tools for problem solving

Taken from:

Introductory and Intermediate Algebra,By K. Elayn Martin-GayCopyright © 1999 by Prentice-Hall, Inc.A Pearson Education CompanyUpper Saddle River, New Jersey 07458

All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Printed in the United States of America

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CONTENTS

Chapter R Prealgebra Review 133

R.1 Factors and the Least Common Multiple 135

R.2 Fractions 139

R.3 Decimals and Percents 147

Chapter 1 Real Numbers and Introduction to Algebra 161

1.1 Symbols and Sets of Numbers 163

1.2 Introduction to Variable Expressions and Equations 169

1.3 Adding Real Numbers 177

1.4 Subtracting Real Numbers 183

1.5 Multiplying Real Numbers 189

1.6 Dividing Real Numbers 193

1.7 Properties of Real Numbers 199

1.8 Reading Graphs 205

Chapter 2 Equations, Inequalities, and Problem Solving 217

2.1 Simplifying Expressions 219

2.2 The Addition Property of Equality 225

2.3 The Multiplication Property of Equality 233

2.4 Further Solving Linear Equations 241

2.5 An Introduction to Problem Solving 247

2.6 Formulas and Problem Solving 253

2.7 Percent, Ratio, and Proportion 259

2.8 Linear Inequalities and Problem Solving 267

Chapter 3 Graphing Equations and Inequalities 279

3.1 The Rectangular Coordinate System 281

3.2 Graphing Linear Equations 289

3.3 Intercepts 299

3.4 Slope 309

3.5 Graphing Linear Inequalities in Two Variables 319

Chapter 4 Exponents and Polynomials 329

4.1 Exponents 331

4.2 Negative Exponents and Scientific Notation 339

iv

4.3 Introduction to Polynomials 347

4.4 Adding and Subtracting Polynomials 355

4.5 Multiplying Polynomials 359

4.6 Special Products 363

4.7 Dividing Polynomials 369

Chapter 5 Factoring Polynomials 379

5.1 The Greatest Common Factor 381

5.2 Factoring Trinomials of the Form x2 + bx + c 387

5.3 Factoring Trinomials of the Form ax2 + bx + c 393

5.4 Factoring Trinomials of the Form ax2 + bx + c by grouping 399

5.5 Factoring Perfect Square Trinomials and theDifference of Two Squares 403

5.6 Solving Quadratic Equations by Factoring 411

5.7 Quadratic Equations and Problem Solving 419

Chapter 6 Rational Expressions 429

6.1 Simplifying Rational Expressions 431

6.2 Multiplying and Dividing Rational Expressions 437

6.3 Adding and Subtracting Rational Expressions withthe Same Denominator and Least Common Denominator 445

6.4 Adding and Subtracting Rational Expressions withDifferent Denominators 451

6.5 Solving Equations Containing Rational Expressions 457

6.6 Rational Equations and Problem Solving 463

6.7 Simplifying Complex Fractions 473

Chapter 7 Graphs and Functions 485

7.1 The Slope-Intercept Form 487

7.2 More Equations of Lines 493

7.3 Introduction to Functions 501

7.4 Polynomial and Rational Functions 509

7.5 An Introduction to Graphing Polynomial Functions 515

Chapter 8 Systems of Equations and Inequalities 531

8.1 Solving Systems of Linear Equations by Graphing 533

8.2 Solving Systems of Linear Equations by Substitution 539

8.3 Solving Systems of Linear Equations by Addition 547

8.4 Systems of Linear Equations and Problem Solving 553

8.5 Solving Systems of Linear Equations in Three Variables 561

8.6 Solving Systems of Equations Using Matrices 567

8.7 Solving Systems of Equations Using Determinants 571

8.8 Systems of Linear Inequalities 577

Contents

v

Chapter 9 Rational Exponents, Radicals, and Complex Numbers 589

9.1 Radical Expressions 591

9.2 Rational Exponents 597

9.3 Simplifying Radical Expressions 601

9.4 Adding, Subtracting, and Multiplying Radical Expressions 607

9.5 Rationalizing Numerators and Denominators ofRadical Expressions 613

9.6 Radical Equations and Problem Solving 617

9.7 Complex Numbers 623

Chapter 10 Quadratic Equations and Functions 633

10.1 Solving Quadratic Equations by Completing the Square 635

10.2 Solving Quadratic Equations by the Quadratic Formula 645

10.3 Solving Equations by Using Quadratic Methods 653

10.4 Nonlinear Inequalities in One Variable 661

10.5 Quadratic Functions and Their Graphs 665

10.6 Further Graphing of Quadratic Functions 675

Chapter 11 Exponential and Logarithmic 685

11.1 The Algebra of Functions 687

11.2 Inverse Functions 691

11.3 Exponential Functions 697

11.4 Logarithmic Functions 705

11.5 Properties of Logarithms 711

11.6 Common Logarithms, Natural Logarithms, andChange of Base 715

11.7 Exponential and Logarithmic Equations andProblem Solving 721

Appendix A Geometry 729

A.1 Lines and Angles 731

A.2 Plane Figures and Solids 737

A.3 Perimeter 741

A.4 Area 749

A.5 Volume 757

A.6 Square Roots and the Pythagorean Theorem 763

A.7 Congruent and Similar Triangles 771

Answers A–1

Index I–1

Contents

133

RMathematics is an important tool for everyday life. Knowing basicmathematical skills can simplify many tasks. For example, we use frac-tions to represent parts of a whole, such as “half an hour” or “third of acup.” Understanding decimals helps us work efficiently in our moneysystem. Percent is a concept used virtually every day in ordinary andbusiness life.

This optional review chapter covers basic topics and skills from preal-gebra. Knowledge of these topics is needed for success in algebra.

R.1 Factors and the LeastCommon Multiple

R.2 Fractions

R.3 Decimals and Percents

Donald and Doris Fisher opened the first The Gap store(named after the “generation gap”) in 1969 near the campus ofwhat is now the San Francisco State University. This singlestore, which sold mostly Levi’s blue jeans, has evolved into aninternational clothing giant. The Gap, Inc., now sells a variety ofclothing through its store chains that include babyGap, BananaRepublic, The Gap, GapKids, and Old Navy Clothing. In 1997,The Gap, Inc., had approximately $5.3 billion in revenue.

Prealgebra Review C H A P T E R

Problem Solving Notes

R.1 FACTORS AND THE LEAST COMMON MULTIPLE

FACTORING NUMBERS

In arithmetic we factor numbers, and in algebra we factor expressions con-taining variables. Throughout this text, you will encounter the word factoroften. Always remember that factoring means writing as a product

Since , we say that 2 and 3 are factors of 6. Also, is a fac-torization of 6.

Example 1 List the factors of 6.

Solution: First we write the different factorizations of 6.

,

The factors of 6 are 1, 2, 3, and 6.

Example 2 List the factors of 20.

Solution:

The factors of 20 are 1, 2, 4, 5, 10, and 20.

In this section, we will concentrate on natural numbers only. The naturalnumbers (also called counting numbers) are

Natural Numbers: 1, 2, 3, 4, 5, 6, 7, and so on

Every natural number except 1 is either a prime number or a compositenumber.

Example 3 Identify each number as prime or composite:

3, 20, 7, 4

Solution: 3 is a prime number. Its factors are 1 and 3 only.20 is a composite number. Its factors are 1, 2, 4, 5, 10, and

20.7 is a prime number. Its factors are 1 and 7 only.4 is a composite number. Its factors are 1, 2, and 4.

WRITING PRIME FACTORIZATIONS

When a number is written as a product of primes, this product is called theprime factorization of the number. For example, the prime factorization of12 is since

and all the factors are prime.

12 = 2 # 2 # 3

2 # 2 # 3

B

PRIME AND COMPOSITE NUMBERS

A prime number is a natural number greater than 1 whose only factorsare 1 and itself. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17,19, 23, 29, . . .A composite number is a natural number greater than 1 that is notprime.

20 = 1 # 20, 20 = 2 # 10, 20 = 4 # 5, 20 = 2 # 2 # 5

6 = 2 # 36 = 1 # 6

2 # 32 # 3 = 6

To factor means to write as a product.

A

Factors and the Least Common Multiple SECTION R.1 135

Objectives

Write the factors of a number.

Write the prime factorization of anumber.

Find the LCM of a list of numbers.C

BA

Practice Problem 1List the factors of 4.

Practice Problem 2List the factors of 18.

Practice Problem 3Identify each number as prime or com-posite:

5, 18, 11, 6

Answers1. 1, 2, 4, 2. 1, 2, 3, 6, 9, 18, 3. 5, 11 prime;6, 18 composite

SSM CD-ROM VideoR.1

136 CHAPTER R Prealgebra Review

HELPFUL HINT

Recall that order is not important when multiplying numbers. Forexample,

For this reason, any of the products shown can be called the primefactorization of 45.

3 # 3 # 5 = 3 # 5 # 3 = 5 # 3 # 3 = 45

HELPFUL HINT

There are a few quick divisibility tests to determine if a number isdivisible by the primes 2, 3, or 5.A whole number is divisible by

j 2 if the ones digit is 0, 2, 4, 6, or 8.T

132 is divisible by 2

j 3 if the sum of the digits is divisible by 3.T

144 is divisible by 3 since is divisible by 3.

j 5 if the ones digit is 0 or 5.T

1115 is divisible by 5

1 + 4 + 4 = 9

Practice Problem 4Write the prime factorization of 28.


Answers4. , 5.

✓ Concept Check: yes; answers may vary

60 = 2 # 2 # 3 # 528 = 2 # 2 # 7

✓ CONCEPT CHECK

Suppose that you choose as your first step in Example 5 andanother student chooses .Will both end up with the same primefactorization as in Example 5?Explain.

80 = 5 # 16

80 = 4 # 20

Example 4 Write the prime factorization of 45.

Solution: We can begin by writing 45 as the product of two num-bers, say 9 and 5.

The number 5 is prime, but 9 is not. So we write 9 as.

T R R

Each factor is now a prime number, so the prime factor-ization of 45 is .


Solution: We first write 80 as a product of two numbers. We con-tinue this process until all factors are prime.

All factors are now prime, so the prime factorization of80 is

TRY THE CONCEPT CHECK IN THE MARGIN.

2 # 2 # 2 # 2 # 5

80 = 8 # 10

4 # 2 # 2 # 5

= 2 # 2 # 2 # 2 # 5

3 # 3 # 5

= 3 # 3 # 5

45 = 9 # 5

3 # 3

45 = 9 # 5

Factors and the Least Common Multiple SECTION R.1 137


When finding the prime factorization of larger numbers, you may wantto use the procedure shown in Example 6.


Solution: Since the ones digit of 252 is 2, we know that 252 is divisi-ble by 2.

1262��252�

126 is divisible by 2 also.

632��126�2��252�

63 is not divisible by 2 but is divisible by 3. We divide63 by 3 and continue in this same manner until the quo-tient is a prime number.

73��213��632��126�2��252�

The prime factorization of 252 is .

FINDING THE LEAST COMMON MULTIPLE

A multiple of a number is the product of that number and any natural num-ber. For example, the multiples of 3 are

3, 6, 9, 12, 15, 18, 21, and so on

The multiples of 2 are

2, 4, 6, 8, 10, 12, 14, and so on

Notice that 2 and 3 have multiples that are common to both.

The least or smallest common multiple of 2 and 3 is 6. The number 6 iscalled the least common multiple or LCM of 2 and 3. It is the smallest num-ber that is a multiple of both 2 and 3.

Finding the LCM by the method above can sometimes be time-consum-ing. Let’s look at another method that uses prime factorization.

To find the LCM of 4 and 10, for example, we write the prime factoriza-tion of each.

If the LCM is to be a multiple of 4, it must contain the factors . Ifthe LCM is to be a multiple of 10, it must contain the factors . Sincewe decide whether the LCM is a multiple of 4 and 10 separately, the LCMdoes not need to contain three factors of 2. The LCM only needs to contain

2 # 52 # 2

10 = 2 # 5

4 = 2 # 2

Multiples of 2: 2, 4, 6 , 8, 10, 12 , 14, 16, 18

Multiples of 3: 3, 6 , 9, 12 , 15, 18 , 21,, and so on

, and so on

2 # 1 2 # 2 2 # 3 2 # 4 2 # 5 2 # 6 2 # 7rrrrrrr

3 # 1 3 # 2 3 # 3 3 # 4 3 # 5 3 # 6 3 # 7rrrrrrr

C

2 # 2 # 3 # 3 # 7

Answer6. 3 # 3 # 3 # 11


Answers7. 70, 8. 45, 9. 60

Practice Problem 7Find the LCM of 14 and 35.

Practice Problem 8Find the LCM of 5 and 9.

a factor the greatest number of times that the factor appears in any oneprime factorization.

Example 7 Find the LCM of 18 and 24.

Solution: First we write the prime factorization of each number.

Now we write each factor the greatest number of timesthat it appears in any one prime factorization.

The greatest number of times that 2 appears is 3 times.The greatest number of times that 3 appears is 2 times.

Example 8 Find the LCM of 11 and 6.

Solution: 11 is a prime number, so we simply rewrite it. Then wewrite the prime factorization of 6.

.

Example 9 Find the LCM of 5, 6, and 12.

Solution:

.LCM = 2 # 2 # 3 # 5 = 60

12 = 2 # 2 # 3

6 = 2 # 3

5 = 5

LCM = 2 # 3 # 11 = 66

6 = 2 # 3

11 = 11

LCM = 2 # 2 # 2 # 3 # 3 = 72rs

2 is a factor3 times.

3 is a factor2 times.

24 = 2 # 2 # 2 # 3

18 = 2 # 3 # 3

TO FIND THE LCM OF A LIST OF NUMBERS

Step 1. Write the prime factorization of each number.

Step 2. Write the product containing each different prime factor (fromStep 1) the greatest number of times that it appears in any onefactorization. This product is the LCM.

The number 2 is a factor twice since that is thegreatest number of times that 2 is a factor ineither of the prime factorizations.

LCM = 2 # 2 # 5 = 20rrThe LCM is a multiple of 4.

The LCM is a multiple of 10.

Practice Problem 9Find the LCM of 4, 15, and 10.

R.2 FRACTIONS

A quotient of two numbers such as is called a fraction. The parts of a frac-

tion are:

A fraction may be used to refer to part of a whole. For example, of the

circle in the figure is shaded. The denominator 9 tells us how many equalparts the whole circle is divided into and the numerator 2 tells us how manyequal parts are shaded.

In this section, we will use whole numbers. The whole numbers consistof 0 and the natural numbers.

Whole Numbers: 0, 1, 2, 3, 4, 5, and so on

WRITING EQUIVALENT FRACTIONS

More than one fraction can be used to name the same part of a whole. Suchfractions are called equivalent fractions.

To write equivalent fractions, we use the fundamental principle of frac-tions. This principle guarantees that, if we multiply both the numerator andthe denominator by the same nonzero number, the result is an equivalent

fraction. For example, if we multiply the numerator and denominator of

by the same number, 2, the result is the equivalent fraction .

1 # 23 # 2

= 26

26

13

EQUIVALENT FRACTIONS

Fractions that represent the same portion of a whole are calledequivalent fractions.

a Qequivalent fractions

62=3

1

A

29

92 of the circle

is shaded.

d Numeratord Denominator

29

Fraction bar S

29

Fractions SECTION R.2 139

Objectives

Write equivalent fractions.

Write fractions in simplest form.

Multiply and divide fractions.

Add and subtract fractions.DCBA

FUNDAMENTAL PRINCIPLE OF FRACTIONS

If a, b, and c are numbers, then

or

as long as b and c are not 0.

a # cb # c

= ab

ab

= a # cb # c

SSM CD-ROM VideoR.2


Practice Problem 1

Write as an equivalent fraction with

a denominator of 20.

14

Practice Problem 2

Simplify: 2035

✓ CONCEPT CHECK

Explain the error in the followingsteps.

a.

b.5 + 15 + 2

=12

67

=

1555

=1 55 5

=15

Practice Problems 3–4Simplify each fraction.

3.

4.1240

720

Example 1 Write as an equivalent fraction with a denominator

of 15.

Solution: Since , we use the fundamental principle offractions and multiply the numerator and denominator of

by 3.

Then is equivalent to . They both represent the same

part of a whole.

SIMPLIFYING FRACTIONS

A fraction is said to be simplified or in lowest terms when the numeratorand the denominator have no factors in common other than 1. For exam-

ple, the fraction is in lowest terms since 5 and 11 have no common fac-

tors other than 1.One way to simplify fractions is to write both the numerator and the

denominator as a product of primes and then apply the fundamental prin-ciple of fractions.

Example 2 Simplify:

Solution: We write the numerator and the denominator as prod-ucts of primes. Then we apply the fundamental principleof fractions to the common factor 7.


Examples Simplify each fraction.

3. There are no common factors other

than 1, so is already simplified.

4.

The improper fraction from Example 4 may be written as the mixed

number , but in this text, we will not do so.

Some fractions may be simplified by recalling that the fraction barmeans division.

and31

= 3 ,1 = 366

= 6 ,6 = 1

425

225

8820

=2 # 2 # 2 # 11

2 # 2 # 5=

225

1127

1127

= 113 # 3 # 3

4249

=2 # 3 # 7

7 # 7=

2 # 37

=67

4249

511

B

615

25

25

= 2 # 35 # 3

= 615

25

5 # 3 = 15

25

Answers

1. , 2. , 3. , 4.

✓ Concept Check: answers may vary

310

720

47

520

Examples Simplify by dividing the numerator by the denominator.

5.

6.

7.

8.

In general, if the numerator and the denominator are the same, the frac-tion is equivalent to 1. Also, if the denominator of a fraction is 1, the frac-tion is equivalent to the numerator.

MULTIPLYING AND DIVIDING FRACTIONS

To multiply two fractions, we multiply numerator times numerator toobtain the numerator of the product. Then we multiply denominator timesdenominator to obtain the denominator of the product.

Example 9 Multiply: . Simplify the product if possible.

Solution:

To simplify the product, we divide the numerator and thedenominator by any common factors.

= 239

=2 # 5

3 # 5 # 13

Multiply numerators.

Multiply denominators.215

# 513

= 2 # 515 # 13

215

# 513

C

81

= 8 ,1 = 8

77

= 7 ,7 = 1

42

= 4 ,2 = 2

33

= 3 ,3 = 1


If a is any number other than 0, then .

Also, if a is any number, .a1

= a

aa

= 1

Practice Problems 5–8Simplify by dividing the numerator bythe denominator.

5. 6.

7. 8.51

1010

93

44

Practice Problem 9

Multiply: . Simplify the product if

possible.

37

# 35

Answers5. 1, 6. 3, 7. 1, 8. 5, 9. 9

35

MULTIPLYING FRACTIONS

, if b Z 0 and d Z 0ab

# cd

= a # cb # d

Before we divide fractions, we first define reciprocals. Two numbers arereciprocals of each other if their product is 1.

The reciprocal of is because .

The reciprocal of 5 is because .

To divide fractions, we multiply the first fraction by the reciprocal of thesecond fraction. For example,

a c

Examples Divide and simplify.

10.

11.

12.

ADDING AND SUBTRACTING FRACTIONS

To add or subtract fractions with the same denominator, we combinenumerators and place the sum or difference over the common denomi-nator.

D

38

,310

=38

# 103

=3 # 2 # 5

2 # 2 # 2 # 3=

54

710

, 14 =710

,141

=710

# 114

=7 # 1

2 # 5 # 2 # 7=

120

45

, 516

= 45

# 165

= 4 # 165 # 5

= 6425

HELPFUL HINT To divide, multiply by the reciprocal.

ff

12

,57

= 12

# 75

= 1 # 72 # 5

= 710

5 # 15

= 51

# 15

= 55

= 115

23

# 32

= 66

= 132

23


ADDING AND SUBTRACTING FRACTIONS WITH THE

SAME DENOMINATOR

, if

, if b Z 0ab

- cb

= a - cb

b Z 0ab

+ cb

= a + cb

Practice Problems 10–12Divide and simplify.

10.

11.

12.54

,58

811

,24

29

,34

Answers

10. , 11. , 12. 21

33827

DIVIDING FRACTIONS

, if b Z 0, d Z 0, and c Z 0ab

,cd

= ab

# dc


Practice Problems 13–16Add or subtract as indicated. Thensimplify if possible.

13.

14.

15.

16.76

- 26

1310

- 310

18

+ 38

211

+ 511

Practice Problem 17

Add: 38

+ 120

Practice Problem 18

Subtract and simplify: 815

- 13

Examples Add or subtract as indicated. Then simplify if possible.

13.

14.

15.

16.

To add or subtract with different denominators, we first write the frac-tions as equivalent fractions with the same denominator. We will use thesmallest or least common denominator, the LCD. The LCD is the same asthe least common multiple we reviewed in Section R.1.

Example 17 Add:

Solution: We first must find the least common denominator beforethe fractions can be added. The least common multiplefor the denominators 5 and 4 is 20. This is the LCD wewill use.

We write both fractions as equivalent fractions withdenominators of 20. Since

and

then

Example 18 Subtract and simplify:

Solution: The LCD is 12. We write both fractions as equivalentfractions with denominators of 12.

=1512

=3 # 5

2 # 2 # 3=

54

= 3812

- 2312

196

- 2312

= 19 # 26 # 2

- 2312

196

- 2312

25

+ 14

= 820

+ 520

= 1320

14

= 1 # 54 # 5

= 520

25

= 2 # 45 # 4

= 820

25

+ 14

53

- 13

= 5 - 13

= 43

97

-27

=9 - 2

7=

77

= 1

310

+2

10=

3 + 210

=5

10=

52 # 5

=12

27

+ 47

= 2 + 47

= 67

Answers

13. , 14. , 15. 1, 16. , 17. ,

18. 15

1740

56

12

711


Focus On Study SkillsCRITICAL THINKING

What Is Critical Thinking?

Although exact definitions often vary, thinking critically usuallyrefers to evaluating, analyzing, and interpreting information inorder to make a decision, draw a conclusion, reach a goal, make aprediction, or form an opinion. Critical thinking often involvesproblem solving, communication, and reasoning skills. Criticalthinking is more than a technique that helps students pass theircourses. Critical thinking skills are life skills. Developing theseskills can help you solve problems in your workplace and in every-day life. For instance, well-developed critical thinking skills wouldbe useful in the following situation:

Suppose you work as a medical lab technician. Your lab super-visor has decided that some lab equipment should be replaced.She asks you to collect information on several different modelsfrom equipment manufacturers. Your assignment is to study thedata and then make a recommendation on which model the labshould buy.

How Can Critical Thinking Be Developed?

Just as physical exercise can help to develop and strengthen cer-tain muscles of the body, mental exercise can help to develop crit-ical thinking skills. Mathematics is ideal for helping to developcritical thinking skills because it requires using logic and reason-ing, recognizing patterns, making conjectures and educatedguesses, and drawing conclusions. You will find many opportuni-ties to build your critical thinking skills throughout this course:

j In real-life application problems (see Exercise 24 in Section 2.5)j In conceptual and writing exercises marked with the icon

(see Exercise 43 in Section 1.2)j In the Combining Concepts subsection of exercise sets (see Exer-

cise 57 in Section 4.4)j In the Chapter Activities (see the Chapter 8 Activity)j In the Critical Thinking and Group Activities found in Focus

On features like this one throughout the book (see page 46).

EXERCISE SET R.2

Write each fraction as an equivalent fraction with the given denominator. See Example 1.

1. with a denominator of 20 2. with a denominator of 25

Simplify each fraction. See Examples 2 through 8.

3. 4. 5.

6. 7. 8.

9. 10. 11.

12. 13. 14.

15. 16.

Multiply or divide as indicated. See Examples 9 through 12.

17. 18. 19.

20. 21. 2536

,10920

,12

335

# 1063

710

# 521

78

# 321

C

360700

120244

6424

840

1620

4245

1830

426

357

2020

59

37

1520

36

B

45

45

A

145

Name ____________________________________ Section ________ Date ___________ ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

Add or subtract as indicated. See Examples 13 through 18.

22. 23. 24.

25. 26. 27.

28. 29. 30.

31. 32. 33.

34. 35. 36. 811

- 14

+ 12

23

- 59

+ 56

2 - 38

125

- 1710

- 815

522

- 533

1135

+ 335

1021

+ 521

117

- 335

103

- 521

34

+ 16

23

+ 37

1835

- 1135

1721

- 1021

13132

+ 35132

D

Name _________________________________________________________________________

146

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

R.3 DECIMALS AND PERCENTS

WRITING DECIMALS AS FRACTIONS

Like fractional notation, decimal notation is used to denote a part of awhole. Below is a place value chart that shows the value of each place.

TRY THE CONCEPT CHECK IN THE MARGIN.The next chart shows decimals written as fractions.

Decimal Form Fractional Form

Examples Write each decimal as a fraction. Do not simplify.

1.

2 zeros

2.

1 zero

3.3 zeros3 decimal

places

2.649 = 26491000

1 decimalplace

1.3 = 1310

2 decimalplaces

0.37 = 37100

110

ten-thousandths

986210,000

0.9862 r

231100

hundredths

hundredths

7100

tenths

2.31

0.07

0.1

UU

E

110

1100

11000

110,000

1101001000

Ten-

Thou

sand

ths

Thou

sand

ths

Hun

dred

ths

Tent

hs

One

s

Tens

Hun

dred

s

Thou

sand

s

.82 7 6 1

decimalpoint

A

Decimals and Percents SECTION R.3 147

Objectives

Write decimals as fractions.

Add, subtract, multiply, and dividedecimals.

Round decimals to a given decimalplace.

Write fractions as decimals.

Write percents as decimals anddecimals as percents.

ED

C

BA

SSM CD-ROM VideoR.3

✓ CONCEPT CHECK

Fill in the blank: In the number52.634, the 3 is in the ——— place.

a. Tens

b. Ones

c. Tenths

d. Hundredths

e. Thousandths

Practice Problems 1–3Write each decimal as a fraction. Donot simplify.

1. 0.27

2. 5.1

3. 7.685

Answers✓Concept Check: d

1. , 2. , 3. 76851000

5110

27100


Practice Problem 4Add.

a.

b. 12 + 0.79 + 0.03

7.19 + 19.782 + 1.006

Practice Problem 5Subtract.

a.

b. 90 - 0.19

84.23 - 26.982

Practice Problem 6Multiply.

a.

b. 1.26 * 0.03

0.31 * 4.6

ADDING, SUBTRACTING, MULTIPLYING, AND DIVIDING DECIMALS

To add or subtract decimals, we write the numbers vertically with decimalpoints lined up. We then add the like place values from right to left. Weplace the decimal point in the answer directly below the decimal points inthe problem.

Example 4 Add.

a. b.

Solution: a. b.

Example 5 Subtract.

a. b.

Solution: a. 1 11 4 10 b. 6 9 9 10

Now let’s study the following product of decimals. Notice the pattern inthe decimal points.

In general, to multiply decimals we multiply the numbers as if they werewhole numbers. The decimal point in the product is placed so that the num-ber of decimal places in the product is the same as the sum of the numberof decimal places in the factors.

Example 6 Multiply.

a. b.

Solution: a. b.

To divide a decimal by a whole number using long division, we place thedecimal point in the quotient directly above the decimal point in the divi-dend. For example,

To check, see that

2.47 * 3 = 7.41

2 .473 @ 7 .41- 6

1 4 - 1 2

21 - 21

0

2 decimal places

2 decimal places

4 decimal places

0.17* 0.020.0034

3 decimal places

1 decimal place

4 decimal places

0.072* 3.5

360 216

0.2520

0.17 * 0.020.072 * 3.5

0.03 * 0.6 = 3100

* 610

= 181000

or 0.018E U

2 decimalplaces

1 decimalplace

3 decimalplaces

7 0 . 0 0- 0 . 4 8

6 9 . 5 2

3 2 . 1 5 0- 1 1 . 2 3 7

2 0 . 9 1 3

70 - 0.4832.15 - 11.237

7. 0.23

+ 0.6 7.83

5.87 23.279

+ 0.00329.152

7 + 0.23 + 0.65.87 + 23.279 + 0.003

B

Answers4. a. 27.978, b. 12.82, 5. a. 57.248,b. 89.81, 6. a. 1.426, b. 0.0378


Practice Problem 7Divide.

a.

b. 15.6 ,0.006

21.75 ,0.5

Practice Problem 8Round 12.9187 to the nearest hun-dredth.

Answers7. a. 43.5, b. 2600, 8. 12.92, 9. 245.3

Practice Problem 9Round 245.348 to the nearest tenth.

In general, to divide decimals we move the decimal point in the divisorto the right until the divisor is a whole number. Then we move the decimalpoint in the dividend the same number of places that the decimal point inthe divisor was moved. The decimal point in the quotient lies directly abovethe decimal point in the dividend.

Example 7 Divide.

a. b.

Solution: a. b.

ROUNDING DECIMALS

We round the decimal part of a decimal number in nearly the same way aswe round the whole numbers. The only difference is that we drop digits tothe right of the rounding place, instead of replacing these digits by 0 s. Forexample,

24.954 rounded to the nearest hundredth is 24.95c

Example 8 Round 7.8265 to the nearest hundredth.

Solution: 7.8265

Thus, 7.8265 rounded to the nearest hundredth is 7.83.

Example 9 Round 19.329 to the nearest tenth.

Solution: 19.329

Thus, 19.329 rounded to the nearest tenth is 19.3.

Step 1. Locate the digit to the right of the tenths place.

Step 2. This digit is less than 5, so we drop this digit and alldigits to its right.

c——

T— tenths place

Step 1. Locate the digit to the right of the hundredthsplace.

Step 2. This digit is 5 or greater, so we add 1 to the hun-dredths place digit and drop all digits to its right.

c——

T— hundredths place

TO ROUND DECIMALS TO A PLACE VALUE TO THE RIGHT OF THE

DECIMAL POINT

Step 1. Locate the digit to the right of the given place value.

Step 2. j If this digit is 5 or greater, add 1 to the digit in the given placevalue and drop all digits to its right.

j If this digit is less than 5, drop all digits to the right of thegiven place.

C

4500 .0007 . @ 31500 .

- 28 35

- 35 0

2 36.504 . 2 946.0

- 8 14

- 12 26

- 24 20

- 20 0

31.5 ,0.0079.46 ,0.04


Practice Problem 11

Write as a decimal.56

Answers10. 0.4, 11. , 12. 0.111

✓Concept Check: c

0.83

✓ CONCEPT CHECK

The notation is the same as

a.

b.

c. 0.52222222 . . .

52 . . .100

52100

0.52

Practice Problem 12

Write as a decimal. Round to the

nearest thousandth.

19

WRITING FRACTIONS AS DECIMALS

To write fractions as decimals, interpret the fraction bar as division and findthe quotient.

Example 10 Write as a decimal.

Solution:

Example 11 Write as a decimal.

Solution:

This pattern will continue so that

A bar can be placed over the digit 6 to indicate that itrepeats.

We can also write a decimal approximation for . For example,

rounded to the nearest hundredth is 0.67. This can be written as .

The sign means “is approximately equal to.”


Example 12 Write as a decimal. Round to the nearest hundredth.

(The fraction is an approximation for .)p227

227

L

23

L 0.67

23

23

23

= 0.666 . . . = 0.6

0.6663 2 2.000-1 8

20 -18

20 -18

2

23

= 0.6666 . . . .

23

14

= 0.25

0.254 2 1.00

-8 20

-200

14

WRITING FRACTIONS AS DECIMALS

To write fractions as decimals, divide the numerator by the denomi-nator.

D

Practice Problem 10

Write as a decimal.25

Solution:

The fraction in decimal form is approximately 3.14.

WRITING PERCENTS AS DECIMALS AND DECIMALS AS PERCENTS

The word percent comes from the Latin phrase per centum, which means“per 100.” Thus, 53% means 53 per 100, or

When solving problems containing percents, it is often necessary to writea percent as a decimal. To see how this is done, study the chart below.

Percent Fraction Decimal

7% 0.07

63% 0.63

109% 1.09

To convert directly from a percent to a decimal, notice that

Example 13 Write each percent as a decimal.

a. 25% b. 2.6% c. 195%

Solution: We drop the % and move the decimal point two places tothe left. Recall that the decimal point of a whole numberis to the right of the ones place digit.

a.

b.

c. 195% = 195.% = 1.95

2.6% = 02.6% = 0.026

25% = 25.% = 0.25

TO WRITE A PERCENT AS A DECIMAL

Drop the percent symbol and move the decimal point two places to theleft.

7% = 0.07

109100

63100

7100

53% = 53100

E

227

3.1427 2 22.000-21

10 -7

30 -28

20 -14

6

L 3.14 If rounding to the nearest hundredth,carry the division process out to one moredecimal place, the thousandths place.


Answers13. a. 0.20, b. 0.12, c. 4.65

Practice Problem 13Write each percent as a decimal.

a. 20%

b. 1.2%

c. 465%


To write a decimal as a percent, we simply reverse the preceding steps.That is, we move the decimal point two places to the right and attach thepercent symbol, %.

Example 14 Write each decimal as a percent.

a. 0.85 b. 1.25 c. 0.012 d. 0.6

Solution: We move the decimal point two places to the right andattach the percent symbol, %.

a.

b.

c.

d. 0.6 = 0.60 = 60%

0.012 = 0.012 = 1.2%

1.25 = 1.25 = 125%

0.85 = 0.85 = 85%

TO WRITE A DECIMAL AS A PERCENT

Move the decimal point two places to the right and attach the percentsymbol,%.

Answers14. a. 42%, b. 0.3%, c. 236%, d. 70%

Practice Problem 14Write each decimal as a percent.

a. 0.42

b. 0.003

c. 2.36

d. 0.7

153

Name ____________________________________ Section ________ Date ___________

EXERCISE SET R.3

Write each decimal as a fraction. Do not simplify. See Examples 1 through 3.

1. 7.23 2. 0.114 3. 0.239 4. 123.1

Add or subtract as indicated. See Examples 4 and 5.

5. 6. 7. 8.

9. 10. 11.

12. 13. 14.

Multiply or divide as indicated. See Examples 6 and 7.

15. 16. 17. 18.

19. 20. 21.

22. 23. 24.

25. 26. 27.

28. 29. 30. 0.054 ƒ 51.84 0.063 ƒ 52.92 0.92 ƒ 3.312

0.82 ƒ 4.7560.9 ƒ 360.6 ƒ 42

2 ƒ 11.7 5 ƒ 0.47 31.006* 3.71

16.003* 5.31

8.03* 5.5

5.62* 7.7

8.91* 100

6.75* 10

0.7: 0.9

0.2* 0.6

863.2 - 39.45

654.9 - 56.67

65.00285.0903

+ 6.9

45.02 3.006

+ 8.405

28 - 3.3118 - 2.78

7.6 - 2.18.8 - 2.332.4 + 1.58 + 0.09342.31 + 6.4

B

A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

Name _________________________________________________________________________

154

Round each decimal to the given place value. See Examples 8 and 9.

31. 0.57, nearest tenth 32. 0.58, nearest tenth

33. 0.234, nearest hundredth 34. 0.452, nearest hundredth

35. 0.5942, nearest thousandth 36. 63.4523, nearest thousandth

37. 98,207.23, nearest tenth 38. 68,936.543, nearest tenth

39. 12.347, nearest tenth 40. 42.9878, nearest thousandth

Write each fraction as a decimal. If the decimal is a repeating decimal, write using the barnotation and then round to the nearest hundredth. See Examples 10 through 12.

41. 42. 43.

44. 45.

Write each percent as a decimal. See Example 13.

46. 36% 47. 3.1% 48. 2.2%

49. 135% 50. 417% 51. 81.49%

E

16

611

58

716

13

D

C

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

53.

52. In a recent telephone survey, approxi-mately 61% of the respondents saidthat they are better off financiallythan their parents were at the sameage. Write this percent as a decimal.(Source: Reader’s Digest, December1996)

53. The average one-year survival rate fora heart transplant recipient is 82.3%.The average one-year survival rate fora liver transplant patient is 81.6%.Write each percent as a decimal.(Source: Bureau of Health ResourcesDevelopment)

58. The estimated life expectancy at birthfor female Canadians was 82.65 yearsin 1996. The estimated life expectancyat birth for male Canadians was only75.67 years in 1996. How much longeris a female Canadian born in 1996expected to live than a male Canadianborn in the same year? (Source: TheCentral Intelligence Agency, 1996World Factbook)

59. The chart shows the average numberof pounds of meats consumed by eachUnited States citizen in 1995. (Source:National Agricultural Statistics Ser-vice)

Meat Pounds

Chicken 71.3

Turkey 17.9

Beef 67.3

Pork 52.5

Veal 1.0

Lamb/Mutton 1.2

a. How much more beef than pork didthe average U.S. citizen consume in1995?

b. How much poultry (chicken andturkey) did the average U.S. citizenconsume in 1995?

c. What was the total amount of meatconsumed by the average U.S. citi-zen in 1995?

60. An estimated of Americans own at

least one credit card. What percent ofAmericans own credit cards?(Sources: Bank Advertising News andThe Gallup Organization, 1996)

1625

155

Name _________________________________________________________________________

Write each decimal as a percent. See Example 14.

54. 0.876 55. 0.521 56. 0.5 57. 0.1

54.

55.

56.

57.

58.

59. a.

b.

c.

60.

COMBINING CONCEPTS

Focus On HistoryFACTORING MACHINE

Small numbers can be broken down into their prime factorsrelatively easily. However, factoring larger numbers can bedifficult and time-consuming. The first known successful attemptto automate the process of factoring whole numbers is credited toa French infantry officer and mathematics enthusiast, EugèneOlivier Carissan. In 1919, he designed and built a machine thatuses gears and a hand crank to factor numbers.

Carissan’s factoring machine had been all but forgotten afterhis death in 1925. In 1989, a Canadian researcher came across adescription of the machine in an article printed in an obscureFrench journal in 1920. This led to a five-year search for traces ofthe machine. Eventually, the factoring machine was found in aFrench astronomical observatory which had received the inven-tion from Carissan’s family after his death.

Mathematical historians agree that the factoring machine was aremarkable achievement in its precomputer era. Up to 40 num-bers per second could be processed by the machine while its oper-ator turned the crank at two revolutions per minute. Carissan wasable to use his machine to prove that the number 708,158,977 wasprime in under 10 minutes. He could also find the prime factoriza-tions of up to 13-digit numbers with the machine.


Chapter R Highlights 157

CHAPTER R HIGHLIGHTSDEFINITIONS AND CONCEPTS EXAMPLES

SECTION R.1 FACTORS AND THE LEAST COMMON MULTIPLE

To factor means to write as a product.

When a number is written as a product of primes,this product is called the prime factorization of anumber.

The factors of 12 are

1, 2, 3, 4, 6, 12

Write the prime factorization of 60.

The prime factorization of 60 is .2 # 2 # 3 # 5

60 = 6 # 10

2 # 3 # 2 # 5

The least common multiple (LCM) of a list of num-bers is the smallest number that is a multiple of allthe numbers in the list.

TO FIND THE LCM OF A LIST OF NUMBERS

Step 1. Write the prime factorization of eachnumber.

Step 2. Write the product containing each differ-ent prime factor (from Step 1) the great-est number of times that it appears inany one factorization. This product is theLCM.

Find the LCM of 12 and 40.

LCM = 2 # 2 # 2 # 3 # 5 = 120

40 = 2 # 2 # 2 # 5

12 = 2 # 2 # 3

SECTION R.2 FRACTIONS

Fractions that represent the same quantity are calledequivalent fractions.

and are equivalent fractions.420

15

15

= 1 # 45 # 4

= 420

FUNDAMENTAL PRINCIPLE OF FRACTIONS

If a, b, and c are numbers, then

or

as long as b and c are not 0.

A fraction is simplified when the numerator andthe denominator have no factors in common otherthan 1.

a # cb # c

= ab

ab

= a # cb # c

is simplified.1317

To simplify a fraction, factor the numerator and thedenominator; then apply the fundamental principleof fractions.

Simplify.

614

=2 # 32 # 7

=37


SECTION R.3 DECIMALS AND PERCENTS

SECTION R.2 (CONTINUED)

Two fractions are reciprocals if their product is 1.

The reciprocal of is , as long as a and b are

not 0.

ba

ab

The reciprocal of is .256

625

To multiply fractions, numerator times numerator isthe numerator of the product and denominatortimes denominator is the denominator of the prod-uct.

25

# 37

= 635

To divide fractions, multiply the first fraction by thereciprocal of the second fraction.

59

,27

= 59

# 72

= 3518

To add fractions with the same denominator, add thenumerators and place the sum over the commondenominator.

511

+ 311

= 811

To subtract fractions with the same denominator,subtract the numerators and place the differenceover the common denominator.

1315

- 315

= 1015

= 23

To add or subtract fractions with different denomi-nators, first write each fraction as an equivalentfraction with the LCD as denominator.

29

+ 36

= 2 # 29 # 2

+ 3 # 36 # 3

= 4 + 918

= 1318

To write decimals as fractions, use place values. 0.12 = 12100

TO ADD OR SUBTRACT DECIMALS

Step 1. Write the decimals so that the decimalpoints line up vertically.

Step 2. Add or subtract as for whole numbers.

Step 3. Place the decimal point in the sum or dif-ference so that it lines up vertically withthe decimal points in the problem.

Subtract: Add: 7 10

25. + 0.02

25.02

2.80-1.04

1.76

25 + 0.022.8 - 1.04

TO MULTIPLY DECIMALS

Step 1. Multiply the decimals as though they arewhole numbers.

Step 2. The decimal point in the product isplaced so that the number of decimalplaces in the product is equal to the sumof the number of decimal places in thefactors.

Multiply:

d2 decimal places

d1 decimal place

d3 decimal places

1.4 8: 5.9

1 3 3 2 7 4 0

8.7 3 2

1.48 * 5.9

Chapter R Highlights 159

SECTION R.3 (CONTINUED)

TO DIVIDE DECIMALS

Step 1. Move the decimal point in the divisor tothe right until the divisor is a whole num-ber.

Step 2. Move the decimal point in the dividendto the right the same number of places asthe decimal point was moved in Step 1.

Step 3. Divide. The decimal point in the quo-tient is directly over the moved decimalpoint in the dividend.

Divide:

0.432.6 2 1.118

104 78

�780

�

1.118 ,2.6

To write fractions as decimals, divide the numeratorby the denominator. Write as a decimal.

0

0.3758 2 3.000-2 4

60 -56

40 -40

38

To write a percent as a decimal, drop the % symboland move the decimal point two places to the left.

To write a decimal as a percent, move the decimalpoint two places to the right and attach the % sym-bol.

25% = 25.% = 0.25

0.7 = 0.70 = 70%

1

161

The power of mathematics is its flexibility. We apply numbers toalmost every aspect of our lives. The power of algebra is its generality. Inalgebra, we use letters to represent numbers.

In this chapter, we begin with a review of the basic symbols—the lan-guage—of arithmetic. We then introduce the use of a variable in place ofa number. From there, we translate phrases to algebraic expressions andsentences to equations. This is the beginning of problem solving, whichwe formally study in Chapter 2.

1.1 Symbols and Sets of Numbers

1.2 Introduction to VariableExpressions and Equations

1.3 Adding Real Numbers

1.4 Subtracting Real Numbers

1.5 Multiplying Real Numbers

1.6 Dividing Real Numbers

1.7 Properties of Real Numbers

1.8 Reading Graphs

The stars have been a source of interest to different cultures forcenturies. Polaris, the North Star, guided ancient sailors. TheEgyptians honored Sirius, the brightest star in the sky, in tem-ples. Around 150 B.C., a Greek astronomer, Hipparchus, de-vised a system of classifying the brightness of stars. He called thebrightest stars “first magnitude” and the faintest stars “sixthmagnitude.” Hipparchus’s system is the basis of the apparentmagnitude scale used by modern astronomers. This modernscale has been modified to include negative numbers.

Real Numbers and Introduction to Algebra C H A P T E R

1.1 SYMBOLS AND SETS OF NUMBERS

We begin with a review of the set of natural numbers and the set of wholenumbers and how we use symbols to compare these numbers. A set is a col-lection of objects, each of which is called a member or element of the set.A pair of brace symbols encloses the list of elements and is translatedas “the set of” or “the set containing.”

E F

Symbols and Sets of Numbers SECTION 1.1 163

These numbers can be pictured on a number line. To draw a numberline, first draw a line. Choose a point on the line and label it 0. To the rightof 0, label any other point 1. Being careful to use the same distance as from0 to 1, mark off equally spaced distances. Label these points 2, 3, 4, 5, andso on. Since the whole numbers continue indefinitely, it is not possible toshow every whole number on the number line. The arrow at the right endof the line indicates that the pattern continues indefinitely.

EQUALITY AND INEQUALITY SYMBOLS

Picturing natural numbers and whole numbers on a number line helps us tosee the order of the numbers. Symbols can be used to describe in writingthe order of two quantities. We will use equality symbols and inequalitysymbols to compare quantities.

Below is a review of these symbols. The letters and are used to rep-resent quantities. Letters such as and that are used to represent num-bers or quantities are called variables.

baba

A

10 2 3 4 5

These symbols may be used to form mathematical statements such as

and 2 Z 62 = 2

MEANING

Equality symbol: a is equal to b.Inequality symbols: a is not equal to b.

a is less than b.a is greater than b.a is less than or equal to b.a is greater than or equal to b. a ≥ b

a ≤ b a 7 b a 6 b a Z ba = b

NATURAL NUMBERS

1, 2, 3, 4, 5, 6, . . .FE

WHOLE NUMBERS

E0, 1, 2, 3, 4, 5, 6, . . .F

HELPFUL HINT

The three dots (an ellipsis) at the end of the list of elements of a setmeans that the list continues in the same manner indefinitely.

SSM CD-ROM Video1.1

Objectives

Define the meaning of the symbols, , , , , and .

Translate sentences into mathe-matical statements.

Identify integers, rational numbers,irrational numbers, and real num-bers.

Find the absolute value of a realnumber.

D

C

B≥≤76Z=

A

164 CHAPTER 1 Real Numbers and Introduction to Algebra

On the number line, we see that a number to the right of another num-ber is larger. Similarly, a number to the left of another number is smaller.For example, 3 is to the left of 5 on the number line, which means that 3 isless than 5, or . Similarly, 2 is to the right of 0 on the number line,which means 2 is greater than 0, or . Since 0 is to the left of 2, we canalso say that 0 is less than 2, or .

0 4 52 312 > 0 or 0 < 2

0 4 52 313 < 5

0 6 22 7 0

3 6 5

Examples Determine whether each statement is true or false.

1. True. Since 2 is to the left of 3 on the number line

2. True. Since 72 is to the right of 27 on the number line

3. True. Since is true

4. True. Since is true

5. False. Since neither nor is true

6. True. Since is true0 6 230 ≤ 2323 = 023 6 023 ≤ 0

8 = 88 ≤ 88 = 88 ≥ 8

72 7 272 6 3

Practice Problems 1–6Determine whether each statement istrue or false.

1.

2.

3.

4.

5.

6. 25 ≥ 22

0 ≤ 5

21 ≥ 21

21 ≤ 21

100 7 10

8 6 6

TRANSLATING SENTENCES INTO MATHEMATICAL STATEMENTS

Now, let’s use the symbols discussed above to translate sentences intomathematical statements.

Example 7 Translate each sentence into a mathematical statement.

a. Nine is less than or equal to eleven.b. Eight is greater than one.c. Three is not equal to four.

Solution: a.

T T T9 11

b.

T T T8 17

oneis greater thaneight

≤

elevenis less thanor equal to

nine

B

Practice Problem 7Translate each sentence into a mathe-matical statement.

a. Fourteen is greater than or equal to fourteen.

b. Zero is less than five.

c. Nine is not equal to ten.

HELPFUL HINTNotice that has exactly the same meaning as Switchingthe order of the numbers and reversing the “direction of the inequalitysymbol” does not change the meaning of the statement.

has the same meaning as

Also notice that when the statement is true, the inequality arrowpoints to the smaller number.

3 6 5.5 7 3

0 6 2.2 7 0

HELPFUL HINTIf either or is true, then is true.3 ≤ 33 = 33 6 3

Answers1. false, 2. true, 3. true, 4. true, 5. true,6. true, 7. a. , b. , c. 9 Z 100 6 514 ≥ 14

c.

T T T3 4

IDENTIFYING COMMON SETS OF NUMBERS

Whole numbers are not sufficient to describe many situations in the realworld. For example, quantities smaller than zero must sometimes be repre-sented, such as temperatures, less than 0 degrees.

We can place numbers less than zero on the number line as follows:Numbers less than 0 are to the left of 0 and are labeled , , , and soon. The numbers we have labeled on the number line below are called theset of integers.

Integers to the left of 0 are called negative integers; integers to the right of0 are called positive integers. The integer 0 is neither positive nor negative.

1

Negative NumbersZero

Positive Numbers

0 32 4 5−5 −4 −3 −2 −1

-3-2-1

C

Z

fouris not

equal tothree


Example 8 Use an integer to express the number in the following.“Pole of Inaccessibility, Antarctica, is the coldest locationin the world, with an average annual temperature of 72degrees below zero.” (Source: The Guinness Book ofRecords)

Solution: The integer represents 72 degrees below zero.

A problem with integers in real-life settings arises when quantities aresmaller than some integer but greater than the next smallest integer. On thenumber line, these quantities may be visualized by points between integers.Some of these quantities between integers can be represented as a quotientof integers. For example,

The point on the number line halfway between 0 and 1 can be repre-

sented by , a quotient of integers.12

-72

Practice Problem 8Use an integer to express the numberin the following. “The lowest altitudein North America is found in DeathValley, California. Its altitude is 282feet below sea level.” (Source: TheWorld Almanac, 1997 )

INTEGERS

E . . . , -3, -2, -1, 0, 1, 2, 3, . . .F

HELPFUL HINTA sign, such as the one in ,tells us that the number is to theleft of 0 on the number line.

is read “negative one,”

A sign or no sign tells us that anumber lies to the right of 0 on thenumber line. For example, 3 and

both mean positive three.+3

+-1

-1-

Answer8. -282

The point on the number line halfway between 0 and can be repre-

sented by . Other quotients of integers and their graphs are shown

in the margin.

These numbers, each of which can be represented as a quotient of integers,are examples of rational numbers. It’s not possible to list the set of rationalnumbers using the notation that we have been using. For this reason, wewill use a different notation.

We read this set as “the set of numbers such that a and b are integers and

b is not 0.”Notice that every integer is also a rational number since each integer can

be written as a quotient of integers. For example, the integer 5 is also a

rational number since . In this rational number, , recall that the top

number, 5, is called the numerator and the bottom number, 1, is called thedenominator.

Let’s practice graphing numbers on a number line.

Example 9 Graph the numbers on the number line.

Solution: To help graph the improper fractions in the list, we firstwrite them as mixed numbers.

Every rational number has a point on the number line that correspondsto it. But not every point on the number line corresponds to a rational num-ber. Those points that do not correspond to rational numbers correspondinstead to irrational numbers.

An irrational number that you have probably seen is . Also, , thelength of the diagonal of the square shown, is an irrational number.

Both rational and irrational numbers can be written as decimal numbers.The decimal equivalent of a rational number will either terminate or repeatin a pattern. For example, upon dividing we find that

(decimal number terminates or ends)

. . . (decimal number repeats in a pattern)23

= 0.66666

34

= 0.75

12p

−2 −1 0

− −

1 4

3.521oror

3

43

2

14

1211

332

18

- 43

, 14

, 32

, 218

, 3.5

51

5 = 51

ab

- 12

-1


HELPFUL HINT

We commonly refer to rationalnumbers as fractions.

IRRATIONAL NUMBERS

nonrational numbers that correspond to points on the number lineFE

Answer9. −2

315

54 2.25−2.5

−4 −3−5 −2 −1 0 21 3 4 5

Practice Problem 9Graph the numbers on the numberline.

−4 −3−5 −2 −1 0 21 3 4 5

-2.5, - 23

, 15

, 54

, 2.25

−3 −2 −1

− −

0 32

74

1

12

12

43

52

RATIONAL NUMBERS

and b are integers and b Z 0 fe ab

` a

1 unit

units2


The decimal representation of an irrational number will neither terminatenor repeat. (For further review of decimals, see Section R.3.)

The set of numbers, each of which corresponds to a point on the numberline, is called the set of real numbers. One and only one point on the num-ber line corresponds to each real number.

Several different sets of numbers have been discussed in this section.The following diagram shows the relationships among these sets of realnumbers. Notice that, together, the rational numbers and the irrationalnumbers make up the real numbers.

REAL NUMBERS

{numbers that correspond to points on the number line}

Practice Problem 10Given the set ,list the numbers in this set that belongto the set of:

a. Natural numbers

b. Whole numbers

c. Integers

d. Rational numbers

e. Irrational numbers

f. Real numbers

E-100, - 25, 0, p, 6, 913F

Answers10. a. 6, 913, b. 0, 6, 913, c. , 0, 6, 913,d. 0, 6, 913, e. , f. all numbersin the given set

p-100, - 25,

-100

Real Numbers

Irrational Numbers Rational Numbers

–18, – 1, 0, √2, π, 47102

–35, –7, 0, 5, 27118

π, √7

Noninteger RationalNumbers

Integers–10, 0, 8

Negative Integers–20, –13, –1

Whole Numbers0, 2, 56, 198

Zero0

Natural Numbers orPositive Integers

1, 16, 170

514

109

1330, ,–

Common Sets of Numbers

Example 10 Given the set , list the numbersin this set that belong to the set of:

a. Natural numbers b. Whole numbersc. Integers d. Rational numberse. Irrational numbers f. Real numbers

Solution: a. The natural numbers are 11 and 112.b. The whole numbers are 0, 11, and 112.c. The integers are , 0, 11, and 112.d. Recall that integers are rational numbers also. The

rational numbers are 11, and 112.e. The irrational number is .f. The real numbers are all numbers in the given set.

12-3, -2, 0, 14,

-3, -2

E-2, 0, 14, 112, -3, 11, 12 F


For example, and since both 3 and are a distanceof 3 units from 0 on the number line.

0 2−2 3−3 1

3 units 3 units

−1

-3@ -3 @ = 3@3 @ = 3

Example 11 Find the absolute value of each number.

a. b. c.

Solution: a. since 4 is 4 units from 0 on the number line.b. since is 5 units from 0 on the number

line.c. since 0 is 0 units from 0 on the number line.

Example 12 Insert , or in the appropriate space to makeeach statement true.

a. 2 b. 5 c.

d. e.

Solution: a. since and .b. .c. since .d. since .e. since .7 7 6@ -7 @ 7 @6 @

5 6 6@5 @ 6 @6 @3 7 2@ -3 @ 7 @ -2 @

@ -5 @ = 50 6 2@0 @ = 0@0 @ 6 2

@6 @@ -7 @@6 @@5 @@ -2 @@ -3 @@ -5 @@0 @

=6 , 7

@0 @ = 0

-5@ -5 @ = 5@4 @ = 4

@0 @@ -5 @@4 @

ABSOLUTE VALUE

The absolute value of a real number a, denoted by , is the distancebetween a and 0 on a number line.

@a @

HELPFUL HINT

Since is a distance, is always either positive or 0, never negative.That is, for any real number a, @a @ ≥ 0.

@a @@a @

FINDING THE ABSOLUTE VALUE OF A NUMBER

The number line not only gives us a picture of the real numbers, it alsohelps us visualize the distance between numbers. The distance between areal number a and 0 is given a special name called the absolute value of a.“The absolute value of a” is written in symbols as .@a @

D

Practice Problem 11Find the absolute value of each num-ber.

a. , b. , c. @ - 23 @@ -8 @@7 @

Practice Problem 12Insert , or in the appropri-ate space to make each statement true.

a. 4, b. ,

c. , d. ,

e. @ -6 @ @ -16 @@6 @ @16 @@ -2.7 @ @ -2 @

-3 @0 @@ -4 @=6 , 7

Answers11. a. 7, b. 8, c. , 12. a. , b. ,c. , d. , e. 667

6=23

Using symbols for mathematical operations is a great convenience. Themore operation symbols presented in an expression, the more careful wemust be when performing the indicated operation. For example, in theexpression , do we add first or multiply first? To eliminate confu-sion, grouping symbols are used. Examples of grouping symbols are paren-theses , brackets , braces , and the fraction bar. If we wish to be simplified by adding first, we enclose in parentheses.

A2 + 3 B # 7 = 5 # 7 = 35

2 + 32 + 3 # 7E FC DA B

2 + 3 # 7

1.2 INTRODUCTION TO VARIABLE EXPRESSIONS

AND EQUATIONS

EXPONENTS AND THE ORDER OF OPERATIONS

Frequently in algebra, products occur that contain repeated multiplicationof the same factor. For example, the volume of a cube whose sides eachmeasure 2 centimeters is cubic centimeters. We may use expo-nential notation to write such products in a more compact form. Forexample.

may be written as 23

The 2 in 23 is called the base; it is the repeated factor. The 3 in 23 is calledthe exponent and is the number of times the base is used as a factor. Theexpression 23 is called an exponential expression.

—— exponentT

base——c 2 is a factor 3 times.

Example 1 Evaluate (find the value of) each expression.

a. [read as “3 squared” or as “3 to the second power”]b. [read as “5 cubed” or as “5 to the third power”]c. [read as “2 to the fourth power”]

d. e.

Solution: a.b.c.d.

e. a 37b 2

= a 37b a 3

7b = 3 # 3

7 # 7= 9

49

71 = 724 = 2 # 2 # 2 # 2 = 1653 = 5 # 5 # 5 = 12532 = 3 # 3 = 9

a 37b 2

71

245332

23 = 2 # 2 # 2 = 8

2cm

Volume is (2 . 2 . 2)cubic centimeters.

2 # 2 # 2

A2 # 2 # 2 B

A

Introduction to Variable Expressions and Equations SECTION 1.2 169

HELPFUL HINT

since indicates repeated multiplication of the same factor.

, whereas 2 # 3 = 623 = 2 # 2 # 2 = 8

2323 Z 2 # 3

Practice Problem 1Evaluate each expression.

a.

b.

c.

d.

e. a 25b 2

91

34

22

42

SSM CD-ROM Video1.2

Objectives

Define and use exponents and theorder of operations.

Evaluate algebraic expressions,given replacement values for vari-ables.

Determine whether a number is asolution of a given equation.

Translate phrases into expressionsand sentences into equations.

D

C

B

A

Answers

1. a. 16, b. 4, c. 81, d. 9, e. 425


Practice Problems 2–4Simplify each expression.

2.

3.

4.

Practice Problem 5Simplify.

1 + @7 - 4 @ + 32

8 - 5

8 C2 16 + 3 2 - 9 D95

# 13

- 13

3 + 2 # 42

If we wish to multiply first, may be enclosed in parentheses.

To eliminate confusion when no grouping symbols are present, we usethe following agreed-upon order of operations.

Using this order of operations, we now simplify . There are nogrouping symbols and no exponents, so we multiply and then add.

Multiply.

Add.

Examples Simplify each expression.

2. Evaluate .

Divide.

Add.

3. Multiply.

The least common denominator is 4.

Subtract.

4.

Multiply.

In the next example, the fraction bar serves as a grouping symbol and sep-arates the numerator and denominator. Simplify each separately.

Example 5 Simplify:

Solution:

Evaluate the exponential expression.

Simplify the numerator. = 83

= 3 + 1 + 43

Find the absolute value and simplifythe denominator. = 3 + 1 + 22

3

Simplify the expression inside theabsolute value bars.

3 + @4 - 3 @ + 22

6 - 3=

3 + @1 @ + 22

6 - 3

3 + @4 - 3 @ + 22

6 - 3

= 54

= 3 C18 D = 3 C28 - 10 D

Simplify the expression in parentheses.They are the innermost grouping symbols.

Multiply 4 and 7.

Subtract inside the brackets.

3 C4 A5 + 2 B - 10 D = 3 C4 A7 B - 10 D = 1

4

= 34

- 24

32

# 12

- 12

= 34

- 12

= 27 = 2 + 25

526 ,3 + 52 = 6 ,3 + 25

= 23

2 + 3 # 7 = 2 + 21

2 + 3 # 7

ORDER OF OPERATIONS

Simplify expressions using the following order. If grouping symbolssuch as parentheses are present, simplify expressions within those first,starting with the innermost set. If fraction bars are present, simplify thenumerator and the denominator separately.

1. Evaluate exponential expressions.2. Perform multiplications or divisions in order from left to right.3. Perform additions or subtractions in order from left to right.

2 + A3 # 7 B = 2 + 21 = 23

3 # 7

Answers

2. 35, 3. , 4. 72, 5. 133

415


Practice Problem 6Evaluate each expression when and .

a.

b.

c.

d. y2 - x2

xy

+ 5y

8x3y

2y - x

y = 4x = 1

EVALUATING ALGEBRAIC EXPRESSIONS

An algebraic expression is a collection of numbers, variables, operationsymbols, and grouping symbols. For example,

2x, , , and

are algebraic expressions. The expression 2x means . Also, means and means . If we give a specific value to a vari-able, we can evaluate an algebraic expression. To evaluate an algebraicexpression means to find its numerical value once we know the value of thevariables.

Algebraic expressions often occur during problem solving. For example,the expression

gives the distance in feet (neglecting air resistance) that an object will fallin t seconds. (See Exercise 63 in this section.)

Example 6 Evaluate each expression when and .

a. b. c. d.

Solution: a. Replace x with 3 and y with 2.

Let and .

Multiply.

Subtract.

b. Let and .

c. Replace x with 3 and y with 2. Then simplify.

d. Replace x with 3 and y with 2.

x2 - y2 = 32 - 22 = 9 - 4 = 5

xy

+ y2

= 32

+ 22

= 52

y = 2x = 33x2y

= 3 # 32 # 2

= 94

= 4

= 6 - 2

y = 2x = 32x - y = 2 13 2 - 2

x2 - y2xy

+ y2

3x2y

2x - y

y = 2x = 3

16t2

3 # y23y25 # 1p2 + 1 2 5 Ap2 + 1 B2 # x

3y2 - 6y + 15

5 Ap2 + 1 B2x - 10-3,

B

Answers

6. a. 7, b. , c. , d. 1532

23

HELPFUL HINTBe careful when evaluating an exponential expression.

c cBase is 4. Base is .3 # 4

A3 # 4 B 2 = A12 B 2 = 1443 # 42 = 3 # 16 = 48

SOLUTIONS OF EQUATIONS

Many times a problem-solving situation is modeled by an equation. Anequation is a mathematical statement that two expressions have equalvalue. The equal symbol “=” is used to equate the two expressions. For

example, , , , and are all equa-

tions.

I = PRT2 Ax - 1 B

3= 07x = 353 + 2 = 5

C


HELPFUL HINTAn equation contains the equal symbol “=”. An algebraic expressiondoes not.

✓ CONCEPT CHECK

Which of the following are equations?Which are expressions?

a.b.c.d. 12y = 3x

12y + 3x5x - 85x = 8

Practice Problem 7Decide whether 3 is a solution of

.5x - 10 = x + 2

TRY THE CONCEPT CHECK IN THE MARGIN.When an equation contains a variable, deciding which values of the vari-able make an equation a true statement is called solving an equation for thevariable. A solution of an equation is a value for the variable that makesthe equation true. For example, 3 is a solution of the equation ,because if x is replaced with 3 the statement is true.

TReplace x with 3.

True.

Similarly, 1 is not a solution of the equation , because is not a true statement.

Example 7 Decide whether 2 is a solution of .

Solution: Replace with and see if a true statement results.

Original equation

Replace x with 2.

Simplify each side.

True.

Since we arrived at a true statement after replacing xwith 2 and simplifying both sides of the equation, 2 is asolution of the equation.

TRANSLATING WORDS TO SYMBOLS

Now that we know how to represent an unknown number by a variable,let’s practice translating phrases into algebraic expressions and sentencesinto equations. Oftentimes solving problems involves the ability to trans-late word phrases and sentences into symbols. Below is a list of key wordsand phrases to help us translate.

Addition Subtraction Multiplication Division Equality

Sum Difference of Product Quotient EqualsPlus Minus Times Divide GivesAdded to Subtracted Multiply Into Is/was/

from should beMore than Less than Twice Ratio YieldsIncreased by Decreased by Of Divided by Amounts toTotal Less Represents

Is the same as

A = BA,BA # BA- BA+ B

D

16 = 166 + 10 � 16

3 12 2 + 10 � 8 12 23x + 10 = 8x

2x

3x + 10 = 8x

1 + 4 = 7x + 4 = 7

7 = 73 + 4 � 7

x + 4 = 7

x + 4 = 7

Answers

7. It is a solution.

✓ Concept Checkequations: a, d; expressions: b, c

Now let’s practice translating sentences into equations.

Example 9 Write each sentence as an equation. Let x represent theunknown number.

a. The quotient of 15 and a number is 4.b. Three subtracted from 12 is a number.c. Four times a number added to 17 is 21.

Solution: a. In words:

T T T

Translate: 4

b. In words:

T T TTranslate:

Care must be taken when the operation is subtraction.The expression would be incorrect. Notice that

.

c. In words:

T T T T T

Translate: 21=17+4x

21is17added tofour times a number

3 - 12 Z 12 - 33 - 12

x=12 - 3

a numberisthree subtracted from 12

=15x

4isthe quotient of 15

and a number

Example 8 Write an algebraic expression that represents each phrase.Let the variable x represent the unknown number.

a. The sum of a number and 3b. The product of 3 and a numberc. Twice a numberd. 10 decreased by a numbere. 5 times a number increased by 7

Solution: a. since “sum” means to addb. 3 and 3x are both ways to denote the product of 3

and xc. or 2xd. 10 because “decreased by” means to subtracte.

5 times a number

5x + 7- x

2 # x

# xx + 3


HELPFUL HINTMake sure you understand the difference when translating phrases con-taining “decreased by,” “subtracted from” and “less than.”

Phrase TranslationA number decreased by 10A number subtracted from 10 v10 less than a number x - 10

Notice the order.10 - x

x - 10

Practice Problem 8Write an algebraic expression that rep-resents each phrase. Let the variable xrepresent the unknown number.

a. The product of a number and 5

b. A number added to 7

c. Three times a number

d. A number subtracted from 8

e. Twice a number plus 1

Answers8. a. , b. , c. , d. ,e. , 9. a. , b. ,c. 2x - 1 = 99

10 - x = 186x = 242x + 18 - x3x7 + x5x

Practice Problem 9Write each sentence as an equation.Let x represent the unknown number.

a. The product of a number and 6 is24.

b. The difference of 10 and a numberis 18.

c. Twice a number decreased by 1 is99.


CALCULATOR EXPLORATIONSEXPONENTS

To evaluate exponential expressions on a calculator, find the key

marked or . To evaluate, for example, , press the

following keys: or . D or

The display should read .

ORDER OF OPERATIONS

Some calculators follow the order of operations, and others donot. To see whether or not your calculator has the order of opera-tions built in, use your calculator to find . To do this,press the following sequence of keys:

.D or

The correct answer is 14 because the order of operations is to

multiply before we add. If the calculator displays ,

then it has the order of operations built in.

Even if the order of operations is built in, parentheses must

sometimes be inserted. For example, to simplify , press the

keys

.D or

The display should read

Use a calculator to evaluate each expression.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10. 956 - 45289 - 86

4623 + 12936 - 34

99 + 1401 + 962 224 1862 - 455 2 + 89

3 114 - 7 2 + 212 120 - 5 28695

7453

1

ENTER

=)7-21(,5

512 - 7

14

ENTER

=4*3+2

2 + 3 # 4

243

ENTER

=5^3=5yx3

35^yx

175


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

EXERCISE SET 1.2

Evaluate each expression when , , and . See Example 6.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12. 2z25y2yz - x

xy + z@5z - 2y @@2x + 3y @

6y - 83x - 2y2z

z5x

4x3y

z = 5y = 3x = 1B

1.3 ADDING REAL NUMBERS

Real numbers can be added, subtracted, multiplied, divided, and raised topowers, just as whole numbers can.

ADDING REAL NUMBERS

To begin, we will use the number line to help picture the addition of realnumbers.

Example 1 Add:

Solution: We start at 0 on a number line, and draw an arrow repre-senting 3. This arrow is three units long and points to theright since 3 is positive. From the tip of this arrow, we drawanother arrow representing 2. The number below the tipof this arrow is the sum, 5.

Example 2 Add:

Solution: We start at 0 on a number line, and draw an arrow repre-senting . This arrow is one unit long and points to theleft since is negative. From the tip of this arrow, wedraw another arrow representing . The number belowthe tip of this arrow is the sum, .

Thinking of integers as money earned or lost might help make additionmore meaningful. Earnings can be thought of as positive numbers. If $1 isearned and later another $3 is earned, the total amount earned is $4. Inother words, .

On the other hand, losses can be thought of as negative numbers. If $1 islost and later another $3 is lost, a total of $4 is lost. In other words,

.In Examples 1 and 2, we added numbers with the same sign. Adding

numbers whose signs are not the same can be pictured on a number linealso.

Example 3 Add:

Solution:

−1 0 1 2 3 4−4−5 −3 −2

StartEnd

−4 + 6 = 2

− 46

-4 + 6

A-1 B + A-3 B = -4

1 + 3 = 4

−1 0 1 2 3 4 5−5 −4 −3 −2

StartEnd−2 −1

−1 + (−2) = −3

-3-2

-1-1

-1 + A-2 B

−1 0 1 2 3 4 5−5 −4 −3 −2

Start End3 2

3 + 2 = 5

3 + 2

A

Adding Real Numbers SECTION 1.3 177

Objectives

Add real numbers.

Solve problems that involve addi-tion of real numbers.

Find the opposite of a number.C

BA

Practice Problem 1Add using a number line:

−4 −3−5 −2 −1 0 21 3 4 5

1 + 5

SSM CD-ROM Video1.3


−4 −3−5 −2 −1 0 21 3 4 5

-2 + A-4 B

Answers1. 6, 2. , 3. 3-6


−4 −3−5 −2 −1 0 21 3 4 5

-5 + 8



−4 −3−5 −2 −1 0 21 3 4 5

5 + A-4 B

Practice Problem 5Add without using a number line:A-8 B + A-5 BPractice Problem 6Add without using a number line:A-14 B + 6

Practice Problems 7–12Add without using a number line.

7.

8.

9.

10.

11.

12. - 45

+ 23

12.8 + A-3.6 B- 6

11+ a - 3

11b

1.5 + A-3.2 BA-4 B + 12

A-17 B + A-10 B

Using temperature as an example, if the thermometer registers 4 degreesbelow 0 degrees and then rises 6 degrees, the new temperature is 2 degreesabove 0 degrees. Thus, it is reasonable that .

Example 4 Add:

Solution:

Using a number line each time we add two numbers can be time con-suming. Instead, we can notice patterns in the previous examples and writerules for adding real numbers.

Example 5 Add without using a number line:

Solution: Here, we are adding two numbers with the same sign.

c a sum of absolute valuesƒ same sign

Example 6 Add without using a number line:

Solution: Here, we are adding two numbers with different signs.

c a difference of absolute valuesƒ sign of number with larger absolute value,

Examples Add without using a number line.

7.8.9.

10.

11.

12.

Example 13 Find each sum.

a.b. C7 + A-10 B D + C-2 + A-4 B D

3 + A-7 B + A-8 B

- 38

+ 25

= - 1540

+ 1640

= 140

11.4 + A-4.7 B = 6.7

- 710

+ a - 110b = - 8

10= - 4

5

0.2 + A-0.5 B = -0.3

A-2 B + 10 = 8

A-8 B + A-11 B = -19

A-10 BA-10 B + 4 = -6

A-10 B + 4

A-7 B + A-6 B = -13

A-7 B + A-6 B

ADDING REAL NUMBERS

To add two real numbers

1. with the same sign, add their absolute values. Use their common signas the sign of the answer.

2. with different signs, subtract their absolute values. Give the answerthe same sign as the number with the larger absolute value.

−1 0 1 2 3 4−4−5 −3 −2

StartEnd

4 + (−6) = −2

4− 6

4 + A-6 B-4 + 6 = 2

Answers4. 1, 5. , 6. , 7. , 8. 8,

9. , 10. , 11. 9.2, 12. ,

13. a. , b. -21-2

- 215

- 911-1.7

-27-8-13

Practice Problem 13Find each sum.

a.

b. C3 + A-13 B D + C-4 + A-7 B D16 + A-9 B + A-9 B

Rises 6Degrees

2°0°

–4°

Adding Real Numbers SECTION 1.3 179

Answer

14. a gain of $3

Solution: a. Perform the additions from left to right.

b. Simplify inside the brackets first.

Add.

SOLVING PROBLEMS THAT INVOLVE ADDITION

Positive and negative numbers are used in everyday life. Stock marketreturns show gains and losses as positive and negative numbers. Tempera-tures in cold climates often dip into the negative range, commonly referredto as “below zero” temperatures. Bank statements report deposits andwithdrawals as positive and negative numbers.

Example 14 Calculating Gain or LossDuring a three-day period, a share of Lamplighter’s Inter-national stock recorded the following gains and losses:

Monday Tuesday Wednesdaya gain of $2 a loss of $1 a loss of $3

Find the overall gain or loss for the stock for the threedays.

Solution: Gains can be represented by positive numbers. Losses canbe represented by negative numbers. The overall gain orloss is the sum of the gains and losses.

In words:

T T T T T

Translate: 2

The overall loss is $2.

FINDING OPPOSITES

To help us subtract real numbers in the next section, we first review whatwe mean by opposites. To help us, the graphs of 4 and are shown on thenumber line below.

Notice that the graph of 4 and lie on opposite sides of 0, and each is 4units away from 0. Such numbers are known as opposites or additiveinverses of each other.

-4

−1 0 1 2 3 4 5−5 −4 −3 −2

4 units4 units

-4

C

= -2A-3 B+A-1 B+

losspluslossplusgain

B

= -9

C7 + A-10 B D + C-2 + A-4 B D = C-3 D + C-6 D

HELPFUL HINT:Don’t forgetthat bracketsare groupingsymbols. We simplifywithin themfirst.

= -12

Adding numberswith different signsAdding numberswith like signs

3 + A-7 B + A-8 B = -4 + A-8 B

Practice Problem 14During a four-day period, a share ofWalco stock recorded the followinggains and losses:

Tuesday Wednesdaya loss of $2 a loss of $1

Thursday Fridaya gain of $3 a gain of $3

Find the overall gain or loss for thestock for the four days.


The sum of a number a and its opposite is 0.

a + A-a B = 0

-a

Practice Problems 15–18Find the opposite of each number.

15.

16. 12

17.

18. 1.9

- 311

-35

Practice Problem 19Simplify each expression.

a.

b.

c.

d. - @ -14 @- A-x B- a - 2

7b

- A-22 B

Examples Find the opposite of each number.

15. 10 The opposite of 10 is .

16. The opposite of is 3.

17. The opposite of is .

18. The opposite of is 4.5.

We use the symbol “ ” to represent the phrase “the opposite of” or“the additive inverse of.” In general, if a is a number, we write the oppositeor additive inverse of a as . We know that the opposite of is 3. Noticethat this translates as

T T T T3

This is true in general.

Example 19 Simplify each expression.

a. b.

c. d.

Solution: a. b.

c.

d. Since , then .ƒ——————c

Let’s discover another characteristic about opposites. Notice that thesum of a number and its opposite is 0.

In general, we can write the following:

Notice that this means that the opposite of 0 is then 0 since .0 + 0 = 0

12

+ a - 12b = 0

-3 + 3 = 0

10 + A-10 B = 0

- @ -6 @ = -6@ -6 @ = 6

- A-2x B = 2x

- a- 12b = 1

2- A-10 B = 10

- @ -6 @- A-2x B- a- 1

2b- A-10 B

=A-3 B-

3is-3the opposite of

-3-a

-

-4.5-4.5

- 12

12

12

-3-3-10

OPPOSITE OR ADDITIVE INVERSE

Two numbers that are the same distance from 0 but lie on oppositesides of 0 are called opposites or additive inverses of each other.

Answers

15. 35, 16. , 17. , 18. ,

19. a. 22, b. , c. , d. -14x27

-1.9311

-12

If a is a number, then .- A-a B = a

181


EXERCISE SET 1.3

Add. See Examples 1 through 13.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. 23 + A-23 B-16 + 16

3 + A-6 B5 + A-7 B8 + A-6 B

10 + A-3 B-5 + 9-7 + 3

7 + A-5 B-9 + A-3 B-7 + A-4 B

-2 + A-3 B-10 + 5-14 + 2

6 + A-4 B8 + A-7 B-6 + A-14 B

-6 + A-8 B9 + A-12 B6 + 3A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

1.4 SUBTRACTING REAL NUMBERS

SUBTRACTING REAL NUMBERS

Now that addition of real numbers has been discussed, we can explore sub-traction. We know that . Notice that , also. Thismeans that

Notice that the difference of 9 and 7 is the same as the sum of 9 and theopposite of 7. This is how we can subtract real numbers.

9 - 7 = 9 + A-7 B9 + A-7 B = 29 - 7 = 2

A

Subtracting Real Numbers SECTION 1.4 183

SUBTRACTING REAL NUMBERS

If a and b are real numbers, then .a - b = a + A-b BSSM CD-ROM Video

1.4In other words, to find the difference of two numbers, we add the oppositeof the number being subtracted.

Example 1 Subtract.

a. b. c. d.

Solution:ƒ——— add ———T

a. Add to the opposite of 4, ƒ——————————c which is .

opposite

ƒ——— add ———Tb. Add 5 to the opposite of ,

ƒ————————c which is 6.opposite

c. Add 3 to the opposite of 6, which is .

d. -1 - A-7 B = -1 + A7 B = 6

= -3-63 - 6 = 3 + A-6 B

= 11

-65 - A-6 B = 5 + A6 B= -17

-4-13-13 - 4 = -13 + A-4 B

-1 - A-7 B3 - 65 - A-6 B-13 - 4

Objectives

Subtract real numbers.

Evaluate algebraic expressions us-ing real numbers.


Solve problems that involve sub-traction of real numbers.

Find complementary and supple-mentary angles.

E

D

C

BA

Practice Problem 1Subtract.

a.

b.

c.

d. -4 - A-9 B7 - 17

3 - A-5 B-20 - 6

Practice Problems 2–4Subtract.

2. 3.

4. - 14

- a - 25b

- 49

- 29

9.6 - A-5.7 BExamples Subtract.

2.

3.

4. - 23

- a - 45b = - 2

3+ a 4

5b = - 10

15+ 12

15= 2

15

- 310

- 510

= - 310

+ a - 510b = - 8

10= - 4

5

5.3 - A-4.6 B = 5.3 + A4.6 B = 9.9

Answers1. a. , b. 8, c. , d. 5,

2. 15.3, 3. , 4. 320

- 23

-10-26

HELPFUL HINT

Study the patterns indicated.

No ƒ——Change to addition.change—T T T—Change to opposite.

7 - A-1 B = 7 + A1 B = 8

-3 - 4 = -3 + A-4 B = -7

5 - 11 = 5 + A-11 B = -6


Practice Problem 5Subtract 7 from .-11


a.

b. 5.2 - A-4.4 B + A-8.8 B-20 - 5 + 12 - A-3 B


a.

b. 52 - 20 + C-11 - A-3 B D-9 + C A-4 - 1 B - 10 D

Example 5 Subtract 8 from .

Solution: Be careful when interpreting this: The order of numbers insubtraction is important. 8 is to be subtracted from .

If an expression contains additions and subtractions, just write the sub-tractions as equivalent additions. Then simplify from left to right.


a.

b.

Solution: a.

b.

When an expression contains parentheses and brackets, remember theorder of operations. Start with the innermost set of parentheses or bracketsand work your way outward.


a.

b.

Solution: a. Start with the innermost set of parentheses. Rewriteas an addition.

Add: .Write as anaddition.Add.

Add.

b. Start simplifying the expression inside the brackets bywriting as an addition.

Add.

Evaluate .

Write as an addition.

Add.

Add.


It is important to be able to evaluate expressions for given replacement val-ues. This helps, for example, when checking solutions of equations.

B

= -3

= -2 + A-1 B8 - 10 = 8 + A-10 B + A-1 B

23 = 8 - 10 + A-1 B = 23 - 10 + C-1 D

23 - 10 + C-6 - A-5 B D = 23 - 10 + C-6 + 5 D-6 - A-5 B

= -12

= -3 + C-9 D-7 - 2 = -3 + C-7 + A-2 B D

-2 + A-5 B = -3 + C A-7 B - 2 D-3 + C A-2 - 5 B - 2 D = -3 + C A-2 + A-5 B B - 2 D

-2 - 5

23 - 10 + C-6 - A-5 B D-3 + C A-2 - 5 B - 2 D

1.6 + 10.3 + A-5.6 B = 6.31.6 - A-10.3 B + A-5.6 B =

-14 + A-8 B + 10 + 6 = -6

-14 - 8 + 10 - A-6 B =

1.6 - A-10.3 B + A-5.6 B-14 - 8 + 10 - A-6 B

-4 - 8 = -4 + A-8 B = -12

-4

-4

Answers

5. , 6. a. , b. 0.8,7. a. , b. -3-24

-10-18

Subtracting Real Numbers SECTION 1.4 185

Practice Problem 8Find the value of each expressionwhen and .

a.

b. x2 - y

x - y14 + x

y = -4x = 1

Practice Problem 9Determine whether is a solution of

.-1 + x = 1-2

Answers

8. a. , b. 5, 9. is not a solution,

10. °-37

-213

Example 8 Find the value of each expression when and.

a. b.

Solution: a. Replace x with 2 and y with . Be sure to put paren-theses around to separate signs. Then simplify theresulting expression.

b. Replace the x with 2 and y with and simplify.


Recall from Section 1.2 that a solution of an equation is a value for the vari-able that makes the equation true.

Example 9 Determine whether is a solution of .

Solution: Replace x with and see if a true statement results.

Original equation

Replace x with .

True.

Thus is a solution of .

SOLVING PROBLEMS THAT INVOLVE SUBTRACTION

Another use of real numbers is in recording altitudes above and below sealevel, as shown in the next example.

Example 10 Finding the Difference in ElevationsThe lowest point in North America is in Death Valley, atan elevation of 282 feet below sea level. Nearby, MountWhitney reaches 14,494 feet, the highest point in theUnited States outside Alaska. How much of a variation inelevation is there between these two extremes? (Source:U.S. Geological Survey)

Solution: To find the variation in elevation between the two heights,find the difference of the high point and the low point.

14,494 feetabove sea

level(+14,494)

Sea level(0)

282 feet belowsea level(−282)

DeathValley

Mt. Whitney

D

x - 5 = -9-4

-9 = -9-4 + A-5 B � -9

-4 -4 - 5 � - 9 x - 5 = -9

-4

x - 5 = -9-4

C

x2 - y = 22 - A-5 B = 4 - A-5 B = 4 + 5 = 9

-5

x - y12 + x

=2 - A-5 B

12 + 2= 2 + 5

14= 7

14= 1

2

-5-5

x2 - yx - y12 + x

y = -5x = 2

Practice Problem 10At 6.00 P.M., the temperature at theWinter Olympics was 14°; by morningthe temperature dropped to °.Find the overall change in tempera-ture.

-23


COMPLEMENTARY AND SUPPLEMENTARY ANGLES

Two angles are complementary if their sum is 90°.

Two angles are supplementary if their sum is 180°.

x + y = 180°x y

x + y = 90°

x y

In words:

T T TTranslate: 14,494

feet

Thus, the variation in elevation is 14,776 feet.

FINDING COMPLEMENTARY AND SUPPLEMENTARY ANGLES

A knowledge of geometric concepts is needed by many professionals, suchas doctors, carpenters, electronic technicians, gardeners, machinists, andpilots, just to name a few. With this in mind, we review the geometric con-cepts of complementary and supplementary angles.

E

= 14,776

= 14,494 + 282A-282 B-

low pointminushigh point

Practice Problem 11Find each unknown complementary orsupplementary angle.

a.

b.

81°

y

78°x

Example 11 Find each unknown complementary or supplementaryangle.

a. b.

Solution: a. These angles are complementary, so their sum is 90°.This means that x is 90° 38°.

b. These angles are supplementary, so their sum is 180°.This means that y is 180° 62°.

y = 180° - 62° = 118°

-

x = 90° - 38° = 52°

-

62° y38°x

Answer11. a. 102°, b. 9°

187


EXERCISE SET 1.4

Subtract. See Examples 1 through 4.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. 8.3 - 11.29.7 - 16.1

15 - A-33 B16 - A-21 B-4 - A-16 B

-6 - A-11 B3 - A-6 B7 - A-4 B

-8 - 4-6 - 5-20 - A-48 B

-16 - A-18 B34

- 78

12

- 13

12 - A-5 B16 - A-3 B8 - 11

4 - 9-12 - 8-6 - 4A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

188

Focus On Study SkillsSTUDY TIPS

Have you wondered what you can do to be successful in your algebra course? If so, that maywell be your first step to success in algebra! Here are some tips on how to use this text and howto study mathematics in general.

Using this Text1. Each example in the section has a parallel Practice Problem. As you read a section, try

each Practice Problem after you’ve finished the corresponding example. This “learn-by-doing” approach will help you grasp ideas before you move on to other concepts.

2. The main section of exercises in an exercise set are referenced by an objective, such as or and also an example(s). Use this referencing in case you have trouble complet-ing an assignment from the exercise set.

3. If you need extra help in a particular section, check at the beginning of the section to seewhat videotapes and software are available.

4. Integrated Reviews in each chapter offer you a chance to practice—in one place—themany concepts that you have learned separately over several sections.

5. There are many opportunities at the end of each chapter to help you understand the con-cepts of the chapter.

Chapter Highlights contain chapter summaries with examples.Chapter Reviews contain review problems organized by section.Chapter Tests are sample tests to help you prepare for an exam.Cumulative Reviews are reviews consisting of material from the beginning of the book to

the end of the particular chapter.

General Tips1. Choose to attend all class periods. If possible, sit near the front of the classroom. This way,

you will see and hear the presentation better. It may also be easier for you to participatein classroom activities.

2. Do your homework. You’ve probably heard the phrase “practice makes perfect” in rela-tion to music and sports. It also applies to mathematics. You will find that the more timeyou spend solving mathematics problems, the easier the process becomes. Be sure to blockout enough time to complete your assignments.

3. Check your work. Review the steps you made while working a problem. Learn to checkyour answers in the original problems. You can also compare your answers to the answersto selected exercises listed in the back of the book. If you have made a mistake, figure outwhat went wrong. Then correct your mistake.

4. Learn from your mistakes. Everyone, even your instructors, makes mistakes. You can useyour mistakes to become a better math student. The key is finding and understanding yourmistakes. Was your mistake a careless mistake? If so, you can try to work more slowly andmake a conscious effort to carefully check your work. Did you make a mistake becauseyou don’t understand a concept? If so, take the time to review the concept or ask ques-tions to better understand the concept.

5. Know how to get help if you need it. It’s OK to ask for help. In fact, it’s a good idea to askfor help whenever there is something that you don’t understand. Make sure you knowwhen your instructor has office hours and how to find his or her office. Find out if mathtutoring services are available on your campus. Check out the hours, location, andrequirements of the tutoring service. You might also want to find another student in yourclass that you can call to discuss your assignment.

6. Organize your class materials, including homework assignments, graded quizzes and tests,and notes from your class or lab. All of these items will make valuable references through-out your course and as you study for upcoming tests and your final exam. Make sure youcan locate any of these materials when you need them.

BA

Multiplying Real Numbers SECTION 1.5 189

1.5 MULTIPLYING REAL NUMBERS

MULTIPLYING REAL NUMBERS

Multiplication of real numbers is similar to multiplication of whole num-bers. We just need to determine when the answer is positive, when it is neg-ative, and when it is zero. To discover sign patterns for multiplication, recallthat multiplication is repeated addition. For example, 3(2) means that 2 isadded to itself three times, or

Also,

Since , this suggests that the product of a positive number anda negative number is a negative number.

What about the product of two negative numbers? To find out, considerthe following pattern.

Factor decreases by 1 each time.T———————

Product increases by 3 each time

This suggests that the product of two negative numbers is a positive num-ber. Our results are given below.

Examples Multiply.

1.2.3.

4.

5.6.

We already know that the product of 0 and any whole number is 0. This istrue of all real numbers.

-18 A-3 B = 54

5 A-1.7 B = -8.5

- 23

# 47

= - 2 # 43 # 7

= - 821

-5 A-10 B = 50

2 A-10 B = -20

-6 A4 B = -24


1. The product of two numbers with the same sign is a positive number.2. The product of two numbers with different signs is a negative number.

-3 # -2 = 6

-3 # -1 = 3

-3 # 0 = 0

-3 # 1 = -3

-3 # 2 = -6

3 A-2 B = -6

3 A-2 B = A-2 B + A-2 B + A-2 B = -6

3 A2 B = 2 + 2 + 2 = 6

A

Practice Problems 1–6Multiply.

1. 2.

3. 4.

5. 6. -15 A-2 B6 A-2.3 B- 5

6# 14

-4 A-12 B5 A-30 B-8 A3 B

PRODUCTS INVOLVING ZERO

If b is a real number, then . Also .0 # b = 0b # 0 = 0

SSM CD-ROM Video1.5

Objectives

Multiply real numbers.



C

BA

Answers

1. , 2. , 3. 48, 4. ,

5. , 6. 30-13.8

- 524

-150-24


HELPFUL HINT

You may have noticed from theexample that if we multiply:

j an even number of negative num-bers, the product is positive.

j an odd number of negative num-bers, the product is negative.

Example 7 Use the order of operations and simplify each expression.

a. b. c.d. e.

Solution: a. By the order of operations, we multiply from left toright. Notice that because one of the factors is 0, theproduct is 0.

b. Multiply two factors at a time, from left to right.

Multiply .

c. Multiply from left to right.

Multiply .

d. The exponent 3 means 3 factors of the base .

Multiply.

e. Follow the order of operations.

Find the products.

Add 44 to the opposite of .

Add.


Now that we know how to multiply positive and negative numbers, we con-tinue to practice evaluating algebraic expressions.

Example 8 Evaluate each expression when and .

a. b.

Solution: a. Replace x with and y with and simplify.

b. Replace x with and y with .

Evaluate and .

Write as a sum.

Add.


To prepare for solving equations, we continue to check possible solutionsfor an equation.


Solution: Replace x with and see if a true statement results.

Original equation

Replace x with .

Multiply.

True.

Thus, is a solution of -7x + 2 = 23.-3

23 = 2321 + 2 � 23

-3-7 1-3 2 + 2 � 23 -7x + 2 = 23

-3

-7x + 2 = 23-3

C

= -24

= -8 + A-16 BA-4 2 2A-2 B 3 = -8 - A16 B

Substitute the given values forthe variables.x3 - y2 = A-2 B 3 - A-4 B 2-4-2

5x - y = 5 A-2 B - A-4 B = -10 - A-4 B = -10 + 4 = -6

-4-2

x3 - y25x - y

y = -4x = -2

B

= 54

-10 = 44 + 10

A-4 B A-11 B - 5 A-2 B = 44 - A-10 B

= -8

A-2 B 3 = A-2 B A-2 B A-2 B-2

= 45

A-1 B A5 BA-1 B A5 B A-9 B = A-5 B A-9 B

= -24

A-2 B A-3 BA-2 B A-3 B A-4 B = A6 B A-4 B

7 A0 B A-6 B = 0 A-6 B = 0

A-4 B A-11 B - 5 A-2 BA-2 B 3A-1 B A5 B A-9 BA-2 B A-3 B A-4 B7 A0 B A-6 B

Practice Problem 7Use the order of operations and sim-plify each expression.

a. b.

c. d.

e. -3 A-9 B - 4 A-4 BA-2 B 2A-2 B A4 B A-8 BA-1 B A-6 B A-7 B5 A0 B A-3 B

Practice Problem 8Evaluate each expression when

and .

a. b. x2 - y33x - y

y = -5x = -1

Practice Problem 9Determine whether is a solutionof .3x + 4 = -26

-10

Answers

7. a. 0, b. , c. 64, d. 4, e. 43,

8. a. 2 b. 126, 9. is a solution.-10

-42

191


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

EXERCISE SET 1.5

Multiply. See Examples 1 through 6.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. -0.5 A-0.3 B-0.2 A-0.7 B

6 A-2.5 B5 A-1.4 B- 56a- 3

10b

- 34a- 8

9b- 1

8a- 1

3b- 1

2a- 3

5b

3 A-5 B2 A-9 B-6 A-9 B

-6 A-7 B-2 # 8-3 # 4

-6 A-11 B-5 A-10 B7 A-4 B

2 A-1 B-8 A5 B-6 A4 BA

192

Focus On HistoryMULTICULTURALISM

Numbers have a long history. The numbers we are accustomed to using probably originated inIndia in the 3rd century and were later adapted by Arabic cultures. Many other ancient civiliza-tions developed their own unique number systems.

Roman NumeralsI V X L C D M1 5 10 50 100 500 1000

If numerals decrease in value from left to right, the values are added. If a smaller numeralappears to the left of a larger numeral, the smaller value is subtracted. For example:

but

Chinese Numerals

Numerals are written vertically. If a digit representing 2–9 appears before a digit representing10, 100, 1000, 10,000, or 100,000, multiplication is indicated.

is

Egyptian Hieroglyphic Numerals

1 10 100 1000 10,000 100,000

The Egyptian system is also multiplicative. For example, 3 hieroglyphs represents .

GROUP ACTIVITIES

j Write several numbers using each of the Roman, Chinese, and Egyptian hieroglyphic sys-tems. Trade your numbers with another student in your group to translate into our numer-als. Check one another’s work.

j Research the number system of another ancient culture (such as Babylonian, MayanIndian, or Ionic Greek). Write the numbers 712, 4690, 5113, and 208 using that system.Demonstrate the system to the rest of your group.

= 12,524

= 4 * 1 + 2 * 10 + 5 * 100 + 2 * 1000 + 1 * 10,000

3 * 10

7000 + 300 + 80 + 5 = 7385f5

f A8 * 10 B

f A3 * 100 Bf A7 * 1000 B

1 2 3 4 5 6 7 8 9 10 100 1000 10,000 100,000

XLIV = 50 - 10 + 5 - 1 = 44XVII = 10 + 5 + 1 + 1 = 17

Dividing Real Numbers SECTION 1.6 193

1.6 DIVIDING REAL NUMBERS

FINDING RECIPROCALS

Addition and subtraction are related. Every difference of two numberscan be written as the sum . Multiplication and division are

related also. For example, the quotient can be written as the prod-

uct . Recall that the pair of numbers 3 and has a special relationship.

Their product is 1 and they are called reciprocals or multiplicative inversesof each other.

Examples Find the reciprocal of each number.

1. 22 The reciprocal of 22 is since .

2. The reciprocal of is since .



5. The reciprocal of 1.7 is since .

Does the number 0 have a reciprocal? If it does, it is a number n suchthat . Notice that this can never be true since . This meansthat 0 has no reciprocal.

DIVIDING REAL NUMBERS

We may now write a quotient as an equivalent product.

In other words, the quotient of two real numbers is the product of the firstnumber and the multiplicative inverse or reciprocal of the second number.

Example 6 Use the definition of the quotient of two numbers to findeach quotient.

a. b. c. 20-4

-14-2

-18 ,3

B

QUOTIENTS INVOLVING ZERO

The number 0 does not have a reciprocal.

0 # n = 00 # n = 1

1.7 # 11.7

= 11

1.71.7

- 913

# - 139

= 1-139

- 913

- 913

-10 # - 110

= 1- 110

-10-10

316

# 163

= 1163

316

316

22 # 122

= 11

22

RECIPROCAL OR MULTIPLICATIVE INVERSE

Two numbers whose product is 1 are called reciprocals or multiplicativeinverses of each other.

13

6 # 13

6 ,3a + A-b Ba - b

A

Practice Problems 1–5Find the reciprocal of each number.

1. 13 2. 3.

4. 5. 7.9- 811

-5715

QUOTIENT OF TWO REAL NUMBERS

If a and b are real numbers and b is not 0, then

a ,b = ab

= a # 1b

Answers

1. , 2. , 3. , 4. , 5. ,

6. a. , b. 2, c. -9-3

17.9

- 118

- 15

157

113

SSM CD-ROM Video1.6

Objectives

Find the reciprocal of a realnumber.

Divide real numbers.



D

CB

A

Practice Problem 6Use the definition of the quotient oftwo numbers to find each quotient.

a. b.

c.36-4

-20-10

-12 ,4


Solution: a.

b.

c.

Since the quotient can be written as the product , it follows that

sign patterns for dividing two real numbers are the same as sign patterns formultiplying two real numbers.

Examples Divide.

7. Same sign, so the quotient is positive.

8.

9. ∂ Unlike signs, so the quotient is negative.

10.0.6 ƒ

� �

In the examples above, we divided mentally or by long division. When wedivide by a fraction, it is usually easier to multiply by its reciprocal.

Examples Divide.

11.

12.

Our definition of the quotient of two real numbers does not allow for divi-sion by 0 because 0 does not have a reciprocal. How then do we interpret

? We say that an expression such as this one is undefined. Can we divide

0 by a number other than 0? Yes; for example,

03

= 0 # 13

= 0

30

- 16

, A- 23B = - 1

6# A- 3

2B = 3

12= 1

4

23

, A- 54B = 2

3# A- 4

5B = - 8

15

70.42.0

42-0.6

= -70

20-2

= -10

-1005

= -20

-30-10

= 3


1. The quotient of two numbers with the same sign is a positivenumber.

2. The quotient of two numbers with different signs is a negativenumber.

a # 1b

a ,b

20-4

= 20 # - 14

= -5

-14-2

= -14 # - 12

= 7

= -613

-18 ,3 = -18 #

DIVISION INVOLVING ZERO

Division by 0 is undefined. For example, is undefined.

0 divided by a nonzero number is 0. For example, .0

-5= 0

-50

Practice Problems 7–10Divide.

7. 8.

9. 10.-720.2

50-2

-48-6

-255

Practice Problems 11–12Divide.

11.

12. - 27

, A- 15B

- 59

,23

Answers

7. , 8. 8, 9. , 10. , 11. ,

12. 107

- 56

-360-25-5

Examples Perform each indicated operation.

13. is undefined. 14.

15.

Notice that , , and . This means that

In other words, a single negative sign in a fraction can be written in thedenominator, in the numerator, or in front of the fraction without changingthe value of the fraction.

Examples combining basic arithmetic operations along with the princi-ples of the order of operations help us to review these concepts.


a.

b.

Solution: a. First, simplify the numerator and denominator sepa-rately, then divide.

Divide.

b. Simplify the numerator and denominator separately,then divide.


Using what we have learned about dividing real numbers, we continue topractice evaluating algebraic expressions.

Example 17 Evaluate when and .

Solution: Replace x with and y with and simplify.

3x2y

=3 A-2 B2 A-4 B = -6

-8= 3

4

-4-2

y = -4x = -23x2y

C

2 1-3 2 2 - 20-5 + 4

= 2 # 9 - 20-5 + 4

= 18 - 20-5 + 4

= -2-1

= 2

= -8

= 40-5

A-12 B A-3 B + 4-7 - A-2 B = 36 + 4

-7 + 2

2 A-3 B 2 - 20-5 + 4

A-12 B A-3 B + 4

-7 - A-2 B

12-2

= - 122

= -122

-122

= -6- 122

= -612-2

= -6

0 A-8 22

= 02

= 0

0-3

= 010

Dividing Real Numbers SECTION 1.6 195

In general, if a and b are real numbers, , .a

-b= -a

b= - a

bb Z 0

Practice Problems 13–15Perform each indicated operation.

13.

14.

15.0 A-5 B

3

0-2

-70


a.

b.5 A-2 B 3 + 52

-4 + 1

-7 A-4 B + 2

-10 - A-5 B

Answers13. undefined, 14. 0, 15. 0,16. a. , b. , 17. 2-4-6

Practice Problem 17

Evaluate when and

.y = -5

x = -1x + y

3x


Practice Problem 18Determine whether is a solution of

.x4

- 3 = x + 3

-8


We use our skills in dividing real numbers to check possible solutions for anequation.


Solution: Original equation

Replace x with .

Divide.

False.

Since we have a false statement, is not a solution ofthe equation.

-10

7 = -10 2 + 5 � -10

-10-20-10

+ 5 � -10

-20x

+ 5 = x

-20x

+ 5 = x-10

D

Answer18. is a solution.-8

CALCULATOR EXPLORATIONSENTERING NEGATIVE NUMBERS ON A

SCIENTIFIC CALCULATOR

To enter a negative number on a scientific calculator, find a key

marked . (On some calculators, this key is marked for

“change sign.”) To enter , for example, press the keys .

The display will read .

ENTERING NEGATIVE NUMBERS ON A GRAPHING CALCULATOR

To enter a negative number on a graphing calculator, find a key

marked . Do not confuse this key with the key , which is used

for subtraction. To enter , for example, press the keys .

The display will read .

OPERATIONS WITH REAL NUMBERS

To evaluate on a calculator, press the keys

, or

.

The display will read or

Use a calculator to simplify each expression.1. 2.3. 4.

5. 6.

7. 8.9. (Be careful.)10. (Be careful.)-1252A-125 B 2 58 - 625995 - 4550

-444 - 444.8-181 - 324

-50 A294 B175 - 265

45 A32 B - 8 A218 B134 + 25 A68 - 91 B -59 A-8 B + 1726-38 A26 - 27 B

-2(7-9) - 20-16-16

ENTER02-)9-7(2(-)

=02-)9-7(*+/-2

-2 A7 - 9 B - 20

-8

8(-)-8

-(-)

-8

+/-8-8

CHS+/-

197


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

EXERCISE SET 1.6

Divide. See Examples 6 through 15.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. 21. - 49

,49

- 110

, a- 811b- 5

9, a - 3

4b

45

, a - 12b6

7, a - 1

3b14

-2

30-2

-45-9

-12-4

30

50

- 248

- 153

0-9

0-4

-605

-4812

-18-6

-16-4

20-10

18-2

B

Properties of Real Numbers SECTION 1.7 199

1.7 PROPERTIES OF REAL NUMBERS

USING THE COMMUTATIVE AND ASSOCIATIVE PROPERTIES

In this section we give names to properties of real numbers with which weare already familiar. Throughout this section, the variables a, b, and c rep-resent real numbers.

We know that order does not matter when adding numbers. For exam-ple, we know that is the same as . This property is given a spe-cial name—the commutative property of addition. We also know that orderdoes not matter when multiplying numbers. For example, we know that

. This property means that multiplication is commutativealso and is called the commutative property of multiplication.

These properties state that the order in which any two real numbers areadded or multiplied does not change their sum or product. For example, ifwe let and , then the commutative properties guarantee that

and

Example 1 Use a commutative property to complete each statement.

a. ——— b. ———

Solution: a.b.

TRY THE CONCEPT CHECK IN THE MARGIN.Let’s now discuss grouping numbers. We know that when we add three

numbers, the way in which they are grouped or associated does not changetheir sum. For example, we know that . Thisresult is the same if we group the numbers differently. In other words,

, also. Thus, . Thisproperty is called the associative property of addition.

We also know that changing the grouping of numbers when multiplyingdoes not change their product. For example, (checkit). This is the associative property of multiplication.

2 # A3 # 4 B = A2 # 3 B # 4

2 + A3 + 4 B = A2 + 3 B + 4A2 + 3 B + 4 = 5 + 4 = 9

2 + A3 + 4 B = 2 + 7 = 9

3 # x = x # 3By the commutative property of addition

By the commutative property of multiplicationx + 5 = 5 + x

3 # x =x + 5 =

3 # 5 = 5 # 33 + 5 = 5 + 3

b = 5a = 3

-5 A6 B = 6 A-5 B

5 + 77 + 5

A

COMMUTATIVE PROPERTIES

Addition:

Multiplication: a # b = b # a

a + b = b + a

HELPFUL HINT

Is subtraction also commutative? Try an example. Is ?No! The left side of this statement equals 1; the right side equals .There is no commutative property of subtraction. Similarly, there is nocommutative property for division. For example, does notequal .2 ,10

10 ,2

-13 - 2 = 2 - 3

Answers1. a. , b.✓ Concept Check: b, d

x + 4y # 7

SSM CD-ROM Video1.7

Objectives

Use the commutative and associa-tive properties.

Use the distributive property.

Use the identity and inverse prop-erties.

CB

A

Practice Problem 1Use a commutative property to com-plete each statement.

a. ———b. ———4 + x =

7 # y =

✓ CONCEPT CHECK

Which of the following pairs ofactions are commutative?a. “taking a test” and “studying for

the test”

b. “putting on your left shoe” and“putting on your right shoe”

c. “putting on your shoes” and“putting on your socks”

d. “reading the sports section” and“reading the comics section”


These properties state that the way in which three numbers are groupeddoes not change their sum or their product.

Example 2 Use an associative property to complete each statement.

a. ———b. ———

Solution: a.

b.

Examples Determine whether each statement is true by an associa-tive property or a commutative property.

3.

4.

Let’s now illustrate how these properties can help us simplify expres-sions.


5.

6. By the associative property of multiplication

Multiply.

USING THE DISTRIBUTIVE PROPERTY

The distributive property of multiplication over addition is used repeatedlythroughout algebra. It is useful because it allows us to write a product as asum or a sum as a product.

We know that . Compare that with . Since both original expressions equal 42,

they must equal each other, or

This is an example of the distributive property. The product on the left sideof the equal sign is equal to the sum on the right side. We can think of the7 as being distributed to each number inside the parentheses.

7 A2 + 4 B = 7 A2 B + 7 A4 B14 + 28 = 427 A4 B =7 A2 B + 7 A2 + 4 B = 7 A6 B = 42

B = -21x

-3 A7x B = A-3 # 7 Bx = 22 + x

= A10 + 12 B + x

By the commutative property of addition

By the associative property of addition

Add.

10 + Ax + 12 B = 10 + A12 + x B

2 # A3 # 1 B = A2 # 3 B # 1

Since the order of two numbers was changedand their grouping was not, this is true by thecommutative property of addition.

Since the grouping of the numbers waschanged and their order was not, this is trueby the associative property of multiplication.

A7 + 10 B + 4 = A10 + 7 B + 4

A-1 # 2 B # 5 = -1 # A2 # 5 BBy the associative propertyof addition

By the associative propertyof multiplication

5 + A4 + 6 B = A5 + 4 B + 6

A-1 # 2 B # 5 =5 + A4 + 6 B =

ASSOCIATIVE PROPERTIES

Addition:

Multiplication: Aa # b B # c = a # Ab # c BAa + b B + c = a + Ab + c B

HELPFUL HINTRemember the difference betweenthe commutative properties and theassociative properties. The commu-tative properties have to do withthe order of numbers and the asso-ciative properties have to do withthe grouping of numbers.

DISTRIBUTIVE PROPERTY OF MULTIPLICATION OVER ADDITION

a Ab + c B = ab + ac

Practice Problem 2Use an associative property to com-plete each statement.

a. ———

b. ———A-2 + 7 B + 3 =

5 # A-3 # 6 B =

Practice Problems 3–4Determine whether each statement istrue by an associative property or acommutative property.

3.

4. -2 + A4 + 9 B = A-2 + 4 B + 9

5 # A4 # 7 B = 5 # A7 # 4 B

Answers2. a. , b. , 3. com-mutative, 4. associative, 5. , 6. 20x14 + x

-2 + A7 + 3 BA5 # -3 B # 6


5.

6. 4 A5x BA-3 + x B + 17

Properties of Real Numbers SECTION 1.7 201

Since multiplication is commutative, this property can also be written as

The distributive property can also be extended to more than two num-bers inside the parentheses. For example,

Since we define subtraction in terms of addition, the distributive propertyis also true for subtraction. For example,

Examples Use the distributive property to write each expressionwithout parentheses. Then simplify the result.

7.

8.

9.

10.

11.

12. Apply the distributive property.

Multiply.

Add.

The distributive property can also be used to write a sum as a product.

Examples Use the distributive property to write each sum as aproduct.

13.14.

USING THE IDENTITY AND INVERSE PROPERTIES

Next, we look at the identity properties.The number 0 is called the identity for addition because when 0 is added

to any real number, the result is the same real number. In other words, theidentity of the real number is not changed.

The number 1 is called the identity for multiplication because when areal number is multiplied by 1, the result is the same real number. In otherwords, the identity of the real number is not changed.

C7s + 7t = 7 As + t B8 # 2 + 8 # x = 8 A2 + x B

= 12x + 38 = 12x + 28 + 10

4 A3x + 7 B + 10 = 4 A3x B + 4 A7 B + 10 = -3 - x + w = A-1 B A3 B + A-1 B Ax B - A-1 B Aw B

- A3 + x - w B = -1 A3 + x - w B = -2 + y-1 A2 - y B = A-1 B A2 B - A-1 B Ay B = 5x + 15y - 5z5 Ax + 3y - z B = 5 Ax B + 5 A3y B - 5 Az B = 15 - 10z-5 A-3 + 2z B = -5 A-3 B + A-5 B A2z B = 2x + 2y2 Ax + y B = 2 Ax B + 2 Ay B

= 2x - 2y

2 Ax - yB = 2 Ax B - 2 Ay B

= 3x + 3y + 3z

3 Ax + y + z B = 3 Ax B + 3 Ay B + 3 Az B

Ab + c Ba = ba + ca

Answers7. , 8. ,9. , 10. ,11. , 12. ,13. , 14. 4 Ax + y B9 A3 + y B 18x + 45-8 - a + b

-3 + a4x + 24y - 8z-6 - 21x5x + 5y

Practice Problems 7–12Use the distributive property to writeeach expression without parentheses.Then simplify the result.

7.

8.

9.

10.

11.

12. 9 A2x + 4 B + 9

- A8 + a - b B-1 A3 - a B4 Ax + 6y - 2z B-3 A2 + 7x B5 Ax + y B

HELPFUL HINT

Notice in Example 11that isfirst rewritten as

.-1 A3 + x - w B- A3 + x - w B

IDENTITIES FOR ADDITION AND MULTIPLICATION

0 is the identity element for addition.

and

1 is the identity element for multiplication.

and 1 # a = aa # 1 = a

0 + a = aa + 0 = a

Practice Problems 13–14Use the distributive property to writeeach sum as a product.

13.

14. 4x + 4y

9 # 3 + 9 # y



Examples Name the property illustrated by each true statement.

15.

16.

17.

18.

19.

20.

21. -6 # Ay # 2 B = A-6 # 2 B # y

-2 + 2 = 0

-2 # a - 12b = 1

2 # Az # 5 B = 2 # A5 # z BAb + 0 B + 3 = b + 3

Ax + 7 B + 9 = x + A7 + 9 BCommutative property of multiplication (orderchanged)Associative property of addition (groupingchanged)

Identity element for addition

Commutative property of multiplication (orderchanged)

Multiplicative inverse property

Additive inverse property

Commutative and associative properties ofmultiplication (order and grouping changed)

3 # y = y # 3

Notice that 0 is the only number that can be added to any real numberwith the result that the sum is the same real number. Also, 1 is the onlynumber that can be multiplied by any real number with the result that theproduct is the same real number.

Additive inverses or opposites were introduced in Section 1.3. Two num-bers are called additive inverses or opposites if their sum is 0. The additiveinverse or opposite of 6 is because . The additive inverseor opposite of is 5 because .

Reciprocals or multiplicative inverses were introduced in Section R.2.Two nonzero numbers are called reciprocals or multiplicative inverses if

their product is 1. The reciprocal or multiplicative inverse of is because

. Likewise, the reciprocal of is because .-5 a - 15b = 1- 1

5-5

23

# 32

= 1

32

23

-5 + 5 = 0-56 + A-6 B = 0-6

Practice Problems 15–21Name the property illustrated by eachtrue statement.

15.

16.

17.

18.

19.

20.

21. A7 # y B # 10 = y # A7 # 10 BAx + 0 B + 23 = x + 23

3 a 13b = 1

6 + Az + 2 B = 6 + A2 + z B-4 # A6 # x B = A-4 # 6 B # x

12 + y = y + 12

5 + A-5 B = 0

Answers

✓ Concept Check: a. , b.

15. additive inverse property, 16. commuta-tive property of addition, 17. associativeproperty of multiplication, 18. commutativeproperty of addition, 19. multiplicativeinverse property, 20. identity element foraddition, 21. commutative and associativeproperties of multiplication

- 103

310

ADDITIVE OR MULTIPLICATIVE INVERSES

The numbers a and are additive inverses or opposites of each otherbecause their sum is 0; that is,

The numbers b and (for ) are reciprocals or multiplicative

inverses of each other because their product is 1; that is,

b # 1b

= 1

b Z 01b

a + A-a B = 0

-a✓ CONCEPT CHECK

Which of the following is the

a. opposite of , and which is the

b. reciprocal of ?

1, , , 0, , - 310

103

310

- 103

- 310

- 310

203


EXERCISE SET 1.7

Use the distributive property to write each expression without parentheses. Then simplifythe result. See Examples 7 through 12.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. 21. - Ar - 3 - 7p B- A9r + 5 B- A5x + 2 B

-4 A4 + 2p + 5 B-4 A1 - 2m + n B8 A3y + z - 6 B

5 Ax + 4m + 2 B-5 A2r + 11 B-7 A3y + 5 B

-3 Az - y B-2 Ay - z B2 Ax + 5 B

3 A6 + x B3 A8x - 1 B7 A4x - 3 B

5 A7 + 8y B2 A3x + 5 B11 Ay - 4 B

9 Ax - 6 B7 Aa + b B4 Ax + y BB

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

Reading Graphs SECTION 1.8 205

1.8 READING GRAPHS

In today’s world, where the exchange of information must be fast andentertaining, graphs are becoming increasingly popular. They provide aquick way of making comparisons, drawing conclusions, and approximatingquantities.

READING BAR GRAPHS

A bar graph consists of a series of bars arranged vertically or horizontally.The bar graph in Example 1 shows a comparison of the rates charged byselected electricity companies. The names of the companies are listed hor-izontally and a bar is shown for each company. Corresponding to the heightof the bar for each company is a number along a vertical axis. These verti-cal numbers are cents charged for each kilowatt-hour of electricity used.

Example 1 The following bar graph shows the cents charged perkilowatt-hour for selected electricity companies.

a. Which company charges the highest rate?b. Which company charges the lowest rate?c. Approximate the electricity rate charged by the first

four companies listed.d. Approximate the difference in the rates charged by

the companies in parts (a) and (b).

Solution: a. The tallest bar corresponds to the company thatcharges the highest rate. Alaska Village ElectricCooperative charges the highest rate.

b. The shortest bar corresponds to the company thatcharges the lowest rate. Waterville, Washingtoncharges the lowest rate.

c. To approximate the rate charged by Kentucky Power,we go to the top of the bar that corresponds to thiscompany. From the top of the bar, we move horizon-tally to the left until the vertical axis is reached.

50

40

30

20

10

0Kentucky

PowerAlaskaVillage

Electric Coop.

East Miss.Electric

CentralVermont

SouthernCal. Edison

Waterville,Wash.

Cen

ts p

er K

ilow

att-

hour

Source: USA Today Research

A

Objectives

Read bar graphs.

Read line graphs.BA

SSM CD-ROM Video1.8

Practice Problem 1Use the bar graph from Example 1 toanswer the following.

a. Approximate the rate charged byEast Mississippi Electric.

b. Approximate the rate charged bySouthern California Edison.

c. Find the difference in rates chargedby Southern California Edison andEast Mississippi Electric.

Answers1.a. 8¢ per kilowatt-hour, b. 12¢ per kilowatt-hour, c. 4¢


The height of the bar is approximately halfway betweenthe 0 and 10 marks. We therefore conclude that

Kentucky Power charges approximately 5¢ per kilo-watt-hour.

Central Vermont charges approximately 11¢ per kilo-watt-hour.

Waterville, Washington charges approximately 2¢ perkilowatt-hour.

Alaska Village Electric charges approximately 42¢ perkilowatt-hour.

d. The difference in rates for Alaska Village ElectricCooperative and Waterville, Washington is approxi-mately or 40¢.

Example 2 The following bar graph shows Disney’s top eight ani-mated films and the amount of money they generated attheaters.

100 120 140 160 180 200 220

Dollars (in millions)

Disney’s Top 8 Animated Feature Films

Bambi

The Hunchback of Notre Dame

The Jungle Book

Pocahontas

Beauty and the Beast

Aladdin

Snow White and the Seven Dwarfs

The Lion King

240 260 280 300 320

1994

1992

1937

1991

1967

1995

1942

1996

Yea

r Fi

rst R

elea

sed

Source: Walt Disney Co. (includes re-releases, re-adjusted for inflation)

42¢ - 2¢

50

40

30

20

10

0Kentucky

PowerEast Miss.

ElectricWaterville,

Wash.

Cen

ts p

er K

ilow

att-

hour

42

115

2

AlaskaVillage

Electric Coop.

CentralVermont

SouthernCal. Edison

Source: USA Today Research

Practice Problem 2Use the graph from Example 2 toanswer the following.

a. How much money did the film Snow White and the Seven Dwarfsgenerate?

b. How much more money did the filmThe Jungle Book make than the filmBambi?

Answers2. a. 175 million, b. 20 million

a. Find the film shown that generated the most incomefor Disney and approximate the income.

b. How much more money did the film Aladdin makethan the film Beauty and the Beast?

Solution: a. Since these bars are arranged horizontally, we look forthe longest bar, which is the bar representing the filmThe Lion King. To approximate the income from thisfilm, we move from the right edge of this bar verticallydownward to the dollars axis. This film generatedapproximately 315 million dollars, or $315,000,000, themost income for Disney.

b. Aladdin generated approximately 215 million dollars.Beauty and the Beast generated approximately145 million dollars. To find how much more moneyAladdin generated than Beauty and the Beast, we sub-tract million dollars, or $70,000,000.

READING LINE GRAPHS

A line graph consists of a series of points connected by a line. The graph inExample 3 is a line graph.

Example 3 The line graph below shows the relationship between thedistance driven in a 14-foot U-Haul truck in one day andthe total cost of renting this truck for that day. Noticethat the horizontal axis is labeled Distance and the verti-cal axis is labeled Total Cost.

B

215 - 145 = 70

100 120 140 160 180 200 220

Dollars (in millions)

Disney’s Top 8 Animated Feature Films

Bambi

The Hunchback of Notre Dame

The Jungle Book

Pocahontas

Beauty and the Beast

Aladdin

Snow White and the Seven Dwarfs

The Lion King

240 260 280 300 320

1994

1992

1937

1991

1967

1995

1942

1996

Yea

r Fi

rst R

elea

sed

Source: Walt Disney Co. (includes re-releases, re-adjusted for inflation)

145 215 315



a. Find the total cost of renting thetruck if 50 miles are driven.

b. Find the total number of milesdriven if the total cost of renting is$100.

Answers3. a. $50, b. 180 miles


a. Find the total cost of renting the truck if 100 miles aredriven.

b. Find the number of miles driven if the total cost ofrenting is $140.

Solution: a. Find the number 100 on the horizontal scale and movevertically upward until the line is reached. From thispoint on the line, we move horizontally to the leftuntil the vertical scale is reached. We find that thetotal cost of renting the truck if 100 miles are driven isapproximately $70.

b. We find the number 140 on the vertical scale andmove horizontally to the right until the line is reached.From this point on the line, we move vertically down-ward until the horizontal scale is reached. We find thatthe truck is driven approximately 280 miles.

From the previous example, we can see that graphing provides a quickway to approximate quantities. In Chapter 6 we show how we can use equa-tions to find exact answers to the questions posed in Example 3. The next

160

140

120

100

80

60

40

20

0

Tot

al C

ost (

Dol

lars

)

Distance (Miles)100 200 3002500 50

70

280

150

One Day 14-foot Truck Rental

160

140

120

100

80

60

40

20

0

Tot

al C

ost (

Dol

lars

)

Distance (Miles)

One Day 14-foot Truck Rental

3001000 50 250150 200


graph is another example of a line graph. It is also sometimes called abroken line graph.

Example 4 The line graph shows the relationship between time spentsmoking a cigarette and pulse rate. Time is recordedalong the horizontal axis in minutes, with 0 minutes beingthe moment a smoker lights a cigarette. Pulse is recordedalong the vertical axis in heartbeats per minute.

a. What is the pulse rate 15 minutes after lighting a ciga-rette?

b. When is the pulse rate the lowest?c. When does the pulse rate show the greatest

change?

Solution: a. We locate the number 15 along the time axis andmove vertically upward until the line is reached. Fromthis point on the line, we move horizontally to the leftuntil the pulse rate axis is reached. Reading the num-ber of beats per minute, we find that the pulse rate is80 beats per minute 15 minutes after lighting a ciga-rette.

100

90

80

70

60

50

40

30

20

10

0

Puls

e ra

te (

hear

tbea

ts p

er m

inut

e)

Time (minutes)

10 20 30 35 4025−5 0 5 15

100

90

80

70

60

50

40

30

20

10

0

Puls

e ra

te (

hear

tbea

ts p

er m

inut

e)

Time (minutes)

10 20 30 35 4025−5 0 5 15


a. What is the pulse rate 40 minutesafter lighting a cigarette?

b. What is the pulse rate when the cig-arette is being lit?

c. When is the pulse rate the highest?

Answers4. a. 70, b. 60, c. 5 min. after lighting


b. We find the lowest point of the line graph, which rep-resents the lowest pulse rate. From this point, wemove vertically downward to the time axis. We findthat the pulse rate is the lowest at minutes, whichmeans 5 minutes before lighting a cigarette.

c. The pulse rate shows the greatest change during the5 minutes between 0 and 5. Notice that the line graph issteepest between 0 and 5 minutes.

-5

211


EXERCISE SET 1.8

The following bar graph shows the number of teenagers expected to use the Internet forthe years shown. Use this graph to answer Exercises 1–4. See Example 1.

1614121086420T

eena

gers

(in

mill

ions

)

2002(estimated)

Teenagers Using the Internet

1999(estimated)

YearsSource: 1997 Jupiter Communications/College and Teen Report

19971995

A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

1. Approximate the number of teenagersexpected to use the Internet in 1999.

2. Approximate the number of teenagerswho use the Internet in 1995.

3. What year shows the greatest increasein number of teenagers using theInternet?

4. How many more teenagers areexpected to use the Internet in 2002than in 1999?

Many fires are deliberately set. An increasing number of those arrested for arson are juve-niles (age 17 and under). The following line graph shows the percent of deliberately setfires started by juveniles. Use this graph to answer Exercises 5–12. See Example 3 and 4.

60%

55%

50%

45%

40%

35%

Perc

ent

1987 19931992

Juveniles Starting Fires

1990

Year

19911989 19940

1988 19951986

Source: National Fire Protection Association and the FBI Uniform Crime Report.

B

5. What year shows the highest percentof arson fires started by juveniles?

6. What year since 1990 shows adecrease in the percent of fires startedby juveniles?

7. Name two consecutive years wherethe percent remained the same.

8. What year(s) shows the lowest percentof arson fires started by juveniles?

9. Estimate the percent of arson firesstarted by juveniles in 1995.

10. Estimate the percent of arson firesstarted by juveniles in 1992.

11. What year shows the greatest increasein the percent of fires started by juve-niles?

12. What trend do you notice from thisgraph?


CHAPTER 1 HIGHLIGHTSDEFINITIONS AND CONCEPTS EXAMPLES

SECTION 1.1 SYMBOLS AND SETS OF NUMBERS

A set is a collection of objects, called elements,enclosed in braces.

Ea, c, eF

Natural numbers:

Whole numbers:

Integers:

Rational numbers: {real numbers that can be ex-pressed as a quotient of integers}

Irrational numbers: {real numbers that cannot be ex-pressed as a quotient of integers}

Real numbers: {all numbers that correspond to apoint on the number line}

E . . . , -3, -2, -1, 0, 1, 2, 3, . . .FE0, 1, 2, 3, 4, . . .FE1, 2, 3, 4, . . .F Given the set list the num-

bers that belong to the set ofNatural numbers 5Whole numbers 0, 5IntegersRational numbersIrrational numbersReal numbers -3.4, 13, 0, 23, 5, -4

13-3.4, 0, 23, 5,-4

-4, 0, 5

E-3.4, 13, 0, 23, 5, -4F

A line used to picture numbers is called a numberline. 0 2−2 3−3 1−1

The absolute value of a real number a denoted by is the distance between a and 0 on the number line.

@a @ @ -2 @ = 2@0 @ = 0@5 @ = 5

SYMBOLS:

≥ is greater than or equal to ≤ is less than or equal to

6 is less than 7 is greater than Z is not equal to= is equal to

18 ≥ - 13

6 ≤ 61 6 44 7 13 Z -3

-7 = -7

ORDER PROPERTY FOR REAL NUMBERS

For any two real numbers a and b, a is less than b ifa is to the left of b on the number line.

0 2−2 3−3−3 < 0 0 > −3 0 < 2.5 2.5 > 0

1−1

SECTION 1.2 INTRODUCTION TO VARIABLE EXPRESSIONS AND EQUATIONS

The expression is an exponential expression. Thenumber a is called the base; it is the repeated factor.The number n is called the exponent; it is the num-ber of times that the base is a factor.

an

72 = 7 # 7 = 4943 = 4 # 4 # 4 = 64

Chapter 1 Highlights 213

SECTION 1.2 (CONTINUED)

ORDER OF OPERATIONS

Simplify expressions in the following order. If group-ing symbols are present, simplify expressions withinthose first, starting with the innermost set. Also,simplify the numerator and the denominator of afraction separately.

1. Simplify exponential expressions.2. Multiply or divide in order from left to right.3. Add or subtract in order from left to right. = 4

= 8421

= 64 + 2021

=64 + 5 A4 B

21

82 + 5 17 - 3 23 # 7

=82 + 5 14 2

21

A symbol used to represent a number is called a vari-able.

Examples of variables areq, x, z

An algebraic expression is a collection of numbers,variables, operation symbols, and grouping sym-bols.

Examples of algebraic expressions are

5x, 2 Ay - 6 B , q2 - 3q + 16

To evaluate an algebraic expression containing avariable, substitute a given number for the variableand simplify.

Evaluate when and .

= 16 = 25 - 9

x2 - y2 = A5 B 2 - 32

y = 3x = 5x2 - y2

A mathematical statement that two expressions areequal is called an equation.

Equations:

A = pr23x - 9 = 20

A solution of an equation is a value for the variablethat makes the equation a true statement.

Determine whether 4 is a solution of .

True.

4 is a solution.

27 = 27

20 + 7 � 27

5 A4 B + 7 � 27

5x + 7 = 27

5x + 7 = 27

SECTION 1.3 ADDING REAL NUMBERS

TO ADD TWO NUMBERS WITH THE SAME SIGN

1. Add their absolute values.2. Use their common sign as the sign of the sum.

Add.

-3 + A-8 B = -1110 + 7 = 17

TO ADD TWO NUMBERS WITH DIFFERENT SIGNS

1. Subtract their absolute values.2. Use the sign of the number whose absolute value

is larger as the sign of the sum.

14 + A-9 B = 5-25 + 5 = -20


SECTION 1.5 MULTIPLYING REAL NUMBERS


The product of two numbers with the same sign is apositive number. The product of two numbers withdifferent signs is a negative number.

Multiply.

2 # A-4 B = -8 -2 # 4 = -8

-7 # A-8 B = 567 # 8 = 56

PRODUCTS INVOLVING ZERO

The product of 0 and any number is 0.

and 0 # b = 0b # 0 = 0 0 # a - 34b = 0-4 # 0 = 0

SECTION 1.6 DIVIDING REAL NUMBERS

QUOTIENT OF TWO REAL NUMBERS

ab

= a # 1b

Divide.

422

= 42 # 12

= 21


The quotient of two numbers with the same sign is apositive number. The quotient of two numbers withdifferent signs is a negative number.

-42

6= -7

42-6

= -7

-90-10

= 99010

= 9

SECTION 1.4 SUBTRACTING REAL NUMBERS

To subtract two numbers a and b, add the first num-ber a to the opposite of the second number b.

a - b = a + A-b BSubtract.

-30 - A-30 B = -30 + 30 = 0

-5 - 22 = -5 + A-22 B = -27

3 - A-44 B = 3 + 44 = 47


Two numbers that are the same distance from 0 butlie on opposite sides of 0 are called opposites oradditive inverses. The opposite of a number a isdenoted by .-a

The opposite of is 7.The opposite of 123 is .-123

-7


QUOTIENTS INVOLVING ZERO

The quotient of a nonzero number and 0 is unde-fined.

is undefined.b0

is undefined.-85

0

The quotient of 0 and any nonzero number is 0.

0b

= 00

-47= 0

018

= 0

SECTION 1.7 PROPERTIES OF REAL NUMBERS

COMMUTATIVE PROPERTIES

Addition:Multiplication: a # b = b # a

a + b = b + a

-8 # 5 = 5 # A-8 B3 + A-7 B = -7 + 3

ASSOCIATIVE PROPERTIES

Addition:Multiplication: Aa # b B # c = a # Ab # c B

Aa + b B + c = a + Ab + c B A-3 # 2 B # 11 = -3 # A2 # 11 BA5 + 10 B + 20 = 5 + A10 + 20 B


Two numbers whose product is 1 are called multi-plicative inverses or reciprocals. The reciprocal of anonzero number a is because .a # 1

a = 11a

The reciprocal of 3 is .

The reciprocal of is .- 52

- 25

13

DISTRIBUTIVE PROPERTY

IDENTITIES

1 # a = aa # 1 = a0 + a = aa + 0 = a

a Ab + c B = a # b + a # c

1 # 27 = 27-14 # 1 = -140 + A-2 B = -25 + 0 = 5

-2 A3 + x B = -2 # 3 + A-2 B Ax B5 A6 + 10 B = 5 # 6 + 5 # 10

INVERSES

Additive or opposite:

Multiplicative or reciprocal: b # 1b

= 1

a + A-a B = 0

3 # 13

= 1

7 + A-7 B = 0


SECTION 1.8 READING GRAPHS

To find the value on the vertical axis representing alocation on a graph, move horizontally from thelocation on the graph until the vertical axis isreached. To find the value on the horizontal axisrepresenting a location on a graph, move verticallyfrom the location on the graph until the horizontalaxis is reached.

This broken line graph shows the average publicclassroom teachers’ salaries for the school yearending in the years shown.

Estimate the average public teacher’s salary for theschool year ending in 1989. The average salary isapproximately $29,500.

Find the earliest year that the average salary roseabove $32,000. The year was 1991.

36,000

35,000

34,000

33,000

32,000

31,000

30,000

29,000

28,000

Ave

rage

Yea

rly

Sala

ry(i

n do

llars

)

19921988 199419931989 1990 1991Source: U.S. Bureau of the Census, Statistical Abstract of theUnited States: 1994 (114th edition) Washington, D.C., 1994

217

2In this chapter, we solve equations and inequalities. Once we knowhow to solve equations and inequalities, we may solve problems. Ofcourse, problem solving is an integral topic in algebra and its discussionis continued throughout this text.

2.1 Simplifying Expressions

2.2 The Addition Property ofEquality

2.3 The Multiplication Property ofEquality

2.4 Further Solving LinearEquations

2.5 An Introduction to ProblemSolving

2.6 Formulas and Problem Solving

2.7 Percent, Ratio, and Proportion

2.8 Linear Inequalities andProblem Solving

A glacier is a giant mass of rocks and ice that flows downhilllike a river. Alaska alone has an estimated 100,000 glaciers.They form high in the mountains where snow does not melt.After years of accumulated snowfall, the weight of the com-pacted ice causes it to slide downhill, and a glacier is formed. InExample 1 on page 93, we will find the time it takes for ice atthe beginning of a glacier to reach its face, or ending wall.

Glaciers account for approximately 77 percent of the world’sfresh water. They also contain records of past environments be-tween their layers of snow. Scientists study glaciers to learnabout past global climates and then use this information to pre-dict future environmental changes.(Source: Blue Ice in Motion—The Story of Alaska’s Glaciers, by Sally D. Wiley, published in 1990by The Alaska Natural History Association and Judith Ann Rose)

Equations, Inequalities, and Problem Solving C H A P T E R

2.1 SIMPLIFYING EXPRESSIONS

As we explore in this section, we will see that an expression such asis not as simple as possible. This is because—even without replac-

ing x by a value—we can perform the indicated addition.

IDENTIFYING TERMS, LIKE TERMS, AND UNLIKE TERMS

Before we practice simplifying expressions, some new language is pre-sented. A term is a number or the product of a number and variables raisedto powers.

Terms

The numerical coefficient of a term is the numerical factor. The numer-ical coefficient of 3x is 3. Recall that 3x means .

Term Numerical Coefficient

3x 3

since means

z 1

Example 1 Identify the numerical coefficient in each term.

a. b. c. y d. e.

Solution: a. The numerical coefficient of is .b. The numerical coefficient of is 22.c. The numerical coefficient of y is 1, since y is 1y.d. The numerical coefficient of is , since is

.

e. The numerical coefficient of is , since is

.

Terms with the same variables raised to exactly the same powers arecalled like terms. Terms that aren’t like terms are called unlike terms.

Like Terms Unlike Terms

3x, 2x 5x,

6abc3, 6ab22ab2c3, ac3b2

7y, 3z, 8x2-6x2y, 2x2y, 4x2y

5x2

17

# x

x7

17

x7

-1x-x-1-x

22z4-3-3y

x7

-x22z4-3y

-5-5-1-y

-0.7-0.7ab3c5

15

# y3y3

515

y3

5

3 # x

0.8z2y

,3xz2,-5,2x3,-y,

A

3x + 2x

Simplifying Expressions SECTION 2.1 219

Why? Same variable x, but different powers xand Why? Different variables

Why? Different variables and different powers

x2

Practice Problem 1Identify the numerical coefficient ineach term.

a. b. c. x d. e.z4

-y15y3-4x

HELPFUL HINT

The term means and thus has a numerical coefficient of .The term z means 1z and thus has a numerical coefficient of 1.

-1-1y-y

Objectives

Identify terms, like terms, and un-like terms.

Combine like terms.

Simplify expressions containingparentheses.

Write word phrases as algebraicexpressions.

D

CB

A

Answers

1. a. , b. 15, c. 1, d. , e. 14

-1-4

SSM CD-ROM Video2.1

Example 2 Determine whether the terms are like or unlike.

a. b. c.d.

Solution: a. Unlike terms, since the exponents on x are not thesame.

b. Like terms, since each variable and its exponent match.c. Like terms, since by the commutative property.d. Like terms.

COMBINING LIKE TERMS

An algebraic expression containing the sum or difference of like terms canbe simplified by applying the distributive property. For example, by the dis-tributive property, we rewrite the sum of the like terms as

Also,

Simplifying the sum or difference of like terms is called combining liketerms.

Example 3 Simplify each expression by combining like terms.

a. b. c.

Solution: a.b.c.

Examples Simplify each expression by combining like terms.

4.

5.

6. These two terms cannot be combined be-cause they are unlike terms.

7.

The examples above suggest the following.

= 7.3x - 62.3x + 5x - 6 = 12.3 + 5 2x - 6

4y - 3y2 = -4a - 1 = 1-5 + 1 2a + 1-3 + 2 2-5a - 3 + a + 2 = -5a + 1a + 1-3 + 2 2 = 5x + 7

2x + 3x + 5 + 2 = 12 + 3 2x + 15 + 2 2

8x2 + 2x - 3x = 8x2 + 12 - 3 2x = 8x2 - x10y2 + y2 = 110 + 1 2y2 = 11y27x - 3x = 17 - 3 2x = 4x

8x2 + 2x - 3x10y2 + y27x - 3x

-y2 + 5y2 = 1-1 + 5 2y2 = 4y2

3x + 2x = 13 + 2 2x = 5x

3x + 2x

B

zy = yz

-x4, x4-2yz, -3zy4x2y, x2y, -2x2y2x, 3x2

220 CHAPTER 2 Equations, Inequalities, and Problem Solving

HELPFUL HINT

In like terms, each variable and its exponent must match exactly, butthese factors don’t need to be in the same order.

and are like terms.3yx22x2y

Practice Problem 2Determine whether the terms are likeor unlike.

a. b.

c. -5ab, 3ba

3x2y2, -x2y2, 4x2y27x, -6x

Practice Problem 3Simplify each expression by combininglike terms.

a. b.

c. 5y - 3x + 6x

11x2 + x29y - 4y

Practice Problems 4–7Simplify each expression by combininglike terms.

4.

5.

6.

7.

Answers2. a. like, b. like, c. like,3. a. , b. , c. ,4. , 5. , 6. ,7. 13.1y - 3

3z - 3z2-x - 79y + 165y + 3x12x25y

8.9y + 4.2y - 3

3z - 3z2

-2x + 4 + x - 11

7y + 2y + 6 + 10

SIMPLIFYING EXPRESSIONS CONTAINING PARENTHESES

In simplifying expressions we make frequent use of the distributive prop-erty to remove parentheses.

Examples Find each product by using the distributive property toremove parentheses.

8.

9.

10.

= -x - y + 2z - 6 - 1 16 2 = -1 1x 2 - 1 1y 2 - 1 1-2z 2- 1x + y - 2z + 6 2 = -1 1x + y - 2z + 6 2

= -2y - 0.6z + 2 + 1-2 2 1-1 2-2 1y + 0.3z - 1 2 = -2 1y 2 + 1-2 2 10.3z 2

= 5x + 10Apply the distributive property.Multiply.

5 1x + 2 2 = 5 1x 2 + 5 12 2

C

COMBINING LIKE TERMS

To combine like terms, combine the numerical coefficients and multi-ply the result by the common variable factors.

Simplifying Expressions SECTION 2.1 221

Apply the distributive property.

Multiply.

Distribute over each term.

-1

HELPFUL HINT

If a “ ” sign precedes parentheses, the sign of each term inside theparentheses is changed when the distributive property is applied toremove parentheses.

Examples:

- 1-3x - 4y - 1 2 = 3x + 4y + 1- 1-5x + y - z 2 = 5x - y + z- 1x - 2y 2 = -x + 2y- 12x + 1 2 = -2x - 1

-

Answers8. , 9. ,10. , 11. ,12. , 13. -25x - 1-3x - 4

4x - 4-3x - 2y - z + 1-4x - 0.8y + 123y + 18

Practice Problems 8–10Find each product by using the distrib-utive property to remove parentheses.

8.

9.

10. - 13x + 2y + z - 1 2-4 1x + 0.2y - 3 23 1y + 6 2


11.

12.

13. -3 17x + 1 2 - 14x - 2 25 - 13x + 9 24 1x - 6 2 + 20

When simplifying an expression containing parentheses, we often usethe distributive property first to remove parentheses and then again tocombine any like terms.


11.

12.

= -4x + 6

= -7x + 3x + 8 - 2

Apply the distributiveproperty.

Combine like terms.

8 - 17x + 2 2 + 3x = 8 - 7x - 2 + 3x

= 6x - 14


Combine like terms.

3 12x - 5 2 + 1 = 6x - 15 + 1

13.

Example 14 Subtract from .

Solution: We first note that “subtract from ”translates to . Next, we simplify thealgebraic expression.

WRITING ALGEBRAIC EXPRESSIONS

To prepare for problem solving, we next practice writing word phrases asalgebraic expressions.

Examples Write each phrase as an algebraic expression and simplifyif possible. Let x represent the unknown number.

15. Twice a number, plus 6

6

This expression cannot be simplified.

16. The difference of a number and 4, divided by 7

This expression cannot be simplified.

17. Five plus the sum of a number and 1

5 +

Next, we simplify this expression.

= 6 + x

5 + 1x + 1 2 = 5 + x + 1

1x + 1 2T T T

fd

7,1x - 4 2T T T

df+2x

T T Tdd

D

= -2x - 1


Combine like terms.

12x - 3 2 - 14x - 2 2 = 2x - 3 - 4x + 2

12x - 3 2 - 14x - 2 2 2x - 34x - 2

2x - 34x - 2

= -11x - 13

Apply the distribu-tive property.Combine like terms.

-2 14x + 7 2 - 13x - 1 2 = -8x - 14 - 3x + 1


Answers

14. , 15. , 16. ,

17. 5x + 6

1x + 2 25

10 - 3x-5x + 7

Practice Problems 15–17Write each phrase as an algebraic ex-pression and simplify if possible. Let xrepresent the unknown number.

15. Three times a number, subtractedfrom 10

16. The sum of a number and 2, di-vided by 5

17. Three times a number, added tothe sum of twice a number and 6

Practice Problem 14Subtract from .4x - 39x - 10

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

223


EXERCISE SET 2.1

Identify the numerical coefficient of each term. See Example 1.

1. 2. x

Indicate whether each list of terms are like or unlike. See Example 2.

3. 4.

Simplify each expression by combining any like terms. See Examples 3 through 7.

5. 6. 7.

8. 9. 10.

11. 12. 13.

14.

Simplify each expression. Use the distributive property to remove any parentheses. SeeExamples 8 through 10.

15. 16.

Remove parentheses and simplify each expression. See Examples 11 through 13.

17.

Write each phrase as an algebraic expression. Simplify if possible. See Example 14.

18. Add to 19. Add to

20. Subtract from 3x - 87x + 1

y + 163y - 54x - 106x + 7

7 1d - 3 2 + 10

7 1r + 3 25 1y + 4 2C

8p + 4 - 8p - 15

5g - 3 - 5 - 5ga + 3a - 2 - 7am - 4m + 2m - 6

6g + 5 - 3g - 73b - 5 - 10b - 4c - 7c + 2c

8w - w + 6w3x + 2x7y + 8y

B

-2x2y, 6xy5y, -y

-7y

A

The Addition Property of Equality SECTION 2.2 225

LINEAR EQUATION IN ONE VARIABLE

A linear equation in one variable can be written in the form

where A, B, and C are real numbers and .A Z 0

Ax + B = C

ADDITION PROPERTY OF EQUALITY

If a, b, and c are real numbers, then

and

are equivalent equations.

a + c = b + ca = b

2.2 THE ADDITION PROPERTY OF EQUALITY

USING THE ADDITION PROPERTY

Recall from Section 1.2 that an equation is a statement that two expressionshave the same value. Also, a value of the variable that makes an equationa true statement is called a solution or root of the equation. The process offinding the solution of an equation is called solving the equation for thevariable. In this section, we concentrate on solving linear equations in onevariable.

A

SSM CD-ROM Video2.2

Evaluating a linear equation for a given value of the variable, as we didin Section 1.2, can tell us whether that value is a solution. But we can’t relyon evaluating an equation as our method of solving it—with what valuewould we start?

Instead, to solve a linear equation in x, we write a series of simpler equa-tions, all equivalent to the original equation, so that the final equation hasthe form

or

Equivalent equations are equations that have the same solution. Thismeans that the “number” above is the solution to the original equation.

The first property of equality that helps us write simpler equivalentequations is the addition property of equality.

number = xx = number

This property guarantees that adding the same number to both sides of anequation does not change the solution of the equation. Since subtraction isdefined in terms of addition, we may also subtract the same number fromboth sides without changing the solution.

A good way to picture a true equation is as a balanced scale. Since it isbalanced, each side of the scale weighs the same amount.

x – 2 5

Objectives

Use the addition property ofequality to solve linear equations.

Simplify an equation and then usethe addition property of equality.


C

B

A


✓ CONCEPT CHECK

Use the addition property to fill in theblank so that the middle equationsimplifies to the last equation.

— —x = 8

= 3 +x - 5 +x - 5 = 3

Practice Problem 1Solve for x.x - 5 = 8

If the same weight is added to or subtracted from each side, the scaleremains balanced.

We use the addition property of equality to write equivalent equationsuntil the variable is alone (by itself on one side of the equation) and theequation looks like “ ” or “ .”


Example 1 Solve for x.

Solution: To solve for x, we first get x alone on one side of theequation. To do this, we add 7 to both sides of theequation.

Add 7 to both sides.

Simplify.

The solution of the equation is obviously 17.Since we are writing equivalent equations, the solution ofthe equation is also 17.

Check: To check, replace x with 17 in the original equation.

Original equation.

Replace x with 17.

True.

Since the statement is true, 17 is the solution.

Example 2 Solve:

Solution: To solve for y, we subtract 0.6 from both sides of theequation.

Subtract 0.6 from both sides.

Combine like terms.

Check: Original equation.

Replace y with .

True.

The solution is .-1.6

-1.0 = -1.0

-1.6-1.6 + 0.6 � -1.0

y + 0.6 = -1.0

y = -1.6

y + 0.6 - 0.6 = -1.0 - 0.6

y + 0.6 = -1.0

y + 0.6 = -1.0

10 = 10

17 - 7 � 10

x - 7 = 10

x - 7 = 10

x = 17

x = 17x - 7 + 7 = 10 + 7

x - 7 = 10

x - 7 = 10

number = xx = number

x – 2 + 2

5

x – 2 + 2 5 + 2

Practice Problem 2Solve: y + 1.7 = 0.3

Answers✓ Concept Check: 51. , 2. y = -1.4x = 13


HELPFUL HINT

We may solve an equation so that the variable is alone on either side ofthe equation. For example, is equivalent to .x = 5

454 = x

Example 3 Solve:

Solution: To get x alone, we add to both sides.

Add to both sides.

The LCD is 4.

Add the fractions.


Replace x with .

Subtract.

True.

The solution is .54

12

= 12

12

�24

54

12

�54

- 34

12

= x - 34

54

= x

24

+ 34

= x

12

# 22

+ 34

= x

34

12

+ 34

= x - 34

+ 34

12

= x - 34

34

12

= x - 34

Practice Problem 3

Solve: 78

= y - 13

Example 4 Solve:

Solution: To solve for t, we first want all terms containing t on oneside of the equation. To do this, we subtract 5t from bothsides of the equation.

Subtract 5t from both sides.

Combine like terms.


Replace t with .

True.

The solution is .-5

-30 = -30

-25 - 5 � -30

-55 A- 5 B - 5 � 6 A-5 B5t - 5 = 6t

-5 = t

5t - 5 - 5t = 6t - 5t

5t - 5 = 6t

5t - 5 = 6t Practice Problem 4Solve: 3x + 10 = 4x

Answers

3. , 4. x = 10y = 2924


Practice Problem 5Solve:

= -2w + 3 + 7w10w + 3 - 4w + 4

Practice Problem 6Solve: = -33 12w - 5 2 - 15w + 1 2

SIMPLIFYING EQUATIONS

Many times, it is best to simplify one or both sides of an equation beforeapplying the addition property of equality.

Example 5 Solve:

Solution: First we simplify both sides of the equation.

Next, we want all terms with a variable on one side of theequation and all numbers on the other side.

Subtract 4x from both sides.

Combine like terms.Subtract 2 from both sides toget x alone.

Combine like terms.


Replace x with .

Multiply.

True.

The solution is .

If an equation contains parentheses, we use the distributive property toremove them, as before. Then we combine any like terms.

Example 6 Solve:

Solution:

Multiply.

Combine like terms.

Simplify.

Check: Check by replacing a with 19 in the original equation.

Example 7 Solve:

Solution: First we subtract 3 from both sides.

Subtract 3 from both sides.

Simplify. -x = 4

3 - x - 3 = 7 - 3

3 - x = 7

3 - x = 7

a = 19

Add 12 toboth sides.a - 12 + 12 = 7 + 12

a - 12 = 7

12a - 6 - 11a - 6 = 7

Apply the distrib-utive property.6 12a 2 + 6 1-1 2 - 1 111a 2 - 1 16 2 = 7

6 12a - 1 2 - 1 111a + 6 2 = 7

6 12a - 1 2 - 111a + 6 2 = 7

-3

-13 = -13

-6 - 9 - 5 + 7 � -30 + 3 + 18 - 4-3

2 1-3 2 + 3 1-3 2 - 5 + 7 � 10 1-3 2 + 3 - 6 1-3 2 - 4

2x + 3x - 5 + 7 = 10x + 3 - 6x - 4

x = -3

x + 2 - 2 = -1 - 2

x + 2 = -1

5x + 2 - 4x = 4x - 1 - 4x

Combine like terms oneach side of the equation.

5x + 2 = 4x - 1

2x + 3x - 5 + 7 = 10x + 3 - 6x - 4

2x + 3x - 5 + 7 = 10x + 3 - 6x - 4

B


Answers5. , 6. , 7. y = 3w = 13w = -4

12 - y = 9


Answer8. 11 - x

We have not yet solved for x since x is not alone. However,this equation does say that the opposite of x is 4. If theopposite of x is 4, then x is the opposite of 4, or .

If ,

then .


bReplace x with .

Add.

True.

The solution is .


In this section, we continue to practice writing algebraic expressions.

Example 8 a. The sum of two numbers is 8. If one number is 3, findthe other number.

b. The sum of two numbers is 8. If one number is x, writean expression representing the other number.

Solution: a. If the sum of two numbers is 8 and one number is 3, wefind the other number by subtracting 3 from 8. Theother number is , or 5.

b. If the sum of two numbers is 8 and one number is x, wefind the other number by subtracting x from 8. Theother number is represented by .

x 8 − x

8

8 - x

3 8 − 3 or 5

8

8 - 3

C

-4

7 = 7

3 + 4 � 7

-43 - 1-4 2 � 7

3 - x = 7

x = -4

-x = 4

x = -4

Practice Problem 8The sum of two numbers is 11. If onenumber is x, write an expression repre-senting the other number.

b


THE GOLDEN RECTANGLE IN ART

The golden rectangle is a rectangle whose length is approximately 1.6 times its width. The earlyGreeks thought that a rectangle with these dimensions was the most pleasing to the eye. Examples of the golden rectangle are found in many ancient, as well as modern, works ofart. For example, the Parthenon in Athens, Greece, shows the golden rectangle in many aspectsof its design. Modern-era artists, including Piet Mondrian (1872–1944) and Georges Seurat(1859–1891), also frequently used the proportions of a golden rectangle in their paintings.

To test whether a rectangle is a golden rectangle, divide the rectangle’s length by its width. Ifthe result is approximately 1.6, we can consider the rectangle to be a golden rectangle. Forinstance, consider Mondrian’s Composition with Gray and Light Brown, which was painted on

an cm canvas. Because , the dimensions of the canvas form a golden rec-

tangle. In what other ways are golden rectangles connected with this painting?Examples of golden rectangles can be found in the designs of many everyday objects. Visual

artists, from architects to product and package designers, use the golden rectangle shape in suchthings as the face of a building, the floor of a room, the front of a food package, the front coverof a book, and even the shape of a credit card.

GROUP ACTIVITY

Find an example of a golden rectangle in a building or an everyday object. Use a ruler to mea-sure its dimensions and verify that the length is approximately 1.6 times the width.

80.249.9

L 1.680.2 * 49.9

Focus On History

Mondrian

Composition with Grayand Light Brown 1918,Oil on canvas, 80.2 x 49.9cm (31 9/16 x 19 5/8 in);Museum of Fine Arts,Houston, Texas

Name ____________________________________ Section _________ Date _________

231

EXERCISE SET 2.2

Solve each equation. Check each solution. See Examples 1 through 4.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. c + 16

= 38

13

+ f = 34

t - 9.2 = -6.8r - 8.6 = -8.1

8 + z = -83 + x = -11y - 9 = 1

x - 2 = -4x + 14 = 25x + 7 = 10

A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

The Multiplication Property of Equality SECTION 2.3 233

MULTIPLICATION PROPERTY OF EQUALITY

If a, b, and c are real numbers and , then

and

are equivalent equations.

ac = bca = b

c Z 0

SSM CD-ROM Video2.3

Objectives

Use the multiplication property ofequality to solve linear equations.

Use both the addition and multi-plication properties of equality tosolve linear equations.


C

B

A

2.3 THE MULTIPLICATION PROPERTY OF EQUALITY

USING THE MULTIPLICATION PROPERTY

As useful as the addition property of equality is, it cannot help us solveevery type of linear equation in one variable. For example, adding or sub-tracting a value on both sides of the equation does not help solve

Instead, we apply another important property of equality, the multiplica-tion property of equality.

52

x = 15

A

Practice Problem 1

Solve: 37

x = 9

This property guarantees that multiplying both sides of an equation bythe same nonzero number does not change the solution of the equation.Since division is defined in terms of multiplication, we may also divide bothsides of the equation by the same nonzero number without changing thesolution.

Example 1 Solve:

Solution: To get x alone, we multiply both sides of the equation by

the reciprocal of which is .

Multiply both sides by .

Apply the associative property.

Simplify.

or

Check: Replace x with 6 in the original equation.

Original equation.

Replace x with 6.

True.

The solution is 6.

15 = 15

5216 2 � 15

52

x = 15

x = 6

1x = 6

a 25

# 52bx = 2

5# 15

25

25

# a 52

x b = 25

# 15

52

x = 15

25

52

,

52

x = 15

Answer1. x = 21


Practice Problem 2Solve: 7x = 42

Practice Problem 3Solve: -4x = 52

In the equation , is the coefficient of x. When the coefficient

of x is a fraction, we will get x alone by multiplying by the reciprocal. Whenthe coefficient of x is an integer or a decimal, it is usually more convenientto divide both sides by the coefficient. (Dividing by a number is, of course,the same as multiplying by the reciprocal of the number.)

Example 2 Solve:

Solution: To get x alone, we divide both sides of the equation by 5,the coefficient of x.

Divide both sides by 5.

Simplify.


Replace x with 6.

True.

The solution is 6.

Example 3 Solve:

Solution: Recall that means . To get x alone, we divideboth sides by the coefficient of x, that is, .

Divide both sides by .

Simplify.


Replace x with .

True.

The solution is .

Example 4 Solve:

Solution: Recall that . To get y alone, we multiply both sides

of the equation by 7, the reciprocal of .

Multiply both sides by 7.

Simplify.

y = 140

1y = 140

7 # 17

y = 7 # 20

17

y = 20

y7

= 20

17

y7

= 17

y

y7

= 20

-11

33 = 33

-11-3 1-11 2 � 33

-3x = 33

x = -11

1x = -11

-3-3x-3

= 33-3

-3x = 33

-3-3 # x-3x

-3x = 33

30 = 30

5 # 6 � 30

5x = 30

x = 6

1 # x = 6

5x5

= 305

5x = 30

5x = 30

52

52

x = 15

Practice Problem 4

Solve:

Answers2. , 3. , 4. y = 65x = -13x = 6

y5

= 13


Practice Problem 5Solve: 2.6x = 13.52

Practice Problem 6

Solve: - 56

y = - 35

Practice Problem 7Solve: -x + 7 = -12


Replace y with 140.

True.

The solution is 140.

Example 5 Solve:

Solution:

Divide both sides by 3.1.

Simplify.

Check: Check by replacing x with 1.6 in the original equation. Thesolution is 1.6.

Example 6 Solve:

Solution: To get x alone, we multiply both sides of the equation by

, the reciprocal of the coefficient of x.

Multiply both sides by ,

the reciprocal of .

Simplify.

Check: Check by replacing x with in the original equation. The

solution is .

USING BOTH THE ADDITION AND MULTIPLICATION PROPERTIES

We are now ready to combine the skills learned in the last section with theskills learned from this section to solve equations by applying more thanone property.

Example 7 Solve:

Solution: First, to get , the term containing the variable alone, weadd 4 to both sides of the equation.


Simplify.

Next, recall that means . Thus to get z alone, weeither multiply or divide both sides of the equation by

. In this example, we divide.

Divide both sides by the coefficient .

Simplify.

z = -10

1z = -10

-1-z-1

= 10-1

-z = 10

-1

-1 # z-z

-z = 10

-z - 4 + 4 = 6 + 4

-z

-z - 4 = 6

B

154

154

x = 154

- 23

- 32

- 32

# - 23

x = - 32

# - 52

- 23

x = - 52

- 32

- 23

x = - 52

x = 1.6

1x = 1.6

3.1x3.1

= 4.963.1

3.1x = 4.96

3.1x = 4.96

20 = 20

1407

� 20

y7

= 20

Answers5. , 6. , 7. x = 19y = 18

25x = 5.2


Practice Problem 8Solve: -7x + 2x + 3 - 20 = -2

Practice Problem 9If x is the first of two consecutive inte-gers, express the sum of the first andthe second integer in terms of x. Sim-plify if possible.


Replace z with .

True.

The solution is .

Example 8 Solve:

Solution: First, we simplify both sides of the equation by combininglike terms.

Combine like terms.


Simplify.


Simplify.

Check: To check, replace a with in the original equation. Thesolution is .


We continue to sharpen our problem-solving skills by writing algebraicexpressions.

Example 9 Writing an Expression for Consecutive IntegersIf x is the first of three consecutive integers, express thesum of the three integers in terms of x. Simplify if possible.

Solution: An example of three consecutive integers is

The second consecutive integer is always 1 more than thefirst, and the third consecutive integer is 2 more than thefirst. If x is the first of three consecutive integers, the threeconsecutive integers are

Their sum is

In words:

Translate: x

which simplifies to .

Study these examples of consecutive even and odd integers.

Even integers:

x + 2, x + 4

+4+2

x,

10 128

3x + 3

1x + 2 2+1x + 1 2+

thirdinteger+

secondinteger+

firstinteger

x + 1 x + 2

+2+1

x

8 9

+2+1

7

C

-5-5

a = -5

2a2

= -102

2a = -10

2a - 3 + 3 = -13 + 3

2a - 3 = -13

a + a - 10 + 7 = -13

a + a - 10 + 7 = -13

-10

6 = 6

10 - 4 � 6

-10- 1-10 2 - 4 � 6

-z - 4 = 6

Answers8. , 9. 2x + 1x = -3

Odd integers:

x + 2, x + 4

+4+2

x,7 95


HELPFUL HINTIf x is an odd integer, then is the next odd integer. This 2 simplymeans that odd integers are always 2 units from each other.

2units

2units

2units

2units

0 1 5432−3 −2 −1

x + 2T

Focus On The Real WorldSURVEYS

Recall that the golden rectangle is a rectangle whose length is ap-proximately 1.6 times its width. It is thought that for about 75% ofadults, a rectangle in the shape of the golden rectangle is mostpleasing to the eye.

GROUP ACTIVITIES

1. Measure the dimensions of each of the three rectangles shownabove and decide which one best approximates the shape ofthe golden rectangle.

2. Using the three rectangles shown above, conduct a survey ask-ing students which rectangle they prefer. (To avoid bias, takecare not to reveal which rectangle is the golden rectangle.)Tally your results and find the percent of survey respondentswho preferred each rectangle. Do your results agree with thepercent suggested above?

(a) (b) (c)

239

Name ____________________________________ Section ________ Date ___________ MENTAL MATH ANSWERS

1.

2.

3.

4.

5.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

MENTAL MATH

Solve each equation mentally. See Examples 2 and 3.

1. 2. 3.4. 5.

EXERCISE SET 2.3

Solve each equation. Check each solution. See Examples 1 through 6.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10.

Solve each equation. Check each solution. See Examples 7 and 8.

11. 12. 13.

14. 15. 6a + 3 = 3-x + 4 = -24

-x + 2 = 223x - 1 = 262x - 4 = 16

B

18

v = 14

16

d = 12

34

n = -1523

x = -8

-y = 8-x = -122x = 0

3x = 07x = 49-5x = 20

A

6x = -307t = 145b = 109c = 543a = 27

Further Solving Linear Equations SECTION 2.4 241

Practice Problem 1Solve: 5 13x - 1 2 + 2 = 12x + 6

Answer1. x = 3

SSM CD-ROM Video2.4

Objectives

Apply the general strategy forsolving a linear equation.

Solve equations containing frac-tions or decimals.

Recognize identities and equationswith no solution.

C

B

A

2.4 FURTHER SOLVING LINEAR EQUATIONS

SOLVING LINEAR EQUATIONS

We now combine our knowledge from the previous sections into a generalstrategy for solving linear equations. One new piece in this strategy is a sug-gestion to “clear an equation of fractions” as a first step. Doing so makesthe equation more manageable, since working with integers is more conve-nient than working with fractions.

Example 1 Solve:

Solution: There are no fractions, so we begin with Step 2.

Step 2.Step 3.Step 4. Get all variable terms on the same side of the

equation by subtracting 3x from both sides; thenadding 5 to both sides.

Step 5. Use the multiplication property of equality to get xalone.


Simplify.

Step 6. Check.

Original equation

Replace x with 2.

True.

The solution is 2.

11 = 11

4 + 7 � 11

4 11 2 + 7 � 11

4 14 - 3 2 + 7 � 6 + 5

4 32 12 2 - 3 4 + 7 � 3 12 2 + 5

4 12x - 3 2 + 7 = 3x + 5

x = 2

5x5

= 105

5x = 10

5x - 5 + 5 = 5 + 5

5x - 5 = 5

Subtract 3x from bothsides.Simplify.


Simplify.

8x - 5 - 3x = 3x + 5 - 3x

8x - 5 = 3x + 5


Combine like terms.

8x - 12 + 7 = 3x + 5

4 12x - 3 2 + 7 = 3x + 5

4 12x - 3 2 + 7 = 3x + 5

TO SOLVE LINEAR EQUATIONS IN ONE VARIABLE

Step 1. Multiply on both sides to clear the equation of fractions if theyoccur.

Step 2. Use the distributive property to remove parentheses if theyoccur.

Step 3. Simplify each side of the equation by combining like terms.

Step 4. Get all variable terms on one side and all numbers on the otherside by using the addition property of equality.

Step 5. Get the variable alone by using the multiplication property ofequality.

Step 6. Check the solution by substituting it into the original equation.

A


HELPFUL HINTWhen checking solutions, use the original written equation.

Practice Problem 2Solve: 9 15 - x 2 = -3x

Example 2 Solve:

Solution: First, we apply the distributive property.

Step 2.Step 4.

Step 5.

Step 6. Check.

Original equation

Replace t with .

The LCD is 3.

Subtract fractions.

True.

The solution is .

SOLVING EQUATIONS CONTAINING FRACTIONS OR DECIMALS

If an equation contains fractions, we can clear the equation of fractions bymultiplying both sides by the LCD of all denominators. By doing this, weavoid working with time-consuming fractions.

Example 3 Solve:

Solution: We begin by clearing fractions. To do this, we multiplyboth sides of the equation by the LCD of 2 and 3, whichis 6.

Step 1. Multiply bothsides by theLCD, 6.

6 a x2

- 1 b = 6 a 23

x - 3 b

x2

- 1 = 23

x - 3

x2

- 1 = 23

x - 3

B

163

- 803

� - 803

8 a - 103b � - 80

3

8 a 63

- 163b � - 80

3

163

8 a2 - 163b � -5 a 16

3b

8 12 - t 2 = -5t

163

= t

163

= 3t3

16 = 3t

16 - 8t + 8t = -5t + 8t

Use the distributiveproperty.

Add 8t to both sides.

Combine like terms.


Simplify.

16 - 8t = -5t

8 12 - t 2 = -5t

8 12 - t 2 = -5t

Practice Problem 3

Solve: 52

x - 1 = 32

x - 4

Answers

2. , 3. x = -3x = 152


Step 2.

ƒ ƒ ƒ ƒSimplify.

There are no longer grouping symbols and no like termson either side of the equation, so we continue with Step 4.

Step 4.

Step 5. The variable is now alone, so there is no need toapply the multiplication property of equality.

Step 6. Check.

Original equation

Replace x with 12.

Simplify.

True.

The solution is 12.

Example 4 Solve:

Solution: We clear the equation of fractions first.

Step 1.

Step 2. Next, we use the distributive property and removeparentheses.

Step 4.

Step 5. Divide both sides by .

Step 6. To check, replace a with 0 in the original equation.The solution is 0.

Write the fraction in sim-plest form. a = 0

-16 -16a- 16

= 0-16

-16a = 0Subtract 18a fromboth sides.

2a - 18a = 18a - 18a

2a = 18a

Subtract 6 from bothsides.

2a + 6 - 6 = 18a + 6 - 6


2a + 6 = 18a + 6

Clear the fraction by mul-tiplying both sides by theLCD, 3.

3 # 2 1a + 3 23

= 3 16a + 2 2

2 1a + 3 2

3= 6a + 2

2 1a + 3 23

= 6a + 2

5 = 5

6 - 1 � 8 - 3

122

- 1 �23

# 12 - 3

x2

- 1 = 23

x - 3

12 = x

-6 + 18 = x - 18 + 18

-6 = x - 18

Subtract 3x fromboth sides.Simplify.


Simplify.

3x - 6 - 3x = 4x - 18 - 3x

3x - 6 = 4x - 18

HELPFUL HINTDon’t forgetto multiplyeach term bythe LCD.

3x - 6 = 4x - 18

Apply the distrib-utive property.

6 a x2b - 6 11 2 = 6 a 2

3x b - 6 13 2

Practice Problem 4

Solve: 3 1x - 2 2

5= 3x + 6

Answer4. x = -3


When solving a problem about money, you may need to solve an equationcontaining decimals. If you choose, you may multiply to clear the equationof decimals.

Example 5 Solve:

Solution: First we clear this equation of decimals by multiplyingboth sides of the equation by 100. Recall that multiplyinga decimal number by 100 has the effect of moving thedecimal point 2 places to the right.

Step 1.� � �

Step 2.Step 3.

Step 4.

Step 5.

Step 6. To check, replace x with 4 in the original equation.The solution is 4.

RECOGNIZING IDENTITIES AND EQUATIONS WITH NO SOLUTION

So far, each equation that we have solved has had a single solution. How-ever, not every equation in one variable has a single solution. Some equa-tions have no solution, while others have an infinite number of solutions.For example,

has no solution since no matter which real number we replace x with, theequation is false.

FALSE

On the other hand,

has infinitely many solutions since x can be replaced by any real numberand the equation is always true.

TRUE

The equation is called an identity. The next few examplesillustrate special equations like these.

x + 6 = x + 6

real number + 6 = same real number + 6

x + 6 = x + 6

real number + 5 = same real number + 7

x + 5 = x + 7

C

x = 4

35x35

= 14035

35x = 140

Add 30 to bothsides.Combine liketerms.Divide both sidesby 35.

35x - 30 + 30 = 110 + 30

35x - 30 = 110

Apply the distribu-tive property.Combine like terms.

25x + 10x - 30 = 110

25x + 10 1x - 3 2 = 5 122 2Multiply bothsides by 100.

0.25x + 0.10 1x - 3 2 = 0.05 122 20.25x + 0.10 1x - 3 2 = 0.05 122 2

0.25x + 0.10 1x - 3 2 = 0.05 122 2Practice Problem 5Solve:0.06x - 0.10 1x - 2 2 = -0.02 18 2

Answer5. x = 9


Practice Problem 6Solve: 5 12 - x 2 + 8x = 3 1x - 6 2


-10 1x + 2 2 - 2x-6 12x + 1 2 - 14 =

✓ CONCEPT CHECK

Suppose you have simplified severalequations and obtain the followingresults. What can you conclude aboutthe solutions to the original equation?a. b. c. 7 = -4x = 07 = 7

Example 6 Solve:

Solution:

The final equation contains no variable terms, and there is no value for x that makes a trueequation. We conclude that there is no solution to thisequation.

Example 7 Solve:

Solution:


The left side of the equation is now identical to the rightside. Every real number may be substituted for x and atrue statement will result. We arrive at the same conclu-sion if we continue.

Again, one side of the equation is identical to the otherside. Thus, is an identity and everyreal number is a solution.


3 1x - 4 2 = 3x - 12

0 = 0

3x - 3x = 3x - 3x

3x = 3x


Combine like terms.


3x - 12 + 12 = 3x - 12 + 12

3x - 12 = 3x - 12

3x - 12 = 3x - 12

3 1x - 4 2 = 3x - 12

3 1x - 4 2 = 3x - 12

20 = -6

20 = -6

-2x + 20 + 2x = -2x - 6 + 2x

-2x + 20 = -2x - 6

Apply the distributiveproperty on both sides.Combine like terms.

Add 2x to both sides.

Combine like terms.

-2x + 10 + 10 = -3x - 6 + x

-2 1x - 5 2 + 10 = -3 1x + 2 2 + x

-2 1x - 5 2 + 10 = -3 1x + 2 2 + x

Answers6. no solution, 7. Every real number is a solu-tion.

✓ Concept Check:a. Every real number is a solution., b. The solution is 0., c. There is no solution.


CALCULATOR EXPLORATIONSCHECKING EQUATIONS

We can use a calculator to check possible solutions of equations. Todo this, replace the variable by the possible solution and evaluateboth sides of the equation separately.

Equation: Solution:

Original equation

Replace x with 16.

Now evaluate each side with your calculator.

Evaluate left side: Display:or

Evaluate right side: Display:or

Since the left side equals the right side, the equation checks.

Use a calculator to check the possible solutions to each equation.

1.

2.

3.

4.

5.

6. 20 1x - 39 2 = 5x - 432; x = 23.2

564x4

= 200x - 11 1649 2 ; x = 121

-1.6x - 3.9 = -6.9x - 25.6; x = 5

5x - 2.6 = 2 1x + 0.8 2 ; x = 4.4

-3x - 7 = 3x - 1; x = -1

2x = 48 + 6x; x = -12

ENTER

44=)6+16(2

ENTER

44=4-16*3

3 116 2 - 4 � 2 116 + 6 23x - 4 = 2 1x + 6 2

x = 163x - 4 = 2 1x + 6 2

An Introduction to Problem Solving SECTION 2.5 247

Objective

Translate a problem to an equa-tion, then use the equation to solvethe problem.

A

SSM CD-ROM Video2.5

Practice Problem 1Three times the difference of a numberand 5 is the same as twice the numberdecreased by 3. Find the number.

2.5 AN INTRODUCTION TO PROBLEM SOLVING

In the preceding sections, we practiced translating phrases into expressionsand sentences into equations as well as solving linear equations. We arenow ready to put our skills to practical use. To begin, we present a generalstrategy for problem solving.

TRANSLATING AND SOLVING PROBLEMS

Much of problem solving involves a direct translation from a sentence to anequation.

Example 1 Finding an Unknown NumberTwice the sum of a number and 4 is the same as four timesthe number decreased by 12. Find the number.

Solution: 1. UNDERSTAND. Read and reread the problem. If welet

the unknown number, then

“the sum of a number and 4” translates to “ ” and

“four times the number” translates to “ ”

2. TRANSLATE.

T T T T T T2 12-4x=1x + 4 2

12decreasedby

four timesthe number

is thesame as

sum of anumber and 4

twice

4x

x + 4

x =

A

GENERAL STRATEGY FOR PROBLEM SOLVING

1. UNDERSTAND the problem. During this step, become comfort-able with the problem. Some ways of doing this are:

Read and reread the problem.Choose a variable to represent the unknown.Construct a drawing.Propose a solution and check. Pay careful attention to how youcheck your proposed solution. This will help when writing anequation to model the problem.

2. TRANSLATE the problem into an equation.3. SOLVE the equation.4. INTERPRET the results: Check the proposed solution in the stated

problem and state your conclusion.

Answer1. The number is 12.


Practice Problem 2An 18-foot wire is to be cut so that thelonger piece is 5 times longer than theshorter piece. Find the length of eachpiece.

3. SOLVE.

4. INTERPRET.

Check: Check this solution in the problem as it was originallystated. To do so, replace “number” with 10. Twice thesum of “10” and 4 is 28, which is the same as 4 times “10”decreased by 12.

State: The number is 10.

Example 2 Finding the Length of a BoardA 10-foot board is to be cut into two pieces so that thelonger piece is 4 times the shorter. Find the length of eachpiece.

Solution: 1. UNDERSTAND the problem. To do so, read and re-read the problem. You may also want to propose asolution. For example, if 3 feet represents the length ofthe shorter piece, then feet is the length ofthe longer piece, since it is 4 times the length of theshorter piece. This guess gives a total board length of3 feet 12 feet 15 feet, too long. However, thepurpose of proposing a solution is not to guess cor-rectly, but to help better understand the problem andhow to model it.

In general, if we let

2. TRANSLATE the problem. First, we write the equa-tion in words.

T T T T T10=4x+x

totallengthof board

equalslength oflongerpiece

added tolength ofshorterpiece

x feet 4x feet

10 feet

4x = length of longer piece

x = length of shorter piece, then

=+

4 13 2 = 12

x = 10

Divide both sides by -2. -2x-2

= -20-2

-2x = -20

Subtract 8 from both sides.-2x + 8 - 8 = -12 - 8

-2x + 8 = -12

Subtract 4x from both sides. 2x + 8 - 4x = 4x - 12 - 4x

Apply the distributive property. 2x + 8 = 4x - 12

2 1x + 4 2 = 4x - 12

Answer2. shorter piece 3 ft; longer piece 15 ft==


3. SOLVE.

4. INTERPRET.

Check: Check the solution in the stated problem. If the shorterpiece of board is 2 feet, the longer piece is

and the sum of the two pieces is.

State: The shorter piece of board is 2 feet and the longer piece of board is 8 feet.

Example 3 Finding the Number of Republican and Democratic SenatorsIn a recent year, Congress had 8 more Republican sena-tors than Democratic. If the total number of senators is100, how many senators of each party were there?

Solution: 1. UNDERSTAND the problem. Read and reread theproblem. Let’s suppose that there are 40 Democraticsenators. Since there are 8 more Republicans thanDemocrats, there must be Republicans.The total number of Democrats and Republicans isthen . This is incorrect since the totalshould be 100, but we now have a betterunderstanding of the problem.

In general, if we let

2. TRANSLATE the problem. First, we write the equa-tion in words.

T T T T T100=1x + 8 2+x

100equalsnumber ofRepublicans

added tonumber ofDemocrats

x + 8 = number of Republicans

x = number of Democrats, then

40 + 48 = 88

40 + 8 = 48

HELPFUL HINT:

Make sure thatunits are includedin your answer, ifappropriate.

2 feet + 8 feet = 10 feet4 # 12 feet 2 = 8 feet

x = 2

5x5

= 105

Combine like terms.


5x = 10

x + 4x = 10

Practice Problem 3In a recent year, the total number ofDemocrats and Republicans in theU.S. House of Representatives was433. There were 39 more Republicansthan Democrats. Find the number ofrepresentatives from each party.

Answer3.

Republicans = 236 representativesDemocrats = 197 representatives;


3. SOLVE.

4. INTERPRET.

Check: If there are 46 Democratic senators, then there are Republican senators. The total number of

senators is then . The results check.

State: There were 46 Democratic and 54 Republican senators.

Example 4 Calculating Cellular Phone UsageA local cellular phone company charges Elaine Chapoton$50 per month and $0.36 per minute of phone use in herusage category. If Elaine was charged $99.68 for a month’scellular phone use, determine the number of whole min-utes of phone use.

Solution: 1. UNDERSTAND. Read and reread the problem. Let’s propose that Elaine uses the phone for 70minutes. Pay careful attention as to how we calcu-late her bill. For 70 minutes of use, Elaine’s phone bill will be $50 plus $0.36 per minute of use. This is

, less than $99.68. We nowunderstand the problem and know that the number ofminutes is greater than 70.

If we let

2. TRANSLATE.

T T T T T50 99.68

3. SOLVE.

x = 138

0.36x0.36

= 49.680.36

0.36x = 49.68

Subtract 50 from bothsides.

Simplify.

Divide both sides by0.36.

Simplify.

50 + 0.36x - 50 = 99.68 - 50

50 + 0.36x = 99.68

=0.36x+

$99.68is equal tominute chargeadded to$50

0.36x = charge per minute of phone use

x = number of minutes, then

$50 + 0.36 170 2 = $75.20

46 + 54 = 10046 + 8 = 54

x = 46

2x2

= 922

2x = 92

2x + 8 - 8 = 100 - 8

Combine like terms.



2x + 8 = 100

x + 1x + 8 2 = 100

Practice Problem 4Enterprise Car Rental charges a dailyrate of $34 plus $0.20 per mile. Sup-pose that you rent a car for a day andyour bill (before taxes) is $104. Howmany miles did you drive?

Answer4. 350 miles


Answer5. smallest = 30°; second = 60°; third = 90°

4. INTERPRET.

Check: If Elaine spends 138 minutes on her cellular phone, herbill is .

State: Elaine spent 138 minutes on her cellular phone thismonth.

Example 5 Finding Angle MeasuresIf the two walls of the Vietnam Veterans Memorial inWashington D.C. were connected, an isosceles trianglewould be formed. The measure of the third angle is 97.5°more than the measure of either of the other two equalangles. Find the measure of the third angle. (Source:National Park Service)

Solution: 1. UNDERSTAND. Read and reread the problem. Wethen draw a diagram (recall that an isosceles trianglehas two angles with the same measure) and let

(x + 97.5)°

x° x°

x + 97.5 = degree measure of the third angle

x = degree measure of the second equal angle

x = degree measure of one angle

$50 + $0.36 1138 2 = $99.68

Practice Problem 5The measure of the second angle of atriangle is twice the measure of thesmallest angle. The measure of thethird angle of the triangle is threetimes the measure of the smallestangle. Find the measures of the angles.


2. TRANSLATE. Recall that the sum of the measures ofthe angles of a triangle equals 180.

T T T T T180

3. SOLVE.

Combine like terms.

4. INTERPRET.

Check: If , then the measure of the third angle is. The sum of the angles is then

, the correct sum.

State: The third angle measures 125°.*

(*This is rounded to the nearest whole degree. The twowalls actually meet at an angle of 125 degrees 12 minutes.)

27.5 + 27.5 + 125 = 180x + 97.5 = 125

x = 27.5

x = 27.5

3x3

= 82.53

3x = 82.5

Subtract 97.5 fromboth sides.

Divide both sidesby 3.

3x + 97.5 - 97.5 = 180 - 97.5

3x + 97.5 = 180

x + x + 1x + 97.5 2 = 180

=1x + 97.5 2+x+x

180equalsmeasureof thirdangle

measureof secondangle

measureof firstangle

Formulas and Problem Solving SECTION 2.6 253

Practice Problem 1A family is planning their vacation toDisney World. They will drive from asmall town outside New Orleans,Louisiana, to Orlando, Florida, a dis-tance of 700 miles. They plan to aver-age a rate of 55 miles per hour. Howlong will this trip take?

Answer

1. approximately hours128

11

SSM CD-ROM Video2.6

Objectives

Use formulas to solve problems.

Solve a formula or equation for oneof its variables.

BA

2.6 FORMULAS AND PROBLEM SOLVING

USING FORMULAS TO SOLVE PROBLEMS

A formula describes a known relationship among quantities. Many formu-las are given as equations. For example, the formula

stands for the relationship

Let’s look at one way that we can use this formula.If we know we traveled a distance of 100 miles at a rate of 40 miles per

hour, we can replace the variables d and r in the formula and findour time, t.

To solve for t, we divide both sides of the equation by 40.


Simplify.

The time traveled is hours, or hours.In this section, we solve problems that can be modeled by known for-

mulas. We use the same problem-solving strategy that was introduced inthe previous section.

Example 1 Finding Time Given Rate and DistanceA glacier is a giant mass of rocks and ice that flows down-hill like a river. Portage Glacier in Alaska is about 6 miles,or 31,680 feet, long and moves 400 feet per year. Icebergsare created when the front end of the glacier flows intoPortage Lake. How long does it take for ice at the head(beginning) of the glacier to reach the lake?

2 12

52

52

= t

10040

= 40t40

Replace d with 100 and r with 40.100 = 40t

Formula. d = rt

d = rt

distance = rate # time

d = r # t

A

Solution: 1. UNDERSTAND. Read and reread the problem. Theappropriate formula needed to solve this problem isthe distance formula, . To become familiar withthis formula, let’s find the distance that ice traveling ata rate of 400 feet per year travels in 100 years. To doso, we let time t be 100 years and rate r be the given400 feet per year, and substitute these values into theformula . We then have that distance

feet. Since we are interestedin finding how long it takes ice to travel 31,680 feet,we now know that it is less than 100 years.

Since we are using the formula , we let

2. TRANSLATE. To translate to an equation, we use theformula and let distance feet andrate feet per year.

b RLet and .

3. SOLVE. Solve the equation for t. To solve for t, divideboth sides by 400.


Simplify.

4. INTERPRET.

Check: To check, substitute 79.2 for t and 400 for r in thedistance formula and check to see that the distance is31,680 feet.

State: It takes 79.2 years for the ice at the head of PortageGlacier to reach the lake.

Example 2 Calculating the Length of a GardenCharles Pecot can afford enough fencing to enclose arectangular garden with a perimeter of 140 feet. If thewidth of his garden is to be 30 feet, find the length.

Solution: 1. UNDERSTAND. Read and reread the problem. Theformula needed to solve this problem is the formulafor the perimeter of a rectangle, . Beforecontinuing, let’s become familar with this formula.

P = perimeter of the garden

w = the width of the rectangular garden

l = the length of the rectangular garden

P = 2l + 2w

HELPFUL HINT

Don’t forget to include units, if appropriate.

79.2 = t

31,680400

= 400 # t400

r = 400d = 31,68031,680 = 400 # t

d = r # t

r = 400d = 31,680d = rt

d = distance from beginning of glacier to lake

r = rate or speed of ice

t = the time in years for ice to reach the lake

d = rt

d = 400 1100 2 = 40,000d = rt

d = rt


Practice Problem 2A wood deck is being built behind ahouse. The width of the deck must be18 feet because of the shape of thehouse. If there is 450 square feet ofdecking material, find the length of thedeck.

Answer2. 25 feet

18 ft?

w = 30 feet

l

Formulas and Problem Solving SECTION 2.6 255

2. TRANSLATE. To translate to an equation, we use theformula and let perimeter feetand width feet.

Let and .b

3. SOLVE.

Multiply 2(30).


Combine like terms.


4. INTERPRET.

Check: Substitute 40 for l and 30 for w in the perimeter formulaand check to see that the perimeter is 140 feet.

State: The length of the rectangular garden is 40 feet.

SOLVING A FORMULA FOR A VARIABLE

We say that the formula

is solved for d because d is alone on one side of the equation and the otherside contains no d’s. Suppose that we have a large number of problems tosolve where we are given distance d and rate r and asked to find time t. Inthis case, it may be easier to first solve the formula for t. To solvefor t, we divide both sides of the equation by r.

Divide both sides by r.

Simplify.

To solve a formula or an equation for a specified variable, we use thesame steps as for solving a linear equation. These steps are listed next.

TO SOLVE EQUATIONS FOR A SPECIFIED VARIABLE

Step 1. Multiply on both sides to clear the equation of fractions if theyoccur.

Step 2. Use the distributive property to remove parentheses if theyoccur.

Step 3. Simplify each side of the equation by combining like terms.

Step 4. Get all terms containing the specified variable on one side andall other terms on the other side by using the addition propertyof equality.

Step 5. Get the specified variable alone by using the multiplicationproperty of equality.

dr

= t

dr

= rtr

d = rt

d = rt

d = rt

B

40 = l

80 = 2l

140 - 60 = 2l + 60 - 60

140 = 2l + 60

140 = 2l + 2 130 2

140 = 2l + 2 130 2w = 30P = 140 P = 2l + 2w

w = 30P = 140P = 2l + 2w

R


Practice Problem 3Solve for r. (This formula isused to find the circumference C of acircle given its radius r.)

rC

C = 2pr

Practice Problem 4Solve for w.P = 2l + 2w

Example 3 Solve for l.

Solution: This formula is used to find the volume of a box. Tosolve for l, we divide both sides by wh.

Divide both sides by wh.

Simplify.

Since we have l alone on one side of the equation, wehave solved for l in terms of V, w, and h. Remember thatit does not matter on which side of the equation we getthe variable alone.

Example 4 Solve for x.

Solution: First we get mx alone by subtracting b from both sides.

Subtract b from both sides.

Combine like terms.

Next we solve for x by dividing both sides by m.

Simplify.

Example 5 Solve for h.

Solution: First let’s clear the equation of fractions by multiplyingboth sides by 2.

Multiply both sides by 2 to clear fractions.

Divide both sides by b to get h alone.

Simplify. 2Ab

= h

2Ab

= bhb

2A = bh

2 # A = 2 a bh2b

A = bh2

A = bh2

y - bm

= x

y - bm

= mxm

y - b = mx

y - b = mx + b - b

y = mx + b

y = mx + b

Vwh

= l

Vwh

= lwhwh

V = lwhh

wl

V = lwh

Practice Problem 5

Solve for b.

Answers

3. , 4. ,

5. b = 2A - a

w = P - 2l2

r = C2p

A = a + b2

257


EXERCISE SET 2.6

Substitute the given values into each given formula and solve for the unknown variable.See Examples 1 and 2.

A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

1. ; , (Area of a parallelogram)

b = 15A = 45A = bh 2. ; , (Distance formula)

t = 3d = 195d = rt

3. ; , ,

(Surface area of a special rectangularbox)

w = 3l = 7S = 102S = 4lw + 2wh 4. ; , ,

(Volume of a rectangular box)h = 3w = 8l = 14V = lwh

5. ; , ,

(Area of a trapezoid)b = 7

B = 11A = 180A = 121B + b 2h 6. ; , ,

(Area of a trapezoid)b = 3

B = 7A = 60A = 121B + h 2h

7. ; , ,

(Perimeter of a triangle)b = 10

a = 8P = 30P = a + b + c 8. ; ,

(Volume of a pyramid)

h = 5V = 45V = 13

Ah

9. ; (use the approx-imation 3.14 for )(Circumference of a circle)

p

C = 15.7C = 2pr 10. ; (use the approxi-mation 3.14 for )(Area of a circle)

p

r = 4.5A = pr2

11. ; , ,

(Simple interest formula)R = 0.05

P = 25,000I = 3750I = PRT 12. ; , ,

(Simple interest formula)T = 6

R = 0.055I = 1,056,000I = PRT

13. ; , (use

the approximation 3.14 for )(Volume of a cone)

p

r = 6V = 565.2V = 13pr2h 14. ; (use the approxi-

mation 3.14 for )(Volume of a sphere)

p

r = 3V = 43pr3

Solve each formula for the specified variable. See Examples 3 through 5.

15. for h 16. for r

17. for W 18. for n

19. for y 20. for y-x + y = 133x + y = 7

T = mnrV = LWH

C = 2prf = 5gh

B

2.7 PERCENT, RATIO, AND PROPORTION

SOLVING PERCENT EQUATIONS

Much of today’s statistics is given in terms of percent: a basketball player’sfree throw percent, current interest rates, stock market trends, and nutri-tion labeling, just to name a few. In this section, we first explore percent,percent equations, and applications involving percents. See Section R.3 if afurther review of percents is needed.

Example 1 The number 63 is what percent of 72?

Solution: 1. UNDERSTAND. Read and reread the problem. Next,let’s suppose that the percent is 80%. To check, wefind 80% of 72.

80% of

Close, but not 63. At this point, though, we have abetter understanding of the problem, we know the cor-rect answer is close to and greater than 80%, and weknow how to check our proposed solution later.

Let the unknown percent.

2. TRANSLATE. Recall that “is” means “equals” and“of” signifies multiplying. Let’s translate the sentencedirectly.

T T T T T63 72

3. SOLVE.


Write as a percent.

4. INTERPRET.

Check: Verify that 87.5% of 72 is 63.

State: The number 63 is 87.5% of 72.

Example 2 The number 120 is 15% of what number?

Solution: 1. UNDERSTAND. Read and reread the problem.

Let the unknown number.

2. TRANSLATE.

T T T T T120 15%

3. SOLVE.

Write 15% as 0.15.

Divide both sides by 0.15.

4. INTERPRET.

800 = x

120 = 0.15x

x#=

what numberof15%isthe number 120

x =

87.5% = x

0.875 = x

63 = 72x

#x=

72ofwhat percentisthe number 63

x =

72 = 0.80 172 2 = 57.6

A

Practice Problem 1The number 22 is what percent of 40?

Percent, Ratio, and Proportion SECTION 2.7 259

Practice Problem 2The number 150 is 40% of what num-ber?

Answers1. 55%, 2. 375

SSM CD-ROM Video2.7

Objectives

Solve percent equations.

Solve problems involving percents.

Write ratios as fractions.

Solve proportions.

Solve problems modeled by pro-portions.

EDCBA


Practice Problem 3Use the circle graph and part (c) ofExample 3 to answer each question.

a. What percent of homeowners spend$250–$4999 on yearly home mainte-nance?

b. What percent of homeowners spend$250 or more on yearly home main-tenance?

c. How many of the homeowners inthe town of Fairview might we ex-pect to spend from $250–$999 peryear on home maintenance.

Check: Check the proposed solution by finding 15% of 800 andverifying that the result is 120.

State: Thus, 120 is 15% of 800.

SOLVING PROBLEMS INVOLVING PERCENT

As mentioned earlier, percents are often used in statistics. Recall that thegraph below is called a circle graph or a pie chart. The circle or pie repre-sents a whole, or 100%. Each circle is divided into sectors (shaped likepieces of a pie) that represent various parts of the whole 100%.

Example 3 The circle graph below shows how much moneyhomeowners in the United States spend annually onmaintaining their homes. Use this graph to answer thequestions below.

a. What percent of homeowners spend under $250 onyearly home maintenance?

b. What percent of homeowners spend less than $1000per year on home maintenance?

c. How many of the 22,000 homeowners in a town calledFairview might we expect to spend under $250 a yearon home maintenance?

Solution: a. From the circle graph, we see that 45% ofhomeowners spend under $250 per year on homemaintenance.

b. From the circle graph, we know that 45% of home-owners spend under $250 per year and 38% of home-owners spend $250–$999 per year, so that the sum

or 83% of homeowners spend less than$1000 per year.

c. Since 45% of homeowners spend under $250 per yearon maintenance, we find 45% of 22,000.

We might then expect that 9900 homeowners inFairview spend under $250 per year on home mainte-nance.

WRITING RATIOS AS FRACTIONS

A ratio is the quotient of two numbers or two quantities. For example,a percent can be thought of as a ratio, since it is the quotient of a numberand 100.

C

= 9900

45% of 22,000 = 0.45 122,000 2

45% + 38%

45%Under $250

38%$250–$999

16%$1000–$4999

1%$5000

or more

Yearly Home Maintenance in the U.S.

B

Answers3. a. 54%, b. 55%, c. 8360


RATIO

The ratio of a number a to a number b is their quotient. Ways of writ-ing ratios are

a to b, a : b,ab

or the ratio of 53 to 10053% = 53100

Practice Problem 4Write a ratio for each phrase. Use frac-tional notation.

a. The ratio of 3 parts oil to 7 partsgasoline

b. The ratio of 40 minutes to 3 hours

Example 4 Write a ratio for each phrase. Use fractional notation.

a. The ratio of 2 parts salt to 5 parts waterb. The ratio of 18 inches to 2 feet

Solution: a. The ratio of 2 parts salt to 5 parts water is .b. First we convert to the same unit of mea-

surement. For example,

The ratio of 18 inches to 2 feet is then

, or in lowest terms.

SOLVING PROPORTIONS

Ratios can be used to form proportions. A proportion is a mathematicalstatement that two ratios are equal.

For example, the equation

is a proportion that says that the ratios and are equal.

Notice that a proportion contains four numbers. If any three numbersare known, we can solve and find the fourth number. One way to do so is to use cross products. To understand cross products, let’s start with theproportion

and multiply both sides by the LCD, bd.

Multiply both sides by the LCD, bd.

Simplify.

Q across product cross product

ad = bc

bd a abb = bd a c

db

ab

= cd

ab

= cd

48

12

12

= 48

D

34

1824

2 feet = 2 # 12 inches = 24 inches

25

Answers

4. a. , b. 29

37

F F


CROSS PRODUCTS

If , then .ad = bcab

= cd

Notice why ad and bc are called cross products.

ad

bc

=a

dbc

Practice Problem 5

Solve for x: 38

= 63x

Example 5 Solve for x:

Solution: To solve, we set cross products equal.

Set cross products equal.

Multiply.


Simplify.

Check: To check, substitute 63 for x in the original proportion.The solution is 63.

Example 6 Solve for x:

Solution:


Multiply.




Check: Verify that is the solution.312

x = 312

2x2

= 312

2x = 31

5x = 3x + 31

5x - 25 = 3x + 6

5 1x - 5 2 = 3 1x + 2 2

=x – 5

53x + 2

x - 53

= x + 25

63 = x

3155

= 5x5

315 = 5x

45 # 7 = x # 5

=457x5

45x

= 57

Practice Problem 6

Solve for x: 2x + 1

7= x - 3

5

Answers

5. , 6. x = - 263

x = 168


✓ CONCEPT CHECK

For which of the following equationscan we immediately use cross prod-ucts to solve for x?

a.

b.25

- x = 1 + x3

2 - x5

= 1 + x3

Practice Problem 7To estimate the number of people inJackson, population 50,000, who haveno health insurance, 250 people werepolled. Of those polled, 39 had noinsurance. How many people in thecity might we expect to be uninsured?


SOLVING PROBLEMS MODELED BY PROPORTIONS

Proportions can be used to model and solve many real-life problems. Whenusing proportions in this way, it is important to judge whether the solutionis reasonable. Doing so helps us to decide if the proportion has beenformed correctly, We use the same problem-solving strategy that was intro-duced in Section 2.5.

Example 7 Calculating Cost with a ProportionThree boxes of 3.5-inch high-density diskettes cost $37.47.How much should 5 boxes cost?

Solution: 1. UNDERSTAND. Read and reread the problem. Weknow that the cost of 5 boxes is more than the cost of3 boxes, or $37.47, and less than the cost of 6 boxes,which is double the cost of 3 boxes, or

. Let’s suppose that 5 boxes cost$60.00. To check, we see if 3 boxes is to 5 boxes as theprice of 3 boxes is to the price of 5 boxes. In otherwords, we see if

or


or

Not a true statement.

Thus, $60 is not correct but we now have a better un-derstanding of the problem.

2. TRANSLATE.

3. SOLVE.



4. INTERPRET.

x = 62.45

3x = 187.35

3x = 5 137.47 2 35

= 37.47x

35

= 37.47x

3 boxes5 boxes

= price of 3 boxesprice of 5 boxes

Let x = price of 5 boxes of diskettes.

180.00 = 187.35

3 160.00 2 = 5 137.47 223

= 812

3 boxes5 boxes

=price of 3 boxesprice of 5 boxes

2 1$37.47 2 = $74.94

E

Answers✓ Concept Check: a

7. 7800 people


HELPFUL HINT

The proportion could also have been used

to solve the problem above. Notice that the cross products are the same.

5 boxes3 boxes

= price of 5 boxesprice of 3 boxes

Check: Verify that 3 boxes is to 5 boxes as $37.47 is to $62.45.Also, notice that our solution is a reasonable one asdiscussed in Step 1.

State: Five boxes of high-density diskettes cost $62.45.

When shopping for an item offered in many different sizes, it is impor-tant to be able to determine the best buy, or the best price per unit. To findthe unit price of an item, divide the total price of the item by the total num-ber of units.

For example, if a 16-ounce can of green beans is priced at $0.88, its unitprice is

Example 8 Finding the Better BuyA supermarket offers a 14-ounce box of cereal for $3.79and an 18-ounce box of the same brand of cereal for $4.99.Which is the better buy?

Solution: To find the better buy, we compare unit prices. The fol-lowing unit prices were rounded to three decimal places.

Size Price Unit Price

14 ounce $3.79

18 ounce $4.99

The 14-ounce box of cereal has the lower unit price so it isthe better buy.

$4.9918

L $0.277

$3.7914

L $0.271

unit price = $0.8816

= $0.055

unit price = total pricenumber of units

Practice Problem 8Which is the better buy for the samebrand of toothpaste?

8 ounces for $2.59

10 ounces for $3.11

Answer8. 10 ounces

265


EXERCISE SET 2.7

Find each number described. See Examples 1 and 2.A1. What number is 16% of 70? 2. What number is 88% of 1000?

3. The number 28.6 is what percent of 52?

4. The number 87.2 is what percent of 436?

5. The number 45 is 25% of whatnumber?


7. Find 23% of 20. 8. Find 140% of 86.


10. The number 56.25 is 45% of whatnumber?

11. The number 144 is what percent of 480?

12. The number 42 is what percent of 35?

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Solve each proportion. See Examples 5 and 6.

13. 14. 15.

16. 17. 18.

19. 20. yy - 16

= 53

x - 3x

= 47

a5

= 32

4x6

= 72

94x

= 62

x10

= 59

x2

= 166

23

= x6

D

2.8 LINEAR INEQUALITIES AND PROBLEM SOLVING

Relationships among measurable quantities are not always described byequations. For example, suppose that a salesperson earns a base of $600 permonth plus a commission of 20% of sales. Find the minimum amount ofsales needed to receive a total income of at least $1500 per month. Here, thephrase “at least” implies that an income of $1500 or more is acceptable. Insymbols, we can write

This is an example of an inequality, which we will solve in Example 11.A linear inequality is similar to a linear equation except that the equal-

ity symbol is replaced with an inequality symbol, such as , , , or .

In this section, when we make definitions, state properties, or list stepsabout an inequality containing the symbol , we mean that the definition,property, or steps apply to an inequality containing the symbols , and

, also.

USING INTERVAL NOTATION

A solution of an inequality is a value of the variable that makes the inequal-ity a true statement. The solution set of an inequality is the set of all solu-tions. For example, the solution set of the inequality contains allnumbers greater than 2. In set notation, this solution set is written as

. Its graph is an interval on the number line since an infinite num-ber of values satisfy the variable. If we use open/closed-circle notation, thegraph of looks like the following:

In this text, a different graphing notation will be used to help us under-stand interval notation. Instead of an open circle, we use a parenthesis;instead of a closed circle, we use a bracket. With this new notation, thegraph of now looks like

and can be represented in interval notation as . The symbol isread “infinity” and indicates that the interval includes all numbers greater

qA2, q B2

Ex @x 7 2F

2{x | x > 2}

Ex @x 7 2F5x|x 7 26

x 7 2

A

≥7 , ≤

6

≥≤76

income ≥ 1500

Linear Inequalities and Problem Solving SECTION 2.8 267

Objective

Use interval notation.

Solve linear inequalities using theaddition property of inequality.

Solve linear inequalities using themultiplication property of inequal-ity.

Solve linear inequalities using bothproperties of inequality.

Solve problems that can be mod-eled by linear inequalities.

E

D

C

BA

LINEAR INEQUALITY IN ONE VARIABLE

A linear inequality in one variable is an inequality that can be writtenin the form

where a, b, and c are real numbers and . For example,

x3

≤ 53 Ax - 4 B 6 5x

4n ≥ n - 32y 6 03x + 5 ≥ 4

a Z 0

ax + b 6 c

SSM CD-ROM Video2.8


Answers1. ,

2. ,

3.

✓ Concept Check: Should be since aparenthesis is always used to enclose .q

A5, q B]

−0.5

−3−4 −2 −1−5 210 3 4 5

C-0.5, 2 B]

−3−4 −2 −1−5 210 3 4 5

A- q, 0 D−3−4 −2 −1−5 210 3 4 5

A-3, q B

Practice Problems 1–3Graph each set on a number line andthen write it in interval notation.

1.

2.

3.

−3−4 −2 −1−5 210 3 4 5

Ex @ -0.5 ≤ x 6 2F−3−4 −2 −1−5 210 3 4 5

Ex @x ≤ 0F−3−4 −2 −1−5 210 3 4 5

Ex @x 7 -3F

than 2. The left parenthesis indicates that 2 is not included in the interval.Using a left bracket, , would indicate that 2 is included in the interval. Thefollowing table shows three equivalent ways to describe an interval: in setnotation, as a graph, and in interval notation.

Set Notation Graph Interval Notation

Examples Graph each set on a number line and then write it ininterval notation.

1.

2.

3.


USING THE ADDITION PROPERTY OF INEQUALITY

Interval notation can be used to write solutions of linear inequalities. Tosolve a linear inequality, we use a process similar to the one used to solve alinear equation. We use properties of inequalities to write equivalentinequalities until the variable is alone on one side of the inequality.

B

A0.5, 3 D) ]−1 0 321

0.5Ex @0.5 6 x ≤ 3FA- q, -1 B(

−3 −2 −1 3210Ex @x 6 -1F

C2, q B]−2 −1 0 4321

Ex @x ≥ 2F

HELPFUL HINT

Notice that a parenthesis is always used to enclose q and -q.

Ca, b Bb(

a]Ex @a ≤ x 6 bF

Aa, b Db[

a)Ex @a 6 x ≤ bF

Ca, b Db[

a]Ex @a ≤ x ≤ bF

Aa, b Bb(

a)Ex @a 6 x 6 bF

Ca, q Ba]Ex @x ≥ aF

A- q, a Da[Ex @x ≤ aF

Aa, q Ba(Ex @x 7 aF

A- q, a Ba(Ex @x 6 aF

C

✓ CONCEPT CHECK

Explain what is wrong with writing theinterval .A5, q D


Answers4. ,

5. ,

−3−4 −2 −1−5 210 3 4 5

Ex @x 7 -4F , A-4, q B−3−4 −2 −1−5 210 3 4 5

Ex @x 6 -2F , A- q, -2 B

Practice Problem 4Solve: . Graph the solutionset.

−3−4 −2 −1−5 210 3 4 5

x + 3 6 1

In other words, we may add the same real number to both sides of aninequality, and the resulting inequality will have the same solution set. Thisproperty also allows us to subtract the same real number from both sides.

Example 4 Solve: . Graph the solution set.

Solution:


Simplify.

The solution set is , which in interval notationis . The graph of the solution set is


Solution: To get x alone on one side of the inequality, we subtract4x from both sides.

Simplify.

The solution set is , which in interval nota-tion is . The graph is

(−2−3 −1 0 21

A-2, q B Ex @x 7 -2F

-2 6 x or x 7 -2

Subtract 4x from bothsides.4x - 2 - 4x 6 5x - 4x

4x - 2 6 5x

4x - 2 6 5x

(4 8765

A- q, 7 BEx @x 6 7F

x 6 7

x - 2 + 2 6 5 + 2

x - 2 6 5

x - 2 6 5

Practice Problem 5Solve: . Graph the solu-tion set.

−3−4 −2 −1−5 210 3 4 5

3x - 4 6 4x

HELPFUL HINT

Don’t forget that means the same as .x 7 -2-2 6 x

ADDITION PROPERTY OF INEQUALITY

If a, b, and c are real numbers, then

and

are equivalent inequalities.

a + c 6 b + ca 6 b

HELPFUL HINT

In Example 4, the solution set is . This means that all num-bers less than 7 are solutions. For example, 6.9, 0, , 1, and are solutions, just to name a few. To see this, replace x in with each of these numbers and see that the result is a true inequality.

x - 2 6 5-56.7-p

Ex @x 6 7F



Solution:


Combine like terms.


Simplify.

The solution set is , which in interval nota-tion is . The graph of the solution set is

USING THE MULTIPLICATION PROPERTY OF INEQUALITY

Next, we introduce and use the multiplication property of inequality tosolve linear inequalities. To understand this property, let’s start with thetrue statement and multiply both sides by 2.


True.

The statement remains true.Notice what happens if both sides of are multiplied by .

False.

The inequality is a false statement. However, if the direction of theinequality sign is reversed, the result is

True.

These examples suggest the following property.

In other words, we may multiply both sides of an inequality by the samepositive real number, and the result is an equivalent inequality. We mayalso multiply both sides of an inequality by the same negative number andreverse the direction of the inequality symbol, and the result is an equivalentinequality. The multiplication property holds for division also, since divi-sion is defined in terms of multiplication.

HELPFUL HINT

Whenever both sides of an inequality are multiplied or divided by anegative number, the direction of the inequality symbol must bereversed to form an equivalent inequality.

6 7 -14

6 6 -14

6 6 -14

-3 A-2 B 6 7 A-2 B -3 6 7

-2-3 6 7

-6 6 14

-3 A2 B 6 7 A2 B -3 6 7

-3 6 7

C

]−7−8−9−10−11 −6

C-10, q B Ex @x ≥ -10F x ≥ -10

x + 4 - 4 ≥ -6 - 4

x + 4 ≥ -6

3x + 4 - 2x ≥ 2x - 6 - 2x

3x + 4 ≥ 2x - 6

3x + 4 ≥ 2x - 6

Answer6.

]−3−4 −2 −1−5 210 3 4 5

Ex @x ≥ 5F , C5, q B

MULTIPLICATION PROPERTY OF INEQUALITY

If a, b, and c are real numbers and c is positive, then andare equivalent inequalities.

If a, b, and c are real numbers and c is negative, then andare equivalent inequalities.ac 7 bc

a 6 bac 6 bc

a 6 b

Practice Problem 6Solve: . Graph thesolution set.

−3−4 −2 −1−5 210 3 4 5

5x - 1 ≥ 4x + 4


Answers7. ,

8. ,

9.

✓ Concept Check: a, d

C2, q B(

−3−4 −2 −1−5 210 3 4 5

Ex @x 7 -5F , A-5, q B]

−3−4 −2 −1−5 210 3 4 5

Ex @x ≤ 4F , A- q, 4 D

Practice Problem 7

Solve: . Graph the solution set.

−3−4 −2 −1−5 210 3 4 5

16

x ≤ 23


Solution:


Simplify.

The solution set is , which in interval notationis . The graph of this solution set is


Solution:

Simplify.

The solution set is , which is ininterval notation. The graph of the solution set is

TRY THE CONCEPT CHECK IN THE MARGIN.To solve linear inequalities in general, we follow steps similar to those

for solving linear equations.

USING BOTH PROPERTIES OF INEQUALITY

Many problems require us to use both properties of inequality.

Example 9 Solve: . Write the solution set ininterval notation.

5 - x ≤ 4x - 15

D

SOLVING A LINEAR INEQUALITY IN ONE VARIABLE

Step 1. Clear the equation of fractions by multiplying both sides of theinequality by the least common denominator (LCD) of all frac-tions in the inequality.

Step 2. Use the distributive property to remove grouping symbols suchas parentheses.

Step 3. Combine like terms on each side of the inequality.

Step 4. Use the addition property of inequality to write the inequality asan equivalent inequality with variable terms on one side andnumbers on the other side.

Step 5. Use the multiplication property of inequality to get the variablealone on one side of the inequality.

(−1−2−3−4−5 0

A-3, q BEx @x 7 -3F x 7 -3

Divide both sides by -2.3 andreverse the inequality symbol.

-2.3x-2.3

7 6.9-2.3

-2.3x 6 6.9

-2.3x 6 6.9

]6 7 8 9 102 3 4 5

A- q, 6 D Ex @x ≤ 6F x ≤ 6

4 # 14

x ≤ 4 # 32

14

x ≤ 32

14

x ≤ 32

Practice Problem 8Solve: . Graph the solu-tion set.

−3−4 −2 −1−5 210 3 4 5

-1.1x 6 5.5

HELPFUL HINT

The inequalitysymbol is thesame since we aremultiplying by apositive number.

ƒƒ

HELPFUL HINT

The inequal-ity symbol isreversedsince wedivided by anegativenumber.

ƒ

✓ CONCEPT CHECK

In which of the following inequalitiesmust the inequality symbol bereversed during the solution process?a.

b.

c.

d. -x + 4 6 5

-x + 4 + 3x 6 5

2x - 3 7 10

-2x 7 7

Practice Problem 9Solve: . Write thesolution set in interval notation.

6 - 2x ≤ 8x - 14


Solution:

Add x to both sides.

Combine like terms.


Combine like terms.


Simplify.

The solution set is .

Example 10 Solve: . Write the solution set in

interval notation.

Solution:


LINEAR INEQUALITIES AND PROBLEM SOLVING

Problems containing words such as “at least,” “at most,” “between,” “nomore than,” and “no less than” usually indicate that an inequality is to besolved instead of an equation. In solving applications involving linearinequalities, we use the same four-step strategy as when we solved applica-tions involving linear equations.

Example 11 Calculating Income with CommissionA salesperson earns $600 per month plus a commissionof 20% of sales. Find the minimum amount of salesneeded to receive a total income of at least $1500 permonth.

Solution: 1. UNDERSTAND. Read and reread the problem. Let

2. TRANSLATE. As stated in the beginning of this sec-tion, we want the income to be greater than or equalto $1500. To write an inequality, notice that the sales-person’s income consists of $600 plus a commission(20% of sales).

x = amount of sales

E

a- q, - 73d

x ≤ - 73

-3x-3

≤ 7-3

-3x ≥ 7

-3x - 12 ≥ -5

2x - 12 ≥ 5x - 5

Multiply both sides by 5 to elimi-nate fractions.




Divide both sides by -3 andreverse the inequality symbol.

Simplify.

5 c 25

Ax - 6 B d ≥ 5 Ax - 1 B

25

Ax - 6 B ≥ x - 1

25

Ax - 6 B ≥ x - 1

C4, q B 4 ≤ x or x ≥ 4

205

≤ 5x5

20 ≤ 5x

5 + 15 ≤ 5x - 15 + 15

5 ≤ 5x - 15

5 - x + x ≤ 4x - 15 + x

5 - x ≤ 4x - 15

Practice Problem 11A salesperson earns $1000 a monthplus a commission of 15% of sales.Find the minimum amount of salesneeded to receive a total income of atleast $4000 per month.

Answers10. , 11. $20,000A- q, 30 D

Practice Problem 10

Solve: . Write the 34

Ax + 2 B ≥ x - 6

solution set in interval notation.


Answer12. after the year 2020

In words:

T T TTranslate: 600 0.20x 1500

3. SOLVE the inequality for x.

4. INTERPRET.

Check: The income for sales of $4500 is

Thus, if sales are greater than or equal to $4500, incomeis greater than or equal to $1500.

State: The minimum amount of sales needed for the salesper-son to earn at least $1500 per month is $4500.

Example 12 Finding the Annual ConsumptionIn the United States, the annual consumption of ciga-rettes is declining. The consumption c in billions of ciga-rettes per year since the year 1985 can be approximatedby the formula

where t is the number of years after 1985. Use this for-mula to predict the years that the consumption of ciga-rettes will be less than 200 billion per year.

Solution: 1. UNDERSTAND. Read and reread the problem. Tobecome familiar with the given formula, let’s find thecigarette consumption after 20 years, which would bethe year , or 2005. To do so, we substitute20 for t in the given formula.

Thus, in 2005, we predict cigarette consumption to beabout 313.69 billion.

Variables have already been assigned in the givenformula. For review, they are

the annual consumption of cigarettes in theUnited States in billions of cigarettes

the number of years after 1985

2. TRANSLATE. We are looking for the years that theconsumption of cigarettes c is less than 200. Since weare finding years t, we substitute the expression in theformula given for c, or

-14.25t + 598.69 6 200

t =

c =

c = -14.25 A20 B + 598.69 = 313.69

1985 + 20

c = -14.25t + 598.69

600 + 0.20 A4500 B , or 1500

x ≥ 4500

0.20x ≥ 900

600 + 0.20x - 600 ≥ 1500 - 600

600 + 0.20x ≥ 1500

≥+

1500≥commission(20% of sales)+600

Practice Problem 12Use the formula given in Example 12to predict when the consumption ofcigarettes will be less than 100 billionper year.


3. SOLVE the inequality.

4. INTERPRET.

Check: We substitute a number greater than 27.98 and see thatc is less than 200.

State: The annual consumption of cigarettes will be less than200 billion for the years more than 27.98 years after 1985,or in approximately .28 + 1985 = 2013

t 7 27.98

-14.25t 6 -398.69

Subtract 598.69 from bothsides.

Divide both sides by -14.25 and round theresult.

-14.25t + 598.69 6 200

Focus On Study SkillsSTUDYING FOR AND TAKING A MATH EXAM

Remember that one of the best ways to start preparing for an exam is to keep current with your assign-ments as they are made. Make an effort to clear up any confusion on topics as you cover them.Begin reviewing for your exam a few days in advance. If you find a topic during your review that you stilldon’t understand, you’ll have plenty of time to ask your instructor, another student in your class, or amath tutor for help. Don’t wait until the last minute to “cram” for the test.

j Reread your notes and carefully review the Chapter Highlights at the end of each chapter to be covered.j Try solving a few exercises from each section.j Pay special attention to any new terminology or definitions in the chapter. Be sure you can state the

meanings of definitions in your own words.j Find a quiet place to take the Chapter Test found at the end of the chapter to be covered. This gives you

a chance to practice taking the real exam, so try the Chapter Test without referring to your notes or look-ing up anything in your book. Give yourself the same amount of time to take the Chapter Test as you willhave to take the exam for which you are preparing. If your exam covers more than one chapter, youshould try taking the Chapter Tests for each chapter covered. You may also find working through theCumulative Reviews helpful when preparing for a multi-chapter test.

j When you have finished taking the Chapter Test, check your answers against those in the back of thebook. Redo any of the problems you missed. Then spend extra time solving similar problems.

j If you tend to get anxious while taking an exam, try to visualize yourself taking the exam in advance. Pic-ture yourself being calm, clearheaded, and successful. Picture yourself remembering concepts and defin-itions with no trouble. When you are well prepared for an exam, a lot of nervousness can be avoidedthrough positive thinking.

j Get lots of rest the night before the exam. It’s hard to show how well you know the material if your brainis foggy from lack of sleep.

When it’s time to take your exam, remember these hints:

j Make sure you have all the tools you will need to take the exam, including an extra pencil and eraser,paper (if needed), and calculator (if allowed).

j Try to relax. Taking a few deep breaths, inhaling and then exhaling slowly before you begin, might help.j Are there any special definitions or solution steps that you’ll need to remember during the exam? As

soon as you get your exam, write these down at the top, bottom, or on the back of your paper.j Scan the entire test to get an idea of what questions are being asked.j Start with the questions that are easiest for you. This will help build your confidence. Then return to the

harder ones.j Read all directions carefully. Make sure that your final result answers the question being asked.j Show all of your work. Try to work neatly.j Don’t spend too much time on a single problem. If you get stuck, try moving on to other problems so you

can increase your chances of finishing the test. If you have time, you can return to the problem giving youtrouble.

j Before turning in your exam, check your work carefully if time allows. Be on the lookout for careless mis-takes.

275


MENTAL MATH

Solve each inequality mentally.

1. 2. 3. 4.

5. 6. 7. 8.

EXERCISE SET 2.8

Solve. Write the solution set using interval notation. See Examples 9 and 10.

1. 2. 3.

4. 5.

6. 7.

8. 9.

10. 11.

12. 35

Ax + 1 B ≤ x + 1

14

Ax - 7 B ≥ x + 2-6x + 2 6 -3 Ax + 4 B

-5x + 4 ≤ -4 Ax - 1 B34

- 23

7 x6

12

+ 23

≥ x6

5 Ax + 4 B ≤ 4 A2x + 3 B

3 Ax - 5 B 6 2 A2x - 1 B20 + x 6 6x

15 + 2x ≥ 4x - 78 - 5x ≤ 23-2x + 7 ≥ 9D

x4

≥ 2x2

≤ 15x 6 203x 7 12

x + 1 ≤ 8x + 5 ≥ 15x - 1 7 6x - 2 6 4

MENTAL MATH ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

276


SECTION 2.1 SIMPLIFYING EXPRESSIONS

The numerical coefficient of a term is its numericalfactor.

Terms with the same variables raised to exactly thesame powers are like terms.

To combine like terms, add the numerical coeffi-cients and multiply the result by the common vari-able factor.

To remove parentheses, apply the distributive prop-erty.

TERM NUMERICAL COEFFICIENT

x 1

LIKE TERMS UNLIKE TERMS

= 26x - 38= -4x - 28 + 30x - 10

-4 1x + 7 2 + 10 13x - 1 2-4z2 + 5z2 - 6z2 = -5z29y + 3y = 12y

7a2b, -2ab2-2xy, 5yx3y, 3y212x, -x

15

15

a2b

-7-7y

SECTION 2.2 THE ADDITION PROPERTY OF EQUALITY

A linear equation in one variable can be written inthe form where A, B, and C are realnumbers and .

Equivalent equations are equations that have thesame solution.

ADDITION PROPERTY OF EQUALITY

Adding the same number to or subtracting the samenumber from both sides of an equation does notchange its solution.

A Z 0Ax + B = C

and are equivalent equations.

y = -6y + 9 - 9 = 3 - 9

y + 9 = 3

x = 17x - 7 = 10

3 1x - 1 2 = -8 1x + 5 2 + 4-3x + 7 = 2

SECTION 2.3 THE MULTIPLICATION PROPERTY OF EQUALITY

MULTIPLICATION PROPERTY OF EQUALITY

Multiplying both sides or dividing both sides of anequation by the same nonzero number does notchange its solution.

(18)

a = 27

32

a 23

a b =32

23

a = 18

SECTION 2.4 FURTHER SOLVING LINEAR EQUATIONS

TO SOLVE LINEAR EQUATIONS

1. Clear the equation of fractions.

2. Remove any grouping symbols such as parenthe-ses.

3. Simplify each side by combining like terms.

Solve:

1.

2.

3. -10x + 63 = 3 -10x + 45 + 18 = 3

Apply the distribu-tive property.

Combine like terms.

5 1-2x + 9 2 + 18 = 3

6 # 5 1-2x + 9 26

+ 6 # 3 = 6 # 12

5 1-2x + 9 26

+ 3 = 12



4. Get all variable terms on one side and all num-bers on the other side by using the addition prop-erty of equality.

5. Get the variable alone by using the multiplica-tion property of equality.

6. Check the solution by substituting it into theoriginal equation.

4. Subtract 63.

5. Divide by .

x = 6

-10 -10x-10

= -60-10

-10x = -60 -10x + 63 - 63 = 3 - 63

SECTION 2.5 AN INTRODUCTION TO PROBLEM SOLVING

PROBLEM-SOLVING STEPS

1. UNDERSTAND the problem.

2. TRANSLATE the problem.

3. SOLVE the equation.

4. INTERPRET the results.

The height of the Hudson volcano in Chili is twicethe height of the Kiska volcano in the AleutianIslands. If the sum of their heights is 12,870 feet,find the height of each.

1. Read and reread the problem. Guess a solutionand check your guess.

Let x be the height of the Kiska volcano. Then 2xis the height of the Hudson volcano.

x 2xKiska Hudson

2.

T T T T Tx + 2x = 12,870

3.

4. Check: If x is 4290 then 2x is 2(4290) or 8580.Their sum is or 12,870, therequired amount.

State: The Kiska volcano is 4290 feet high andthe Hudson volcano is 8580 feet high.

4290 + 8580

x = 42903x = 12,870

x + 2x = 12,870

12,870isheight ofHudson

addedto

heightof Kiska

SECTION 2.6 FORMULAS AND PROBLEM SOLVING

An equation that describes a known relationshipamong quantities is called a formula.

To solve a formula for a specified variable, use thesame steps as for solving a linear equation. Treatthe specified variable as the only variable of theequation.

(area of a rectangle)(simple interest)

Solve for l.

Subtract 2w.

Divide by 2.

Simplify.P - 2w

2= l

P - 2w2

= 2l2

P - 2w = 2lP - 2w = 2l + 2w - 2w

P = 2l + 2w

P = 2l + 2w

I = PRTA = lw


SECTION 2.7 PERCENT, RATIO, AND PROPORTION

Use the same problem-solving steps to solve a prob-lem containing percents.

1. UNDERSTAND.

2. TRANSLATE.

3. SOLVE.

4. INTERPRET.

A ratio is the quotient of two numbers or two quan-tities.

The ratio of a to b can also be written as

or a :b

A proportion is a mathematical statement that tworatios are equal.

In the proportion , the products ad and bc are

called cross products.

If , then .ad = bcab

= cd

ab

= cd

ab

32% of what number is 36.8?

1. Read and reread. Propose a solution and check.Let x = the unknown number.

2.

T T T T T32% x = 36.8

3. Solve

Divide by 0.32.

Simplify.

4. 32% of 115 is 36.8.

Write the ratio of 5 hours to 1 day using fractionalnotation.

Solve:


-3 = x3x - 3 = 4x

3 1x - 1 2 = 4x

34

= xx - 1

34

= xx - 1

2 # 12 or 24¡3 # 8 or 24¡2

3= 8

12

x7

= 1535

23

= 812

5 hours1 day

= 5 hours24 hours

= 524

x = 115

0.32x0.32

= 36.80.32

0.32x = 36.832% # x = 36.8

#

36.8iswhat

numberof32%

SECTION 2.8 LINEAR INEQUALITIES AND PROBLEM SOLVING

Properties of inequalities are similar to properties ofequations. Don’t forget that if you multiply ordivide both sides of an inequality by the same neg-ative number, you must reverse the direction of theinequality symbol.

TO SOLVE LINEAR INEQUALITIES

1. Clear the inequality of fractions.2. Remove grouping symbols.3. Simplify each side by combining like terms.4. Write all variable terms on one side and all num-

bers on the other side using the addition prop-erty of inequality.

5. Get the variable alone by using the multiplica-tion property of inequality.

Divide by , reverse the inequality symbol.

Solve: 1. No fractions to clear.

2. Apply the distributive property.

3. Combine like terms.

4. Subtract 6.

5. Divide by 3.

]0 2−2 1−1

x ≤ 0

3x3

≤ 03

3x ≤ 03x + 6 - 6 ≤ 6 - 63x + 6 ≤ 63x + 6 ≤ -2 + 83 1x + 2 2 ≤ -2 + 8

3 1x + 2 2 ≤ -2 + 8

[0 2−2−3 1−1

x ≥ -2

-2-2x-2

≥ 4

-2

-2x ≤ 4

279

3In Chapter 2 we learned to solve and graph the solutions of linear equa-tions and inequalities in one variable on number lines. Now we defineand present techniques for solving and graphing linear equations andinequalities in two variables on grids.

3.1 The Rectangular CoordinateSystem

3.2 Graphing Linear Equations

3.3 Intercepts

3.4 Slope

3.5 Graphing Linear Inequalitiesin Two Variables

OPEC, the Organization of Petroleum Exporting Countries,was established in 1960. OPEC’s goal is to control the price ofcrude oil worldwide by controlling oil production. For example,if OPEC countries agree to limit the amount of oil they produce,an oil shortage is created and oil prices rise. In 1997, the mem-bers of OPEC were Algeria, Indonesia, Iran, Iraq, Kuwait, Libya,Nigeria, Qatar, Saudi Arabia, United Arab Emirates, andVenezuela. OPEC’s headquarters is in Vienna, Austria.

Graphing Equations and Inequalities C H A P T E R

3.1 THE RECTANGULAR COORDINATE SYSTEM

In Section 1.8, we learned how to read graphs. Example 4 in Section 1.8presented the broken line graph below showing the relationship betweentime spent smoking a cigarette and pulse rate. The horizontal line or axisshows time in minutes and the vertical line or axes shows the pulse rate inheartbeats per minute. Notice in this graph that there are two numbersassociated with each point of the graph. For example, we discussed earlierthat 15 minutes after “lighting up,” the pulse rate is 80 beats per minute. Ifwe agree to write the time first and the pulse rate second, we can say thereis a point on the graph corresponding to the ordered pair of numbers (15, 80). A few more ordered pairs are shown alongside their correspond-ing points.

PLOTTING ORDERED PAIRS OF NUMBERS

In general, we use the idea of ordered pairs to describe the location of apoint in a plane (such as a piece of paper). We start with a horizontal anda vertical axis. Each axis is a number line, and for the sake of consistencywe construct our axes to intersect at the 0 coordinate of both. This point ofintersection is called the origin. Notice that these two number lines or axesdivide the plane into four regions called quadrants. The quadrants are usu-ally numbered with Roman numerals as shown. The axes are not consid-ered to be in any quadrant.

It is helpful to label axes, so we label the horizontal axis the x-axis andthe vertical axis the y-axis. We call the system described above the rectan-gular coordinate system, or the coordinate plane. Just as with other graphsshown, we can then describe the locations of points by ordered pairs ofnumbers. We list the horizontal x-axis measurement first and the vertical y-axis measurement second.

quadrant IV

quadrant Iquadrant II

quadrant III

x-axis

y-axis

origin

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

A

1009080706050403020100

Puls

e R

ate

(hea

rtbe

ats

per

min

ute)

Time (minutes)

40−5 0

(0, 60)

(5, 95)

(15, 80)

(40, 70)

5 10 15 20 25 30 35

The Rectangular Coordinate System SECTION 3.1 281

Objectives

Plot ordered pairs of numbers onthe rectangular coordinate system.

Graph paired data to create a scat-ter diagram.

Find the missing coordinate of anordered pair solution, given onecoordinate of the pair.

C

B

A

SSM CD-ROM Video3.1

282 CHAPTER 3 Graphing Equations and Inequalities

Answers1.

Point lies in quadrant I.

Point lies in quadrant II.

Point lies in quadrant III.

Point lies in quadrant IV.

Points , , , and

lie on axes, so they are not in any quadrant.

✓ Concept Check:The graph of point lies in quadrant IIand the graph of point lies in quadrantIV. They are not in the same location.

A1, -5 BA-5, 1 B

a-212

, 0bA0, -4 BA3, 0 BA0, 3 BA2, -2 BA-1, -3 BA-5, 1 BA4, 2 B

(−5, 1)

(−1, −3) (0, −4)

(2, −2)

(0, 3)

(3, 0)

(4, 2)

(−212, 0)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

✓ CONCEPT CHECK

Is the graph of the point inthe same location as the graph of thepoint ? Explain.A1, -5 B

A-5, 1 B

Practice Problem 1On a single coordinate system, ploteach ordered pair. State in which quad-rant, if any, each point lies.

a. b.

c. d.

e. f.

g. h. a-212

, 0 bA0, -4 BA3, 0 BA0, 3 BA-5, 1 BA2, -2 BA-1, -3 BA4, 2 B

To plot or graph the point corresponding to the ordered pair

We start at the origin. We then move a units left or right (right if a is posi-tive, left if a is negative). From there, we move b units up or down (up if bis positive, down if b is negative). For example, to plot the point corre-sponding to the ordered pair , we start at the origin, move 3 unitsright, and from there move 2 units up. (See the figure below.) The x-value,3, is also called the x-coordinate and the y-value, 2, is also called the y-coor-dinate. From now on, we will call the point with coordinates simplythe point . The point is graphed below also.


Example 1 On a single coordinate system, plot each ordered pair.State in which quadrant, if any, each point lies.

a. b. c. d.

e. f. g. h.

Solution:

Point lies in quadrant I.Point lies in quadrant II.Point lies in quadrant III.Point lies in quadrant IV.Points , ,

and lie on axes, so

they are not in any quadrant.

a0, -512b

A-5, 0 BA0, 0 B , A0, 2 BA1, -2 BA-2, -4 BA-5, 3 BA5, 3 B(−5, 3)

(−5, 0)

(−2, −4)

(1, −2)

(0, 0)

(0, 2)(5, 3)

0, − )( 512

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

a0, -512bA-5, 0 BA0, 2 BA0, 0 B

A-5, 3 BA1, -2 BA-2, -4 BA5, 3 B

HELPFUL HINT

Don’t forget that each ordered pair corresponds to exactly one point inthe plane and that each point in the plane corresponds to exactly oneordered pair.

x-axis

y-axis

(3, 2)

(−2, 5)

5 units up3 unitsright

2 unitsleft

2 units up

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

A-2, 5 BA3, 2 B A3, 2 B

A3, 2 B

Aa, b B

TRY THE CONCEPT CHECK IN THE MARGIN.From Example 1, notice that the y-coordinate of any point on the x-axis

is 0. For example, the point lies on the x-axis. Also, the x-coordi-nate of any point on the y-axis is 0. For example, the point lies on they-axis.

CREATING SCATTER DIAGRAMS

Data that can be represented as an ordered pair is called paired data. Manytypes of data collected from the real world are paired data. For instance,the annual measurement of a child’s height can be written as an orderedpair of the form (year, height in inches) and is paired data. The graph ofpaired data as points in the rectangular coordinate system is called a scat-ter diagram. Scatter diagrams can be used to look for patterns and trendsin paired data.

Example 2 The table gives the an-nual revenues for Wal-Mart Stores for the yearsshown. (Source: Wal-Mart Stores, Inc.)

a. Write this paired dataas a set of orderedpairs of the form (year,revenue in billions of dollars).

b. Create a scatter diagram of the paired data.c. What trend in the paired data does the scatter diagram

show?

Solution: a. The ordered pairs are , and .

b. We begin by plotting the ordered pairs. Because the x-coordinate in each ordered pair is a year, we labelthe x-axis “Year” and mark the horizontal axis withthe years given. Then we label the y-axis or verticalaxis “Wal-Mart Revenue (in billions of dollars).” It isconvenient to mark the vertical axis in multiples of 20,starting with 0.

c. The scatter diagram shows that Wal-Mart revenuesteadily increased over the years 1993–1997.

120

100

80

60

40

20

0

Wal

-Mar

t Rev

enue

(in

billi

ons

of d

olla

rs)

1994 19971993 1995 1996

Year

A1997, 106 BA1996, 95 BA1995, 83 B ,A1994, 68 B ,A1993, 56 B ,

B

A0, 2 BA-5, 0 B


✓ Concept Check: a. , b. A0, -6 BA-3, 5 B

Wal-Mart RevenueYear (in billions of dollars)

1993 561994 681995 831996 951997 106

✓ CONCEPT CHECK

For each description of a point in therectangular coordinate system, writean ordered pair that represents it.a. Point A is located three units to

the left of the y-axis and five unitsabove the x-axis.

b. Point B is located six units belowthe origin.

Practice Problem 2The table gives the number of cableTV subscribers (in millions) for theyears shown. (Source: Television andCable Factbook, Warren Publishing,Inc., Washington, D.C.)

a. Write this paired data as a set ofordered pairs of the form (year,number of cable TV subscribers inmillions).

b. Create a scatter diagram of thepaired data.

c. What trend in the paired data doesthe scatter diagram show?

70

60

50

40

30

20

10

0

Cab

le T

V S

ubsc

ribe

rs(i

n m

illio

ns)

1988 1994 19961986 1990 1992

Year

Cable TV SubscribersYear (in millions)

1986 381988 441990 501992 531994 571996 62

Answers2. a.

,b.

c. The number of cable TV subscribers hassteadily increased.

1990 19961992 199419881986

Year

Cab

le T

V S

ubsc

ribe

rs(i

n m

illio

ns)

A1996, 62 BA1994, 57 B ,A1992, 53 B ,A1990, 50 B ,A1988, 44 B ,A1986, 38 B ,


Answers3. a. , b. , c. A-4, 6 BA2, 3 BA0, 4 B

COMPLETING ORDERED PAIRS SOLUTIONS

Let’s see how we can use ordered pairs to record solutions of equationscontaining two variables. An equation in one variable such as has one solution, which is 4: the number 4 is the value of the variable x thatmakes the equation true.

An equation in two variables, such as , has solutions con-sisting of two values, one for x and one for y. For example, and

is a solution of because, if x is replaced with 3 and y with2, we get a true statement.

True.

The solution and can be written as , an ordered pairof numbers.

In general, an ordered pair is a solution of an equation in two variablesif replacing the variables by the values of the ordered pair results in a truestatement. For example, another ordered pair solution of is

. Replacing x with 5 and y with results in a true statement.

Replace x with 5 and y with .

True.

Example 3 Complete each ordered pair so that it is a solution to theequation .

a. b. c.

Solution: a. In the ordered pair , the x-value is 0. We letin the equation and solve for y.

Replace x with 0.

The completed ordered pair is .b. In the ordered pair , the y-value is 6. We let

in the equation and solve for x.

Replace y with 6.



The ordered pair is .c. In the ordered pair , the x-value is . We let

in the equation and solve for y.

Replace x with .


The ordered pair is .A-1, 15 B y = 15

-3 + y = 12

-13 A-1 B + y = 12

3x + y = 12

x = -1-1A-1, B

A2, 6 B x = 2

3x = 63x + 6 = 123x + y = 12

y = 6A , 6 B

A0, 12 B y = 12

0 + y = 123 A0 B + y = 12

3x + y = 12

x = 0A0, B

A-1, BA , 6 BA0, B3x + y = 12

8 = 8 10 - 2 � 8

-22 A5 B + A-2 B � 8 2x + y = 8

-2A5, -2 B 2x + y = 8

A3, 2 By = 2x = 3

8 = 82 A3 B + 2 � 8

2x + y = 8

2x + y = 8y = 2x = 3

2x + y = 8

x + 1 = 5

C

Practice Problem 3Complete each ordered pair so that itis a solution to the equation

.

a.

b.

c. A-4, BA , 3 BA0, B

x + 2y = 8

Solutions of equations in two variables can also be recorded in a table ofpaired values, as shown in the next example.

Example 4 Complete the table for the equation .

x y

a.b. 0c.

Solution: a. We replace x with in the equation and solve for y.

Let .

The ordered pair is

b. We replace y with 0 in the equation and solve for x.

Let .


The ordered pair is .

c. We replace y with in the equation and solve for x.

Let .


The ordered pair is . The completed table is shown to the right.

Example 5 Complete the table for the equation.

Solution: The equation is the same as. No matter what value we

replace x by, y always equals 3. Thecompleted table is shown to the right.

By now, you have noticed that equations in two variables often havemore than one solution. We discuss this more in the next section.

A table showing ordered pair solutions may be written vertically, or hor-izontally as shown in the next example.

0x + y = 3y = 3

y = 3

A-3, -9 B-3 = x

y = -9-9 = 3x

y = 3x

-9

A0, 0 B0 = x

y = 00 = 3x

y = 3x

A-1, -3 By = -3

x = -1y = 3 A-1 By = 3x

-1

-9

-1

y = 3x


Practice Problem 4Complete the table for the equation

.

x y

a.b. 0c. 10

-3

y = -2x

x y

0-5

-2

x y

0 0-9-3

-3-1

Practice Problem 5Complete the table for the equation

.

x y

04

-2

x = 5

Answers

4. x y

a. 6b. 0 0c. 10

5. x y55 05 4

-2

-5

-3

x y

30 3

3-5

-2

Example 6 A small business purchased a computer for $2000. Thebusiness predicts that the computer will be used for 5years and the value in dollars y of the computer in xyears is . Complete the table.

x 0 1 2 3 4 5

y

Solution: To find the value of y when x is 0, we replace x with 0 inthe equation. We use this same procedure to find y whenx is 1 and when x is 2.

WHEN x � 0, WHEN x � 1, WHEN x � 2,

We have the ordered pairs , , and. This means that in 0 years the value of the

computer is $2000, in 1 year the value of the computer is$1700, and in 2 years the value is $1400. To complete thetable of values, we continue the procedure for

.

WHEN x � 3, WHEN x � 4, WHEN x � 5,

The completed table is

x 0 1 2 3 4 5

y 2000 1700 1400 1100 800 500

The ordered pair solutions recorded in the completed table for Example6 are another set of paired data. They are graphed next. Notice that thisscatter diagram gives a visual picture of the decrease in value of the com-puter.

2000

1800

1600

1400

1200

1000

800

600

400

200

0

Val

ue o

f C

ompu

ter

(dol

lars

)

3 540 1 2

Time (years)

y = 500y = 800y = 1100y = -1500 + 2000y = -1200 + 2000y = -900 + 2000y = -300 # 5 + 2000y = -300 # 4 + 2000y = -300 # 3 + 2000y = -300x + 2000y = -300x + 2000y = -300x + 2000

x = 3, x = 4, and x = 5

A2, 1400 BA1, 1700 BA0, 2000 B

y = 1400y = 1700y = 2000y = -600 + 2000y = -300 + 2000y = 0 + 2000y = -300 # 2 + 2000y = -300 # 1 + 2000y = -300 # 0 + 2000y = -300x + 2000y = -300x + 2000y = -300x + 2000

y = -300x + 2000

Practice Problem 6A company purchased a fax machinefor $400. The business manager of thecompany predicts that the fax machinewill be used for 7 years and the value indollars y of the machine in x years is

. Complete the table.

x 1 2 3 4 5 6 7y

y = -50x + 400


x y

0 20001 17002 14003 11004 8005 500

Answer6.

x 1 2 3 4 5 6 7y 350 300 250 200 150 100 50

287


MENTAL MATH

Give two ordered pair solutions for each linear equation.

1. 2.

EXERCISE SET 3.1

Plot each ordered pair. State in which quadrant, if any, each point lies. See Example 1.A

x + y = 6x + y = 10

1.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

a-1, 412bA2, -4 BA0,-1 B

A-3, 0 BA-5,-2 BA1, 5 B 2.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

A5, -4 Ba334

, 0bA-3, -3 BA-2, 1 BA0, 2 BA2, 4 B

3. When is the graph of the ordered pair(a, b) the same as the graph of theordered pair (b, a)?

4. In your own words, describe how toplot an ordered pair.

Find the x- and y-coordinates of each labeled point. See Example 1.

5. A

6. B

7. C

8. D

9. E

10. F

11. G

12. A

13. B

14. C

15. D

16. E

17. F

18. G

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5

C D

F

G A

E

B

x

y

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5

DC

B

EF G

Ax

y

MENTAL MATH ANSWERS

1.

2.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

3.2 GRAPHING LINEAR EQUATIONS

In the previous section, we found that equations in two variables may havemore than one solution. For example, both are solutions ofthe equation . In fact, this equation has an infinite number ofsolutions. Other solutions include . Notice thepattern that appears in the graph of these solutions.

These solutions all appear to lie on the same line, as seen in the secondgraph. It can be shown that every ordered pair solution of the equation cor-responds to a point on this line, and every point on this line corresponds toan ordered pair solution. Thus, we say that this line is the graph of the equa-tion .

The equation is called a linear equation in two variables andthe graph of every linear equation in two variables is a straight line.

The form is called standard form. Following are exam-ples of linear equation in two variables.

y = 7y = 13

x + 2-2x = 7y2x + y = 8

Ax + By = C

x + y = 4

(4, 0)(2, 2)

(0, 4)(−2, 6)

(6, −2)x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

x + y = 4

(4, 0)(2, 2)

(0, 4)(−2, 6)

(6, −2)x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

A-2, 6 B , A4, 0 B , and A6, -2 Bx + y = 4A2, 2 B and A0, 4 B

Graphing Linear Equations SECTION 3.2 289

Objectives

Graph a linear equation by findingand plotting ordered pair solutions.

A

LINEAR EQUATION IN TWO VARIABLES

A linear equation in two variables is an equation that can be written inthe form

where A, B, and C are real numbers and A and B are not both 0. Thegraph of a linear equation in two variables is a straight line.

Ax + By = C

SSM CD-ROM Video3.2


Answer1.

(3, 1)(6, 0)(0, 2)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

GRAPHING LINEAR EQUATIONS

From geometry, we know that a straight line is determined by just twopoints. Thus, to graph a linear equation in two variables we need to find justtwo of its infinitely many solutions. Once we do so, we plot the solutionpoints and draw the line connecting the points. Usually, we find a thirdsolution as well, as a check.

Example 1 Graph the linear equation .

Solution: To graph this equation, we find three ordered pair solu-tions of . To do this, we choose a value forone variable, x or y, and solve for the other variable. Forexample, if we let , then becomes

Replace x with 1.

Multiply.


Since , the ordered pair is a solu-tion of . Next, we let .

Replace x with 0.

The ordered pair is a second solution.

The two solutions found so far allow us to draw thestraight line that is the graph of all solutions of

. However, we will find a third ordered pairas a check. Let .

Replace y with .



The third solution is . These three ordered pairsolutions are listed in the table and plotted on the coordi-nate plane. The graph of is the line throughthe three points.

x y

1 30 53

2x + y = 5

(0, 5)

(1, 3)

(3, −1)x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

-1

2x + y = 5

A3, -1 B x = 3

2x = 6 2x - 1 = 5

-12x + A-1 B = 5 2x + y = 5

y = -12x + y = 5

A0, 5 B y = 5

0 + y = 52 A0 B + y = 5

2x + y = 5

x = 02x + y = 5A1, 3 By = 3 when x = 1

y = 3 2 + y = 5

2 A1 B + y = 5 2x + y = 5

2x + y = 5x = 1

2x + y = 5

2x + y = 5

A

Practice Problem 1Graph the linear equation .

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

x + 3y = 6


Solution: We find three ordered pair solutions of .

Let x � 0. Let y � 0. Let x � �2.

The ordered pairs are , and . The

graph of is the line through the threepoints.

x y

0 50


Solution: We find three ordered pair solutions. Since this equationis solved for y, we’ll choose three x values.

If x = -1, y = 3 # -1 = -3.

If x = 0, y = 3 # 0 = 0.

If x = 2, y = 3 # 2 = 6.

y = 3x

67

54321

1−1

−2−3−4

−2−3−4−5−6−7 2 3 4

−5x + 3y = 15

(0, 5)

(−3, 0)x

y

(−2, )132

53 = 12

3-2-3

-5x + 3y = 15

a-2, 53bA0, 5 B , A-3, 0 B

y = 53

x = -3 y = 5

3y = 5 -5x = 15 3y = 15

10 + 3y = 15 -5x + 0 = 15 0 + 3y = 15

-5 # -2 + 3y = 15-5x + 3 # 0 = 15-5 # 0 + 3y = 15

-5x + 3y = 15 -5x + 3y = 15 -5x + 3y = 15

-5x + 3y = 15

-5x + 3y = 15

Helpful Hint

All three points should fall on the same straight line. If not, check yourordered pair solutions for a mistake.


Practice Problem 2Graph the linear equation

.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

-2x + 4y = 8


x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

y = 2x

Answers2.

3.

(−2, −4)

(0, 0)

(3, 6)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

(0, 2)(2, 3)

(−2, 1)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

Next, we plot the ordered pair solutions and draw a linethrough the plotted points. The line is the graph of

. Every point on the graph represents an orderedpair solution of the equation and every ordered pair solu-tion is a point on this line.

x y

2 60 0


Solution We find three ordered pair solutions, plot the solutions,and draw a line through the plotted solutions. To avoidfractions, we’ll choose x values that are multiples of 3 tosubstitute into the equation.

x y

60 0

1

Let’s compare the graphs in Examples 3 and 4. The graph of tilts

upward (as we follow the line from left to right) and the graph of

tilts downward (as we follow the line from left to right). Also notice thatboth lines go through the origin or that is an ordered pair solution ofboth equations. We will learn more about the tilt, or slope, of a line in Sec-tion 3.4.

A0, 0 B

y = - 13

x

y = 3x

(6, −2)

(−3, 1)(0, 0)

y = x31−

108642

2−2

−4−6−8

−10

−4−6−8−10 4 6 8 10 x

y

-3

-2

If x = -3, then y = - 13

# -3 = 1.

If x = 0, then y = - 13

# 0 = 0.

If x = 6, then y = - 13

# 6 = -2.

y = - 13

x

108642

2−2

−4−6−8

−10

(−1, −3)

y = 3x

(0, 0)

(2, 6)

−4−6−8−10 4 6 8 10 x

y

-3-1

y = 3x


Answer4.

(−6, 3)(0, 0)

(4, −2)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

Practice Problem 4

Graph the linear equation .

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

y = - 12

x

Example 5 Graph the linear equation and compare thisgraph with the graph of in Example 3.

Solution We find three ordered pair solutions, plot the solutions,and draw a line through the plotted solutions. We choosex values and substitute into the equation .

x y

0 61 9

The most startling similarity is that both graphs appear tohave the same upward tilt as we move from left to right.Also, the graph of crosses the y-axis at the origin,while the graph of crosses the y-axis at 6. Itappears that the graph of is the same as thegraph of except that the graph of ismoved 6 units upward.


Solution: Recall from Section 3.1 that the equation is thesame as . No matter what value we replacex with, y is .

x y

03

Notice that the graph of is a horizontal line.y = -2

(−2, −2) (3, −2)(0, −2)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

-2-2-2-2

-20x + y = -2

y = -2

y = -2

y = 3x + 6y = 3xy = 3x + 6

y = 3x + 6y = 3x

y = 3x + 6 y = 3x

(0, 6)

(−3, −3)

108642

2−2

−4−6−8

−10

−4−6−8−10 4 6 8 10 x

y

6 units up

(1, 9)-3-3

If x = 1, then y = 3 A1 B + 6 = 9.If x = 0, then y = 3 A0 B + 6 = 6.If x = -3, then y = 3 A-3 B + 6 = -3.

y = 3x + 6

y = 3xy = 3x + 6


Practice Problem 5Graph the linear equation and compare this graph with the graphof in Practice Problem 3.

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

y = 2x

y = 2x + 3


x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x = 3

Answers5.

Same as the graph of except that thegraph of is moved 3 units upward.

6.

(3, −4)

(3, 0)

(3, 2)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

y = 2x + 3y = 2x

(−2, −1)(0, 3)

(2, 7)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10


GRAPHING CALCULATOR EXPLORATIONSIn this section, we begin an optional study of graphing calculators and graphing software packages forcomputers. These graphers use the same point plotting technique that was introduced in this section. The

advantage of this graphing technology is, of course, that graphing calculators and computers can find and plotordered pair solutions much faster than we can. Note, however, that the features described in these boxes maynot be available on all graphing calculators.

The rectangular screen where a portion of the rectangular coordinate system is displayed is called awindow. We call it a standard window for graphing when both the x- and y-axes show coordinates between

and 10. This information is often displayed in the window menu on a graphing calculator as

The scale on the x-axis is one unit per tick mark.

The scale on the y-axis is one unit per tick mark.

To use a graphing calculator to graph the equation , press the key and enter the keystrokes

. The top row should now read . Next press the key, and the display should

look like this:

Use a standard window and graph the following linear equations. (Unless otherwise stated, use a standard windowwhen graphing.)1. 2. 3.

4. 5. 6. y = 29

x - 223

y = - 310

x + 325

y = -1.3x + 5.2

y = 2.5x - 7.9y = -x + 5y = -3x + 7

y = 2x + 3

−10

−10

10

10

GRAPHY1 = 2x + 33+X2

Y=y = 2x + 3

Yscl = 1 Ymax = 10 Ymin = -10

Xscl = 1Xmax = 10 Xmin = -10

-10

EXERCISE SET 3.2

For each equation, find three ordered pair solutions by completing the table. Then use the ordered pairs to graph theequation. See Examples 1 through 6.

1. 2. 3.

4. 5. 6.

7. 8.

x y x y

0 01 12 2

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

y = -5x + 2y = -4x + 3

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

y = 12

xy = 13

xy = -5x

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

y = -4xx - y = 4x - y = 6

A

295


x y

04

-1

x y

02

-1

x y

10

-1

x y

10

-1

x y

06

-3

x y

0

2-4

Name _________________________________________________________________________

296

Graph each linear equation. See Examples 1 through 6.

9. 10. 11.

12. 13. 14.

15. 16. 17.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x = -4y = -2x + 7y = 6x + 3

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

-x + 5y = 5x - 2y = 6-x + y = 6

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x - y = -2x + y = 7x + y = 1

297

Name _________________________________________________________________________

18. 19. 20.

21. 22. 23.

24. 25.

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x + 3y = 9y = 4x

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

y = 5xy = -xy = x

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x = -1y = 3y = 5

3.3 INTERCEPTS

IDENTIFYING INTERCEPTS

The graph of is shown below. Notice that this graph crosses they-axis at the point . This point is called the y-intercept point and is called the y-intercept. Likewise the graph crosses the x-axis at . Thispoint is called the x-intercept point and 2 is the x-intercept.

The intercept points are and .

Examples Identify the x- and y-intercepts and the intercept points.

1.

Solution: x-intercept: y-intercept: 2intercept points: and A0, 2 BA-3, 0 B

-3

4321

1−1

−2−3−4

−2−3−4 2 3 x

y

A0, -8 BA2, 0 B

(2, 0)

y = 4x − 8

(0, −8)

The y-interceptis −8.

The x-interceptis 2.

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

A2, 0 B -8A0, -8 By = 4x - 8

A

Intercepts SECTION 3.3 299

Objectives

Identify intercepts of a graph.

Graph a linear equation by findingand plotting intercept points.

Identify and graph vertical andhorizontal lines.

C

BA

HELPFUL HINT

If a graph crosses the x-axis at and the y-axis at , then

Intercept points

c cx-intercept y-intercept

Notice that if y is 0, the corresponding x-value is the x-intercept. Like-wise, if x is 0, the corresponding y-value is the y-intercept.

A0, 7 BA-3, 0 Bd A0, 7 BA-3, 0 B

Practice Problems 1–3Identify the x- and y-intercepts and theintercept points.

1.

2.

3.

x

y

−4 −2 2

2

2

4

4

4

x

y

−4 −2 2

2

2

4

4

4

x

y

−4 −2 2

2

2

4

4

4

SSM CD-ROM Video3.3

Answers1. x-intercept: 2; y-intercept: ; interceptpoints: and , 2. x-intercepts:

; y-intercept: 3; intercept points: ,, and , 3. x-intercept:

none; y-intercept: 3; intercept point: A0, 3 B

A0, 3 BA2, 0 BA-4, 0 B-4, 2

A0, -4 BA2, 0 B-4

2.

Solution: x-intercepts: y-intercept: 1intercept points: , ,

3.

Solution: x-intercept: 0y-intercept: 0intercept point:

FINDING AND PLOTTING INTERCEPT POINTS

Given an equation of a line, intercept points are usually easy to find sinceone coordinate is 0.

One way to find the y-intercept of a line, given its equation, is to let, since a point on the y-axis has an x-coordinate of 0. To find the

x-intercept of a line, let , since a point on the x-axis has a y-coordi-nate of 0.

Example 4 Graph by finding and plotting its interceptpoints.

Solution: We let to find the x-intercept and to findthe y-intercept.

. .

The x-intercept is 6 and the y-intercept is . We finda third ordered pair solution to check our work. If we let

, then . We plot the points , ,A0, -2 BA6, 0 Bx = 3y = -1

-2

y = -2 x = 6

-3y = 6 x - 0 = 6

0 - 3y = 6x - 3 A0 B = 6

x - 3y = 6 x - 3y = 6

Let x = 0 Let y = 0

x = 0y = 0

x - 3y = 6

y = 0x = 0

B

A0, 0 B

321

1−1

−2−3

−2−3 2 3 x

y

A0, 1 BA-1,0 BA-4, 0 B-4, -1

54321

1−1

−2−3

−2−3−4−5 2 x

y


FINDING x- AND y-INTERCEPTS

To find the x-intercept, let and solve for x.To find the y-intercept, let and solve for y.x = 0

y = 0

Answer4.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2

(2, 0)

(0, −4)

3 4 5

Practice Problem 4Graph by finding andplotting its intercept points.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

2x - y = 4

and . The graph of is the line drawnthrough these points as shown.

x y

6 003

Example 5 Graph by finding and plotting its interceptpoints.

Solution: We let to find the x-intercept and to findthe y-intercept.

Both the x-intercept and y-intercept are 0. In otherwords, when , then , which gives the orderedpair . Also, when , then , which givesthe same ordered pair . This happens when thegraph passes through the origin. Since two points areneeded to determine a line, we must find at least onemore ordered pair that satisfies . We let

to find a second ordered pair solution and letas a checkpoint.

The ordered pairs are , , and . Weplot these points to graph .

x y

0 02

1

GRAPHING VERTICAL AND HORIZONTAL LINES

The equation , for example, is a linear equation in two variablesbecause it can be written in the form . The graph of this equa-tion is a vertical line, as shown in the next example.

x + 0y = 2x = 2

C

321

1−1

−2−3

−2−3−4−5 2 3 4 5

x = −2y

(0, 0)(−2, 1)

(2, −1)

x

y

-2-1

x = -2yA-2, 1 BA2, -1 BA0, 0 B

x = -2 x = 2

x = -2 A1 B x = -2 A-1 BLet y = 1.Let y = -1.

y = 1y = -1

x = -2y

A0, 0 B x = 0y = 0A0, 0 B y = 0x = 0

0 = y x = 0

0 = -2y x = -2 A0 B x = -2y x = -2y

Let x = 0.Let y = 0.

x = 0y = 0

x = -2y

x − 3y = 6

(6, 0)

(0, −2)(3, −1)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

-1-2

x - 3y = 6A3, -1 B


Practice Problem 5Graph by finding and plottingits intercept points.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

y = 3x

Answer5.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2

(0, 0)

3 4 5

Example 6 Graph: .

Solution: The equation can be written as . Forany y-value chosen, notice that x is 2. No other value forx satisfies . Any ordered pair whose x-coordi-nate is 2 is a solution of . We will use theordered pair solutions , , and to graph

.

x y

2 32 02

The graph is a vertical line with x-intercept 2. Note thatthis graph has no y-intercept because x is never 0.

In general, we have the following.

Example 7 Graph:

Solution: The equation can be written as .For any x-value chosen, y is . If we chose 4, 1, and as x-values, the ordered pair solutions are ,

, and . We use these ordered pairs tograph . The graph is a horizontal line with y-intercept and no x-intercept.

x y

41 2

1

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5

y = −3

(1, −3)(−2, −3) (4, −3)

x

y

-3-2-3-3

-3y = -3

A-2, -3 BA1, -3 BA4, -3 B

-2-30x + y = -3y = -3

y = -3

4321

1−1

−2−3−4

−2−3 2 3 4

x = 2

(2, 3)

(2, 0)

(2, −3)

x

y

-3

x = 2A2, -3 BA2, 0 BA2, 3 Bx + 0y = 2

x + 0y = 2

x + 0y = 2x = 2

x = 2


Answers6.

7.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2

(0, 4)

3 4 5

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2

(−3, 0)

3 4 5

VERTICAL LINES

The graph of x = c, where c is a real number, is a vertical line with x-intercept c. (c, 0)

x = c

x

y

Practice Problem 6Graph:

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x = -3


x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

y = 4


In general, we have the following.

GRAPHING CALCULATOR EXPLORATIONSYou may have noticed that to use the key on a grapherto graph an equation, the equation must be solved for y. For

example, to graph , we solve this equation for y.



Simplify.

To graph or , press the key and

enter

Graph each linear equation.

1. 2.

3. 4. 5.9x - 0.8y = -10.4-2.2x + 6.8y = 15.5

-2.61y = xx = 3.78y

−10

−10

10

10

2x + 3y = 7 or y = − + x23

73

Y1 = - 23

x + 73

Y=y = - 23

x + 73

2x + 3y = 7

y = - 23

x + 73

3y3

= - 2x3

+ 73

3y = -2x + 7

2x + 3y = 7

2x + 3y = 7

Y=

HORIZONTAL LINES

The graph of , where c is a real number, is a horizontal line with y-intercept c. (0, c)

y = c

x

y

y = c


Focus On The Real WorldREADING A MAP

How do you find a location on a map? Most maps we use todayhave a grid that is based on the rectangular coordinate system weuse in algebra. After finding the coordinates of cities and otherlandmarks from the map index, the grid can help us find places on the map. To eliminate confusion, many maps use letters to label the grid along one edge and numbers along the other. How-ever, the coordinates are still pairs of numbers and letters. Forinstance, the coordinates for Toledo on the map are A-2.

CRITICAL THINKING

1. Find the coordinates of the following cities: Hamilton, Colum-bus, Youngstown, and Cincinnati.

2. What cities correspond to the following coordinates: F-3, A-3,B-4, and D-2?

3. How are the map’s coordinate system and the rectangular coor-dinate system we use in algebra the same? How are they differ-ent? What are the advantages of each?

F

E

D

C

B

A

F

E

D

C

B

A

1

Toledo

Springfield

Lakewood

Akron

Canton

Youngstown

Xenia

Massillon

Orange

Sandusky

East Liverpool

Steubenville

Portsmouth

Dayton

Cincinnati

OHIO

Cleveland

KENTUCKY

WEST V

IRGIN

IAHamilton

CitiesandTowns

0 to 50,000

0 10 20 40 80 Mi.

1,000,000 and overCapital50,000 to 500,000

500,000 to 1,000,000

Columbus

LAKE ERIE

PEN

NSY

LV

AN

IA

IND

IAN

A

MICHIGAN

Lima

2 3 4

N

5

1 2 3 4 5


EXERCISE SET 3.3

Graph each linear equation by finding and plotting its intercept points. See Examples 4 and 5.

1. 2.

3. 4. 5.

6. 7. 8.

9. 10. 11.

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

y = 3x + 6y = -2xx = 2y

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

2x + 3y = 62x - 4y = 8x - 2y = -8

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

-x + 2y = 62x = yx = 5y

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x - y = -4x - y = 3

B

305

Name _________________________________________________________________________

306

12 13. 14.

Graph each linear equation. See Examples 6 and 7.

15. 16. 17.

18.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x = 0

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

y = 0y = 5x = -1

C

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

x = -yx = yy = 2x + 10

307

Focus On The Real WorldROAD GRADES

Have you ever driven on a hilly highway and seen a sign like theone below? The 7% on the sign refers to the grade of the road.The grade of a road is the same as its slope given as a percent. A7 percent grade means that for every rise of 7 units there is a run of100 units. The type of units doesn’t matter as long as they are thesame. For instance, we could say that a 7 percent grade representsa rise of 7 feet for every run of 100 feet or a rise of 7 meters forevery run of 100 meters, and so on.

Most highways are designed to have grades of 6 percent or less. Ifa portion of a highway has a grade that is steeper than 6 percent, asign is usually posted giving the grade and the number of miles forthe grade. Truck drivers need to know when the road is particu-larly steep. They may need to take precautions such as using a dif-ferent gear, reducing their speed, or testing their brakes.

Here is a sampling of road grades:j A portion of the John Scott Highway in Steubenville, Ohio, has

a 10% grade. (Source: Ohio Department of Transportation)j Joaquin Road in Portola Valley, California, has an average

grade of 15%. (Source: Western Wheelers Bicycle Club)j The steepest grade in Seattle, Washington, is 26% on East Roy

Street between 25th Avenue and 26th Avenue. (Source: SeattlePost-Intelligencer, Nov. 21, 1994)

j The steepest street in Pittsburgh, Pennsylvania, is CantonAvenue with a 37% grade. (Source: Pittsburgh Department ofPublic Works)

j The steepest street in the world is Baldwin Street in Dunedin,New Zealand. Its maximum grade is 79%. (Source: GuinnessBook of Records, 1996)

COOPERATIVE LEARNING ACTIVITY

Try to find a road sign with a percent grade warning or the nameand grade of a steep road in your area. Describe its slope andmake a scale drawing to represent its grade.

7%

NEXT3 MILES

7% grade = rise of 7run of 100

7% = 0.07 = 7100

3.4 SLOPE

FINDING THE SLOPE OF A LINE GIVEN TWO POINTS

Thus far, much of this chapter has been devoted to graphing lines. Youhave probably noticed by now that a key feature of a line is its slant orsteepness. In mathematics, the slant or steepness of a line is formally knownas its slope. We measure the slope of a line by the ratio of vertical change(rise) to the corresponding horizontal change (run) as we move along theline.

On the line below, for example, suppose that we begin at the point and move to the point . The vertical change is the change in y-coordi-nates: or 4 units. The corresponding horizontal change is the changein x-coordinates: units. The ratio of these changes is

The slope of this line, then, is . This means that for every 4 units of change

in y-coordinates, there is a corresponding change of 3 units in x-coordi-nates.

HELPFUL HINT

It makes no difference which two points of a line are chosen to find itsslope. The slope of a line is the same everywhere on the line.

43

67

54321

1−1

−2−3

−2−3−4 2 3 4 5 6 7

Vertical change is6 − 2 = 4 units.

Horizontal change is4 − 1 = 3 units.

(4, 6)

(1, 2)

x

y

slope =change in y Avertical change B

change in x Ahorizontal change B = 43

4 - 1 = 36 - 2

A4, 6 B A1, 2 B

A

Slope SECTION 3.4 309

Objectives

Find the slope of a line given twopoints of the line.

Find the slope of a line given itsequation.

Find the slopes of horizontal andvertical lines.

Compare the slopes of parallel andperpendicular lines.

D

C

B

A

SLOPE OF A LINE

The slope m of the line containing the points and isgiven by

, as long as x2 Z x1m = riserun

=change in ychange in x

=y2 - y1

x2 - x1

Ax2, y2 BAx1, y1 B

SSM CD-ROM Video3.4

Example 1 Find the slope of the line through and .Graph the line.

Solution: Let be and be . Then, bythe definition of slope,

The slope of the line is .

In Example 1, we could just as well have identified with and with . It makes no difference which point is called

or .


Ax2, y2 BAx1, y1 BA-1, 5 BAx2, y2 B

A2, -3 BAx1, y1 B

- 83

rise: −8

run: 3

(−1, 5)

( 2, −3)

x

y

= -83

= - 83

= -3 - 52 - A-1 B

m =y2 - y1

x2 - x1

A2, -3 BAx2, y2 BA-1, 5 BAx1, y1 B

A2, -3 BA-1, 5 B


HELPFUL HINT

When finding the slope of a line through two given points, it makesno difference which given point is called and which is called

. However, once an x-coordinate is called , make sure itscorresponding y-coordinate is called .y1

x1Ax2, y2 BAx1, y1 B

✓ CONCEPT CHECK

The points ,and all lie on the same line.Work with a partner and verify thatthe slope is the same no matter whichpoints are used to find slope.

A10, 13 BA-2, -5 B , A0, -2 B , A4, 4 B

Practice Problem 1Find the slope of the line through

and . Graph the line.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

A4, -1 BA-2, 3 B

Answers

1.

✓ Concept Check: m = 32

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5(4, −1)

(−2, 3)

- 23

Example 2 Find the slope of the line through and .Graph the line.

Solution: Let be and be .

The slope is 2.

TRY THE CONCEPT CHECK IN THE MARGIN.Notice that the slope of the line in Example 1 is negative, and the slope

of the line in Example 2 is positive. Let your eye follow the line with nega-tive slope from left to right and notice that the line “goes down.” If you fol-low the line with positive slope from left to right, you will notice that theline “goes up.” This is true in general.

FINDING THE SLOPE OF A LINE GIVEN ITS EQUATION

As we have seen, the slope of a line is defined by two points on the line.Thus, if we know the equation of a line, we can find its slope by finding twoof its points. For example, let’s find the slope of the line

To find two points, we can choose two values for x and substitute to findcorresponding y-values. If , for example, or .If or . This gives the ordered pairs and

. Using the definition for slope, we have

The slope is 3.

Notice that the slope, 3, is the same as the coefficient of x in the equa-tion . This is true in general.y = 3x - 2

m =1 - A-2 B

1 - 0= 3

1= 3

A1, 1 BA0, -2 By = 1x = 1, y = 3 # 1 - 2

y = -2y = 3 # 0 - 2x = 0

y = 3x - 2

B

Goes up

Positive slope

x

y

Goes down

Negative slope

x

y

54321

1−1

−2−3−4

−2−3−4−5 2 3

(2, 4)

(−1, −2)

x

y

= -6-3

= 2

= -2 - 4-1 - 2

m =y2 - y1

x2 - x1

A-1, -2 BAx2, y2 BA2, 4 BAx1, y1 BA2, 4 BA-1, -2 B


If a linear equation is solved for y, the coefficient of x is its slope. Inother words, the slope of the line given by is m, the coef-ficient of x.

y = mx + b

Practice Problem 2Find the slope of the line through

and . Graph the line.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

A3, 5 BA-2, 1 B

✓ CONCEPT CHECK

What is wrong with the followingslope calculation?

and

m = 5 - 6-2 - 3

= -1-5

= 15

A-2, 6 BA3, 5 B

Answers

2.

✓ Concept Check:

m = 5 - 63 - A-2 B = -1

5= - 1

5

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

45


Answers

3. , 4. 0- 54

Practice Problem 4Find the slope of .y = 3

Example 3 Find the slope of the line .

Solution: When we solve for y, the coefficient of x is the slope.



cThe slope is . d

INTERPRETING SLOPE

The slope of a line can be thought of as a rate of change. For example, theslope of the line given by the equation is 2. Thinking of a slope of

2 as a rate of change means that for every change of 1 unit in the

value of x, the value of y will change similarly by 2 units. A slope of means that for every increase of 1 unit in the value of x, the value of ydecreases by 3 units.

Interpreting slope as a rate of change is also meaningful in real-life appli-cations. For example, the cost y (in cents) of an in-state long-distance tele-phone call in Massachusetts is given by the linear equation ,where x is the length of the call in minutes (Source: Based on data from BellAtlantic). By looking at the equation, we can see that the

slope of this line is 11 . This slope means that the cost of such a tele-

phone call increases 11 cents for each additional 1 minute spent on the tele-phone. Another way to say this is that the cost of the telephone call isincreasing at a rate of 11 cents per minute.

FINDING SLOPES OF HORIZONTAL AND VERTICAL LINES


Solution: Recall that is a horizontal line with y-intercept. To find the slope, we find two ordered pair solutions

of , knowing that solutions of must havea y-value of . We will use and . Welet be and be .

The slope of the line is 0. Since the y-values willhave a difference of 0 for all horizontal lines, we can saythat all horizontal lines have a slope of 0.

y = -1

321

1−1

−2−3−4

−2−3−4−5 2 3 4 5y = −1

x

y

m =y2 - y1

x2 - x1=

-1 - A-1 B-3 - 2

=0

-5 = 0

A-3, -1 BAx2, y2 BA2,-1 BAx1, y1 BA-3, -1 BA2, -1 B-1

y = -1y = -1-1

y = -1

y = -1

C

aor 111b

y = 11x + 27

-3

aor 21b

y = 2x

23

y = 23

x + 113

3y = 2x + 11-2x + 3y = 11

-2x + 3y = 11Practice Problem 3Find the slope of the line

.5x + 4y = 10


Practice Problem 5Find the slope of the line .x = -2


Solution: Recall that the graph of is a vertical line with x-intercept 5. To find the slope, we find two ordered pairsolutions of . Ordered pair solutions of must have an x-value of 5. We will use and . We let

and .

Since is undefined, we say the slope of the vertical line

is undefined. Since the x-values will have a differ-ence of 0 for all vertical lines, we can say that all verticallines have undefined slope.

SLOPES OF PARALLEL AND PERPENDICULAR LINES

Two lines in the same plane are parallel if they do not intersect. Slopes oflines can help us determine whether lines are parallel. Since parallel lineshave the same steepness, it follows that they have the same slope.

For example, the graphs of

and

are shown. These lines have the same slope, -2. They also have different y-intercepts, so the lines are parallel. (If the y-intercepts are the same also,the lines are the same.)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

y = −2x + 4m = −2

y = −2x − 3m = −2

y = -2x - 3

y = -2x + 4

D

HELPFUL HINT

Slope of 0 and undefined slope are not the same. Vertical lines haveundefined slope or no slope, while horizontal lines have a slope of 0.

x = 5

40

654321

1−1

−2−3

−2−3 2 3 4 5 6

x = 5

x

y

m =y2 - y1

x2 - x1= 4 - 0

5 - 5= 4

0

Ax2, y2 B = A5, 4 BAx1, y1 B = A5, 0 B A5, 4 BA5, 0 B x = 5x = 5

x = 5

x = 5

Answer

5. undefined slope


Two lines are perpendicular if they lie in the same plane and meet at a90° (right) angle. How do the slopes of perpendicular lines compare? Theproduct of the slopes of two perpendicular lines is .

For example, the graphs of

and

are shown. The slopes of the lines are 4 and . Their product is

, so the lines are perpendicular.

HELPFUL HINT

Here are a few reminders about vertical and horizontal lines.

j Two distinct vertical lines are parallel.j Two distinct horizontal lines are parallel.j A horizontal line and a vertical line are always perpendicular.

PERPENDICULAR LINES

If the product of the slopes of two lines is -1, then the lines are per-pendicular. (Two nonvertical lines are perpendicular if the slope of oneis the negative reciprocal of the slope of the other.)

4 a- 14b = -1

- 14

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

y = 4x + 1m = 4

x − 3

m = − 14

14y = −

y = - 14

x - 3

y = 4x + 1

-1

PARALLEL LINES

Nonvertical parallel lines have the same slope and different y-intercepts.

HELPFUL HINT

Here are examples of numbers that are negative reciprocals.

Number Negative ReciprocalT

dƒƒ

15

-5 or - 51

or -3- 31

13


✓ Concept CheckFor example, y = 2x + 3, y = 2x - 1, y = 2x

Answers6. a. neither, b. perpendicular, c. parallel

Practice Problem 6Determine whether each pair of linesis parallel, perpendicular, or neither.

a.

b.

c.

4x - 2y = 8

y = 2x + 1

5x + 2y = 1

5y = 2x - 3

2x + y = 5

x + y = 5

✓ CONCEPT CHECK

Write the equations of three parallellines.

Example 6 Determine whether each pair of lines is parallel,perpendicular, or neither.

a. b. c.

Solution: a. The slope of the line is . We find

the slope of the second line by solving it for y.



Simplify.

ƒ———TThe slope of this line is also. Since the lines have the

same slope and different y-intercepts, they are parallel, asshown in the figure.b. To find each slope, we solve each equation for y.

c cThe slope is -1. The slope is 1.

The slopes are not the same, so the lines are not parallel.Next we check the product of the slopes: .Since the product is , the lines are perpendicular, asshown in the figure.

(a) (b)

c. We solve each equation for y to find each slope. The

slopes are and . The slopes are not the same

and their product is not . Thus, the lines are neitherparallel nor perpendicular.


-1

- 23

-3

x

y

1−1

1234

−2−3−4

−2−3−4−5 2 3 4 5

x + y = 3

−x + y = 4x

y

1−1

1234

−2−3−4

−2−3−4−5 2 3 4 52x + 10y = 3

x + 115y = −

-1A-1 B A1 B = -1

y = x + 4 y = -x + 3

-x + y = 4x + y = 3

- 15

y = - 15

x + 310

y = -210

x + 310

10y = -2x + 3

2x + 10y = 3

- 15

y = - 15

x + 1

2x + 3y = 6-x + y = 42x + 10y = 3

3x + y = 5x + y = 3y = - 15

x + 1

GRAPHING CALCULATOR EXPLORATIONSIt is possible to use a grapher and sketch the graph of more than one equation on the same set of axes.This feature can be used to see parallel lines with the same slope. For example, graph the equations

, and on the same set of axes. To do so, press the key and enter the

equations on the first three lines.

The displayed equations should look like:

These lines are parallel as expected since they all have a slope of . The graph of is the graph of

moved 7 units upward with a y-intercept of 7. Also, the graph of is the graph of

moved 4 units downward with a y-intercept of .

Graph the parallel lines on the same set of axes. Describe the similarities and differences in their graphs.

1. 2.

3. 4. y = - 34

x, y = - 34

x - 5, y = - 34

x + 6y = 14

x, y = 14

x + 5, y = 14

x - 8

y = -4.9x, y = -4.9x + 1, y = -4.9x + 8y = 3.8x, y = 3.8x - 3, y = 3.8x + 9

-4

y = 25

xy = 25

x - 4y = 25

x

y = 25

x + 725

y2 = x + 725

y1 = x25

y3 = x − 425

10

−10

−10 10

Y3 = a 25b x - 4

Y2 = a 25b x + 7

Y1 = a 25b x

Y=y = 25

x - 4y = 25

x, y = 25

x + 7

MENTAL MATH

Decide whether a line with the given slope is upward-sloping, downward-sloping, horizontal,or vertical.

1. 2. 3. 4. m is undefined

EXERCISE SET 3.4

Use the points shown on each graph to find the slope of each line. See Examples 1 and 2.

1. 2.

3. 4.

Find the slope of the line that passes through the given points. See Examples 1 and 2.

5. 6. 7.

8. 9. 10.

11.

Find the slope of each line. See Example 3.

12. 13. 14.

15. 16. 2x - 3y = 10-5x + y = 10

2x + y = 7y = -2x + 6y = 5x - 2

B

A-2, 8 B and A1, 6 B

A3, 1 B and A2, 6 BA1, 4 B and A5, 3 BA-1, 9 B and A-3, 4 B

A-1, 5 B and A6, -2 BA-1, 5 B and A0, 0 BA0, 0 B and A7, 8 B

x

y

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5

x

y

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5

A

m = 0m = -3m = 76

317


1.

2.

3.

4.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

3.5 GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES

Recall that a linear equation in two variables is an equation that can bewritten in the form where A, B, and C are real numbers andA and B are not both 0. A linear inequality in two variables is an inequal-ity that can be written in one of the forms

where A, B, and C are real numbers and A and B are not both 0.

DETERMINING SOLUTIONS OF LINEAR INEQUALITIES IN TWO VARIABLES

Just as for linear equations in x and y, an ordered pair is a solution of aninequality in x and y if replacing the variables with the coordinates of theordered pair results in a true statement.

Example 1 Determine whether each ordered pair is a solution of theequation .

a. b.

Solution: a. We replace x with 5 and y with and see if a truestatement results.

Replace x with 5 and y with .

False.

The ordered pair is not a solution sinceis a false statement.

b. We replace x with 2 and y with 7 and see if a truestatement results.

Replace x with 2 and y with 7.

True.

The ordered pair is a solution since is atrue statement.

GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES

The linear equation is graphed next. Recall that all points onthe line correspond to ordered pairs that satisfy the equation . Itcan be shown that all the points above the line have coordinatesthat satisfy the inequality . Similarly, all points below the linehave coordinates that satisfy the inequality .x - y 7 1

x - y 6 1x - y = 1

x - y = 1x - y = 1

B

-3 6 6A2, 7 B -3 6 6

4 - 7 6 6

2 A2 B - 7 6 6

2x - y 6 6

11 6 6A5, -1 B

11 6 6

10 + 1 6 6

-12 A5 B - A-1 B 6 6

2x - y 6 6

-1

A2, 7 BA5, -1 B2x - y 6 6

A

Ax + By ≥ CAx + By 7 CAx + By ≤ CAx + By 6 C

Ax + By = C

Graphing Linear Inequalities in Two Variables SECTION 3.5 319

Objectives

Determine whether an orderedpair is a solution of a linear in-equality in two variables.

Graph a linear inequality in twovariables.

B

A

Practice Problem 1Determine whether each ordered pairis a solution of .

a. b. A9, 0 BA-3, 2 Bx - 4y 7 8

Answers1. a. no, b. yes

SSM CD-ROM Video3.5



x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x - y 7 3

Answer2.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

The region above the line and the region below the line are called half-planes. Every line divides the plane (similar to a sheet of paper extendingindefinitely in all directions) into two half-planes; the line is called theboundary.

Recall that the inequality means

Thus, the graph of is the half-plane along with theboundary line .

Example 2 Graph:

Solution: First we graph the boundary line by graphing the equa-tion . We graph this boundary as a dashed linebecause the inequality sign is , and thus the points onthe line are not solutions of the inequality .

(0, 0)

108642

2−2

−4−6−8

−10

−4−6−8−10 4 6 8 10

x + y = 7

(0, 7)(2, 5)

(7, 0)x

y

x + y 6 76

x + y = 7

x + y 6 7

TO GRAPH A LINEAR INEQUALITY IN TWO VARIABLES

Step 1. Graph the boundary line found by replacing the inequality signwith an equal sign. If the inequality sign is 7 or 6 , graph adashed boundary line (indicating that the points on the line arenot solutions of the inequality). If the inequality sign is ≥ or ≤ ,graph a solid boundary line (indicating that the points on theline are solutions of the inequality).

Step 2. Choose a point, not on the boundary line, as a test point. Sub-stitute the coordinates of this test point into the original in-equality.

Step 3. If a true statement is obtained in Step 2, shade the half-planethat contains the test point. If a false statement is obtained,shade the half-plane that does not contain the test point.

x - y = 1x - y 6 1x - y ≤ 1

x - y = 1 or x - y 6 1

x - y ≤ 1

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5

x − y = 1x − y < 1

x − y > 1

(1, 0)

(0, −1)

x

y


Next we choose a test point, being careful not to choose apoint on the boundary line. We choose , and substi-tute the coordinates of into .

Original inequality

Replace x with 0 and y with 0.

True.

Since the result is a true statement, is a solutuion of, and every point in the same half-plane as

is also a solution. To indicate this, we shade theentire half-plane containing , as shown.

Example 3 Graph:

Solution: We graph the boundary line by graphing .We draw this line as a solid line because the inequalitysign is , and thus the points on the line are solutions of

. Once again, is a convenient test pointsince it is not on the boundary line.

We substitute 0 for x and 0 for y into the originalinequality.

Let and .

False.

Since the statement is false, no point in the half-planecontaining is a solution. Therefore, we shade thehalf-plane that does not contain . Every point in theshaded half-plane and every point on the boundary line isa solution of .

(0, 0)

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5

2x − y = 3

2x − y ≥ 3(0, −3)

x

y

( , 0)121

2x - y ≥ 3

A0, 0 BA0, 0 B 0 ≥ 3

y = 0x = 0 2 A0 B - 0 ≥ 3

2x - y ≥ 3

A0, 0 B2x - y ≥ 3≥

2x - y = 3

2x - y ≥ 3

(0, 0)

108642

2−2

−4−6−8

−10

−4−6−8−10 4 6 8 10

x + y < 7

(0, 7)(2, 5)

(7, 0)x

y

A0, 0 BA0, 0 Bx + y 6 7A0, 0 B

0 6 7

0 + 0 6 7

x + y 6 7

x + y 6 7A0, 0 B A0, 0 B


x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x - 4y ≤ 4

Answer3.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5


Answers4.

5.

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5


x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

y 6 3x


x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

3x + 2y ≥ 12

Example 4 Graph:

Solution: We find the boundary line by graphing . Theboundary line is a dashed line since the inequality symbolis . We cannot use as a test point because it is apoint on the boundary line. We choose instead .

Let and .

False.

Since the statement is false, we shade the half-plane thatdoes not contain the test point , as shown.

Example 5 Graph:

Solution: We graph the solid boundary line andchoose as the test point.

Let and .

True.

We shade the half-plane that contains , as shown.

(0, 0)

654321

1−1

−2−3−4

−2−3−4−5 2 3 4 5

5x + 4y ≤ 20 5x + 4y = 20

(0, 5)

(4, 0)x

y

A0, 0 B 0 ≤ 20

y = 0x = 05 A0 B + 4 A0 B ≤ 20

5x + 4y ≤ 20

A0, 0 B 5x + 4y = 20

5x + 4y ≤ 20

(0, 2)

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5x = 2y

x > 2y

(0, 0) (2, 1)x

y

A0, 2 B0 7 4

y = 2x = 00 7 2 A2 Bx 7 2y

A0, 2 BA0, 0 B7

x = 2y

x 7 2y

HELPFUL HINT

When graphing an inequality, make sure the test point is substitutedinto the original inequality. For Example 3, we substituted the testpoint into the original inequality , not .2x - y = 32x - y ≥ 3A0, 0 B


Answer6.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

Example 6 Graph:

Solution: We graph the dashed boundary line and chooseas the test point. (Recall that the graph of is

a horizontal line with y-intercept 3.)

Let .

False.

We shade the half-plane that does not contain , asshown.

(0, 0)

y > 3

y = 3

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5 x

y

A0, 0 B0 7 3

y = 00 7 3

y 7 3

y = 3A0, 0 B y = 3

y 7 3 Practice Problem 6Graph:

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x 6 2


Focus On The Real WorldMISLEADING GRAPHS

Graphs are very common in magazines and in newspapers such as USA Today. Graphs can bea convenient way to get an idea across because, as the old saying goes, “a picture is worth athousand words.” However, some graphs can be deceptive, which may or may not be inten-tional. It is important to know some of the ways that graphs can be misleading.

Beware of graphs like the one at the right. Noticethat the graph shows a company’s profit for vari-ous months. It appears that profit is growing quiterapidly. However, this impressive picture tells uslittle without knowing what units of profit arebeing graphed. Does the graph show profit in dol-lars or millions of dollars? An unethical companywith profit increases of only a few pennies coulduse a graph like this one to make the profitincrease seem much more substantial than it reallyis. A truthful graph describes the size of the unitsused along the vertical axis.

Another type of graph to watch for is one that mis-represents relationships. For example, the bargraph at the right shows the number of men andwomen employees in the accounting and shippingdepartments of a certain company. In the account-ing department, the bar representing the numberof women is shown twice as tall as the bar repre-senting the number of men. However, the numberof women (13) is not twice the number of men(10). This set of bars misrepresents the relationshipbetween the number of men and women. Do you see how the relationship between the number of men and women in the shipping department isdistorted by the heights of the bars used? A truthful graph will use bar heights that are propor-tional to the numbers they represent.

The impression a graph can give also depends on its vertical scale. The two graphs below rep-resent exactly the same data. The only difference between the two graphs is the vertical scale—one shows enrollments from 246 to 260 students and the other shows enrollments between 0and 300 students. If you were trying to convince readers that algebra enrollment at UPH hadchanged drastically over the period 1996–2000, which graph would you use? Which graph doyou think gives the more honest representation?

300

250

200

150

100

50

0

Num

ber

of S

tude

nts

Algebra Enrollment at UPH

Year

200019991996 1997 1998

260258256254252250248246

Num

ber

of S

tude

nts

Algebra Enrollment at UPH

Year

200019991996 1997 1998

Num

ber

of E

mpl

oyee

s

Department

10

40

9

13

Accounting Shipping

Men

WomenPr

ofit

Month

121086421 3 5 7 9 11



SECTION 3.1 THE RECTANGULAR COORDINATE SYSTEM

The rectangular coordinate system consists of aplane and a vertical and a horizontal number lineintersecting at their 0 coordinate. The vertical num-ber line is called the y-axis and the horizontal num-ber line is called the x-axis. The point of intersec-tion of the axes is called the origin.

To plot or graph an ordered pair means to find itscorresponding point on a rectangular coordinatesystem.

To plot or graph an ordered pair such as ,start at the origin. Move 3 units to the right andfrom there, 2 units down.

To plot or graph ; start at the origin. Move 3units to the left and from there, 4 units up.

An ordered pair is a solution of an equation in twovariables if replacing the variables with the coordi-nates of the ordered pair results in a true statement.

A-3, 4 B

A3, -2 B

5432

1

4 units

2 units

3 units3 units

(3, −2)

(−3, 4)

−1

−2−3−4−5

−2−3−4−5 2 3 4 5 x

y

54321

1

Origin

−1

−2−3−4−5

−2−3−4

quadrant II quadrant I

quadrant III quadrant IV

−5 2 3 4 5 x

y

If one coordinate of an ordered pair solution isknown, the other value can be determined by sub-stitution.

Complete the ordered pair for the equation.

Let .

Divide by .

The ordered pair solution is .A0, -2 B y = -2

-6 -6y-6

= 12-6

x = 00 - 6y = 12

x - 6y = 12

x - 6y = 12A0, B

SECTION 3.2 GRAPHING LINEAR EQUATIONS

A linear equation in two variables is an equation thatcan be written in the form , where Aand B are not both 0. The form iscalled standard form.

Ax + By = CAx + By = C

is in standard form.x + y = 10

y = -x + 10 y = 3x = -53x + 2y = -6



To graph a linear equation in two variables, findthree ordered pair solutions. Plot the solutionpoints and draw the line connecting the points.

Graph: x y

5 01

(1, −2)(−1, −3)

(5, 0)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

-3-1-2

x - 2y = 5

SECTION 3.3 INTERCEPTS

An intercept point of a graph is a point where thegraph intersects an axis. If a graph intersects the x-axis at a, then a is the x-intercept and the corre-sponding intercept point is . If a graph inter-sects the y-axis at b, then b is the y-intercept and thecorresponding intercept point is .A0, b B

Aa, 0 B54321

1

The x-intercept is 5.Intercept point: (5, 0)

The y-intercept is 3.Intercept point: (0, 3)

−1

−2−3−4−5

−2−3−4−5 2 3 4 5 x

y

To find the x-intercept, let and solve for x.To find the y-intercept, let and solve for y.x = 0

y = 0 Find the intercepts for

If , then If , then

The x-intercept is . The y-intercept is 2.Intercept point: . Intercept point: .

(−5, 0)2x − 5y = −10

(0, 2)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

A0, 2 BA-5, 0 B-5y = 2 x = -5

-5y-5

= -10-5

2x2

= -102

-5y = -10 2x = -102 # 0 - 5y = -102x - 5 # 0 = -10

x = 0y = 0

2x - 5y = -10.

The graph of is a vertical line with x-inter-cept c.

The graph of is a horizontal line with y-inter-cept c.

y = c

x = c

y = −1x

y

x = 3

x

y


SECTION 3.4 SLOPE

The slope m of the line through points andis given by

m =y2 - y1

x2 - x1 as long as x2 Z x1

Ax2, y2 BAx1, y1 B The slope of the line through points and

is

m =y2 - y1

x2 - x1= 8 - 6

-5 - A-1 B = 2-4

= - 12

A-5, 8 BA-1, 6 B

A horizontal line has slope 0.The slope of a vertical line is undefined.Nonvertical parallel lines have the same slope.Two nonvertical lines are perpendicular if the slope

of one is the negative reciprocal of the slope of theother.

The slope of the line is 0.The line has undefined slope.

Perpendicularlines

x

ym = 3

2

m = 23−

Parallellines

x

y m = 54

m = 54

x = 3y = -5

SECTION 3.5 GRAPHING LINEAR INEQUALITIES IN TWO VARIABLESE

A linear inequality in two variables is an inequalitythat can be written in one of these forms:

Ax + By ≥ CAx + By 7 CAx + By ≤ CAx + By 6 C

y ≤ 2 y 7 -8xx ≥ -52x - 5y 6 6

TO GRAPH A LINEAR INEQUALITY

1. Graph the boundary line by graphing the relatedequation. Draw the line solid if the inequalitysymbol is ≤ or ≥. Draw the line dashed if theinequality symbol is .

2. Choose a test point not on the line. Substitute itscoordinates into the original inequality.

3. If the resulting inequality is true, shade the half-plane that contains the test point. If the inequal-ity is not true, shade the half-plane that does notcontain the test point.

6 or 7

Graph: 1. Graph . Draw a solid line because

the inequality symbol is ≤.2. Check the test point in the original

inequality, .

Let and .

True.

3. The inequality is true, so shade the half-planecontaining as shown.

2x − y = 4

2x − y ≤ 4

x

y

A0, 0 B

0 ≤ 4

y = 0x = 02 # 0 - 0 ≤ 4

2x - y ≤ 4A0, 0 B

2x - y = 42x - y ≤ 4

329

4Recall from Chapter 1 that an exponent is a shorthand notation forrepeated factors. This chapter explores additional concepts about expo-nents and exponential expressions. An especially useful type of expo-nential expression is a polynomial. Polynomials model many real-worldphenomena. Our goal in this chapter is to become proficient with oper-ations on polynomials.

4.1 Exponents

4.2 Negative Exponents andScientific Notation

4.3 Introduction to Polynomials

4.4 Adding and SubtractingPolynomials

4.5 Multiplying Polynomials

4.6 Special Products4.7 Dividing Polynomials

One way exponents are used is to express very large or verysmall numbers in scientific notation.The U.S. Bureau of theCensus conducts a census every 10 years to count U.S. citizens.However, government agencies and private businesses needmore up-to-date population figures between censuses. From itsresearch, the Census Bureau devised a way to estimate the U.S.population based on a few simple observations. There is a birthin the U.S. every 8 seconds and there is a death in the U.S. every14 seconds. The estimate also takes into account the arrival ofimmigrants (one approximately every 42 seconds) and U.S. cit-izens returning from abroad (one approximately every 4108seconds). These facts along with the most recent census figurescan be used to estimate the current population. Visit the Bureauof the Census Web site, at http://www.census.gov/cgi-bin/popclock.

Exponents and Polynomials C H A P T E R

4.1 EXPONENTS

EVALUATING EXPONENTIAL EXPRESSIONS

In this section, we continue our work with integer exponents. As wereviewed in Section 1.2, for example,

The exponent 5 tells us how many times that 2 is a factor. The expres-sion is called an exponential expression. It is also called the fifth powerof 2, or we can say that 2 is raised to the fifth power.

and

6 factors; each factor is 5 4 factors; each factor is

The base of an exponential expression is the repeated factor. Theexponent is the number of times that the base is used as a factor.

T—exponent or power

c base n factors of a

Examples Evaluate (find the value of) each expression.

1.2. . To raise 3 to the first power means to use 3 as

a factor only once. When no exponent is shown, theexponent is assumed to be 1.

3.4.

5.

6.

Notice how similar is to in the examples above. The differ-ence between the two is the parentheses. In , the parentheses tell usthat the base, or the repeated factor, is . In , only 4 is the base.

An exponent has the same meaning whether the base is a number or avariable. If is a real number and is a positive integer, then is the prod-uct of factors, each of which is .

n factors; each factor is x

fxn=x # x # x # x # x . . . x

xnxnnx

-42-4A-4 B 2A-4 B 2-42

4 # 32 = 4 # 9 = 36

a 12b 4

= 12

# 12

# 12

# 12

= 116

-42 = - A4 # 4 B = -16A-4 B 2 = A-4 B A-4 B = 16

31 = 323 = 2 # 2 # 2 = 8

dan = a # a # a . . . a

-3

fd A-3 B 4 = A-3 B # A-3 B # A-3 B # A-3 B 56 = 5 # 5 # 5 # 5 # 5 # 5

25

2 # 2 # 2 # 2 # 2 = 25

A

Exponents SECTION 4.1 331

Objectives

Evaluate exponential expressions.

Use the product rule for expo-nents.

Use the power rule for exponents.

Use the power rules for productsand quotients.

Use the quotient rule for expo-nents, and define a number raisedto the 0 power.

Decide which rule(s) to use to sim-plify an expression.

F

E

DC

BA

Practice Problems 1–6Evaluate (find the value of) each ex-pression.

1. 2.

3. 4.

5. 6. 5 # 62a 23b 2

-23A-2 B 37134

HELPFUL HINT

Be careful when identifying the base of an exponential expression. Payclose attention to the use of parentheses.

The base is . The base is 3. The base is 3.2 # 32 = 2 # 3 # 3 = 18-32 = - A3 # 3 B = -9A-3 B 2 = A-3 B A-3 B = 9

-32 # 32-32A-3 B 2

Answers1. 81, 2. 7, 3. , 4. , 5. , 6. 180

49

-8-8

SSM CD-ROM Video4.1

332 CHAPTER 4 Exponents and Polynomials

PRODUCT RULE FOR EXPONENTS

If m and n are positive integers and a is a real number, then

For example:

35 # 37 = 35 + 7 = 312

am # an = am + n

Example 7 Evaluate each expression for the given value of .

a. when is 5 b. when is

Solution: a. When x is 5,

b. When x is

USING THE PRODUCT RULE

Exponential expressions can be multiplied, divided, added, subtracted, andthemselves raised to powers. By our definition of an exponent,

4 factors of 5 3 factors of 5

7 factors of 5

Also,

In both cases, notice that the result is exactly the same if the exponents areadded.

and

This suggests the following rule.

In other words, to multiply two exponential expressions with the same base,we keep the base and add the exponents. We call this simplifying the expo-nential expression.

Examples Use the product rule to simplify each expression.

8.9.

10. ——

= y4 = y3 + 1

y3 # y = y3 # y1HELPFUL HINT

Don’t forget that if no exponent is written, it isassumed to be 1.

x2 # x5 = x2 + 5 = x742 # 45 = 42 + 5 = 47

x2 # x3 = x2 + 3 = x554 # 53 = 54 + 3 = 57

= x5

= x # x # x # x # x

x2 # x3 = Ax # x B # Ax # x # x B

= 57

= 5 # 5 # 5 # 5 # 5 # 5 # 5

c b54 # 53 = A5 # 5 # 5 # 5 B # A5 # 5 # 5 B

B

= 99

= 1

= 9A-3 B A-3 B

-3, 9x2 = 9

-32

= 250 = 2 # 125 = 2 # A5 # 5 # 5 B

2x3 = 2 # 53

-3x9x2x2x3

xPractice Problem 7Evaluate each expression for the givenvalue of .

a. when is 4

b. when is -2xx4

-8

x3x2

x

Answers7. a. 48, b. , 8. , 9. , 10. ,11. , 12. A-3 B 10s11

r6x1375-2

Practice Problems 8–12Use the product rule to simplify eachexpression.

8. 9.

10. 11.

12. A-3 B 9 # A-3 Bs6 # s2 # s3r5 # r

x4 # x973 # 72

11.

12.

Example 13 Use the product rule to simplify .

Solution: Recall that means and means .

Remove parentheses.

Group factors with common bases.

Simplify.

USING THE POWER RULE

Exponential expressions can themselves be raised to powers. Let’s try todiscover a rule that simplifies an expression like . By the definitionof ,

means 3 factors of .

which can be simplified by the product rule for exponents.

Notice that the result is exactly the same if we multiply the exponents.

The following rule states this result.

In other words, to raise an exponential expression to a power, we keep thebase and multiply the exponents.

Examples Use the power rule to simplify each expression.

14.15. Ay8 B 2 = y8 # 2 = y16

A53 B 6 = 53 # 6 = 518

Ax2 B 3 = x2 # 3 = x6

Ax2 B 3 = Ax2 B Ax2 B Ax2 B = x2 + 2 + 2 = x6

Ax2 BAx2 B 3Ax2 B 3 = Ax2 B Ax2 B Ax2 Ban

Ax2 B 3C

= -6x7

= 2 # -3 # x2 # x5

A2x2 B A-3x5 B = 2 # x2 # -3 # x5

-3 # x5-3x52 # x22x2

A2x2 B A-3x5 BA-5 B 7 # A-5 B 8 = A-5 B 7 + 8 = A-5 B 15

y3 # y2 # y7 = y3 + 2 + 7 = y12


HELPFUL HINTThese examples will remind you of the difference between adding andmultiplying terms.

Addition

Multiplication

A7x B A4x2 B = 7 # 4 # x # x2 = 28x1 + 2 = 28x3

A5x3 B A3x3 B = 5 # 3 # x3 # x3 = 15x3 + 3 = 15x6

7x + 4x2 = 7x + 4x2

5x3 + 3x3 = A5 + 3 Bx3 = 8x3

Practice Problem 13Use the product rule to simplify

.A6x3 B A-2x9 B

POWER RULE FOR EXPONENTS

If and are positive integers and is a real number, then

For example:

A72 B 5 = 72 # 5 = 710

Aam Bn = amn

anm

Practice Problems 14–15Use the power rule to simplify eachexpression.

14. 15. Az6 B 3A94 B 10

Answers13. , 14. , 15. z18940-12x12

USING THE POWER RULES FOR PRODUCTS AND QUOTIENTS

When the base of an exponential expression is a product, the definition ofstill applies. To simplify , for example,


Group factors with common bases.

Simplify.

Notice that to simplify the expression , we raise each factor within theparentheses to a power of 3.

In general, we have the following rule.

In other words, to raise a product to a power, we raise each factor to thepower.


16. Use the power of a product rule.



Use the power rule for exponents.

Let’s see what happens when we raise a quotient to a power. To simplify

, for example,


Multiply fractions.

Simplify.

Notice that to simplify the expression, , we raise both the numerator

and the denominator to a power of 3.

In general, we have the rule shown in the margin.

In other words, to raise a quotient to a power, we raise both the numeratorand the denominator to the power.

a xyb 3

= x3

y3

a xyb 3

= x3

y3

= x # x # xy # y # y

a xyba x

yb 3a x

yb 3

= a xyb a x

yb a x

yb

a xyb 3

= 25x4y6z2A-5x2y3z B 2 = A-5 B 2 # Ax2 B 2 # Ay3 B 2 # Az1 B 2A2a B 3 = 23 # a3 = 8a3Ast B 4 = s4 # t4 = s4t4

Axy B 3 = x3y3

Axy B 3 = x3y3

= x # x # x # y # y # y

Axy BAxy B 3Axy B 3 = Axy B Axy B Axy BAxy B 3an

D


HELPFUL HINT

Take a moment to make sure thatyou understand when to apply theproduct rule and when to apply thepower rule.

Product Rule S Add Exponents

Power Rule S Multiply Exponents

Ay6 B 2 = y6 # 2 = y12Ax5 B 7 = x5 # 7 = x35

y6 # y2 = y6 + 2 = y8

x5 # x7 = x5 + 7 = x12

POWER OF A PRODUCT RULE

If n is a positive integer and a and b are real numbers, then

For example:

A3x B 5 = 35x5

Aab Bn = anbn


16. 17.

18. A-2p4q2r B 3A3y B 4Axy B 7

Answers16. , 17. , 18. -8p12q6r381y4x7y7

POWER OF A QUOTIENT RULE

If n is a positive integer and a andc are real numbers, then

,

For example:

a y7b 3

= y3

73

c Z 0a acb n

= an

cn



19. , Use the power of a quotient rule.

20. Use the power of a quotient rule.

, Use the power rule for exponents.

USING THE QUOTIENT RULE AND DEFINING THE ZERO EXPONENT

Another pattern for simplifying exponential expressions involves quo-tients.

Notice that the result is exactly the same if we subtract exponents of thecommon bases.

The following rule states this result in a general way.

In other words, to divide one exponential expression by another with acommon base, we keep the base and subtract the exponents.

Examples Simplify each quotient.

21. Use the quotient rule.



24.

Use the quotient rule.

= 2x4y1 or 2x4y

= 2 # Ax5-1 B # Ay2-1 B2x5y2

xy= 2 # x5

x1# y2

y1

A-3 B 5A-3 B 2 = A-3 B 3 = -27

47

43 = 47 - 3 = 44 = 256

x5

x2 = x5 - 2 = x3

x5

x3 = x5-3 = x2

= x2 = x # x

= x # x # x # x # x

x # x # x

x5

x3 = x # x # x # x # xx # x # x

E

y Z 0 = 16x16

81y20

a 2x4

3y5 b 4

=24 # Ax4 B 434 # Ay5 B 4

n Z 0amnb 7

= m7

n7


19. 20. a 5x6

9y3 b 2a rsb 6

QUOTIENT RULE FOR EXPONENTS

If m and n are positive integers and a is a real number, then

,

For example:

, x Z 0x6

x2 = x6-2 = x4

a Z 0am

an = am-n

Answers

19. , , 20. , , 21. ,

22. 125, 23. 16, 24. 7a3b10

y4y Z 025x12

81y6s Z 0r6

s6

Practice Problems 21–24Simplify each quotient.

21. 22.

23. 24.7a4b11

abA-2 B 14

A-2 B 10

59

56y7

y3

Let’s now give meaning to an expression such as . To do so, we will

simplify in two ways and compare the results.

Apply the quotient rule.

Apply the fundamental principle for fractions.

Since and , we define that as long as is not 0.

In other words, a base raised to the 0 power is 1, as long as the base is not 0.


25.26.27.28.


DECIDING WHICH RULE TO USE

Let’s practice deciding which rule to use to simplify.


a.

b.

c.

Solution: a. Here, we have a product, so we use the product rule tosimplify.

b. This is a quotient raised to a power, so we use thepower of a quotient rule.

c. This is a product raised to a power, so we use thepower of a product rule.

A9y5 B 2 = 92 Ay5 B 2 = 81y10

a 12b 4

= 14

24 = 116

x7 # x4 = x7 + 4 = x11

A9y5 B 2a 1

2b 4

x7 # x4

F

-40 = -1 # 40 = -1 # 1 = -1A-4 B 0 = 1A5x3y2 B 0 = 130 = 1

xx0 = 1x3

x3 = 1x3

x3 = x0

x3

x3 = x # x # x x # x # x

= 1

x3

x3 = x3 - 3 = x0

x3

x3

x0


ZERO EXPONENT

, as long as is not 0.

For example: .50 = 1

aa0 = 1


25. 26.

27. 28. -50A-5 B 0A2r2s B 080

✓ CONCEPT CHECK

To simplify each expression, tellwhether you would add the expo-nents, subtract the exponents, multiplythe exponents, or divide the expo-nents, or none of these.

a. b.

c. d. w45 # w9z16 + z8

y15

y3Ax63 B 21

Answers25. 1, 26. 1, 27. 1, 28. , 29. a. ,

b. , c.

✓Concept Checka. multiply, b. subtract, c. none of these,d. add

x3

6481y16

x3-1


a. b. c. a x4b 3A3y4 B 4x7

x4

MENTAL MATH

State the bases and the exponents for each expression.

1. 2. 3. 4.

5.

EXERCISE SET 4.1

Evaluate each expression. See Examples 1 through 6.

1. 2. 3. 4.

5.

Evaluate each expression with the given replacement values. See Example 7.

6. when 7. when 8. when

9. when 10. when and

Use the product rule to simplify each expression. Write the results using exponents. SeeExamples 8 through 13.

11. 12.

Use the power rule and the power of a product or quotient rule to simplify each expres-sion. See Examples 14 through 20.

13. 14. 15.

Use the quotient rule and simplify each expression. See Examples 21 through 24.

16. 17. 18. 19.

20. 7x2y6

14x2y3

x8y6

xy5p7q20

pq15y10

y9x3

x

E

Apq B 7Ay7 B 5Ax9 B 4C

y2 # yx2 # x5

B

y = 5x = 32xy2x = -14x2

x = 35x3x = -2x3x = -2x2

-24

A-3 B 2A-5 B 1-3272

A

-42

-37A-3 B 65432

337


1.

2.

3.

4.

5.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

338

STUDYING FOR A MATH EXAM

Remember that one of the best ways to start preparing for anexam is to keep current with your assignments as they are made.Make an effort to clear up any confusion on topics as you coverthem.

Begin reviewing for your exam a few days in advance. This way,if you find a topic that you still don’t understand, you’ll haveplenty of time to ask your instructor, another student in your class,or a math tutor for help. Don’t wait until the last minute to “cram”for an exam.

j Reread your notes and carefully review the Chapter Highlights atthe end of each chapter to be covered.

j Pay special attention to any new terminology or definitions inthe chapter. Be sure you can state the meanings of definitions inyour own words.

j If you tend to get anxious while taking an exam, try to visualizeyourself taking the exam in advance. Picture yourself beingcalm, clearheaded, and successful. Picture yourself rememberingformulas and definitions with no trouble. When you are wellprepared for an exam, a lot of nervousness can be avoidedthrough positive thinking.

j Get lots of rest the night before the exam. It’s hard to show howwell you know the material if your brain is foggy from lack ofsleep.

j Give yourself enough time so that you will arrive early for theexam. This way, if you run into difficulties on the way, you shouldstill arrive on time.

Focus On Study Skills

4.2 NEGATIVE EXPONENTS AND SCIENTIFIC NOTATION

SIMPLIFYING EXPRESSIONS CONTAINING NEGATIVE EXPONENTS

Our work with exponential expressions so far has been limited to expo-nents that are positive integers or 0. Here we expand to give meaning to anexpression like .

Suppose that we wish to simplify the expression . If we use the quo-

tient rule for exponents, we subtract exponents:

,

But what does mean? Let’s simplify using the definition of .

If the quotient rule is to hold true for negative exponents, then must

equal .

From this example, we state the definition for negative exponents.

In other words, another way to write is to take its reciprocal and changethe sign of its exponent.

Examples Simplify by writing each expression with positiveexponents only.

1. Use the definition of negative exponent.

2. Use the definition of negative exponent.

ƒ

3.

4. A-2 B-4 = 1A-2 B 4 = 1

A-2 B A-2 B A-2 B A-2 B = 116

2-1 + 4-1 = 12

+ 14

= 24

+ 14

= 34

2x-3 = 2 # 1x3 = 2

x3

3-2 = 132 = 1

9

a-n

1x3

x-3

= 1x3

Divide numerator and denominator by commonfactors by applying the fundamental principle forfractions.

= x # xx # x # x # x # x

x2

x5 = x # xx # x # x # x # x

anx2

x5x-3

x Z 0x2

x5 = x2-5 = x-3

x2

x5

x-3

A

Negative Exponents and Scientific Notation SECTION 4.2 339

Objectives

Simplify expressions containingnegative exponents.

Use the rules and definitions forexponents to simplify exponentialexpressions.

Write numbers in scientific nota-tion.

Convert numbers in scientific nota-tion to standard form.

D

C

B

A

NEGATIVE EXPONENTS

If a is a real number other than 0 and n is an integer, then

For example,

x-3 = 1x3

a-n = 1an

HELPFUL HINT

Don’t forget that since there are no parentheses, only x is the base forthe exponent .-3

Answers

1. , 2. , 3. , 4. 181

815

7x4

1125

SSM CD-ROM Video4.2

Practice Problems 1–4Simplify by writing each expressionwith positive exponents only.

1. 2.

3. 4. A-3 B-45-1 + 3-1

7x-45-3

Examples Simplify each expression. Write each result using positiveexponents only.

T¬¬¬¬¬¬ƒ

5. Use the negative exponent rule.

T¬¬¬¬¬¬ƒ6. Use the quotient rule.

7. Use the negative exponent rule.

8.

SIMPLIFYING EXPONENTIAL EXPRESSIONS

All the previously stated rules for exponents apply for negative exponentsalso. Here is a summary of the rules and definitions for exponents.

Examples Simplify each expression. Write each result using positiveexponents only.

T—————————————————————ƒ

9. Use the power rule.Ax3 B 4xx7 = x12 # x

x7 = x12+1

x7 = x13

x7 = x13-7 = x6

B

x-5

x7 = x-5-7 = x-12 = 1x12

p-4

q-9 = q9

p4

yy-2 = y1

y-2 = y1 - 1-22 = y3

a 23b-3

= 2-3

3-3 = 33

23 = 278


HELPFUL HINT

A negative exponent does not affect the sign of its base.Remember: Another way to write is to take its reciprocal and

change the sign of its exponent, . For example,

, or

, or 251

5-2 = 521y-4 = 1

1y4

= y4

18

2-3 = 123x-2 = 1

x2

a-n = 1an

a-n

Practice Problems 5–8Simplify each expression. Write eachresult using positive exponents only.

5. 6.

7. 8.y-4

y6y-9

z-5

xx-4a 6

7b-2

Answers

5. , 6. , 7. , 8. 1y10

z5

y9x54936

SUMMARY OF EXPONENT RULES

If m and n are integers and a, b, and c are real numbers, then:

Product rule for exponents:

Power rule for exponents:

Power of a product:

Power of a quotient: ,

Quotient rule for exponents: ,

Zero exponent: ,

Negative exponent: , a Z 0a-n = 1an

a Z 0a0 = 1

a Z 0am

an = am-n

c Z 0a acb n

= an

cn

Aab Bn = anbn

Aam Bn = am # nam # an = am+n


Practice Problems 9–14Simplify each expression. Write eachresult using positive exponents only.

9. 10.

11. 12.

13. 14. A4a2 B-3y-10

Ay5 B 4

A2x B 4x8Aa-4b7 B-5

a 9x3

yb -2Ax5 B 3x

x4

10. Raise each factor in the numerator and the denominatorto the power.

Use the power rule.

Use the negative exponent rule.

Write as 27.

11. Raise each factor to the power.

T¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬ƒ

12.

13.

14. Raise each factor to the power.

WRITING NUMBERS IN SCIENTIFIC NOTATION

Both very large and very small numbers frequently occur in many fields ofscience. For example, the distance between the sun and the planet Pluto isapproximately 5,906,000,000 kilometers, and the mass of a proton isapproximately 0.00000000000000000000000165 gram. It can be tedious towrite these numbers in this standard decimal notation, so scientific notationis used as a convenient shorthand for expressing very large and very smallnumbers.

The following numbers are written in scientific notation. The signfor multiplication is used as part of the notation.

(Distance between the sun andPluto)(Mass of a proton)1.65 * 10-248.1 * 10-51 * 10-3

5.906 * 1097.362 * 1072.03 * 102

*

5,906,000,000kilometers

Sun

Pluto

C

= 5-2y-6 = 152y6 = 1

25y6

-2A5y3 B-2 = 5-2 Ay3 B-2

x-7

Ax4 B 3 = x-7

x12 = x-7 - 12 = x-19 = 1x19

Raise each factor in the numeratorto the fifth power.A2x B 5

x3 = 25 # x5

x3 = 25 # x5-3 = 32x2

u

= y18z-36 = y18

z36

-6Ay-3z6 B - 6= Ay-3 B-6 Az6 B-6

33 = b3

27a6

= b3

33a6

= 3-3a-6

b-3

-3a 3a2

bb -3

=3-3 Aa2 B-3

b-3

Answers

9. , 10. , 11. , 12. , 13. ,

14. 164a6

1y30

16x4

a20

b35

y2

81x6x12

SCIENTIFIC NOTATION

A positive number is written in scientific notation if it is written as theproduct of a number a, where , and an integer power r of 10:

a * 10r

1 ≤ a 6 10

proton

Mass of proton is approximately0.000 000 000 000 000 000 000 001 65 gram


The following steps are useful when writing numbers in scientific no-tation.

Example 15 Write each number in scientific notation.

a. 367,000,000 b. 0.000003c. 20,520,000,000 d. 0.00085

Solution: a. Step 1. Move the decimal point until the number isbetween 1 and 10.

8 places

Step 2. The decimal point is moved to the left 8 places,so the count is positive 8.

Step 3. .

b. Step 1. Move the decimal point until the number isbetween 1 and 10.

6 places

Step 2. The decimal point is moved 6 places to theright, so the count is

Step 3.

c.

d.

CONVERTING NUMBERS TO STANDARD FORM

A number written in scientific notation can be rewritten in standard form.For example, to write in standard form, recall that .

Notice that the exponent on the 10 is positive 3, and we moved the decimalpoint 3 places to the right.

To write in standard form, recall that .

The exponent on the 10 is negative 3, and we moved the decimal to the left3 places.

7.29 * 10-3 = 7.29 a 11000

b = 7.291000

= 0.00729

10-3 = 1103 = 1

10007.29 * 10-3

8.63 * 103 = 8.63 11000 2 = 8630

103 = 10008.63 * 103

D

0.00085 = 8.5 * 10-4

20,520,000,000 = 2.052 * 1010

0.000003 = 3.0 * 10-6

-6.

0.00 0030

367,000,000 = 3.67 * 108

367,000,000

TO WRITE A NUMBER IN SCIENTIFIC NOTATION

Step 1. Move the decimal point in the original number so that the newnumber has a value between 1 and 10.

Step 2. Count the number of decimal places the decimal point is movedin Step 1. If the decimal point is moved to the left, the count ispositive. If the decimal point is moved to the right, the count isnegative.

Step 3. Multiply the new number in Step 1 by 10 raised to an exponentequal to the count found in Step 2.

Practice Problem 15Write each number in scientific nota-tion.

a. 420,000 b. 0.00017

c. 9,060,000,000 d. 0.000007

Answers15. a. , b. ,c. , d. 7 * 10-69.06 * 109

1.7 * 10-44.2 * 105


✓ CONCEPT CHECK

Which number in each pair is larger?a. or

b. or

c. or 6.3 * 10-55.6 * 10-4

2.7 * 1049.2 * 10-2

2.1 * 1057.8 * 103

Practice Problem 16Write the numbers in standard nota-tion, without exponents.

a. b.

c. d. 6.002 * 1069.6 * 10-5

5.21 * 1043.062 * 10-4

Practice Problem 17Perform each indicated operation.Write each result in standard decimalnotation.

a.

b.8 * 104

2 * 10-3

A9 * 107 B A4 * 10-9 B

In general, to write a scientific notation number in standard form, movethe decimal point the same number of places as the exponent on 10. If theexponent is positive, move the decimal point to the right; if the exponent isnegative, move the decimal point to the left.


Example 16 Write each number in standard notation, withoutexponents.

a. b.

c. d.

Solution: a. Move the decimal point 5 places to the right.

b. Move the decimal point 3 places to the left.

c. 7 places to the right

d. 5 places to the left

Performing operations on numbers written in scientific notation makesuse of the rules and definitions for exponents.

Example 17 Perform each indicated operation. Write each result instandard decimal notation.

a.

b.

Solution: a.

b. 12 * 102

6 * 10-3 = 126

* 102-A- 3B = 2 * 105 = 200,000

= 0.056

= 56 * 10-3

A8 * 10-6 B A7 * 103 B = 8 # 7 # 10-6 # 103

12 * 102

6 * 10-3

A8 * 10-6 B A7 * 103 B

3.007 * 10-5 = 0.00003007

8.4 * 107 = 84,000,000.

7.358 * 10-3 = 0.007358

1.02 * 105 = 102,000.

3.007 * 10-58.4 * 107

7.358 * 10-31.02 * 105

✓ Concept Checka. b. c.Answers16. a. 0.0003062, b. 52,100, c. 0.000096,d. 6,002,000, 17. a. 0.36, b. 40,000,000

5.6 * 10-42.7 * 104,2.1 * 105,


CALCULATOR EXPLORATIONSSCIENTIFIC NOTATION

To enter a number written in scientific notation on a scientific cal-

culator, locate the scientific notation key, which may be marked

or . To enter , press . The display

should read .

Enter each number written in scientific notation on your calculator.

1.

2.

3.

4.

Multiply each of the following on your calculator. Notice the form ofthe result.

5.

6.

Multiply each of the following on your calculator. Write the productin scientific notation.

7.

8. A8.76 * 10-4 B A1.237 * 109 BA3.26 * 106 B A2.5 * 1013 B

230,000 * 1,000

3,000,000 * 5,000,000

-9.9811 * 10-2

6.6 * 10-9

-4.8 * 1014

5.31 * 103

3.1 07

7EE3.13.1 * 107EXPEE

345


MENTAL MATH

State each expression using positive exponents only.

1. 2. 3.

4. 5. 6.

EXERCISE SET 4.2

Simplify each expression. Write each result using positive exponents only. See Examples1 through 8.

1. 2. 3. 4.

Simplify each expression. Write each result using positive exponents only. See Examples9 through 14.

5. 6. 7. 8.

Write each number in scientific notation. See Example 15.

9. 78,000 10. 9,300,000,000 11. 0.00000167 12. 0.00000017

13. 0.00635 14. 0.00194

C

y3yy-2

p2pp-1

y4y5

y6x2x5

x3

B

A7x B-37x-36-24-3

A

16y-7

4y-3

1x-3

1y-63x-35x-2

MENTAL MATH ANSWERS

1.

2.

3.

4.

5.

6.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

Expression Terms

, , ,

The numerical coefficient of a term, or simply the coefficient, is thenumerical factor of each term. If no numerical factor appears in the term,then the coefficient is understood to be 1. If the term is a number only, it iscalled a constant term or simply a constant.

Term Coefficient

13

3 (constant) 3

Example 1 Complete the table for the expression.

Term Coefficient

75

Solution: The completed table is

Term Coefficient

1

75 5

7x5

-3-3x

-8-8x4

x2

-3x

-8x2

3x + 5x2 -7x5 - 8x4 +

-1-x2y-4-4x

3x2x5

7y37y3-1-7x9x49x4 - 7x - 1

3x4x24x2 + 3x

4.3 INTRODUCTION TO POLYNOMIALS

DEFINING TERM AND COEFFICIENT

In this section, we introduce a special algebraic expression called a polyno-mial. Let’s first review some definitions presented in Section 2.1.

Recall that a term is a number or the product of a number and variablesraised to powers. The terms of the expression are and . Theterms of the expression are , , and .-1-7x9x49x4 - 7x - 1

3x4x24x2 + 3x

A

Introduction to Polynomials SECTION 4.3 347

Objectives

Define term and coefficient of aterm.

Define polynomial, monomial, bi-nomial, trinomial, and degree.

Evaluate polynomials for given re-placement values.

Simplify a polynomial by combin-ing like terms.

Simplify a polynomial in severalvariables.

E

D

C

B

A

SSM CD-ROM Video4.3

Practice Problem 1Complete the table for the expression

.

Term Coefficient

4-1

-6x6

-97x3

7x3 - 9x2 - 14x5 +-6x6 +

Answers1. term: , coefficient: 7; ; -1-6-9x2; 4x5

DEFINING POLYNOMIAL, MONOMIAL, BINOMIAL, TRINOMIAL, AND DEGREE

Now we are ready to define what we mean by a polynomial.

For example,

is a polynomial in x. Notice that this polynomial is written in descendingpowers of x because the powers of x decrease from left to right. (Recall thatthe term 1 can be thought of as .)

On the other hand,

is not a polynomial because one of its terms contains a variable with anexponent, , that is not a whole number.

The following are examples of monomials, binomials, and trinomials.Each of these examples is also a polynomial.

POLYNOMIALS

Monomials Binomials Trinomials None of These

4

Each term of a polynomial has a degree. The degree of a term in onevariable is the exponent on the variable.

Example 2 Identify the degree of each term of the trinomial.

Solution: The term has degree 4.The term has degree 1 since is .The term 3 has degree 0 since 3 is .

Each polynomial also has a degree.

Example 3 Find the degree of each polynomial and tell whether thepolynomial is a monomial, binomial, trinomial, or noneof these.

a. b. c. 7x + 3x3 + 2x2 - 115x - 10-2t2 + 3t + 6

DEGREE OF A POLYNOMIAL

The degree of a polynomial is the greatest degree of any term of thepolynomial.

3x0-7x1-7x- 7x

12x4

12x4 - 7x + 3

x6 + x4 - x3 + 1-q4 + q3 - 2q4x2 - 7-y5 + y4 - 3y3 - y2 + yx5 + 7x2 - x3p + 2-3z

5x3 - 6x2 + 3x - 6x2 + 4xy + y2x + yax2

TYPES OF POLYNOMIALS

A monomial is a polynomial with exactly one term.A binomial is a polynomial with exactly two terms.A trinomial is a polynomial with exactly three terms.

-5

x-5 + 2x - 3

1x0

x5 - 3x3 + 2x2 - 5x + 1

B


POLYNOMIAL

A polynomial in x is a finite sum of terms of the form , where a is areal number and n is a whole number.

axn

Practice Problem 2Identify the degree of each term of thetrinomial .-15x3 + 2x2 - 5

Answers2. 3; 2; 0, 3. a. binomial, 1,b. none of these, 6, c. trinomial, 2

Practice Problem 3Find the degree of each polynomialand tell whether the polynomial isa monomial, binomial, trinomial, ornone of these.

a.

b.

c. 10x2 - 6x - 6

9x - 3x6 + 5x4 + 2

-6x + 14


Practice Problem 4Evaluate each polynomial when

.

a.

b. 6x2 + 11x - 20

-2x + 10

x = -1

Practice Problem 5From Example 5, find the height of theobject when seconds and when

seconds.t = 7t = 3

Answers4. a. 12, b. , 5. 1677 feet; 1037 feet-25

Solution: ƒ¬¬¬¬¬¬¬¬¬Ta. The degree of the trinomial is 2, the greatest degree of

any of its terms.

ƒ¬¬¬¬¬¬Tb. The degree of the binomial or is 1.

ƒ¬¬¬¬¬¬¬¬¬Tc. The degree of the polynomial is 3.

EVALUATING POLYNOMIALS

Polynomials have different values depending on replacement values for thevariables. When we find the value of a polynomial for a given replacementvalue, we are evaluating the polynomial for that value.

Example 4 Evaluate each polynomial when .a.b.

Solution: a. Replace x with .

b. Replace x with

.

Many physical phenomena can be modeled by polynomials.

Example 5 Finding Free-Fall TimeThe CN Tower in Toronto, Ontario, is 1821 feet tall andis the world’s tallest self-supporting structure. An objectis dropped from the top of this building. Neglecting airresistance, the height of the object at time t seconds isgiven by the polynomial . Find the heightof the object when second and when sec-onds. (Source: World Almanac)

Solution: To find each height, we evaluate the polynomial whenand when .

Replace t with 1.

The height of the object at 1 second is 1805 feet.

Replace t with 10.

The height of the object at 10 seconds is 221 feet.

= 221

= -1600 + 1821

= -16 A100 B + 1821

-16t2 + 1821 = -16 A10 B 2 + 1821

= 1805

= -16 + 1821

= -16 A1 B + 1821

-16t2 + 1821 = -16 A1 B 2 + 1821

t = 10t = 1

t = 10t = 1-16t2 + 1821

= 17 = 12 + 4 + 1

-2 = 3 A4 B + 4 + 13x2-2x + 1 = 3 A-2 B 2 - 2 A-2 B + 1

= 16 = 10 + 6

-2-5x + 6 = -5 A-2 B + 6

3x2 - 2x + 1-5x + 6

x = -2

C

7x + 3x3 + 2x2 - 1

15x1 - 1015x - 10

-2t2 + 3t + 6

SIMPLIFYING POLYNOMIALS BY COMBINING LIKE TERMS

Polynomials with like terms can be simplified by combining like terms.Recall that like terms are terms that contain exactly the same variablesraised to exactly the same powers.

Like Terms Unlike Terms

, , , ,

, ,

Only like terms can be combined. We combine like terms by applying thedistributive property.

Examples Simplify each polynomial by combining any like terms.

6.7.

8. Write as .

9. Combine like terms and .

10.

Example 11 Write a polynomial that describes the total area of thesquares and rectangles shown on the next page. Thensimplify the polynomial.

= 12

x4 + 12

x3 - x2

= 510

x4 + 36

x3 - x2

= a 410

+ 110bx4 + a 4

6- 1

6bx3 - x2

= a 25

+ 110bx4 + a 2

3- 1

6bx3 - x2

25

x4 + 23

x3 - x2 + 110

x4 - 16

x3

-9x6x5x2 + 6x - 9x - 3 = 5x2 - 3x - 3 = 10x3

1x3x39x3 + x3 = 9x3 + 1x3 = 13x2 - 2

11x2 + 5 + 2x2 - 7 = 11x2 + 2x2 + 5 - 7-3x + 7x = A-3 + 7 Bx = 4x

4s2t6st2-a2b12

a2b

-5x-2x22yy3y3x-7x25x2

D

t = 1

t = 10

1821 ft

221 ft

1805 ft


Practice Problems 6–10Simplify each polynomial by combin-ing any like terms.

6.

7.

8.

9.

10.27

x3 - 14

x + 2 - 12

x3 + 38

x

23x2 - 6x - x - 15

7x3 + x3

14y2 + 3 - 10y2 - 9

-6y + 8y

Answers6. , 7. , 8. ,

9. , 10. - 314

x3 + 18

x + 223x2 - 7x - 15

8x34y2 - 62y

Solution:

Area:+ + + +

SIMPLIFYING POLYNOMIALS CONTAINING SEVERAL VARIABLES

A polynomial may contain more than one variable, such as

We call this expression a polynomial in several variables.The degree of a term with more than one variable is the sum of the expo-

nents on the variables. The degree of the polynomial in several variables isstill the greatest degree of the terms of the polynomial.

Example 12 Identify the degrees of the terms and the degree of thepolynomial .

Solution: To organize our work, we use a table.

Terms Degreeof Polynomial of Term Degree of Polynomial

1or 3or 4 4 (highest degree)or 3

11 0

To simplify a polynomial containing several variables, we combine anylike terms.

Examples Simplify each polynomial by combining any like terms.

13.

c

14.17a2 + 7b2= 10a2b -

9a2b - 6a2 + 5b2 + a2b - 11a2 + 2b2

= 10xy - 5y2 - 9x23xy - 5y2 + 7xy - 9x2 = A3 + 7 Bxy - 5y2 - 9x2

-2y2 + 1x2y2 + 2-6x2y21 + 23xy2

5x

5x + 3xy2 - 6x2y2 + x2y - 2y + 1

5x + 3xy2 - 6x2y2 + x 2y - 2y + 1

E

= 3x2 + 7x + 9

Recall that the area of a rectangle is lengthtimes width.Combine like terms.

= x2 + 3x + 9 + 4x + 2x2

x # 2x4 # x3 # 33 # xx # x


Practice Problem 11Write a polynomial that describes thetotal area of the squares and rectanglesshown below. Then simplify the poly-nomial.

x 45

5

x

x

x

x

8

x

HELPFUL HINT

This term can be written as or 10yx10xy

Practice Problem 12Identify the degrees of the termsand the degree of the polynomial

.-2x3y2 + 4 - 8xy + 3x3y + 5xy2

Practice Problems 13–14Simplify each polynomial by combin-ing any like terms.

13.

14.5x2x2 - y2 +

4y2x2 +2y2 -7x2y2 +

11ab - 6a2 - ba + 8b2

Answers11. , 12. 5, 0, 2, 4, 3; 513. , 14. 3x2y2 + y2 + 6x210ab - 6a2 + 8b2

2x2 + 13x + 20

2x

x

x

4

3

3

x

3

x

x


Focus On HistoryEXPONENTIAL NOTATION

The French mathematician and philosopher René Descartes(1596–1650) is generally credited with devising the system of expo-nents that we use in math today. His book La Géométrie was thefirst to show successive powers of an unknown quantity x as x, xx,x3, x4, x5, and so on. No one knows why Descartes preferred towrite xx instead of x2. However, the use of xx for the square of thequantity x continued to be popular. Those who used the notationdefended it by saying that xx takes up no more space when writtenthan x2 does.

Before Descartes popularized the use of exponents to indicatepowers, other less convenient methods were used. Some mathe-maticians preferred to write out the Latin words quadratus andcubus whenever they wanted to indicate that a quantity was to beraised to the second power or the third power. Other mathemati-cians used the abbreviations of quadratus and cubus, Q and C, toindicate second and third powers of a quantity.

3.

Term Coefficient

x-5

3.2x2-5x4

-5x4 + 3.2x2 + x - 5 4.

Term Coefficient

- 14

x2

x3-3x59.7x7

9.7x7 - 3x5 + x3 - 14

x2

2.

Term Coefficient

2

4-x

2x3 - x + 41.

Term Coefficient

5-3

x2

x2 - 3x + 5

353


EXERCISE SET 4.3

Complete each table for each polynomial. See Example 1.A

ANSWERS

1. see table

2. see table

3. see table

4. see table

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Evaluate each polynomial when (a) and (b) . See Examples 4 and 5.

5. 6. 7.

8. 9. 10.

Simplify each expression by combining like terms. See Examples 6 through 10.

11. 12. 13.

14. 15.

Simplify each polynomial by combining any like terms. See Examples 13 and 14.

16. 17.

18. 19.

20. 5x2y + 6xy2 - 5yx2 + 4 - 9y2x

3a2 - 9ab + 4b2 - 7ab4x2 - 6xy + 3y2 - xy

-9xy + 7y - xy - 6y3ab - 4a + 6ab - 7a

E

8s - 5s + 4s12k3 - 9k3 + 11

15x2 - 3x2 - y18x3 - 4x314x2 + 9x2

D

-2x3 + 3x2 - 6x3 - 15x2 - 4

x2 - 5x - 22x - 10x + 6

x = -1x = 0C

4.4 ADDING AND SUBTRACTING POLYNOMIALS

ADDING POLYNOMIALS

To add polynomials, we use commutative and associative properties andthen combine like terms. To see if you are ready to add polynomials,


Examples Add.

1.Remove parentheses.

Combine like terms.

Simplify.

2. and translates to

Remove parentheses.

Combine like terms.

Simplify.

Polynomials can be added vertically if we line up like terms underneathone another.

Example 3 Add and using a verticalformat.

Solution: Vertically line up like terms and add.

SUBTRACTING POLYNOMIALS

To subtract one polynomial from another, recall the definition of subtrac-tion. To subtract a number, we add its opposite: . Tosubtract a polynomial, we also add its opposite. Just as is the oppositeof b, is the opposite of .

Example 4 Subtract:

Solution: From the definition of subtraction, we have

Add the opposite.

= 3x + 8

Apply the distribu-tive property.Combine liketerms.

= A5x - 3 B + A-2x + 11 BA5x - 3 B - A2x - 11 B = A5x - 3 B + C- A2x - 11 B D

A5x - 3 B - A2x - 11 BAx2 + 5 B- Ax2 + 5 B -b

a - b = a + A-b BB

7y3 - 2y2 + 7 6y2 + 17y3 + 4y2 + 8

A6y2 + 1 BA7y3 - 2y2 + 7 B

= -4x2 + 6x + 2

= A-2x2 - 2x2 B + A5x + 1x B + A-1 + 3 B= -2x2 + 5x - 1 - 2x2 + x + 3

A-2x2 + 5x - 1 B + A-2x2 + x + 3 BA-2x2 + x + 3 BA-2x2 + 5x - 1 B

= 4x3 - x2 + 7

= 4x3 + A-6x2 + 5x2 B + A2x - 2x B + 7

= 4x3 - 6x2 + 2x + 7 + 5x2 - 2x

A4x3 - 6x2 + 2x + 7 B + A5x2 - 2x B

TO ADD POLYNOMIALS

To add polynomials, combine all like terms.

A

Adding and Subtracting Polynomials SECTION 4.4 355

Objectives

Add polynomials.

Subtract polynomials.

Add or subtract polynomials in onevariable.

Add or subtract polynomials inseveral variables.

D

CBA

Practice Problem 4Subtract: A9x + 5 B - A4x - 3 B

Answers1. , 2. ,3. , 4.

✓ Concept Check: b

5x + 89y2 - 2y + 8-x2 - x3x5 - 4x3 - 1

SSM CD-ROM Video4.4

✓ CONCEPT CHECK

When combining like terms in theexpression , which ofthe following is the proper result?

a. b.c. d. -11x4-11x

-3x - 8x2-11x2

5x - 8x2 - 8x

Practice Problems 1–2

Add. 1.

2. andA-6x2 + x - 1 BA5x2 - 2x + 1 BA3x3 - 2x BA3x5 - 7x3 + 2x - 1 B +

Practice Problem 3Add: and using a vertical format.

A4y + 3 BA9y2 - 6y + 5 B

TO SUBTRACT POLYNOMIALS

To subtract two polynomials,change the signs of the terms ofthe polynomial being subtractedand then add.

Example 5 Subtract:

Solution: First, we change the sign of each term of the secondpolynomial, then we add.

Combine like terms.

Just as polynomials can be added vertically, so can they be subtractedvertically.

Example 6 Subtract from usinga vertical format.

Solution: Arrange the polynomials in a vertical format, lining up liketerms.

ADDING AND SUBTRACTING POLYNOMIALS IN ONE VARIABLE

Let’s practice adding and subtracting polynomials in one variable.

Example 7 Subtract from the sum of and .

Solution: Notice that is to be subtracted from a sum. Thetranslation is

Remove grouping symbols.

Group like terms.

Combine like terms.

ADDING AND SUBTRACTING POLYNOMIALS IN SEVERAL VARIABLES

Now that we know how to add or subtract polynomials in one variable, wecan also add and subtract polynomials in several variables.

Examples Add or subtract as indicated.

8.

Combine like terms.

9.

= 9a2b2 + 4ab - 8ab2 + 3 + 9b2

= 9a2b2 + 6ab - 3ab2 - 5b2a - 2ab + 3 + 9b2Change the sign ofeach term of thepolynomial beingsubtracted.

Combine like terms.

A9a2b2 + 6ab - 3ab2 B - A5b2a + 2ab - 3 - 9b2 B= x2 + 2xy + 4y2= 3x2 - 6xy + 5y2 - 2x2 + 8xy - y2

A3x2 - 6xy + 5y2 B + A-2x2 + 8xy - y2 B

D

= 12z + 16= 8z + 9z - 5z + 11 - 2 + 7= 8z + 11 + 9z - 2 - 5z + 7

C A8z + 11 B + A9z - 2 B D - A5z - 7 BA5z - 7 B

A9z - 2 BA8z + 11 BA5z - 7 B

C

-8y2 - 4y + 17-5y2 - 2y + 6- A5y2 + 2y - 6 B

HELPFUL HINT

Don’t forget to change thesign of each term in thepolynomial being subtracted.

-3y2 - 2y + 11-3y2 - 2y + 11

A-3y2 - 2y + 11 BA5y2 + 2y - 6 B

= 9x2 - 6x - 1= 2x3 - 2x3 + 8x2 + x2 - 6x - 1

A2x3 + 8x2 - 6x B + A-2x3 + x2 - 1 BA2x3 + 8x2 - 6x B - A2x3 - x2 + 1 B =

A2x3 + 8x2 - 6x B - A2x3 - x2 + 1 B


Practice Problem 5Subtract: A-4x3 + x2 - 11 B

A4x3 - 10x2 + 1 B -

Practice Problem 6Subtract from

using a verticalformat.A2y2 - 2y + 7 BA6y2 - 3y + 2 B

Practice Problem 7Subtract from the sum of

and .A12x - 5 BA4x - 3 BA3x + 1 B

Answers5. , 6. ,7. , 8. ,9. 6x2y2 - 4 - 9x2y + 8xy2 - y2

5a2 - 2ab + 13b213x - 9-4y2 + y + 58x3 - 11x2 + 12

Practice Problems 8–9Add or subtract as indicated.

8.

9. A-x2y2 + 7 - 8xy2 + 2y2 BA5x2y2 + 3 - 9x2y + y2 B -

A-3a2 + ab - 7b2 BA2a2 - ab + 6b2 B -

357


EXERCISE SET 4.4

Add. See Examples 1 through 3.

1. 2.

3. 4.

5.

Subtract. See Examples 4 and 5.

6. 7.

8. 9.

10.

Add or subtract as indicated. See Example 7.

11. 12.

13. 14.

15.

Add or subtract as indicated. See Examples 8 and 9.

16. 17.

18.

19.

20. Ax2 + 2xy - y2 B + A5x2 - 4xy + 20y2 B

A7a2 - 3b2 + 10 B - A-2a2 + b2 - 12 B

A4x2 + y2 + 3 B - Ax2 + y2 - 2 B

A3x - 2 + 6y B + A7x - 2 - y BA9a + 6b - 5 B + A-11a - 7b + 6 BD

Ax2 + 2x + 1 B - A3x2 - 6x + 2 B

A14y + 12 B + A-3y - 5 BA7y + 7 B - Ay - 6 B

A9x - 1 B - A5x + 2 BA3x + 5 B + A2x - 14 BC

A2x2 + 3x - 9 B - A-4x + 7 B

4 - A-y - 4 B3x - A5x - 9 B

A5x2 + 4 B - A-2y2 + 4 BA2x + 5 B - A3x - 9 BB

A-5x2 + 3 B + A2x2 + 1 B

A3x - 8 B + A4x2 - 3x + 3 BA-7x + 5 B + A-3x2 + 7x + 5 B

A3x2 + 7 B + A3x2 + 9 BA3x + 7 B + A9x + 5 BA

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

4.5 MULTIPLYING POLYNOMIALS

MULTIPLYING MONOMIALS

Recall from Section 4.1 that to multiply two monomials such as and, we use the associative and commutative properties and regroup.

Remember also that to multiply exponential expressions with a commonbase, we add exponents.

Examples Multiply.

1.

2.

3.

MULTIPLYING MONOMIALS BY POLYNOMIALS

To multiply a monomial such as by a trinomial such as , weuse the distributive property.

Examples Multiply.

4.

5.

6.

MULTIPLYING TWO POLYNOMIALS

We also use the distributive property to multiply two binomials.

Example 7 Multiply:

Solution:

.

Multiply.Combine liketerms.

This idea can be expanded so that we can multiply any two polynomials.

= 6x2 - 11x - 10

= 6x2 - 15x + 4x - 10

= 3x A2x B + 3x A-5 B + 2 A2x B + 2 A-5 BUse the dis-tributiveproperty.

A3x + 2 B A2x - 5 B = 3x A2x - 5 B + 2 A2x - 5 B

A3x + 2 B A2x - 5 B

C

= -15x4 - 18x3 + 3x2


Multiply.

= A-3x2 B A5x2 B + A-3x2 B A6x B + A-3x2 B A-1 B-3x2 A5x2 + 6x - 1 B

= 10x4 + 30x


Multiply.

5x A2x3 + 6 B = 5x A2x3 + 5x A6 BB = 7x3 + 14x2 + 35x


Multiply.

7x Ax2 + 2x + 5 B = 7x Ax2 B + 7x A2x B + 7x A5 B

x3 + 2x + 57x

B

= 12x6

= A-12 B A-1 B Ax5 # x BA-12x5 B A-x B = A-12x5 B A-1x B = -14x7

-7x2 # 2x5 = A-7 # 2 B Ax2 # x5 B = 24x2

Use the commutative andassociative properties.

Multiply.

6x # 4x = A6 # 4 B Ax # x B

= 10x7

Use the commutative andassociative properties.

Multiply.

A-5x3 B A-2x4 B = A-5 B A-2 B Ax3 # x4 B

A-2x4 B A-5x3 BA

Multiplying Polynomials SECTION 4.5 359

Objectives

Multiply monomials.

Multiply a monomial by a polyno-mial.

Multiply two polynomials.

Multiply polynomials vertically.DC

BA


1.

2.

3.


4.

5.

6. -2x3 A3x2 - x + 2 B8x A7x4 + 1 B4x Ax2 + 4x + 3 B

A-5x4 B A-x B8x3 A-11x7 B10x # 9x

Answers1. , 2. , 3. ,4. , 5. ,6. , 7. 12x2 - x - 20-6x5 + 2x4 - 4x3

56x5 + 8x4x3 + 16x2 + 12x5x5-88x1090x2

SSM CD-ROM Video4.5

Practice Problem 7Multiply: A4x + 5 B A3x - 4 B

Examples Multiply.

8.

Multiply.

Combine like terms.

9.

Combine like terms.

MULTIPLYING POLYNOMIALS VERTICALLY

Another convenient method for multiplying polynomials is to multiply ver-tically, similar to how we multiply real numbers. This method is shown inthe next examples.

Example 10 Multiply vertically:

Solution:

Multiply by 5

Multiply by

Combine like terms.

Example 11 Find the product of and using a vertical format.

Solution: First, we arrange the polynomials in a vertical format.Then we multiply each term of the second polynomial byeach term of the first polynomial.

2x4 - 3x3 + 4x2

2x4 + 7x3 - 15x2 + 26x - 8

2x2 - 3x + 4x2 + 5x - 2

-4x2 + 6x - 810x3 - 15x2 + 20x

Ax2 + 5x - 2 BA2x2 - 3x + 4 B

2y2y2 - 3y + 42y4 - 6y3 + 8y2

2y4 - 6y3 + 13y2 - 15y + 20

y2 - 3y + 4

y2 - 3y + 4 2y2 + 5

5y2 - 15y + 20

A2y2 + 5 B Ay2 - 3y + 4 B

D

= 3t3 + 2t2 - 6t + 4

= 3t3 - 4t2 + 2t + 6t2 - 8t + 4

= t A3t2 B + t A-4t B + t A2 B + 2 A3t2 B + 2 A-4t B + 2 A2 BA t + 2 B A3t2 - 4t + 2 B = 4x2 - 4xy + y2

= 4x2 - 2xy - 2xy + y2

= 2x A2x B + 2x A-y B + A-y B A2x B + A-y B A-y B = A2x - y B A2x - y BA2x - y B 2

TO MULTIPLY TWO POLYNOMIALS

Multiply each term of the first polynomial by each term of the secondpolynomial, and then combine like terms.


Multiply by .

Multiply by .Multiply by .

Combine like terms.

x22x2 - 3x + 45x2x2 - 3x + 4

-22x2 - 3x + 4


8.

9. Ax + 3 B A2x2 - 5x + 4 BA3x - 2y B 2

Practice Problem 10Multiply vertically:A3y2 + 1 B Ay2 - 4y + 5 B

Answers8. ,9. ,10. ,11. 12x4 + 21x3 - 17x2 - 4x + 2

3y4 - 12y3 + 16y2 - 4y + 52x3 + x2 - 11x + 129x2 - 12xy + 4y2

Practice Problem 11Find the product of and using a verticalformat.

A3x2 + 6x - 2 B A4x2 - x - 1 B

361


MENTAL MATH

Find each product.

1. 2. 3.

4. 5.

EXERCISE SET 4.5


1. 2. 3.

4. 5.


6. 7.

8. 9.

10. 11.

12. 13.

14. 15. Ax2 - 4 B 2A7xy - y B 2

A3 + b B A2 - 5b - 3b2 BA2a - 3 B A5a2 - 6a + 4 B

Aa + 2 B Aa3 - 3a2 + 7 BAx + 5 B Ax3 - 3x + 4 B

Ax + 3 B Ax2 + 5x - 8 BAx - 2 B Ax2 - 3x + 7 B

Ax + 2 B Ax + 9 BAx + 4 B Ax + 3 BC

A-x3 B A-x BA-5.2x4 B A3x4 B

A-3.1x3 B A4x9 B6x # 3x28x2 # 3x

A

x7 # x7y9 # y

y4 # yx2 # x6x3 # x5

MENTAL MATH ANSWERS

1.

2.

3.

4.

5.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

4.6 SPECIAL PRODUCTS

USING THE FOIL METHOD

In this section, we multiply binomials using special products. First, we intro-duce a special order for multiplying binomials called the FOIL order ormethod. We demonstrate by multiplying by .

F O I L

Combine like terms.

Let’s practice multiplying binomials using the FOIL method.

Example 1 Multiply:

Solution:

Combine like terms.

Example 2 Multiply:

Solution:

Combine like terms. = 5x2 - 17x + 14

= 5x2 - 10x - 7x + 14

8 12 - t 2 = -5t

A5x - 7 B Ax - 2 B

= x2 + x - 12

= x2 + 4x - 3x - 12

Ax - 3 B Ax + 4 B = Ax B Ax B + Ax B A4 B + A-3 B Ax B + A-3 B A4 BF F O I L

I

L

O

Ax - 3 B Ax + 4 B

= 6x2 + 17x + 5

A3x + 1 B A2x + 5 B = 6x2 + 15x + 2x + 5

A2x + 5 BA3x + 1 B

A

Special Products SECTION 4.6 363

Objectives

Multiply two binomials using theFOIL method.

Square a binomial.

Multiply the sum and difference oftwo terms.

Use special products to multiplybinomials.

D

CB

A

THE FOIL METHOD

F stands for the product of the First terms.

F

O stands for the product of the Outer terms.

O

I stands for the product of the Inner terms.

I

L stands for the product of the Last terms.

LA1 B A5 B = 5A3x + 1 B A2x + 5 B

A1 B A2x B = 2x

A3x + 1 B A2x + 5 BA3x B A5 B = 15x

A3x + 1 B A2x + 5 BA3x B A2x B = 6x2

A3x + 1 B A2x + 5 B

Practice Problem 1Multiply: Ax + 7 B Ax - 5 B

Practice Problem 2Multiply: A6x - 1 B Ax - 4 B

SSM CD-ROM Video4.6

Answers1. , 2. 6x2 - 25x + 4x2 + 2x - 35


SQUARING A BINOMIAL

A binomial squared is equal to the square of the first term plus orminus twice the product of both terms plus the square of the secondterm.

Aa - b B 2 = a2 - 2ab + b2

Aa + b B 2 = a2 + 2ab + b2

Practice Problem 3Multiply: A2y2 + 3 B Ay - 4 B

Practice Problem 4Multiply: A2x + 9 B 2

Example 3 Multiply:

F O I L

Solution:

Notice in this example that there are no like terms thatcan be combined, so the product is .

SQUARING BINOMIALS

An expression such as is called the square of a binomial. Since, we can use the FOIL method to find this

product.

Example 4 Multiply:

Solution:F O I L

Notice the pattern that appears in Example 4.

is the first term of the binomialsquared. .

is 2 times the product of bothterms of the binomial.

.1 is the second term of the binomialsquared. .

This pattern leads to the following, which can be used when squaring abinomial. We call these special products.

This product can be visualized geometrically.

The area of the large square is side side.

The area of the large square is also the sum ofthe areas of the smaller rectangles.

Thus, .Aa + b B 2 = a2 + 2ab + b2

Area = a2 + ab + ab + b2 = a2 + 2ab + b2

Area = Aa + b B Aa + b B = Aa + b B 2#

a

a

b

b

b2

a2

ab

ab

a + b

a + b

A1 B 2 = 1

A2 B A3y B A1 B = 6y

6y

A3y B 2 = 9y29y2

A3y + 1 B 2 = 9y2 + 6y + 1

= 9y2 + 6y + 1

= 9y2 + 3y + 3y + 1

= A3y B A3y B + A3y B A1 B + 1 A3y B + 1 A1 BA3y + 1 B 2 = A3y + 1 B A3y + 1 B

A3y + 1 B 2A3y + 1 B 2 = A3y + 1 B A3y + 1 BA3y + 1 B 2B

2y3 - y2 + 12y - 6

Ay2 + 6 B A2y - 1 B = 2y3 - 1y2 + 12y - 6

Ay2 + 6 B A2y - 1 B

Answers3. , 4. 4x2 + 36x + 812y3 - 8y2 + 3y - 12

Examples Use a special product to square each binomial.

R T T b b5.6.7.8.

MULTIPLYING THE SUM AND DIFFERENCE OF TWO TERMS

Another special product is the product of the sum and difference of thesame two terms, such as . Finding this product by theFOIL method, we see a pattern emerge.

Notice that the middle two terms subtract out. This is because the Outerproduct is the opposite of the Inner product. Only the difference of squaresremains.

Examples Use a special product to multiply.

9.10. 72 = 36t2 - 49-A6t + 7 B A6t - 7 B = A6t B 2

42 = x2 - 16-Ax + 4 B Ax - 4 B = x2

firstterm

squared

minus secondterm

squared

= x2 - y2

Ax + y B Ax - y B = x2 - xy + xy - y2F F O I L

L

I

O

Ax + y B Ax - y BC

A7y B 2 = x4 - 14x2y + 49y2+2 1x2 B A7y B-Ax2 B 2 Ax2 - 7y B 2 = 52 = 4x2 + 20x + 25+2 A2x B A5 B+A2x B 2 A2x + 5 B 2 = q2 = p2 - 2pq + q2+2 Ap B Aq B-p2 Ap - q B 2 = 22 = t2 + 4t + 4+2 A t B A2 B+t2 A t + 2 B 2 =

secondtermsquared

plus

twice theproduct ofthe terms

plusorminus

firsttermsquared

Special Products SECTION 4.6 365

HELPFUL HINT

Notice that

The middle term is missing.

Likewise,

Aa - b B 2 = Aa - b B Aa - b B = a2 - 2ab + b2

Aa - b B 2 Z a2 - b2

Aa + b B 2 = Aa + b B Aa + b B = a2 + 2ab + b2

2abAa + b B 2 Z a2 + b2

Practice Problems 5–8Use a special product to square eachbinomial.

5. 6.

7. 8. Ax2 - 3y B 2A6x + 5 B 2Ar - s B 2Ay + 3 B 2

Answers5. , 6. ,7. , 8. ,

9 . , 1 0 . , 1 1 . ,

12. , 13. 4x4 - 36y29a2 - b2

x2 - 19

16y2 - 25x2 - 49

x4 - 6x2y + 9y236x2 + 60x + 25r2 - 2rs + s2y2 + 6y + 9

MULTIPLYING THE SUM AND DIFFERENCE OF TWO TERMS

The product of the sum and difference of two terms is the square ofthe first term minus the square of the second term.

Aa + b B Aa - b B = a2 - b2

Practice Problems 9–13Use a special product to multiply.

9.

10.

11.

12.

13. A2x2 - 6y B A2x2 + 6y BA3a - b B A3a + b Bax - 1

3b ax + 1

3b

A4y + 5 B A4y - 5 BAx + 7 B Ax - 7 B


Answers

14. , 15. ,16.

✓ Concept Check: a or e, b

4a2 - 110y2 - 19y - 1549x2 - 14x + 1

✓ CONCEPT CHECK

Match each expression on the left tothe equivalent expression or expres-sions in the list on the right.

a.b.

c.d.e. a2 + 2ab + b2

a2 - 2ab + b2a2 + b2

a2 - b2Aa + b B Aa - b B Aa + b B Aa + b BAa + b B 2

11.

12.

13.


USING SPECIAL PRODUCTS

Let’s now practice using our special products on a variety of multiplica-tions. This practice will help us recognize when to apply what special prod-uct.

Examples Use a special product to multiply.

14.

15. This is a binomial squared.

16.

F O I L

= 6a2 - 41a - 7

= 6a2 - 42a + a - 7

= 6a # a + 6a A-7 B + 1 # a + 1 A-7 BNo special product applies.

Use the FOIL method.

A6a + 1 B Aa - 7 B = 9y2 + 12y + 4

= A3y B 2 + 2 A3y B A2 B + 22

A3y + 2 B 2= x2 - 81= x2 - 92

This is the sum and difference

of the same two terms.Ax - 9 B Ax + 9 B

D

A5y B 2 = 9x4 - 25y2-A3x2 - 5y B A3x2 + 5y B = A3x2 B 2-q2 = 4p2 - q2A2p - q B A2p + q B = A2p B 2

- a 14b 2

= x2 - 116

ax - 14b ax + 1

4b = x2

Practice Problems 14–16Use a special product to multiply.

14.

15.

16. A2a - 1 B A2a + 1 BA5y + 3 B A2y - 5 BA7x - 1 B 2

367


EXERCISE SET 4.6

Multiply using the FOIL method. See Examples 1 through 3.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10.


11. 12. 13.

14. 15. 16.

17. 18. 19.

20. A3x2 + 1 B A3x2 - 1 B

A2y2 - 1 B A2y2 + 1 BAa2 + 6 B Aa2 - 6 BAx2 + 5 B Ax2 - 5 B

A4x - 5 B A4x + 5 BA3x - 1 B A3x + 1 BAx - 8 B Ax + 8 B

Ax + 6 B Ax - 6 BAb + 3 B Ab - 3 BAa - 7 B Aa + 7 BC

A6x + 2 B Ax - 2 B

A2x + 5 B A3x - 1 BA2x - 9 B Ax - 11 BAy - 6 B A4y - 1 B

A3y - 5 B A2y - 7 BA5x - 6 B Ax + 2 BAy - 12 B Ay + 4 B

Ax - 5 B Ax + 10 BAx + 5 B Ax + 1 BAx + 3 B Ax + 4 BA

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Focus On The Real WorldSPACE EXPLORATION

From scientific observations on Earth, we know that Saturn, the second largest planet in our solarsystem, is a giant ball of gas surrounded by rings and orbited by 19 moons. We also know that Sat-urn has a diameter of 120,000 kilometers and a mass of 569,000,000,000,000,000,000,000,000 kilo-grams. But what is Saturn like below its outer layer of clouds? What are Saturn’s rings made of?What is the surface of Saturn’s largest moon, Titan, like? Could life ever be supported on Titan?

NASA is hoping to answer these questions and more about the sixth planet from the sun inour solar system with its $3,400,000,000 Cassini mission. The Cassini spacecraft is scheduled toarrive in orbit around Saturn in June 2004. The goal of this mission is to study Saturn, its rings,and its moons. The Cassini spacecraft will also launch the Huygens probe to study Titan.

The Cassini mission began on October 15, 1997, with the launch of a mighty Titan IV boosterrocket. The entire launch vehicle including rocket fuel weighed more than 2,000,000 poundsbefore launch. The Titan IV flung Cassini into space at a speed of 14,400 kilometers per hour. Totake advantage of something called gravity assist, Cassini is taking a roundabout path to Saturnpast Venus (twice), Earth, and Jupiter. Altogether, Cassini will travel 3,540,000,000 kilometersbefore reaching Saturn, a planet that is only 1,430,000,000 kilometers from the sun.

Once Cassini reaches Saturn, it will begin collecting all kinds of data about the planet and itsmoons. Over the course of Cassini’s mission, it will collect over 2,000,000,000,000 bits of scientificinformation, or about the same amount of data in 800 sets of the Encyclopedia Britannica. Aboutonce per day, Cassini will use its 4-meter antenna to transmit the latest data that it has collectedback to Earth at a frequency of 8,400,000,000 cycles per second. For comparison, the FM bandon a radio is centered around 100,000,000 cycles per second. It will take from 70 to 90 minutesfor Cassini’s transmissions to reach Earth and by then the signals are very weak. The power ofthe signal transmitted by the spacecraft is 20 watts but, even with the huge antennas used onEarth, only 0.0000000000000001 watt can be received. (Source: based on data from NationalAeronautics and Space Administration)

CRITICAL THINKING

1. Make a list of the numbers (other than those in dates) used in the article. Rewrite each num-ber in scientific notation.

2. What are the advantages of scientific notation?3. What are the disadvantages of scientific notation?4. In your opinion, how large or small should a number be to make using scientific notation

worthwhile?

368

4.7 DIVIDING POLYNOMIALS

DIVIDING BY A MONOMIAL

To divide a polynomial by a monomial, recall addition of fractions. Frac-tions that have a common denominator are added by adding the numera-tors:

If we read this equation from right to left and let a, b, and c be monomials,, we have the following:

Throughout this section, we assume that denominators are not 0.

Example 1 Divide: by 2m

Solution: We begin by writing the quotient in fraction form. Thenwe divide each term of the polynomial by themonomial 2m.

Simplify.

Check: To check, we multiply.

The quotient checks.


Example 2 Divide:

Solution:

Divide each term by .

Simplify.

Notice that the quotient is not a polynomial because of the

term . This expression is called a rational expression—

we will study rational expressions further in Chapter 5.Although the quotient of two polynomials is not always apolynomial, we may still check by multiplying.

Check:

= 9x5 - 12x2 + 3x

3x2 a3x3 - 4 + 1xb = 3x2 A3x3 B - 3x2 A4 B + 3x2 a 1

xb

1x

= 3x3 - 4 + 1x

3x29x5 - 12x2 + 3x3x2 = 9x5

3x2 - 12x2

3x2 + 3x3x2

9x5 - 12x2 + 3x3x2

3m + 1

2m A3m + 1 B = 2m A3m B + 2m A1 B = 6m2 + 2m

= 3m + 1

6m2 + 2m2m

= 6m2

2m+ 2m

2m

6m2 + 2m

6m2 + 2m

c Z 0

ac

+ bc

= a + bc

A

Dividing Polynomials SECTION 4.7 369

Objectives

Divide a polynomial by a mono-mial.

Use long division to divide a poly-nomial by a polynomial other thana monomial.

Use synthetic division.C

B

A

TO DIVIDE A POLYNOMIAL BY A MONOMIAL

Divide each term of the polynomial by the monomial.

, c Z 0a + b

c= a

c+ b

c

Practice Problem 1Divide: by 5x225x3 + 5x2

SSM CD-ROM Video4.7

✓ CONCEPT CHECK

In which of the following is

simplified correctly?

a. b. c. x + 1xx5

+ 1

x + 55

Answers1. , 2.

✓ Concept Check: a

6x5 + 2 - 1x

5x + 1

Practice Problem 2

Divide: 30x7 + 10x2 - 5x

5x2

Example 3 Divide:

Solution:

Divide each term by 4xy.

Simplify.

Check:

DIVIDING BY A POLYNOMIAL OTHER THAN A MONOMIAL

To divide a polynomial by a polynomial other than a monomial, we use aprocess known as long division. Polynomial long division is similar to num-ber long division, so we review long division by dividing 13 into 3660.

281

13��3660��26T

106 Subtract and bring down the next digit in the dividend.

10420 Subtract and bring down the next digit in the dividend.

13

7 Subtract. There are no more digits to bring down, so the remainder is 7.

The quotient is 281 R 7, which can be written as .

Recall that division can be checked by multiplication. To check a divisionproblem such as this one, we see that

Now we demonstrate long division of polynomials.

Example 4 Divide by using long division.

Solution:

How many times does x divide ?

.Multiply: .Subtract and bring down the nextterm.

Now we repeat this process. How many times does x divide

4x? .

Multiply: .

Subtract. The remainder is 0.

The quotient is .x + 4

4 Ax + 3 B

x + 4x + 3 �x2 + 7x + 12

x2 + 3x

4x + 124x + 12

0

- -

To subtract, changethe signs of theseterms and add.

4xx

= 4

x Ax + 3 Bx + 3 �x2 + 7x + 12

x2 + 3x4x + 12

- - ¡

xTo subtract, changethe signs of theseterms and add.

x2

x= x

x2

x + 3x2 + 7x + 12

13 # 281 + 7 = 3660

d remainderd divisor

281 713

1 # 13 = 13

8 # 13 = 104

2 # 13 = 26

B

= 8x2y2 - 16xy + 2x

4xy a2xy - 4 + 12yb = 4xy A2xy B - 4xy A4 B + 4xy a 1

2yb

= 2xy - 4 + 12y

8x2y2 - 16xy + 2x4xy

= 8x2y2

4xy- 16xy

4xy+ 2x

4xy

8x2y2 - 16xy + 2x4xy


Practice Problem 3

Divide: 12x3y3 - 18xy + 6y

3xy

Practice Problem 4Divide: by x + 5x2 + 12x + 35

Answers

3. , 4. x + 74x2y2 - 6 + 2x


.Subtract and bring down the next term.

Subtract. The remainder is .-1

12x3x

= 4, 4 A3x - 1 B2x A3x - 1 B

Practice Problem 5Divide: by 2x - 16x2 + 7x - 5

Practice Problem 6

Divide: 5 - x + 9x3

3x + 2

Check: We check by multiplying.

or T T T T


Example 5 Divide by using long division.

Solution:

, so is a term of the quotient.

Thus divided by is with a remainder of . This can be written as

Check: To check, we multiply . Then we addthe remainder, , to this product.


Notice that the division process is continued until the degree of theremainder polynomial is less than the degree of the divisor polynomial.

Example 6 Divide:

Solution: Before we begin the division process, we rewriteas . Notice that we

have written the polynomial in descending order andhave represented the missing x term by 0x.

Thus, .4x2 + 7 + 8x3

2x + 3= 4x2 - 4x + 6 + -11

2x + 3

Remainder.

4x2 - 4x + 62x + 3�8x3 + 4x2 + 0x + 7

8x3 + 12x2

- 8x2 + 0x- 8x2 - 12x

12x + 712x + 18

- 11

+ +

- -

--

8x3 + 4x2 + 0x + 74x2 + 7 + 8x3

4x2 + 7 + 8x3

2x + 3

= 6x2 + 10x - 5

A3x - 1 B A2x + 4 B + A-1 B = A6x2 + 12x - 2x - 4 B - 1

-1A3x - 1 B A2x + 4 B

d remainderd divisor

6x2 + 10x - 53x - 1

= 2x + 4 + -13x - 1

-1A2x + 4 BA3x - 1 BA6x2 + 10x - 5 B

2x6x2

3x= 2x

2x + 43x - 1�6x2 + 10x - 5

6x2 - 2x 12x - 512x - 4

- 1

+-

+-

¡

3x - 16x2 + 10x - 5

= x2 + 7x + 120Ax + 4 B +Ax + 3 B #

dividend=remainder+quotient#divisor

Answers

5. , 6. 3x2 - 2x + 1 + 33x + 2

3x + 5

USING SYNTHETIC DIVISION

When a polynomial is to be divided by a binomial of the form , ashortcut process called synthetic division may be used. On the left is anexample of long division, and on the right is the same example showing thecoefficients of the variables only.

Notice that as long as we keep coefficients of powers of x in the same col-umn, we can perform division of polynomials by performing algebraic oper-ations on the coefficients only. This shortest process of dividing with coef-ficients only in a special format is called synthetic division. To find

by synthetic division, follow the nextexample.

Example 7 Use synthetic division to divide by.

Solution: To use synthetic division, the divisor must be in the form. Since we are dividing by , c is 3. We write

down 3 and the coefficients of the dividend.

2 -1 -13 1

2

2 -1 -13 1

6

2

2 -1 -13 1

6

2 5

2 -1 -13 1

6 15

2 5 2

2 -1 -13 1

6 15 6

2 5 2 7

3

3

3

3

3

c

Next, draw a line and bring downthe first coefficient of the dividend.

Multiply 3 . 2 and write down theproduct, 6.

Add -1 + 6. Write down the sum, 5.

3 . 5 = 15.-13 + 15 = 2.

3 . 2 = 6.1 + 6 = 7.

x - 3x - c

x - 32x3 - x2 - 13x + 1

A2x3 - x2 - 13x + 1 B, Ax - 3 B

2 5 21 - 3 2 2 - 1 - 13 + 1

2 - 6 5 - 13 5 - 15

2 + 12 - 6

7

2x2 + 5x + 2x - 3 2 2x3 - x2 - 13x + 1

2x3 - 6x2 5x2 - 13x 5x2 - 15x

2x + 12x - 6

7

x - c

C


Practice Problem 7Use synthetic division to divide

by .x - 23x3 - 2x2 + 5x + 4

Answers7. 3x2 + 4x + 13 + 30

x - 2

The quotient is found in the bottom row. The numbers 2,5, and 2 are the coefficients of the quotient polynomial,and the number 7 is the remainder. The degree of thequotient polynomial is one less than the degree of thedividend. In our example, the degree of the dividend is 3,so the degree of the quotient polynomial is 2. As wefound when we performed the long division, the quotient is

, remainder 7

or

When using synthetic division, if there are missing powers of the vari-able, insert 0s as coefficients.

Example 8 Use synthetic division to divide by .

Solution: The divisior is , which in the form is. Thus, c is . There is no x-term in the divi-

dend, so we insert a coefficient of 0. The dividend coeffi-cients are 1, , , 0, and 34.

The dividend is a fourth-degree polynomial, so the quo-tient polynomial is a third-degree polynomial. The quo-tient is with a remainder of 22. Thus,


HELPFUL HINT

Before dividing by synthetic division, write the dividend in descendingorder of variable exponents. Any “missing powers” of the variablemust be represented by 0 times the variable raised to the missingpower.

x4 - 2x3 - 11x2 + 34x + 2

= x3 - 4x2 - 3x + 6 + 22x + 2

x3 - 4x2 - 3x + 6

1 -2 -11 0 34

-2 8 6 -12

1 -4 -3 6 22

-2

c

-11-2

-2x - A-2 B x - cx + 2

x + 2x4 - 2x3 - 11x2 + 34

2x2 + 5x + 2 + 7x - 3

2x2 + 5x + 2


✓ CONCEPT CHECK

Which division problems are candi-dates for the synthetic divisionprocess?a.

b.

c.

d. x5, Ax - 5 BAy4 + y - 3 B, Ax2 + 1 BAx3 - x2 + 2 B, A3x3 - 2 BA3x2 + 5 B, Ax + 4 B

Answers

8.

✓ Concept Check: a and d

x3 + 2x2 - 2x - 3 + 7x + 1

Practice Problem 8Use synthetic division to divide

by .x + 1x4 + 3x3 - 5x + 4


Focus On Business and CareerBUSINESS TERMS

For most businesses, a financial goal is to “make money.” But what does that mean from a mathematicalpoint of view? To find out, we must first discuss some common business terms.

j Revenue is the amount of money a business takes in. A company’s annual revenue is the amount ofmoney it collects during its fiscal, or business, year. For most companies, the largest source of revenue isfrom the sales of their products or services. For instance, a computer manufacturer’s annual revenue isthe amount of money it collects during the year from selling computers to customers. Large companiesmay also have revenues from investment interest or leases. When revenue can be expressed as a functionof another variable, it is often denoted as .

j Expenses are the costs of doing business. For instance, a large part of a computer manufacturer’sexpenses include the cost of the computer components it buys from wholesalers to use in the manufac-turing or assembling process. Other expenses include salaries, mortgage payments, equipment, taxes,advertising, and so on. Some businesses refer to their expenses simply as cost. When cost can beexpressed as a function of another variable, it is often denoted as .

j Net income/loss is the difference between a company’s annual revenues and expenses. This differencemay also be referred to as net earnings. Positive net earnings—that is, a positive difference—result in anet income or net profit. Posting a net income can be interpreted as “making money.” Negative net earn-ings—that is, a negative difference—result in a net loss. Posting a net loss can be interpreted as “losingmoney.” A profit function can be expressed as . In this case, a negative profit isinterpreted as a net loss.

GROUP ACTIVITY

Locate several corporate annual reports. Using the data in the reports, verify that the net income or netearnings given in a report was calculated as the difference between revenue and expenses. If this was notthe case, can you tell what caused the variation? If so, explain.

P Ax B = R Ax B - C Ax B

C Ax B

R Ax B

375


MENTAL MATH

Simplify each expression.

1. 2. 3.

4. 5. 6.

EXERCISE SET 4.7

Perform each division. See Examples 1 through 3.

1. 2. 3.

4. 5. 6.

7. 8. m3n2 - mn4

mna2b2 - ab3

ab

6x5 + 3x4

3x414m2 - 27m3

7m15p3 + 18p2

3p

15x2 - 9x5

x12x4 + 3x2

x8x3 - 4x2 + 6x + 2

2

A

k7

k5k5

k2p8

p3

a3

ay2

ya6

a4

MENTAL MATH ANSWERS

1.

2.

3.

4.

5.

6.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.



SECTION 4.1 EXPONENTS

means the product of n factors, each of which is a.an

a 12b 4

= 12

# 12

# 12

# 12

= 116

A-5 B 3 = A-5 B A-5 B A-5 B = -12532 = 3 # 3 = 9

If m and n are integers and no denominators are 0,Product Rule:Power Rule:Power of a Product Rule:

Power of a Quotient Rule:

Quotient Rule:

Zero Exponent: a0 = 1, a Z 0

am

an = am-n

a abb n

= an

bn

Aab Bn = anbnAam Bn = amn

am # an = am+n

, , x Z 0x0 = 150 = 1

x9

x4 = x9-4 = x5

a x8b 3

= x3

83

A7y B 4 = 74y4A53 B 8 = 53 #8 = 524x2 # x7 = x2+7 = x9

SECTION 4.2 NEGATIVE EXPONENTS AND SCIENTIFIC NOTATION

If and n is an integer,

a-n = 1an

a Z 0

Simplify:

= x14

y2

= x4-1-10By-2

a x-2yx5 b -2

= x4y-2

x-10

3-2 = 132 = 1

9; 5x-2 = 5

x2

A positive number is written in scientific notation if it is written as the product of a number a,

, and an integer power r of 10.

a * 10r

1 ≤ a 6 10 0.000 0568 = 5.68 * 10-60

12000 = 1.2 * 104

SECTION 4.3 INTRODUCTION TO POLYNOMIALS

A term is a number or the product of a number andvariables raised to powers.

-5x, 7a2b, 14

y4, 0.2

The numerical coefficient or coefficient of a term isits numerical factor.

TERM COEFFICIENT

7y 1

-1-a2b

7x2

A polynomial is a finite sum of terms of the form where a is a real number and n is a whole number.

axn(Polynomial)5x3 - 6x2 + 3x - 6

A monomial is a polynomial with exactly 1 term.(Monomial)

56

y3

A binomial is a polynomial with exactly 2 terms. (Binomial)-0.2a2b - 5b2


SECTION 4.3 CONTINUED

A trinomial is a polynomial with exactly 3 terms. (Trinomial)3x2 - 2x + 1

The degree of a polynomial is the greatest degree ofany term of the polynomial.

POLYNOMIAL DEGREE

22 + 3 = 57y + 8y2z3 - 12

5x2 - 3x + 2

SECTION 4.4 ADDING AND SUBTRACTING POLYNOMIALS

To add polynomials, combine like terms. Add.

= 7x2 - 8x - 4 = 7x2 - 3x + 2 - 5x - 6

A7x2 - 3x + 2 B + A-5x - 6 B

To subtract two polynomials, change the signs of theterms of the second polynomial, then add.

Subtract.

= 3y3 + 17y2 - 7y + 7 = 17y2 - 2y + 1 + 3y3 - 5y + 6

= A17y2 - 2y + 1 B + A3y3 - 5y + 6 BA17y2 - 2y + 1 B - A-3y3 + 5y - 6 B

SECTION 4.5 MULTIPLYING POLYNOMIALS

To multiply two polynomials, multiply each term ofone polynomial by each term of the other polyno-mial, and then combine like terms.

Multiply.

= 10x3 - 7x2 - 2x + 2= 10x3 - 12x2 + 4x + 5x2 - 6x + 2= 2x A5x2 - 6x + 2 B + 1 A5x2 - 6x + 2 B

A2x + 1 B A5x2 - 6x + 2 B

SECTION 4.6 SPECIAL PRODUCTS

The FOIL method may be used when multiplyingtwo binomials.

Multiply:

F O I L

= 10x2 + 9x - 9= 10x2 + 15x - 6x - 9

= A5x B A2x B + A5x B A3 B + A-3 B A2x B + A-3 B A3 B

A5x - 3 B A2x + 3 B

FirstLast

InnerOuter

A5x - 3 B A2x + 3 B

Squaring a BinomialAa + b B 2 = a2 + 2ab + b2

Square each binomial.

= x2 + 10x + 25Ax + 5 B 2 = x2 + 2 Ax B A5 B + 52

Aa - b B 2 = a2 - 2ab + b2

= 9x2 - 12xy + 4y2A3x - 2y B 2 = A3x B 2 - 2 A3x B A2y B + A2y B 2

Multiplying the Sum and Difference of Two TermsAa + b B Aa - b B = a2 - b2

Multiply.

= 36y2 - 25A6y + 5 B A6y - 5 B = A6y B 2 - 52


To divide a polynomial by a monomial:a + b

c= a

c+ b

c, c Z 0

Divide.

= 3x3 - 2x + 1 - 25x

= 15x5

5x2 - 10x3

5x2 + 5x2

5x2 - 2x5x2

15x5 - 10x3 + 5x2 - 2x5x2

To divide a polynomial by a polynomial other than amonomial, use long division.

or

5x - 1 - 42x + 3

2x + 3 � 10x2 + 13x - 710x2 + 15x

-2x - 7-2x - 3

- 4

5x - 1 + -42x + 3

SECTION 4.7 DIVIDING POLYNOMIALS

A shortcut method called synthetic division may beused to divide a polynomial by a binomial of theform .x - c

Use synthetic division to divide by .

The quotient is .2x2 + 3x - 2 - 5x - 2

2 -1 -8 -1

4 6 -4

2 3 -2 -5

2

x - 22x3 - x2 - 8x - 1

379

5In Chapter 4, we learned how to multiply polynomials. Now we willdeal with an operation that is the reverse process of multiplying—fac-toring. Factoring is an important algebraic skill because it allows us towrite a sum as a product. As we will see in Sections 5.6 and 5.7, factor-ing can be used to solve equations other than linear equations. In Chap-ter 6, we will also use factoring to simplify and perform arithmetic oper-ations on rational expressions.

5.1 The Greatest Common Factor

5.2 Factoring Trinomials of theForm

5.3 Factoring Trinomials of theForm

5.4 Factoring Trinomials of theForm byGrouping

5.5 Factoring Perfect SquareTrinomials and the Differenceof Two Squares

5.6 Solving Quadratic Equationsby Factoring

5.7 Quadratic Equations andProblem Solving

ax2 + bx + c

ax2 + bx + c

x2 + bx + c

The America’s Cup sailing competition is held roughly everythree years. This international yacht race dates from 1851. Thesailboats were originally built from wood and later from alu-minum. Today’s International America’s Cup Class sailboats aremade of carbon fiber. This strong but lightweight materialreduced the boats’ weight by 50% and made them much faster.

Factoring Polynomials C H A P T E R

5.1 THE GREATEST COMMON FACTOR

In the product , 2 and 3 are called factors of 6 and is a fac-tored form of 6. This is true of polynomials also. Since

, then and are factors of , andis a factored form of the polynomial.

Study the examples below and look for a pattern.


Multiplying:

Factoring:

Do you see that factoring is the reverse process of multiplying?

FINDING THE GREATEST COMMON FACTOR

The first step in factoring a polynomial is to see whether the terms of thepolynomial have a common factor. If there is one, we can write the poly-nomial as a product by factoring out the common factor. We will usuallyfactor out the greatest common factor (GCF).

The greatest common factor (GCF) of a list of terms is the product of theGCF of the numerical coefficients and the GCF of the variable factors.

Notice that the GCF of a list of variables raised to powers is the variablesraised to the smallest exponent in the list. For example,

the GCF of , , and is ,

because is the largest common factor.

Example 1 Find the greatest common factor of each list of terms.

a. , , and b. , and c. , , and

Solution: a.The GCF of , , ¶Sand x is x.

orb.

The GCF of , , ¶Sand is .

or 9y2GCF = 3 # 3 # y2

y2y4 27y4 = 3 # 3 # 3 # y4

y3y2 -63y3 = -1 # 3 # 3 # 7 # y3 -18y2 = -1 # 2 # 3 # 3 # y2

2xGCF = 2 # x1 -8x = -1 # 2 # 2 # 2 # x1

x3x2

10x3 = 2 # 5 # x3 6x2 = 2 # 3 # x2

a6b2a5ba3b227y4-18y2,-63y3

-8x10x36x2

x2

x2x3x5x2

GCF = 2 # x # y # y = 2xy2

6xy3 = 2 # 3 # x # y # y # y

20x2y2 = 2 # 2 # 5 # x # x # y # y

A

x2 + 5x + 6 = Ax + 2 B Ax + 3 Bfactoring

multiplying

2x2 - 14x = 2x Ax - 7 B5x2 + 15 = 5 Ax2 + 3 B

2x Ax - 7 B = 2x2 - 14x5 Ax2 + 3 B = 5x2 + 15

The process of writing a polynomial as a product is called factoring thepolynomial.

Ax + 2 B Ax + 3 B x2 + 5x + 6Ax + 3 BAx + 2 Bx2 + 5x + 6Ax + 2 B Ax + 3 B =

2 # 32 # 3 = 6

The Greatest Common Factor SECTION 5.1 381

Objectives

Find the greatest common factor ofa list of terms.

Factor out the greatest commonfactor from the terms of a polyno-mial.

Factor by grouping.C

B

A

✓ CONCEPT CHECK

Multiply: What do you think the result of fac-toring would be? Why?2x - 8

2 Ax - 4 B

Answers1. a. , b. , c.

✓ Concept Check; The result would be because

factoring is the reverse process of multiplying.2 Ax - 4 B2x - 8

ab24y3x2

Practice Problem 1Find the greatest common factor ofeach list of terms.

a. , , and

b. , , and

c. , , and a3b2ab3a5b4

40y4-20y6-16y

-12x59x46x2

SSM CD-ROM Video5.1

382 CHAPTER 5 Factoring Polynomials

c. The GCF of , , and is .The GCF of , b, and is . Thus,the GCF of , , and is .

FACTORING OUT THE GREATEST COMMON FACTOR

To factor a polynomial such as , we first see whether the termshave a greatest common factor other than 1. In this case, they do: The GCFof and 14 is 2.

We factor out 2 from each term by writing each term as a product of 2and the term’s remaining factors.

Using the distributive property, we can write

Thus, a factored form of is . We can check by multiply-

ing: .


Example 2 Factor each polynomial by factoring out the greatestcommon factor (GCF).

a. b.

Solution: a. The GCF of terms and 18 is 6. Thus,


We can check our work by multiplying 6 and .

, the originalpolynomial.

b. The GCF of and is . Thus,

ƒHELPFUL HINT Don’t forget the 1.

= y5 A1 - y2 By5 - y7 = Ay5 B1 - Ay5 By2

y5y7y5

6 A t + 3 B = 6 # t + 6 # 3 = 6t + 18

A t + 3 B = 6 A t + 3 B

6t + 18 = 6 # t + 6 # 3

6t

y5 - y76t + 18

2 A4x + 7 B = 2 # 4x + 2 # 7 = 8x + 14

2 A4x + 7 B8x + 14

= 2 A4x + 7 B8x + 14 = 2 # 4x + 2 # 7

8x + 14 = 2 # 4x + 2 # 7

8x

8x + 14

B

a3ba6b2a5ba3b2bb2b2a3a6a5a3

HELPFUL HINT

A factored form of is not

Although the terms have been factored (written as a product), thepolynomial has not been factored. A factored form of

is the product .2 A4x + 7 B8x + 148x + 14

2 # 4x + 2 # 7

8x + 14

✓ CONCEPT CHECK

Which of the following is/are factoredform(s) of ?a. 6

b.

c.

d. 3 A t + 6 B6 A t + 3 B6 # t + 6 # 3

6t + 18

Practice Problem 2Factor each polynomial by factoringout the greatest common factor(GCF).

a.

b. x4 - x9

10y + 25

Answers

2. a. , b.

✓ Concept Check: c

x4 A1 - x5 B5 A2y + 5 B

The Greatest Common Factor SECTION 5.1 383

Practice Problem 3Factor: -10x3 + 8x2 - 2x

Practice Problems 4–6Factor.

4.

5.

6. 6a3b + 3a3b2 + 9a2b4

25

a5 - 45

a3 + 15

a2

4x3 + 12x

Practice Problem 7Factor: 7 Ap + 2 B + q Ap + 2 B

Practice Problem 8Factor by group-ing.

ab + 7a + 2b + 14

Answers3. , 4. ,

5. ,

6. , 7. ,

8. Ab + 7 B Aa + 2 BAp + 2 B A7 + q B3a2b A2a + ab + 3b3 B

15

a2 A2a3 - 4a + 1 B4x Ax2 + 3 B-2x A5x2 - 4x + 1 B

Example 3 Factor:

Solution:

ƒ

In Example 3, we could have chosen to factor out instead of . Ifwe factor out , we have

ƒ ƒ ƒ

Examples Factor.

4.

5.

6.

Example 7 Factor:

Solution: The binomial is present in both terms and is thegreatest common factor. We use the distributive propertyto factor out .

FACTORING BY GROUPING

Once the GCF is factored out, we can often continue to factor the poly-nomial, using a variety of techniques. We discuss here a technique calledfactoring by grouping. This technique can be used to factor some polyno-mials with four terms.

Example 8 Factor by grouping.

Solution: The GCF of the first two terms is x, and the GCF of thelast two terms is 3.

Next we factor out the common binomial factor, .

x Ay + 2 B + 3 Ay + 2 B = Ay + 2 B Ax + 3 BAy + 2 B

HELPFUL HINT

Notice that this is not a factored form of the origi-nal polynomial. It is a sum, not a product.

f = x Ay + 2 B + 3 Ay + 2 Bxy + 2x + 3y + 6 = Axy + 2x B + A3y + 6 B

xy + 2x + 3y + 6

C

5 Ax + 3 B + y Ax + 3 B = Ax + 3 B A5 + y BAx + 3 BAx + 3 B

5 Ax + 3 B + y Ax + 3 B

15p2q4 + 20p3q5 + 5p3q3 = 5p2q3 A3q + 4pq2 + p B37

x4 + 17

x3 - 57

x2 = 17

x2 A3x2 + x - 5 B6a4 - 12a = 6a Aa3 - 2 B

HELPFUL HINT Notice the changes in signs when factoring out -3a.

= -3a A3a4 - 6a + 1 B-9a5 + 18a2 - 3a = A-3a B A3a4 B + A-3a B A-6a B + A-3a B A1 B

-3a3a-3a

HELPFUL HINT Don’t forget the -1.

= 3a A-3a4 + 6a - 1 B-9a5 + 18a2 - 3a = A3a B A-3a4 B + A3a B A6a B + A3a B A-1 B

-9a5 + 18a2 - 3a


Practice Problems 9–10Factor by grouping.

9.

10. 2xy + 5y2 - 4x - 10y

28x3 - 7x2 + 12x - 3

Practice Problems 11–13Factor by grouping.

11.

12.

13. 3xy - 4 + x - 12y

2x - 2 + x3 - 3x2

4x3 + x - 20x2 - 5

Check: Multiply by .

,

the original polynomial.

Thus, the factored form of is theproduct

Examples Factor by grouping.

9.

Factor each group.

Factor out the common factor, .

10.

Examples Factor by grouping.

11.

12.

There is no common binomial factor that can now be factored out. Nomatter how we rearrange the terms, no grouping will lead to a com-mon factor. Thus, this polynomial is not factorable by grouping.

13.

Notice that the first two terms have no common factor other than 1.However, if we rearrange these terms, a grouping emerges that doeslead to a common factor.

Factor from the second group.

Factor out the common factor, .

HELPFUL HINT

Throughout this chapter, we will be factoring polynomials. Even whenthe instructions do not so state, it is always a good idea to check youranswers by multiplying.

Ay - 1 B = Ay - 1 B A3x - 2 B-2 = 3x Ay - 1 B - 2 Ay - 1 B

= A3xy - 3x B + A-2y + 2 B3xy + 2 - 3x - 2y

3xy + 2 - 3x - 2y

5x - 10 + x3 - x2 = 5 Ax - 2 B + x2 Ax - 1 B = A3x2 - 2 B Ax - 3 B = x A3x2 - 2 B - 3 A3x2 - 2 B

3x3 - 2x - 9x2 + 6

= A3x + 4y B Ax - 1 B = x A3x + 4y B - 1 A3x + 4y B = A3x2 + 4xy B + A-3x - 4y B

3x2 + 4xy - 3x - 4y

A3x - 2 B = A3x - 2 B A5x2 + 2 B = 5x2 A3x - 2 B + 2 A3x - 2 B = A15x3 - 10x2 B + A6x - 4 B

15x3 - 10x2 + 6x - 4

Ay + 2 B Ax + 3 B xy + 2x + 3y + 6

Ay + 2 B Ax + 3 B = xy + 2x + 3y + 6

Ax + 3 BAy + 2 B

Answers9. , 10. ,11. , 12. Can’t be factored,13. A3y + 1 B Ax - 4 BA4x2 + 1 B Ax - 5 B A2x + 5y B Ay - 2 BA4x - 1 B A7x2 + 3 B

Factor each group. A is factored from the second pair of terms so that thereis a common factor, .Factor out the common factor, .A3x + 4y BA3x + 4y B

-1

Factor each group. A is factored fromthe second pair of terms so that there is acommon factor, . Factor out the common factor, .A3x2 - 2 BA3x2 - 2 B

-3

EXERCISE SET 5.1

Factor out the GCF from each polynomial. See Examples 2 through 7.

1. 2. 3. 4.

5. 6. 7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20. z Ay + 4 B + 3 Ay + 4 By Ax + 2 B + 3 Ax + 2 B

25

y7 - 45

y5 + 35

y2 - 25

y13

x4 + 23

x3 - 43

x5 + 13

x

9y6 - 27y4 + 18y2 + 68x5 + 16x4 - 20x3 + 12

14x3y + 7x2y - 7xy5x3y - 15x2y + 10xy

x9y6 + x3y5 - x4y3 + x3y3a7b6 - a3b2 + a2b5 - a2b2

10xy - 15x232xy - 18x2

5x2 + 10x66y4 - 2yy5 - 6y4x3 + 5x2

42x - 730x - 1518a + 123a + 6

B

385


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

5.2 FACTORING TRINOMIALS OF THE FORM x2 � bx � cFACTORING TRINOMIALS OF THE FORM x2 � bx � c

In this section, we factor trinomials of the form , such as

, , ,

Notice that for these trinomials, the coefficient of the squared variable is 1.Recall that factoring means to write as a product and that factoring and

multiplying are reverse processes. Using the FOIL method of multiplyingbinomials, we have that

F O I L

Thus, a factored form of is .Notice that the product of the first terms of the binomials is ,

the first term of the trinomial. Also, the product of the last two terms of thebinomials is , the third term of the trinomial. The sum of thesesame terms is , the coefficient of the middle, x, term of the tri-nomial.

The product of these numbers is 3.

The sum of these numbers is 4.

Many trinomials, such as the one above, factor into two binomials. Tofactor , let’s assume that it factors into two binomials andbegin by writing two pairs of parentheses. The first term of the trinomial is

, so we use x and x as the first terms of the binomial factors.

n n

To determine the last term of each binomial factor, we look for two inte-gers whose product is 10 and whose sum is 7. The integers are 2 and 5. Thus,

To see if we have factored correctly, we multiply.

Combine like terms. = x2 + 7x + 10

Ax + 2 B Ax + 5 B = x2 + 5x + 2x + 10

x2 + 7x + 10 = Ax + 2 B Ax + 5 B

BB Ax +x2 + 7x + 10 = Ax +

x2

x2 + 7x + 10

x2 + 4x + 3 = Ax + 3 B Ax + 1 B

3 + 1 = 43 # 1 = 3

x # x = x2Ax + 3 B Ax + 1 Bx2 + 4x + 3

= x2 + 4x + 3

Ax + 3 B Ax + 1 B = x2 + 1x + 3x + 3

r2 - r - 42x2 + 4x - 12x2 - 8x + 15x2 + 4x + 3

x2 + bx + c

A

Factoring Trinomials of the Form x2 � bx � c SECTION 5.2 387

Objectives

Factor trinomials of the form.

Factor out the greatest commonfactor and then factor a trinomialof the form .x2 + bx + c

Bx2 + bx + c

A

HELPFUL HINT

Since multiplication is commutative, the factored form ofcan be written as either or

.Ax + 5 B Ax + 2 B Ax + 2 B Ax + 5 Bx2 + 7x + 10

SSM CD-ROM Video5.2

Example 1 Factor:

Solution: We begin by writing the first terms of the binomial fac-tors.

Next we look for two numbers whose product is 12 andwhose sum is 7. Since our numbers must have a positiveproduct and a positive sum, we look at pairs of positivefactors of 12 only.

Factors of 12 Sum of Factors

1, 12 132, 6 8

Check: Multiply by .

Example 2 Factor:

Solution: Again, we begin by writing the first terms of the binomi-als.

Now we look for two numbers whose product is 15 andwhose sum is . Since our numbers must have a posi-tive product and a negative sum, we look at pairs of neg-ative factors of 15 only.


,

Example 3 Factor:

Solution: x2 + 4x - 12 = Ax + n B Ax + n Bx2 + 4x - 12

x2 - 8x + 15 = Ax - 3 B Ax - 5 B

Correct sum, so thenumbers are -3 and -5.-8-3, -5

-16-15-1

-8

Ax + n B Ax + n B

x2 - 8x + 15

Ax + 4 BAx + 3 Bx2 + 7x + 12 = Ax + 3 B Ax + 4 B

Correct sum, so thenumbers are 3 and 4.

73, 4

Ax + n B Ax + n B

x2 + 7x + 12


TO FACTOR A TRINOMIAL OF THE FORM x2 � bx � c

The product of these numbers is c.

The sum of these numbers is b.

x2 + bx + c = Ax + n B Ax + n B

Practice Problem 1Factor: x2 + 9x + 20

Practice Problem 2Factor each trinomial.

a.

b. x2 - 27x + 50

x2 - 13x + 22

Practice Problem 3Factor: x2 + 5x - 36

Answers1. , 2. a. ,b. , 3. Ax + 9 B Ax - 4 BAx - 2 B Ax - 25 B Ax - 2 B Ax - 11 BAx + 4 B Ax + 5 B

We look for two numbers whose product is andwhose sum is 4. Since our numbers must have a negativeproduct, we look at pairs of factors with opposite signs.

Factors of Sum of Factors

, 12 111,

2, , 4 1

3,

Example 4 Factor:

Solution: Because the variable in this trinomial is r, the first termof each binomial factor is r.

Now we look for two numbers whose product is andwhose sum is , the numerical coefficient of r. Thenumbers are 6 and . Therefore,

Example 5 Factor:

Solution: Look for two numbers whose product is 10 and whosesum is 2. Neither 1 and 10 nor 2 and 5 give the requiredsum, 2. We conclude that is not factorablewith integers. A polynomial such as iscalled a prime polynomial.

Example 6 Factor:

Solution:

Recall that the middle term is the same as .Notice that is the “coefficient” of x. We then look fortwo terms whose product is and whose sum is . Theterms are and because and

. Therefore,

Example 7 Factor:

Solution: As usual, we begin by writing the first terms of the bino-mials. Since the greatest power of x in this polynomial is

, we write

since

Now we look for two factors of 6 whose sum is 5. Thenumbers are 2 and 3. Thus,

x4 + 5x2 + 6 = Ax2 + 2 B Ax2 + 3 B

x2 # x2 = x4Ax2 + n B Ax2 + n Bx4

x4 + 5x2 + 6

x2 + 5xy + 6y2 = Ax + 2y B Ax + 3y B2y + 3y = 5y

2y # 3y = 6y23y2y5y6y2

5y5yx5xy

x2 + 5xy + 6y2 = Ax + n B Ax + n Bx2 + 5xy + 6y2

a2 + 2a + 10a2 + 2a + 10

a2 + 2a + 10

r2 - r - 42 = Ar + 6 B Ar - 7 B-7

-1-42

r2 - r - 42 = Ar + n B Ar + n B

r2 - r - 42

x2 + 4x - 12 = Ax - 2 B Ax + 6 B

-1-4-3

-4-6

Correct sum, so thenumbers are -2 and 6.

4-2, 6

-11-12-1

-12

-12

Factoring Trinomials of the Form x2 � bx � c SECTION 5.2 389


a.

b. y2 + 2y - 48

q2 - 3q - 40



a.

b. a2 - 13ab + 30b2

x2 + 6xy + 8y2

Practice Problem 7Factor: x4 + 8x2 + 12

Answers4. a. , b. ,5. prime polynomial, 6. a. ,b. , 7. Ax2 + 6 B Ax2 + 2 BAa - 3b B Aa - 10b B Ax + 2y B Ax + 4y BAy + 8 B Ay - 6 BAq - 8 B Aq + 5 B

The following sign patterns may be useful when factoring trinomials.


Remember that the first step in factoring any polynomial is to factor out thegreatest common factor (if there is one other than 1 or ).

Example 8 Factor:

Solution: First we factor out the greatest common factor, 3, fromeach term.

Now we factor by looking for two factorsof whose sum is . The factors are and 2.Therefore, the complete factored form is

ƒ——ƒHELPFUL HINTRemember to write the common factor 3 as part of the factored form.

3m2 - 24m - 60 = 3 Am + 2 B Am - 10 B

-10-8-20m2 - 8m - 20

3m2 - 24m - 60 = 3 Am2 - 8m - 20 B

3m2 - 24m - 60

-1

B


HELPFUL HINT—SIGN PATTERNS

A positive constant in a trinomial tells us to look for two numbers withthe same sign. The sign of the coefficient of the middle term tells uswhether the signs are both positive or both negative.

T T

T T

A negative constant in a trinomial tells us to look for two numbers withopposite signs.

T Tx2 - 6x - 16 = Ax - 8 B Ax + 2 Bx2 + 6x - 16 = Ax + 8 B Ax - 2 B

oppositesigns

oppositesigns

x2 - 10x + 16 = Ax - 2 B Ax - 8 B

samesign

bothnegative

x2 + 10x + 16 = Ax + 2 B Ax + 8 B

samesign

bothpositive


a.

b. 4x2 - 24x + 36

x3 + 3x2 - 4x

Answers8. a. , b. 4 Ax - 3 B Ax - 3 Bx Ax + 4 B Ax - 1 B

EXERCISE SET 5.2

Factor each trinomial completely. If a polynomial can’t be factored, write “prime.” SeeExamples 1 through 7.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10.

Factor each trinomial completely. See Examples 1 through 8.

11. 12. 13.

14. 15. 16.

17. 18. 19.

20. x2 - 5x - 14

x2 - x - 2x2 + 19x + 60x2 + 15x + 36

x2 - 4xy - 77y2x2 - 3xy - 4y2x3 - x2 - 56x

2x3 - 18x2 + 40x3x2 + 30x + 632z2 + 20z + 32B

x2 - 7x + 5

x2 + 5x + 2x2 + 4x - 32x2 + 3x - 70

x2 - x - 30x2 - 3x - 18x2 - 6x + 9

x2 - 10x + 9x2 + 6x + 8x2 + 7x + 6

A

391


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

5.3 FACTORING TRINOMIALS OF THE FORM ax2 � bx � cFACTORING TRINOMIALS OF THE FORM ax2 � bx � c

In this section, we factor trinomials of the form , such as

, ,

Notice that the coefficient of the squared variable in these trinomials is anumber other than 1. We will factor these trinomials using a trial-and-checkmethod based on our work in the last section.

To begin, let’s review the relationship between the numerical coeffi-cients of the trinomial and the numerical coefficients of its factored form.For example, since , the factored formof is

Notice that and x are factors of , the first term of the trinomial. Also,6 and 1 are factors of 6, the last term of the trinomial, as shown:

ƒ——————————————T— —T

ƒ————————c ——c

Also notice that 13x, the middle term, is the sum of the following products:

Let’s use this pattern to factor . First, we find factors of .Since all numerical coefficients in this trinomial are positive, we will usefactors with positive numerical coefficients only. Thus, the factors of are

and x. Let’s try these factors as first terms of the binomials. Thus far, wehave

Next, we need to find positive factors of 2. Positive factors of 2 are 1 and2. Now we try possible combinations of these factors as second terms of thebinomials until we obtain a middle term of .

Let’s try switching factors 2 and 1.

Thus the factored form of is . To check,we multiply and . The product is .5x2 + 7x + 2Ax + 1 BA5x + 2 B A5x + 2 B Ax + 1 B5x2 + 7x + 2

2x+ 5x

7x

A5x + 2 B Ax + 1 B = 5x2 + 7x + 2Uu

Correct middle term

1x+ 10x

11x

A5x + 1 B Ax + 2 B = 5x2 + 11x + 2Uu

Incorrect middle term

7x

5x2 + 7x + 2 = A5x + n B Ax + n B5x

5x2

5x25x2 + 7x + 2

1x+ 12x

13x

2x2 + 13x + 6 = A2x + 1 B Ax + 6 BUuMiddle term

1 # 6

2x2 + 13x + 6 = A2x + 1 B Ax + 6 B2x # x

2x22x

2x2 + 13x + 6 = A2x + 1 B Ax + 6 B2x2 + 13x + 6

A2x + 1 B Ax + 6 B = 2x2 + 13x + 6

2x2 + 13x - 78x2 - 22x + 53x2 + 11x + 6

ax2 + bx + c

A

Factoring Trinomials of the Form ax2 � bx � c SECTION 5.3 393

Objectives

Factor trinomials of the form, where .

Factor out the GCF before factor-ing a trinomial of the form

.ax2 + bx + c

Ba Z 1ax2 + bx + c

A

SSM CD-ROM Video5.3



a.

b. 5x2 + 27x + 10

4x2 + 12x + 5

✓ CONCEPT CHECK

Do the terms of havea common factor? Without multiply-ing, decide which of the following fac-tored forms could not be a factoredform of .a.

b.

c.

d. A3x + 9 B Ax + 2 BA3x + 6 B Ax + 3 BA3x + 2 B Ax + 9 BA3x + 18 B Ax + 1 B3x2 + 29x + 18

3x2 + 29x + 18


a.

b. 2x2 - 11x + 12

6x2 - 5x + 1

Example 1 Factor:

Solution: Since all numerical coefficients are positive, we use fac-tors with positive numerical coefficients. We first findfactors of .

Factors of :

If factorable, the trinomial will be of the form

Next we factor 6.

Factors of 6: ,

Now we try combinations of factors of 6 until a middleterm of is obtained. Let’s try 1 and 6 first.

Now let’s next try 6 and 1.

Before multiplying, notice that the terms of the factorhave a common factor of 3. The terms of the

original trinomial have no common factorother than 1, so the terms of the factored form of

can contain no common factor otherthan 1. This means that is not a fac-tored form.

Next let’s try 2 and 3 as last terms.

Thus the factored form of is.


Example 2 Factor:

Solution: Factors of : ,

We’ll try and x.8x

8x2 = 4x # 2x8x2 = 8x # x8x2

8x2 - 22x + 5

HELPFUL HINT

If the terms of a trinomial have no common factor (other than 1), thenthe terms of neither of its binomial factors will contain a common fac-tor (other than 1).

A3x + 2 B Ax + 3 B 3x2 + 11x + 6

2x+ 9x

11x

A3x + 2 B Ax + 3 B = 3x2 + 11x + 6Uu

Correct middle term

A3x + 6 B Ax + 1 B3x2 + 11x + 6

3x2 + 11x + 63x + 6

A3x + 6 B Ax + 1 B

1x+ 18x

19x

A3x + 1 B Ax + 6 B = 3x2 + 19x + 6UuIncorrect middle term

11x

6 = 2 # 36 = 1 # 6

3x2 + 11x + 6 = A3x + n B Ax + n B

3x2 = 3x # x3x2

3x2

3x2 + 11x + 6

Answers1. a. , b. ,

2. a. , b.

✓ Concept Check: no; a, c, d

A2x - 3 B Ax - 4 BA3x - 1 B A2x - 1 BA5x + 2 B Ax + 5 BA2x + 5 B A2x + 1 B

Factoring Trinomials of the Form ax2 � bx � c SECTION 5.3 395


a.

b. 4x2 + 3x - 7

35x2 + 4x - 4


a.

b. 12a2 - 16ab - 3b2

14x2 - 3xy - 2y2

Since the middle term, , has a negative numericalcoefficient, we factor 5 into negative factors.

Factors of 5:

Let’s try and .

Now let’s try and .

Don’t give up yet! We can still try other factors of .Let’s try and with and .

The factored form of is.

Example 3 Factor:

Solution: Factors of :

Factors of : ,

We try possible combinations of these factors:

The factored form of is .

Example 4 Factor:

Solution: Factors of : ,

Factors of : ,

We try some combinations of these factors:

Correct middle term

A2x - 3y B A5x + y B = 10x2 - 13xy - 3y2

A5x + 3y B A2x - y B = 10x2 + xy - 3y2

Ax + 3y B A10x - y B = 10x2 + 29xy - 3y2

A10x - 3y B Ax + y B = 10x2 + 7xy - 3y2

-3y2 = 3y # -y-3y2 = -3y # y-3y2

10x2 = 2x # 5x10x2 = 10x # x10x2

10x2 - 13xy - 3y2

A2x - 1 B Ax + 7 B2x2 + 13x - 7

A2x - 1 B Ax + 7 B = 2x2 + 13x - 7


Correct middle term

A2x + 1 B Ax - 7 B = 2x2 - 13x - 7

-7 = 1 # -7-7 = -1 # 7-7

2x2 = 2x # x2x2

2x2 + 13x - 7

A4x - 1 B A2x - 5 B 8x2 - 22x + 5

-2x+ A-20x B

-22x

A4x - 1 B A2x - 5 B = 8x2 - 22x + 5UuCorrect middle term

-5-12x4x8x2

-5x+ A-8x B

-13x

A8x - 5 B Ax - 1 B = 8x2 - 13x + 5Uu


-1-5

-1x+ A-40x B

-41x

A8x - 1 B Ax - 5 B = 8x2 - 41x + 5Uu


-5-1

5 = -1 # -5

-22x

8x2 - 22x + 5 = A8x + n B Ax + n B

Answers3. a. , b. ,4. a. ,b. A6a + b B A2a - 3b BA7x + 2y B A2x - y B A4x + 7 B Ax - 1 BA5x + 2 B A7x - 2 B



a.

b. 6xy2 + 33xy - 18x

3x3 + 17x2 + 10x

Answers5. a. ,b. 3x A2y - 1 B Ay + 6 Bx A3x + 2 B Ax + 5 B



Don’t forget that the first step in factoring any polynomial is to look for acommon factor to factor out.

Example 5 Factor:

Solution: Notice that all three terms have a common factor of .Thus we factor out first.

Next we factor .

Factors of : , ,

Since all terms in the trinomial have positive numericalcoefficients, we factor 3 using positive factors only.

Factors of 3:

We try some combinations of the factors.


ƒƒƒ

HELPFUL HINT

Don’t forget to include the common factor in the factored form.

2x2 A2x + 3 B A6x + 1 B24x4 + 40x3 + 6x2

Correct middle term2x2 A2x + 3 B A6x + 1 B = 2x2 A12x2 + 20x + 3 B2x2 A12x + 1 B Ax + 3 B = 2x2 A12x2 + 37x + 3 B2x2 A4x + 3 B A3x + 1 B = 2x2 A12x2 + 13x + 3 B

3 = 1 # 3

12x2 = 6x # 2x12x2 = 12x # x12x2 = 4x # 3x12x2

12x2 + 20x + 3

24x4 + 40x3 + 6x2 = 2x2 A12x2 + 20x + 3 B2x2

2x2

24x4 + 40x3 + 6x2

B

A2x - 3y B A5x + y B 10x2 - 13xy - 3y2

EXERCISE SET 5.3

Complete each factored form.

1. ( ) 2. ( )

3. ( ) 4. ( )

5. ( )


6. 7. 8.

9. 10. 11.

12. 13. 14.

15.


16. 17. 18.

19. 20. 12x2 + 7x - 1212x2 - 14x - 10

21x2 - 48x - 458a3 + 14a2 + 3a12x3 + 11x2 + 2x

B

2x2 + 7x + 5

10x2 + 17x + 33x2 + 20x - 6320r2 + 27r - 8

36r2 - 5r - 242x2 - 9x - 521x2 - 41x + 10

8y2 - 17y + 93x2 + 8x + 42x2 + 13x + 15

20x2 - 7x - 6 = A5x + 2 B

6y2 + 11y - 10 = A2y + 5 B50x2 + 15x - 2 = A5x + 2 B

2y2 + 15y + 25 = A2y + 5 B5x2 + 22x + 8 = A5x + 2 BA

397


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

5.4 FACTORING TRINOMIALS OF THE FORM ax2 � bx � cBY GROUPING

USING THE GROUPING METHOD

There is an alternative method that can be used to factor trinomials of theform , . This method is called the grouping methodbecause it uses factoring by grouping as we learned in Section 5.1.

To see how this method works, let’s multiply the following:

Notice that the product of the coefficients of the first and last terms is. This is the same as the product of the coefficients of the two

middle terms, .Let’s use this pattern to write as a four-term polyno-

mial. We will then factor the polynomial by grouping.

Find two numbers whose product is and whose sum is 11.

Since we want a positive product and a positive sum, we consider pairs ofpositive factors of 24 only.


1, 24 252, 12 14

The factors are 3 and 8. Now we use these factors to write the middle termas (or ). We replace with in the original

trinomial and then we can factor by grouping.

Group the terms.

Factor each group.

Factor out .

In general, we have the following procedure.

A2x + 3 B = A2x + 3 B Ax + 4 B = x A2x + 3 B + 4 A2x + 3 B = A2x2 + 3x B + A8x + 12 B

2x2 + 11x + 12 = 2x2 + 3x + 8x + 12

3x + 8x11x8x + 3x3x + 8x11x

Correct sum113, 8

= 2x2 + nx + nx + 122 # 12 = 242x2 + 11x + 12

2x2 + 11x + 1210 # 3 = 30

6 # 5 = 30

A2x + 1 B A3x + 5 B = 6x2 + 10x + 3x + 5

10 # 3 = 30

6 # 5 = 30

= 6x2 + 13x + 5

6 # 5 30

a Z 1ax2 + bx + c

A

Factoring TRINOMIALS OF THE FORM ax2 � bx � c BY GROUPING SECTION 5.4 399

Objectives

Use the grouping method to factor trinomials of the form

, .a Z 1ax2 + bx + c

A

TO FACTOR TRINOMIALS OF THE FORM ax2 � bx � c BY

GROUPING

Step 1. Factor out a greatest common factor, if there is one other than1 (or ).

Step 2. Find two numbers whose product is and whose sum is b.

Step 3. Write the middle term, bx using the factors found in Step 2.

Step 4. Factor by grouping.

a # c

-1

SSM CD-ROM Video5.4


Practice Problem 1Factor each trinomial by grouping.

a.

b. 12x2 + 19x + 5

3x2 + 14x + 8

Practice Problem 2Factor each trinomial by grouping.

a.

b. 30x2 - 26x + 4

6x2y - 7xy - 5y

Answers1. a. , b. ,2. a. ,b. 2 A5x - 1 B A3x - 2 By A2x + 1 B A3x - 5 B A4x + 5 B A3x + 1 BAx + 4 B A3x + 2 B


Solution: Step 1. The terms of this trinomial contain no greatestcommon factor other than 1 (or ).

Step 2. This trinomial is of the form with, , and . Find two numbers

whose product is or , and whosesum is or . The numbers are and .

Step 3. Write as so that



Solution: Step 1. First factor out the greatest common factor, 2.

Step 2. Next notice that , , and inthis trinomial. Find two numbers whose product is

or and whose sum is , .The numbers are and 5.

Step 3.

Step 4.

The factored form of .c— Don’t forget to include the common

factor of 2.

2 Ax - 2 B A3x + 5 B6x2 - 2x - 20 =

= Ax - 2 B A3x + 5 B = 3x Ax - 2 B + 5 Ax - 2 B

3x2 - x - 10 = 3x2 - 6x + 5x - 10

-6-1b3 A-10 B = -30a # c

c = -10b = -1a = 3

6x2 - 2x - 20 = 2 A3x2 - x - 10 B

6x2 - 2x - 20

= A2x - 1 B A4x - 5 B8x2 - 4x - 10x + 5 = 4x A2x - 1 B - 5 A2x - 1 B

8x2 - 14x + 5 = 8x2 - 4x - 10x + 5

-4x - 10x-14x

-10-4-14b8 # 5 = 40a # c

c = 5b = -14a = 8ax2 + bx + c

-1

8x2 - 14x + 5

401


EXERCISE SET 5.4

Factor each polynomial by grouping. Notice that Step 3 has already been done in theseexercises. See Examples 1 and 2.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10. 4x2 - 9x - 32x + 728x2 - 5x - 24x + 15

2x2 + 5x + 14x + 353x2 + 4x + 12x + 16

z2 + 10z - 7z - 70y2 + 8y - 2y - 16

x2 - 6x + 2x - 12x2 - 4x + 7x - 28

x2 + 5x + 3x + 15x2 + 3x + 2x + 6

A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

5.5 FACTORING PERFECT SQUARE TRINOMIALS AND THE

DIFFERENCE OF TWO SQUARES

RECOGNIZING PERFECT SQUARE TRINOMIALS

A trinomial that is the square of a binomial is called a perfect square trino-mial. For example,

Thus is a perfect square trinomial.In Chapter 4, we discovered special product formulas for squaring bino-

mials, recognizing that

Because multiplication and factoring are reverse processes, we can now usethese special products to help us factor perfect square trinomials. If wereverse these equations, we have the following.

To use these equations to help us factor, we must first be able to recog-nize a perfect square trinomial. A trinomial is a perfect square when

1. Two terms, and , are squares, and2. another term is or . That is, this term is twice the

product of a and b, or its opposite.

Example 1 Decide whether is a perfect squaretrinomial.

Solution: 1. Two terms, and 16, are squares .ƒ—————T T————————ƒ

2. Twice the product of x and 4 is the other term of the tri-nomial.

Thus, is a perfect square trinomial.


Solution: 1. Two terms, and 9, are squares.

ƒ———————Tƒ——T

2. Twice the product of and 3 is not the other term ofthe trinomial.

The trinomial is not a perfect square trinomial.

2 # 2x # 3 = 12x, not 10x

2x

9 = 32and4x2 = A2x B 24x2

4x2 + 10x + 9

x2 + 8x + 16

2 # x # 4 = 8x

A16 = 42 Bx2

x2 + 8x + 16

-2 # a # b2 # a # bb2a2

Aa - b B 2 = a2 - 2ab + b2andAa + b B 2 = a2 + 2ab + b2

x2 + 6x + 9

= x2 + 6x + 9

Ax + 3 B 2 = Ax + 3 B Ax + 3 B

A

Factoring Perfect Square Trinomials and the Difference of Two Squares SECTION 5.5 403

Objectives

Recognize perfect square trino-mials.

Factor perfect square trinomials.

Factor the difference of twosquares.

CB

A

FACTORING PERFECT SQUARE TRINOMIALS

a2 - 2ab + b2 = Aa - b B 2a2 + 2ab + b2 = Aa + b B 2

Practice Problem 1Decide whether each trinomial is aperfect square trinomial.

a.

b. x2 + 20x + 100

x2 + 12x + 36

SSM CD-ROM Video5.5

Answers1. a. yes, b. yes, 2. a. no, b. no


a.

b. 4x2 + 8x + 11

9x2 + 20x + 25



a.

b. 9x2 - 42x + 49

25x2 - 10x + 1


Practice Problem 5Factor: 9r2 + 24rs + 16s2

Answers3. a. yes, b. yes, 4. ,5. , 6. ,7. A3x + 1 B A3x + 4 BA3n - 1 B 2A3r + 4s B 2 Ax + 8 B 2

Practice Problem 6Factor: 9n2 - 6n + 1


Solution: 1. Two terms, and 4, are squares.

ƒ———————Tƒ——T

2. Twice the product of and 2 is the opposite of theother term of the trinomial.

Thus, is a perfect square trinomial.

FACTORING PERFECT SQUARE TRINOMIALS

Now that we can recognize perfect square trinomials, we are ready to fac-tor them.

Example 4 Factor:

Solution: and

Example 5 Factor:

Solution:

Example 6 Factor:

Solution:

Example 7 Factor:

Solution: Notice that this trinomial is not a perfect square trino-mial.

but b¬ƒb¬¬¬ƒ

and is not the middle term .

Although is not a perfect square trino-mial, it is factorable. Using techniques we learned in Sec-tion 4.3, we find that

25x2 + 50x + 9 = A5x + 9 B A5x + 1 B

25x2 + 50x + 9

50x30x

2 # 5x # 3 = 30x

25x2 = A5x B 2, 9 = 32

25x2 + 50x + 9

Aa - b B 2 = A2m - 1 B 2

a2 - 2 # a # b + b2

4m2 - 4m + 1 = A2m B 2 - 2 # 2m # 1 + 12

4m2 - 4m + 1

= A5x + 2y B 225x2 + 20xy + 4y2 = A5x B 2 + 2 # 5x # 2y + A2y B 2

25x2 + 20xy + 4y2

Aa + b B 2 = Ax + 6 B 2

12x = 2 # x # 6

a2 + 2 # a # b + b2

36 = 62x2 + 12x + 36 = x2 + 2 # x # 6 + 62

x2 + 12x + 36

B

9x2 - 12x + 4

2 # 3x # 2 = 12x, the opposite of -12x

3x

4 = 22and9x2 = A3x B 29x2

9x2 - 12x + 4

Practice Problem 7Factor: 9x2 + 15x + 4

Example 8 Factor:

Solution: Don’t forget to first look for a common factor. There is agreatest common factor of in this trinomial.

FACTORING THE DIFFERENCE OF TWO SQUARES

In Chapter 3, we discovered another special product, the product of thesum and difference of two terms a and b:

Reversing this equation gives us another factoring pattern, which we use tofactor the difference of two squares.

Let’s practice using this pattern.

Examples Factor each binomial.

9.

10.

11.

12.

Note that the binomial is the sum of two squaressince we can write as . We might try tofactor using or . Butwhen we multiply to check, we find that neither factoringis correct.

In both cases, the product is a trinomial, not the requiredbinomial. In fact, is a prime polynomial.

HELPFUL HINT

After the greatest common factor has been removed, the sum of twosquares cannot be factored further using real numbers.

x2 + 4

Ax - 2 B Ax - 2 B = x2 - 4x + 4

Ax + 2 B Ax + 2 B = x2 + 4x + 4

Ax - 2 B Ax - 2 BAx + 2 B Ax + 2 B x2 + 22x2 + 4

x2 + 4

x2 + 4

y2 - 49

= y2 - a 23b 2

= ay + 23b ay - 2

3b

y2 - 25 = y2 - 52 = Ay + 5 B Ay - 5 B a2 - b2 = Aa + b B Aa - b B

x2 - 4 = x2 - 22 = Ax + 2 B Ax - 2 B

Aa + b B Aa - b B = a2 - b2

C

= 2x A9x - 4 B 2 = 2x C A9x B 2 - 2 # 9x # 4 + 42 D

162x3 - 144x2 + 32x = 2x A81x2 - 72x + 16 B2x

162x3 - 144x2 + 32x

HELPFUL HINT

A perfect square trinomial can also be factored by the methods foundin Sections 4.2 through 4.4.


FACTORING THE DIFFERENCE OF TWO SQUARES

a2 - b2 = Aa + b B Aa - b B

Practice Problem 8Factor: 12x3 - 84x2 + 147x

Practice Problems 9–12Factor each binomial.

9. 10.

11. 12. s2 + 9c2 - 925

a2 - 16x2 - 9

Answers

8. , 9. ,

10. , 11. ,

12. prime polynomial

a c - 35b a c + 3

5bAa - 4 B Aa + 4 B

Ax - 3 B Ax + 3 B3x A2x - 7 B 2


Practice Problems 13–15Factor each difference of two squares.

13.14.15. p4 - 81

16x2 - 49y29s2 - 1

Practice Problems 16–17Factor each difference of two squares.

16.

17. 48x4 - 3

9x3 - 25x

Answers13. ,14. ,15. ,16. ,17. 3 A4x2 + 1 B A2x + 1 B A2x - 1 Bx A3x - 5 B A3x + 5 BAp2 + 9 B Ap + 3 B Ap - 3 BA4x - 7y B A4x + 7y BA3s - 1 B A3s + 1 B

Examples Factor each difference of two squares.

13.

14.

15.Factor the difference of two squares.

Factor the difference of two squares.

Examples Factor each difference of two squares.

16. Factor out the common factor x.


17. Factor out the common factor, 2.


Factor the difference of two

squares.

= 2 A9x2 + 1 B A3x + 1 B A3x - 1 B = 2 A9x2 + 1 B A9x2 - 1 B

162x4 - 2 = 2 A81x4 - 1 B = x A2x + 7 B A2x - 7 B = x C A2x B 2 - 72 D

4x3 - 49x = x A4x2 - 49 B

HELPFUL HINTS1. Don’t forget to first see whether there’s a greatest common factor

(other than 1 or -1) that can be factored out.2. Factor completely. In other words, check to see whether any factors

can be factored further (as in Example 15).

= Ay2 + 4 B Ay + 2 B Ay - 2 B = Ay2 + 4 B Ay2 - 4 B

y4 - 16 = Ay2 B 2 - 42

25a2 - 9b2 = A5a B 2 - A3b B 2 = A5a + 3b B A5a - 3b B4x2 - 1 = A2x B 2 - 12 = A2x + 1 B A2x - 1 B


CALCULATOR EXPLORATIONGRAPHING

A graphing calculator is a convenient tool for evaluating anexpression at a given replacement value. For example, let’s evalu-ate when . To do so, store the value 2 in the vari-able x and then enter and evaluate the algebraic expression.

The value of when is . You may want to use thismethod for evaluating expressions as you explore the following.

We can use a graphing calculator to explore factoring patternsnumerically. Use your calculator to evaluate

, and for each value of x givenin the table. What do you observe?

Notice in each case that . Because foreach x in the table the value of and the value of

are the same, we might guess that . We can verify our guess algebraically with multiplica-

tion:

Ax - 1 B Ax - 1 B = x2 - x - x + 1 = x2 - 2x + 1

Ax - 1 B 2 x2 - 2x + 1 =Ax - 1 B 2 x2 - 2x + 1x2 - 2x - 1 Z Ax - 1 B 2

x = 0

x = -12.1

x = 2.7

x = -3

x = 5

Ax - 1 B 2x2 - 2x - 1x2 - 2x + 1

Ax - 1 B 2x2 - 2x + 1, x2 - 2x - 1

-8x = 2x2 - 6x

2 S X2

X2 - 6X-8

x = 2x2 - 6x


Focus On Mathematical ConnectionsGEOMETRY

Factoring polynomials can be visualized using areas of rectangles. To see this, let’s first find theareas of the following squares and rectangles. (Recall that Area )

To use these areas to visualize factoring the polynomial , for example, use theshapes below to form a rectangle. The factored form is found by reading the length and thewidth of the rectangle as shown below.

Thus, .Try using this method to visualize the factored form of each polynomial below.

GROUP ACTIVITY

Work in a group and use tiles to find the factored form of the polynomials below. (Tiles can behand made from index cards.)

1. 4.

2. 5.

3. 6. x2 + 4x + 4x2 + 5x + 4

x2 + 6x + 9x2 + 5x + 6

x2 + 4x + 3x2 + 6x + 5

x2 + 3x + 2 = Ax + 2 B Ax + 1 B

x

1

x2 + 3x + 2

x + 1

x

x + 2 1x

x2

1 x

x1

1

x2 + 3x + 2

x

x

1 1

1

Area: 1 . x = x square units Area: 1 . 1 = 1 square unitArea: x . x = x2 square units

x

= Length # Width

409


MENTAL MATH

State each number as a square.

1. 1 2. 25

State each term as a square.

3. 4.

EXERCISE SET 5.5

Determine whether each trinomial is a perfect square trinomial. See Examples 1 through 3.

1. 2. 3.

4.


5. 6. 7.

8. 9. 10.

11. 12.

Factor each binomial completely. See Examples 9 through 17.

13. 14. 15. 16.

17. 18. 19. 20.

21. 22. 23. y2 - 116

x2 - 14

16 - a2b2

x3y - 4xy3n4 - 169t2 - 14r2 - 1

100 - t281 - p2x2 - 36x2 - 4C

m4 + 10m2 + 25x4 + 4x2 + 4

25x2 + 20x + 416a2 - 24a + 9x2 - 12x + 36

x2 - 16x + 64x2 + 18x + 81x2 + 22x + 121B

y2 + 4y + 16

y2 + 5y + 25x2 + 22x + 121x2 + 16x + 64

A

16y29x2

MENTAL MATH ANSWERS

1.

2.

3.

4.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

5.6 SOLVING QUADRATIC EQUATIONS BY FACTORING

In this section, we introduce a new type of equation—the quadratic equa-tion.

Some examples of quadratic equations are

The form is called the standard form of a quadraticequation. The quadratic equation is the only equationabove that is in standard form.

Quadratic equations model many real-life situations. For example, let’ssuppose an object is dropped from the top of a 256-foot cliff and we wantto know how long before the object strikes the ground. The answer to thisquestion is found by solving the quadratic equation . (SeeExample 1 in Section 5.7.)

SOLVING QUADRATIC EQUATIONS BY FACTORING

Some quadratic equations can be solved by making use of factoring and thezero factor property.

In other words, if the product of two numbers is 0, then at least one of thenumbers must be 0.

Example 1 Solve:

Solution: If this equation is to be a true statement, then either thefactor must be 0 or the factor must be 0. Inother words, either

or

If we solve these two linear equations, we have

or x = -1x = 3

x + 1 = 0x - 3 = 0

x + 1x - 3

Ax - 3 B Ax + 1 B = 0

A

256 feet

-16t2 + 256 = 0

3x2 + 5x + 6 = 0ax2 + bx + c = 0

y2 + y = 1x2 = 93x2 + 5x + 6 = 0

Solving Quadratic Equations by Factoring SECTION 5.6 411

Objectives

Solve quadratic equations by fac-toring.

Solve equations with degreegreater than 2 by factoring.

B

A

QUADRATIC EQUATION

A quadratic equation is one that can be written in the form

where a, b, and c are real numbers and .a Z 0

ax2 + bx + c = 0

ZERO FACTOR PROPERTY

If a and b are real numbers and if , then , or .b = 0a = 0ab = 0

SSM CD-ROM Video5.6

Answer

1. 7 and -2

Practice Problem 1Solve: Ax - 7 B Ax + 2 B = 0

Thus, 3 and are both solutions of the equation. To check, we replace x with 3 in

the original equation. Then we replace x with in theoriginal equation.

Check:

Replace x

with .

True. True.

The solutions are 3 and .

Example 2 Solve:

Solution: The product is 0. By the zero factor property, this is trueonly when a factor is 0. To solve, we set each factor equalto 0 and solve the resulting linear equations.

or

Check: Let .

Replace x with 5.

True.

Let .

Replace x with .

True.

The solutions are 5 and .- 72

0 = 0

a- 172b # 0 � 0

a- 172b A-7 + 7 B � 0

- 72

a -72

- 5 b a2 a -72b + 7 b � 0

Ax - 5 B A2x + 7 B = 0

x = - 72

0 = 0

0 # 17 � 0

A5 - 5 B A2 # 5 + 7 B � 0

A x - 5 B A2x + 7 B = 0

x = 5

x = - 72

2x = -7 x = 5

2x + 7 = 0 x - 5 = 0

Ax - 5 B A2x + 7 B = 0

Ax - 5 B A2x + 7 B = 0

-1

A-4 B A0 B = 0 0 A4 B = 0

-1A-1 - 3 B A-1 + 1 B = 0Replace x

with 3.A3 - 3 B A3 + 1 B = 0

Ax - 3 B Ax + 1 B = 0Ax - 3 B Ax + 1 B = 0

-1Ax - 3 B Ax + 1 B = 0

-1


HELPFUL HINT

The zero factor property says that if a product is 0, then a factor is 0.

If , then or .If , then or .If , then or .

Use this property only when the product is 0. For example, if , we do not know the value of a or b. The values may be

, or , , or any other two numbers whose prod-uct is 8.

b = 1a = 8b = 4a = 2a # b = 8

2x - 3 = 0x + 7 = 0Ax + 7 B A2x - 3 B = 0x + 5 = 0x = 0x Ax + 5 B = 0

b = 0a = 0a # b = 0

Practice Problem 2Solve: Ax - 10 B A3x + 1 B = 0

Answer

2. 10 and - 13


Practice Problem 3Solve each equation.

a.

b. x A4x - 3 B = 0

y Ay + 3 B = 0

Practice Problem 4Solve: x2 - 3x - 18 = 0

Practice Problem 5Solve: x2 - 14x = -24

Answers

3. a. 0 and , b. 0 and , 4. 6 and ,

5. 12 and 2

-334

-3

Example 3 Solve:

Solution:

Use the zero factor property.

Check these solutions in the original equation. The solu-

tions are 0 and .

Example 4 Solve:

Solution: One side of the equation is 0. However, to use the zerofactor property, one side of the equation must be 0 andthe other side must be written as a product (must befactored). Thus, we must first factor this polynomial.

Factor.

Now we can apply the zero factor property.

or

Check:

Let . Let .

True. True.

The solutions are 11 and .

Example 5 Solve:

Solution: First we rewrite the equation in standard form so thatone side is 0. Then we factor the polynomial.

Next we use the zero factor property and set each factorequal to 0.

or

Check: Check these solutions in the original equation. Thesolutions are 4 and 5.

x = 5 x = 4

Set each factor equal to 0.

Solve.

x - 5 = 0x - 4 = 0

Ax - 4 B Ax - 5 B = 0

Write in standard form by adding20 to both sides.

Factor.

x2 - 9x + 20 = 0

x2 - 9x = -20

x2 - 9x = -20

-2

0 = 0 0 = 0

22 - 22 � 0 22 - 22 � 0

4 + 18 - 22 � 0 121 - 99 - 22 � 0

A-2 B 2 - 9 A-2 B - 22 � 0112 - 9 # 11 - 22 � 0

x2 - 9x - 22 = 0 x2 - 9x - 22 = 0

x = -2x = 11

x = -2 x = 11

x + 2 = 0x - 11 = 0

Ax - 11 B Ax + 2 B = 0

x2 - 9x - 22 = 0

x2 - 9x - 22 = 0

25

x = 25

5x = 2

x = 0 or 5x - 2 = 0

x A5x - 2 B = 0

x A5x - 2 B = 0

The following steps may be used to solve a quadratic equation by fac-toring.

Since it is not always possible to factor a quadratic polynomial, not allquadratic equations can be solved by factoring. Other methods of solvingquadratic equations are presented in Chapter 10.

Example 6 Solve:

Solution: First we write the equation in standard form; then wefactor.

Multiply.

Write in standard form.

Factor.

Check the solutions in the original equation. The solu-

tions are and 4.

SOLVING EQUATIONS WITH DEGREE GREATER THAN TWO

BY FACTORING

Some equations with degree greater than 2 can be solved by factoring andthen using the zero factor property.

Example 7 Solve:

Solution: To factor the left side of the equation, we begin byfactoring out the greatest common factor, .3x

3x3 - 12x = 0

B

- 12

x = - 12

x = 4 2x = -1

Set each factorequal to zero.Solve.

2x + 1 = 0 or x - 4 = 0

A2x + 1 B Ax - 4 B = 0

2x2 - 7x - 4 = 0

2x2 - 7x = 4

x A2x - 7 B = 4

x A2x - 7 B = 4

TO SOLVE QUADRATIC EQUATIONS BY FACTORING

Step 1. Write the equation in standard form so that one side of theequation is 0.

Step 2. Factor the quadratic equation completely.

Step 3. Set each factor containing a variable equal to 0.

Step 4. Solve the resulting equations.

Step 5. Check each solution in the original equation.


HELPFUL HINT

To solve the equation , do not set each factor equal to4. Remember that to apply the zero factor property, one side of theequation must be 0 and the other side of the equation must be in fac-tored form.

x A2x - 7 B = 4

Practice Problem 6Solve each equation.

a.

b. x A3x + 7 B = 6

x Ax - 4 B = 5

Answers

6. a. 5 and , b. and , 7. 0, 3, and -3-323

-1

Practice Problem 7Solve: 2x3 - 18x = 0


Practice Problem 8Solve: Ax + 3 B A3x2 - 20x - 7 B = 0

Answer

8. , , and 7- 13

-3

Thus, the equation has three solutions: 0,, and 2.

Check: Replace x with each solution in the original equation.

Let . Let . Let .

True.True. True.

The solutions are 0, , and 2.

Example 8 Solve:

Solution:

Factor the trinomial.

or or

Check each solution in the original equation. The solu-

tions are , , and .-6- 32

15

x = - 32

x = 15

x = -6 2x = -3 5x = 1

Set each factorequal to 0.Solve.

x + 6 = 02x + 3 = 05x - 1 = 0

A5x - 1 B A2x + 3 B Ax + 6 B = 0

A5x - 1 B A2x2 + 15x + 18 B = 0

A5x - 1 B A2x2 + 15x + 18 B = 0

-2

0 = 0 0 = 0

3 A8 B - 24 � 0 3 A-8 B + 24 � 0 0 = 0

3 A2 B 3 - 12 A2 B � 03 A-2 B 3 - 12 A-2 B � 03 A0 B 3 - 12 A0 B � 0

x = 2x = -2x = 0

-23x3 - 12x = 0

x = 2x = -2 x = 0

3x = 0 or x + 2 = 0 or x - 2 = 0

3x Ax + 2 B Ax - 2 B = 0

3x Ax2 - 4 B = 0 Factor out the GCF, 3x.Factor x2 - 4, a differ-ence of two squares.Set each factor equal to 0.Solve.

3x3 - 12x = 0

Focus On Mathematical ConnectionsNUMBER THEORY

By now, you have realized that being able to write a number as the productof prime numbers is very useful in the process of factoring polynomials. Youprobably also know at least a few numbers that are prime (such as 2, 3, and5). But what about the other prime numbers? When we come across a num-ber, how will we know if it is a prime number? Apparently, the ancientGreek mathematician Eratosthenes had a similar question because in thethird century B.C. he devised a simple method for identifying primes. Themethod is called the Sieve of Eratosthenes because it “sifts out” the primesin a list of numbers. The Sieve of Eratosthenes is generally considered to bethe most useful for identifying primes less than 1,000,000.

Here’s how the sieve works: suppose you want to find the prime numbers inthe first n natural numbers. Write the numbers, in order, from 2 to n. Weknow that 2 is prime, so circle it. Now, cross out each number greater than 2that is a multiple of 2. Consider the next number in the list that is notcrossed out. This number is 3, which we know is prime. Circle 3 and thencross out all multiples of 3 in the remainder of the list. Continue consideringeach uncircled number in the list. Once you have reached the largest primenumber less than or equal to , and you have eliminated all its multiples inthe remainder of the list, you can stop. Circle all the numbers left in the listthat have not yet been circled. Now all of the circled numbers in the list areprime numbers. The list below demonstrates the Sieve of Eratosthenes onthe numbers 2 through 30. Because , we need only check andeliminate the multiples of primes up to and including 5, which is the largestprime less than or equal to the square root of 30.

2 3 4 5 6 7 8 9 10~ ~ π ~ π ~ π π π11 12 13 14 15 16 17 18 19 20~ π ~ π π π ~ π ~ π21 22 23 24 25 26 27 28 29 30π π ~ π π π π π ~ π

We can see that the prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19,23, and 29.

GROUP ACTIVITY

Work with your group to identify the prime numbers less than 300 using theSieve of Eratosthenes. What is the largest prime number that you will needto check in this process?

130 L 5.477

1n

416

EXERCISE SET 5.6

Solve each equation. See Examples 1 through 3.

1. 2. 3.

4. 5.

Solve each equation. See Examples 4 through 8.

6. 7. 8.

9. 10. 11.

12. 13. 14.

15. 16. 17.

18. 19. 20. 3x2 + 19x - 72 = 0x A4x - 11 B = 3x A3x - 1 B = 14

x2 - 5x = 24x2 - 4x = 32x2 = 9

x2 = 16x2 + 15x = 0x2 + 20x = 0

x2 - 3x = 0x2 - 7x = 0x2 - 5x + 6 = 0

x2 + 2x - 8 = 0x2 + 2x - 63 = 0x2 - 13x + 36 = 0

B

Ax + 9 B Ax + 17 B = 0Ax + 4 B Ax - 10 B = 0

Ax - 6 B Ax - 7 B = 0Ax + 3 B Ax + 2 B = 0Ax - 2 B Ax + 1 B = 0

A

417


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

5.7 QUADRATIC EQUATIONS AND PROBLEM SOLVING

SOLVING PROBLEMS MODELED BY QUADRATIC EQUATIONS

Some problems may be modeled by quadratic equations. To solve theseproblems, we use the same problem-solving steps that were introduced inSection 2.5. When solving these problems, keep in mind that a solution ofan equation that models a problem may not be a solution to the problem.For example, a person’s age or the length of a rectangle is always a positivenumber. Thus we discard solutions that do not make sense as solutions ofthe problem.

Example 1 Finding Free-Fall TimeFor a TV commercial, a piece of luggage is dropped froma cliff 256 feet above the ground to show the durability ofthe luggage. Neglecting air resistance, the height h in feetof the luggage above the ground after t seconds is givenby the quadratic equation

Find how long it takes for the luggage to hit the ground.

Solution: 1. UNDERSTAND. Read and reread the problem. Thendraw a picture of the problem.

The equation models the height ofthe falling luggage at time t. Familiarize yourself withthis equation by finding the height of the luggage at

second and seconds.

When second, the height of the suitcase isfeet.

When seconds, the height of the suitcase isfeet.

2. TRANSLATE. To find how long it takes the luggageto hit the ground, we want to know the value of t forwhich the height .

3. SOLVE. We solve the quadratic equation byfactoring.

t = -4 t = 4

t + 4 = 0ort - 4 = 0

0 = -16 A t - 4 B A t + 4 B 0 = -16 A t2 - 16 B 0 = -16t2 + 256

0 = -16t2 + 256

h = 0

h = -16 A2 B 2 + 256 = 192t = 2

h = -16 A1 B 2 + 256 = 240t = 1

t = 2t = 1

h = -16t2 + 256

256 feet

h = -16t2 + 256

A

Quadratic Equations and Problem Solving SECTION 5.7 419

Objectives

Solve problems that can be mod-eled by quadratic equations.

A

Answer1. 3 seconds

SSM CD-ROM Video5.7

Practice Problem 1An object is dropped from the roof ofa 144-foot-tall building. Neglecting airresistance, the height h in feet of theobject above ground after t seconds isgiven by the quadratic equation

Find how long it takes the object to hitthe ground.

144feet

h = -16t2 + 144


Practice Problem 2The square of a number minus twicethe number is 63. Find the number.

Answers2. 9 and , 3. length 16 ft; width 11 ft=

=-7

4. INTERPRET. Since the time t cannot be negative, theproposed solution is 4 seconds.

Check: Verify that the height of the luggage when t is 4 secondsis 0.

When seconds, feet.

State: The solution checks and the luggage hits the ground4 seconds after it is dropped.

Example 2 Finding a NumberThe square of a number plus three times the number is70. Find the number.

Solution: 1. UNDERSTAND. Read and reread the problem. Sup-pose that the number is 5. The square of 5 is or 25.Three times 5 is 15. Then , not 70, so thenumber must be greater than 5. Remember, the pur-pose of proposing a number, such as 5, is to betterunderstand the problem. Now that we do, we will let

the number.2. TRANSLATE.

3. SOLVE.


Factor.

Solve.

4. INTERPRET.

Check: The square of is , or 100. Three times isor . Then , the correct

sum, so checks.

The square of 7 is or 49. Three times 7 is , or 21.Then , the correct sum, so 7 checks.

State: There are two numbers. They are and 7.

Example 3 Finding the Dimensions of a SailThe height of a triangular sail is 2 meters less than twicethe length of the base. If the sail has an area of 30 squaremeters, find the length of its base and the height.

Solution: 1. UNDERSTAND. Read and reread the problem. Sincewe are finding the length of the base and the height,we let

x = the length of the base

-10

49 + 21 = 703 A7 B72

-10100 + A-30 B = 70-303 A-10 B -10A-10 B 2-10

x = 7 x = -10

Set each factor equalto 0.x - 7 = 0orx + 10 = 0

Ax + 10 B Ax - 7 B = 0

x2 + 3x - 70 = 0

x2 + 3x = 70

70=3x+x2

the square of anumber

plus three timesthe number

is 70

x =

25 + 15 = 4052

-256 + 256 = 0h = -16 A4 B 2 + 256 =t = 4

Practice Problem 3The length of a rectangle is 5 feet morethan its width. The area of the rectan-gle is 176 square feet. Find the lengthand the width of the rectangle.


and since the height is 2 meters less than twice thebase,

An illustration is shown to the left.

2. TRANSLATE. We are given that the area of the tri-angle is 30 square meters, so we use the formula forarea of a triangle.

T T T T

3. SOLVE. Now we solve the quadratic equation.

Multiply.


Factor.

4. INTERPRET. Since x represents the length of thebase, we discard the solution . The base of a trian-gle cannot be negative. The base is then 6 feet and theheight is feet.

Check: To check this problem, we recall that base height

area, or

The required area

State: The base of the triangular sail is 6 meters and the heightis 10 meters.

The next example makes use of the Pythagorean theorem and consecu-tive integers. Before we review this theorem, recall that a right triangle is atriangle that contains a 90° or right angle. The hypotenuse of a right trian-gle is the side opposite the right angle and is the longest side of the trian-gle. The legs of a right triangle are the other sides of the triangle.

12

A6 B A10 B = 30

=#12

2 A6 B - 2 = 10

-5

x = -5 x = 6

Set each factorequal to 0.x + 5 = 0orx - 6 = 0

Ax - 6 B Ax + 5 B = 0

x2 - x - 30 = 0

30 = x2 - x

30 = 12

x A2x - 2 B

A2x - 2 B#x#12

=30

height#base#12=

area of triangle

2x - 2 = the height

Height= 2x − 2

Base = x

PYTHAGOREAN THEOREM

In a right triangle, the sum of the squares of the lengths of the two legsis equal to the square of the length of the hypotenuse.

Leg b

Leg a

Hypotenuse c

a2 + b2 = c2orA leg B 2 + A leg B 2 = Ahypotenuse B 2


Practice Problem 4Solve.

a. Find two consecutive odd integerswhose product is 23 more than theirsum.

b. The length of one leg of a right tri-angle is 7 meters less than thelength of the other leg. The lengthof the hypotenuse is 13 meters.Find the lengths of the legs.

Answers4. a. 5 and 7 or , b. 5 meters, 12 meters

-5 and -3

Study the following diagrams for a review of consecutive integers.

Consecutive integers:

If x is the first integer: x, ,

Consecutive even integers:

If x is the first even integer: x, ,

Consecutive odd integers:

If x is the first odd integer: x, ,

Example 4 Finding the Dimensions of a TriangleFind the lengths of the sides of a right triangle if thelengths can be expressed as three consecutive evenintegers.

Solution: 1. UNDERSTAND. Read and reread the problem. Let’ssuppose that the length of one leg of the right triangleis 4 units. Then the other leg is the next even integer,or 6 units, and the hypotenuse of the triangle is thenext even integer, or 8 units. Remember that thehypotenuse is the longest side. Let’s see if a trianglewith sides of these lengths forms a right triangle. Todo this, we check to see whether the Pythagorean the-orem holds true.

False.

Our proposed numbers do not check, but we now have abetter understanding of the problem.

We let x, , and be three consecu-tive even integers. Since these integers repre-sent lengths of the sides of a right triangle, wehave

x + 4 = hypotenuse A longest side B x + 2 = other leg

x = one leg

x + 4x + 2

x

x + 2

x + 4

52 = 64

16 + 36 � 644 units

6 units

8 units 42 + 62 � 82

x + 4x + 2

11

+2

139

+4

x + 4x + 2

12

+2

1410

+4

x + 2x + 1

6

+1

75

+2


2. TRANSLATE. By the Pythagorean theorem, we havethat

3. SOLVE. Now we solve the equation.

Multiply.

Combine like terms.


Factor.


4. INTERPRET. We discard since length cannot be negative. If , then and

.

Check: Verify that (hypotenuse) (leg) (leg) , or, or .

State: The sides of the right triangle have lengths 6 units,8 units, and 10 units.

10 units

8 units

6 units

100 = 36 + 64102 = 62 + 82

22 +2 =

x + 4 = 10x + 2 = 8x = 6

x = -2

x = -2 x = 6

x + 2 = 0orx - 6 = 0

Ax - 6 B Ax + 2 B = 0

x2 - 4x - 12 = 0

x2 + 8x + 16 = 2x2 + 4x + 4

x2 + 8x + 16 = x2 + x2 + 4x + 4

Ax + 4 B 2 = x2 + Ax + 2 B 2

Ax + 4 B 2 = Ax B 2 + Ax + 2 B 2 Ahypotenuse B 2 = A leg B 2 + A leg B 2

Focus On The Real WorldPRIME NUMBERS IN THE NEWS

Now that you have discovered a way to identify relatively small prime numbers with the Sieveof Eratosthenes, perhaps you are wondering whether there are very large primes. The answer isyes. Another ancient Greek mathematician, Euclid, proved that there are an infinite number ofprimes. Thus, there do exist huge numbers that are prime. Researchers call prime numbers withmore than 1000 digits, titanic primes.

Knowing that very large prime numbers exist and finding them and proving that they are infact prime are two very different things. David Slowinski, a research scientist at Silicon Graph-ics Inc.’s Cray Research unit and a discoverer or co-discoverer of seven titanic primes, com-pares the hunt for large primes with looking for a needle in a haystack. At the time that thisbook was written, the largest-known prime number, discovered November 13, 1996, had420,921 digits which would fill up about 13.5 newspaper pages in standard-size type. The follow-ing newspaper article from the San Jose Mercury News describes how this prime number wasfound.

San Jose Mercury NewsPublished: Nov. 23, 1996

“Move Over, Supercomputers”Net-linked band of off-the shelf PCusers finds largest example ofMersenne prime number

By Dan GillmorMercury News Computing Editor

In an undertaking previously re-served for powerful supercomputers,a worldwide band of personal com-puter users has found the largestknown example of a special kind ofprime number. Combining the Inter-net’s global reach with hundreds ofoff-the-shelf PCs running Intel micro-processors, they divided an enormousproblem into megabyte-sized tasks—and showed again how PCs are mov-ing rapidly up computing’sevolutionary scale.

Ten days ago in Paris, a PC oper-ated by 29-year-old programmer JoelArmengaud found what is now thebiggest-known Mersenne prime num-ber. Mersenne primes are namedafter a 17th-century French monk,Father Marin Mersenne, who was fas-cinated by mathematics.

To test whether the nearly421,000-digit number was such aprime, Armengaud was running aprogram written by Florida-basedprogrammer George Woltman. Earlythis year, Woltman organized what hedubbed “The Great InternetMersenne Prime Search,” a hunt thathas attracted more than 750 peoplefrom around the globe, some ofwhom have devoted more than onemachine to the task.

A prime number is an integergreater than zero whose divisors areonly itself and 1. (The number 2 isprime because it can only be dividedevenly by 1 and 2, for example.)Mersenne primes take the form 2 tosome power, minus 1—in otherwords, 2 multiplied by itself a certainnumber of times with 1 subtractedfrom the result.

The smallest Mersenne prime is 3,or 2 to the 2nd power minus 1. The next largest “regular”prime number is 5, and the nextlargest Mersenne prime is 7, or 2 tothe 3rd power minus 1.

The latest discovery—only the35th known Mersenne prime—is 2 to

the 1,398,269th power (2 multipliedby itself 1,398,269 times) minus 1. It’s420,921 digits long, or more than 150pages of single-spaced text.

The previous largest-knownMersenne prime, discovered earlierthis year, is 2 to the 1,257,787th powerminus 1. It was found by DavidSlowinski and Paul Gage, computerscientists at Silicon Graphics Inc.’sCray Research unit, using a Craysupercomputer.

While Armengaud, Woltman andothers in the search are in it mainlyfor fun, finding more efficient ways tocrunch huge numbers is more than anabstract exercise. Related techniqueshave helped people handle otherimportant computing tasks—such asmodeling complex weather patternsand exploring for oil. And cryptogra-phy, the art of scrambling messages toensure privacy, relies on the difficultyof performing complex operations onextremely large numbers.

And the way the latest Mersenneprime was found represents anotherstep in computing’s evolution.A2 * 2 * 2 = 8 B

A2 * 2 = 4 B

GROUP ACTIVITIES

1. The following World Wide Web site contains information on the current status of the largest-known prime numbers: http://www.utm.edu/research/primes/largest.html. Visit this site andreport on the five largest-known prime numbers. Be sure to include information about thenumbers themselves, who found them, when they were found, and how they were found (ifpossible). How many numbers larger than that found by Armengaud, Woltman, et. al. inNovember 1996 have been found?

2. Explore some of the links from the Web site listed in Activity 1 to find information on thehistory of the search for the largest-known primes (look for information on the largest knownprime by year). Summarize the trends in the methods used to hunt for titanic primes. Whenis a prime number with over 1,000,000 digits expected to be discovered?

424

425


EXERCISE SET 5.7

See Examples 1 through 4 for all exercises. Represent each given condition using a singlevariable, x.

A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

1. The length and width of a rectanglewhose length is 4 centimeters morethan its width

x

2. The length and width of a rectanglewhose length is twice its width

x

3. Two consecutive odd integers 4. Two consecutive even integers

5. The base and height of a trianglewhose height is one more than fourtimes its base

x

6. The base and height of a trapezoidwhose base is three less than fivetimes its height

base

x

Use the information given to find the dimensions of each figure.

7.

The area of the square is 121 squareunits. Find the length of its sides.

x

8.

The area of the rectangle is 84 squareinches. Find its length and width.

x + 3

x − 2

9.

The perimeter of the quadrilateral is120 centimeters. Find the lengths ofthe sides.

x + 5

x + 3

3x − 8

x2 − 3x

10.

The perimeter of the triangle is 85 feet.Find the lengths of its sides.

2x 2x + 5

x2 + 3


SECTION 5.1 THE GREATEST COMMON FACTOR

Factoring is the process of writing an expression as aproduct.

Factor: Factor: x2 + 5x + 6 = Ax + 2 B Ax + 3 B6 = 2 # 3

The GCF of a list of common variables raised topowers is the variable raised to the smallest expo-nent in the list.

The GCF of , and is .z3z10z5, z3

The GCF of a list of terms is the product of all com-mon factors.

Find the GCF of , and .

or 2x2y GCF = 2 # x2 # y50x2y3 = 2 # 5 # 5 # x2 # y310x3y2 = 2 # 5 # x3 # y2

8x2y = 2 # 2 # 2 # x2 # y

50x2y38x2y, 10x3y2

TO FACTOR BY GROUPING Factor: 10ax + 15a - 6xy - 9y

Step 1. Group the terms into two groups of twoterms.

Step 2. Factor out the GCF from each group.

Step 3. If there is a common binomial factor, factorit out.

Step 4. If not, rearrange the terms and try Steps 1–3again.

Step 1.Step 2.Step 3. A2x + 3 B A5a - 3y B

5a A2x + 3 B - 3y A2x + 3 BA10ax + 15a B + A-6xy - 9y B

SECTION 5.2 FACTORING TRINOMIALS OF THE FORM x2 � bx � c

The sum of these numbers is b.

The product of these numbers is c.

x2 + bx + c = Ax + n B Ax + n BFactor:

x2 + 7x + 12 = Ax + 3 B Ax + 4 B3 + 4 = 7 3 # 4 = 12

x2 + 7x + 12

SECTION 5.3 FACTORING TRINOMIALS OF THE FORM ax2 � bx � c

To factor , try various combinations offactors of and c until a middle term of bx isobtained when checking.

ax2ax2 + bx + c Factor:

Factors of : ,

Factors of : , 5 and 1, .

-1x+ 15x

14x

A3x - 1 B Ax + 5 BUu

Correct middle term

-5-1-5

x3x3x2

3x2 + 14x - 5


SECTION 5.4 FACTORING TRINOMIALS OF THE FORM ax2 � bx � c BY GROUPING

TO FACTOR ax2 + bx + c BY GROUPING

Step 1. Find two numbers whose product is and whose sum is b.

a # c

Factor:

Step 1. Find two numbers whose product isor and whose sum is 14. They

are 15 and .-1-153 # A-5 B

3x2 + 14x - 5

Step 2. Rewrite bx, using the factors found in step 1. Step 2. 3x2 + 14x - 5


Step 3.= Ax + 5 B A3x - 1 B= 3x Ax + 5 B - 1 Ax + 5 B= 3x2 + 15x - 1x - 5

SECTION 5.5 FACTORING PERFECT SQUARE TRINOMIALS AND THE DIFFERENCE OF TWO SQUARES

A perfect square trinomial is a trinomial that is thesquare of some binomial.

PERFECT SQUARE TRINOMIAL SQUARE OF BINO-MIAL

25x2 - 10x + 1 = A5x - 1 B 2x2 + 4x + 4 = Ax + 2 B 2

=

Factoring Perfect Square Trinomials

a2 - 2ab + b2 = Aa - b B 2a2 + 2ab + b2 = Aa + b B 2

Factor:

4x2 - 12x + 9 = A2x B 2 - 2 A2x # 3 B + 32 = A2x - 3 B 2

x2 + 6x + 9 = x2 + 2 Ax # 3 B + 32 = Ax + 3 B 2

Difference of Two Squares

a2 - b2 = Aa + b B Aa - b BFactor:

x2 - 9 = x2 - 32 = Ax + 3 B Ax - 3 B

SECTION 5.6 SOLVING QUADRATIC EQUATIONS BY FACTORING

A quadratic equation is an equation that can be writ-ten in the form with a not 0.

The form is called the standardform of a quadratic equation.

ax2 + bx + c = 0ax2 + bx + c = 0

Quadratic Equation Standard Form

2y2 + y - 5 = 0y = -2y2 + 5x2 - 16 = 0x2 = 16

Zero Factor PropertyIf a and b are real numbers and if , then

or .b = 0a = 0ab = 0

If , then or.x - 1 = 0

x + 3 = 0Ax + 3 B Ax - 1 B = 0

TO SOLVE QUADRATIC EQUATIONS BY FACTORING

Step 1. Write the equation in standard form so thatone side of the equation is 0.

Step 2. Factor completely.

Step 3. Set each factor containing a variable equalto 0.

Solve:

Step 1.Step 2.Step 3. or x - 4 = 03x - 1 = 0

A3x - 1 B Ax - 4 B = 0

3x2 - 13x + 4 = 0

3x2 = 13x - 4

Step 4. Solve the resulting equations. Step 4.

x = 13

x = 4 3x = 1

Step 5. Check in the original equation. Step 5. Check both and 4 in the original equation.13


SECTION 5.7 QUADRATIC EQUATIONS AND PROBLEM SOLVING

PROBLEM-SOLVING STEPS A garden is in the shape of a rectangle whose lengthis two feet more than its width. If the area of thegarden is 35 square feet, find its dimensions.

1. UNDERSTAND the problem. 1. Read and reread the problem. Guess a solutionand check your guess. Draw a diagram.

Let x be the width of the rectangular garden.Then is the length.

x + 2

x

x + 2

2. TRANSLATE. 2.

T T T35=x#Ax + 2 B

area=width#length

3. SOLVE. 3.

x = -7 x = 5

x - 5 = 0 or x + 7 = 0

Ax - 5 B Ax + 7 B = 0

x2 + 2x - 35 = 0

Ax + 2 Bx = 35

4. INTERPRET. 4. Discard the solution of since x representswidth.

Check: If x is 5 feet then feet. The area of a rectangle whose width is 5feet and whose length is 7 feet is (5 feet)(7 feet)or 35 square feet.

State: The garden is 5 feet by 7 feet.

x + 2 = 5 + 2 = 7

-7


429

6In this chapter, we expand our knowledge of algebraic expressions toinclude algebraic fractions, called rational expressions. We explore theoperations of addition, subtraction, multiplication, and division usingprinciples similar to the principles for number fractions.

6.1 Simplifying RationalExpressions

6.2 Multiplying and DividingRational Expressions

6.3 Adding and SubtractingRational Expressions with theSame Denominator and LeastCommon Denominator

6.4 Adding and SubtractingRational Expressions withDifferent Denominators

6.5 Solving Equations ContainingRational Expressions

6.6 Rational Equations andProblem Solving

6.7 Simplifying ComplexFractions

Rational expressions can be found in formulas for many real-world situations. Rainfall intensity describes the depth of rainthat falls during a certain time period. According to the NationalGeographic Society, the rainiest location in the world is MountWaialeale, Hawaii. Mount Waialeale receives an average of 460inches of rain per year. Most of the United States only receivesan average of 10 to 60 inches of rain per year.

Rainfall also varies in intensity from storm to storm. Mostrainfall can be classified on a scale ranging from 0.10 inches perhour (light rain) to 0.30 inches per hour (heavy rain). However,the intensities on this scale are occasionally exceeded by partic-ularly strong storms.

Rational Expressions C H A P T E R

6.1 SIMPLIFYING RATIONAL EXPRESSIONS

EVALUATING RATIONAL EXPRESSIONS

A rational number is a number that can be written as a quotient of integers.A rational expression is also a quotient; it is a quotient of polynomials.Examples are

, ,

Rational expressions have different numerical values depending on whatvalue replaces the variable.

Example 1 Find the numerical value of for each replacementvalue.

a. b.

Solution: a. We replace each in the expression with 5 and thensimplify.

b. We replace each in the expression with and thensimplify.

or

In the example above, we wrote as . For a negative fraction such

as , recall from Section 1.6 that

In general, for any fraction

This is also true for rational expressions. For example,

cNotice the parentheses.

IDENTIFYING WHEN A RATIONAL EXPRESSION IS UNDEFINED

In the preceding box, notice that we wrote for the denominator .The denominator of a rational expression must not equal 0 since division by

bb Z 0

B

d

- Ax + 2 Bx

= x + 2-x

= - x + 2x

2-7

= -27

= - 27

2-7

- 27

2-7

- 27

x + 42x - 3

= -2 + 42 A-2 B - 3

= 2-7

-2x

x + 42x - 3

= 5 + 42 A5 B - 3

= 910 - 3

= 97

x

x = -2x = 5

x + 42x - 3

5x2 - 3x + 23x + 7

-4pp3 + 2p + 1

3y3

8

A

Simplifying Rational Expressions SECTION 6.1 431

Objectives

Find the value of a rational expres-sion given a replacement number.

Identify values for which a rationalexpression is undefined.

Simplify, or write rational expres-sions in lowest terms.

C

B

A

RATIONAL EXPRESSION

A rational expression is an expression that can be written in the form

where P and Q are polynomials and .Q Z 0

PQ

Practice Problem 1

Find the value of for each

replacement value.

a.

b. x = -3

x = 4

x - 35x + 1

-ab

= a-b

= - ab

, b Z 0

Answers

1. a. , b. -6-14

= 37

121

SSM CD-ROM Video6.1

432 CHAPTER 6 Rational Expressions

Practice Problem 2Are there any values for x for whicheach rational expression is undefined?

a. b.

c.x2 - 3x + 2

5

x - 3x2 + 5x + 4

xx + 2

0 is not defined. This means we must be careful when replacing the variablein a rational expression by a number. For example, suppose we replace

x with 5 in the rational expression . The expression becomes

But division by 0 is undefined. Therefore, in this expression we can allowx to be any real number except 5. A rational expression is undefined for val-ues that make the denominator 0.

Example 2 Are there any values for x for which each expression isundefined?

a. b.

c.

Solution: To find values for which a rational expression is un-defined, we find values that make the denominator 0.

a. The denominator of is 0 when or

when . Thus, when , the expression

is undefined.b. We set the denominator equal to 0.

Factor.

or Set each factor equal to 0.

Solve.

Thus, when or , the denominatoris 0. So the rational expression

is undefined when or when .

c. The denominator of is never 0, so

there are no values of for which this expression isundefined.

SIMPLIFYING RATIONAL EXPRESSIONS

A fraction is said to be written in lowest terms or simplest form when the numerator and denominator have no common factors other than 1

. For example, the fraction is in lowest terms since the numera-tor and denominator have no common factors other than 1 .

The process of writing a rational expression in lowest terms or simplestform is called simplifying a rational expression. The following fundamentalprinciple of rational expressions is used to simplify a rational expression.

Aor -1 B710Aor -1 B

C

x

x3 - 6x2 - 10x3

x = 1x = 2x2 + 2

x2 - 3x + 2

x2 - 3x + 2x = 1x = 2

x = 1 x = 2

x - 1 = 0 x - 2 = 0

Ax - 2 B Ax - 1 B = 0

x2 - 3x + 2 = 0

xx - 3

x = 3x = 3

x - 3 = 0x

x - 3

x3 - 6x2 - 10x3

x2 + 2x2 - 3x + 2

xx - 3

2 + xx - 5

= 2 + 55 - 5

= 70

2 + xx - 5

Answers2. a. , b. , c. nox = -4, x = -1x = -2

FUNDAMENTAL PRINCIPLE OF RATIONAL EXPRESSIONS

If is a rational expression and R is a nonzero polynomial, then

PRQR

= PQ

PQ

Simplifying Rational Expressions SECTION 6.1 433

Simplifying a rational expression is similar to simplifying a fraction.

Simplify:

Simplify:

Thus, the rational expression has the same value as the ratio-

nal expression for all values of except 2 and . (Remember that

when is 2, the denominator of both rational expressions is 0 and when is , the original rational expression has a denominator of 0.)

As we simplify rational expressions, we will assume that the simplifiedrational expression is equal to the original rational expression for all realnumbers except those for which either denominator is 0. The followingsteps may be used to simplify rational expressions.

Example 3 Simplify:

Solution: To begin, we factor the numerator and denominator ifpossible. Then we apply the fundamental principle.

Example 4 Simplify:

Solution: We factor the numerator and denominator and thenapply the fundamental principle.

x2 + 8x + 7x2 - 4x - 5

=Ax + 7 B Ax + 1 BAx - 5 B Ax + 1 B =

x + 7x - 5

x2 + 8x + 7x2 - 4x - 5

5x - 5x3 - x2 =

5 Ax - 1 Bx2 Ax - 1 B =

5x2

5x - 5x3 - x2

TO SIMPLIFY A RATIONAL EXPRESSION

Step 1. Completely factor the numerator and denominator.

Step 2. Apply the fundamental principle of rational expressions todivide out common factors.

-3xx

-3xx - 3x - 2

x2 - 9x2 + x - 6

= x - 3x - 2

=Ax - 3 B Ax + 3 BAx - 2 B Ax + 3 B

Factor the numerator and the denominator.

Look for common factors.

Apply the fundamental principle.

x2 - 9x2 + x - 6

=Ax - 3 B Ax + 3 BAx - 2 B Ax + 3 B

x2 - 9x2 + x - 6

= 32 # 2

= 34

= 3 # 52 # 2 # 5

Factor the numerator and the denominator.

Look for common factors.


1520

= 3 # 52 # 2 # 5

1520

Practice Problem 3

Simplify: x4 + x3

5x + 5

Practice Problem 4

Simplify: x2 + 11x + 18

x2 + x - 2

Answers

3. , 4. x + 9x - 1

x3

5

Example 5 Simplify:

Solution: We factor the numerator and denominator and thenapply the fundamental principle.


Example 6 Simplify:

Solution: We factor and then apply the fundamental principle.

Example 7 Simplify each rational expression.

a. b.

Solution: a. The expression can be simplified by using the

commutative property of addition to rewrite thedenominator as .

b. The expression can be simplified by recognizing

that and are opposites. In other words,. We proceed as follows:

x - y

y - x=

1 # Ax - y BA-1 B Ax - y B =

1-1

= -1

y - x = -1 Ax - y Bx - yy - x

x - yy - x

x + yy + x

= x + yx + y

= 1

x + yy + x

x + yy + x

x - yy - x

x + yy + x

x + 9x2 - 81

= x + 9Ax + 9 B Ax - 9 B = 1

x - 9

x + 9x2 - 81

x2 + 4x + 4x2 + 2x

=Ax + 2 B Ax + 2 B

x Ax + 2 B =x + 2

x

x2 + 4x + 4x2 + 2x


HELPFUL HINT

When simplifying a rational expression, the fundamental principleapples to common factors, not common terms.

Common factors. These can be Common terms. Fundamental principledivided out. does not apply. This is in simplest form.

x + 2x

x # Ax + 2 Bx # x

=x + 2

x

Practice Problem 5

Simplify: x2 + 10x + 25

x2 + 5x

✓ CONCEPT CHECK

Recall that the fundamental principleapplies to common factors only.Which of the following are not true?Explain why.

a.

b.

c.

d.2x + 3

2= x + 3

3772

= 32

2x + 102

= x + 5

3 - 13 + 5

= - 15

Practice Problem 6

Simplify: x + 5

x2 - 25

Practice Problem 7Simplify each rational expression.

a. b.x - 44 - x

x + 44 + x

Answers

5. , 6. , 7. a. 1, b.

✓ Concept Check: a, c, d

-11

x - 5x + 5

x

435


EXERCISE SET 6.1

Simplify each expression. See Examples 3 through 7.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. 4x2 + 24xx + 6

2x2 + 3x - 22x - 1

x - 10x2 - 17x + 70

x + 7x2 + 5x - 14

12x2 + 4x - 12x + 1

5x2 + 11x + 2x + 2

x - 3x2 - 6x + 9

x + 5x2 - 4x - 45

7x + 35x2 + 5x

-5a - 5ba + b

y - 99 - y

x - 77 - x

y + 99 + y

x + 77 + x

3x - 124x - 16

2x - 103x - 30

x + 5x2 - 25

x - 2x2 - 4

39x + 6

28x + 16

C

ANSWERS

1.

2.

3.

4.

5.

6.

7.

7.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

6.2 MULTIPLYING AND DIVIDING RATIONAL EXPRESSIONS

MULTIPLYING RATIONAL EXPRESSIONS

Just as simplifying rational expressions is similar to simplifying numberfractions, multiplying and dividing rational expressions is similar to multi-plying and dividing number fractions.

Multiply: Multiply:

Multiply numerators and then multiply denominators.

Simplify by factoring numerators and denominators.


or

Example 1 Multiply.

a. b.

Solution: To multiply rational expressions, we first multiply thenumerators and then multiply the denominators of bothexpressions. Then we write the product in lowest terms.

a.

The expression is in lowest terms.

b. Multiply.

The expression is not in lowest terms, so we

factor the numerator and the denominator and apply thefundamental principle.

-7x2 # 3y5

5y # 14x2

-7x2

5y #

3y5

14x2 = -7x2 # 3y5

5y # 14x2

25x2y3

25x2

# 1y3 = 25x # 1

2 # y3 = 25x2y3

-7x2

5y #

3y5

14x225x

2 #

1y3

= 2x + 3

611

= 3 # 211

=Ax - 3 B # 2 Ax + 5 BAx + 5 B Ax + 3 B Ax - 3 B= 3 # 2 # 5

5 # 11

x - 3x + 5

# 2x + 10x2 - 9

=Ax - 3 B # A2x + 10 BAx + 5 B # Ax2 - 9 B

35

# 1011

= 3 # 105 # 11

x - 3x + 5

# 2x + 10x2 - 9

35

# 1011

A

Multiplying and Dividing Rational Expressions SECTION 6.2 437

Objectives

Multiply rational expressions.

Divide rational expressions.

Multiply and divide rational ex-pressions.

Converting between units of mea-sure.

D

CBA

MULTIPLYING RATIONAL EXPRESSIONS

If and are rational expressions, then

To multiply rational expressions, multiply the numerators and thenmultiply the denominators.

PQ

# RS

= PRQS

RS

PQ

Practice Problem 1Multiply.

a. b.-5a3

3b3 # 2b2

15a16y

3 #

1x2

Answers

1. a. , b. - 2a2

9b16y3x2

SSM CD-ROM Video6.2

When multiplying rational expressions, it is usually best to first factoreach numerator and denominator. This will help us when we apply the fun-damental principle to write the product in lowest terms.

Example 2 Multiply:

Solution:

Multiply.


The following steps may be used to multiply rational expressions.


Example 3 Multiply:

Solution:

Factor.

Multiply.

Simplify. =3 Ax + 1 B

5x A2x - 3 B

=3 Ax + 1 B A2x + 3 B Ax - 1 B

5x Ax - 1 B A2x - 3 B A2x + 3 B

3x + 35x2 - 5x

# 2x2 + x - 3

4x2 - 9=

3 Ax + 1 B5x Ax - 1 B #

A2x + 3 B Ax - 1 BA2x - 3 B A2x + 3 B

3x + 35x2 - 5x

# 2x2 + x - 3

4x2 - 9

TO MULTIPLY RATIONAL EXPRESSIONS

Step 1. Completely factor numerators and denominators.

Step 2. Multiply numerators and multiply denominators.

Step 3. Simplify or write the product in lowest terms by applying thefundamental principle to all common factors.

= 25

=x Ax + 1 B # 2 # 3

3x # 5 Ax + 1 B

Factor numerators anddenominators.

x2 + x3x

# 6

5x + 5=

x Ax + 1 B3x

# 2 # 3

5 Ax + 1 B

x2 + x3x

# 6

5x + 5

= - 3y4

10

= -1 # 7 # 3 # x2 # y # y4

5 # 2 # 7 # x2 # y


Practice Problem 2

Multiply: 6x + 6

7 #

14x2 - 1

✓ CONCEPT CHECK

Which of the following is a true state-ment?

a. b.

c.

d.x7

# x + 5

4= 2x + 5

28

3x

# 12

= 32x

2x

# 5 x

= 10x

13

# 12

= 15

Practice Problem 3

Multiply: 4x + 8

7x2 - 14x #

3x2 - 5x - 29x2 - 1

Answers

2. , 3.


4 Ax + 2 B7x A3x - 1 B

12x - 1

DIVIDING RATIONAL EXPRESSIONS

We can divide by a rational expression in the same way we divide by a num-ber fraction. Recall that to divide by a fraction, we multiply by its reciprocal.

For example, to divide by , we multiply by .

Example 4 Divide:

Solution: Multiply by the reciprocal of .

Simplify.

Example 5 Divide:

Solution:

of .

Simplify. = x - 14

Factor andmultiply. = Ax - 1 B Ax + 2 B # 5

5 # 2 # 2 # Ax + 2 B

2x + 45

Ax - 1 B Ax + 2 B10

,2x + 45

=Ax - 1 B Ax + 2 B

10 #

52x + 4

Multiply by thereciprocal

Ax - 1 B Ax + 2 B10

,2x + 45

= 3y2

160

= 3 x3y2

160 x3

4x3

y23x3

40,4x3

y2 = 3x3

40 #

y2

4x3

3x3

40,4x3

y2

32

,78

= 32

# 87

= 3 # 4 # 22 # 7

= 127

87

32

78

32

B


HELPFUL HINT

Don’t forget how to find reciprocals. The reciprocal of ,

.a Z 0, b Z 0

ab

is ba

DIVIDING RATIONAL EXPRESSIONS

If and are rational expressions and is not 0, then

To divide two rational expressions, multiply the first rational expres-sion by the reciprocal of the second rational expression.

PQ

,RS

= PQ

# SR

= PSQR

RS

RS

PQ

Practice Problem 4

Divide: 7x2

6, x

2y

Practice Problem 5

Divide: A2x + 3 B Ax - 4 B

6,3x - 12

2

Answers

4. , 5. 2x + 39

7xy3

The following may be used to divide by a rational expression.

Example 6 Divide:

Solution:

Factor and multiply.

Simplify.

Example 7 Divide:

Solution:

Factor and multiply.

or 1 Simplify.

MULTIPLYING AND DIVIDING RATIONAL EXPRESSIONS

Let’s make sure that we understand the difference between multiplying anddividing rational expressions.

Rational Expressions

Multiplication Multiply the numerators and multiply the denominators.Division Multiply by the reciprocal of the divisor.

Example 8 Multiply or divide as indicated.

a. b.

Solution: a.

b. x - 45

, xx - 4

= x - 45

# x - 4

x=Ax - 4 B 2

5x

x - 45

# x

x - 4=Ax - 4 B # x

5 # Ax - 4 B = x5

x - 45

, xx - 4

x - 45

# x

x - 4

C

= 11

=A2x - 1 B Ax - 5 B # 2 # 5

5 Ax - 5 B # 2 A2x - 1 B

Multiply by thereciprocal.

2x2 - 11x + 55x - 25

,4x - 210

= 2x2 - 11x + 55x - 25

# 10

4x - 2

2x2 - 11x + 55x - 25

,4x - 210

= 2x Ax + 1 B

=2 A3x + 1 B Ax - 1 B

Ax + 1 B Ax - 1 B # x A3x + 1 B

Multiply by thereciprocal.

6x + 2x2 - 1

,3x2 + xx - 1

= 6x + 2x2 - 1

# x - 1

3x2 + x

6x + 2x2 - 1

,3x2 + xx - 1

TO DIVIDE BY A RATIONAL EXPRESSION

Multiply by its reciprocal.

Practice Problem 8Multiply or divide as indicated.

a.

b.x + 3

x, 7

x + 3

x + 3x

# 7

x + 3


Answers

6. , 7. 1, 8. a. , b.Ax + 3 B 2

7x7x

2x2 Ax - 2 B

Practice Problem 7

Divide: 3x2 - 10x + 8

7x - 14,9x - 12

21

Practice Problem 6

Divide: 10x + 4x2 - 4

,5x3 + 2x2

x + 2

CONVERTING BETWEEN UNITS OF MEASURE

Now that we know how to multiply fractions and rational expressions, wecan use this knowledge to help us convert between units of measure. To doso, we will use unit fractions. A unit fraction is a fraction that equals 1. Forexample, since 12 in. 1 ft, we have the unit fractions

and

Example 9 Converting from Square Yards to Square FeetThe largest casino in the world is the Foxwoods ResortCasino in Ledyard, CT. The gaming area for this casino is21,444 square yards. Find the size of the gaming area insquare feet. (Source: The Guinness Book of Records,1996)

Solution: There are 9 square feet in 1 square yard.

Unit fraction

L 193,000 square feet

= 192,996 square feet

21,444 square yards = 21,444 sq. yd # 9 sq. ft1 sq. yd

1 square yard= 9 square feet

1 ydor 3 ft

1 yd or3 ft

d

1 ft12 in.

= 112 in.1 ft

= 1

=

D


HELPFUL HINT

When converting among units of measurement, if possible, write theunit fraction so that the numerator contains units converting to andthe denominator contains original units.

Unit fraction

dd

= 4812

ft = 4 ft

Units converting toOriginal units

48 ft = 48 in.1

# 1 ft

12 in.

d

Answer

9. 4444.44 sq. yards

Practice Problem 9Convert 40,000 square feet to squareyards.

Example 10 Converting from Feet per Second to Miles per HourFlorence Griffith Joyner holds the women’s record forspeed. She has been timed at 36 feet per second. Convertthis to miles per hour. (Source: The Guinness Book ofRecords, 1996)

Solution: Recall that 1 mile 5280 feet and 1 hour 3600 seconds .

Unit fractionsb R

L 24.5 miles>hour Arounded to the nearest tenth B = 36 # 3600

5280 miles>hour

36 feet>second = 36 feet1 second

# 3600 seconds

1 hour #

1 mile5280 feet

ddA60 # 60 B=

=


Practice Problem 10Carl Lewis holds the men’s record forspeed. He has been timed at 39.5 feetper second. Convert this to miles perhour. (Source: The Guinness Book ofRecords, 1996)

Answer

10. 26.9 miles/hour

443


MENTAL MATH

Find each product. See Example 1.

1. 2. 3.

4. 5.

EXERCISE SET 6.2

Find each product and simplify if possible. See Examples 1 through 3.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10.

Find each quotient and simplify. See Examples 4 through 7.

11. 12. 13. 14.

15.

Multiply or divide as indicated. See Example 8.

16. 17.

18. 19.

20. 76p2 + q

, 1418p2 + 3q

3x2 + 12x6

# 9

2x + 8x2 + 5x

8 #

93x + 15

6x + 65

,3x + 310

5x - 1012

,4x - 88

C

Ax - 6 B Ax + 4 B4x

,2x - 128x2

7a2b3ab2 ,21a2b2

14ab8x2

y3 ,4x2y3

69y4

6y,y2

35x7

2x5 ,10x4x3

B

m2 - n2

m + n #

mm2 - mn

x2 + x8

# 16

x + 16x + 6

5 #

1036x + 36

4x - 2420x

# 5

x - 6

x2x - 14

# x2 - 7x

5- 5a2b

30a2b2 # b3 6x2

10x3 # 5x12

8x2

# x5

4x29x2

y #

4y3x2

3xy2 #

7y4x

A

9x

# x5

x5

11 #

4z3

57

# y2

x23x4

# 1y

2y

# x3

MENTAL MATH ANSWERS

1.

2.

3.

4.

5.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

6.3 ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH

THE SAME DENOMINATOR AND LEAST COMMON DENOMINATOR

ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH THE

SAME DENOMINATOR

Like multiplication and division, addition and subtraction of rationalexpressions is similar to addition and subtraction of rational numbers. Inthis section, we add and subtract rational expressions with a commondenominator.

Add: Add:

Add the numerators and place the sum over the common denominator.

Simplify. Simplify.

Example 1 Add:

Solution: Add the numerators.

Example 2 Subtract:

Solution: Subtract the numerators.

Simplify.

Example 3 Subtract: 3x2 + 2x

x - 1- 10x - 5

x - 1

= 11

or 1

2y2y - 7

- 72y - 7

= 2y - 72y - 7

2y2y - 7

- 72y - 7

= 3mn

Simplify the numerator by combininglike terms.

Simplify by applying the fundamentalprinciple.

= 6m2n

5m2n

+ m2n

= 5m + m2n

5m2n

+ m2n

= 12x + 2

= 85

9x + 2

+ 3x + 2

= 9 + 3x + 2

65

+ 25

= 6 + 25

9x + 2

+ 3x + 2

65

+ 25

A

Adding and Subtracting Rational Expressions SECTION 6.3 445

Objectives

Add and subtract rational expres-sions with common denominators.

Find the least common denomina-tor of a list of rational expressions.

Write a rational expression as anequivalent expression whose de-nominator is given.

C

B

A

SSM CD-ROM Video6.3

Practice Problem 1

Add: 8x3y

+ x3y

Practice Problem 2

Subtract: 3x

3x - 7- 7

3x - 7

Answers

1. , 2. 1, 3. 2x - 33xy

Practice Problem 3

Subtract: 2x2 + 5x

x + 2- 4x + 6

x + 2

ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH

COMMON DENOMINATORS

If and are rational expressions, then

To add or subtract rational expressions, add or subtract numeratorsand place the sum or difference over the common denominator.

PR

- QR

= P - QR

andPR

+ QR

= P + QR

QR

PR


Solution:

Combine like terms.

Factor.

Simplify.

FINDING THE LEAST COMMON DENOMINATOR

Recall from Chapter R that to add and subtract fractions with differentdenominators, we first find a least common denominator (LCD). Then wewrite all fractions as equivalent fractions with the LCD.

For example, suppose we add and . The LCD of denominators 3 and5 is 15, since 15 is the smallest number that both 3 and 5 divide into evenly.So we rewrite each fraction so that its denominator is 15. (Notice how weapply the fundamental principle.)

To add or subtract rational expressions with different denominators, wealso first find an LCD and then write all rational expressions as equivalentexpressions with the LCD. The least common denominator (LCD) of a listof rational expressions is a polynomial of least degree whose factors includeall the factors of the denominators in the list.

Example 4 Find the LCD for each pair.

a. b. 75x

, 6

15x218

, 322

TO FIND THE LEAST COMMON DENOMINATOR (LCD)

Step 1. Factor each denominator completely.

Step 2. The least common denominator (LCD) is the product of allunique factors found in Step 1, each raised to a power equal tothe greatest number of times that the factor appears in any onefactored denominator.

83

+ 25

=8 A5 B3 A5 B +

2 A3 B5 A3 B = 40

15+ 6

15= 40 + 6

15= 46

15

25

83

B

= 3x - 5

=Ax - 1 B A3x - 5 B

x - 1

= 3x2 - 8x + 5x - 1

Use the distributiveproperty. = 3x2 + 2x - 10x + 5

x - 1

Subtract the numerators.Notice the parentheses.

3x2 + 2xx - 1

- 10x - 5x - 1

=3x2 + 2x - A10x - 5 B

x - 1

HELPFUL HINT

Notice how the numerator has been subtracted in Example 3.

This sign applies to the So parentheses are insertedentire numerator of . here to indicate this.

T T T

3x2 + 2xx - 1

- 10x - 5x - 1

=3x2 + 2x - A10x - 5 B

x - 1

10x - 5-

10x - 5

Practice Problem 4Find the LCD for each pair.

a. b.5

6x3, 118x5

29

, 715

Answers4. a. 45, b. 24x5

Adding and Subtracting Rational Expressions SECTION 6.3 447

Practice Problem 5

Find the LCD of and .7a

a - 53a

a + 5

Practice Problem 6

Find the LCD of and

.5x

3x - 12

7x2

Ax - 4 B 2

Solution: a. We start by finding the prime factorization of eachdenominator.

Next we write the product of all the unique factors, eachraised to a power equal to the greatest number of timesthat the factor appears.

The greatest number of times that the factor 2appears is 3.


b. We factor each denominator.



The greatest number of times that the factor xappears is 2.

Example 5 Find the LCD of and .

Solution: The denominators and are completelyfactored already. The factor appears once and thefactor appears once.


Solution: We factor each denominator.

This denominator is already factored.

The greatest number of times that the factor 3 appearsis 1.

The greatest number of times that the factor appears in any one denominator is 2.



Solution:

LCD = A t - 3 B A t + 2 B A t + 1 B t2 + 3t + 2 = A t + 1 B A t + 2 B

t2 - t - 6 = A t - 3 B A t + 2 Bt + 5

t2 + 3t + 2t - 10

t2 - t - 6

LCD = 3 Am + 5 B 2m + 5

Am + 5 B 2 = Am + 5 B 23m + 15 = 3 Am + 5 B

2Am + 5 B 2

6m2

3m + 15

LCD = Ax + 2 B Ax - 2 Bx - 2

x + 2x - 2x + 2

5x2

x - 27x

x + 2

LCD = 31 # 51 # x2 = 15x2

15x2 = 3 # 5 # x2and5x = 5 # x

LCD = 23 # 111 = 8 # 11 = 88

22 = 2 # 11and8 = 2 # 2 # 2 = 23

✓ CONCEPT CHECK

Choose the correct LCD of

and .

a. b.

c. d. 5x Ax + 1 B 2Ax + 1 B 3Ax + 1 B 2x + 1

5x + 1

xAx + 1 B 2

Practice Problem 7

Find the LCD of and

.y + 4

y2 - 3y + 2

y + 5y2 + 2y - 3

Answers5. , 6. ,7.


Ay + 3 B Ay - 2 B Ay - 1 B3 Ax - 4 B 2Aa + 5 B Aa - 5 B


Practice Problem 8

Find the LCD of and .9

4 - x6

x - 4

Practice Problem 9Write the rational expression as anequivalent rational expression with thegiven denominator.

2x5y

=20x2y2


Solution: The denominators and are opposites. Thatis, . We can use either or

as the LCD.

WRITING EQUIVALENT RATIONAL EXPRESSIONS

Next we practice writing a rational expression as an equivalent rationalexpression with a given denominator. To do this, we apply the fundamen-

tal principle, which says that , or equivalently that . This

can be seen by recalling that multiplying an expression by 1 produces anequivalent expression. In other words,

Example 9 Write the rational expression as an equivalent rationalexpression with the given denominator.

Solution: We can ask ourselves: “What do we multiply by to get?” The answer is , since . So we

multiply the numerator and denominator by .

Example 10 Write the rational expression as an equivalent rationalexpression with the given denominator.

Solution: First we factor the denominator as. If we multiply the original denominatorby , the result is the new

denominator . Thus, we multiplythe numerator and the denominator by .

= 5x - 20Ax - 2 B Ax + 2 B Ax - 4 B

5x2 - 4

= 5Ax - 2 B Ax + 2 B =

5 Ax - 4 BAx - 2 B Ax + 2 B Ax - 4 B

x - 4Ax - 2 B Ax + 2 B Ax - 4 Bx - 4Ax - 2 B Ax + 2 BAx - 2 B Ax + 2 B x2 - 4

5x2 - 4

= Ax - 2 B Ax + 2 B Ax - 4 B

4b9a

=4b A3ab B9a A3ab B = 12ab2

27a2b

3ab9a A3ab B = 27a2b3ab27a2b

9a

4b9a

=27a2b

PQ

= PQ

# 1 = PQ

# RR

= PRQR

PQ

= PRQR

PRQR

= PQ

C

LCD = x - 2 or LCD = 2 - x

2 - xx - 22 - x = -1 Ax - 2 B 2 - xx - 2

102 - x

2x - 2

Practice Problem 10Write the rational expression as anequivalent rational expression with thegiven denominator.

3x2 - 25

= Ax + 5 B Ax - 5 B Ax - 3 B

Answers

8. or , 9. ,

10. 3x - 9Ax + 5 B Ax - 5 B Ax - 3 B

8x3y20x2y2A4 - x BAx - 4 B

MENTAL MATH

Perform each indicated operation.

1. 2. 3.

4. 5. 6.

EXERCISE SET 6.3

Add or subtract as indicated. Simplify the result if possible. See Examples 1 through 3.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11.

12. 13.

14. 3x - 1x2 + 5x - 6

- 2x - 7x2 + 5x - 6

2x + 3x2 - x - 30

- x - 2x2 - x - 30

3yy2 + 3y - 10

- 6y2 + 3y - 10

aa2 + 2a - 15

- 3a2 + 2a - 15

x2 + 9xx + 7

- 4x + 14x + 7

5x + 4x - 1

- 2x + 7x - 1

9y + 9

+ yy + 9

93 + y

+ y + 13 + y

8yy - 2

- 16y - 2

4mm - 6

- 24m - 6

3p2

+ 11p2

4m3n

+ 5m3n

x + 17

+ 67

a13

+ 913

A

7y5

+ 10y5

89

- 79

3y8

+ 2y8

3x9

+ 4x9

511

+ 111

23

+ 13

449


1.

2.

3.

4.

5.

6.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

6.4 ADDING AND SUBTRACTING RATIONAL EXPRESSIONS

WITH DIFFERENT DENOMINATORS

ADDING AND SUBTRACTING RATIONAL EXPRESSIONS


In the previous section, we practiced all the skills we need to add and sub-tract rational expressions with different denominators. The steps are as fol-lows:

Example 1 Perform each indicated operation.

a. b.

Solution: a. First, we must find the LCD. Since and ,the . Next we write each fraction as anequivalent fraction with the denominator 8, then wesubtract.

ƒ———————T

ƒ————————c

b. Since and , theLCD . We write each fraction asan equivalent fraction with a denominator of .

Example 2 Subtract:

Solution: Since , the . We write equivalent

expressions with the LCD as denominators.LCD = Ax - 2 B Ax + 2 Bx2 - 4 = Ax + 2 B Ax - 2 B

6xx2 - 4

- 3x + 2

Add numerators. Write the sumover the common denominator. = 15 + 14x

50x2

= 1550x2 + 14x

50x2

310x2 + 7

25x=

3 A5 B10x2 A5 B +

7 A2x B25x A2x B

50x2= 2 # 52 # x2 = 50x2

25x = 5 # 5 # x10x2 = 2 # 5 # x # x

a4

- 2a8

=a A2 B4 A2 B - 2a

8= 2a

8- 2a

8= 2a - 2a

8= 0

8= 0

LCD = 23 = 88 = 234 = 22

310x2 + 7

25xa4

- 2a8

TO ADD OR SUBTRACT RATIONAL EXPRESSIONS


Step 1. Find the LCD of the rational expressions.

Step 2. Rewrite each rational expression as an equivalent expressionwhose denominator is the LCD found in Step 1.

Step 3. Add or subtract numerators and write the sum or differenceover the common denominator.

Step 4. Simplify or write the rational expression in lowest terms.

A

Adding and Subtracting Rational Expressions with Different Denominators SECTION 6.4 451

Objective

Add and subtract rational expres-sions with different denominators.

A

Practice Problem 1Perform each indicated operation.

a. b.5

8x+ 11

10x2y5

- 3y15

Answers

1. a. 0, b. , 2. 5x - 3

25x + 4440x2

SSM CD-ROM Video6.4

Practice Problem 2

Subtract: 10x

x2 - 9- 5

x + 3


Practice Problem 3

Add: 5

7x+ 2

x + 1

Practice Problem 4

Subtract: 10

x - 6- 15

6 - x

Next we factor the numerator to see if this rationalexpression can be simplified.

Factor.

Apply the fundamental principle to simplify.

Example 3 Add:

Solution: The LCD is . We write each rational expressionas an equivalent rational expression with a denominatorof .

Example 4 Subtract:

Solution: To find a common denominator, we notice that and are opposites. That is, .We write the denominator as andsimplify.

Apply .

Example 5 Add:

Solution: Recall that 1 is the same as . The LCD of and

is .m + 1

mm + 1

11

11

1 + mm + 1

= 16x - 3

Subtract numerators. Write the dif-ference over the common denominator. = 7 - A-9 B

x - 3

a-b

= -ab

= 7x - 3

- - 9x - 3

7x - 3

- 93 - x

= 7x - 3

- 9- Ax - 3 B

- Ax - 3 B3 - x3 - x = - Ax - 3 B3 - x

x - 3

7x - 3

- 93 - x

= 17t + 23t A t + 1 B

= 2t + 2 + 15t3t A t + 1 B

Add numerators. Write the sum overthe common denominator.

Apply the distributive property in thenumerator.

Combine like terms in the numerator.

=2 A t + 1 B + 5 A3t B

3t A t + 1 B

23t

+ 5t + 1

=2 A t + 1 B3t A t + 1 B +

5 A3t BA t + 1 B A3t B

3t A t + 1 B3t A t + 1 B

23t

+ 5t + 1

= 3x - 2

=3 Ax + 2 B

Ax + 2 B Ax - 2 B

= 3x + 6Ax + 2 B Ax - 2 B

= 6x - 3x + 6Ax + 2 B Ax - 2 B

Subtract numerators. Write the differenceover the common denominator.

Apply the distributive property in thenumerator.


=6x - 3 Ax - 2 BAx + 2 B Ax - 2 B

6xx2 - 4

- 3x + 2

= 6xAx - 2 B Ax + 2 B -

3 Ax - 2 BAx + 2 B Ax - 2 B

Practice Problem 5

Add: 2 + xx + 5

Answers

3. , 4. , 5. 3x + 10x + 5

25x - 6

19x + 57x Ax + 1 B

Adding and Subtracting Rational Expressions with Different Denominators SECTION 6.4 453

Practice Problem 6

Subtract: 4

3x2 + 2x- 3x

12x + 8

Practice Problem 7

Add: 6x

x2 + 4x + 4+ x

x2 - 4

Write 1 as .

Multiply both the numerator and the denominator of by .


Example 6 Subtract:

Solution: First, we factor the denominators.

The LCD is . We write equivalent expres-sions with denominators of .

Example 7 Add:

Solution: First we factor the denominators.

Now we write the rational expressions as equivalent expressions withdenominators of , the LCD.

or

The numerator was factored as a last step to see if therational expression could be simplified further. Sincethere are no factors common to the numerator and thedenominator, we can’t simplify further.

x A3x - 1 BAx + 1 B 2 Ax - 1 B= 3x2 - x

Ax + 1 B 2 Ax - 1 B

Apply the distributive property in the numerator.= 2x2 - 2x + x2 + xAx + 1 B 2 Ax - 1 B

Add numerators. Write the sum overthe common denominator.

=2x Ax - 1 B + x Ax + 1 BAx + 1 B 2 Ax - 1 B

=2x Ax - 1 B

Ax + 1 B Ax + 1 B Ax - 1 B +x Ax + 1 B

Ax + 1 B Ax - 1 B Ax + 1 B

Ax + 1 B Ax + 1 B Ax - 1 B

2xx2 + 2x + 1

+ xx2 - 1

= 2xAx + 1 B Ax + 1 B + x

Ax + 1 B Ax - 1 B

2xx2 + 2x + 1

+ xx2 - 1

Subtract numerators. Write the differ-ence over the common denominator.= 9 - 2x2

3x A2x + 1 B

=3 A3 B

x A2x + 1 B A3 B -2x Ax B

3 A2x + 1 B Ax B

3x A2x + 1 B3x A2x + 1 B

32x2 + x

- 2x6x + 3

= 3x A2x + 1 B - 2x

3 A2x + 1 B

32x2 + x

- 2x6x + 3

= 2m + 1m + 1

Add numerators. Write the sum over thecommon denominator. = m + 1 + m

m + 1

m + 111

=1 Am + 1 B1 Am + 1 B + m

m + 1

111 + m

m + 1= 1

1+ m

m + 1

Answers

6. , 7.x A7x - 10 B

Ax + 2 B 2 Ax - 2 B16 - 3x2

4x A3x + 2 B

Focus On Business and CareerFAST-GROWING CAREERS

According to U.S. Bureau of Labor Statistics projections, the careers listed below will have thelargest job growth into the next century.

Employment[Numbers in thousands]

Occupation 1994 2005 Change

1. Cashiers 3,005 3,567

2. Janitors and cleaners, including maids andhousekeeping cleaners 3,043 3,602

3. Salespersons, retail 3,842 4,374

4. Waiters and waitresses 1,847 2,326

5. Registered nurses 1,906 2,379

6. General managers and top executives 3,046 3,512

7. Systems analysts 483 928

8. Home health aides 420 848

9. Guards 867 1,282

10. Nursing aides, orderlies, and attendants 1,265 1,652

11. Teachers, secondary 1,340 1,726

12. Marketing and sales worker supervisors 2,293 2,673

13. Teacher’s aides and educational associates 932 1,296

14. Receptionists and information clerks 1,019 1,337

15. Truck drivers, light and heavy 2,565 2,837

16. Secretaries, except legal and medical 2,842 3,109

17. Clerical supervisors and managers 1,340 1,600

18. Child care workers 757 1,005

19. Maintenance repairers, general utility 1,273 1,505

20. Teachers, elementary 1,419 1,639

Source: Bureau of Labor Statistics, Office of Employment Projections, November 1995

What do all of these in-demand occupations have in common? They all require a knowledge ofmath! For some careers like cashiers, salespersons, waiters and waitresses, financial managers,and computer engineers, the ways math is used on the job may be obvious. For other occupa-tions, the use of math may not be quite as obvious. However, tasks common to many jobs likefilling in a time sheet, writing up an expense or mileage report, planning a budget, figuring abill, ordering supplies, completing a packing list, and even making a work schedule all requiremath.

CRITICAL THINKING

Suppose that your college placement office is planning to publish an occupational handbook onmath in popular occupations. Choose one of the occupations from the list above that interestsyou. Research the occupation. Then write a brief entry for the occupational handbook thatdescribes how a person in that career would use math in his or her job. Include an example ifpossible.

+220

+231

+248

+261

+267

+271

+318

+364

+380

+386

+387

+415

+428

+445

+466

+473

+479

+532

+559

+562

454

EXERCISE SET 6.4

Perform each indicated operation. Simplify if possible. See Example 1.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10.

Perform each indicated operation. Simplify if possible. See Examples 4 through 7.

11. 12. 13.

14. 15. 16.

17. 18. 19.

20. 6x

- 1

2m

+ 11

y + 5+ 2

3y4

5b+ 1

b - 1

5xAx - 2 B 2 - 3

x - 21

x + 3- 1Ax + 3 B 2

5x6

+ 15x2

2

3x4

x- 4x2

x27x

x - 3- 4x + 9

x - 35x

x + 2- 3x - 4

x + 2

9x - 3

+ 93 - x

6x - 3

+ 83 - x

5y2 - y

2y + 13

4x+ 8

x - 2

3x + 2

- 1x2 - 4

3x

+ 52x2

4cd

- 8x5

15ab

+ 6b5

157a

+ 86a

42x

+ 93x

A

455


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

6.5 SOLVING EQUATIONS CONTAINING RATIONAL EXPRESSIONS

SOLVING EQUATIONS CONTAINING RATIONAL EXPRESSIONS

In Chapter 2, we solved equations containing fractions. In this section, wecontinue the work we began in Chapter 2 by solving equations containingrational expressions. For example,

and

are equations containing rational expressions. To solve equations such asthese, we use the multiplication property of equality to clear the equationof fractions by multiplying both sides of the equation by the LCD.

Example 1 Solve:

Solution: The LCD of denominators 2, 3, and 6 is 6, so we multiplyboth sides of the equation by 6.


ƒ ƒ ƒMultiply and simplify.



Check: To check, we replace x with in the original equation.

Replace x with .

True.

This number checks, so the solution is .

Example 2 Solve:

Solution: The LCD of denominators 2, 9, and 18 is 18, so wemultiply both sides of the equation by 18.

ƒ ƒ ƒ

Solve for t. t = 5

7t = 35

7t - 30 = 5

9t - 36 - 2t + 6 = 5

9 A t - 4 B - 2 A t - 3 B = 5HELPFUL HINT

Multiplyeach termby 18.


Simplify.Use the distributiveproperty.

Combine like terms.

18 a t - 42b - 18 a t - 3

9b = 18 a 5

18b

18a t - 42

- t - 39b = 18 a 5

18b

t - 42

- t - 39

= 518

-5

16

= 16

-5-52

+ 83

�16

-5

x = -5

3x = -15

3 # x + 16 = 1

6 a x2b + 6 a 8

3b = 6 a 1

6b

6 a x2

+ 83b = 6 a 1

6b

HELPFUL HINT

Make surethat eachterm is mul-tiplied bythe LCD.

x2

+ 83

= 16

x + 19x - 5

= 23x

x5

+ x + 29

= 8

A

Solving Equations Containing Rational Expressions SECTION 6.5 457

Objectives

Solve equations containing rationalexpressions.

Solve equations containing rationalexpressions for a specified vari-able.

B

A

Practice Problem 1

Solve: x4

+ 45

= 120

Answers1. , 2. x = -6x = -3

SSM CD-ROM Video6.5

Practice Problem 2

Solve: x + 2

3- x - 1

5= 1

15


Practice Problem 3

Solve: 2 + 6x

= x + 7

Check:

Replace t with 5.

Simplify.

True.

The solution is 5.

Recall from Section 6.1 that a rational expression is defined for all realnumbers except those that make the denominator of the expression 0. Thismeans that if an equation contains rational expressions with variables in thedenominator, we must be certain that the proposed solution does not makethe denominator 0. If replacing the variable with the proposed solutionmakes the denominator 0, the rational expression is undefined and this pro-posed solution must be rejected.

Example 3 Solve:

Solution: In this equation, 0 cannot be a solution because if x is 0,

the rational expression is undefined. The LCD is x, so

we multiply both sides of the equation by x.

ƒ ƒ ƒ ƒ ƒ

Simplify.

Now we write the quadratic equation in standard formand solve for x.

Factor.

Notice that neither nor makes the denominator inthe original equation equal to 0.

Check: To check these solutions, we replace x in the originalequation by , and then by .

If : If :

True. True.

Both and are solutions.-2-3

6 = 6 5 = 5

3 - A-3 B � 6 3 - A-2 B � 5

3 - 6-2

� -2 + 83 - 6-3

� -3 + 8

3 - 6x

= x + 8 3 - 6x

= x + 8

x = -2x = -3

-2-3

-2-3

x = -2 x = -3

Set each factor equalto 0 and solve.x + 3 = 0 or x + 2 = 0

0 = Ax + 3 B Ax + 2 B 0 = x2 + 5x + 6

3x - 6 = x2 + 8x

HELPFUL HINT

Multiplyeach termby x.


x A3 B - x a 6xb = x # x + x # 8

xa3 - 6xb = x Ax + 8 B

6x

3 - 6x

= x + 8

518

= 518

12

- 29

�518

5 - 42

- 5 - 39

�518

t - 4

2- t - 3

9= 5

18

Answer3. x = -6, x = 1

Solving Equations Containing Rational Expressions SECTION 6.5 459

Practice Problem 4

Solve: 2

x + 3+ 3

x - 3= -2

x2 - 9

Practice Problem 5

Solve: 5x

x - 1= 5

x - 1+ 3

The following steps may be used to solve an equation containing ratio-nal expressions.

Example 4 Solve:

Solution: The denominator factors as .The LCD is then , so we multiply bothsides of the equation by this LCD.

Multiply by the LCD.

Simplify.


Combine like terms.


Check: Check by replacing x with in the original equation.The solution is .

Example 5 Solve:

Solution: Multiply both sides by the LCD, .

Multiply by the LCD.

Simplify.

Notice that 4 makes the denominator 0 in the originalequation. Therefore, 4 is not a solution and this equationhas no solution.


HELPFUL HINT

As we can see from Example 5, it is important to check the proposedsolution(s) in the original equation.

x = 4

2x = 4 + x

2x = 8 + Ax - 4 BUse the distributiveproperty. Ax - 4 B #

2xx - 4

= Ax - 4 B # 8

x - 4+ Ax - 4 B # 1

Ax - 4 B a 2xx - 4

b = Ax - 4 B a 8x - 4

+ 1 bx - 4

2xx - 4

= 8x - 4

+ 1

-3-3

x = -3

5x = -15

6x + 10 = x - 5

4x + 2x + 10 = x - 5

4x + 2 Ax + 5 B = x - 5

= Ax + 5 B Ax - 5 B # 1

x + 5

Use the distribu-tive property.

Ax + 5 B Ax - 5 B # 4x

x2 - 25+ Ax + 5 B Ax - 5 B #

2x - 5

Ax + 5 B Ax - 5 B a 4xx2 - 25

+ 2x - 5

b = Ax + 5 B Ax - 5 B a 1x + 5

b

Ax + 5 B Ax - 5 B Ax + 5 B Ax - 5 Bx2 - 25

4xx2 - 25

+ 2x - 5

= 1x + 5

TO SOLVE AN EQUATION CONTAINING RATIONAL EXPRESSIONS

Step 1. Multiply both sides of the equation by the LCD of all rationalexpressions in the equation.

Step 2. Remove any grouping symbols and solve the resulting equation.Step 3. Check the solution in the original equation.

Answers4. , 5. No solution


x = -1

✓ CONCEPT CHECK

When can fractions be cleared bymultiplying through by the LCD?a. When adding or subtracting ratio-

nal expressionsb. When solving an equation contain-

ing rational expressionsc. Both of thesed. Neither of these


Practice Problem 6

Solve: x - 6x + 3

= 2xx + 3

+ 2

Practice Problem 7

Solve for a.1a

+ 1b

= 1x

Answers

6. , 7. a = bxb - x

x = 4

Example 6 Solve:

Solution: Notice the denominators in this equation. We can seethat 2 can’t be a solution. The LCD is , so wemultiply both sides of the equation by .

Simplify.

Combine like terms.

Solve.

As we have already noted, 2 can’t be a solution of theoriginal equation. So we need only replace x with 8 in theoriginal equation. We find that 8 is a solution; the onlysolution is 8.

SOLVING EQUATIONS FOR A SPECIFIED VARIABLE

The last example in this section is an equation containing several variables,and we are directed to solve for one of the variables. The steps used in thepreceding examples can be applied to solve equations for a specified vari-able as well.

Example 7 Solve for x

Solution: (This type of equation often models a work problem, aswe shall see in the next section.) The LCD is abx, so wemultiply both sides by abx.

Simplify.

Factor out x from each term on the left side.


Simplify.

This equation is now solved for x.

x = abb + a

b + a x Ab + a B

b + a= ab

b + a

x Ab + a B = ab

bx + ax = ab

abx a 1ab + abx a 1

bb = abx #

1x

abx a 1a

+ 1bb = abx a 1

xb

1a

+ 1b

= 1x

B

x = 2 x = 8

x - 8 = 0 or x - 2 = 0

Ax - 8 B Ax - 2 B = 0

Write the quadratic equa-tion in standard form.Factor.


x2 - 10x + 16 = 0

x2 - 2x + 14 = 8x - 2

x2 - 2x + 14 = 7x + x - 2

Ax - 2 B Ax B + Ax - 2 B a 14x - 2

b = Ax - 2 B a 7xx - 2

b + Ax - 2 B A1 B Ax - 2 B ax + 14

x - 2b = Ax - 2 B a 7x

x - 2+ 1 bx - 2x - 2

x + 14x - 2

= 7xx - 2

+ 1

MENTAL MATH

Solve each equation for the variable.

1. 2. 3. 4.

EXERCISE SET 6.5

Solve each equation and check each solution. See Examples 1 and 2.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11.

Solve each equation and check each answer. See Examples 3 through 6.

12. 13. 14.

15. 16. 2 + 3a - 3

= aa - 3

53

- 32x

= 32

112x

+ 23

= 72x

63y

+ 3y

= 12y

+ 12

= 52y

x - 35

+ x - 22

= 12

b5

= b + 26

a5

= a - 32

6 + 5y

= y - 2y

2 + 10x

= x + 5

5 + 4x

= 12 - 8x

= 6x6

+ 4x3

= x18

x2

+ 5x4

= x12

x5

- 2 = 9x5

+ 3 = 9

A

y7

= 8z6

= 6x8

= 4x5

= 2

461


1.

2.

3.

4.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

6.6 RATIONAL EQUATIONS AND PROBLEM SOLVING

SOLVING PROBLEMS ABOUT NUMBERS

In this section, we solve problems that can be modeled by equations con-taining rational expressions. To solve these problems, we use the sameproblem-solving steps that were first introduced in Section 2.5. In our firstexample, our goal is to find an unknown number.

Example 1 Finding an Unknown NumberThe quotient of a number and 6 minus is the quotient

of the number and 2. Find the number.

Solution: 1. UNDERSTAND. Read and reread the problem.Suppose that the unknown number is 2, then we see if

the quotient of 2 and 6, or , minus is equal to the

quotient of 2 and 2, or .

Don’t forget that the purpose of a proposed solution isto better understand the problem.

2. TRANSLATE.

In words:

T T T T T

Translate:

3. SOLVE. Here, we solve the equation . We

begin by multiplying both sides of the equation by the

LCD 6.


Simplify.

Subtract x from both sides.


Simplify. -5 = x

- 102

= 2x2

-10 = 2x

x - 10 = 3x

6 a x6b - 6 a 5

3b = 6 a x

2b

6 a x6

- 53b = 6 a x

2b

x6

- 53

= x2

x2

=53

-x6

the quotientof x and 2

is53

minusthe quotient

of x and 6

Let x = the unknown number.

26

- 53

= 13

- 53

= - 43

, not 22

22

53

26

53

A

Rational Equations and Problem Solving SECTION 6.6 463

Objectives

Solve problems about numbers.

Solve problems about work.

Solve problems about distance,rate, and time.

Solve problems about similar trian-gles.

D

CBA

SSM CD-ROM Video6.6

Practice Problem 1The quotient of a number and 2 minus

is the quotient of the number and 6.13

Answer1. x = 1

4. INTERPRET.

Check: To check, we verify that “the quotient of and 6 minus

is the quotient of and 2, or .

State: The unknown number is .

SOLVING PROBLEMS ABOUT WORK

The next example is often called a work problem. Work problems usuallyinvolve people or machines doing a certain task.

Example 2 Finding Work RatesSam Waterton and Frank Schaffer work in a plant thatmanufactures automobiles. Sam can complete a qualitycontrol tour of the plant in 3 hours while his assistant,Frank, needs 7 hours to complete the same job. Theregional manager is coming to inspect the plant facilities,so both Sam and Frank are directed to complete a qualitycontrol tour together. How long will this take?

Solution: 1. UNDERSTAND. Read and reread the problem. Thekey idea here is the relationship between the time(hours) it takes to complete the job and the part of thejob completed in 1 unit of time (hour). For example, ifthe time it takes Sam to complete the job is 3 hours,

the part of the job he can complete in 1 hour is .

Similarly, Frank can complete of the job in 1 hour.

Let the time in hours it takes Sam and Frank tocomplete the job together.

Then the part of the job they complete in 1 hour.

Hours to Part of JobComplete Total Job Completed in 1 Hour

Sam 3

Frank 7

Together x

2. TRANSLATE.

In words:

T T T T T

Translate:1x

=17

+13

part of jobthey com-

pletedtogether in

1 hour

isequal

to

part of jobFrank

completedin 1 hour

addedto

part of job Sam

completedin 1 hour

1x

17

13

1x

=

x =

17

13

B

-5

- 56

- 53

= - 52

-553

-5


Practice Problem 2

Answer

2. hours115

Andrew and Timothy Larson volun-teer at a local recycling plant. Andrewcan sort a batch of recyclables in2 hours alone while his brother Timo-thy needs 3 hours to complete thesame job. If they work together, howlong will it take them to sort onebatch?


Practice Problem 3

3. SOLVE. Here, we solve the equation . We

begin by multiplying both sides of the equation by the LCD, 21x.

Simplify.

4. INTERPRET.

Check: Our proposed solution is hours. This proposed

solution is reasonable since hours is more than half

of Sam’s time and less than half of Frank’s time. Checkthis solution in the originally stated problem.

State: Sam and Frank can complete the quality control tour in

hours.

SOLVING PROBLEMS ABOUT DISTANCE, RATE, AND TIME

Next we look at a problem solved by the distance formula.

Example 3 Finding Speeds of VehiclesA car travels 180 miles in the same time that a truck trav-els 120 miles. If the car’s speed is 20 miles per hour fasterthan the truck’s, find the car’s speed and the truck’sspeed.

Solution: 1. UNDERSTAND. Read and reread the problem.Suppose that the truck’s speed is 45 miles per hour.Then the car’s speed is 20 miles per hour more, or65 miles per hour.

We are given that the car travels 180 miles in the sametime that the truck travels 120 miles. To find the timeit takes the car to travel 180 miles, remember that

since , we know that .

Car’s Time Truck’s Time

hours hours

Since the times are not the same, our proposed solu-tion is not correct. But we have a better understandingof the problem.

t = dr

= 12045

= 23045

= 223

t = dr

= 18065

= 25065

= 21013

dr

= td = rt

C

2110

2110

2110

x = 2110

or 2110

hours

10x = 21

7x + 3x = 21

21x a 13b + 21x a 1

7b = 21x a 1

xb

13

+ 17

= 1x

Answer3. car: 70 mph; motorcycle: 60 mph

A car travels 280 miles in the sametime that a motorcycle travels 240miles. If the car’s speed is 10 miles perhour more than the motorcycle’s, findthe speed of the car and the speed ofthe motorcycle.

Let the speed of the truck.

Since the car’s speed is 20 miles per hour faster thanthe truck’s, then

Use the formula or distance rate time.Prepare a chart to organize the information in theproblem.

Distance Rate Time

Truck 120 x d distanced rate

Car 180d distanced rate

2. TRANSLATE. Since the car and the truck traveledthe same amount of time, we have that

In words:T T

Translate:

3. SOLVE. We begin by multiplying both sides of theequation by the LCD, , or cross multiplying.




4. INTERPRET. The speed of the truck is 40 miles perhour. The speed of the car must then be or60 miles per hour.

Check: Find the time it takes the car to travel 180 miles and thetime it takes the truck to travel 120 miles.

Car’s Time Truck’s Time

hours hours

Since both travel the same amount of time, the proposedsolution is correct.

t = dr

= 12040

= 3t = dr

= 18060

= 3

x + 20

x = 40

60x = 2400

180x = 120x + 2400

180x = 120 Ax + 20 B

180x + 20

= 120x

x Ax + 20 B

120x

=180x + 20

truck’s time=car’s time

180x + 20

x + 20

120x

#=

#=d = r # t

x + 20 = the speed of the car

x =


HELPFUL HINT

If ,

then

or time

.= distancerate

t = dr

d = r # t—

—


Practice Problem 4If the following two triangles are simi-lar, find the missing length x.

12 unitsx units

9 units 15 units

Answer4. unitsx = 20

State: The car’s speed is 60 miles per hour and the truck’s speedis 40 miles per hour.

SOLVING PROBLEMS ABOUT SIMILAR TRIANGLES

Similar triangles have the same shape but not necessarily the same size. Insimilar triangles, the measures of corresponding angles are equal, and cor-responding sides are in proportion.

If triangle ABC and triangle XYZ shown are similar, then we know thatthe measure of angle A the measure of angle X, the measure of angleB the measure of angle Y, and the measure of angle C the measureof angle Z. We also know that corresponding sides are in proportion:

.

In this section, we will position similar triangles so that they have the sameorientation.

To show that corresponding sides are in proportion for the trianglesabove, we write the ratios of the corresponding sides.

Example 4 Finding the Length of a Side of a TriangleIf the following two triangles are similar, find the missinglength x.

Solution: Since the triangles are similar, their corresponding sidesare in proportion and we have

23

= 10x

2 yards 10 yards

7 yards

3 yards x yards

cz

= 155

= 3by

= 124

= 3ax

= 186

= 3

(15 in.) c

b (12 in.)

a (18 in.)

A

B

C(5 in.) z

y (4 in.)

x (6 in.)

X

Y

Z

ax

= by

= cz

===

D

To solve, we multiply both sides by the LCD, 3x, or crossmultiply.


The missing length is 15 yards.

x = 15

2x = 30


469


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

EXERCISE SET 6.6

Solve. See Example 1.A1. Three times the reciprocal of a num-

ber equals 9 times the reciprocal of 6.Find the number.

2. Twelve divided by the sum of a num-ber and 2 equals the quotient of 4 andthe difference of the number and 2.Find the number.

3. If twice a number added to 3 isdivided by the number plus 1, theresult is three halves. Find the num-ber.

4. A number added to the product of 6and the reciprocal of the numberequals . Find the number.-5

5. Two divided by the difference of anumber and 3, minus 4 divided by thenumber plus 3, equals 8 times the re-ciprocal of the difference of the num-ber squared and 9. What is the num-ber?

6. If 15 times the reciprocal of a numberis added to the ratio of 9 times thenumber minus 7 and the number plus2, the result is 9. What is the number?

7. One-fourth equals the quotient of anumber and 8. Find the number.

8. Four times a number added to 5 isdivided by 6. The result is . Find thenumber.

72

Solve. See Example 2.B9. Smith Engineering found that an

experienced surveyor surveys aroadbed in 4 hours. An apprenticesurveyor needs 5 hours to survey thesame stretch of road. If the two worktogether, find how long it takes themto complete the job.

10. An experienced bricklayer constructsa small wall in 3 hours. The apprenticecompletes the job in 6 hours. Findhow long it takes if they worktogether.

11. In 2 minutes, a conveyor belt moves300 pounds of recyclable aluminumfrom the delivery truck to a storagearea. A smaller belt moves the samequantity of cans the same distance in6 minutes. If both belts are used, findhow long it takes to move the cans tothe storage area.

12. Find how long it takes the conveyorbelts described in Exercise 11 to move1200 pounds of cans. (Hint: Think of1200 pounds as four 300-pound jobs.)

19. A jogger begins her workout byjogging to the park, a distance of3 miles. She then jogs home at thesame speed but along a differentroute. This return trip is 9 miles andher time is one hour longer. Completethe accompanying chart and use it tofind her jogging speed.

Distance Rate Time

Trip to park 13 x

Return trip 9 x + 1

#=

Name _________________________________________________________________________

470

13. Marcus and Tony work for Lom-bardo’s Pipe and Concrete. Mr. Lom-bardo is preparing an estimate for acustomer. He knows that Marcus laysa slab of concrete in 6 hours. Tonylays the same size slab in 4 hours. Ifboth work on the job and the cost oflabor is $45.00 per hour, decide whatthe labor estimate should be.

14. Mr. Dodson can paint his house byhimself in 4 days. His son needs anadditional day to complete the job ifhe works by himself. If they worktogether, find how long it takes topaint the house.

13.

14.

15.

16.

17.

18.

19.

20.

15. One custodian cleans a suite of officesin 3 hours. When a second worker isasked to join the regular custodian,the job takes only hours. How longdoes it take the second worker to dothe same job alone?

112

16. One person proofreads copy for asmall newspaper in 4 hours. If a sec-ond proofreader is also employed, thejob can be done in hours. How longdoes it take for the second proof-reader to do the same job alone?

212

17. One pipe fills a storage pond in20 hours. A second pipe fills the samepond in 15 hours. When a third pipe isadded and all three are used to fill thepond, it takes only 6 hours. Find howlong it takes the third pipe to do thejob.

18. One pump fills a tank 3 times as fastas another pump. If the pumps worktogether, they fill the tank in 21 min-utes. How long does it take for eachpump to fill the tank?

Solve. See Example 3.C20. A marketing manager travels

1080 miles in a corporate jet and thenan additional 240 miles by car. If thecar ride takes one hour longer thanthe jet ride takes, and if the rate of thejet is 6 times the rate of the car, findthe time the manager travels by jetand find the time the manager travelsby car.

471

Focus On HistoryEPIGRAM OF DIOPHANTES

One of the great algebraists of ancient times was a man named Diophantus. Little is known ofhis life other than that he lived and worked in Alexandria. Some historians believe he lived dur-ing the first century of the Christian era, about the time of Nero. The only clue to his personallife is the following epigram found in a collection called the Palatine Anthology.

God granted him youth for a sixth of his life and added a twelfth part to this. He clothed hischeeks in down. He lit him the light of wedlock after a seventh part, and five years after hismarriage, He granted him a son. Alas, lateborn wretched child. After attaining the measureof half his father’s life, cruel fate overtook him, thus leaving Diophantus during the last fouryears of his life only such consolation as the science of numbers. How old was Diophantus athis death?*

*From The Nature and Growth of Modern Mathematics, Edna Kramer, 1970, Fawcett Premier Books, Vol. 1, pages 107–108.

We are looking for Diophantus’ age when he died, so let x represent that age. If we sum theparts of his life, we should get the total age.

is the time of his youth.

is the time between his youth and when he married.Parts of

5 years is the time between his marriage and the birth of his son.his life

µis the time Diophantus had with his son.

4 years is the time between his son’s death and his own.

The sum of these parts should equal Diophantus’ age when he died.

CRITICAL THINKING

1. Solve the epigram.2. How old was Diophantus when his son was born? How old was the son when he died?

3. Solve the following epigram:I was four when my mother packed my lunch and sent me off to school. Half my life wasspent in school and another sixth was spent on a farm. Alas, hard times befell me. My cropsand cattle fared poorly and my land was sold. I returned to school for 3 years and havespent one tenth of my life teaching. How old am I?

GROUP ACTIVITY

4. Write an epigram describing your life. Be sure that none of the time periods in your epi-gram overlap. Exchange epigrams with a partner to solve and check.

16

# x + 112

# x + 17

# x + 5 + 12

# x + 4 = x

12

# x

17

# x

16

# x + 112

# x

6.7 SIMPLIFYING COMPLEX FRACTIONS

A rational expression whose numerator or denominator or both numeratorand denominator contain fractions is called a complex rational expressionor a complex fraction. Some examples are

Our goal in this section is to write complex fractions in simplest form. A

complex fraction is in simplest form when it is in the form , where P and

Q are polynomials that have no common factors.

SIMPLIFYING COMPLEX FRACTIONS—METHOD 1In this section, two methods of simplifying complex fractions are repre-sented. The first method presented uses the fact that the main fraction barindicates division.

Example 1 Simplify the complex fraction .

Solution: Since the numerator and denominator of the complexfraction are already single fractions, we proceed to step 2:perform the indicated division by multiplying the

numerator by the reciprocal of the denominator .

c ———c

The reciprocal of is .

Example 2 Simplify:

Solution: We simplify above and below the main fraction bar

separately. First we add and to obtain a single 15

23

23

+ 15

23

- 29

32

23

5823

= 58

# 32

= 1516

23

58

5823

METHOD 1: TO SIMPLIFY A COMPLEX FRACTION

Step 1. Add or subtract fractions in the numerator or denominator sothat the numerator is a single fraction and the denominator is asingle fraction.

Step 2. Perform the indicated division by multiplying the numerator ofthe complex fraction by the reciprocal of the denominator of thecomplex fraction.

Step 3. Write the rational expression in lowest terms.

A

PQ

d Numerator of complex fraction

d Main fraction bar

d Denominator of complex fraction

1x + 2

x + 2 - 1x

ff

32

47

- x

4

2 - 12

Simplifying Complex Fractions SECTION 6.7 473

Objectives

Simplify complex fractions usingmethod 1.

Simplify complex fractions usingmethod 2.

B

A

Practice Problem 1

Simplify the complex fraction .

3759

SSM CD-ROM Video6.7

Answers

1. , 2. 221

2735

Practice Problem 2

Simplify:

34

- 23

12

+ 38


Practice Problem 3

Simplify:

25

- 1x

x10

- 13

fraction in the numerator. Then we subtract from to

obtain a single fraction in the denominator.

Simplify.

Next we perform the indicated division by multiplyingthe numerator of the complex fraction by the reciprocalof the denominator of the complex fraction.

The reciprocal of is .

Example 3 Simplify:

Solution: Subtract to get a single fraction in the numerator and asingle fraction in the denominator of the complex frac-tion.

Multiply by the reciprocal of .

Factor.

Write in lowest terms. = 3z

=2 # 3 # A2 - z B2 # z # A2 - z B

2 - z6

= 2 - z2z

# 6

2 - z

=

2 - z2z

2 - z6

The LCD of the numerator’s frac-tions is 2z.

The LCD of the denominator’sfractions is 6.

1z

- 12

13

- z6

=

22z

- z2z

26

- z6

1z

- 12

13

- z6

= 13 # 3 # 33 # 5 # 4

= 3920

94

49

131549

= 1315

# 94

Add the numerator’s fractions.

Subtract the denominator’s fractions. =

131549

=

1015

+ 315

69

- 29

The LCD of the numerator’s fractions is 15.

The LCD of the denominator’s fractions is 9.

23

+ 15

23

- 29

=

2 A5 B3 A5 B +

1 A3 B5 A3 B

2 A3 B3 A3 B - 2

9

23

29

Answer

3.6 A2x - 5 B

x A3x - 10 B

Simplifying Complex Fractions SECTION 6.7 475

Practice Problem 4Use method 2 to simplify the complexfraction in Practice Problem 2:

34

- 23

12

+ 38

Answer

4. 221

SIMPLIFYING COMPLEX FRACTIONS—METHOD 2

Next we study a second method for simplifying complex fractions. In thismethod, we multiply the numerator and the denominator of the complexfraction by the LCD of all fractions in the complex fraction.

We use method 2 to rework Example 2.

Example 4 Simplify:

Solution: The LCD of is 45, so we multiply the

numerator and the denominator of the complex fractionby 45. Then we perform the indicated operations, andwrite in lowest terms.

Simplify.

HELPFUL HINT

The same complex fraction was simplified using two different methodsin Examples 2 and 4. Notice that each time the simplified result is thesame.

= 30 + 930 - 10

= 3920


=45 a 2

3b + 45 a 1

5b

45 a 23b - 45 a 2

9b

23

+ 15

23

- 29

=45 a 2

3+ 1

5b

45 a 23

- 29b

23

, 15

, 23

and 29

23

+ 15

23

- 29


Step 1. Find the LCD of all the fractions in the complex fraction.

Step 2. Multiply both the numerator and the denominator of the com-plex fraction by the LCD from Step 1.

Step 3. Perform the indicated operations and write the result in lowestterms.

B


Practice Problem 6

Simplify:

56y

+ yx

y3

- x

Example 5 Simplify:

Solution: The LCD of and is y, so we multiply the

numerator and the denominator of the complex fractionby y.

Simplify.

Example 6 Simplify:

Solution: The LCD of , so we multiply both

the numerator and the denominator of the complex frac-tion by .

or 2x2 + 3y

xy Ax + 2y B

= 2x2 + 3yx2y + 2xy2

Apply the distribu-tive property. =

2xy a xyb + 2xy a 3

2xb

2xy a x2b + 2xy Ay B

xy

+ 32x

x2

+ y=

2xy a xy

+ 32xb

2xy a x2

+ yb

2xy

xy

, 3

2x,

x2

, and y1

is 2xy

xy

+ 32x

x2

+ y

= x + 1x + 2y

Apply the distributive propertyin the denominator.

=y a x + 1

yb

y a xyb + y # 2

x + 1y

xy

+ 2=

y a x + 1yb

y a xy

+ 2 b

xy

x + 1y

x + 1y

xy

+ 2

Answers

5. , 6. 5x + 6y2

2yx Ay - 3x By + x2x + 1

Practice Problem 5

Simplify: 1 + x

y2x + 1

y

EXERCISE SET 6.7

Simplify each complex fraction. See Examples 1 through 6.

1. 2. 3. 4.

5. 6. 7. 8.

9. 10. 11. 12.

13. 14. 15. 16.

17. 18. 19. 20.

35y

+ 8

35y

- 8

xy

+ 1

xy

- 1

7y + 213

3y + 98

4y - 816

6y - 124

x - 12x + 1

1 - x2x + 1

1 + 1y - 2

y + 1y - 2

1y2 + 2

31y

- 56

15

- 1x

710

+ 1x2

x2

+ 2

x2

- 2

mn

- 1

mn

+ 1

4 - 1112

5 + 14

2 + 710

1 + 35

710

- 35

12

13

12

- 14

- 78y

214y

- 512x2

2516x3

- 6y11

4y9

- 4x9

- 2x3

18

- 512

1234

BA

477


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.



SECTION 6.1 SIMPLIFYING RATIONAL EXPRESSIONS

A rational expression is an expression that can be

written in the form , where P and Q are polyno-

mials and Q does not equal 0.

PQ

7y3

4,

x2 + 6x + 1x - 3

, -5

s3 + 8

To find values for which a rational expression isundefined, find values for which the denominatoris 0.

Find any values for which the expression

is undefined.

The expression is undefined when y is 3 and when yis 1.

y = 1y = 3

y - 3 = 0 or y - 1 = 0

Ay - 3 B Ay - 1 B = 0

Set the denominatorequal to 0.

Factor.

Set each factor equalto 0.Solve.

y2 - 4y + 3 = 0

5yy2 - 4y + 3

FUNDAMENTAL PRINCIPLE OF RATIONAL

EXPRESSIONS

If P, Q, and R are polynomials, and Q and R are not0, then

PRQR

= PQ

By the fundamental principle,

as long as and .x Z -1x Z 0

Ax - 3 B Ax + 1 Bx Ax + 1 B =

x - 3x

TO SIMPLIFY A RATIONAL EXPRESSION

Step 1. Factor the numerator and denominator.

Step 2. Apply the fundamental principle to divideout common factors.

Simplify:

4x + 20x2 - 25

=4 Ax + 5 B

Ax + 5 B Ax - 5 B =4

x - 5

4x + 20x2 - 25

SECTION 6.2 MULTIPLYING AND DIVIDING RATIONAL EXPRESSIONS

TO MULTIPLY RATIONAL EXPRESSIONS

Step 1. Factor numerators and denominators.

Step 2. Multiply numerators and multiply denomi-nators.

Step 3. Write the product in lowest terms.

PQ

# RS

= PRQS

Multiply:

=4 Ax + 2 B

x - 1

= 4 Ax + 1 B A2x - 3 B Ax + 2 BA2x - 3 B Ax + 1 B Ax - 1 B

=4 Ax + 1 B2x - 3

# A2x - 3 B Ax + 2 BAx + 1 B Ax - 1 B

4x + 42x - 3

# 2x2 + x - 6

x2 - 1

4x + 42x - 3

# 2x2 + x - 6

x2 - 1



To divide by a rational expression, multiply by thereciprocal. Divide:

15x + 53x2 - 14x - 5

, 153x - 12

PQ

,RS

= PQ

# SR

= PSQR

SECTION 6.3 ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH THE SAME DENOMINATOR AND LEAST COMMON DENOMINATOR

To add or subtract rational expressions with thesame denominator, add or subtract numerators,and place the sum or difference over the commondenominator.


5x + 1

+ xx + 1

= 5 + xx + 1

PR

- QR

= P - QR

PR

+ QR

= P + QR

= 1y - 3

= y + 3Ay + 3 B Ay - 3 B

= 2y + 7 - y - 4y2 - 9

=A2y + 7 B - Ay + 4 B

y2 - 9

2y + 7y2 - 9

- y + 4y2 - 9

TO FIND THE LEAST COMMON DENOMINATOR (LCD)

Step 1. Factor the denominators.

Step 2. The LCD is the product of all unique fac-tors, each raised to a power equal to thegreatest number of times that it appears inany one factored denominator.

Find the LCD for

and

LCD = 3x Ax + 5 B Ax + 5 B or 3x Ax + 5 B 2

3x2 + 15x = 3x Ax + 5 Bx2 + 10x + 25 = Ax + 5 B Ax + 5 B

113x2 + 15x

7xx2 + 10x + 25

SECTION 6.4 ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH DIFFERENT DENOMINATORS

TO ADD OR SUBTRACT RATIONAL EXPRESSIONS WITH

DIFFERENT DENOMINATORS

Step 1. Find the LCD.

Perform the indicated operation.

LCD is .Ax + 3 B Ax - 3 B= 9x + 3Ax + 3 B Ax - 3 B - 5

x - 3

9x + 3x2 - 9

- 5x - 3

= x - 4x - 5

=5 A3x + 1 B

A3x + 1 B Ax - 5 B # 3 A - 4 B

3 # 5x

15x + 53x2 - 14x - 5

, 153x - 12



Step 2. Rewrite each rational expression as anequivalent expression whose denominator isthe LCD.

Step 3. Add or subtract numerators and place thesum or difference over the common denom-inator.

Step 4. Write the result in lowest terms.

=4 Ax - 3 B

Ax + 3 B Ax - 3 B =4

x + 3

= 4x - 12Ax + 3 B Ax - 3 B

= 9x + 3 - 5x - 15Ax + 3 B Ax - 3 B

=9x + 3 - 5 Ax + 3 BAx + 3 B Ax - 3 B

= 9x + 3Ax + 3 B Ax - 3 B -

5 Ax + 3 BAx - 3 B Ax + 3 B

SECTION 6.5 SOLVING EQUATIONS CONTAINING RATIONAL EXPRESSIONS

TO SOLVE AN EQUATION CONTAINING RATIONAL

EXPRESSIONS

Step 1. Multiply both sides of the equation by theLCD of all rational expressions in the equa-tion.

Solve: The LCD is x + 2.5x

x + 2+ 3 = 4x - 6

x + 2

Step 2. Remove any grouping symbols and solve theresulting equation.

Step 3. Check the solution in the original equation.

The solution checks; the solution is .-3

x = -3 4x = -12

5x + 3x + 6 = 4x - 6

= Ax + 2 B a 4x - 6x + 2

bAx + 2 B a 5x

x + 2b + Ax + 2 B A3 B

= Ax + 2 B a 4x - 6x + 2

bAx + 2 B a 5xx + 2

+ 3b


SECTION 6.6 RATIONAL EQUATIONS AND PROBLEM SOLVING


1. UNDERSTAND. Read and reread the problem.

A small plane and a car leave Kansas City, Missouri,and head for Minneapolis, Minnesota, a distance of450 miles. The speed of the plane is 3 times thespeed of the car, and the plane arrives 6 hoursahead of the car. Find the speed of the car.

the speed of the car. the speed of the plane.

Distance Rate Time

Car 450 x

Plane 450 3x4503xa distance

rateb

450xa distance

rateb

#=

Then 3x = Let x =

2. TRANSLATE.In words: + =

T T T

Translate: 6450x

=+4503x

car’stime

6 hoursplane’s

time

3. SOLVE.

x = 50 18x = 900

450 + 18x = 1350

3x a 4503xb + 3x A6 B = 3x a 450

xb

4503x

+ 6 = 450x

4. INTERPRET. Check the solution by replacing x with 50 in the orig-inal equation. State the conclusion: The speed ofthe car is 50 miles per hour.

SECTION 6.7 SIMPLIFYING COMPLEX FRACTIONS


Step 1. Add or subtract fractions in the numeratorand the denominator of the complex frac-tion.

Step 2. Perform the indicated division.

Step 3. Write the result in lowest terms.

Simplify:

=y A1 + 2x B

y - x

= 1 + 2xx

# xy

y - x

=

1 + 2xx

y - xxy

1x

+ 2

1x

- 1y

=

1x

+ 2xx

yxy

- xxy



METHOD 2. TO SIMPLIFY A COMPLEX FRACTION

Step 1. Find the LCD of all fractions in the complexfraction.

Step 2. Multiply the numerator and the denomina-tor of the complex fraction by the LCD.

Step 3. Perform the indicated operations and writethe result in lowest terms.

= y + 2xyy - x

or y A1 + 2x B

y - x

=xy a 1

xb + xy A2 B

xy a 1xb - xy a 1

yb

1x

+ 2

1x

- 1y

=xy a 1

x+ 2 b

xya 1x

- 1yb

483

Focus On Business and CareerMORTGAGES

A loan for which the purpose is to buy a house or other property iscalled a mortgage. When you are thinking of getting a mortgage tobuy a house, it is helpful to know how much your monthly mort-gage payment will be. One way to calculate the monthly paymentis to use the formula

where is the amount of the mortgage, annual interestrate (written as a decimal), and loan term in years. Try theexercises below.

CRITICAL THINKING

1. The average mortgage rate in the United States in 1996 was7.93% (Source: National Association of Realtors®). Supposeyou had borrowed $80,000 to buy a house in 1996. If your loanterm is 30 years, calculate your monthly mortgage payment.

2. The average mortgage interest rate in the United States in1989 was 10.11% (Source: National Association of Realtors®).Suppose you had borrowed $71,000 to buy a house in 1989. Ifyour loan term is 20 years, calculate your monthly mortgagepayment.

Another way to calculate a monthly mortgage payment is to useone of the many sites on the World Wide Web that offer an inter-active mortgage calculator. For instance, by visiting the givenWorld Wide Web address, you will be able to access the FleetBank Web site, or a related site, where you can calculate amonthly mortgage payment by entering the amount to be bor-rowed, the interest rate as a percent, and the term of the loan inyears. Use the site below to solve the given exercises.

Go to http://www.prenhall.com/martin-gay

Internet Excursions

3. Suppose you would like to borrow $64,000 to buy a house. Ifthe interest rate is 8.2% and you plan to take out a 25-yearloan, what will be your monthly mortgage payment?

4. Suppose you would like to borrow $100,000 to buy a house. Ifthe interest rate is 7.5% and you plan to take out a 20-yearloan, what will be your monthly mortgage payment?

t =r =A =

P =

Ar12

1 - 1

a1 + r12b 12t

,

485

7Ithis chapter, we continue our investigation of graphs of equations intwo variables, which began in Chapter 3. These equations and theirgraphs lead to the notion of relation and to the notion of function, per-haps the single most important and useful concept in all of mathematics.

7.1 The Slope-Intercept Form

7.2 More Equations of Lines

7.3 Introduction to Functions

7.4 Polynomial and RationalFunctions

7.5 An Introduction to GraphingPolynomial Functions

At the beginning of the 20th century, there were approxi-mately 237,600 students enrolled in the 977 institutions ofhigher education in the United States. At that time, only 19% ofbachelor’s degree recipients were women. By the year 2000, theprojected 3800 colleges and universities in the United Stateswill have an estimated 14,800,000 students. Roughly 56% ofbachelor’s degree recipients are expected to be women. Thephenomenal growth of colleges and universities can also be seenin the average tuition costs at these institutions of higher learn-ing. For instance, the average annual tuition at a private four-year college or university has increased from $1809 in 1970 to$13,664 in 1998, an increase of about 655%!

Graphs and Functions C H A P T E R

7.1 THE SLOPE-INTERCEPT FORM

GRAPHING A LINE USING SLOPE AND Y-INTERCEPT

From Section 3.4, we know that when a linear equation is solved for y, thecoefficient of x is the slope of the line. For example, the slope of the linewhose equation is is 3. In this equation, , whatdoes 1 represent? To find out, let and watch what happens.

Let .

We now have the ordered pair , which means that 1 is the y-intercept.This is true in general. To see this, let and solve for y in

.

Let .

We obtain the ordered pair , which means that b is the y-intercept.The form is appropriately called the slope-intercept form

of a linear equation.

We can use the slope-intercept form to graph a linear equation.

Example 1 Graph:

Solution: Recall that the slope of the graph of is and

the y-intercept is . To graph the line, we first plot they-intercept point . To find another point on the

line, we recall that slope is . Another point may

then be plotted by starting at , rising 1 unit up,and then running 4 units to the right. We are now at the

point . The graph of is the line

through points and , as shown.

Rise = 1Run = 4

(0, −3)(4, −2)

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

m = 14

A4, -2 BA0, -3 By = 1

4x - 3A4, -2 B

A0, -3 Briserun

= 14

A0, -3 B-3

14

y = 14

x - 3

y = 14

x - 3

SLOPE-INTERCEPT FORM

When a linear equation in two variables is written in slope-interceptform,

y = mx + bthen m is the slope of the line and b is the y-intercept of the line.

y-interceptslope

y = mx + bA0, b B

y = b

x = 0y = m # 0 + b

y = mx + bx = 0

A0, 1 By = 1

x = 0y = 3 # 0 + 1y = 3x + 1

x = 0y = 3x + 1y = 3x + 1

A

The Slope-Intercept Form SECTION 7.1 487

Objectives

Graph a line using its slope and y-intercept.

Use the slope-intercept form towrite an equation of the line.

Interpret the slope-intercept formin an application.

C

B

A

Practice Problem 1

Graph:

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

y = 23

x + 1

Answer1.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

SSM CD-ROM Video7.1

488 CHAPTER 7 Graphs and Functions

Example 2 Graph:

Solution: First, we solve the equation for y to write it in slope-intercept form. In slope-intercept form, the equation is

. Next we plot the y-intercept point, .To find another point on the line, we use the slope ,

which can be written as . We start at and

move vertically 2 units down, since the numerator of theslope is ; then we move horizontally 1 unit to the rightsince the denominator of the slope is 1. We arrive at thepoint . The line through and will havethe required slope of .

USING THE SLOPE-INTERCEPT FORM TO WRITE AN EQUATION

Given the slope and y-intercept of a line, we may write its equation as wellas graph the line.

Example 3 Write an equation of the line with y-intercept and

slope of .

Solution: We let and , and write the equation in

slope-intercept form, .

Let and .

Simplify.

Notice that the graph of this equation has slope and

y-intercept , as desired.


INTERPRETING THE SLOPE-INTERCEPT FORM

On the next page is a graph of an adult one-day pass price for Disney Worldover time. Often, businesses depend on linear equations that “closely fit”data to model the data and predict future trends. For example, by a methodcalled least squares regression, the linear equation approximates the data shown, where x is the number of years since 1988and y is the ticket price for that year.

y = 1.462x + 29.35

C

-3

14

y = 14

x - 3

b = -3m = 14

y = 14

x + A-3 B y = mx + b

y = mx + b

b = -3m = 14

14

-3

B

54321

1

(0, 3)

(1, 1)

−1

−2−3−4−5

−2−3

Rise = −2

Run = +1

−4−5 2 3 4 5 x

y

-2A0, 3 BA1, 1 BA1, 1 B

-2

A0, 3 Briserun

= -21

-2A0, 3 By = -2x + 3

2x + y = 3

Practice Problem 3Write an equation of the line with

slope and y-intercept 1.23

✓ CONCEPT CHECK

What is wrong with the followingequation of a line with y-intercept 4 and slope 2?y = 4x + 2

Answers2.

3.

✓ Concept Check: The y-intercept and slopewere switched. It should be .y = 2x + 4

y = 23

x + 1

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5


x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

3x + y = -2

The Slope-Intercept Form SECTION 7.1 489

Answers4. a. $12,477.20, b. The yearly average incomeincreases by $356.50 every year, c. At year

, or 1991, the yearly average income was$8912.20.x = 0

Practice Problem 4

a. Predict the income for the year2001.

b. What does the slope of this equa-tion mean?

c. What does the y-intercept of thisequation mean?

Example 4 Predicting Future PricesThe adult one-day pass price y for Disney World is givenby

where x is the number of years since 1988.

a. Use this equation to predict the ticket price for theyear 2002.

b. What does the slope of this equation mean?c. What does the y-intercept of this equation mean?

Solution: a. To predict the price of a pass in 2002, we need to findy when x is 14. (Since year 1988 corresponds to ,year 2002 corresponds to .

Let .

We predict that in the year 2002, the price of an adultone-day pass to Disney World will be about $49.82.

b. The slope of is 1.462. We can

think of this number as or . This means that

the ticket price increases on the average by $1.462

every 1 year.

c. The y-intercept of is 29.35. Noticethat it corresponds to the point of the graph

c ayear price

This means that at year or 1988, the ticket pricewas $29.35.

x = 0

A0, 29.35 By = 1.462x + 29.35

1.4621

riserun

y = 1.462x + 29.35

= 49.818

x = 14 = 1.462 A14 B + 29.35

y = 1.462x + 29.35

x = 2002 - 1988 = 14x = 0

y = 1.462x + 29.35

Pric

e of

Adu

lt O

ne-D

ay P

ass

(in

dolla

rs)

0 1 2 3 4 5Years

0 1988

6 7 8 9 10

Ticket Price at Disney World

44.00

42.00

40.00

38.00

36.00

34.00

32.00

30.00

28.00

26.00

24.00

22.00

20.00

0

Source: The Walt Disney Company

y

x

y = 1.462x + 29.35

HELPFUL HINT

The notation 0 4 1988 meansthat the number 0 correspondsto the year 1988, 1 correspondsto the year 1989, and so on.

The yearly average income y of anAmerican woman with some highschool education but no diploma isgiven by the equation

,where x is the number of years since1991. (Source: Based on data from theU.S. Bureau of the Census, 1991–1996)

8912.2y = 356.5x +


GRAPHING CALCULATOR EXPLORATIONSYou may have noticed by now that to use the key on a grapher to graph anequation, the equation must be solved for y.

Graph each equation by first solving the equation for y.

1. 2.

3. 4.

5. 6.

7. 8. y + 2.6x = -3.2y - 5.6x = 7.7x + 1.5

3y - 5x = 6x - 4y - x = 3.78

-7.22x + 3.89y = 12.575.78x + 2.31y = 10.98

-2.7y = xx = 3.5y

Y =

491


MENTAL MATH

Find the slope and the y-intercept of each line.

1. 2. 3.

4. 5.

EXERCISE SET 7.1

Graph each line passing through the given point with the given slope. See Examples 1 and 2.

1. Through with slope 2. Through with slope

3. Through with slope 5 4. Through with slope 2

5. Through with slope 6. Through with slope

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

-3A3, 0 B-1A0, 7 B

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

A- 5, 2 BA0, 0 B

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

25

A-2, -4 B32

A1, 3 B

A

y = 12

x + 6y = -x

y = 5xy = 23

x - 72

y = -4x + 12

MENTAL MATH ANSWERS

1.

2.

3.

4.

5.

ANSWERS

1. see graph

2. see graph

3. see graph

4. see graph

5. see graph

6. see graph

Name _________________________________________________________________________

Graph each linear equation using the slope and y-intercept. See Examples 1 and 2.

7. 8. 9. 10.

11.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = 12

x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = 2x + 6y = -2x + 3y = 2xy = -2x

Use the slope-intercept form of a linear equation to write the equation of each line withthe given slope and y-intercept. See Example 3.

12. Slope ; y-intercept 1 13. Slope ; y-intercept

14. Slope 2; y-intercept 15. Slope ; y-intercept - 15

-334

-612

-1

B

492

12.

13.

14.

15.

7.2 MORE EQUATIONS OF LINES

USING THE POINT-SLOPE FORM TO WRITE AN EQUATION

When the slope of a line and a point on the line are known, the equation ofthe line can also be found. To do this, we use the slope formula to write theslope of a line that passes through points , and . We have

We multiply both sides of this equation by to obtain

This form is called the point-slope form of the equation of a line.

Example 1 Write an equation of the line with slope containingthe point .

Solution: Because we know the slope and a point on the line, weuse the point-slope form with and

.

Point-slope form

Let and .


The equation is .

Example 2 Write an equation of the line through points and.

Solution: First we find the slope of the line.

Next we make use of the point-slope form. We replaceby either or in the point-slope

equation. We will choose the point . The line

through with slope is58

A4, 0 BA4, 0 B

A-4, -5 BA4, 0 BAx1, y1 B

m = -5 - 0-4 - 4

= -5-8

= 58

A-4, -5 B A4, 0 B

y = -3x - 2

y = -3x - 2

y + 5 = -3x + 3

Ax1, y1 B = A1, -5 Bm = -3y - A-5 B = -3 Ax - 1 B y - y1 = m Ax - x1 B

Ax1, y1 B = A1, -5 B m = -3

A1, -5 B -3

y - y1 = m Ax - x1 Bx - x1

m =y - y1

x - x1

Ax1, y1 BAx, y B

A

More Equations of Lines SECTION 7.2 493

Objectives

Use the point-slope form to writethe equation of a line.

Write equations of vertical andhorizontal lines.

Write equations of parallel andperpendicular lines.

Use the point-slope form in real-world applications.

D

C

B

A

POINT-SLOPE FORM OF THE EQUATION OF A LINE

The point-slope form of the equation of a line is

slopeT

a Qpoint

where m is the slope of the line and is a point on the line.Ax1, y1 B

y - y1 = m Ax - x1 B

Practice Problem 1

Answers

1. , 2. y = - 45

x + 125

y = -2x

SSM CD-ROM Video7.2

Practice Problem 2

Write an equation of the line withslope containing the point .Write the equation in slope-interceptform, .y = mx + b

A2, -4 B-2

Write an equation of the line throughpoints and . Write theequation in slope-intercept form,

.y = mx + b

A-2, 4 BA3, 0 B


Answers3. , 4. x = 4y = 6

Practice Problem 3

Point-slope form

Let and .


Simplify.

The equation is . If we choose to use the

point , we have ,

which also simplifies to .

WRITING EQUATIONS OF VERTICAL AND HORIZONTAL LINES

A few special types of linear equations are those whose graphs are verticaland horizontal lines.

Example 3 Write an equation of the horizontal line containing thepoint .

Solution: Recall from Section 3.1, that a horizontal line has anequation of the form . Since the line contains thepoint , the equation is .

Example 4 Write an equation of the line containing the point with undefined slope.

Solution: Since the line has undefined slope, the line must bevertical. A vertical line has an equation of the form

, and since the line contains the point , the equation is .

WRITING EQUATIONS OF PARALLEL AND PERPENDICULAR LINES

Next, we write equations of parallel and perpendicular lines.

C

(2, 3)

x = 2

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

x = 2A2, 3 Bx = c

A2, 3 By = 3A2, 3 B

y = b

A2, 3 B

B

HELPFUL HINT

If two points of a line are given, either one may be used with the slope-intercept form to write an equation of the line.

y = 58

x - 52

y - A-5 B = 58

Cx - A-4 B DA-4, -5 By = 5

8 x - 5

2

y = 58

x - 52

y = 58

x - 58

# 4

Ax1, y1 B = A4, 0 Bm = 58

y - 0 = 58Ax - 4 B

y - y1 = m Ax - x1 B

Practice Problem 4Write an equation of the line contain-ing the point with undefinedslope.

A4, 7 B

Write an equation of the horizontalline containing the point .A-1, 6 B


Answers5. , 6. 2x - y = 2y = -3x - 1

Practice Problem 5Example 5 Write an equation of the line containing the point and parallel to the line .

Solution: Because the line we want to find is parallel to the line, the two lines must have equal slopes. So

we first find the slope of by solving it for yto write it in the form . Here so the slope is .

Now we use the point-slope form to write the equationof a line through with slope .

Let , and .


The equation, , and the new equation,, have the same slope but different

y-intercepts so their graphs are parallel. Also, the graph of contains the point , as desired.

Example 6 Write an equation of the line containing the pointand perpendicular to the line .

Solution: First we find the slope of by solving it for y.

The slope of the given line is . A line perpendicular to

this line will have a slope that is the negative reciprocal

of , or . We use the point-slope form to write an

equation of a new line through with slope .

(−2, 1)

y = x +53

x

y

133

y = − x +35

45

53

A-2, 1 B53

- 35

- 35

y = - 35

x + 45

5y = -3x + 4

3x + 5y = 4

3x + 5y = 4A-2, 1 B

A4, 4 By = -2x + 12

y = -2x + 12y = -2x - 6

(4, 4)

y = −2x + 12

y = −2x − 6 x

y

y = -2x + 12

y - 4 = -2x + 8

y1 = 4m = -2, x1 = 4 y - 4 = -2 Ax - 4 By - y1 = m Ax - x1 B

-2A4, 4 B-2

y = -2x - 6y = mx + b2x + y = -6

2x + y = -6

2x + y = -6A4, 4 B

Practice Problem 6

Write an equation of the line contain-ing the point and parallel tothe line . Write the equa-tion in the form .y = mx + b

3x + y = 5A-1, 2 B

Write an equation of the line contain-ing the point and perpendicularto the line . Write theequation in standard form.

2x + 4y = 5A3, 4 B


Answer7. 57 new contracts

Simplify.



The equation and the new equation

have negative reciprocal slopes so their

graphs are perpendicular. Also, the graph of

contains the point , as desired.

USING THE POINT-SLOPE FORM IN APPLICATIONS

The point-slope form of an equation is very useful for solving real-worldproblems.

Example 7 Predicting SalesSouthern Star Realty is an established real estate com-pany that has enjoyed constant growth in sales since1990. In 1992 the company sold 200 houses, and in 1997the company sold 275 houses. Use these figures to predictthe number of houses this company will sell in the year2002.

Solution: 1. UNDERSTAND. Read and reread the problem. Thenlet

the number of years after 1990 and

the number of houses sold in the year corre-sponding to x.

The information provided then gives the ordered pairsand . To better visualize the sales of

Southern Star Realty, we graph the linear equationthat passes through the points and .

2. TRANSLATE. We write a linear equation that passesthrough the points and . To do so, wefirst find the slope of the line.

A7, 275 BA2, 200 B

4 6 7 8 9 10 11 1251 2 3 x

y

380350320290260230200170

Years after 1990

(2, 200)

(7, 275)

A7, 275 BA2, 200 BA7, 275 BA2, 200 B

y =x =

D

A-2, 1 By = 53

x + 133

y = 53

x + 133

y = -35

x + 45

y = 53

x + 133

y - 1 = 53

x + 103

y - 1 = 53Ax + 2 B

y - 1 = 53Cx - A-2 B D

Practice Problem 7Southwest Regional is an establishedoffice product maintenance companythat has enjoyed constant growth innew maintenance contracts since 1985.In 1990, the company obtained 15 newcontracts and in 1997, the companyobtained 36 new contracts. Use thesefigures to predict the number of newcontracts this company can expect in2004.


Then, using the point-slope form to write the equation,we have

Let and .

Multiply.


3. SOLVE. To predict the number of houses sold in theyear 2002, we use and complete theordered pair , since .

Let .

4. INTERPRET.

Check: Verify that the point is a point on the linegraphed in step 1.

State: Southern Star Realty should expect to sell 350 houses inthe year 2002.

A12, 350 B

y = 350

x = 12y = 15 A12 B + 170

2002 - 1990 = 12A12, By = 15x + 170

y = 15x + 170

y - 200 = 15x - 30

Ax1, y1 B = A2, 200 Bm = 15y - 200 = 15 Ax - 2 B y - y1 = m Ax - x1 B

m = 275 - 2007 - 2

= 755

= 15

GRAPHING CALCULATOR EXPLORATIONSMany graphing calculators have a TRACE feature. This fea-ture allows you to trace along a graph and see the corre-

sponding x- and y-coordinates appear on the screen. Use thisfeature for the following exercises.

Graph each equation and then use the TRACE feature to completeeach ordered pair solution. (Many times the tracer will not show anexact x- or y-value asked for. In each case, trace as closely as you canto the given x- or y-coordinate and approximate the other, unknown,coordinate to one decimal place.)

1. 2.

3. 4.

5. 6.

x = ?, y = 12x = ?, y = 36x = 3.2, y = ?x = 2.3, y = ?y = -6.2x - 8.3y = 5.2x - 3.3

x = ?, y = -4.4x = ?, y = 7.2y = 0.4x - 8.6y = -5.9x - 1.6

x = -1.8, y = ?x = 5.1, y = ?y = -4.8x + 2.9y = 2.3x + 6.7


Focus On Business and CareerLINEAR MODELING

As we saw in Section 3.3, businesses often depend on equations that “closely fit” data. Tomodel the data means to find an equation that describes the relationship between the paireddata of two variables, such as time in years and profit. A model that accurately summarizes therelationship between two variables can be used to replace a potentially lengthy listing of theraw data. An accurate model might also be used to predict future trends by answering ques-tions such as “If the trend seen in our company’s performance in the last several years contin-ues, what level of profit can we reasonably expect in 3 years?”

There are several ways to find a linear equation that models a set of data. If only twoordered pair data points are involved, an exact equation that contains both points can be foundusing the methods of Section 3.4. When more than two ordered pair data points are involved, itmay be impossible to find a linear equation that contains all of the data points. In this case, thegraph of the best fit equation should have a majority of the plotted ordered pair data points onthe graph or close to it. In statistics, a technique called least squares regression is used to deter-mine an equation that best fits a set of data. Various graphing utilities have built-in capabilitiesfor finding an equation (called a regression equation) that best fits a set of ordered pair datapoints. Regression capabilities are often found with a graphing utility’s statistics features.* Abest fit equation can also be estimated using an algebraic method, which is outlined in theGroup Activity below. In either case, a useful first step when finding a linear equation thatmodels a set of data is creating a scatter diagram of the ordered pair data points to verify that alinear equation is an appropriate model.

GROUP ACTIVITY

Coca-Cola Company is the world’slargest producer of soft drinks andjuices. The table shows Coca-Cola’snet profit (in billions of dollars) forthe years 1993–1997. Use the tablealong with your answers to the questions below to find a linear function that represents netprofit (in billions of dollars) as a function of the number of years after 1993.

1. Create a scatter diagram of the paired data given in the table. Does a linear model seemappropriate for the data?

2. Use a straight edge to draw on your graph what appears to be the line that “best fits” the datayou plotted.

3. Estimate the coordinates of two points that fall on your best fit line. Use these points to finda linear function for the line.

4. Use your linear function to find , and interpret its meaning in context.

5. Compare your group’s linear function with other groups’ functions. Are they the same or dif-ferent? Explain why.

6. (Optional) Enter the data from the table into a graphing utility and use the linear regressionfeature to find a linear function that models the data. Compare this function with the one youfound in Question 3. How are they alike or different?

7. (Optional) Using corporation annual reports or articles from magazines or newspapers,search for a set of business-related data that could be modeled with a linear function. Explainhow modeling this data could be useful to a business. Then find the best fit equation for thedata.

*To find out more about using a graphing utility to find a regression equation, consult the user’s manual for your graphing utility.

f A8 Bf Ax B

f Ax B

Year 1993 1994 1995 1996 1997

Net Profit(in billions of dollars) 2.2 2.6 3.0 3.5 4.1

(Source: The Coca-Cola Company)

499


MENTAL MATH

Find the slope and a point of the graph of each equation.

1. 2. 3.

4. 5.

EXERCISE SET 7.2

Write an equation of each line with the given slope and containing the given point. Writethe equation in the form . See Example 1.

1. Slope 3; through 2. Slope 4; through

3. Slope ; through 4. Slope ; through

5. Slope ; through

Write an equation of the line passing through the given points. Write the equation in the form. See Example 2.

6. and 7. and

8. and 9. and

10. and

Write an equation of each line. See Examples 3 and 4.

11. Vertical; through 12. Slope 0; through

13. Horizontal; through 14. Vertical; through

15. Undefined slope; through A0, 5 BA4, 7 BA-3, 1 BA-2, -4 BA2, 6 B

B

A-4, -3 BA-2, -4 BA2, 6 BA7, -4 BA-6, 13 BA-2, 5 BA7, 8 BA3, 0 BA4, 6 BA2, 0 B

y = mx + b

A-6, 2 B12

A2, -4 B-4A1, -3 B-2

A5, 1 BA1, 2 By = mx + b

A

y + 2 = 5 Ax - 3 By - 1 = - 23Ax - 0 B

y - 0 = 14Ax - 2 By - 6 = -3 Ax - 4 By - 4 = -2 Ax - 1 B

MENTAL MATH ANSWERS

1.

2.

3.

4.

5.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Write an equation of each line. Write the equation in the form . See Exam-ples 5 and 6.

y = mx + bC

16. Through ; parallel toy = 4x - 2

A3, 8 B 17. Through ; parallel toy = 3x - 4

A1, 5 B

18. Through ; perpendicular toy = -2x - 6

A2, -5 B 19. Through ; perpendicular toy = -4x - 1

A-4, 8 B

20. Through ; parallel to3x + 2y = 5

A-2, -3 B

7.3 INTRODUCTION TO FUNCTIONS

DEFINING RELATION, DOMAIN, AND RANGE

Equations in two variables, such as , describe relations betweenx-values and y-values. For example, if , then this equation describeshow to find the y-value related to . In words, the equation

says that twice the x-value increased by 1 gives the corre-sponding y-value. The x-value of 1 corresponds to the y-value of

for this equation, and we have the ordered pair .There are other ways of describing relations or correspondences

between two numbers or, in general, a first set (sometimes called the set ofinputs) and a second set (sometimes called the set of outputs). For example,

First Set: Input Correspondence Second Set: Output

People in a certain city Each person’s age The set of nonnegativeintegers

A few examples of ordered pairs from this relation might be (Ana, 4);(Bob, 36); (Trey, 21); and so on.

Below are just a few other ways of describing relations between two setsand the ordered pairs that they generate.

TOrdered Pairs

T TOrdered Pairs Some Ordered Pairs

RELATION, DOMAIN, AND RANGE

A relation is a set of ordered pairs.The domain of the relation is the set of all first components of theordered pairs.The range of the relation is the set of all second components of theordered pairs.

A1, 3 B , A0, 1 B , and so onA-3, -1 B , A1, 1 B , A2, 3 B , A3, -2 B

y = 2x + 1

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

Aa, 3 B , Ac, 3 B , Ae, 1 B

First Set: Second Set:Input Output

Correspondence

a 1c 2e 3

A1, 3 B2 A1 B + 1 = 3

y = 2x + 1x = 1

x = 1y = 2x + 1

A

Introduction to Functions SECTION 7.3 501

Objectives

Define relation, domain, andrange.

Identify functions.

Use the vertical line test for func-tions.

Find the domain and range of arelation from its graph.

Use function notation.

Graph a linear function.FE

D

CB

A

SSM CD-ROM Video7.3

Practice Problems 1–3Determine the domain and range ofeach relation.

1.

2.

3. Output:Input: Number ofStates Representatives

Arkansas 4Texas 30

Oklahoma 10 8South Carolina 6

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

E A1, 6 B , A2, 8 B , A0, 3 B , A0, -2 B F


Answers

1. domain: , range: ,2. domain: , range: ,3. domain: Arkansas, Texas, Oklahoma,South Carolina , range: E4, 30, 6FF

EE-1, 0, 1, 2, 3FE1F

E6, 8, 3, -2FE1, 2, 0F

For example, the domain for our middle relation above is andthe range is . Notice that the range does not include the element 2 ofthe second set. This is because no element of the first set is assigned to thiselement. If a relation is defined in terms of x- and y-values, we will agreethat the domain corresponds to x-values and that the range corresponds toy-values.

Examples Determine the domain and range of each relation.

1.The domain is the set of all first coordinates of theordered pairs, .The range is the set of all second coordinates,

.2.

The relation is .

The domain is .The range is .

3.

The domain is the first set, Erie, Escondido, Gary,Miami, Waco .The range is the numbers in the second set that corre-spond to elements in the first set, .

IDENTIFYING FUNCTIONS

Now we consider a special kind of relation called a function.

FUNCTION

A function is a relation in which each first component in the orderedpairs corresponds to exactly one second component.

B

E104, 109, 117, 359FF E

Output:Input: PopulationCities (in thousands)

Erie 109Miami 359

200

Escondido 11752

Waco 182 104Gary

E1FE-3, -2, -1, 0, 1, 2F

A2, 1 B FA1, 1 B , A0, 1 B ,A-1, 1 B ,A-2, 1 B ,E A-3, 1 B ,

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

E3, 4, -1FE2, 0, 3F

E A2, 3 B , A2, 4 B , A0, -1 B , A3, -1 B F

E1, 3F Ea, c, eF

Practice Problems 4–6Determine whether each relation isalso a function.

4.

5.

6. Input Correspondence Output

People in County/ Counties ofa certain Parish that state

state that a personlives in

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(3, 4)

(1, 2)

(1, −2)

(−3, 0) (3, 0)

E A-3, 7 B , A1, 7 B , A2, 2 B F


Answers4. function, 5. not a function, 6. function7. yes, 8. no

✓ Concept Check: Two different ordered pairscan have the same y-value, but not the same x-value in a function.

✓ CONCEPT CHECK

Explain why a function can containboth the ordered pairs and

but not both and .A3, 2 BA3, 1 BA2, 3 BA1, 3 B

Practice Problem 7Determine whether the relation

is also a function.y = 3x + 2

Examples Determine whether each relation is also a function.

4.Although the ordered pairs and havethe same y-value, each x-value is assigned to only oney-value, so this set of ordered pairs is a function.

5.

The x-value 0 is assigned to two y-values, and 3, inthis graph so this relation is not a function.

6. Input Correspondence OutputPeople in a Each person’s age The set ofcertain city nonnegative

integers

This relation is a function because although two differentpeople may have the same age, each person has only oneage. This means that each element in the first set isassigned to only one element in the second set.

TRY THE CONCEPT CHECK IN THE MARGIN.We will call an equation such as a relation since this equa-

tion defines a set of ordered pair solutions.

Example 7 Determine whether the relation is also afunction.

Solution: The relation is a function if each x-value cor-responds to just one y-value. For each x-value substitutedin the equation , the multiplication and addi-tion performed gives a single result, so only one y-value will be associated with each x-value. Thus,

is a function.

Example 8 Determine whether the relation is also a function.

Solution: In , if , then . Also, if , then. In other words, we have the ordered pairs

and . Since the x-value 9 corresponds to two y-values, 3 and , is not a function.x = y2-3A9, -3 B

A9, 3 Bx = 9y = -3x = 9y = 3x = y2

x = y2

y = 2x + 1

y = 2x + 1

y = 2x + 1

y = 2x + 1

y = 2x + 1

-2

654321

1−1

−2

−2−3 2 3 4 5 6

(−3, 2)

(0, −2)

(0, 3)

(1, 5)(6, 6)

x

y

A-3, 5 BA-2, 5 BE A-2, 5 B , A2, 7 B , A-3, 5 B , A9, 9 B F

HELPFUL HINT

A function is a special type of relation, so all functions are relations,but not all relations are functions.

Practice Problem 8Determine whether the relation

is also a function.x = y2 + 1

Practice Problems 9–13Use the vertical line test to determinewhich are graphs of functions.

9.

10.

11.

x

y

x

y

x

y


USING THE VERTICAL LINE TEST

As we have seen, not all relations are functions. Consider the graphs ofand shown next. On the graph of , notice

that each x-value corresponds to only one y-value. Recall from Example 7that is a function.

On the graph of , the x-value 9, for example, corresponds to twoy-values, 3 and , as shown by the vertical line. Recall from Example 8that is not a function.

Graphs can be used to help determine whether a relation is also a func-tion by the following vertical line test.

Examples Use the vertical line test to determine which are graphsof functions.

9. 10. 11.

This is the graph of This is the graph of This is not the grapha function since no a function. of a function. Notevertical line will in- that vertical linestersect this graph can be drawn thatmore than once. intersect the graph

in two points.

x

y

x

y

x

y

VERTICAL LINE TEST

If no vertical line can be drawn so that it intersects a graph more thanonce, the graph is the graph of a function.

x = y2-3

x = y2

456

321

1

−2−3−4

−2−3 2 3 4

(x, y)

y = 2x + 1

x

y

8642

2−2

−4−6−8

−4−6 4 6 8 10

(9, 3)

(9, −3)

12 14

x = y2

x

y

y = 2x + 1

y = 2x + 1x = y2y = 2x + 1

C

12.

13.

x

y

x

y

Answers9. function, 10. function, 11. not a function,12. function, 13. not a function


Answers14. domain: , range: ,15. domain: , range: ,16. domain: , range: ,17. domain: , range:

✓ Concept Check: a, c

C-2, 2 DC-2, 2 DA- q, q BA- q, q B

A- q, q BC0, q BC-3, 4 DC-2, 4 D

✓ CONCEPT CHECK

Determine which equations representfunctions. Explain your answer.a.b.c. x + y = 6

x = -5y = 14

Practice Problems 14–17Find the domain and range of eachrelation.

14.

15.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

x

y

21

3

(4, 4)

(−2, 1)

45

−2−3−4−5

−2−1−3−4−5 21 3 4 5

12. 13.

This is the graph This is not the graph of a function. of a function. A vertical line

can be drawn that intersects this line at every point.


FINDING DOMAIN AND RANGE FROM A GRAPH

Next we practice finding the domain and range of a relation from its graph.

Examples Find the domain and range of each relation.

Solutions:

14.

15.

Range:[0, ∞)

Domain: (−∞, ∞)

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(−3, 1)

Domain:The x-values graphed are

from −3 to 5, or[−3, 5]

(5, −2)

(2, 4)

The y-valuesgraphed are

from −2 to 4, or[−2, 4]

Range:

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(−3, 1)

(5, −2)

(2, 4)

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

D

x

y

x

y

16.

17.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5


16.

17.

USING FUNCTION NOTATION

Many times letters such as f, g, and h are used to name functions. To denotethat y is a function of x, we can write

This means that y is a function of x or that y depends on x. For this reason,y is called the dependent variable and x the independent variable. Thenotation is read “f of x” and is called function notation.

For example, to use function notation with the function , wewrite . The notation means to replace x with 1 and findthe resulting y or function value. Since

then

This means that when , y or . The corresponding orderedpair is . Here, the input is 1 and the output is or 7. Now let’s find

, and .

Ordered Pairs: A-1, -1 BA0, 3 BA2, 11 B

= -1 = 3 = 11 = -4 + 3 = 0 + 3 = 8 + 3

f A-1 B = 4 A-1 B + 3f A0 B = 4 A0 B + 3f A2 B = 4 A2 B + 3 f Ax B = 4x + 3f Ax B = 4x + 3f Ax B = 4x + 3

f A-1 Bf A2 B , f A0 Bf A1 BA1, 7 B

f Ax B = 7x = 1

f A1 B = 4 A1 B + 3 = 7

f Ax B = 4x + 3

f A1 Bf Ax B = 4x + 3y = 4x + 3

f Ax B

y = f Ax B

E

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

Domain:(−∞, ∞)

Range:(−∞, ∞)

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

Domain:[−4, 4]

Range:[−2, 2]x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5


Practice Problems 18–21Find each function value.

18. If , find .

19. If , find .

20. If , find .

21. If , find.f A-1 B

f Ax B = 3x2 - x + 2

f A2 Bf Ax B = 3x2 - x + 2

g A-5 Bg Ax B = 4x + 5

g A0 Bg Ax B = 4x + 5

Examples Find each function value.

18. If , find .

19. If , find .

20. If , find .

21. If , find .


GRAPHING LINEAR FUNCTIONS

Recall that the graph of a linear equation in two variables is a line, and aline that is not vertical will always pass the vertical line test. Thus, all linearequations are functions except those whose graph is a vertical line. We callsuch functions linear functions.

Example 22 Graph the function .

Solution: Since , we could replace with y and graphas usual. The graph of has slope 2 and y-intercept 1. Its graph is shown.

x

y

21

3

y = 2x + 1or

f(x) = 2x + 1

(0, 1)

45

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = 2x + 1f Ax By = f Ax B

f Ax B = 2x + 1

F

f A-2 B = 7 A-2 B 2 - 3 A-2 B + 1 = 35

f A-2 Bf Ax B = 7x2 - 3x + 1

f A1 B = 7 A1 B 2 - 3 A1 B + 1 = 5

f A1 Bf Ax B = 7x2 - 3x + 1

g A0 B = 3 A0 B - 2 = -2

g A0 Bg Ax B = 3x - 2

g A1 B = 3 A1 B - 2 = 1

g A1 Bg Ax B = 3x - 2

✓ CONCEPT CHECK

Suppose and we are toldthat . Which is not true?a. When

b. A possible function is .

c. A point on the graph of the func-tion is .

d. A possible function isf Ax B = 2x + 4.

A3, 9 Bf Ax B = x2

x = 3, y = 9.f A3 B = 9

y = f Ax B

HELPFUL HINT

Note that is a special symbol in mathematics used to denote afunction. The symbol is read “f of x.” It does not mean

.f # x Af times x Bf Ax B

f Ax B

LINEAR FUNCTION

A linear function is a function that can be written in the form

f Ax B = mx + b

Practice Problem 22Graph the function .

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

f Ax B = 3x - 2

Answers

18. , 19. ,

20. , 21.22.

✓ Concept Check: d

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(0, −2)

f(x) = 3x − 2or

y = 3x − 2

f A-1 B = 6f A2 B = 12

g A-5 B = -15g A0 B = 5


Focus On Mathematical ConnectionsPERPENDICULAR BISECTORS

A perpendicular bisector is a line that is perpendicular to a givenline segment and divides the segment into two equal lengths. Aperpendicular bisector crosses the line segment at the point that islocated exactly halfway between the two endpoints of the line segment. That point is called the midpoint of the line segment. If a line segment has the endpoints and , then the midpoint of this line segment is the point with coordinates

. An example of a line segment and its

perpendicular bisector is shown in the figure.

To find the equation of a line segment’s perpendicular bisector,follow these steps:

Step 1. Find the midpoint of the line segment.Step 2. Find the slope of the line segment.Step 3. Find the slope of a line that is perpendicular to the line seg-

ment.Step 4. Use the midpoint and the slope of the perpendicular line to

find the equation of the perpendicular bisector.

CRITICAL THINKING

Use the steps given above and what you have learned in this chapterto find the equation of the perpendicular bisector of each linesegment whose endpoints are given.

1. 2.

3. 4.

5. 6. A-6, 8 B ; A-4, -2 BA2, 3 B ; A-4, 7 BA5, 8 B ; A7, 2 BA-2, 6 B ; A-22, -4 BA-6, -3 B ; A-8, -1 BA3, -1 B ; A-5, 1 B

67

54321

1−1

−2−3

−2−3−4 2 3 4 5 6 7 8 9 10

Line segment

Perpendicularbisector

x + 4y = 9

x

y

(0, −2)

(2, 6)

a x1 + x2

2,

y1 + y2

2b

Ax2, y2 BAx1, y1 B

7.4 POLYNOMIAL AND RATIONAL FUNCTIONS

EVALUATING POLYNOMIAL FUNCTIONS

At times it is convenient to use function notation to represent polyno-mials. For example, we may write to represent the polynomial

. In symbols, we would write

This function is called a polynomial function because the expressionis a polynomial.

Examples If , find each function value.

1. Let in the function P(x).

2. Let in the function P(x).

Many real-world phenomena are modeled by polynomial functions. Ifthe polynomial function model is given, we can often find the solution of aproblem by evaluating the function at a certain value.

Example 3 Finding the Height of an ObjectThe world’s highest bridge, Royal Gorge suspensionbridge in Colorado, is 1053 feet above the ArkansasRiver. An object is dropped from the top of this bridge.Neglecting air resistance, the height of the object at timet seconds is given by the polynomial function

. Find the height of the objectwhen second and when seconds.

Solution: To find the height of the object at 1 second, we find .

When second, the height of the object is 1037 feet.

t = 1

P A1 B = 1037

P A1 B = -16 A1 B 2 + 1053

P A t B = -16t2 + 1053

P A1 Bt = 8t = 1

P A t B = -16t2 + 1053

x = -2P A-2 B = 3 A-2 B 2 - 2 A-2 B - 5 = 11x = 1P A1 B = 3 A1 B 2 - 2 A1 B - 5 = -4

P Ax B = 3x2 - 2x - 5

HELPFUL HINT

Recall that the symbol P(x) does not mean P times x. It is a specialsymbol used to denote a function.

3x2 - 2x - 5

P Ax B = 3x2 - 2x - 5

3x2 - 2x - 5P Ax B

A

Polynomial and Rational Functions SECTION 7.4 509

Objectives

Evaluate polynomial functions.

Evaluate rational functions.BA

SSM CD-ROM Video7.4

Practice Problems 1–2If , find eachfunction value.

1.

2. P A-1 BP A2 BP Ax B = 5x2 - 3x + 7

Practice Problem 3Use the polynomial function in Ex-ample 3 to find the height of the objectwhen seconds and sec-onds.

t = 7t = 3

Answers1. , 2. , 3. At 3 seconds, height is 909 feet; at 7 seconds,height is 269 feet

P A-1 B = 15P A2 B = 21


To find the height of the object at 8 seconds, we find.

When seconds, the height of the object is 13 feet.Notice that as time t increases, the height of the objectdecreases.

EVALUATING RATIONAL FUNCTIONS

Functions can also represent rational expressions. For example, we call the

function a rational function since is a rational

expression in one variable.

The domain of a rational function such as is the set of all

possible replacement values for x. In other words, since the rational ex-

pression is not defined when , we say that the domain of

is all real numbers except 3. We can write the domain as:

is a real number and

Example 4 Finding Unit Cost

For the ICL Production Company, the rational function

describes the company’s cost per

disc of pressing x compact discs. Find the cost per disc forpressing:

a. 100 compact discsb. 1000 compact discs

Solution: a.

The cost per disc for pressing 100 compact discs is$102.60.

b.

The cost per disc for pressing 1000 compact discs is$12.60. Notice that as more compact discs are produced,the cost per disc decreases.

C A1000 B =2.6 A1000 B + 10,000

1000= 12,600

1000= 12.6

C A100 B =2.6 A100 B + 10,000

100= 10,260

100= 102.6

C Ax B = 2.6x + 10,000x

x Z 3FEx @xf Ax B = x2 + 2

x - 3

x = 3x2 + 2x - 3

f Ax B = x2 + 2x - 3

x2 + 2x - 3

f Ax B = x2 + 2x - 3

B

t = 8

P A8 B = 13

P A8 B = -1024 + 1037

P A8 B = -16 A8 B 2 + 1037

P A t B = -16t2 + 1037

P A8 B

Practice Problem 4A company’s cost per book for print-ing x particular books is given by the

rational function .

Find the cost per book for printing:

a. 100 books

b. 1000 books

C Ax B = 0.8x + 5000x

Answers4. a. $50.80, b. $5.80

GRAPHING CALCULATOR EXPLORATIONS

Recall that since the rational expression is not defined when or when , we

say that the domain of the rational function is all real numbers except 2 and .

This domain can be written as is a real number and , . This means that the graph of should not cross the vertical lines and . The graph of in connected mode follows. In con-nected mode the grapher tries to connect all dots of the graph so that the result is a smooth curve. This is whathas happened in the graph. Notice that the graph appears to contain vertical lines at and at .We know that this cannot happen because the function is not defined at and at . We also knowthat this cannot happen because the graph of this function would not pass the vertical line test.

If we graph in dot mode, the graph appears as follows. In dot mode the grapher will not connect dotswith a smooth curve. Notice that the vertical lines have disappeared, and we have a better picture of thegraph. The graph, however, actually appears more like the hand-drawn graph to its right. By using a Tablefeature, a Calculate Value feature, or by tracing, we can see that the function is not defined at and at

.

Find the domain of each rational function. Then graph each rational function and use the graph to confirm thedomain.

1. 2.

3. 4. f Ax B = 3x + 24x2 - 19x - 5

h Ax B = x2

2x2 + 7x - 4

g Ax B = 5xx2 - 9

f Ax B = x + 1x2 - 4

32−2 −1−1−2−3−4−5

−3−4−5−6−7−8−9−10 1

12345

4 5 6 x

y

−10

−10

10

10

x = -5x = 2

f Ax B

10

10

−10

−10

x = -5x = 2x = -5x = 2

f Ax Bx = -5x = 2f Ax Bx Z -5Fx Z 2Ex @x

-5f Ax B = 7x - 2Ax - 2 B Ax + 5 B

x = -5x = 27x - 2

Ax - 2 B Ax + 5 B

Polynomial and Rational Functions SECTION 7.4 511


Focus On Mathematical ConnectionsFINITE DIFFERENCES

When polynomial functions are evaluated at successive integer values, a list of values called asequence is generated. The differences between successive pairs of numbers in such a sequencehave special properties. Let’s investigate these properties, beginning with a first-degree polyno-mial function, the linear function.

Notice in the table below on the left that first differences are the differences between the suc-cessive pairs of numbers in the original sequence. Find the first differences for any other linearfunction and fill in the table on the right. What do you notice? (Note: You may wish to try sev-eral different linear functions.)

Now let’s look at differences for a second-degree polynomial. Notice in the table below onthe left that second differences are the differences between successive pairs of first differences.Find first and second differences for any other second-degree polynomial function and fill inthe table on the right. What do you notice? (Note: You may wish to try several differentsecond-degree polynomial functions.)

CRITICAL THINKING

1. As you might guess, third differences are the differences between successive pairs of seconddifferences. Find the first, second, and third differences for any two third-degree polynomialfunctions. What do you notice?

2. What would you expect to be true about the differences for a fourth-degree polynomial func-tion?

3. What would you expect to be true about the differences for an nth-degree polynomial func-tion?

Original Sequence Firstx Differences

8 287 25 3

6 22 3

5 19 3

4 16 3

3 13 3

2 10 3

1 7 3

f Ax B = 3x + 4Original Sequence First

x Differencesf Ax B =

Original Sequence First Secondx Differences Differencesf Ax B =

Original Sequence First Secondx Differences Differences

8 108277 81 4

6 58 234

5 39 194

4 24 154

3 13 114

2 6 74

1 3 3

f Ax B = 2x2 - 3x + 4

513


EXERCISE SET 7.4

If and , find each function value. See Examples 1and 2.

1. 2. 3.

4. 5. 6.

If and , find each function value. See Examples 1 and 2.

7. 8. 9.

10.

If and , find each function value. See Example 4.

11. 12. 13.

14. 15. 16.

If and , find each function value. See Example 4.

17. 18. 19.

20. f A-2 B

g A-6 Bg A3 Bf A1 Bg Ax B = x2 + 2x

x + 3f Ax B = x2 + 5

x

g A-5 Bf A-1 Bf A0 B

g A0 Bg A10 Bf A2 Bg Ax B = x - 2

x - 5f Ax B = x + 8

2x - 1B

P A-1 B

Q A-3 BQ A6 BP A2 BQ Ax B = 7x + 5P Ax B = x3 + 2x - 3

Q A0 BP A0 BP A-4 B

Q A-10 BQ A4 BP A7 BQ Ax B = 5x2 - 1P Ax B = x2 + x + 1A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

7.5 AN INTRODUCTION TO GRAPHING POLYNOMIAL FUNCTIONS

ANALYZING GRAPHS OF POLYNOMIAL FUNCTIONS

Some polynomial functions are given special names according to theirdegree. For example,

is called a linear function; its degree is one.is called a quadratic function; its degree is two.

is called a cubic function; its degree is three.

is called a quartic function; its degreeis four.

All the above functions are also polynomial functions.

Example 1 Given the graph of the function below to the left:

a. Find the domain and the range of the function.b. List the x- and y-intercept points.c. Find the coordinates of the point with the greatest

y-value.d. Find the coordinates of the point with the least

y-value.

Solution: a. The domain is the set of all real numbers, or in inter-val notation, . The range is .

b. The x-intercept points are , , , and. The y-intercept point is .

c. The point with the greatest y-value corresponds to the“highest” point. This is the point with coordinates

. (This means that for all real number values forx, the greatest y-value, or value, is 4.)

d. The point with the least y-value corresponds to the“lowest” point. This graph contains no “lowest” point,so there is no point with the least y-value.

The graph of any polynomial function (linear, quadratic, cubic, and soon) can be sketched by plotting a sufficient number of ordered pairs thatsatisfy the function and connecting them to form a smooth curve. Thegraphs of all polynomial functions will pass the vertical line test since theyare graphs of functions.

g Ax BA3, 4 B

A0, -2 BA5, 0 BA1, 0 BA-1, 0 BA-3, 0 B

A- q, 4 DA- q, q B

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

g Ax B

f Ax B = -8x4 - 3x3 + 2x2 + 20

f Ax B = 7x3 + 3x2 - 1f Ax B = 5x2 - x + 3f Ax B = 2x - 6

A

An Introduction to Graphing Polynomial Functions SECTION 7.5 515

Objectives

Analyze the graph of a polynomialfunction.

Graph quadratic functions.

Find the vertex of a parabola byusing the vertex formula.

Graph cubic functions.D

CB

A

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4

Range:(−∞, 4]


(−1, 0)

(−2, 2)

Highest point(3, 4)

No lowestpoint

(1, 0)

(0, −2)

(5, 0)

(−3, 0)

5

Practice Problem 1Given the graph of the function :

a. Find the domain and the range ofthe function.

b. List the x- and y-intercept points.c. Find the coordinates of the point

with the greatest y-value.d. Find the coordinates of the point

with the least y-value.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

f Ax B

Answers1. a. domain: ; range: ,b. , c. no greatest y-value point, d. A3, -3 B

A2, 0 B , A4, 0 BA-2, 0 B , A0, 0 B ,C-3, q BA- q, q B

SSM CD-ROM Video7.5


Answer2.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

GRAPHING QUADRATIC FUNCTIONS

Since we know how to graph linear functions (see Section 7.3), we will nowgraph quadratic functions and discuss special characteristics of their graphs.

We know that an equation of the form may be written as . Thus, both and

define quadratic functions as long as a is not 0.

Example 2 Graph the function by plotting points.

Solution: This function is not linear, and its graph is not a line. Webegin by finding ordered pair solutions. Then we plot thepoints and draw a smooth curve through them.

If , then , or 9.If , then , or 4.If , then , or 1.If , then , or 0.If , then , or 1.If , then , or 4.

If , then , or 9.

Notice that the graph of Example 2 passes the vertical line test, as itshould since it is a function. This curve is called a parabola. The highestpoint on a parabola that opens downward or the lowest point on a parabolathat opens upward is called the vertex of the parabola. The vertex of thisparabola is , the lowest point on the graph. If we fold the graph alongthe y-axis, we can see that the two sides of the graph coincide. This meansthat this curve is symmetric about the y-axis, and the y-axis, or the line

, is called the axis of symmetry. The graph of every quadratic func-tion is a parabola and has an axis of symmetry: the vertical line that passesthrough the vertex of the parabola.

x = 0

A0, 0 B

x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8

(3, 9)(−3, 9)

(2, 4)(−2, 4)

(1, 1)

(0, 0)

(−1, 1)

10

f A3 B = 32x = 3

f A2 B = 22x = 2f A1 B = 12x = 1f A0 B = 02x = 0

f A-1 B = A-1 B 2x = -1f A-2 B = A-2 B 2x = -2f A-3 B = A-3 B 2x = -3

f Ax B = x2

y = ax2 + bx + cf Ax B = ax2 + bx + cy = ax2 + bx + c

f Ax B = ax2 + bx + c

B

Practice Problem 2Graph the function .

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f Ax B = -x2

QUADRATIC FUNCTION

A quadratic function is a function that can be written in the form

where a, b, and c are real numbers and .a Z 0


x

941

0 01 12 43 9

-1-2-3

y = f Ax B


Answer3.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

Practice Problem 3Graph the quadratic function

.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

-x2 - 2x - 3f Ax B =

Example 3 Graph the quadratic function byplotting points.

Solution: To graph, we choose values for x and find correspondingor y-values. Then we plot the points and draw a

smooth curve through them.

The vertex of this parabola is , the highestpoint on the graph. The vertical line is the axis ofsymmetry. Recall that to find the x-intercepts of a graph,we let . Using function notation, this is the same asletting . Since this graph has no x-intercepts, itmeans that has no real number solu-tions.

Notice that the parabola opens downward,whereas opens upward. When the equation of a quadratic func-tion is written in the form , the coefficient of thesquared variable, a, determines whether the parabola opens downward orupward. If , the parabola opens upward, and if , the parabolaopens downward.

FINDING THE VERTEX OF A PARABOLA

In both and , the vertex happens to beone of the points we chose to plot. Since this is not always the case, andsince plotting the vertex allows us to draw the graph quickly, we need a con-sistent method for finding the vertex. One method is to use the followingformula, which we shall derive in Chapter 10.

f Ax B = -x2 + 2x - 3f Ax B = x2

C

x

y

f(x) = ax2 + bx + c,a > 0, opens upward

f(x) = ax2 + bx + c,a < 0, opens downward

x

y

a 6 0a 7 0

f Ax B = ax2 + bx + cf Ax B = x2

f Ax B = -x2 + 2x - 3

0 = -x2 + 2x - 3f Ax B = 0y = 0

x = 1A1, -2 B

x

y

2

64

108

−4−6−8

−10

−4−8 −2−6−10 4 82 6 10

x = 1

f(x) = −x2 + 2x − 3

f Ax B

f Ax B = -x2 + 2x - 3

x

0123 -6

-3-2-3-6-1-11-2

y = f Ax B


Answer4.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4

(1, −4)

5

Practice Problem 4Graph . Find thevertex and any intercepts.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f Ax B = x2 - 2x - 3

We can also find the x- and y-intercepts of a parabola to aid in graphing.Recall that x-intercepts of the graph of any equation may be found by let-ting or in the equation and solving for x. Also, y-interceptsmay be found by letting in the equation and solving for y or .

Example 4 Graph . Find the vertex and anyintercepts.

Solution: To find the vertex, we use the vertex formula. For thefunction , and . Thus,

Find .

The vertex is , and since is greater than 0,this parabola opens upward. Graph the vertex and noticethat this parabola will have two x-intercepts because itsvertex lies below the x-axis and it opens upward. To findthe x-intercepts, we let y or and solve for x.

To find the y-intercept, let .

or

The x-intercepts are and 1 and the correspondingpoints are . The y-intercept is andthe corresponding point is .

Now we plot these points and connect them with asmooth curve.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

(−1, −4)

(−3, 0) (1, 0)

(0, −3)f(x) = x2 + 2x − 3

A0, -3 B-3A-3, 0 B and A1, 0 B

-3

x = 1 x = -3

x - 1 = 0x + 3 = 0

f A0 B = -3 0 = Ax + 3 B Ax - 1 Bf A0 B = 02 + 2 A0 B - 3 0 = x2 + 2x - 3

f Ax B = x2 + 2x - 3 f Ax B = x2 + 2x - 3

x = 0f Ax B = 0

a = 1A-1, -4 B = -4

= 1 - 2 - 3

f A-1 Bf A-1 B = A-1 B 2 + 2 A-1 B - 3x = -b2a

= -22 A1 B = -1

b = 2a = 1f Ax B = x2 + 2x - 3

f Ax B = x2 + 2x - 3

f Ax Bx = 0f Ax B = 0y = 0

VERTEX FORMULA

The graph of , , is a parabola with vertex

a -b2a

, f a -b2ab b

a Z 0f Ax B = ax2 + bx + c


Answer5.

x

y

1

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

2

Practice Problem 5

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

Example 5 Graph . Find the vertex and anyintercepts.

Solution: To find the vertex, we use the vertex formula. For thefunction , and . Thus

Find .

The vertex is . Also, this parabola opens upwardsince is greater than 0. Graph the vertex andnotice that this parabola has no x-intercepts: Its vertexlies above the x-axis, and it opens upward.

To find the y-intercept, let .

The y-intercept is 13. Use this information along withsymmetry of a parabola to sketch the graph of

.

In Section 10.5 we study the graphing of quadratic functions further.

GRAPHING CUBIC FUNCTIONS

To sketch the graph of a cubic function, we again plot points and then con-nect the points with a smooth curve. The general shapes of cubic graphs aregiven below.

D

x

y

42

68

101214

−4−6

−4−2−6−8−10 42 6 8 10

(0, 13)

(2, 1)

f(x) = 3x2 − 12x + 13

3x2 - 12x + 13f Ax B =

= 13

= 0 - 0 + 13

f A0 B = 3 A0 B 2 - 12 A0 B + 13

x = 0

a = 3A2, 1 B

= 1

= 3 A4 B - 24 + 13

f A2 Bf A2 B = 3 A2 B 2 - 12 A2 B + 13x = -b2a

=- A-12 B

2 A3 B = 126

= 2

b = -12a = 3y = 3x2 - 12x + 13

f Ax B = 3x2 - 12x + 13

HELPFUL HINT

Not all graphs of parabolas have x-intercepts. To see this, first plot thevertex of the parabola and decide whether the parabola opens upwardor downward. Then use this information to decide whether the graphof the parabola has x-intercepts.

Graph . Findthe vertex and any intercepts.

f Ax B = 2x2 + 4x + 4


Answer6.

x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

Example 6 Graph . Find any intercepts.

Solution: To find x-intercepts, we let y or and solve for x.

Let .

Factor.

or or

This graph has three x-intercepts. They are 0, , and 2.To find the y-intercept, we let .

Next let’s select some x-values and find their correspond-ing or y-values.

Finally, we plot the intercepts and points and connectthem with a smooth curve.

HELPFUL HINT

When a graph has an x-intercept of 0, notice that the y-intercept willalso be 0.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3

f(x) = x3 − 4x

4 5

f A3 B = 33 - 4 A3 B = 27 - 12 = 15

f A1 B = 13 - 4 A1 B = 1 - 4 = -3

f A-1 B = A-1 B 3 - 4 A-1 B = -1 + 4 = 3

f A-3 B = A-3 B 3 - 4 A-3 B = -27 + 12 = -15

f Ax B = x3 - 4x

f Ax Bf A0 B = 03 - 4 A0 B = 0

x = 0-2

x = 2 x = -2x = 0

Set each factorequal to 0.Solve.

x - 2 = 0x + 2 = 0x = 0 0 = x Ax + 2 B Ax - 2 B 0 = x Ax2 - 4 B

f Ax B = 0 0 = x3 - 4x

f Ax B = x3 - 4x

f Ax B = 0

f Ax B = x3 - 4x

Graph of a Cubic Function

Coefficient of x3

is a positive number.Coefficient of x3

is a negative number.

x

y

x

y

x

y

x

y

(Degree 3)

Practice Problem 6Graph . Find anyintercepts.

x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

f Ax B = x3 - 9x

x

313 15

-3-1

-15-3

f Ax B


Answer7.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

Example 7 Graph . Find any intercepts.

Solution: To find x-intercepts, we let y or and solve for x.

The only x-intercept is 0. This means that the y-interceptis 0 also.

Next we choose some x-values and find their corre-sponding y-values.

Now we plot the points and sketch the graph of.

x

y

42

68

10

−4−6−8

−10

−4 −2−6−8−10 42 6

f(x) = −x3

8 10

f Ax B = -x3

f A2 B = -23 = -8

f A1 B = - A1 B 3 = -1

f A-1 B = - A-1 B 3 = 1

f A-2 B = - A-2 B 3 = 8

f Ax B = -x3

0 = x

0 = -x3

f Ax B = -x3

f Ax B = 0

f Ax B = -x3

HELPFUL HINT

If you are unsure about the graph of a function, plot more points.

Practice Problem 7Graph . Find any inter-cepts.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f Ax B = 2x3

x

81

12 -8

-1-1-2

f Ax B


GRAPHING CALCULATOR EXPLORATIONSWe can use a grapher to approximate real number solutions of any quadratic equation in standard form,whether the associated polynomial is factorable or not. For example, let’s solve the quadratic equation

. The solutions of this equation will be the x-intercepts of the graph of the function. (Recall that to find x-intercepts, we let , or .) When we use a standard

window, the graph of this function looks like this:

The graph appears to have one x-intercept between and and one between 3 and 4. To find the x-inter-cept between 3 and 4 to the nearest hundredth, we can use a Root feature, a Zoom feature, which magnifies aportion of the graph around the cursor, or we can redefine our window. If we redefine our window to

the resulting screen is

By using the Trace feature, we can now see that one of the intercepts is between 3.21 and 3.25. To approxi-mate to the nearest hundredth, Zoom again or redefine the window to

If we use the Trace feature again, we see that, to the nearest thousandth, the x-intercept is 3.236. By repeat-ing this process, we can approximate the other x-intercept to be .

To check, find and . Both of these values should be close to 0. (They will not be exactly 0since we approximated these solutions.)

and

Solve each of these quadratic equations by graphing a related function and approximating the x-intercepts to thenearest thousandth.

1. 2.

3. 4.

5. 6. x2 + 0.08x - 0.01 = 00.09x2 - 0.13x - 0.08 = 0

0.2x2 + 6.2x + 2.1 = 02.3x2 - 4.4x - 5.6 = 0

5x2 - 7x + 1 = 0x2 + 3x - 2 = 0

f A-1.236 B = -0.000304f A3.236 B = -0.000304

f A-1.236 Bf A3.236 B -1.236

Yscl = 1 Xscl = 1

Ymax = 0.1Xmax = 3.3

Ymin = -0.1Xmin = 3.2

2 5

Yscl = 1 Xscl = 1

Ymax = 1Xmax = 5

Ymin = -1Xmin = 2

-1-2

y = x2 − 2x − 4

10−10

−10

10

y = 0f Ax B = 0f Ax B = x2 - 2x - 4x2 - 2x - 4 = 0


EXERCISE SET 7.5

For the graph of each function answer the following. See Example 1.a. Find the domain and the range of the function.b. List the x- and y-intercept points.c. Find the coordinates of the point with the greatest y-value.d. Find the coordinates of the point with the least y-value.

1. 2.

3. The graph in Example 4 of this section 4. The graph in Example 5 of this section

5. The graph in Example 6 of this section 6. The graph in Example 7 of this section

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

f Ax BA

ANSWERS

1. a.

b.

c.

d.

2. a.

b.

c.

d.

3. a.

b.

c.

d.

4. a.

b.

c.

d.

5. a.

b.

c.

d.

6. a.

b.

c.

d.

523

10. 11. 12.

Graph each quadratic function. Find and label the vertex and intercepts. See Examples 4 and 5.

13. 14. 15. 16.

x

y

42

68

10

−4−6−8

−10

−2−4−6−8−10 42 6 8 10x

y

84

121620

−8−12−16−20

−4−12−20 4 12 20x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

f Ax B = x2 - 12x + 35f Ax B = x2 - 2x - 24f Ax B = x2 + 6x + 5f Ax B = x2 + 8x + 7C

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

f Ax B = 12

x2f Ax B = -x2f Ax B = x2 - 2

Name _________________________________________________________________________

524

Graph each quadratic function by plotting points. See Examples 2 and 3.

7. 8. 9.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

f Ax B = x2 + 1f Ax B = -3x2f Ax B = 2x2

B

17. 18. 19. 20.

x

y

21

345

−2−3−4−5

−1−2−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−1−2−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−1−2−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−1−2−3−4−5 21 3 4 5

f Ax B = x2 + 4f Ax B = x2 + 1f Ax B = -3x2 + 6xf Ax B = 2x2 - 6x

525

Name _________________________________________________________________________



SECTION 7.1 THE SLOPE-INTERCEPT FORM

We can use the slope-intercept form to write anequation of a line given its slope and y-intercept.

Write an equation of the line with y-intercept

and slope .

y = 23

x - 1

y = mx + b

23

-1

SECTION 7.2 MORE EQUATIONS OF LINES

The point-slope form of the equation of a line is

where m is the slope of the line and is a pointon the line.

Ax1, y1 By - y1 = m Ax - x1 B

Find an equation of the line with slope 2 containingthe point . Write the equation in standardform: .

Standard form -2x + y = -6

y + 4 = 2x - 2y - A-4 B = 2 Ax - 1 B

y - y1 = m Ax - x1 BAx + By = CA1, -4 B

SECTION 7.3 INTRODUCTION TO FUNCTIONS

A relation is a set of ordered pairs. The domain ofthe relation is the set of all first components of theordered pairs. The range of the relation is the set ofall second components of the ordered pairs.

Domain: {cat, dog, too, give}Range: {1, 2}

Output:Input: Number ofWords Correspondence Vowels

cat 1dog

3too

2give

A function is a relation in which each element of thefirst set corresponds to exactly one element of thesecond set.

The previous relation is a function. Each word con-tains one exact number of vowels.

VERTICAL LINE TEST

If no vertical line can be drawn so that it intersects agraph more than once, the graph is the graph of afunction.

Find the domain and the range of the relation. Alsodetermine whether the relation is a function.

By the vertical line test, this is the graph of a func-tion.

x

y


Range:(−∞, 0]


If , find .

f A-3 B = 2 A-3 B 2 - 5 = 2 A9 B - 5 = 13

f A-3 Bf Ax B = 2x2 - 5

A linear function is a function that can be written inthe form

f Ax B = mx + b

f Ax B = -3, g Ax B = 5x, h Ax B = - 13

x - 7

To graph a linear function, use the slope and y-inter-cept.

Graph: (or )

The slope is .

y-intercept is 0, or the point (0, 0).

2-1

(0, 0)

f(x) = −2x

x

yy = -2x + 0

f Ax B = -2x


The symbol means function of x and is calledfunction notation.

f Ax B

SECTION 7.4 POLYNOMIAL AND RATIONAL FUNCTIONS

A function P is a polynomial function if is apolynomial.

A rational function is a function described by a ratio-nal expression.

P Ax B For the polynomial function

find .

, h 1 t 2 = t2 - 3t + 5t - 1

f 1x 2 = 2x - 67

P A-2 B = - A-2 B 2 + 6 A-2 B - 12 = -28

P A-2 BP Ax B = -x2 + 6x - 12

SECTION 7.5 AN INTRODUCTION TO GRAPHING POLYNOMIAL FUNCTIONS

TO GRAPH A POLYNOMIAL FUNCTION

Find and plot x- and y-intercepts and a sufficientnumber of ordered pair solutions. Then connect theplotted points with a smooth curve.

Graph:

or or

The x-intercept points are , , and .

The y-intercept point is .

x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

A0, 0 Bf A0 B = 03 + 2 # 02 - 3 # 0 = 0

A-3, 0 BA1, 0 BA0, 0 Bx = -3x = 1x = 0

0 = x Ax - 1 B Ax + 3 B0 = x3 + 2x2 - 3x

f Ax B = x3 + 2x2 - 3x

x

64

2 10

- 78

12

-1-2

-20-4

f Ax B


A quadratic function is a function that can be writtenin the form

,

The graph of this quadratic function is a parabola

with vertex .a -b2a

, f a -b2ab b


Find the vertex of the graph of the quadratic func-tion

Here and .

The vertex has coordinates .A2, -7 Bf A2 B = 2 # 22 - 8 # A2 B + 1 = -7

-b2a

=- A-8 B2 # 2

= 2

b = -8a = 2

f Ax B = 2x2 - 8x + 1


Focus On HistoryCARTESIAN COORDINATE SYSTEM

The French mathematician and philosopher René Descartes(1596–1650) is generally credited with devising the rectangularcoordinate system that we use in mathematics today. It is said thatDescartes thought of describing the location of a point in a planeusing a fixed frame of reference while watching a fly crawl on hisceiling as he laid in bed one morning meditating. He incorporatedthis idea of defining a point’s position in the plane by giving itsdistances, x and y, to two fixed axes in his text La Géométrie.

Although Descartes is credited with the concept of the rectan-gular coordinate system, nowhere in his written works does anexample of the modern gridlike coordinate system appear. Healso never referred to a point’s location as we do today with

-notation giving an ordered pair of coordinates. In fact,Descartes never even used the term coordinate! Instead, his basicideas were expanded upon by later mathematicians. The Dutchmathematician Frans van Schooten (1615–1660) is credited withmaking Descartes’ concept of a coordinate system widelyaccepted with his text, Geometria a Renato Des Cartes (Geometryby René Descartes). The German mathematician Gottfried Wil-helm Leibniz (1646–1716) later contributed the terms abscissa(the x-axis in the modern rectangular coordinate system), ordinate(the y-axis), and coordinate to the development of the rectangularcoordinate system.

Ax, y B

529

Focus On the Real WorldCENTERS OF MASS

The center of mass, also known as center of gravity, of an object is the point at which the massof the object may be considered to be concentrated. For a two-dimensional object or surface,such as a flat board, the center of mass can be described as the point on which the surfacewould balance.

The idea of center of mass is an important one in many disciplines, especially physics and itsapplications. The following list describes situations in which an object’s center of mass isimportant.

j Geographers are sometimes concerned with pinpointing the geographic center of a county,state, country, or continent. A geographic center is actually the center of mass of a geographicregion if it is considered as a two-dimensional surface. The geographic center of the 48 con-tiguous United States is near Lebanon, Kansas. The geographic center of the North Americancontinent is 6 miles west of Balta, North Dakota.

j A top-loading washing machine is designed so its center of mass is located within its agitatorpost. During the spin cycle, the washer tub spins around its center of mass. If clothes aren’tcarefully distributed within the tub, they can bunch up and throw off the center of mass. Thiscauses the machine to vibrate, sometimes jumping or shaking wildly. Some washing machinemodels will stop operating when the loads become “unbalanced” in this way.

j Single-hulled boats are normally designed so that their centers of mass are below the waterline. This provides a boat with stability. Otherwise, with a center of mass above the water line,the boat would have a tendency to tip over in the water.

j Most small airplanes must be carefully loaded so as not to affect the location of the airplane’scenter of mass. The center of mass of an airplane must be near the center of the wings. Pilotsof small aircraft usually try to balance the weight of their cargo around the airplane’s centerof mass.

GROUP ACTIVITY

Attach a piece of graph paper to a piece of cardboard. Cut out a triangle and label the vertices withtheir coordinates. Lay the triangle on a horizontal table top with the graph paper face down. Slidethe triangle toward the edge of the table until it is balanced on the edge, just about to tip over theside of the table. Firmly hold the triangle in place while another group member uses the straightedge of the table to draw a line on the graph paper side of the triangle marking the position of thetable edge. Rotate the triangle a quarter turn and rebalance the triangle on the edge of the table.Draw a second line on the graph paper side of the triangle marking the position of the table edge.

1. The point where the two lines drawn on the triangle intersect is the center of mass of the tri-angle. Find the coordinates of this point.

2. Verify that the point you have located is roughly the center of mass of the triangle by bal-ancing it at this point on the tip of a pencil or pen.

3. List the coordinates of the vertices of your triangle. What is the relationship between thecoordinates of the center of mass and the coordinates of the triangle's vertices? (Hint: Youmay find it helpful to examine the sum of the x-coordinates and the sum of the y-coordinatesof the vertices of the triangle.)

4. Test your observation in Question 3. Cut out another triangle. Label its vertices and, usingyour observation from Question 3, predict the location of the center of mass of the triangle.Use the balancing procedure to find the center of mass. How close was your prediction?

531

8In this chapter, two or more equations in two or more variables aresolved simultaneously. Such a collection of equations is called a systemof equations. Systems of equations are good mathematical models formany real-world problems because these problems may involve severalrelated patterns. We will study various methods for solving systems ofequations and will conclude with a look at systems of inequalities.

8.1 Solving Systems of LinearEquations by Graphing

8.2 Solving Systems of LinearEquations by Substitution

8.3 Solving Systems of LinearEquations by Addition

8.4 Systems of Linear Equationsand Problem Solving

8.5 Solving Systems of LinearEquations in Three Variables

8.6 Solving Systems of EquationsUsing Matrices

8.7 Solving Systems of EquationsUsing Determinants

8.8 Systems of Linear Inequalities

L ightning, most often produced during thunderstorms, is arapid discharge of high-current electricity into the atmosphere.At any given moment around the world, there are about 2000thunderstorms in progress producing approximately 100 light-ning flashes per second. In the United States, lightning causesan average of 75 fatalities per year. An estimated 5% of all resi-dential insurance claims in the United States are due to light-ning damage, totaling more than $1 billion per year. In addi-tion, roughly 30% of all power outages in the United States arelightning related. Because of lightning’s potentially destructivenature, meteorologists track lightning activity by recording andplotting the positions of lightning strikes.

Systems of Equations and Inequalities C H A P T E R

8.1 SOLVING SYSTEMS OF LINEAR EQUATIONS BY GRAPHING

A system of linear equations consists of two or more linear equations. Inthis section, we focus on solving systems of linear equations containing twoequations in two variables. Examples of such linear systems are

DECIDING WHETHER AN ORDERED PAIR IS A SOLUTION

A solution of a system of two equations in two variables is an ordered pairof numbers that is a solution of both equations in the system.

Example 1 Determine whether (12, 6) is a solution of the system

Solution: To determine whether (12, 6) is a solution of the system,we replace x with 12 and y with 6 in both equations.

First equation Second equation

Let and . Let and .

Simplify. True.

True.

Since (12, 6) is a solution of both equations, it is a solu-tion of the system.

Example 2 Determine whether is a solution of the system

Solution: We replace x with and y with 2 in both equations.


Let Let and . and .Simplify. Simplify.

True. False.

is not a solution of the second equation,, so it is not a solution of the system.

SOLVING SYSTEMS OF EQUATIONS BY GRAPHING

Since a solution of a system of two equations in two variables is a solutioncommon to both equations, it is also a point common to the graphs of bothequations. Let’s practice finding solutions of both equations in a system—that is, solutions of a system—by graphing and identifying points of inter-section.

B

4x - y = 6A-1, 2 B

-6 = 6 3 = 3

-4 - 2 � 6 -1 + 4 � 3y = 2y = 2x = -1 4 A-1 B - 2 � 6x = -1-1 + 2 A2 B � 3

4x - y = 6 x + 2y = 3

-1

ex + 2y = 34x - y = 6

A-1, 2 B

6 = 6

12 = 12 24 - 18 � 6

y = 6x = 1212 � 2 A6 By = 6x = 122 A12 B - 3 A6 B � 6

x = 2y 2x - 3y = 6

e2x - 3y = 6x = 2y

A

ey = 7x - 1y = 4

ex - y = 02x + y = 10

e3x - 3y = 0x = 2y

Solving Systems of Linear Equations by Graphing SECTION 8.1 533

Objectives

Decide whether an ordered pair isa solution of a system of linearequations.

Solve a system of linear equationsby graphing.

Identify special systems: those withno solution and those with an infi-nite number of solutions.

C

B

A

Practice Problem 1Determine whether (3, 9) is a solutionof the system

e5x - 2y = -3y = 3x

Practice Problem 2Determine whether is a solu-tion of the system

e2x - y = 8x + 3y = 4

A3, -2 B

Answers1. (3, 9) is a solution of the system, 2.is not a solution of the system

A3, -2 B

SSM CD-ROM Video8.1

Example 3 Solve the system of equations by graphing.

Solution: On a single set of axes, graph each linear equation.

x y

0

22 4

x y

0 22 01 1

The two lines appear to intersect at the point . Tocheck, we replace x with and y with 3 in both equa-tions.

First equation Second equationLet Letand . and .Simplify. True.

True.

checks, so it is the solution of the system.


Solution: We graph each linear equation on a single set of axes.

The two lines appear to intersect at the point . Todetermine whether is the solution, we replace xwith 2 and y with in both equations.-2

A2, -2 B A2, -2 B

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5(2, −2)

2x + 3y = −2 x = 2

e2x + 3y = -2x = 2

HELPFUL HINT

Neatly drawn graphs can help when “guessing” the solution of a systemof linear equations by graphing.

A-1, 3 B 10 = 10

2 = 2 1 + 9 � 10y = 3y = 3x = -1-1 + 3 � 2x = -1- A-1 B + 3 A3 B � 10

x + y = 2 -x + 3y = 10

-1A-1, 3 B

x + y = 2

-4

103

HELPFUL HINT

The point ofintersectiongives thesolution ofthe system.x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x + y = 2−x + 3y = 10(−1, 3)

-x + 3y = 10

e -x + 3y = 10x + y = 2

Practice Problem 3Solve the system of equations bygraphing

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

e -3x + y = -10x - y = 6

534 CHAPTER 8 Systems of Equations and Inequalities

Practice Problem 4Solve the system of equations bygraphing.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

ex + 3y = -1y = 1

Answers3. please see page 536, 4. please see page 536



x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

e3x - y = 66x = 2y


Let and . Let .

Simplify. True.

True.

Since a true statement results in both equations, is the solution of the system.

IDENTIFYING SPECIAL SYSTEMS OF LINEAR EQUATIONS

Not all systems of linear equations have a single solution. Some systemshave no solution and some have an infinite number of solutions.


Solution: We graph the two equations in the system. The equationsin slope-intercept form are .Notice from the equations that the lines have the sameslope, , and different y-intercepts. This means that thelines are parallel.

Since the lines are parallel, they do not intersect. Thismeans that the system has no solution.


Solution: We graph each equation.

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5

x − y = 3−x + y = −3

x

y

(4, 1)(3, 0)

ex - y = 3-x + y = -3

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4 5

2y = −4x 2x + y = 7

x

y

-2

y = -2x + 7 and y = -2x

e2x + y = 72y = -4x

C

A2, -2 B -2 = -2

2 = 2 4 + A-6 B � -2

x = 22 = 2y = -2x = 22 A2 B + 3 A-2 B � -2

x = 2 2x + 3y = -2


x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

e3x + 4y = 129x + 12y = 36

Answers5. please see page 536, 6. please see page 536.

The graphs of the equations are the same line. To seethis, notice that if both sides of the first equation in thesystem are multiplied by , the result is the secondequation.

First equation

Multiply both sides by .

Simplify. This is the second equation.

This means that the system has an infinite number ofsolutions. Any ordered pair that is a solution of oneequation is a solution of the other and is then a solutionof the system.

Examples 5 and 6 are special cases of systems of linear equations. A sys-tem that has no solution is said to be an inconsistent system. If the graphsof the two equations of a system are identical—we call the equationsdependent equations.

As we have seen, three different situations can occur when graphing thetwo lines associated with the equations in a linear system. These situationsare shown in the figures.

-x + y = -3

-1-1 Ax - y B = -1 A3 B x - y = 3

-1


x

yParallel lines: no solution

x

yOne point of intersection: one solution

Answers3.

4.

5. no solution

6. infinite number of solutions

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 53x + 4y = 129x + 12y = 36

3x − y = 6

6x = 2y

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x + 3y = −1

y = 1(−4, 1)

A-4, 1 B−3x + y = −10

x − y = 6(2, −4)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

A2, -4 B

x

ySame line: infinite number of solutions


GRAPHING CALCULATOR EXPLORATIONSA graphing calculator may be used to approximate solutions

of systems of equations. For example, to approximate the solutionof the system

first graph each equation on the same set of axes. Then use theintersect feature of your calculator to approximate the point ofintersection.

The approximate point of intersection is

Solve each system of equations. Approximate the solutions to twodecimal places.

1.

2.

3.

4. e -3.6x - 8.6y = 10-4.5x + 9.6y = -7.7

e4.3x - 2.9y = 5.68.1x + 7.6y = -14.1

ey = 4.25x + 3.89y = -1.88x + 3.21

ey = -2.68x + 1.21y = 5.22x - 1.68

A-0.82, 1.23 B .

ey = -3.14x - 1.35y = 4.88x + 5.25 ,

8.2 SOLVING SYSTEMS OF LINEAR EQUATIONS BY

SUBSTITUTION

USING THE SUBSTITUTION METHOD

You may have suspected by now that graphing alone is not an accurate way

to solve a system of linear equations. For example, a solution of is

unlikely to be read correctly from a graph. In this section, we discuss a sec-ond, more accurate method for solving systems of equations. This methodis called the substitution method and is introduced in the next example.

Example 1 Solve the system:

Solution: The second equation in this system is . Thistells us that x and have the same value. This meansthat we may substitute for x in the first equation.

First equationb

Substitute for x since

Notice that this equation now has one variable, y. Let’snow solve this equation for y.

—


Combine like terms.



Now we know that the y-value of the ordered pairsolution of the system is 2. To find the corresponding x-value, we replace y with 2 in the equation andsolve for x.

Let .

The solution of the system is the ordered pair (4, 2).Since an ordered pair solution must satisfy both linearequations in the system, we could have chosen the equa-tion to find the corresponding x-value. Theresulting x-value is the same.

Check: We check to see that (4, 2) satisfies both equations of theoriginal system.

First Equation Second Equation

Let and .

True. True.4 = 4 10 = 10

y = 2x = 44 � 2 + 22 A4 B + 2 � 10

x = y + 2 2x + y = 10

2x + y = 10

x = 4

y = 2x = 2 + 2

x = y + 2

x = y + 2

y = 2

3y = 6

3y + 4 = 10

2y + 4 + y = 10

2 Ay + 2 B + y = 10HELPFUL HINT

Don’t forgetthe distri-butiveproperty.

x = y + 2.y + 22 Ay + 2 B + y = 10

d 2x + y = 10

y + 2y + 2

x = y + 2

First equation

Second equatione2x + y = 10

x = y + 2

a 12

, 29b

A

Solving Systems of Linear Equations by Substitution SECTION 8.2 539

Objective

Use the substitution method tosolve a system of linear equations.

A

Practice Problem 1Use the substitution method to solvethe system:

e2x + 3y = 13x = y + 4

Answer1. (5, 1)

SSM CD-ROM Video8.2


Practice Problem 2Use the substitution method to solvethe system:

e4x - y = 2 y = 5x

Answers2. , 3. a 1

3, 4 bA-2, -10 B

The solution of the system is (4, 2).A graph of the two equations shows the two lines

intersecting at the point (4, 2).


Solution: The second equation is solved for y in terms of x. Wesubstitute 3x for y in the first equation.

First equation

T

Now we solve for x.

Combine like terms.


The x-value of the ordered pair solution is . To findthe corresponding y-value, we replace x with in theequation .

Let .

Check to see that the solution of the system is .

To solve a system of equations by substitution, we first need an equationsolved for one of its variables.


Solution: We choose one of the equations and solve for x or y. Wewill solve the first equation for x by subtracting 2y fromboth sides.

First equation

Subtract 2y from both sides. x = 7 - 2y

x + 2y = 7

e x + 2y = 72x + 2y = 13

A-1, -3 By = -3

x = -1y = 3 A-1 By = 3x

y = 3x-1

-1

x = -1

2x = -2

5x - 3x = -2

5x - A3x B = -2

5x - y = -2

e5x - y = -2y = 3x

54321

1−1

−2−3−4−5

−2−3−4−5 2 3 4

(4, 2)

5

2x + y = 10x = y + 2

x

y

Practice Problem 3Solve the system:

e3x + y = 53x - 2y = -7


Answers4.

✓ Concept Check: No, the solution will be anordered pair.

A0, -3 B

✓ CONCEPT CHECK

As you solve the system

you find that . Is this the solu-tion of the system?

y = -5

e2x + y = -5x - y = 5

Since , we now substitute for x in thesecond equation and solve for y.

Second equation

Let .

——ƒ—————ƒ


Simplify.



To find x, we let in the equation .

Let .

Check the solution in both equations of the original sys-

tem. The solution is .

The following steps summarize how to solve a system of equations by thesubstitution method.



Solution: To avoid introducing fractions, we will solve the secondequation for y.

Second equation

Next, we substitute for y in the first equation.3x + 6

y = 3x + 6

-3x + y = 6

e7x - 3y = -14-3x + y = 6

TO SOLVE A SYSTEM OF TWO LINEAR EQUATIONS BY THE

SUBSTITUTION METHOD

Step 1. Solve one of the equations for one of its variables.

Step 2. Substitute the expression for the variable found in Step 1 intothe other equation.

Step 3. Solve the equation from Step 2 to find the value of one variable.

Step 4. Substitute the value found in Step 3 in any equation containingboth variables to find the value of the other variable.

Step 5. Check the proposed solution in the original system.

a6, 12b

x = 6

x = 7 - 1

y = 12

x = 7 - 2 a 12b

x = 7 - 2y

x = 7 - 2yy = 12

-2 y = 12

-2y = -1

14 - 2y = 13

14 - 4y + 2y = 13

x = 7 - 2y2 A7 - 2y B + 2y = 13

2x + 2y = 13HELPFUL HINT

Don’t forgetto insertparentheses.

7 - 2yx = 7 - 2y


e5x - 2y = 6-3x + y = -3



•-x

y =

+13

3y = 6

x + 2

Answer5. infinite number of solutions

First equation

Let .


Simplify.



To find the corresponding y-value, we substitute for xin the equation . Then or

. The solution of the system is . Check thissolution in both equations of the system.


Solution: The second equation is already solved for x in terms of y.Thus we substitute for x in the first equation andsolve for y.


Simplify.

Arriving at a true statement such as indicatesthat the two linear equations in the original system areequivalent. This means that their graphs are identical, asshown in the figure. There is an infinite number of solu-tions to the system, and any solution of one equation isalso a solution of the other.

y

x

x = 6 + 2y

x − y = 312

2

4

−4

6

−6

−2−2−4−6 2 4 6

3 = 3

3 = 3

3 + y - y = 3

12

x - y = 3

12A6 + 2y B - y = 3

•

First equation

Let x � 6 � 2y.

6 + 2y

•12x

x - y = 3

= 6 + 2y

HELPFUL HINT

When solving a system of equations by the substitution method, beginby solving an equation for one of its variables. If possible, solve for avariable that has a coefficient of 1 or -1 to avoid working with time-consuming fractions.

A-2, 0 By = 0y = 3 A-2 B + 6y = 3x + 6

-2

x = -2

-2 -2x-2

= 4-2

-2x = 4

-2x - 18 = -14

7x - 9x - 18 = -14

y = 3x + 67x - 3 A3x + 6 B = -14

7x - 3y = -14



e2x - 3y = 6-4x + 6y = -12

Answers6. infinite number of solutions

✓ Concept Checka. parallel linesb. intersect at one pointc. identical graphs


Solution: We choose the second equation and solve for y.

Second equation



Simplify.

Now we replace y with in the first equation.

First equation

Let .

Simplify.

Combine like terms.

The false statement indicates that this system hasno solution. The graph of the linear equations in the sys-tem is a pair of parallel lines, as shown in the figure.


y

x

−4x − 8y = 0

6x + 12y = 5

2

4

−4

6

−6

−2−2−4−6 2 4 6

0 = 5

0 = 5

6x + A-6x B = 5

y = -12

x6x + 12 a- 12

xb = 5

6x + 12y = 5

- 12

x

y = - 12

x

-8 -8y-8

= 4x-8

-8y = 4x

-4x - 8y = 0

e6x + 12y = 5-4x - 8y = 0

✓ CONCEPT CHECK

Describe how the graphs of the equa-tions in a system appear if the systemhasa. no solutionb. one solutionc. an infinite number of solutions


Focus On Business and CareerBREAK-EVEN POINT

When a business sells a new product, it generally does not start making a profit right away.There are usually many expenses associated with producing a new product. These expensesmight include an advertising blitz to introduce the product to the public. These start-upexpenses might also include the cost of market research and product development or anybrand-new equipment needed to manufacture the product. Start-up costs like these are gener-ally called fixed costs because they don’t depend on the number of items manufactured.Expenses that depend on the number of items manufactured, such as the cost of materials andshipping, are called variable costs. The total cost of manufacturing the new product is given bythe cost equation: Total cost Fixed costs Variable costs.

For instance, suppose a greeting card company is launching a new line of greeting cards. Thecompany spent $7000 doing product research and development for the new line and spent$15,000 on advertising the new line. The company did not need to buy any new equipment tomanufacture the cards, but the paper and ink needed to make each card will cost $0.20 percard. The total cost y in dollars for manufacturing x cards is .

Once a business sets a price for the new product, the company can find the product’sexpected revenue. Revenue is the amount of money the company takes in from the sales of itsproduct. The revenue from selling a product is given by the revenue equation: Revenue Price per item Number of items sold.

For instance, suppose that the card company plans to sell its new cards for $1.50 each. Therevenue y, in dollars, that the company can expect to receive from the sales of x cards is

.If the total cost and revenue equations are graphed on the same coordinate system, the

graphs should intersect. The point of intersection is where total cost equals revenue and iscalled the break-even point. The break-even point gives the number of items x that must be soldfor the company to recover its expenses. If fewer than this number of items is sold, the com-pany loses money. If more than this number of items is sold, the company makes a profit. Inthe case of the greeting card company, approximately 16,923 cards must be sold for the com-pany to break even on this new card line. The total cost and revenue of producing and selling16,923 cards is the same. It is approximately $25,385.

GROUP ACTIVITY

Suppose your group is starting a small business near your campus.a. Choose a business and decide what campus-related product or service you will provide.b. Research the fixed costs of starting up such a business.c. Research the variable costs of producing such a product or providing such a service.d. Decide how much you would charge per unit of your product or service.e. Find a system of equations for the total cost and revenue of your product or service.f. How many units of your product or service must be sold before your business will break

even?

$50,00045,00040,00035,00030,00025,00020,00015,00010,000

50000

Dol

lars

(co

st o

r re

venu

e)

Number of units

y = 1.50x

y = 22,000 + 0.20x

(16,923, 25,385)

20,000 25,000 30,00015,00050000 10,000

y = 1.50x

*=

y = 22,000 + 0.20x

+=

EXERCISE SET 8.2

Solve each system of equations by the substitution method. See Examples 1 and 2.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

Solve each system of equations by the substitution method. See Examples 3 through 6.

13. 14. 15.

16. 17. 18.

19. 20. 21.

22. 23. 24.

25. e3x - y = 12x - 3y = 10

e10x - 5y = -21x + 3y = 0

e6x - 3y = 5x + 2y = 0

e8x - 3y = -47x = y + 3

e2x - 3y = -93x = y + 4

e3x + y = -144x + 3y = -22

e4x + y = 112x + 5y = 1

e3y = x + 64x + 12y = 0

e2y = x + 26x - 12y = 0

e4x + 2y = 52x + y = -4

e2x - 5y = 13x + y = -7

ex + 3y = -52x + 2y = 6

ex + 2y = 62x + 3y = 8

ey = 5x - 3y = 8x + 4

ey = 2x + 9y = 7x + 10

ey = 2x + 35y - 7x = 18

ey = 3x + 14y - 8x = 12

e3x - 4y = 10y = 2x

e3x - 4y = 10x = 2y

e2x + 3y = 18x = 2y - 5

e3x + 2y = 16x = 3y - 2

ex + y = 6y = -4x

ex + y = 6y = -3x

ex + y = 20x = 3y

ex + y = 3x = 2y

A

545


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

8.3 SOLVING SYSTEMS OF LINEAR EQUATIONS BY ADDITION

USING THE ADDITION METHOD

We have seen that substitution is an accurate method for solving a systemof linear equations. Another accurate method is the addition or eliminationmethod. The addition method is based on the addition property of equal-ity: Adding equal quantities to both sides of an equation does not changethe solution of the equation. In symbols,

if and , then


Solution: Since the left side of each equation is equal to its rightside, we are adding equal quantities when we add the leftsides of the equations together and the right sides of theequations together. This adding eliminates the variable yand gives us an equation in one variable, x. We can thensolve for x.

First equation

Second equation

Add the equations to eliminate y.


The x-value of the solution is 6. To find the correspond-ing y-value, we let in either equation of the system.We will use the first equation.

First equation

Let .

Solve for y.

The solution is (6, 1).

Check: Check the solution in both equations.

First Equation

Let and .

True.

Second Equation

Let and .

True.

Thus, the solution of the system is (6, 1) and the graphsof the two equations intersect at the point (6, 1), asshown.

5 = 5

y = 1x = 66 - 1 � 5

x - y = 5

7 = 7

y = 1x = 66 + 1 � 7

x + y = 7

y = 1

x = 66 + y = 7

x + y = 7

x = 6

x = 62x = 12

x - y = 5

x + y = 7

ex + y = 7x - y = 5

A + C = B + DC = DA = B

A

Solving Systems of Linear Equations by Addition SECTION 8.3 547

Objective

Use the addition method to solve asystem of linear equations.

A

Answer1. (9, 4)

Practice Problem 1

Use the addition method to solve the

system: ex + y = 13x - y = 5

SSM CD-ROM Video8.3


Practice Problem 2

Solve the system: e2x - y = -6-x + 4y = 17


Solution: If we simply add these two equations, the result is still anequation in two variables. However, our goal is to elimi-nate one of the variables so that we have an equation inthe other variable. To do this, notice what happens if wemultiply both sides of the first equation by . We areallowed to do this by the multiplication property ofequality. Then the system

simplifies to

When we add the resulting equations, the y variable iseliminated.

To find the corresponding y-value, we let inany of the preceding equations containing both variables.We use the first equation of the original system.

First equation

Let .

Check the ordered pair in both equations ofthe original system. The solution is .

In Example 2, the decision to multiply the first equation by was noaccident. To eliminate a variable when adding two equations, the coeffi-

-3

A-2, -2 BA-2, -2 B y = -2

4 + y = 2

x = -2-2 A-2 B + y = 2

-2x + y = 2

x = -2

Add.


6x - 3y = -6 -x + 3y = -4 5x = -10

x = -2

e6x - 3y = -6-x + 3y = -4

e -3 A-2x + y B = -3 A2 B-x + 3y = -4

-3

e -2x + y = 2-x + 3y = -4

54321

1−1

−2−3−4−5

−2−3−4 2 3 4

(6, 1)

5 6

x + y = 7

x − y = 5

x

y

Answer2. A-1, 4 B


Practice Problem 3

Solve the system: ex - 3y = -2-3x + 9y = 5

Practice Problem 4

Solve the system: e2x + 5y = 1-4x - 10y = -2

Answers3. no solution, 4. infinite number of solutions

cient of the variable in one equation must be the opposite of its coefficientin the other equation.


Solution: When we multiply both sides of the first equation by ,the resulting coefficient of x is . This is the opposite of8, the coefficient of x in the second equation. Then thesystem

———ƒ—————————ƒ

simplifies to

Add the equations.

When we add the equations, both variables are elimi-nated and we have , a false statement. Thismeans that the system has no solution. The equations, ifgraphed, are parallel lines.


Solution: First we multiply both sides of the first equation by 3,then we add the resulting equations.

simplifies to

Both variables are eliminated and we have , a truestatement. This means that the system has an infinitenumber of solutions.

0 = 0

Add theequations.

e 9x - 6y = 6 -9x + 6y = -6

0 = 0

e3 A3x - 2y B = 3 # 2-9x + 6y = -6

e3x - 2y = 2-9x + 6y = -6

0 = -27

e -8x + 4y = -28 8x - 4y = 1 0 = -27

e -4 A2x - y B = -4 A7 B8x - 4y = 1

HELPFUL HINT

Don’t for-get to multi-ply bothsides by -4.

-8-4

e2x - y = 78x - 4y = 1

HELPFUL HINT

Be sure to multiply both sides of an equation by a chosen number whensolving by the addition method. A common mistake is to multiply onlythe side containing the variables.


Answer

5. (1, 2)

✓ Concept Check: ba. Multiply the second equation by 4.b. Possible answer: Multiply the first equation

by and the second equation by 7.-2

✓ CONCEPT CHECK

Suppose you are solving the system

by the addition method.a. What step(s) should you take if

you wish to eliminate x whenadding the equations?

b. What step(s) should you take ifyou wish to eliminate y whenadding the equations?

e -4x + 7y = 6x + 2y = 5



Solution: We can eliminate the variable y by multiplying the firstequation by 9 and the second equation by 4. Then we addthe resulting equations.

simplifies to

To find the corresponding y-value, we let in anyequation in this example containing two variables. Doingso in any of these equations will give . Check to seethat (3, 1) satisfies each equation in the original system.The solution is (3, 1).

If we had decided to eliminate x instead of y in Example 5, the first equa-tion could have been multiplied by 5 and the second by . Try solving theoriginal system this way to check that the solution is (3, 1).

The following steps summarize how to solve a system of linear equationsby the addition method.


TO SOLVE A SYSTEM OF TWO LINEAR EQUATIONS BY THE

ADDITION METHOD

Step 1. Rewrite each equation in standard from Ax + By = C.

Step 2. If necessary, multiply one or both equations by a nonzero num-ber so that the coefficients of a chosen variable in the system areopposites.

Step 3. Add the equations.

Step 4. Find the value of one variable by solving the resulting equationfrom Step 3.

Step 5. Find the value of the second variable by substituting the valuefound in Step 4 into either of the original equations.

Step 6. Check the proposed solution in the orignal system.

-3

y = 1

x = 3

Add theequations.Solve for x.

e27x + 36y = 11720x - 36y = 24

47x = 141 x = 3

e9 A3x + 4y B = 9 A13 B4 A5x - 9y B = 4 A6 B

e3x + 4y = 135x - 9y = 6

✓ CONCEPT CHECK

Suppose you are solving the system

You decide to use the additionmethod by multiplying both sides ofthe second equation by 2. In which ofthe following was the multiplicationperformed correctly? Explain.a.b. 4x - 8y = 6

4x - 8y = 3

e3x + 8y = -52x - 4y = 3

Practice Problem 5

Solve the system: e4x + 5y = 143x - 2y = -1


Solution: We begin by clearing each equation of fractions. To doso, we multiply both sides of the first equation by theLCD 2 and both sides of the second equation by theLCD 4. Then the system

simplifies to

Now we add the resulting equations in the simplified sys-tem.

Add.

To find y, we could replace x with in one of the

equations with two variables. Instead, let’s go back to thesimplified system and multiply by appropriate factors toeliminate the variable x and solve for y. To do this, wemultiply the first equation in the simplified system by .Then the system

simplifies to

Add.

Check the ordered pair in both equations of

the original system. The solution is .a - 54

, - 52b

a - 54

, - 52b

y = - 52

e 2x + y = -5-2x + y = 0- 2y = -5

e -1 A-2x - y B = -1 A5 B-2x + y = 0

-1

- 54

x = - 54

-2x - y = 5-2x + y = 0

-4x = 5

e -2x - y = 5-2x + y = 0

µ 2 a-4 a-

x

x2

-

+

y2y4

bb

= 2 a= 4 A

52

0 Bb

µ -

-

x

x2

-

+

y2y4

=

=

52

0


Practice Problem 6

Solve the system: µ-

x2

x3

-

+

52

y

y

=

=

43

- 12

Answer

6. a- 172

, - 32b

552

Focus On HistoryEVERYDAY MATH IN ANCIENT CHINA

The oldest known arithmetic book is a Chinese textbook calledNine Chapters on the Mathematical Art. No one knows for surewho wrote this text or when it was first written. Experts believethat it was a collection of works written by many different people.It was probably written over the course of several centuries. Eventhough no one knows the original date of the Nine Chapters, wedo know that it existed in 213 BC. In that year, all of the originalcopies of the Nine Chapters, along with many other books, wereburned when the first emperor of the Qin Dynasty (221–206 BC)tried to erase all traces of previous rulers and dynasties.

The Qin Emperor was not quite successful in destroying all ofthe Nine Chapters. Pieces of the text were found and many Chi-nese mathematicians filled in the missing material. In 263 AD, theChinese mathematician Liu Hui wrote a summary of the NineChapters, adding his own solutions to its problems. Liu Hui’s ver-sion was studied in China for over a thousand years. At one point,the Chinese government even adopted Nine Chapters as the offi-cial study aid for university students to use when preparing forcivil service exams.

The Nine Chapters is a guide to everyday math in ancientChina. It contains a total of 246 problems covering widely encoun-tered problems like field measurement, rice exchange, fair taxa-tion, and construction. It includes the earliest known use ofnegative numbers and shows the first development of solving sys-tems of linear equations. The following problem appears in Chap-ter 7, “Excess and Deficiency,” of the Nine Chapters. (Note: Awen is a unit of currency.)

A certain number of people are purchasing some chickenstogether. If each person contributes 9 wen, there is an excess of11 wen. If each person contributes just 6 wen, there is a defi-ciency of 16 wen. Find the number of people and the total priceof the chickens. (Adapted from The History of Mathematics: AnIntroduction, second edition, David M. Burton, 1991, Wm. C.Brown Publishers, p. 164)

CRITICAL THINKING

The information in the excess/deficiency problem from NineChapters can be translated into two equations in two variables. Letc represent the total price of the chickens, and let x represent thenumber of people pooling their money to buy the chickens. In thissituation, an excess of 11 wen can be interpreted as 11 more thanthe price of the chickens. A deficiency of 16 wen can be inter-preted as 16 less than the price of the chickens.

1. Use what you have learned so far in this book about translat-ing sentences into equations to write two equations in twovariables for the excess/deficiency problem.

2. Solve the problem from Nine Chapters by solving the systemof equations you wrote in Question 1. How many peoplepooled their money? What was the price of the chickens?

3. Write a modern-day excess/deficiency problem of your own.

8.4 SYSTEMS OF LINEAR EQUATIONS AND PROBLEM SOLVING

USING A SYSTEM OF EQUATIONS FOR PROBLEM SOLVING

Many of the word problems solved earlier using one-variable equations canalso be solved using two equations in two variables. We use the same prob-lem-solving steps that have been used throughout this text. The only dif-ference is that two variables are assigned to represent the two unknownquantities and that the problem is translated into two equations.

Example 1 Finding Unknown NumbersFind two numbers whose sum is 37 and whose differenceis 21.

Solution: 1. UNDERSTAND. Read and reread the problem.Suppose that one number is 20. If their sum is 37, theother number is 17 because . Is theirdifference 21? No; . Our proposedsolution is incorrect, but we now have a betterunderstanding of the problem.

Since we are looking for two numbers, we let

2. TRANSLATE. Since we have assigned two variablesto this problem, we translate our problem into twoequations.

In words:

T T TTranslate: 37

In words:

T T TTranslate: 21=x - y

21istwo numberswhose difference

=x + y

37istwo numberswhose sum

y = second number

x = first number

20 - 17 = 320 + 17 = 37


1. UNDERSTAND the problem. During this step, become com-fortable with the problem. Some ways of doing this are to

Read and reread the problem.Choose two variables to represent the two unknowns.Construct a drawing.Propose a solution and check. Pay careful attention to how you

check your proposed solution. This will help when writing anequation to model the problem.

2. TRANSLATE the problem into two equations.3. SOLVE the system of equations.4. INTERPRET the results: Check the proposed solution in the

stated problem and state your conclusion.

A

Systems of Linear Equations and Problem Solving SECTION 8.4 553

Objective

Use a system of equations to solveproblems.

A

Practice Problem 1Find two numbers whose sum is 50 andwhose difference is 22.

Answer1. The numbers are 36 and 14.

SSM CD-ROM Video8.4


Practice Problem 2Admission prices at a local week-end fair were $5 for children and $7 for adults. The total money collectedwas $3379, and 587 people attended the fair. How many children and howmany adults attended the fair?

3. SOLVE. Now we solve the system

Notice that the coefficients of the variable y areopposites. Let’s then solve by the addition method andbegin by adding the equations.

Add the equations.


Now we let in the first equation to find y.

First equation

4. INTERPRET. The solution of the system is (29, 8).

Check: Notice that the sum of 29 and 8 is , therequired sum. Their difference is , therequired difference.

State: The numbers are 29 and 8.

Example 2 Solving a Problem about PricesThe Barnum and Bailey Circus is in town. Admission for4 adults and 2 children is $22, while admission for 2adults and 3 children is $16.

a. What is the price of an adult’s ticket?b. What is the price of a child’s ticket?c. A special rate of $60 is charged for groups of

20 persons. Should a group of 4 adults and 16 childrenuse the group rate? Why or why not?

Solution: 1. UNDERSTAND. Read and reread the problem andguess a solution. Let’s suppose that the price of anadult’s ticket is $5 and the price of a child’s ticket is$4. To check our proposed solution, let’s see ifadmission for 4 adults and 2 children is $22. Admissionfor 4 adults is 4($5) or $20 and admission for2 children is 2($4) or $8. This gives a total admissionof , not the required $22. Againthough, we have accomplished the purpose of thisprocess: We have a better understanding of theproblem. To continue, we let

2. TRANSLATE. We translate the problem into twoequations using both variables.

C = the price of a child’s ticket

A = the price of an adult’s ticket

$20 + $8 = $28

29 - 8 = 2129 + 8 = 37

y = 37 - 29 = 8

29 + y = 37

x + y = 37

x = 29

x = 582

= 29

x + y = 37x - y = 212x = 58

ex + y = 37x - y = 21

Answer2. 365 children and 222 adults


In words:

T T T T TTranslate: 4A 2C 22

In words:

T T T T TTranslate: 2A 3C 16

3. SOLVE. We solve the system

Since both equations are written in standard form, wesolve by the addition method. First we multiply thesecond equation by to eliminate the variable A.Then the system

simplifies to

To find A, we replace C with 2.5 in the first equation.

First equation

Let .

4. INTERPRET.

Check: Notice that 4 adults and 2 children will pay, the required

amount. Also, the price for 2 adults and 3 children is, the

required amount.

State: Answer the three original questions.

a. Since , the price of an adult’s ticket is $4.25.b. Since , the price of a child’s ticket is $2.50.c. The regular admission price for 4 adults and 16 chil-

dren is

This is $3 less than the special group rate of $60, sothey should not request the group rate.

= $57.00

4 A$4.25 B + 16 A$2.50 B = $17.00 + $40.00

C = 2.5A = 4.25

2 A$4.25 B + 3 A$2.50 B = $8.50 + $7.50 = $16

4 A$4.25 B + 2 A$2.50 B = $17 + $5 = $22

A = 174

= 4.25 or $4.25, the adult’s ticket price.

4A = 17

4A + 5 = 22

C = 2.54A + 2 A2.5 B = 22

4A + 2C = 22

C = 52

= 2.5 or $2.50, the children’s ticket price.

-4C = -10

Add theequations.

e 4A + 2C = 22-4A - 6C = -32

-4C = -10

e4A + 2C = 22-2 A2A + 3C B = -2 A16 B

-2

e4A + 2C = 222A + 3C = 16

=+

$16isadmission for

3 childrenand

admissionfor 2 adults

=+

$22isadmission for

2 childrenand

admissionfor 4 adults

Practice Problem 3Two cars are 440 miles apart and trav-eling toward each other. They meet in3 hours. If one car’s speed is 10 milesper hour faster than the other car’sspeed, find the speed of each car.

Faster car

Slower car

r # t = d


Example 3 Finding RatesAlbert and Louis live 15 miles away from each other.They decide to meet one day by walking toward oneanother. After 2 hours they meet. If Louis walks one mileper hour faster than Albert, find both walking speeds.

Solution 1. UNDERSTAND. Read and reread the problem. Let’spropose a solution and use the formula tocheck. Suppose that Louis’ rate is 4 miles per hour.Since Louis’ rate is 1 mile per hour faster, Albert’srate is 3 miles per hour. To check, see if they can walka total of 15 miles in 2 hours. Louis’ distance is

miles and Albert’s distance ismiles. Their total distance ismiles, not the required

15 miles. Now that we have a better understanding ofthe problem, let’s model it with a system of equations.

First, we let

Now we use the facts stated in the problem and theformula to fill in the following chart.

Albert x 2 2x

Louis y 2 2y

2. TRANSLATE. We translate the problem into twoequations using both variables.

In words:

T T TTranslate: 2x 2y 15

In words:

T T TTranslate: y

3. SOLVE. The system of equations we are solving is

Let’s use substitution to solve the system since the sec-ond equation is solved for y.

e2x + 2y = 15y = x + 1

x + 1=

1 mile per hourfaster than Albert’s

isLouis’

rate

=+

15=Louis’distance+

Albert’sdistance

r # t = d

d = rt

y = Louis’ rate in miles per hour

x = Albert’s rate in miles per hour

8 miles + 6 miles = 14rate # time = 3 A2 B = 6rate # time = 4 A2 B = 8

d = r # t

15 miles

2y 2x

Answer

3. One car’s speed is mph and the other

car’s speed is mph.7813

6813


4. INTERPRET. Albert’s proposed rate is 3.25 miles perhour and Louis’ proposed rate is 4.25 miles per hour.

Check: Use the formula and find that in 2 hours, Albert’sdistance is (3.25)(2) miles or 6.5 miles. In 2 hours, Louis’distance is (4.25)(2) miles or 8.5 miles. The total distancewalked is or 15 miles, the givendistance.

State: Albert walks at a rate of 3.25 miles per hour and Louiswalks at a rate of 4.25 miles per hour.

Example 4 Finding Amounts of SolutionsEric Daly, a chemistry teaching assistant, needs 10 litersof a 20% saline solution (salt water) for his 2 p.m. labora-tory class. Unfortunately, the only mixtures on hand are a 5% saline solution and a 25% saline solution. Howmuch of each solution should he mix to produce the 20% solution?

Solution: 1. UNDERSTAND. Read and reread the problem.Suppose that we need 4 liters of the 5% solution. Thenwe need liters of the 25% solution. To seeif this gives us 10 liters of a 20% saline solution, let’sfind the amount of pure salt in each solution.

T T T5% solution: 0.05 4 liters 0.2 liters

25% solution: 0.25 6 liters 1.5 liters20% solution: 0.20 10 liters 2 liters

Since , not 2 liters,our proposed solution is incorrect. But we have gainedsome insight into how to model and check thisproblem.

We let

+ =x + y

or10 litersx liters y liters

5% salinesolution

25% salinesolution

20% salinesolution

y = number of liters of 25% solution

x = number of liters of 5% solution

0.2 liters + 1.5 liters = 1.7 liters

=*=*=*

amount ofpure salt

=amount ofsolution

*concentrationrate

10 - 4 = 6

6.5 miles + 8.5 miles

d = rt

y = x + 1 = 3.25 + 1 = 4.25

x = 134

= 3.25

4x = 13

2x + 2x + 2 = 15

2x + 2y = 15

2x + 2 Ax + 1 B = 15

•

First equation

Replace y with x � 1.

Practice Problem 4Barb Hayes, a pharmacist, needs 50 li-ters of a 60% alcohol solution. She currently has available a 20% solutionand a 70% solution. How many litersof each does she need to make theneeded 50 liters of 60% alcohol solu-tion?

Answer4. 10 liters of the 20% alcohol solution and 40

liters of the 70% alcohol solution


Now we use a table to organize the given data.

Concentration Rate Liters of Solution Liters of Pure Salt

Firstsolution

5% x 0.05x

Secondsolution

25% y 0.25y

Mixtureneeded

20% 10 (0.20)(10)

2. TRANSLATE. We translate into two equations usingboth variables.

In words:

T T TTranslate: x y 10

In words:

T T TTranslate: 0.05x 0.25y (0.20)(10)

3. SOLVE. Here we solve the system

To solve by the addition method, we first multiply thefirst equation by and the second equation by 100.Then the system

Add.

To find y, we let in the first equation of theoriginal system.

Let .

4. INTERPRET. Thus, we propose that Eric needs tomix 2.5 liters of 5% saline solution with 7.5 liters of25% saline solution.

Check: Notice that , the required number ofliters. Also, the sum of the liters of salt in the twosolutions equals the liters of salt in the required mixture:

State: Eric needs 2.5 liters of the 5% saline solution and7.5 liters of the 25% solution.


0.125 + 1.875 = 2

0.05 A2.5 B + 0.25 A7.5 B = 0.20 A10 B

2.5 + 7.5 = 10

y = 7.5

x = 2.52.5 + y = 10

x + y = 10

x = 2.5

µ-25x - 25y = -250 5x- + 25y = 200--20x = -50

x = 2.5

simplifiestoe -25 Ax + y B = -25 A10 B

100 A0.05x + 0.25y B = 100 A2 B

-25

ex + y = 100.05x + 0.25y = 2

=+

salt inmixture=

salt in 25%solution+

salt in 5%solution

=+

10=liters of 25%solution

+liters of 5%solution

✓ CONCEPT CHECK

Suppose you mix an amount of 30%acid solution with an amount of50% acid solution. Which of the fol-lowing acid strengths would be possi-ble for the resulting acid mixture?a. 22% b. 44% c. 63%

A


Focus On Business and CareerASSESSING JOB OFFERS

When you finish your present course of study, you will probably look for a job. When your jobsearch has paid off and you receive a job offer, how will you decide whether to take the job?How do you decide between two or more job offers? These decisions are an important part ofthe job search and may not be easy to make. To evaluate the job offer, you should consider thenature of the work involved in the job, the type of company or organization that has offeredthe job, and the salary and benefits offered by the employer. You may also need to comparethe compensation packages of two or more job offers. The following hints on assessing a job’scompensation package were included in an article by Max Carry in the Winter, 1990–91 Occu-pational Outlook Handbook Quarterly.

Most companies will not talk about pay until they have decided to hire you. In order to know if theiroffer is reasonable, you need a rough estimate of what the job should pay. You may have to go to sev-eral sources for this information. Talk to friends who recently were hired in similar jobs. Ask your teach-ers and the staff in the college placement office about starting pay for graduates with your qualifications.Scan the help-wanted ads in newspapers. Check the library or your school’s career center for salary sur-veys, such as the College Council Salary Survey and the Bureau of Labor Statistics wage surveys. If youare considering the salary and benefits for a job in another geographic area, make allowances for dif-ferences in the cost of living. Use the research to come up with a base salary range for yourself, the topbeing the best you can hope to get and the bottom being the least you will take. When negotiating, aimfor the top of your estimated salary range, but be prepared to settle for less. Entry level salaries some-times are not negotiable, particularly in government agencies. If you are not pleased with an offer, how-ever, what harm could come from asking for more?

An employer cannot be specific about the amount of pay if it includes commissions and bonuses. Theway the pay plan works, however, should be explained. The employer also should be able to tell youwhat most people in the job are earning. You should also learn the organization’s policy regarding over-time. Depending on the job, you may or may not be exempt from laws requiring the employer to com-pensate you for overtime. Find out how many hours you will be expected to work each week andwhether evening and weekend work is required or expected. Will you receive overtime pay or time offfor working more than the specified number of hours in a week? Also take into account that the start-ing salary is just that, the start. Your salary should be reviewed on a regular basis—many organizationsdo it every 12 months. If the employer is pleased with your performance, how much can you expect tomake after one year? Two years? Three years and so on?

Don’t think of salary as the only compensation you will receive. Consider benefits. What on the sur-face looks like a great salary could be accompanied by little else. Do you know the value of theemployer’s contribution to your benefits? According to a 1989 Bureau of Labor Statistics study, for eachdollar employers in private industry spent on straight-time wages and salaries, they contributed on aver-age another 38 cents to employee benefits, including contributions required by law. Benefits can add alot to your base pay. Your benefit package probably will consist of health insurance, life insurance, apension plan, and paid vacations, holidays, and sick leave. It also may include items as diverse as profitsharing, moving expenses, parking space, a company car, and on-site day care. Do you know exactlywhat the benefit package includes and how much of the cost must be borne by you? Depending on yourcircumstances, you might want to increase or decrease particular benefits.

CRITICAL THINKING

1. Suppose you have been searching for a position as an electronics sales associate. You havereceived two job offers. The first job pays a monthly salary of $1500 per month plus a com-mission of 4% on all sales made. The second pays a monthly salary of $1800 per monthplus a commission of 2% on all sales made. At what level of monthly sales would the jobspay the same amount? Based only on the given information about the jobs, which jobwould you choose? Why? What other information would you want to have about the jobsbefore making a decision?

2. Suppose you have been searching for an entry-level bookkeeping position. You havereceived two job offers. The first company offers you a starting hourly wage of $6.50 perhour and says that each year entry-level workers receive a raise of $0.75 per hour. The sec-ond company offers you a starting hourly wage of $7.50 per hour and says that you canexpect a $0.50 per hour raise each year. After how many years will the two jobs pay thesame hourly wage? Based only on the given information about the jobs, which job wouldyou choose? Why? What other information would you want to have about the jobs beforemaking a decision?

8.5 SOLVING SYSTEMS OF LINEAR EQUATIONS

IN THREE VARIABLES

In this section, we solve systems of three linear equations in three variables.We call the equation , for example, a linear equation inthree variables since there are three variables and each variable is raisedonly to the power 1. A solution of this equation is an ordered triple (x, y, z)that makes the equation a true statement.

For example, the ordered triple is a solution ofsince replacing x with 2, y with 0, and z with yields

the true statement

The graph of this equation is a plane in three-dimensional space, just as thegraph of a linear equation in two variables is a line in two-dimensionalspace.

Although we will not discuss the techniques for graphing equations inthree variables, visualizing the possible patterns of intersecting planes givesus insight into the possible patterns of solutions of a system of three three-variable linear equations. There are four possible patterns.

1. Three planes have a single point in common. This point represents thesingle solution of the system. This system is consistent.

2. Three planes intersect at no point common to all three. This system hasno solution. A few ways that this can occur are shown. This system isinconsistent.

3. Three planes intersect at all the points of a single line. The system hasinfinitely many solutions. This system is consistent.

3 A2 B - 0 + A-21 B = -15

-213x - y + z = -15A2, 0, -21 B

3x - y + z = -15

Solving Systems of Linear Equations in Three Variables SECTION 8.5 561

Objectives

Solve a system of three linearequations in three variables.

Solve problems that can be mod-eled by a system of three linearequations

B

A

SSM CD-ROM Video8.5


4. Three planes coincide at all points on the plane. The system is consistent,and the equations are dependent.

SOLVING A SYSTEM OF THREE LINEAR EQUATIONS IN THREE VARIABLES

Just as with systems of two equations in two variables, we can use the elim-ination method to solve a system of three equations in three variables. Todo so, we eliminate a variable and obtain a system of two equations in twovariables. Then we use the methods we learned in the previous section tosolve the system of two equations. See the box in the margin for steps.


Solution: We add equations (1) and (2) to eliminate z.

Next we add two other equations and eliminate z again.To do so, we multiply both sides of equation (1) by 2 andadd this resulting equation to equation (3). Then

Now we solve equations (4) and (5) for x and y. To solveby elimination, we multiply both sides of equation (4) by

and add this resulting equation to equation (5). Then

We now replace x with in equation (4) or (5).

Equation (4)

Let .

Solve for y.

Finally, we replace x with and y with 2 in equation(1), (2), or (3).

Equation (2)

Let and .

z = -1

-z = 1

-4 + 4 - z = 1

y = 2x = -4-4 + 2 A2 B - z = 1

x + 2y - z = 1

-4

y = 2

x = -44 A-4 B + y = -14

4x + y = -14

-4

Add theequations.Solve for x.

e -4x - y = 14 8x + y = -30 4x = -16 x = -4

simplifiestoe -1 A4x + y B = -1 A-14 B

8x + y = -30

-1

Equation (5)

e6x - 2y + 2z = -302x + 3y - 2z = 08x + y = -30

simplifiestoe2 A3x - y + z B = 2 A-15 B

2x + 3y - 2z = 0

Equation (4)

3x - y + z = -15 x + 2y - z = 14x + y = -14

Equation (1)

Equation (2)

Equation (3)

•3x - y + z = -15 x + 2y - z = 12x + 3y - 2z = 0

A


•2x - y + 3z = 13 x + y - z = -23x + 2y + 2z = 13

Answer1. A1, 1, 4 B

SOLVING A SYSTEM OF THREE

LINEAR EQUATIONS BY THE

ELIMINATION METHOD

Step 1. Write each equation instandard form, Ax + By + Cz = D.

Step 2. Choose a pair of equationsand use the equations toeliminate a variable.

Step 3. Choose any other pair ofequations and eliminate thesame variable as in Step 2.

Step 4. Two equations in two vari-ables should be obtainedfrom Step 2 and Step 3. Usemethods from Section 4.1to solve this system forboth variables.

Step 5. To solve for the third vari-able, substitute the valuesof the variables found inStep 4 into any of the origi-nal equations containingthe third variable.

Step 6. Check the ordered triplesolution in all three originalequations.


Answers2. ¤, 3.


a 13

, -1, -1 b

The ordered triple solution is . To check, let, and in all three original equa-

tions of the system.

Equation (1) Equation (2)

True. True.

Equation (3)

True.

All three statements are true, so the ordered triple solu-tion is .


Solution: When we add equations (2) and (3) to eliminate x, thenew equation is

(4)

To eliminate x again, we multiply both sides of equation(2) by 2 and add the resulting equation to equation (1).Then

(5)

Next we solve for y and z using equations (4) and (5). Todo so, we multiply both sides of equation (4) by andadd the resulting equation to equation (5).

Since the statement is false, this system is inconsistentand has no solution. The solution set is the empty set

or .



(1)

(2)

(3)•

2x + 4y = 14x - 4z = -1 y - 4z = -3

¤E F

False.

e 10y - 10z = -22-10y + 10z = 24 0 = 2

simplifiestoe -2 A-5y + 5z B = -2 A11 B

-10y + 10z = 24

-2

e 2x - 4y + 8z = 2-2x - 6y + 2z = 22 -10y + 10z = 24

simplifiestoe 2x - 4y + 8z = 2

2 A-x - 3y + z B = 2 A11 B

-5y + 5z = 11

(1)

(2)

(3)•

2x - 4y + 8z = 2-x - 3y + z = 11 x - 2y + 4z = 0

A-4, 2, -1 B

0 = 0 -8 + 6 + 2 = 0

2 A-4 B + 3 A2 B - 2 A-1 B = 0 2x + 3y - 2z = 0

1 = 1 -15 = -15 -4 + 4 + 1 = 1 -12 - 2 - 1 = -15

-4 + 2 A2 B - A-1 B = 13 A-4 B - 2 + A-1 B = -15 x + 2y - z = 1 3x - y + z = -15

z = -1x = -4, y = 2A-4, 2, -1 B


• 2x + 4y - 2z = 3-x + y - z = 6

x + 2y - z = 1

✓ CONCEPT CHECK

In the system

Equations (1) and (2) are used toeliminate y. Which action could beused to finish solving? Why?(a) Use (1) and (2) to eliminate z(b) Use (2) and (3) to eliminate y(c) Use (1) and (3) to eliminate x

Equation (1)

Equation (2)

Equation (3)•

x + y + z = 62x - y + z = 3x + 2y + 3z = 14


•3x + 2y = -16x - 2z = 4 y - 3z = 2


Answer4. E Ax, y, z B @x - 3y + 4z = 2F

Solution: Notice that equation (2) has no term containing the vari-able y. Let us eliminate y using equations (1) and (3). Wemultiply both sides of equation (3) by and add theresulting equation to equation (1). Then

Next we solve for z using equations (4) and (2). We mul-tiply both sides of equation (4) by and add the result-ing equation to equation (2).

Now we replace z with in equation (3) and solve for y.

Let in equation (3).

Finally, we replace y with 0 in equation (1) and solve for x.

Let in equation (1).

The ordered triple solution is . Check to see

that this solution satisfies all three equations of the sys-tem.


Solution: We multiply both sides of equation (3) by 2 to eliminatefractions, and we multiply both sides of equation (2) by

so that the coefficient of x is 1. The resulting system

is then

(1)Multiply (2) by .

Multiply (3) by 2.

All three resulting equations are identical, and thereforeequations (1), (2), and (3) are all equivalent. There areinfinitely many solutions of this system. The equationsare dependent. The solution set can be written as

.E Ax, y, z B @x - 5y - 2z = 6F

-12•

x - 5y - 2z = 6x - 5y - 2z = 6x - 5y - 2z = 6

- 12

(1)

(2)

(3)µ

x - 5y - 2z = 6

-2x + 10y + 4z = -12

12

x -

52

y - z = 3

a 12

, 0, 34b

x = 12

2x = 1

y = 02x + 4 A0 B = 1

y = 0

y - 3 = -3

z = 34

y - 4 a 34b = -3

34

e -4x - 32z = -26 4x - 4z = -1 - 36z = -27 z = 3

4

simplifiestoe -2 A2x + 16z B = -2 A13 B

4x - 4z = -1

-2

(4)•

2x + 4y = 1 -4y + 16z = 122x + 16z = 13

simplifiestoe2x + 4y = 1

-4 A y - 4z B = -4 A-3 B

-4


µ x - 3y + 4z = 2-2x + 6y - 8z = -4

12

x - 32

y + 2z = 1

SOLVING PROBLEMS MODELED BY SYSTEMS OF THREE EQUATIONS

To introduce problem solving by writing a system of three linear equationsin three variables, we solve a problem about triangles.

Example 5 Finding Angle MeasuresThe measure of the largest angle of a triangle is 80° morethan the measure of the smallest angle, and the measureof the remaining angle is 10° more than the measure ofthe smallest angle. Find the measure of each angle.

Solution: 1. UNDERSTAND. Read and reread the problem.Recall that the sum of the measures of the angles of atriangle is 180°. Then guess a solution. If the smallestangle measures 20°, the measure of the largest angle is80° more, or . The measure of theremaining angle is 10° more than the measure of thesmallest angle, or . The sum of thesethree angles is , not therequired 180°. We now know that the measure of thesmallest angle is greater than 20°.

To model this problem we will let

2. TRANSLATE. We translate the given informationinto three equations.

In words:=

T TTranslate: = 180

In words:

T T TTranslate: y =

In words:

T T TTranslate: z =

3. SOLVE. We solve the system

Since y and z are both expressed in terms of x, we willsolve using the subsitution method. We substitute

and in the first equation.Then

z = x + 10y = x + 80

•x + y + z = 180y = x + 80z = x + 10

x + 10

10 more than thesmallest angle

isthe remaining

angle

x + 80

80 more than thesmallest angle

isthe largest

angle

x + y + z

180the sum of

the measures

z

y xz = degree measure of the remaining angle

y = degree measure of the largest angle

x = degree measure of the smallest angle

20° + 100° + 30° = 150°20° + 10° = 30°

20° + 80° = 100°

B


Practice Problem 5The measure of the largest angle of atriangle is 90° more than the measureof the smallest angle, and the measureof the remaining angle is 30° more thanthe measure of the smallest angle. Findthe measure of each angle.

Answer5. 20°, 50°, 110°

First equation

Let and .

Then , and. The ordered triple solu-

tion is .

4. INTERPRET.

Check: Notice that . Also, the measureof the largest angle, 110°, is 80° more than the measure ofthe smallest angle, 30°. The measure of the remainingangle, 40°, is 10° more than the measure of the smallestangle, 30°.

State: The angles measure 30°, 110°, and 40°.

30° + 40° + 110° = 180°

A30, 110, 40 Bz = x + 10 = 30 + 10 = 40

y = x + 80 = 30 + 80 = 110

x = 30

3x = 90

3x + 90 = 180

z = x + 10y = x + 80

x + y + z = 180

x + Ax + 80 B + Ax + 10 B = 180


8.6 SOLVING SYSTEMS OF EQUATIONS USING MATRICES

By now, you may have noticed that the solution of a system of equationsdepends on the coefficients of the equations in the system and not on thevariables. In this section, we introduce how to solve a system of equationsusing a matrix.

USING MATRICES TO SOLVE A SYSTEM OF TWO EQUATIONS

A matrix (plural: matrices) is a rectangular array of numbers. The follow-ing are examples of matrices.

The numbers aligned horizontally in a matrix are in the same row. Thenumbers aligned vertically are in the same column.

c c ccolumn 1 ƒ ƒ

ƒ ƒcolumn 2 ƒ

ƒcolumn 3

To see the relationship between systems of equations and matrices,study the example below.

System of Equations Corresponding Matrix

Notice that the rows of the matrix correspond to the equations in the sys-tem. The coefficients of each variable are placed to the left of a verticaldashed line. The constants are placed to the right. Each of these numbersin the matrix is called an element.

The method of solving systems by matrices is to write this matrix as anequivalent matrix from which we easily identify the solution. Two matricesare equivalent if they represent systems that have the same solution set.The following row operations can be performed on matrices, and the resultis an equivalent matrix.

HELPFUL HINT

Notice that these row operations are the same operations that we canperform on equations in a system.

ELEMENTARY ROW OPERATIONS

1. Any two rows in a matrix may be interchanged.2. The elements of any row may be multiplied (or divided) by the same

nonzero number.3. The elements of any row may be multiplied (or divided) by a

nonzero number and added to their corresponding elements in anyother row.

Row 1

Row 260dc2

1-3

1Equation 1

Equation 2e2x - 3y = 6

x + y = 0

This matrix has 2 rows and 3columns. It is called a 2 * 3(read “two by three”) matrix.

c 2-1

16

02drow 1 S

row 2 S

c ad

be

cfd£ 2

0-6

1-1

2

341

-150§c1

001d

A

Solving Systems of Equations Using Matrices SECTION 8.6 567

Objectives

Use matrices to solve a system oftwo equations.

Use matrices to solve a system ofthree equations.

B

A

SSM CD-ROM Video8.6


Answers1. , 2. no solutionA2, -3 B

Practice Problem 1Use matrices to solve the system:

e x + 2y = -42x - 3y = 13


e -3x + y = 0-6x + 2y = 2

Example 1 Use matrices to solve the system:

Solution: The corresponding matrix is . We use

elementary row operations to write an equivalent matrix

that looks like .

For the matrix given, the element in the first row, firstcolumn is already 1, as desired. Next we write an equiva-lent matrix with a 0 below the 1. To do this, we multiplyrow 1 by and add to row 2. We will change onlyrow 2.

simplifies to

Now we change the to a 1 by use of an elementaryrow operation. We divide row 2 by , then

simplifies to

This last matrix corresponds to the system

To find x, we let in the first equation, .

First equation

Let .

The ordered pair solution is . Check to see thatthis ordered pair satisfies both equations.


Solution: The corresponding matrix is . To get 1 in

the row 1, column 1 position, we divide the elements of row 1 by 2.

simplifies to325§£1

4

- 12

-2

325§£22

4

-12

-2

35dc2

4-1-2

e 2x - y = 34x - 2y = 5

A-1, 2 B x = -1

y = 2 x + 3 A2 B = 5

x + 3y = 5

x + 3y = 5y = 2

ex + 3y = 5y = 2

52dc1

031

5-14-7§£ 1

0-7

3-7-7

-7-7

5-14dc1

03

-7

crow 2

element

crow 1

element

crow 2

element

crow 1

element

crow 2

element

crow 1

element

5-2 A5 B + A-4 B d3

-2 A3 B + A-1 Bc 1-2 A1 B + 2

-2

bcdc1

0a1

5-4dc1

23

-1

e x + 3y = 5 2x - y = -4

Solving Systems of Equations Using Matrices SECTION 8.6 569

Answers3.

✓ Concept Check: Its matrix is � .8- 3dc2

1- 3

5

A1, 2, -2 B

✓ CONCEPT CHECK

Consider the system

What is wrong with its correspondingmatrix shown below?

�83dc2

035

e2x - 3y = 8 x + 5y = -3

To get 0 under the 1, we multiply the elements of row 1by and add the new elements to the elements ofrow 2.

simplifies to

The corresponding system is . The equa-

tion is false for all y or x values; hence the sys-tem is inconsistent and has no solution.


USING MATRICES TO SOLVE A SYSTEM OF THREE EQUATIONS

To solve a system of three equations in three variables using matrices, wewill write the corresponding matrix in the form


Solution: The corresponding matrix is .

Our goal is to write an equivalent matrix with 1s alongthe diagonal (see the numbers in red) and 0s below the1s. The element in row 1, column 1 is already 1. Next weget 0s for each element in the rest of column 1. To dothis, first we multiply the elements of row 1 by 2 and addthe new elements to row 2. Also, we multiply the ele-ments of row 1 by and add the new elements to theelements of row 3. We do not change row 1. Then

simplifies to

29

-10§£10

0

231

14

-3

22 A2 B + 5

-1 A2 B - 8§

12 A1 B + 2

-1 A1 B - 2

22 A2 B - 1

-1 A2 B + 3£

12 A1 B - 2

-1 A1 B + 1

-1

25

-8§£ 1

-21

2-1

3

12

-2

• x + 2y + z = 2

-2x - y + 2z = 5 x + 3y - 2z = -8

def§£10

0

a10

bc1

B

0 = -1

•x -1

y = 32 2

0 = -1

32

-1§£1

0

- 120

¥32

3 -4 a2

b + 5

1- 2

-4 a- 1 b - 22

≥1

-4 A1 B + 4

-4


•x + 3y + z = 5

-3x + y - 3z = 5x + 2y - 2z = 9


We continue down the diagonal and use elementary rowoperations to get 1 where the element 3 is now. To dothis, we interchange rows 2 and 3.

Next we want the new row 3, column 2 element to be 0.We multiply the elements of row 2 by and add theresult to the elements of row 3.

simplifies to

Finally, we divide the elements of row 3 by 13 so that thefinal diagonal element is 1.

simplifies to

This matrix corresponds to the system

We identify the z-coordinate of the solution as 3. Next wereplace z with 3 in the second equation and solve for y.

Second equation

Let .

To find x, we let and in the first equation.

First equation

Let and .

The ordered triple solution is . Check to seethat it satisfies all three equations in the original system.

A1, -1, 3 B x = 1

y = -1z = 3x + 2 A-1 B + 3 = 2

x + 2y + z = 2

y = -1z = 3

y = -1

z = 3y - 3 A3 B = -10

y - 3z = -10

•x + 2y + z = 2

y - 3z = -10z = 3

2-10

3§£10

0

210

1-3

1

2-10

3913

¥≥10

013

21

013

1-3 1313

2-10

39§£10

0

210

1-313

2-10

-3 A-10 B + 9§£ 1

0-3 A0 B + 0

21

-3 A1 B + 3

1-3

-3 A-3 B + 4

-3

£ 100

231

14

-3

29

-10§ is equivalent to

£100

213

1-3

4

2-10

9§

8.7 SOLVING SYSTEMS OF EQUATIONS USING DETERMINANTS

We have solved systems of two linear equations in two variables in four dif-ferent ways: graphically, by substitution, by elimination, and by matrices.Now we analyze another method called Cramer’s rule.

EVALUATING 2 * 2 DETERMINANTS

Recall that a matrix is a rectangular array of numbers. If a matrix has thesame number of rows and columns, it is called a square matrix. Examplesof square matrices are

A determinant is a real number associated with a square matrix. Thedeterminant of a square matrix is denoted by placing vertical bars about thearray of numbers. Thus,

The determinant of the square matrix is .

The determinant of the square matrix is .

We define the determinant of a matrix first. (Recall that isread “two by two.” It means that the matrix has 2 rows and 2 columns.)

Example 1 Evaluate each determinant.

a. b.

Solution: First we identify the values of a, b, c, and d. Then we per-form the evaluation.

a. Here , and .

b. In this example, , and .

` 27

0-5` = ad - bc = 2 A-5 B - A0 B A7 B = -10

d = -5a = 2, b = 0, c = 7

` -13

2-4` = ad - bc = A-1 B A-4 B - A2 B A3 B = -2

d = -4a = -1, b = 2, c = 3

` 27

0-5`` -1

32

-4`

2 * 22 * 2

† 203

456

129†£20

3

456

129§

` 15

62`c1

562d

£203

456

129§c1

562d

A

Solving Systems of Equations Using Determinants SECTION 8.7 571

Objectives

Define and evaluate a deter-minant.

Use Cramer’s rule to solve a sys-tem of two linear equations in twovariables.

Define and evaluate a deter-minant.

Use Cramer’s rule to solve a linearsystem of three equations in threevariables.

D

3 * 3C

B

2 * 2A

DETERMINANT OF A 2 * 2 MATRIX

` ac

bd` = ad - bc

Practice Problem 1Evaluate each determinant.

a.

b. ` 40

5-5`

` -32

61`

Answers1. a. , b. -20-15

SSM CD-ROM Video8.7


USING CRAMER’S RULE TO SOLVE A SYSTEM OF TWO

LINEAR EQUATIONS

To develop Cramer’s rule, we solve the system using

elimination. First, we eliminate y by multiplying both sides of the first equa-tion by d and both sides of the second equation by so that the coeffi-cients of y are opposites. The result is that

simplifies to

We now add the two equations and solve for x.

When we replace x with in the equation and

solve for y, we find that .

Notice that the numerator of the value of x is the determinant of

Also, the numerator of the value of y is the determinant of

Finally, the denominators of the values of x and y are the same and are thedeterminant of

This means that the values of x and y can be written in determinant nota-tion:

and

For convenience, we label the determinants D, , and .

c cx-column replaced y-column replaced

by constants by constants

These determinant formulas for the coordinates of the solution of a sys-tem are known as Cramer’s rule.

` ac

bd` = D ` h

kbd` = Dx ` a

chk` = Dy

x-coefficients

y-coefficients

DyDx

` ac

bd`` a

cbd`

y = x = ` ac

hk`` h

kbd`

` ac

bd` = ad - bc

` ac

hk` = ak - hc

` hk

bd` = hd - kb

y = ak - chad - bc

ax + by = hhd - kbad - bc

Add the equations.

Solve for x.

adx + bdy = hd -bcx - bdy = -kb

adx - bcx = hd - kbAad - bc Bx = hd - kbx = hd - kb

ad - bc

e adx + bdy = hd-bcx - bdy = -kb

e d Aax + by B = d # h-b Acx + dy B = -b # k

-b

eax + by = hcx + dy = k

B


CRAMER’S RULE FOR TWO LINEAR EQUATIONS IN TWO VARIABLES

The solution of the system is given by

as long as is not 0.D = ad - bc

x =` hk

bd`

` ac

bd`

= Dx

D y =

` ac

hk`

` ac

bd`

=Dy

D


When , the system is either inconsistent or the equations are depen-dent. When this happens, we need to use another method to see which isthe case.

Example 2 Use Cramer’s rule to solve the system:

Solution: First we find D, , and .

a b hT T T

c c cc d k

Then and .

The ordered pair solution is .

As always, check the solution in both original equations.


Solution: First we find D, , and .

Dy = ` 5-7

5-7` = 5 A-7 B - 5 A-7 B = 0

Dx = ` 5-7

1-2` = 5 A-2 B - A-7 B A1 B = -3

D = ` 5-7

1-2` = 5 A-2 B - A-7 B A1 B = -3

DyDx

e 5x + y = 5-7x - 2y = -7

A-5, 2 By =

Dy

D= -20

-10= 2x = Dx

D= 50

-10= -5

Dy = ` ac

hk` = ` 3

1-7-9` = 3 A-9 B - A-7 B A1 B = -20

Dx = ` hk

bd` = ` -7

-94

-2` = A-7 B A-2 B - 4 A-9 B = 50

D = ` ac

bd` = ` 3

14

-2` = 3 A-2 B - 4 A1 B = -10

e3x + 4y = -7 x - 2y = -9

DyDx

e3x + 4y = -7 x - 2y = -9

D = 0

Practice Problem 2Use Cramer’s rule to solve the system.

e x - y = -42x + 3y = 2

Answers2. , 3. A0, 3 BA-2, 2 B

Practice Problem 3Use Cramer’s rule to solve the system.

e4x + y = 32x - 3y = -9


Answers4. a. 4, b. 4

✓ Concept Check:+-+-

-+-+

+-+-

-+-+

Then

The ordered pair solution is .

EVALUATING 3 * 3 DETERMINANTS

A determinant can be used to solve a system of three equations inthree variables. The determinant of a matrix, however, is consider-ably more complex than a one.

Notice that the determinant of a matrix is related to the determi-nants of three matrices. Each determinant of these matrices iscalled a minor, and every element of a matrix has a minor associatedwith it. For example, the minor of is the determinant of the matrixfound by deleting the row and column containing .

The minor of is

Also, the minor of element is the determinant of the matrix thathas no row or column containing .

The minor of is

So the determinant of a matrix can be written as

Finding the determinant by using minors of elements in the first columnis called expanding by the minors of the first column. The value of a deter-minant can be found by expanding by the minors of any row or column. Thefollowing array of signs is helpful in determining whether to add or subtractthe product of an element and its minor.

If an element is in a position marked , we add. If marked , we sub-tract.


Example 4 Evaluate by expanding by the minors of the given row orcolumn.

a. First column b. Second row

† 01

-2

532

1-1

4†

-+

+-+

-+-

+-+

a1 # Aminor of a1 B - a2 # Aminor of a2 B + a3 # Aminor of a3 B3 * 3

` b2

b3

c2

c3`a1†

a1 b1 c1a2 b2 c2

a3 b3 c3

a1

2 * 2a1

` a1

a3

b1

b3`c2

a1 b1 c1

a2 b2 c2a3 b3 c3

†c2

2 * 2c2

3 * 32 * 22 * 2

3 * 3

2 * 23 * 3

3 * 3

C

A1, 0 By =

Dy

D= 0

-3= 0x = Dx

D= -3

-3= 1

✓ CONCEPT CHECK

Suppose you are interested in findingthe determinant of a matrix.Study the pattern shown in the arrayof signs for a matrix. Use thepattern to expand the array of signsfor use with a matrix.4 * 4

3 * 3

4 * 4

Practice Problem 4Evaluate by expanding by the minorsof the given row or column.

a. First column b. Third row

† 2-1

5

031

124†

DETERMINANT OF A 3 * 3 MATRIX

† a1

a2

a3

b1

b2

b3

c1

c2

c3

† = a1 # ` b2

b3

c2

c3` - a2 # ` b1

b3

c1

c3` + a3 # ` b1

b2

c1

c2`


Answers5.

✓ Concept Check: Two elements of the secondrow are 0, which makes calculations easier.

A-1, 3, 2 B

Solution: a. The elements of the first column are 0, 1, and . Thefirst column of the array of signs is .

b. The elements of the second row are 1, 3, and . Thistime, the signs begin with and again alternate.

Notice that the determinant of the matrix is thesame regardless of the row or column you select toexpand by.


USING CRAMER’S RULE TO SOLVE A SYSTEM OF THREE

LINEAR EQUATIONS

A system of three equations in three variables may be solved with Cramer’srule also. Using the elimination process to solve a system with unknownconstants as coefficients leads to the following.


• x - 2y + z = 43x + y - 2z = 35x + 5y + 3z = -8

D

3 * 3

= -18 + 6 + 10 = -2

= -1 A20 - 2 B + 3 A0 - A-2 B B - A-1 B A0 - A-10 B B† 0

1-2

532

1-1

4† = -1 # ` 5

214` + 3 # ` 0

-214` - A-1 B # ` 0

-252`

--1

= 0 - 18 + 16 = -2

= 0 A12 - A-2 B B - 1 A20 - 2 B + A-2 B A-5 - 3 B† 0

1-2

532

1-1

4† = 0 # ` 3

2-1

4` - 1 # ` 5

214` + A-2 B # ` 5

31

-1`

+ , - , +-2

✓ CONCEPT CHECK

Why would expanding by minors ofthe second row be a good choice for

the determinant ?† 356

40

-3

-207†

CRAMER’S RULE FOR THREE EQUATIONS IN THREE VARIABLES

The solution of the system is given by

and

where

as long as D is not 0.

Dy = † a1

a2

a3

k1

k2

k3

c1

c2

c3

† Dz = † a1

a2

a3

b1

b2

b3

k1

k2

k3

†

D = † a1

a2

a3

b1

b2

b3

c1

c2

c3

† Dx = † k1

k2

k3

b1

b2

b3

c1

c2

c3

†

z =Dz

Dy =

Dy

Dx =

Dx

D

•a1x + b1y + c1z = k1

a2x + b2y + c2z = k2

a3x + b3y + c3z = k3

Practice Problem 5Use Cramer’s rule to solve the system:

• x + 2y - z = 32x - 3y + z = -9

-x + y - 2z = 0


Solution: First we find D, , , and . Beginning with D, weexpand by the minors of the first column.

From these determinants, we calculate the solution:

The ordered triple solution is . Check thissolution by verifying that it satisfies each equation of thesystem.

A1, -2, -1 Bz =

Dz

D= -61

61= -1y =

Dy

D= -122

61= -2x =

Dx

D= 61

61= 1

= -23 + 12 - 50 = -61

= 1 A-8 - 15 B - 3 A16 - 20 B + 5 A-6 - 4 BDz = † 13

5

-215

43

-8† = 1 # ` 1

53

-8` - 3 # ` -2

54

-8` + 5 # ` -2

143`

= -7 - 60 - 55 = -122

= 1 A9 - 16 B - 3 A12 - A- 8 B B + 5 A- 8 - 3 BDy = † 13

5

43

-8

1-2

3† = 1 # ` 3

-8-2

3` - 3 # ` 4

-813` + 5 # ` 4

31

-2`

= 52 + 33 - 24 = 61

= 4 A3 - A-10 B B - 3 A-6 - 5 B + A-8 B A4 - 1 B= 4 # ` 1

5-2

3` - 3 # ` -2

513` + A-8 B # ` -2

11

-2`Dx = † 4

3-8

-215

1-2

3†

= 13 + 33 + 15 = 61

= 1 A3 - A-10 B B - 3 A-6 - 5 B + 5 A4 - 1 BD = † 13

5

-215

1-2

3† = 1 # ` 1

5-2

3` - 3 # ` -2

513` + 5 # ` -2

11

-2`

DzDyDx

8.8 SYSTEMS OF LINEAR INEQUALITIES

GRAPHING SYSTEMS OF LINEAR INEQUALITIES

In Section 3.5 we solved linear inequalities in two variables. Just as two lin-ear equations make a system of linear equations, two linear inequalitiesmake a system of linear inequalities. Systems of inequalities are veryimportant in a process called linear programming. Many businesses use lin-ear programming to find the most profitable way to use limited resourcessuch as employees, machines, or buildings.

A solution of a system of linear inequalities is an ordered pair that satis-fies each inequality in the system. The set of all such ordered pairs is thesolution set of the system. Graphing this set gives us a picture of the solu-tion set. We can graph a system of inequalities by graphing each inequalityin the system and identifying the region of overlap.

Example 1 Graph the solutions of the system

Solution: We begin by graphing each inequality on the same set ofaxes. The graph of the solutions of the system is theregion contained in the graphs of both inequalities. Inother words, it is their intersection.

First let’s graph . The boundary line is thegraph of . We sketch a solid boundary line sincethe inequality means or . The testpoint satisfies the inequality, so we shade the half-plane that includes .

Next we sketch a solid boundary line onthe same set of axes. The test point satisfies theinequality , so we shade the half-plane thatincludes . (For clarity, the graph of isshown here on a separate set of axes.)

An ordered pair solution of the system must satisfyboth inequalities. These solutions are points that lie in

x + 2y ≤ 8A0, 0 Bx + 2y ≤ 8A0, 0 Bx + 2y = 8

54321

1

−2

−1

−3−4−5

−2−3−5−4 2 3 4 5 x

y

x + 2y ≤ 8

54321

1

−2

−1

−3−4−5

−2−3−5−4 2 3 4 5 x

y

3x ≥ y

A1, 0 BA1, 0 B 3x = y3x 7 y3x ≥ y3x = y

3x ≥ y

e 3x ≥ yx + 2y ≤ 8

GRAPHING THE SOLUTIONS OF A SYSTEM OF LINEAR INEQUALITIES

Step 1. Graph each inequality in the system on the same set of axes.

Step 2. The solutions of the system are the points common to thegraphs of all the inequalities in the system.

A

Systems of Linear Inequalities SECTION 8.8 577

Objective

Graph a system of linear inequali-ties.

A

Practice Problem 1Graph the solutions of the system:

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

e2x ≤ yx + 4y ≥ 4

SSM CD-ROM Video8.8

Answer1.

54Solution

region321

1

−2

−1

−3−4−5

−2−3−5−4 2 3 4 5 x

y


Answer2.

54321

1

−2

−1

−3−4−5

−2−3−5−4 2 3 4 5 x

y

Solutionregion


x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

•-x + y 6 3

y 6 1 2x + y 7 -2

both shaded regions. The solution of the system is thedarkest shaded region. This solution includes parts ofboth boundary lines.

In linear programming, it is sometimes necessary to find the coordinatesof the corner point: the point at which the two boundary lines intersect. Tofind the point of intersection for the system of Example 1, we solve therelated linear system

using either the subsitution or the elimination method. The lines intersect

at , the corner point of the graph.

Example 2 Graph the solutions of the system:

Solution: First we graph all three inequalities on the same set ofaxes. All boundary lines are dashed lines since theinequality symbols are and . The solution of thesystem is the region shown by the darkest shading. In thisexample, the boundary lines are not a part of the solu-tion.

y < 2

67

54321

1−1

−2−3−4−5−6−7

−2−3−4−5−6−7 2 3 4 5 6 7

Solutionregion

x + 2y > −1

x − y < 2

x

y

76

•x - y 6 2x + 2y 7 -1

y 6 2

a 87

, 247b

e 3x = yx + 2y = 8

54321

1

−2

−1

−3−4−5

−2−3−5−4 2 3 4 5 x

y

Solutionregion

87( ), 24

7

x + 2y ≤ 8 3x ≥ y


Answers3.

✓ Concept Check: It is the line x = 2.

54321

1

−2

−1

−3−4−5

−2−3−5−4 2 3 4 5 x

y

✓ CONCEPT CHECK

Describe the solution of the system ofinequalities:

ex ≤ 2x ≥ 2


x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

µ2x - 3y ≤ 6

y ≥ 0y ≤ 4x ≥ 0


Example 3 Graph the solutions of the system

Solution: We graph the inequalities on the same set of axes. Theintersection of the inequalities is the solution region. It isthe only region shaded in this graph and includes por-tions of all four boundary lines.

y = 0

67

54321

1−1

−2−3−4−5−6−7

−2−3−4−5−6−7 2 3 4 5 6 7

Solutionregion

x = 3

x = 0

−3x + 4y = 12

x

y

µ-3x + 4y ≤ 12

x ≤ 3 x ≥ 0 y ≥ 0


Focus On the Real WorldANOTHER MATHEMATICAL MODEL

Sometimes mathematical models other than linear models are appropriate for data. Supposethat an equation of the form is an appropriate model for the ordered pairs

, and . Then it is necessary to find the values of a, b, and c such that thegiven ordered pairs are solutions of the equation . To do so, substitute eachordered pair into the equation. Each time, the result is an equation in three unknowns: a, b,and c. Solving the resulting system of three linear equation in three unknowns will give therequired values of a, b, and c.

GROUP ACTIVITY

1. The table gives the average annual fatalities from lightning ineach of the years listed.a. Write the data as ordered pairs of the form where y

is the number of lightning fatalities in the year x.

b. Find the values of a, b, and c such that the equationmodels this data.

c. Verify that the model you found in part (b) gives each ofthe ordered pair solutions from part (a).

d. According to the model, what was the average annual number of fatalities from light-ning in 1955?

2. The table gives the world production of red meat (in millionsof metric tons) for each of the years listed.a. Write the data as ordered pairs of the form where y

is the red meat production (in millions of metric tons) inthe year x .

b. Find the values of a, b, and c such that the equationmodels this data.

c. According to the model, what is the world production ofred meat in 1999?

3. a. Make up an equation of the form .b. Find three ordered pair solutions of the equation.c. Without revealing your equation from part (a), exchange lists of ordered pair solutions

with another group.d. Use the method described above to find the values of a, b, and c such that the equation

has the ordered pair solutions you received from the other group.e. Check with the other group to see if your equation from part (d) is the correct one.

y = ax2 + bx + c

y = ax2 + bx + c

y = ax2 + bx + c

Ax = 0 represents 1990 BAx, y B

y = ax2 + bx + c

Ax = 0 represents 1900 BAx, y B

y = ax2 + bx + cAx3, y3 BAx1, y1 B , Ax2, y2 B

y = ax2 + bx + c

AVERAGE ANNUAL

LIGHTNING FATALITIES

Year Fatalities

1940 337

1950 184

1960 133

(Source: National Weather Service)

WORLD PRODUCTION OF

RED MEAT

Millions ofYear metric tons

1993 119.3

1996 135.5

1998 140.1

(Source: Economic Research Ser-vice, U.S. Department of Agricul-ture)


Focus On the Real WorldLINEAR MODELING

In Chapter 3, we learned several ways to find a linear model when given either two orderedpairs or an ordered pair and slope. Another way to find a linear model of the form

for two ordered pairs and is to solve the following system of lin-ear equations for m and b:

For example, suppose a researcher wishes to find a linear model for the number of trafficaccidents involving teenagers. The researcher locates statistics stating that there were 5215teenage deaths from motor vehicle accidents in the United States in 1992. By 1996, this numberhad increased to 5805 teenage deaths in motor vehicle accidents. (Source: Insurance Institutefor Highway Safety)

This data gives two ordered pairs: (1992, 5215) and (1996, 5805). Alternatively, the orderedpairs could be written as (2, 5215) and (6, 5805), where the x-coordinate represents the numberof years after 1990. (Adjusting data given as years in this way often simplifies calculations.) Bysubstituting the coordinates of the second set of ordered pairs into the general linear system,we obtain the system

The solution of this system is and . We can use these values to write themodel the researcher wished to find: , where y is the number of teenagedeaths in motor vehicle accidents x years after 1990.

y = 147.5x + 4920b = 4920m = 147.5

e5215 = 2m + b5805 = 6m + b

ey1 = mx1 + by2 = mx2 + b

Ax2, y2 BAx1, y1 By = mx + b



SECTION 8.1 SOLVING SYSTEMS OF LINEAR EQUATIONS BY GRAPHING

A system of linear equations consists of two or morelinear equations. e -3x + 5y = 10

x - 4y = -2e2x + y = 6

x = -3y

A solution of a system of two equations in two vari-ables is an ordered pair of numbers that is a solu-tion of both equations in the system.

Determine whether is a solution of thesystem.

Replace x with and y with 3 in both equations.

True.

True.

is a solution of the system.A-1, 3 B -1 = -1 -1 � 3 A 3 B - 10

x = 3y - 10

-5 = -52 A-1 B - 3 � -5

2x - y = -5

-1

e2x - y = -5x = 3y - 10

A-1, 3 B

Graphically, a solution of a system is a point com-mon to the graphs of both equations. Solve by graphing:

3x − 2y = −3x + y = 4

(1, 3)

y

x

e3x - 2y = -3x + y = 4

Three different situations can occur when graphingthe two lines associated with the equations in a lin-ear system.

One point of inter- Same line; infinite section; one solution number of solutions

Parallel lines; no solution

x

y

x

y

x

y


SECTION 8.2 SOLVING SYSTEMS OF LINEAR EQUATIONS BY SUBSTITUTION

TO SOLVE A SYSTEM OF LINEAR EQUATIONS BY THE

SUBSTITUTION METHOD

Step 1. Solve one equation for a variable.

Step 2. Substitute the expression for the variableinto the other equation.

Step 3. Solve the equation from Step 2 to find thevalue of one variable.

Step 4. Substitute the value from Step 3 in eitheroriginal equation to find the value of theother variable.

Step 5. Check the solution in both equations.

Solve by substitution.

Substitute for x in the first equation

Divide by 5.

To find x, substitute 2 for y in so thator . The solution checks.A-1, 2 B-1x = 2 - 3

x = y - 3

y = 2

5y = 10

3y - 9 + 2y = 1

3 Ay - 3 B + 2y = 1

3x + 2y = 1

y - 3

e3x + 2y = 1x = y - 3

SECTION 8.3 SOLVING SYSTEMS OF LINEAR EQUATIONS BY ADDITION

TO SOLVE A SYSTEM OF LINEAR EQUATIONS BY THE

ADDITION METHOD

Step 1. Rewrite each equation in standard form.Ax + By = C

Solve by addition

ex - 2y = 83x + y = -4

Step 2. Multiply one or both equations by a nonzeronumber so that the coefficients of a variableare opposites.

Step 3. Add the equations.

Step 4. Find the value of one variable by solving theresulting equation.

Step 5. Substitute the value from Step 4 into eitheroriginal equation to find the value of theother variable.

Multiply both sides of the first equation by .

To find x, let in an original equation.

First equation

x = 0

x + 8 = 8

x - 2 A-4 B = 8

y = -4

Add.

Divide by 7.

e -3x + 6y = -24 3x + y = -4

7y = -28y = -4

-3

Step 6. Check the solution in both equations.

If solving a system of linear equations by substitutionor addition yields a true statement such as

, then the graphs of the equations in thesystem are identical and there is an infinite numberof solutions of the system.

-2 = -2

The solution checks.

Solve:

Substitute for x in the first equation.

True.

The system has an infinite number of solutions.

-2 = -2

6y - 2 - 6y = -2

2 A3y - 1 B - 6y = -2

3y - 1

e2x - 6y = -2x = 3y - 1

A0, -4 B

If solving a system of linear equations yields a falsestatement such as , the graphs of the equa-tions in the system are parallel lines and the systemhas no solution.

0 = 3 Solve: False.

The system has no solution.

e 5x - 2y = 6 -5x + 2y = -3

0 = 3


SECTION 8.4 SYSTEMS OF LINEAR EQUATIONS AND PROBLEM SOLVING


1. UNDERSTAND. Read and reread the problem.

Two angles are supplementary if their sum is 180°.The larger of two supplementary angles is threetimes the smaller, decreased by twelve. Find themeasure of each angle. Let

measure of smaller angle

measure of larger angle

x˚y˚

y =x =

2. TRANSLATE. In words:

T T TTranslate: 180

In words:

T T T T TTranslate: y 12-3x=

12decreased

by3 timessmalleris

larger angle

=x + y

180°isthe sum of

supplementary angles

3. SOLVE. Solve the system

Use the substitution method and replace y within the first equation.

x = 48

4x = 192

x + A3x - 12 B = 180

x + y = 180

3x - 12

ex + y = 180y = 3x - 12

4. INTERPRET. Since , then or 132.

The solution checks. The smaller angle measures 48°and the larger angle measures 132°.

y = 3 # 48 - 12y = 3x - 12

SECTION 8.5 SOLVING SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES

A solution of an equation in three variables x, y andz is an ordered triple (x, y, z) that makes the equa-tion a true statement.

Verify that is a solution of.

Replace x with , y with 1, and z with 3.

True.

is a solution.A-2, 1, 3 B -7 = -7

-4 + 3 - 6 = -7

2 A-2 B + 3 A1 B - 2 A3 B = -7

-22x + 3y - 2z = -7

A-2, 1, 3 B



SOLVING A SYSTEM OF THREE LINEAR EQUATIONS

BY THE ELIMINATION METHOD

Step 1. Write each equation in standard form,.

Step 2. Choose a pair of equations and use theequations to eliminate a variable.

Step 3. Choose any other pair of equations andeliminate the same variable.

Step 4. Solve the system of two equations in twovariables from Steps 2 and 3.

Step 5. Solve for the third variable by substitutingthe values of the variables from Step 4 intoany of the original equations.

Step 6. Check the solution in all three originalequations.

Ax + By + Cz = D

Solve:

1. Each equation is written in standard form.2.

3. Eliminate y from equations (1) and (3) also.

4. Solve.

To find z, use equation (5).

5. To find y, use equation (1).

The solution checks.A-12, 14, -10 B y = 14

-24 + y + 10 = 0

2 A-12 B + y - A-10 B = 0

2x + y - z = 0

z = -10

-12 + z = -22

x + z = -22

Divide equation (4) by 3.(5)

x - z = -2 x + z = -222x = -24 x = -12

(4)

(5)e3x - 3z = -6

x + z = -22

Multiply equation (1) by 2.

Add.

(3)

(5)

4x + 2y - 2z = 0-3x - 2y + 3z = -22 x + z = -22

Add.

(1)

(2)

(4)

2x + y - z = 0 x - y - 2z = -6 3x - 3z = -6

(1)

(2)

(3)

• 2x + y - z = 0

x - y - 2z = -6

-3x - 2y + 3z = -22

SECTION 8.6 SOLVING SYSTEMS OF EQUATIONS USING MATRICES

A matrix is a rectangular array of numbers.

£adg

beh

cfi§c -7

102

34d

The corresponding matrix of the system is obtainedby writing a matrix composed of the coefficients ofthe variables and the constants of the system.

The corresponding matrix of the system

is 1

11dc1

2-1

1e x - y = 1

2x + y = 11


Use matrices to solve:

The corresponding matrix is

Use row operations to write an equivalent matrixwith 1s along the diagonal and 0s below each 1 inthe diagonal. Multiply row 1 by and add to row2. Change row 2 only.

simplifies to

Divide row 2 by 3.

simplifies to

This matrix corresponds to the system

Let in the first equation.

The ordered pair solution is .A4, 3 B x = 4

x - 3 = 1

y = 3

ex - y = 1y = 3

13dc1

0-1 1

193§£10

3

-133

19dc1

0-1

3

1-2 A1 B + 11 dc 1

-2 A1 B + 2-1

-2 A-1 B + 1

-2

111dc1

2-1

1

e x - y = 12x + y = 11

SECTION 8.7 SOLVING SYSTEMS OF EQUATIONS USING DETERMINANTS

A square matrix is a matrix with the same number ofrows and columns. £40

1

-121

652§c -2

618d

A determinant is a real number associated with asquare matrix. To denote the determinant, placevertical bars about the array of numbers.

The determinant of is .` -26

18`c -2

618d

The determinant of a matrix is

` ac

bd` = ad - bc

2 * 2 ` -26

18` = -2 # 8 - 1 # 6 = -22


The following row operations can be performed onmatrices, and the result is an equivalent matrix.

Elementary row operations:

1. Interchange any two rows.2. Multiply (or divide) the elements of one row by

the same nonzero number.3. Multiply (or divide) the elements of one row by

the same nonzero number and add to its corre-sponding elements in any other row.



CRAMER’S RULE FOR TWO LINEAR EQUATIONS IN

TWO VARIABLES

The solution of the system is given

by

as long as is not 0.D = ad - bc

x =` hk

bd`

` ac

bd`

= Dx

D y =

` ac

hk`

` ac

bd`

=Dy

D


Use Cramer’s rule to solve

The ordered pair solution is .A-2, 7 By =

Dy

D= -49

-7= 7 x =

Dx

D= 14

-7= -2

Dy = ` 32

8-11

` = 3 A-11 B - 8 A2 B = -49

Dx = ` 8-11

2-1` = 8 A-1 B - 2 A-11 B = 14

D = ` 32

2-1` = 3 A-1 B - 2 A2 B = -7

e3x + 2y = 82x - y = -11

DETERMINANT OF A MATRIX

Each matrix above is called a minor.2 * 2

` b1

b2

c1

c2`` b1

b3

c1

c3` + a3 #

` b2

b3

c2

c3` - a2 #† a1

a2

a3

b1

b2

b3

c1

c2

c3

† = a1 #

3 * 3

= 0 - 40 + 16 = -24- 1 A-10 - 6 B

= 0 A12 - 0 B - 2 A20 - 0 B+ A-1 B ` 5

23

-2`

` 52

04`` 3

-204` - 2† 05

2

23

-2

-104† = 0

CRAMER’S RULE FOR THREE EQUATIONS IN THREE

VARIABLES

The solution of the system

is given by

and

where

as long as D is not 0.

Dz = † a1

a2

a3

b1

b2

b3

k1

k2

k3

†Dy = † a1

a2

a3

k1

k2

k3

c1

c2

c3

†

Dx = † k1

k2

k3

b1

b2

b3

c1

c2

c3

†D = † a1

a2

a3

b1

b2

b3

c1

c2

c3

†

z =Dz

Dy =

Dy

Dx =

Dx

D

•a1x + b1y + c1z = k1

a2x + b2y + c2z = k2

a3x + b3y + c3z = k3

Use Cramer’s rule to solve

The ordered triple solution is .A-1, 0, 4 Bz =

Dz

D= -12

-3= 4

y =Dy

D= 0

-3= 0x =

Dx

D= 3

-3= -1

Dz = † 012

31

-1

832† = -12

Dy = † 012

832

211† = 0

Dx = † 832

31

-1

211† = 3

D = † 012

31

-1

211† = -3

•3y + 2z = 8

x + y + z = 32x - y + z = 2


To graph a system of inequalities, graph eachinequality in the system. The overlapping region isthe solution of the system.

54321

1

−2

−1

−3

−2−3−5−4 2 3 4 5 x

y

y ≤ −2x

Solutionregion

x − y ≥ 3

SECTION 8.8 SYSTEMS OF LINEAR INEQUALITIES

A system of linear inequalities consists of two ormore linear inequalities. ex - y ≥ 3

y ≤ -2x

589

9In this chapter, we define radical notation, and then introduce rationalexponents. As the name implies, rational exponents are exponents thatare rational numbers. We present an interpretation of rational exponentsthat is consistent with the meaning and rules already established for inte-ger exponents, and we present two forms of notation for roots: radicaland exponent. We conclude this chapter with complex numbers, a nat-ural extension of the real number system.

9.1 Radical Expressions

9.2 Rational Exponents

9.3 Simplifying RadicalExpressions

9.4 Adding, Subtracting, andMultiplying RadicalExpressions

9.5 Rationalizing Numerators andDenominators of RadicalExpressions

9.6 Radical Equations andProblem Solving

9.7 Complex Numbers

Mount Vesuvius is the only active volcano on the Europeancontinent. Although the Romans thought the volcano wasextinct, Vesuvius erupted violently in 79 A.D. The eruptionburied the cities of Pompeii, Herculaneum, and Stabiae underup to 60 feet of ash and mud, killing approximately 16,000 peo-ple. Although Pompeii was completely engulfed, the city was farfrom destroyed. The blanket of mud and ash perfectly preservedmuch of the city in a snapshot of daily life of the ancientRomans. Pompeii lay undisturbed for over 1500 years until thefirst excavations were made and its archaeological significancewas proven. The 79 A.D. eruption of Vesuvius seemed to be thevolcano’s renewal. It has erupted with varying degrees of vio-lence more than 30 times since Pompeii’s burial, most recentlyin 1944.

Rational Exponents, Radicals, and Complex Numbers C H A P T E R

9.1 RADICAL EXPRESSIONS

FINDING SQUARE ROOTS

The opposite of squaring a number is taking the square root of a number.For example, since the square of 4, or , is 16, we say that a square root of16 is 4.

Examples Find the square roots of each number.

1. 25 Since and , the squareroots of 25 are 5 and .

2. 49 Since and , the squareroots of 49 are 7 and .

3. There is no real number whose square is .The number has no real number squareroot.

The notation is used to denote the positive, or principal, square root ofa nonnegative number a. We then have in symbols that

We denote the negative square root with the negative radical sign:

An expression containing a radical sign is called a radical expression. Anexpression within, or “under,” a radical sign is called a radicand.

b radical sign

radical expression:a radicand

Examples Find each square root. Assume that all variablesrepresent nonnegative real numbers.

4. because .5. because .

6. because .

7. because .8. because .9. because .

10. . The negative in front of the radicalindicates the negative square root of 81.

11. is not a real number.1-81

-181 = -9

A3x5 B 2 = 9x1029x10 = 3x5

Ax3 B 2 = x62x6 = x3

A0.5 B 2 = 0.2510.25 = 0.5

a 27b 2

= 449A 4

49= 2

7

02 = 010 = 062 = 36136 = 6

1a

-116 = -4

116 = 4

1a

-4-4-4

-7A-7 B 2 = 4972 = 49

-5A-5 B 2 = 2552 = 25

42

A

Radical Expressions SECTION 9.1 591

Objectives

Find square roots.

Approximate roots using a calcula-tor.

Find cube roots.

Find nth roots.

Find when a is any real num-ber.

Find function values of radicalfunctions.

F

n2anEDC

BA

SQUARE ROOT

The number b is a square root of a if .b2 = a

Practice Problems 1–3Find the square roots of each number.

1. 36

2. 81

3. -16

Answers1. , 2. , 3. no real number square

root, 4. 5, 5. 0, 6. , 7. 0.6, 8. ,

9. , 10. , 11. not a real number-56x3

x535

9, -96, -6

PRINCIPAL AND NEGATIVE SQUARE ROOTS

The principal square root of a nonnegative number a is its nonnegativesquare root. The principal square root is written as . The negativesquare root of a is written as .-1a

1a

SSM CD-ROM Video9.1

Practice Problems 4–11Find each square root. Assume that allvariables represent nonnegative realnumbers.

4. 5.

6. 7.

8. 9.

10. 11. 1- 25-125

236x62x10

10.36A 925

10125

592 CHAPTER 9 Rational Exponents, Radicals, and Complex Numbers

\

Answers12. 5.477, 13. 0, 14. , 15. , 16. ,17. -4x2

x314

-2

APPROXIMATING ROOTS

Numbers such as 1, 4, 9, and 25 are called perfect squares, since ,, , and . Square roots of perfect square radicands sim-

plify to rational numbers. What happens when we try to simplify a rootsuch as ? Since 3 is not a perfect square, is not a rational number. Itis called an irrational number, and we can find a decimal approximation ofit. To find decimal approximations, we can use the table in the appendix ora calculator. For example, an approximation for is

capproximation symbol

To see if the approximation is reasonable, notice that since

, then

, or

.

We found , a number between 1 and 2, so our result is rea-sonable.

Example 12 Use a calculator or the appendix to approximate .Round the approximation to 3 decimal places and checkto see that your approximation is reasonable.

Is this reasonable? Since , then, or . The approxima-

tion is between 4 and 5 and is thus reasonable.

FINDING CUBE ROOTS

Finding roots can be extended to other roots such as cube roots. For exam-ple, since , we call 2 the cube root of 8. In symbols, we write

From this definition, we have

since

since

since

Notice that, unlike with square roots, it is possible to have a negative radi-cand when finding a cube root. This is so because the cube of a negativenumber is a negative number. Therefore, the cube root of a negative num-ber is a negative number.

Examples Find each cube root.

13. because .

14. because .A-4 B 3 = -6431-64 = -4

13 = 1311 = 1

x3 = x332x3 = x

A-3 B 3 = -2731-27 = -3

43 = 643164 = 4

318 = 2

23 = 8

C

4 6 120 6 5116 6 120 6 12516 6 20 6 25

120 L 4.472

120

13 L 1.732

1 6 13 6 2

11 6 13 6 14

1 6 3 6 4

13 L 1.732

13

1313

25 = 529 = 324 = 221 = 12

B

Practice Problem 12Use a calculator or the appendix to

Practice Problems 13–17Find each cubic root.

13.

14.

15.

16.

17. 32-64x6

32x9

3A 164

31-8

310

HELPFUL HINT

Don’t forget that the square rootof a negative number is not a realnumber. For example,

it not a realnumber

because there is no real numberthat when multiplied by itselfwould give a product of . InSection 9.7, we will see what kindof a number is.1-9

-9

1-9

CUBE ROOT

The cube root of a real number a is written as , and

only if b3 = a31a = b

31a

approximate . Round the approxi-mation to 3 decimal places and checkto see that your approximation is rea-sonable.

130

Radical Expressions SECTION 9.1 593

Answers18. 2, 19. , 20. , 21. not a real num-ber, 22.


2x2-6-2

15. because .

16. because .

17. because .

FINDING nTH ROOTS

Just as we can raise a real number to powers other than 2 or 3, we can findroots other than square roots and cube roots. In fact, we can find the nthroot of a number, where n is any natural number. In symbols, the nth rootof a is written as , where n is called the index. The index 2 is usuallyomitted for square roots.


Examples Find each root.

18. because and 3 is positive.19. because .20. because is the opposite of .21. is not a real number. There is no real number

that, when raised to the fourth power, is .

22. because .

FINDINGn1—an WHEN a IS ANY REAL NUMBER

Recall that the notation indicates the positive square root of only.For example,

When variables are present in the radicand and it is unclear whether thevariable represents a positive number or a negative number, absolute valuebars are sometimes needed to ensure that the result is a positive number.For example,

This ensures that the result is positive. This same situation may occur whenthe index is any even positive integer. When the index is any odd positiveinteger, absolute value bars are not necessary.

2x2 = @x @

2 A-5 B 2 = 125 = 5

a22a2

E

A4x B 3 = 64x33264x3 = 4x

-81

41-81125-5-125 = -5

A-3 B 5 = -24351-243 = -334 = 814181 = 3

n1a

D

A-2x3 B 3 = -8x932-8x9 = -2x3

Ax2 B 3 = x632x6 = x2

a 25b 3

= 8125

3A 8125

= 25

HELPFUL HINT

If the index is even, such as , , , and so on, the radicandmust be nonnegative for the root to be a real number. For example,

, but is not a real number,

, but is not a real number.

If the index is odd, such as , , and so on, the radicand may beany real number. For example,

and ,and .51-32 = -25132 = 2

31-64 = -43164 = 4

51 31

61-646164 = 2

41-164116 = 2

61 41 1

Practice Problems 18–22Find each root.

18. 19.

20. 21.

22. 328x6

41-16-136

51-324116

✓ CONCEPT CHECK

Which one is not a real number?a.

b.

c.

d. 2 A-15 B 251-15

41-15

31-15


\

Answers23. 5, 24. , 25. , 26. ,

27. , 28. , 29. ,30. 3, 31. 2, 32. , 33. 31912

@x + 3 @6 @x @7x - 1

-3@x + 6 @@x3 @

Examples Simplify. Assume that the variables represent any realnumber.

23.

24.

25.

26.

27.

28.

29.

FINDING FUNCTION VALUES

Functions of the form

are called radical functions. Recall that the domain of a function in x is theset of all possible replacement values of x. This means that if n is even, thedomain is the set of all nonnegative numbers, or . If n is odd,the domain is the set of all real numbers. Keep this in mind as we find func-tion values. In Chapter 9, we will graph these functions.

Examples If and , find eachfunction value.

30.

31.

32.

33. g A1 B = 311 + 2 = 313

g A-1 B = 31-1 + 2 = 311 = 1

f A6 B = 16 - 4 = 12

f A8 B = 18 - 4 = 14 = 2

g Ax B = 31x + 2f Ax B = 1x - 4

Ex @x ≥ 0F

f Ax B = n1x

F

2x2 + 2x + 1 = 2 Ax + 1 B 2 = @x + 1 @225x2 = 5 @x @

52 A2x - 7 B 5 = 2x - 7

Absolute value bars are not needed when the indexis odd.

32 A-5 B 3 = -5

42 Ax - 2 B 4 = @x - 2 @2x2 = @x @

When the index is even, the absolute value barsensure that the result is not negative.2 A-3 B 2 = @ -3 @ = 3

Practice Problems 23–29Simplify. Assume that the variablesrepresent any real number.

23.

24.

25.

26.

27.

28.

29. 2x2 + 6x + 9

236x2

52 A7x - 1 B 532 A-3 B 342 Ax + 6 B 42x6

2 A-5 B 2


If and , find each function

value.

30.

31.

32.

33. g A10 Bf A0 Bg A9 Bf A7 B

31x - 1g Ax B =f Ax B = 1x + 2

FINDING

If n is an even positive integer, then .If n is an odd positive integer, then .

n2an = a

n2an = @a @n2an

HELPFUL HINT

Notice that for the function , the domain includes allreal numbers that make the radicand . To see what numbersthese are, solve and find that . The domain is

.The domain of the cube root function is the set ofreal numbers.

g Ax B = 31x + 2Ex @x ≥ 4F x ≥ 4x - 4 ≥ 0

≥ 0f Ax B = 1x - 4

595


EXERCISE SET 9.1

Find the square roots of each number. See Examples 1 through 3.

1. 4 2. 9

Find each square root. Assume that all variables represent nonnegative real numbers. SeeExamples 4 through 11.

3. 4. 5. 6.

7. 8.

Use a calculator or the appendix to approximate each square root to 3 decimal places.Check to see that each approximation is reasonable. See Example 12.

9. 10.

Find each cube root. See Examples 13 through 17.

11. 12. 13.

14. 15. 16.

Find each root. Assume that all variables represent nonnegative real numbers. See Exam-ples 18 through 22.

17. 18. 19. 20.

21.

If and , find each function value. See Examples 30through 33.

22. 23. 24. 25.

26. g A-19 B

f A-1 Bg A7 Bg A0 Bf A0 Bg Ax B = 31x - 8f Ax B = 12x + 3F

51-32

1-1641-1651-243- 4116

D

31-12531-13A2764

3A18

31273164

C

11117

B

10.0410.0001

A 925A1

414001100

A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

9.2 RATIONAL EXPONENTS

UNDERSTANDING a1/n

So far in this text, we have not defined expressions with rational exponentssuch as , , and . We will define these expressions so that therules for exponents shall apply to these rational exponents as well.

Suppose that . Then

or 5ƒ¬

Since , then x is the number whose cube is 5, or . Notice thatwe also know that . This means that

Notice that the denominator of the rational exponent corresponds to theindex of the radical.

Examples Use radical notation to write each expression. Simplify ifpossible.

1.2.3.4.5.6.

UNDERSTANDING am/n

As we expand our use of exponents to include , we define their meaning

so that rules for exponents still hold true. For example, by properties ofexponents,

or

82>3 = A82 B 1>3 = 3282

82>3 = A81>3 B 2 = A 318 B 2

mn

B

5y1>3 = 5 31y

A81x8 B 1>4 = 4281x8 = 3x2

-91>2 = -19 = -3x1>4 = 41x

641>3 = 3164 = 441>2 = 14 = 2

51>3 = 315

x = 51>3x = 315x3 = 5

using rules cfor exponents

x3 = A51>3 B 3 = 51>3 # 3 = 51

x = 51>3

-9-1>4x2>331>2

A

Rational Exponents SECTION 9.2 597

Objectives

Understand the meaning of .



Use rules for exponents to simplifyexpressions that contain rationalexponents.

Use rational exponents to simplifyradical expressions.

E

Da-m>nCam>nBa1>nA

DEFINITION OF a1/n

If n is a positive integer greater than 1 and is a real number, then

a1>n = n1a

n1a

Practice Problems 1–6Use radical notation to write each ex-pression. Simplify if possible.

1.

2.

3.

4.

5.

6. 7x1>5A-27y6 B 1>3-251>2x1>5271>3251>2

Answers1. 5, 2. 3, 3. , 4. , 5. ,6. 7 51x

-3y2-551x

DEFINITION OF am/n

If m and n are positive integers greater than 1 with in lowest terms,then

as long as is a real number.n1a

am>n = n2am = A n1a Bm

mn

SSM CD-ROM Video9.2


Answers7. 27, 8. , 9. 4, 10. ,

11. , 12. , 13. - 164

19

72 A2x + 1 B 218

-64


7.

8.

9.

10.

11. A2x + 1 B 2>7a 1

4b 3>2

A-32 B 2>5-2563>493>2

Notice that the denominator n of the rational exponent corresponds tothe index of the radical. The numerator m of the rational exponent indi-cates that the base is to be raised to the mth power. This means that

or

Examples Use radical notation to write each expression. Simplify ifpossible.

7.8.9.

10.

11.

UNDERSTANDING a�m/n

The rational exponents we have given meaning to exclude negative ratio-nal numbers. To complete the set of definitions, we define .

Examples Write each expression with a positive exponent. Thensimplify.

12.

13. A-27 B-2>3 = 1A-27 B 2>3 = 1

A 31-27 B 2 = 1A-3 B 2 = 1

9

16-3>4 = 1163>4 = 1

A 4116 B 3 = 123 = 1

8

a-m>n

C

A4x - 1 B 3>5 = 52 A4x - 1 B 3a 1

9b 3>2

= aA19b 3

= a 13b 3

= 127

A-27 B 2>3 = A 31-27 B 2 = A-3 B 2 = 9

-163>4 = - A 4116 B 3 = - A2 B 3 = -8

43>2 = A14 B 3 = 23 = 8

82>3 = A 318 B 2 = 22 = 4

82>3 = 3282 = 3164 = 4

Practice Problems 12–13Write each expression with a positiveexponent. Then simplify.

12.

13. -256-3>427-2>3

HELPFUL HINT

Most of the time, will be easier to calculate than .n2amA n1a Bm

HELPFUL HINT

The denominator of a rational exponent is the index of the corre-sponding radical. For example, and , or

.z2>3 = A 31z B 2z2>3 = 32z2x1>5 = 51x

DEFINITION OF a-m/n

as long as is a nonzero real number.am>n

a-m>n = 1am>n

Use radical notation to write eachexpression. Simplify if possible.


USING RULES FOR EXPONENTS

It can be shown that the properties of integer exponents hold for rationalexponents. By using these properties and definitions, we can now simplifyexpressions that contain rational exponents. These rules are repeated herefor review.

Examples Use the properties of exponents to simplify.

14. Use the product rule.


16. Use the power rule.

Simplify.

Use the quotient rule.

Simplify.

or 32 Substitute 1 for .x0 = 32 # 1 = 32x0 = 32x2 - 2

= 32x2

x2

A2x2>5 B 5x2 =

25 Ax2>5 B 5x2

71>374>3 = 71>3 - 4>3 = 7-3>3 = 7-1 = 1

7

x1>2x1>3 = x1>2 + 1>3 = x3>6 + 2>6 = x5>6

D

Rational Exponents SECTION 9.2 599

HELPFUL HINT

If an expression contains a negative rational exponent, you may wantto first write the expression with a positive exponent, then interpretthe rational exponent. Notice that the sign of the base is not affectedby the sign of its exponent. For example,

Also,

A-27 B-1>3 = 1A-27 B 1>3 = - 1

3

9-3>2 = 193>2 = 1

A19 B 3 = 127

Answers14. , 15. , 16. 27

✓ Concept Check: d

181

x7>12

SUMMARY OF EXPONENT RULES

If m and n are rational numbers, and a, b, and c are numbers for whichthe expressions below exist, then

Product rule for exponents:

Power rule for exponents:

Power rules for products and quotients: and

Quotient rule for exponents:

Zero exponent:

Negative exponent: a-n = 1an, a Z 0

a0 = 1, a Z 0

am

an = am - n, a Z 0

a acb n

= an

cn, c Z 0

Aab Bn = anbn

Aam Bn = am # nam # an = am + n

✓ CONCEPT CHECK

Which one is correct?

a.

b.

c.

d. -8-2>3 = - 14

8-2>3 = -4

8-2>3 = - 14

-82>3 = 14

Practice Problems 14–16Use the properties of exponents tosimplify.

14.

15.

16.A3x2>3 B 3

x2

92>5912>5

x1>3x1>4


Answers

17. , 18. , 19. , 20. ,

21. , 22. 615001 21x

62y532a2b131y

USING RATIONAL EXPONENTS TO SIMPLIFY RADICAL EXPRESSIONS

Some radical expressions are easier to simplify when we first write themwith rational exponents. We can simplify some radical expressions by firstwriting the expression with rational exponents. Use properties of expo-nents to simplify, and then convert back to radical notation.

Examples Use rational exponents to simplify. Assume that allvariables represent positive numbers.

17. Write with rational exponents.

Simplify the exponent.

Write with radical notation.


Write 25 as .

Use the power rule.

Simplify the exponent.



Use the power rule.

Simplify the exponents.

Use .


Examples Use rational exponents to write as a single radical.

20.

21.


Use .


Multiply .32 # 23 = 6172

= 6232 # 23

anbn = Aab Bn = A32 # 23 B 1>6Write the exponents so that they have the samedenominator. = 32>6 # 23>6

313 # 12 = 31>3 # 21>2 = x1>6 = 61x

1x31x

= x1>2x1>3 = x1>2 - 1>3 = x3>6 - 2>6

= x3>4 = 42x3

1x # 41x = x1>2 # x1>4 = x1>2 + 1>4

= 32rs2

anbn = Aab Bn = Ars2 B 1>3 = r1>3s2>3 = r2>6s4>6

62r2s4 = Ar2s4 B 1>6 = 315

= 51>3 = 52>6

52 = A52 B 1>66125 = 251>6

= 1x = x1>2

82x4 = x4>8

E

Practice Problems 17–19Use rational exponents to simplify.Assume that all variables representpositive numbers.

17.

18.

19. 92a6b3

419

1 02y5

Practice Problems 20–22Use rational exponents to write as asingle radical.

20.

21.

22. 15 # 312

31x41x

1y # 31y

9.3 SIMPLIFYING RADICAL EXPRESSIONS

USING THE PRODUCT RULE

It is possible to simplify some radicals that do not evaluate to rational num-bers. To do so, we use a product rule and a quotient rule for radicals. Todiscover the product rule, notice the following pattern:

Since both expressions simplify to 6, it is true that

This pattern suggests the following product rule for exponents.

Notice that the product rule is the relationship statedin radical notation.

Examples Use the product rule to multiply.

1.2.3.4.

5.

USING THE QUOTIENT RULE

To discover the quotient rule for radicals, notice the following pattern:

Since both expressions simplify to , it is true that

This pattern suggests the following quotient rule for radicals.

A49

= 1419

23

1419= 2

3

A49

= 23

B

A 2a

# Bb3

= B2a

# b3

= B2b3a

415 # 422x3 = 425 # 2x3 = 4210x3

314 # 312 = 314 # 2 = 318 = 2121 # 1x = 121x

13 # 15 = 13 # 5 = 115

a1>n # b1>n = Aab B 1>n

19 # 14 = 19 # 4

19 # 4 = 136 = 6

19 # 14 = 3 # 2 = 6

A

Simplifying Radical Expressions SECTION 9.3 601

Objectives

Use the product rule for radicals.

Use the quotient rule for radicals.

Simplify radicals.CBA

PRODUCT RULE FOR RADICALS

If and are real numbers, thenn1a #

n2b = n2ab

n2bn1a

Practice Problems 1–5Use the product rule to multiply.

1.

2.

3.

4.

5. A 3x

# Ay2

416 # 423x2

312 # 3132

117 # 1y

12 # 17

Answers1. , 2. , 3. 4, 4. ,

5. A 3y2x

4218x2117y114

QUOTIENT RULE FOR RADICALS

If and are real numbers and is not zero, then

nA ab

=n1an2b

n2bn2b

n1a

SSM CD-ROM Video9.3


Answers

6. , 7. , 8. , 9. , 10. ,

11. , 12. , 13. 3 4121142 315

312517

2x34

1y6

35

Practice Problems 6–9Use the quotient rule to simplify.

6.

7.

8.

9. 5A 732x5

3A2764

A y36

A 925

Practice Problem 10Simplify: 118

Notice that the quotient rule is the relationship stated in

radical notation. We can use the quotient rule to simplify radical expres-sions by reading the rule from left to right or to divide radicals by readingthe rule from right to left.

For example,

Using

Using

Note: For the remainder of this chapter, we will assume that variables repre-sent positive real numbers. If this is so, we need not insert absolute value barswhen we simplify even roots.

Examples Use the quotient rule to simplify.

6.

7.

8.

9.

SIMPLIFYING RADICALS

Both the product and quotient rules can be used to simplify a radical. If theproduct rule is read from right to left, we have that . Weuse this to simplify the following radicals.

Example 10 Simplify:

Solution: We factor 50 such that one factor is the largest perfectsquare that divides 50. The largest perfect square factorof 50 is 25, so we write 50 as and use the productrule for radicals to simplify.

—c¬

Examples Simplify.

11.c¬

12. The largest perfect square factor of 26 is 1, socannot be simplified further.

13.c¬ The largest 4th power factor of 32.

4132 = 4116 # 2 = 4116 # 412 = 2 412126126

The largest perfect cube factor of 24.

3124 = 318 # 3 = 318 # 313 = 2 313

The largest perfect squarefactor of 50.

150 = 125 # 2 = 125 # 12 = 512

25 # 2

150

n2ab = n1a # n2b

C

4A 316y4 =

4134216y4

=413

2y

3A 827

=318

3127= 2

3

Ax9

= 1x19= 1x

3

A2549

= 125149= 5

7

n1an2b

= nAab

17513= A75

3= 125 = 5

nAab

=n1an2bA x

16= 1x116

= 1x4

a abb 1>n

= a1>nb1>n

Practice Problems 11–13Simplify:

11.

12.

13. 41162

114

3140

HELPFUL HINT

Don’t forgetthat, forexample,

means.5 # 12

512

Simplifying Radical Expressions SECTION 9.3 603

Answers14. , 15. , 16. ,

17. 5, 18. , 19. , 20. 9x3y 413y15x2 312431y

2z2 41z2x3y2 313y7a21a

After simplifying a radical such as a square root, always check the radi-cand to see that it contains no other perfect square factors. It may, if thelargest perfect square factor of the radicand was not originally recognized.For example,

Notice that the radicand 50 still contains the perfect square factor 25. Thisis because 4 is not the largest perfect square factor of 200. We continue asfollows:

The radical is now simplified since 2 contains no perfect square factors(other than 1).

Examples Simplify.

14.

15.

16.

Simplify.

Examples Use the quotient rule to divide. Then simplify if possible.


Simplify.

Simplify.

18. Use the quotient rule.150x212

= 12

# A50x2

= 2 = 14

12015= A20

5

= 3z2 42z3

= 4281 # z8 # 42z3

4281z11 = 4281 # z8 # z3

= 3x2y2 322y2

= 3227 # x6 # y6 # 322y2

= 3227 # x6 # y6 # 2y2

3254x6y8 = 3227 # 2 # x6 # y6 # y2

= 5x1x

= 225 # x2 # 1x

Find the largest perfect square factor.

Use the product rule.

Simplify.

Factor the radicand and identify per-fect cube factors.


Simplify.

Factor the radicand and identify per-fect fourth power factors.


225x3 = 225 # x2 # x

2150 = 2125 # 2 = 2 # 125 # 12 = 2 # 5 # 12 = 1012

1200 = 14 # 50 = 14 # 150 = 2150

Practice Problems 14–16Simplify.

14.

15.

16. 4216z9

3224x9y7

249a5

HELPFUL HINT

To help you recognize largest perfect power factors of a radicand, itwill help if you are familiar with some perfect powers. A few arelisted belowPerfect Squares 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

Perfect Cubes 1, 8, 27, 64, 125

Perfect 4th powers 1, 16, 81, 25644342414

5343332313

122112102928272625242322212

HELPFUL HINTWe say that a radical of the form is simplified when the radicanda contains no factors that are perfect nth powers (other than 1 or ).-1

n1a

Practice Problems 17–20Use the quotient rule to divide. Thensimplify if possible.

17. 18.

19. 20.3 42243x9y6

42x-3y

5 32162x8

323x2

180y315

17513


Simplify.

Factor .

Simplify.

19.

20.


= 2 4216 # a8 # b4 # 412 # a = 2 # 2a2b # 412a = 4a2b 412a

2 4232a8b6

42a-1b2= 2 4A32a8b6

a-1b2 = 2 4232a9b4 = 2 4216 # a8 # b4 # 2 # a

= 7 328 # y3 # 313 = 7 # 2y 313 = 14y 313

7 3248y4

312y= 7 3A48y4

2y= 7 3224y3 = 7 328 # y3 # 3

= 521x

= 12

# 5 # 1x

25x = 12

# 125 # 1x

= 12

# 125x

✓ CONCEPT CHECK

Find and correct the error:312719

= 3A279

= 313

Answer

✓ Concept Check:312719

= 33

= 1

Focus On HistoryDEVELOPMENT OF THE RADICAL SYMBOL

The first mathematician to use the symbol we use today to denotea square root was Christoff Rudolff (1499–1545). In 1525, Rudolffwrote and published the first German algebra text, Die Coss. In it,he used to represent a square root, the symbol to repre-sent a cube root, and the symbol to represent a fourth root,It was another 100 years before the square root symbol wasextended with an overbar called a vinculum, , to indicate theinclusion of several terms under the radical symbol. This innova-tion was introduced by René Descartes (1596–1650) in 1637 in histext La Géométrie. The modern use of a numeral as part of a radi-cal sign to indicate the index of the radical for higher roots did notappear until 1690 when this notation was used by French mathe-matician Michel Rolle (1652–1719) in his text Traité d’Algébre.

1

1

605


EXERCISE SET 9.3

Use the product rule to multiply. See Examples 1 through 5.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

Use the quotient rule to simplify. See Examples 6 through 9.

13. 14. 15.

16. 17. 18.

19. 20. 21.

22. 23. 24.

25. A x2y100

3A 38x6

3A 2x81y12

4A a3

81

4A 8x8

3A 364

3A 427

4A y81x4

4A x3

16A 5121

A 249A 8

81A 649

B

42ab2 # 4227ab424x3 # 415A 6m

# An5

A 7x

# A 2y

13y # 15x12 # 13x

3110 # 315314 # 3194127 # 413

418 # 412111 # 11017 # 12A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

9.4 ADDING, SUBTRACTING, AND MULTIPLYING RADICAL

EXPRESSIONS

ADDING OR SUBTRACTING RADICAL EXPRESSIONS

We have learned that the sum or difference of like terms can be simplified.To simplify these sums or differences, we use the distributive property. Forexample,

The distributive property can also be used to add like radicals.


1.2.3. This expression cannot be simplified

since and do not contain like radicals.

TRY THE CONCEPT CHECK IN THE MARGIN.When adding or subtracting radicals, always check first to see whether

any radicals can be simplified.


4. Factor 20 and 45.


Simplify and .

Add like radicals.

5.Factor and use the product rule.

Simplify and .

Write as 10.

Combine like radicals.= -6 3125 # 2= 3 312 - 10 312 + 312

3183127= 3 # 312 - 5 # 2 # 312 + 312= 3127 # 312 - 5 # 318 # 312 + 312

3154 - 5 3116 + 312

= 815 = 215 + 615

1914 = 2 # 15 + 2 # 3 # 15 = 14 # 15 + 2 # 19 # 15

120 + 2145 = 14 # 5 + 219 # 5

2 317217217 + 2 3175 313x - 7 313x = A5 - 7 B 313x = -2 313x

4111 + 8111 = A4 + 8 B 111 = 12111

2x + 3x = A2 + 3 Bx = 5x

A

Adding, Subtracting, and Multiplying Radical Expressions SECTION 9.4 607

Objectives

Add or subtract radical expres-sions.

Multiply radical expressions.B

A

HELPFUL HINT

The expression

does not contain like radicals and cannot be simplified further.

517 - 316

LIKE RADICALS

Radicals with the same index and the same radicand are like radicals.For example,

2 17 + 317 = A2 + 3 B 17 = 517


1.

2.

3. 6110 - 3 3110

9 312y - 15 312y

5115 + 2115

✓ CONCEPT CHECK

True or false?

Explain.1a + 1b = 1a + b

Answers1. , 2. , 3. ,4. , 5. , 6. ,7. , 8.

✓ Concept Check: false; answers may vary

5y 32y22 412 + 412515x - 241x-13 31320126110 - 3 3110-6 312y7115

SSM CD-ROM Video9.4


4.

5.

6.

7.

8. 328y5 + 3227y5

4132 + 132

120x - 6116x + 145x

3124 - 4 31192 + 313

150 + 5118


Answers

9. , 10. 13 315x3

191318


9.

10. 3A5x27

+ 4 315x

1759

- 132

6.

Simplify and .

Write as 6.ƒ————ƒ————ƒ——

7. Factor and use the product rule.

No further simplification is possible.

8.Simplify and .

Combine like radicals.


9. To subtract, notice that the LCD is 12.

Subtract.

10.

Simplify.

Add.

MULTIPLYING RADICAL EXPRESSIONS

We can multiply radical expressions by using many of the same propertiesused to multiply polynomial expressions. For instance, to multiply

, we use the distributive property and multiply byeach term inside the parentheses.

= 213 - 6

Use the product rule forradicals. = 12 # 2 # 3 - 3 # 2

= 12 # 6 - 312 # 2

Use the distributiveproperty.12 A16 - 312 B = 12 A16 B - 12 A312 B

1212 A16 - 312 B

B

= 5 317x2

=317x2

+ 4 317x2

Write each expression as an equiva-lent expression with a denominatorof 2.

=317x2

+ 2 317x # 22

=317x2

+ 2 317x

Use the quotient rule for radicals.3A7x8

+ 2 317x =317x318

+ 2 317x

= 51512

Multiply factors in the numeratorsand the denominators. = 915

12- 415

12

Write each expression as an equivalentexpression with a denominator of 12. = 315 # 3

4 # 3- 15 # 4

3 # 4

1454

- 153

= 3154

- 153

= 3y 316y

32y3328y3 = 2y 316y + y 316y

Factor and use theproduct rule.

3248y4 + 326y4 = 328y3 # 316y + 32y3 # 316y

= 3198 + 712

3198 + 198 = 3198 + 149 # 12

HELPFUL HINT

None of these terms contain likeradicals. We can simplify no further.

2 # 3= 313x - 61x + 612x

13619= 3 # 13x - 2 # 3 # 1x + 6 # 12x

= 19 # 13x - 2 # 19 # 1x + 136 # 12xFactor and use theproduct rule.

127x - 219x + 172x

Adding, Subtracting, and Multiplying Radical Expressions SECTION 9.4 609

Answers11. ,12. , 13. ,14. , 15. x + 5 + 41x + 152 - 1413

5y - 416 + 713 - 110 - 715612 + 215

Square the binomial: Aa + b B 2 = a2 + 2ab + b2

Practice Problem 11Multiply: 12 A6 + 110 B


12.

13.

14.

15. A1x + 1 + 2 B 2A13 - 7 B 2A15y + 2 B A15y - 2 BA13 - 15 B A12 + 7 B

Example 11 Multiply:

Solution:

Examples Multiply.

First Outer Inner Last12.

13. Multiply the sum and difference oftwo terms:

.

14. Square the binomial:

.

15.

c c c c c c c ca b 2 a b

Simplify.

Combine like terms. = x + 22 + 101x - 3 = x - 3 + 101x - 3 + 25

b2+##+a2

A1x - 3 + 5 B 2 = A1x - 3 B 2 + 2 # 1x - 3 # 5 + 52

= 4 - 213Aa - b B 2 = a2 - 2ab + b2 = 3 - 213 + 1

A13 - 1 B 2 = A13 B 2 - 2 # 13 # 1 + 12

Aa + b B Aa - b B = a2 - b2 = 2x - 25

A12x + 5 B A12x - 5 B = A12x B 2 - 52

Using the FOILorder, simplify. = 135 + 15 - 142 - 16

A15 - 16 B A17 + 1B = 15 # 17 + 15 # 1 - 16 # 17 - 16 # 1

= 513 + 3110

= 513 + 13 # 3 # 10

= 513 + 13 # 30

13 A5 + 130 B = 13 A5 B + 13 A130 B13 A5 + 130 B


Focus On the Real WorldDIFFUSION

Diffusion is the spontaneous movement of the molecules of a substance from a region of higherconcentration to a region of lower concentration until a uniform concentration throughout theregion is reached. For example, if a drop of food coloring is added to a glass of water, the mol-ecules of the coloring are diffused so that the entire glass of water is colored evenly withoutany kind of stirring. Diffusion is also mostly responsible for the spread of the smell of bakingbrownies throughout a house.

Diffusion is used or seen in important aspects of many disciplines. The following listdescribes situations in which diffusion plays a role.j In the commercial production of sugar, sugar can be extracted from sugar cane through a dif-

fusion process.j Solid-state diffusion plays a role in the manufacturing process of silicon computer chips.j In biology, the diffusion phenomenon allows water molecules, nutrient molecules, and dis-

solved gas molecules (such as oxygen and carbon dioxide) to pass through the semipermeablemembranes of cell walls.

j During a human pregnancy, the fetus is nourished from the mother’s blood supply via diffu-sion through the placenta. Waste materials from the fetus are also diffused through the pla-centa to be carried away by the mother’s circulatory system.

j The medical treatment known as kidney dialysis, in which waste materials are removed fromthe blood of a patient without kidney function, is made possible by diffusion.

j A diffusion process is widely used to separate the uranium isotope U-235, which can be usedas a fuel in nuclear power plants, from the uranium isotope U-238, which cannot be used tocreate nuclear energy.In chemistry, Graham’s law states that the diffusion rate of a substance in its gaseous state is

inversely proportional to the square root of its molecular weight. Another useful property ofdiffusion is that the distance a material diffuses over time is directly proportional to the squareroot of the time.

CRITICAL THINKING

1. Write an equation for the relationship described by Graham’s law. Be sure to define the vari-ables and constants that you use.

2. According to Graham’s law, which molecule will diffuse more rapidly: a molecule with a mo-lecular weight of 58.4 or a molecule with a molecular weight of 180.2? Explain your reason-ing.

3. Write an equation for the relationship between the distance that a material diffuses and time.Again, be sure to define the variables and constants that you use.

4. Suppose it takes sugar 1 week to diffuse a distance of 1 cm from its starting point in a par-ticular liquid. How long will it take the sugar to diffuse a total of 3 cm from its starting pointin the liquid?

611


MENTAL MATH

Simplify. Assume that all variables represent positive real numbers.

1. 2. 3. 4.

5. 6. 7. 8.

EXERCISE SET 9.4

Add or subtract as indicated. See Examples 1 through 10.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 719 - 7 + 13

3x175

+ B 7x2

100120x

9+ A5x

9

2 3147

-31414

3A118

-31116

132

+ 4133

5123

+ 2125

24x7 + 9x22x3 - 5x2x529b3 - 225b3 + 249b3

2 323a4 - 3a 3181a3116x - 3154x

4132 - 118 + 211282150 - 31125 + 198

3245x3 + x15x222x3 + 4x18x

127 - 17518 - 132A

A14x + 1 B 2A13 B 28 31z - 2 31z7 31x + 5 31x

31y + 101y81x - 51x517 + 317213 + 413

MENTAL MATH ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

9.5 RATIONALIZING NUMERATORS AND DENOMINATORS

OF RADICAL EXPRESSIONS

RATIONALIZING DENOMINATORS

Often in mathematics, it is helpful to write a radical expression such as either without a radical in the denominator or without a radical in the

numerator. The process of writing this expression as an equivalent ex-pression but without a radical in the denominator is called rationalizing thedenominator. To rationalize the denominator of , we use the funda-mental principle of fractions and multiply the numerator and the denomi-nator by . Recall that this is the same as multiplying by , which sim-plifies to 1.

Example 1 Rationalize the denominator of .

Solution: To rationalize the denominator, we multiply the numera-tor and denominator by a factor that makes the radicandin the denominator a perfect square.


Solution: First we simplify the radicals; then we rationalize thedenominator.

To rationalize the denominator, we multiply the numera-tor and the denominator by .


Solution:

Now we rationalize the denominator. Since is a cuberoot, we want to multiply by a value that will make theradicand 2 a perfect cube. If we multiply by , we get

. Thus,

Multiply numerator and denominator by and thensimplify.


32221 # 3222

312 # 3222=

3143223

=3142

3223 = 2

3222

312

3A12

=311312

= 1312

3A12

831x

= 8 # 1x31x # 1x

= 81x3x

1x

211619x=

2 A4 B19 # 1x= 8

31x

211619x

The denominator isnow rationalized.

215= 2 # 1515 # 15

= 2155

215

1312= 13 # 1212 # 12

= 1614= 16

2

121212

1312

1312

A

Rationalizing Numerators and Denominators of Radical Expressions SECTION 9.5 613

Objectives

Rationalize denominators.

Rationalize numerators.

Rationalize denominators or nu-merators having two terms.

CBA

Practice Problem 1

Rationalize the denominator of .712

Answers

1. , 2. , 3.

✓ Concept Check: a. or , b. 41231493272

31105

31y2y

7122

Practice Problem 2

Rationalize the denominator of .219116y

Practice Problem 3

Rationalize the denominator of .3A 225

SSM CD-ROM Video9.5

✓ CONCEPT CHECK

In each case, determine by whichnumber both the numerator anddenominator should be multiplied torationalize the denominator of theradical expression.

a. b.1418

1317


Answers

4. , 5. , 6. , 7.3a

3263a2b

6516

52a2b3

2b3155mn

11n

Practice Problem 4

Rationalize the denominator of .A 5m11n

Practice Problem 5Rationalize the denominator of

.52a2

5232b12

Practice Problem 6

Rationalize the numerator of .118175


Solution:

Use the quotient rule. No radical may be simplified further.

Multiply numerator and denominator by so that theradicand in the denominator is a perfect square.

Use the product rule in the numerator and denominator.Remember that .


Solution: First, simplify each radical if possible.

Use the product rule in the denominator.

Write as 3y.

Multiply numerator and denominator by so that theradicand in the denominator is a perfect 4th power.

Use the product rule in the numerator and denominator.

In the denominator, and .

RATIONALIZING NUMERATORS

As mentioned earlier, it is also often helpful to write an expression such asas an equivalent expression without a radical in the numerator. This

process is called rationalizing the numerator. To rationalize the numeratorof , we multiply the numerator and the denominator by .

Example 6 Rationalize the numerator of .

Solution: First we simplify .

Next we rationalize the numerator by multiplying thenumerator and the denominator by .

Example 7 Rationalize the numerator of .322x2

315y

17315

= 17 # 17315 # 17

= 7315 # 7

= 73135

17

17145= 1719 # 5

= 17315

145

17145

1312= 13 # 1312 # 13

= 1916= 316

131312

1312

B

3y # y = 3y242y4 = y =42xy3

3y2

=42xy3

3y 42y4

42y3

=41x # 42y3

3y 41y # 42y3

4281y4 =41x

3y 41y

41x4281y5

=41x

4281y4 # 41y

41x4281y5

13y # 13y = 3y= 121xy

3y

13y= 17x # 13y13y # 13y

A7x3y

= 17x13y

A7x3y

Practice Problem 7

Rationalize the numerator of 313a317b

Rationalizing Numerators and Denominators of Radical Expressions SECTION 9.5 615

Answer

8. 3 A215 - 1 B19

Solution:

Multiply the numerator and denominator by so thatthe radicand in the numerator is a perfect cube.

Use the product rule in the numerator and denominator.

Simplify.

RATIONALIZING DENOMINATORS OR NUMERATORS HAVING

TWO TERMS

Remember the product of the sum and difference of two terms?

a QThese two expressions are called conjugates of each other.

To rationalize a numerator or denominator that is a sum or difference oftwo terms, we use conjugates. To see how and why this works, let’s ratio-

nalize the denominator of the expression . To do so, we multiply both

the numerator and the denominator by , the conjugate of the

denominator and see what happens.

Multiply the sum and difference of two terms:.

Notice in the denominator that the product of and its conju-gate, , is . In general, the product of an expression and its con-jugate will contain no radical terms. This is why, when rationalizing adenominator or a numerator containing two terms, we multiply by its con-jugate. Examples of conjugates are

andand


Solution: We multiply the numerator and the denominator by theconjugate of .

Multiply the sum and difference of two terms:.Aa + b B Aa - b B = a2 - b2

=2 A312 - 4 BA312 B 2 - 42

2312 + 4

= 2 A312 - 4 BA312 + 4 B A312 - 4 B

312 + 4

2312 + 4

x - 1y x + 1y1a + 1b1a - 1b

-1A13 + 2 BA13 - 2 B

= -5 A13 + 2 B or -513 - 10

=5 A13 + 2 B

-1

=5 A13 + 2 B

3 - 4

Aa + b B Aa - b B = a2 - b2 =

5 A13 + 2 BA13 B 2 - 22

513 - 2=

5 A13 + 2 BA13 - 2 B A13 + 2 B

13 - 2

13 + 2

513 - 2

Aa + b B Aa - b B = a2 - b2

C

= 2x3120xy

=3223x3

325y # 22x

3222x322x2

315y=

322x2 # 3222x315y # 3222x


.3

215 + 1


Answers

9. ,

10. , 11. x - 253 A1x - 5 B

6 + 31x4 - x

115 + 110 + 313 + 312


.15 + 313 - 12


.3

2 - 1x

Write as or 18 and as 16.

or

As we saw in Example 8, it is often helpful to leave a numerator in factoredform to help determine whether the expression can be simplified.


Solution: We multiply the numerator and the denominator by theconjugate of .


Solution: We multiply by the conjugate of to elimi-nate the radicals from the denominator.

Example 11 Rationalize the numerator of .

Solution: We multiply the numerator and the denominator by theconjugate of , the numerator.

Multiply by , the conjugate of .

.

= x - 45 A1x - 2 B

Aa + b B Aa - b B = a2 - b2 =A1x B 2 - 22

5 A1x - 2 B

1x + 21x - 21x + 2

5=A1x + 2 B A1x - 2 B

5 A1x - 2 B

1x + 2

1x + 25

= 61mx - 2m9x - m

21m31x + 1m

= 21m A31x - 1m BA31x + 1m B A31x - 1m B = 61mx - 2m

A31x B 2 - A1m B 2

31x + 1m

21m31x + 1m

= 130 + 312 + 215 + 2132

= 130 + 118 + 215 + 2135 - 3

= 1615 + 1613 + 215 + 213A15 B 2 - A13 B 2

16 + 215 - 13=A16 + 2 B A15 + 13 BA15 - 13 B A15 + 13 B

15 - 13

16 + 215 - 13

312 - 4 =2 A312 - 4 B

2

429 # 2A312 B 2 =2 A312 - 4 B

18 - 16

Practice Problem 11Rationalize the numerator of

.1x + 5

3

9.6 RADICAL EQUATIONS AND PROBLEM SOLVING

SOLVING EQUATIONS THAT CONTAIN RADICAL EXPRESSIONS

In this section, we present techniques to solve equations containing radicalexpressions such as

We use the power rule to help us solve these radical equations.

This property does not say that raising both sides of an equation to a poweryields an equivalent equation. A solution of the new equation may or maynot be a solution of the original equation. Thus, each solution of the newequation must be checked to make sure it is a solution of the original equa-tion. Recall that a proposed solution that is not a solution of the originalequation is called an extraneous solution.

Example 1 Solve:

Solution: We use the power rule to square both sides of the equa-tion to eliminate the radical.

Now we check the solution in the original equation.

Check:

Let .

True.

The solution checks, so we conclude that the solution setis .

To solve a radical equation, first isolate a radical on one side of the equa-tion.

Example 2 Solve:

Solution: First, isolate the radical on one side of the equation. Todo this, we subtract 3x from both sides.

1-10x - 1 = -3x

1-10x - 1 + 3x - 3x = 0 - 3x

1-10x - 1 + 3x = 0

1-10x - 1 + 3x = 0

E42F 9 = 9

181 � 9

184 - 3 � 9

x = 4222 A42 B - 3 � 9

12x - 3 = 9

x = 42

2x = 84

2x - 3 = 81

A12x - 3 B 2 = 92

12x - 3 = 9

12x - 3 = 9

POWER RULE

If both sides of an equation are raised to the same power, all solutionsof the original equation are among the solutions of the new equation.

12x - 3 = 9

A

Radical Equations and Problem Solving SECTION 9.6 617

Practice Problem 1Solve: 13x - 2 = 5

Answers

1. , 2. e 14

, 2 fE9F

SSM CD-ROM Video9.6

Practice Problem 2Solve: 19x - 2 - 2x = 0

Objectives

Solve equations that contain radi-cal expressions.

Use the Pythagorean theorem tomodel problems.

B

A


Answer3. E4F

Next we use the power rule to eliminate the radical.

Since this is a quadratic equation, we can set the equationequal to 0 and try to solve by factoring.

Factor.


Check: . .

True.

True.

Both solutions check. The solution set is .

The following steps may be used to solve a radical equation.

Example 3 Solve:

Solution: First we isolate the radical by subtracting 5 from bothsides of the equation.

Next we raise both sides of the equation to the thirdpower to eliminate the radical.

31x + 1 = -2

31x + 1 + 5 = 3

31x + 1 + 5 = 3

SOLVING A RADICAL EQUATION

Step 1. Isolate one radical on one side of the equation.

Step 2. Raise each side of the equation to a power equal to the index ofthe radical and simplify.

Step 3. If the equation still contains a radical term, repeat Steps 1 and2. If not, solve the equation.

Step 4. Check all proposed solutions in the original equation.

e - 19

, -1 f

3 - 3 = 0 13

- 13

= 0

19 - 3 � 0 A19

- 13

� 0

110 - 1 - 3 � 0 A109

- 99

- 39

� 0

2-10 A-1 B - 1 + 3 A-1 B � 0A-10 a-19 b - 1 + 3 a-1

9b � 0

1-10x - 1 + 3x = 0 1-10x - 1 + 3x = 0

Let x = -1Let x = - 19

x = -1 x = - 19

9x + 1 = 0 or x + 1 = 0

A9x + 1 B Ax + 1 B = 0

9x2 + 10x + 1 = 0

-10x - 1 = 9x2

A1-10x - 1 B 2 = A-3x B 2

Practice Problem 3Solve: 31x - 5 + 2 = 1


Answers4. , 5.


e 316fE0F

Practice Problem 4Solve: 19 + x = x + 3

✓ CONCEPT CHECK

How can you immediately tell that theequation has no realsolution?

12y + 3 = -4

Practice Problem 5Solve: 13x + 1 + 13x = 2

The solution checks in the original equation, so the solu-tion set is .

Example 4 Solve:

Solution:

Factor.


Check:

Let . Let .

False. True.

The proposed solution 3 checks, but 0 does not. Since 0 isan extraneous solution, the solution set is .


Example 5 Solve:

Solution: We get one radical alone by subtracting from bothsides.

Now we use the power rule to begin eliminating the radi-cals. First we square both sides.

Multiply:

.

There is still a radical in the equation, so we get the radi-cal alone again. Then we square both sides.

A3 - 12x B A3 - 12x B 2x + 5 = 9 - 612x + 2x

A12x + 5 B 2 = A3 - 12x B 2

12x + 5 = 3 - 12x

12x + 5 + 12x = 3

12x

12x + 5 + 12x = 3

E3F 1 = 1 2 = -2

x = 314 - 3 � 3 - 2x = 014 - 0 � 0 - 214 - x = x - 214 - x = x - 2

x = 3

x = 0 or x - 3 = 0

x Ax - 3 B = 0

x2 - 3x = 0 Write the quadratic equa-tion in standard form.

4 - x = x2 - 4x + 4

A14 - x B 2 = Ax - 2 B 2 14 - x = x - 2

14 - x = x - 2

E-9F x = -9

x + 1 = -8

A 31x + 1 B 3 = A-2 B 3

HELPFUL HINT

In Example 4, notice that . Make sure bino-mials are squared correctly.

Ax - 2 B 2 = x2 - 4x + 4


Answers6.


313 cm

Get the radical alone.

Multiply.

Solve.

Simplify.

The proposed solution, checks in the original equation.



USING THE PYTHAGOREAN THEOREM

Recall that the Pythagorean theorem states that in a right triangle, thelength of the hypotenuse squared equals the sum of the lengths of each ofthe legs squared.

Example 6 Find the length of the unknown leg of the right triangle.

Solution: In the formula , c is the hypotenuse. Here,, the length of the hypotenuse, and . We

solve for b. Then becomesa2 + b2 = c2a = 4c = 10

a2 + b2 = c2

b

10 m4 m

B

e 29f

29

x = 29

x = 1672

72x = 16

36 A2x B = 16

Square both sides of the equa-tion to eliminate the radical.A612x B 2 = 42

612x = 4

2x + 5 = 9 - 612x + 2x

✓ CONCEPT CHECK

What is wrong with the followingsolution?

x = 55 x + 9 = 64

A2x + 5 B + A4 - x B = 64A12x + 5 + 14 - x B 2 = 82

12x + 5 + 14 - x = 8

Practice Problem 6Find the length of the unknown leg ofthe right triangle.

b

6 cm3 cm

HELPFUL HINT

Make sure expressions are squared correctly. In Example 5, wesquared as

= 9 - 612x + 2x

= 3 # 3 - 312x - 312x + 12x # 12x

A3 - 12x B 2 = A3 - 12x B A3 - 12x BA3 - 12x B

PYTHAGOREAN THEOREM

If a and b are the lengths of the legs of a right triangle and c is thelength of the hypotenuse, then .

c

b

a

Legs

Hypotenuse

a2 + b2 = c2

Practice Problem 7A furniture upholsterer wishes to cut astrip from a piece of fabric that is 45inches by 45 inches. The strip must becut on the bias of the fabric. What isthe longest strip that can be cut? Givean exact answer and a two decimalplace approximation.

45 inches

45 inches


Answer7. 4512 in. L 63.64 in.


Since b is a length and thus is positive, we have that

The unknown leg of the triangle is meters long.

Example 7 Calculating Placement of a Wire

A 50-foot supporting wire is to be attached to a 75-footantenna. Because of surrounding buildings, sidewalks,and roadways, the wire must be anchored exactly 20 feetfrom the base of the antenna.

a. How high from the base of the antenna is the wireattached?

b. Local regulations require that a supporting wire beattached at a height no less than of the total height ofthe antenna. From part (a), have local regulationsbeen met?

Solution: 1. UNDERSTAND. Read and reread the problem. From the diagram we notice that a right triangle isformed with hypotenuse 50 feet and one leg 20 feet.Let the height from the base of the antenna tothe attached wire.

2. TRANSLATE. Use the Pythagorean theorem.

3. SOLVE.

4. INTERPRET. Check the work and state the solution.a. The wire is attached exactly feet from the

base of the pole, or approximately 45.8 feet.b. The supporting wire must be attached at a height no

less than of the total height of the antenna. Thisheight is (75 feet), or 45 feet. Since we know frompart (a) that the wire is to be attached at a heightof approximately 45.8 feet, local regulations havebeen met.

35

35

10121

= 10121 x = 12100

Subtract 400 from bothsides. x2 = 2100

400 + x2 = 2500A20 B 2 + x2 = A50 B 2

a = 20, c = 50

Aa B 2 + Ab B 2 = Ac B 2A20 B 2 + x2 = A50 B 2

x =x feet

50feet

20 feet

35

75 feet

20 feet

50feet

2121

b = 184 = 14 # 21 = 2121

b2 = 84

16 + b2 = 100

42 + b2 = 102


GRAPHING CALCULATOR EXPLORATIONSWe can use a grapher to solve radical equations. For example, to use a grapher to approximate the solu-tions of the equation solved in Example 4, we graph the following:

and

The x-value of the point of intersection is the solution. Use the Intersect feature or the Zoom and Trace fea-tures of your grapher to see that the solution is 3.

Use a grapher to solve each radical equation. Round all solutions to the nearest hundredth.

1. 2.

3. 4.

5. 6. 21.9x2 - 2.2 = -0.8x + 31.2x = 13.1x + 5

110x - 1 = 1-10x + 10 - 112x + 1 = 12x + 2

13x + 5 = 2x1x + 7 = x

y =

y = x − 2

4 − x

10

10

−10

−10

Y2 = x - 2Y1 = 14 - x

9.7 COMPLEX NUMBERS

WRITING NUMBERS IN THE FORM biOur work with radical expressions has excluded expressions such as because is not a real number; there is no real number whose squareis . In this section, we discuss a number system that includes roots ofnegative numbers. This number system is the complex number system, andit includes the set of real numbers as a subset. The complex number systemallows us to solve equations such as that have no real numbersolutions. The set of complex numbers includes the imaginary unit.

To write the square root of a negative number in terms of i, we use theproperty that if a is a positive number, then

Using i, we can write as

Examples Write using i notation.

1.

2. . Since can easily be con-fused with , we write as .

3.

The product rule for radicals does not necessarily hold true for imagi-nary numbers. To multiply square roots of negative numbers, first we writeeach number in terms of the imaginary unit i. For example, to multiply and , we first write each number in the form bi:

We will also use this method to simplify quotients of square roots of nega-tive numbers.

Examples Multiply or divide as indicated.

4.5.6.

7.

Now that we have practiced working with the imaginary unit, we definecomplex numbers.

1-12515= i112515

= i125 = 5i

18 # 1-2 = 212 A i12 B = 2i A1212 B = 2i A2 B = 4i

1-36 # 1-1 = 6i A i B = 6i 2 = 6 A-1 B = -6

1-3 # 1-5 = i13 A i15 B = i 2115 = -1115 = -115

1-41-9 = 2i A3i B = 6i 2 = 6 A-1 B = -6

1-91-4

-1-20 = -1-1 # 20 = -1-1 # 14 # 5 = - i # 215 = -2i15i1515i15i

15i1-5 = 2-1 A5 B = 1-1 # 15 = i15

1-36 = 1-1 # 36 = 1-1 # 136 = i # 6 or 6i

1-16 = 1-1 # 16 = 1-1 # 116 = i # 4 or 4i

1-16

= i # 1a

1-a = 1-1 # 1a

x2 + 1 = 0

-161-16

1-16

A

Complex Numbers SECTION 9.7 623

Objectives

Write square roots of negativenumbers in the form bi.

Add or subtract complex numbers.

Multiply complex numbers.

Divide complex numbers.

Raise i to powers.EDCB

A

IMAGINARY UNIT

The imaginary unit, written i, is the number whose square is . Thatis,

and i = 1-1i 2 = -1

-1

Practice Problems 1–3Write using i notation.

1.

2.

3. -1-50

1-3

1-25

Answers1. , 2. , 3. , 4. ,5. , 6. , 7. 2i9i-5

-114-5i12i135i

Practice Problems 4–7Multiply or divide as indicated.

4.

5.

6.

7.1-812

127 # 1-3

1-25 # 1-1

1-2 # 1-7

SSM CD-ROM Video9.7


COMPLEX NUMBERS

A complex number is a number that can be written in the form ,where a and b are real numbers.

a + bi

Answer

✓ Concept Check: false

✓ CONCEPT CHECK

True or false? Every complex numberis also a real number.

Notice that the set of real numbers is a subset of the complex numbers sinceany real number can be written in the form of a complex number. Forexample,

In general, a complex number is a real number if . Also, acomplex number is called an imaginary number if . For example,

and

are imaginary numbers.The following diagram shows the relationship between complex num-

bers and their subsets.


ADDING OR SUBTRACTING COMPLEX NUMBERS

Two complex numbers and are equal if and only if and. Complex numbers can be added or subtracted by adding or sub-

tracting their real parts and then adding or subtracting their imaginaryparts.

b = da = cc + dia + bi

B

Complex numbersa + bi:

6, 1 − 2i, i,53 3

√76, 1 − 2i, i,53 3

√7

, π, e, −3√2 √11, π, e, −3√2 √11

Real numbersa + bi, b = 0:

4, −7, 1.8, − , 0119√54, −7, 1.8, − , 0,,

119√5

Complex numbers thatare not real numbers

a + bi, b ≠ 0:

6 − 2i, 3i, + i, −7.2i43

216 − 2i, 3i, + i, −7.2i

43

21

a + bi, a = 0, b ≠ 0:

4i, −2.6i, i874i, −2.6i, i87

a + bi, a ≠ 0, b ≠ 0:

3 + 5i, 6 − 0.2i, i+32

373 + 5i, 6 − 0.2i, i+

32

37

Rational numbers:

6, − , 0, −1.2,54

386, − , 0, −1.2,

54

38

Irrational numbers:

Imaginary numbers

Other complex numbers

i17 = 0 + i173i = 0 + 3i

a = 0b = 0a + bi

16 = 16 + 0i

SUM OR DIFFERENCE OF COMPLEX NUMBERS

If and are complex numbers, then their sum is

Their difference is

Aa + bi B - Ac + di B = a + bi - c - di = Aa - c B + Ab - d B i

Aa + bi B + Ac + di B = Aa + c B + Ab + d B ic + dia + bi


Answers8. , 9. , 10. ,11. 15, 12. , 13. ,14. , 15. 61-3 - 4i

22 - 21i-4 - 12i1 - 4i-2 + 7i9 - i


8.

9.

10. A-2 - 4i B - A-3 B6i - A2 - i BA5 + 2i B + A4 - 3i B


11.

12.

13.

14.

15. A6 + 5i B A6 - 5i BA1 - 2i B 2A3 - 4i B A6 + i B-2i A6 - 2i B-5i # 3i


8.9.

10.

MULTIPLYING COMPLEX NUMBERS

To multiply two complex numbers of the form , we multiply asthough they are binomials. Then use the relationship to simplify.

Examples Multiply.

11.Replace with .

12. Use the distributive property.

Multiply.

Replace with .

Use the FOIL method. (First, Outer, Inner, Last)

13.F O I L

.

14.

.

15.

.

Notice that if you add, subtract, or multiply two complex numbers, theresult is a complex number.

DIVIDING COMPLEX NUMBERS

From Example 15, notice that the product of and is a realnumber. These two complex numbers are called complex conjugates of oneanother. In general, we have the following definition.

7 - 3i7 + 3i

D

= 58 = 49 + 9

i 2 = -1 = 49 - 9 A-1 B = 49 - 21i + 21i - 9i 2A7 + 3i B A7 - 3i B = 7 A7 B - 7 A3i B + 3i A7 B - 3i A3i B

= 3 - 4i

i 2 = -1 = 4 - 4i + A-1 B = 2 A2 B - 2 A i B - 2 A i B + i 2

A2 - i B 2 = A2 - i B A2 - i B = 13 - 18i = 8 - 18i + 5

i 2 = -1 = 8 - 18i - 5 A-1 B = 8 + 2i - 20i - 5i 2

A2 - 5i B A4 + i B = 2 A4 B + 2 A i B - 5i A4 B - 5i A i B

= 3 + 6i = 6i + 3

-1i 2 = 6i - 3 A-1 B = 6i - 3i 23i A2 - i B = 3i # 2 - 3i # i

= 21-1i 2 = -21 A-1 B-7i # 3i = -21i 2

i 2 = -1a + bi

C

= 3 - 7i

= A-3 + 6 B - 7i

A-3 - 7i B - A-6 B = -3 - 7i + 6

= -1 + 6i

= -1 + A5 + 1 B i5i - A1 - i B = 5i - 1 + i

A2 + 3i B + A-3 + 2i B = A2 - 3 B + A3 + 2 B i = -1 + 5i


Answers

16. , 17. - 65

i313

+ 1113

i

Practice Problem 16Divide and write in the form :3 + i2 - 3i

a + bi

To see that the product of a complex number and its conjugateis the real number , we multiply:

We will use complex conjugates to divide by a complex number.

Example 16 Divide and write in the form a + bi:

Solution: We multiply the numerator and the denominator by thecomplex conjugate of to eliminate the imaginarynumber in the denominator.

Example 17 Divide and write in the form :

Solution: We multiply the numerator and the denominator by theconjugate of . Note that , so its conjugate is

or .

FINDING POWERS OF iWe can use the fact that to simplify and .

We continue this process and use the fact that and to sim-plify and .

i 6 = i 4 # i 2 = 1 # A-1 B = -1

i 5 = i 4 # i = 1 # i = i

i 6i 5i 2 = -1i 4 = 1

i 4 = i 2 # i 2 = A-1 B # A-1 B = 1

i 3 = i 2 # i = A-1 B i = - i

i 4i 3i 2 = -1

E

= - 73

i73i

=7 A-3i BA3i B A-3i B = -21i

-9i 2 = -21i

-9 A-1 B = -21i9

= -7i3

-3i0 - 3i3i = 0 + 3i3i

73i

a + bi

= 1 + 3i2

= 12

+ 32

i

= 2 + 3i - 11 + 1

=2 A1 B + 2 A i B + 1 A i B + i 2

12 - i 2

2 + i1 - i

=A2 + i B A1 + i BA1 - i B A1 + i B

1 - i

2 + i1 - i

= a 2 + b 2

= a2 - b 2 A-1 BAa + bi B Aa - bi B = a2 - abi + abi - b 2i 2

a2 + b2a - bia + bi

Practice Problem 17

Divide and write in the form : 65i

a + bi

COMPLEX CONJUGATES

The complex numbers and are called complex con-jugates of each other, and

Aa + bi B Aa - bi B = a2 + b2

Aa - bi BAa + bi B


If we continue finding powers of i, we generate the following pattern.Notice that the values , and 1 repeat as i is raised to higher andhigher powers.

This pattern allows us to find other powers of i. To do so, we will use thefact that and rewrite a power of i in terms of .For example, .

Examples Find each power of i.

18.19.20.

21. i -12 = 1i 12 = 1

A i 4 B 3 = 1A1 B 3 = 1

1= 1

i 46 = i 44 # i 2 = A i 4 B 11 # i 2 = 111 A-1 B = -1i 20 = A i 4 B 5 = 15 = 1i 7 = i 4 # i 3 = 1 A- i B = - i

i22 = i 20 # i 2 = A i 4 B 5 # i 2 = 15 # A-1 B = 1 # A-1 B = -1i 4i 4 = 1

i 12 = 1i 8 = 1i 4 = 1

i 11 = - ii 7 = - ii 3 = - i

i 10 = -1i 6 = -1i 2 = -1

i 9 = ii 5 = ii 1 = i

i, -1, - i

Practice Problems 18–21Find the powers of i.

18.

19.

20.

21. i -10

i 50

i 40

i 11

Answers18. , 19. 1, 20. , 21. -1-1- i


Focus On HistoryHERON OF ALEXANDRIA

Heron (also Hero) was a Greek mathematician and engineer. He lived and worked in Alexan-dria, Egypt, around 75 A.D. During his prolific work life, Heron developed a rotary steamengine called an aeolipile, a surveying tool called a dioptra, as well as a wind organ and a fireengine. As an engineer, he must have had the need to approximate square roots because hedescribed an iterative method for doing so in his work Metrica. Heron’s method for approxi-mating a square root can be summarized as follows:

Suppose that x is not a perfect square and is the nearest perfect square to x. For a rough estimate of

the value of , find the value of . This estimate can be improved by calculating a sec-

ond estimate using the first estimate in place of a: . Repeating this process several

times will give more and more accurate estimates of .

CRITICAL THINKING

1. a. Which perfect square is closest to 80?b. Use Heron’s method for approximating square roots to calculate the first estimate of the

square root of 80.c. Use the first estimate of the square root of 80 to find a more refined second estimate.d. Use a calculator to find the actual value of the square root of 80. List all digits shown on

your calculator’s display.e. Compare the actual value from part (d) to the values of the first and second estimates. What

do you notice?f. How many iterations of this process are necessary to get an estimate that differs no more

than one digit from the actual value recorded in part (d)?

2. Repeat Question 1 for finding an estimate of the square root of 30.

3. Repeat Question 1 for finding an estimate of the square root of 4572.

4. Why would this iterative method have been important to people of Heron’s era? Would yousay that this method is as important today? Why or why not?

1x

y2 = 12ay1 + x

y1by1

y1 = 12aa + x

ab1x

a2



SECTION 9.1 RADICAL EXPRESSIONS

The positive, or principal, square root of a nonnega-tive number a is written as .

only if and b ≥ 0b2 = a1a = b

1a A 9100

= 310

136 = 6

The negative square root of a is written as .-1a 10.04 = 0.2-136 = -6

The cube root of a real number a is written as .

only if

If n is an even positive integer, then .

If n is an odd positive integer, then .n2an = a

n2an = @a @b3 = a31a = b

31a

32 A-7 B 3 = -7

2 A-3 B 2 = @ -3 @ = 3

3264x9 = 4x332y6 = y2

3A- 18

= - 12

3127 = 3

A radical function in x is a function defined by anexpression containing a root of x.

If ,

f A3 B = 13 + 2 L 3.73

f A1 B = 11 + 2 = 1 + 2 = 3

f Ax B = 1x + 2

SECTION 9.2 RATIONAL EXPONENTS

if is a real number.n1aa1>n = n1a

A-8x3 B 1>3 = 32-8x3 = -2x

811>2 = 181 = 9

If m and n are positive integers greater than 1 with

in lowest terms and is a real number, then

am>n = Aa1>n Bm = A n1a Bmn1a

mn

272>3 = A 3127 B 2 = 32 = 9

45>2 = A14 B 5=25 = 32

as long as is a nonzero number.am>na-m>n = 1am>n 16-3>4 = 1

163>4 = 1A 4116 B 3 = 1

23 = 18

Exponent rules are true for rational exponents.

a4>5a-2>5 = a4>5 - A-2>5B = a6>5A814 B 1> 7=82 = 64

x2>3 # x-5>6 = x2>3 - 5>6 = x-1>6 = 1x1>6

SECTION 9.3 SIMPLIFYING RADICAL EXPRESSIONS

PRODUCT AND QUOTIENT RULES

If and are real numbers,

, provided n1b Z 0

n1an1b

= nA ab

n1a # n1b = n1a # b

n1bn1a

Multiply or divide as indicated:

3140x315x

= 318 = 2

111 # 13 = 133

A radical of the form is simplified when a con-tains no factors that are perfect nth powers.

n1a

3224x7y3 = 328x6y3 # 3x = 2x2y 313x

236x5 = 236x4 # x = 6x21x

140 = 14 # 10 = 2110


SECTION 9.4 ADDING, SUBTRACTING, AND MULTIPLYING RADICAL EXPRESSIONS

Radicals with the same index and the same radicandare like radicals.

516 + 216 = A5 + 2 B16 = 716

The distributive property can be used to add likeradicals.

Radical expressions are multiplied by using many ofthe same properties used to multiply polynomials. Multiply:

= 4 A3 B - 8x = 12 - 8xA213 - 18x B A213 + 18x B

= 110 + 110x - 21x - 2x = 110 + 110x - 14x - 2x

A15 - 12x B A12 + 12x B= -11 313x + 3 3110x

= A-1 - 10 B 313x + 3 3110x

- 313x - 10 313x + 3 3110x

SECTION 9.5 RATIONALIZING NUMERATORS AND DENOMINATORS OF RADICAL EXPRESSIONS

The conjugate of is .a - ba + b The conjugate of is .

Rationalize each denominator:

=6 A17 - 13 B

4=

3 A17 - 13 B2

=6 A17 - 13 B

7 - 3

617 + 13=

6 A17 - 13 BA17 + 13 B A17 - 13 B

1513= 15 # 1313 # 13

= 1153

17 - 1317 + 13

The process of writing the denominator of a radicalexpression without a radical is called rationalizingthe denominator.

The process of writing the numerator of a radicalexpression without a radical is called rationalizingthe numerator.

Rationalize each numerator:

=3 A3 - x B

3 # 4 A3 - 13x B = 3 - x4 A3 - 13x B

= 9 - 3x12 A19 - 13x B

19 + 13x12

=A19 + 13x B A19 - 13x B

12 A19 - 13x B

319315

=319 # 313315 # 313

=31273115

= 33115

SECTION 9.6 RADICAL EQUATIONS AND PROBLEM SOLVING

TO SOLVE A RADICAL EQUATION Solve: x = 14x + 9 + 3

Step 1. Write the equation so that one radical is byitself on one side of the equation.

1. x - 3 = 14x + 9

Step 2. Raise each side of the equation to a powerequal to the index of the radical and sim-plify.

2. x2 - 6x + 9 = 4x + 9

Ax - 3 B 2 = A14x + 9 B 2

Step 3. If the equation still contains a radical,repeat Steps 1 and 2. If not, solve the equa-tion.

3.

or x = 10x = 0x Ax - 10 B = 0 x2 - 10x = 0

Step 4. Check all proposed solutions in the originalequation.

4. The proposed solution 10 checks, but 0 does not.The solution is .E10F


SECTION 9.7 COMPLEX NUMBERS

A complex number is a number that can be writtenin the form , where a and b are real numbers.

and i = 1-1i 2 = -1

a + biSimplify:

, or

Complex Numbers Written in Form a + bi

12

Multiply.

= -121

= i 2121

1-3 # 1-7 = i13 # i17

-2 + A-3 B i-2 - 3i

0 + A-5 B i-5i

12 + 0i

3i1-9 = 1-1 # 9 = 1-1 # 19 = i # 3

1-9

To add or subtract complex numbers, add or subtracttheir real parts and then add or subtract their imag-inary parts.


= -10 + 6i

A-3 + 2i B - A7 - 4i B = -3 + 2i - 7 + 4i

To multiply complex numbers, multiply as thoughthey are binomials.

= -40 - 19i

= -42 - 19i + 2

= -42 - 19i - 2 A-1 BA-7 - 2i B A6 + i B = -42 - 7i - 12i - 2i 2

The complex numbers and arecalled complex conjugates.

Aa - bi BAa + bi B The complex conjugate of

is .

Their product is a real number:

= 9 - 36 A-1 B = 9 + 36 = 45

A3 - 6i B A3 + 6i B = 9 - 36i 2

A3 - 6i BA3 + 6i B

To divide complex numbers, multiply the numeratorand the denominator by the conjugate of thedenominator.

Divide:

= 85

+ 45

i = 8 + 4i5

=4 A2 + i B

5

=4 A2 + i B4 - i 2

42 - i

=4 A2 + i B

A2 - i B A2 + i B

633

10An important part of algebra is learning to model and solve problems.Often, the model of a problem is a quadratic equation or a function con-taining a second-degree polynomial. In this chapter, we continue thework from Chapter 5, solving quadratic equations in one variable by fac-toring. Two other methods of solving quadratic equations are analyzedin this chapter, with methods of solving nonlinear inequalities in onevariable and the graphs of quadratic functions.

10.1 Solving Quadratic Equationsby Completing the Square

10.2 Solving Quadratic Equationsby the Quadratic Formula

10.3 Solving Equations by UsingQuadratic Methods

10.4 Nonlinear Inequalities in OneVariable

10.5 Quadratic Functions and TheirGraphs

10.6 Further Graphing of QuadraticFunctions

The surface of the Earth is heated by the sun and then slowlyradiated into outer space. Sometimes, certain gases in theatmosphere reflect the heat radiation back to Earth, preventingit from escaping. The gradual warming of the atmosphere isknown as the greenhouse effect. This effect is compounded bythe increase of certain gases (greenhouse gases) such as carbondioxide, methane, and nitrous oxide. Although these gasesoccur naturally and are needed to keep the surface of the Earthat a temperature that is hospitable to life, the recent buildup ofthese gases is due primarily to human activities. According tothe Natural Resources Defense Council, carbon dioxide concen-trations have increased by 30% globally over the past century.

Quadratic Equations and Functions C H A P T E R

10.1 SOLVING QUADRATIC EQUATIONS BY COMPLETING

THE SQUARE

USING THE SQUARE ROOT PROPERTY

In Chapter 5, we solved quadratic equations by factoring. Recall that a qua-dratic, or second-degree, equation is an equation that can be written in theform , where a, b, and c are real numbers and a is not 0.To solve a quadratic equation such as by factoring, we use the zero-factor theorem. To use the zero-factor theorem, the equation must first bewritten in standard form, .

Subtract 9 from both sides to write in standard form.

Factor.


Solve.

The solution set is , the positive and negative square roots of 9.Not all quadratic equations can be solved by factoring, so we need to

explore other methods. Notice that the solutions of the equation aretwo numbers whose square is 9:

and

Thus, we can solve the equation by taking the square root of bothsides. Be sure to include both and as solutions since both and

are numbers whose square is 9.

The notation (read as plus or minus ) indicates the pair ofnumbers and .

This illustrates the square root property.

Example 1 Use the square root property to solve .

Solution:

Use the square root property.

Simplify the radical. x = ;512

x = ;150

x2 = 50

x2 = 50

-19+19 x = ;3

19;19 x = ;19

x2 = 9

-1919-1919

x2 = 9

A-3 B 2 = 932 = 9

x2 = 9

E-3, 3F x = 3 x = -3

x + 3 = 0 or x - 3 = 0

Ax + 3 B Ax - 3 B = 0

x2 - 9 = 0

x2 = 9

ax2 + bx + c = 0

x2 = 9ax2 + bx + c = 0

A

Solving Quadratic Equations by Completing the Square SECTION 10.1 635

Objectives

Use the square root property tosolve quadratic equations.

Write perfect square trinomials.

Solve quadratic equations by com-pleting the square.

Use quadratic equations to solveproblems.

D

CB

A

HELPFUL HINT

The notation , for example, is read as “plus or minus 3.” It is ashorthand notation for the pair of numbers and .-3+3

;3

SQUARE ROOT PROPERTY

If b is a real number and if , then .a = ;2ba2 = b

SSM CD-ROM Video10.1

Practice Problem 1Use the square root property to solve

.x2 = 20

Answer1. E215, -215F

636 CHAPTER 10 Quadratic Equations and Functions

Answers2. ,3. ,

4.


e 1 - 2i3

, 1 + 2i

3f

E-2 + 312, -2 - 312FE111, -111F


.5x2 = 55


.Ax + 2 B 2 = 18


.A3x - 1 B 2 = -4

Check: . .

True. True.



Solution: First we get the squared variable alone on one side of theequation.



Check: .

True. True.



Solution:


Simplify the radical.


Check: Below is a check for . The check foris almost the same and is left for you to do on

your own.

True.



Solution:





Check each proposed solution in the original equation to

see that the solution set is .


e 5 + 4i2

, 5 - 4i

2f

x = 5 ; 4i2

2x = 5 ; 4i

2x - 5 = ;4i

2x - 5 = ;1-16

A2x - 5 B 2 = -16

A2x - 5 B 2 = -16

E-1 + 213, -1 - 213F 12 = 12

4 # 3 � 12

A213 B 2 � 12

A-1 + 213 + 1 B 2 � 12

Ax + 1 B 2 = 12

-1 - 213-1 + 213

x = -1 ; 213

x + 1 = ;213

x + 1 = ;112

Ax + 1 B 2 = 12

Ax + 1 B 2 = 12

E17, -17F 14 = 14 14 = 14

2 # 7 � 14 2 # 7 � 14

2 A-17 B 2 � 142 A17 B 2 � 14

2x2 = 14 2x2 = 14

Let x = -17 Let x = 17.

x = ;17

x2 = 7

2x2 = 14

2x2 = 14

E512, -512F 50 = 50 50 = 50

25 # 2 � 50 25 # 2 � 50A-512 B 2 � 50 A512 B 2 � 50

x2 = 50 x2 = 50 Let x = -512 Let x = 512

✓ CONCEPT CHECK

How do you know just by looking thathas complex solu-

tions?Ax - 2 B 2 = -4


Answers5. ,

6. ,

7. E-4 - 117, -4 + 117Fy2 - 5y + 25

4= ax - 5

2b 2

x2 + 12x + 36 = Ax + 6 B 2

Practice Problem 5Add the proper constant to so that the result is a perfect square tri-nomial. Then factor.

x2 + 12x

Practice Problem 6Add the proper constant to sothat the result is a perfect square trino-mial. Then factor.

y2 - 5y

WRITING PERFECT SQUARE TRINOMIALS

Notice from Examples 3 and 4 that, if we write a quadratic equation so thatone side is the square of a binomial, we can solve by using the square rootproperty. To write the square of a binomial, we write perfect square trino-mials. Recall that a perfect square trinomial is a trinomial that can be fac-tored into two identical binomial factors, that is, as a binomial squared.

Perfect Square Trinomials Factored Form

Notice that for each perfect square trinomial, the constant term of the trino-mial is the square of half the coefficient of the x-term. For example,

Example 5 Add the proper constant to so that the result is aperfect square trinomial. Then factor.

Solution: We add the square of half the coefficient of x.

Example 6 Add the proper constant to so that the result is aperfect square trinomial. Then factor.

Solution: We add the square of half the coefficient of x.

SOLVING BY COMPLETING THE SQUARE

The process of writing a quadratic equation so that one side is a perfectsquare trinomial is called completing the square. We will use this process inthe next examples.

Example 7 Solve by completing the square.

Solution: First we add the square of half the coefficient of p toboth sides so that the resulting trinomial will be a perfectsquare trinomial. The coefficient of p is 2.

and 12 = 112

A2 B = 1

p2 + 2p = 4

C

In factored formx2 - 3x + 94

= ax - 32b 2

12

A-3 B = -32

and a-32b 2

= 94

x2 - 3x

In factored form= Ax + 3 B 2x2 + 6x + 9

12

A6 B = 3 and 32 = 9

x2 + 6x

x2 + 8x + 16 x2 - 6x + 9

12

A8 B = 4 and 42 = 16 12

A-6 B = -3 and A-3 B 2 = 9

ax + 32b 2

x2 + 3x + 94

Ax - 3 B 2x2 - 6x + 9Ax + 4 B 2x2 + 8x + 16

B

Practice Problem 7Solve by completing thesquare.

x2 + 8x = 1


Answer

8. e 5 - 1172

, 5 + 117

2f

Now we add 1 to both sides of the original equation.


We may now use the square root property and solve for p.



Notice that there are two solutions: and. The solution set is .


Solution: First we add 1 to both sides of the equation so that theleft side has no constant term. We can then add the con-stant term on both sides that will make the left side a per-fect square trinomial.

Now we find the constant term that makes the left side aperfect square trinomial by squaring half the coefficientof m. We add this constant to both sides of the equation.

and

Add to both sides of the equation.


Add to both sides and simplify .

Simplify.


The following steps may be used to solve a quadratic equation such asby completing the square. This method may be used

whether or not the polynomial is factorable.ax2 + bx + cax2 + bx + c = 0

e 7 + 1532

, 7 - 153

2f

m = 7 ; 1532

A534

72

m = 72

; 1532

m - 72

= ;A534

Factor the perfect square trinomial and simplifythe right side. am - 7

2b 2

= 534

494

m2 - 7m + 494

= 1 + 494

a-72b 2

= 494

12

A-7 B = -72

m2 - 7m = 1

m2 - 7m - 1 = 0

m2 - 7m - 1 = 0

E-1 + 15, -1 - 15F-1 - 15-1 + 15

p = -1 ; 15

p + 1 = ;15

Factor the trinomial; simplify theright side. Ap + 1 B 2 = 5

p2 + 2p + 1 = 4 + 1

p2 + 2p = 4

Practice Problem 8Solve by completingthe square.

y2 - 5y + 2 = 0


Answer

9. e 1 + i1132

, 1 - i113

2f

Practice Problem 9Solve by complet-ing the square.

2x2 - 2x + 7 = 0


Solution: First we divide both sides of the equation by 4 so that thecoefficient of is 1.

Step 1. Divide both sides of the equationby 4.

Step 2. Subtract from both sides.

Since , we add 9 to both

sides of the equation.

Step 3. Add 9 to both sides.

Step 4. Factor the perfect squaretrinomial.

Step 5. Use the square root property.



Find a common denominator.

Simplify.

The solution set is .e 6 + i152

, 6 - i15

2f

= 6 ; i152

= 62

; i152

x = 3 ; i152

x - 3 = ; i152

x - 3 = ;A- 54

Ax - 3 B 2 = - 54

Ax - 3 B 2 = - 414

+ 364

x2 - 6x + 9 = - 414

+ 9

12

A-6 B = -3 and A-3 B 2 = 9

414

x2 - 6x = - 414

x2 - 6x + 414

= 0

4x2 - 24x + 41 = 0

x2

4x2 - 24x + 41 = 0

SOLVING A QUADRATIC EQUATION IN x BY COMPLETING

THE SQUARE

Step 1. If the coefficient of x2 is 1, go to Step 2. Otherwide, divide bothsides of the equation by the coefficient of x2.

Step 2. Get all variable terms alone on one side of the equation.

Step 3. Complete the square for the resulting binomial by adding thesquare of half of the coefficient of x to both sides of the equation.

Step 4. Factor the resulting perfect square trinomial and write it as thesquare of a binomial.

Step 5. Use the square root property to solve for x.


Answer10. 5%


Recall the simple interest formula , where I is the interest earned,P is the principal, r is the rate of interest, and t is time. If $100 is invested ata simple interest rate of 5% annually, at the end of 3 years the total inter-est I earned is

or

and the new principal is

Most of the time, the interest computed on money borrowed or moneydeposited is compound interest. Compound interest, unlike simple interest,is computed on original principal and on interest already earned. To see thedifference between simple interest and compound interest, suppose that$100 is invested at a rate of 5% compounded annually. To find the totalamount of money at the end of 3 years, we calculate as follows:

First year:

Second year:

Third year:

At the end of the third year, the total compound interest earned is $15.76,whereas the total simple interest earned is $15.

It is tedious to calculate compound interest as we did above, so we use acompound interest formula. The formula for calculating the total amountof money when interest is compounded annually is

where P is the original investment, r is the interest rate per compoundingperiod, and t is the number of periods. For example, the amount of moneyA at the end of 3 years if $100 is invested at 5% compounded annually is

as we previously calculated.

Example 10 Finding Interest RatesFind the interest rate r if $2000 compounded annuallygrows to $2420 in 2 years.

Solution: 1. UNDERSTAND the problem. For this example,make sure that you understand the formula for com-pounding interest annually.

2. TRANSLATE. We substitute the given values into theformula:

Let .A = 2420, P = 2000, and t = 22420 = 2000 A1 + r B 2 A = P A1 + r B t

A = $100 A1 + 0.05 B 3 L $100 A1.1576 B = $115.76

A = P A1 + r B t

New principal = $110.25 + $5.51 = $115.76Interest = $110.25 # 0.05 # 1 L $5.51New principal = $105.00 + $5.25 = $110.25Interest = $105.00 # 0.05 # 1 = $5.25New principal = $100.00 + $5.00 = $105.00Interest = $100 # 0.05 # 1 = $5.00

t # r#P=I

$100 + $15 = $115

I = 100 # 0.05 # 3 = $15

I = P # r # t

I = Prt

D

Practice Problem 10Use the formula from Example 10 tofind the interest rate r if $1600 com-pounded annually grows to $1764 in 2years.


3. SOLVE. Solve the equation for r.


Simplify the fraction.


Simplify.

4. INTERPRET. The rate cannot be negative, so we

reject .

Check: per year. If we invest $2000 at

10% compounded annually, in 2 years the amount in theaccount would be dollars, thedesired amount.

State: The interest rate is 10% compounded annually.

2000 A1 + 0.10 B 2 = 2420

110

= 0.10 = 10%

- 2110

110

= r or - 2110

= r

- 1010

; 1110

= r

-1 ; 1110

= r

; 1110

= 1 + r

;A121100

= 1 + r

121100

= A1 + r B 2 24202000

= A1 + r B 2 2420 = 2000 A1 + r B 2


GRAPHING CALCULATOR EXPLORATIONSIn Section 7.5, we showed how we can use a grapher to approximate real number solutions of a quadraticequation written in standard form. We can also use a grapher to solve a quadratic equation when it is not

written in standard form. For example, to solve , the quadratic equation in Example 3, wegraph the following on the same set of axes. We use , , , and

.

Use the Intersect feature or the Zoom and Trace features to locate the points of intersection of the graphs.The x-values of these points are the solutions of . The solutions, rounded to two decimal points,are 2.46 and .

Check to see that these numbers are approximations of the exact solutions .

Use a grapher to solve each quadratic equation. Round all solutions to the nearest hundredth.

1. 2.

3. 4.

5. Use a grapher and solve , Example 4 in this section, using the window

Explain the results. Compare your results with the solution found in Example 4.

6. What are the advantages and disadvantages of using a grapher to solve quadratic equations?

Yscl = 1

Ymax = 20

Ymin = -20

Xscl = 1

Xmax = 20

Xmin = -20

A2x - 5 B 2 = -16

x2 - 2.6x = -2.2x + 3x2 + 0.5x = 0.3x + 1

x Ax + 2 B = 5x Ax - 5 B = 8

-1 ; 213-4.46

Ax + 1 B 2 = 12

13

10

y = 12

y = (x + 1)2

−13

−10

Y1 = Ax + 1 B 2 and Y2 = 12

Ymax = 13Ymin = -13Xmax = 10Xmin = -10

Ax + 1 B 2 = 12

643


EXERCISE SET 10.1

Use the square root property to solve each equation. See Examples 1 through 4.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. 21.

22. 23. 24.

25. Ax + 3 B 2 = -8

Ax + 10 B 2 = 11Az + 7 B 2 = 5Ay + 2 B 2 = -25

Ax - 1 B 2 = -163p2 + 36 = 02z2 + 16 = 0

y2 - 10 = 0x2 - 6 = 0x2 + 4 = 0

x2 + 9 = 0A4x + 9 B 2 = 6A2x - 3 B 2 = 8

Ay + 4 B 2 = 27Az - 6 B 2 = 18Ay - 3 B 2 = 4

Ax + 5 B 2 = 92x2 = 43z2 - 30 = 0

y2 = 20x2 = 18x2 - 11 = 0

x2 - 7 = 0x2 = 49x2 = 16A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

644

Focus On Business and CareerFLOWCHARTS

We saw in the Focus On Business and Career feature in Chapter 6 that three of the top 10fastest-growing jobs into the 21st century are computer related. A useful skill in computer-related careers is flowcharting. A flowchart is a diagram showing a sequence of procedures usedto complete a task. Flowcharts are commonly used in computer programming to help a pro-grammer plan the steps and commands needed to write a program. Flowcharts are also used inother types of careers, such as manufacturing or finance, to describe the sequence of eventsneeded in a certain process.

A flowchart usually uses the following symbols to represent certain types of actions.

For instance, suppose we want to write a flowchart for the process of computing a household’smonthly electric bill. The flowchart might look something like the one here.

CRITICAL THINKING

1. Using the given flowchart as a guide,describe in words this utility company’spricing structure for household electricityusage.

2. Make a flowchart for the process of com-puting the human-equivalent age for theage of a dog if 1 dog year is equivalent to7 human years.

3. Make a flowchart for the process ofdetermining the number and type of solu-tions of a quadratic equation of the form

using the discriminant.

4. (Optional) Write a programmable calcu-lator program for your graphing calcula-tor that determines the number and typeof solutions of a quadratic equation of theform .ax2 + bx + c = 0

ax2 + bx + c = 0

Start

Stop

No

Yes

Input total electricalusage U for month

CalculateM = U − 100

Compute bill amountas B = 0.11M + 14

Display billamount B

Compute bill amountas B = 0.14U

Is U ≤ 100 KWH?

Start or end ofprocess

Individual operation ortask description

Conditionor decision

Connector

10.2 SOLVING QUADRATIC EQUATIONS

BY THE QUADRATIC FORMULA

SOLVING EQUATIONS BY USING THE QUADRATIC FORMULA

Any quadratic equation can be solved by completing the square. Since thesame sequence of steps is repeated each time we complete the square, let’scomplete the square for a general quadratic equation, .By doing so, we find a pattern for the solutions of a quadratic equationknown as the quadratic formula.

Recall that to complete the square for an equation such as, , we first divide both sides by the coefficient of .

Divide both sides by a, the coefficient of .

Subtract the constant from both sides.

Next we find the square of half , the coefficient of x.

Now we add this result to both sides of the equation.

Add to both sides.

Subtract from both sides.

Simplify.

The resulting equation identifies the solutions of the general quadraticequation in standard form and is called the quadratic formula. It can beused to solve any equation written in standard form aslong as a is not 0.

ax2 + bx + c = 0

x = -b ; 2b2 - 4ac2a

b2a x = - b

2a; 2b2 - 4ac

2a

x + b2a

= ;2b2 - 4ac2a

x + b2a

= ;Bb2 - 4ac4a2

ax + b2ab 2

= b2 - 4ac4a2

x2 + ba

x + b2

4a2 = b2 - 4ac4a2

Find a common denominatoron the right side.

Simplify the right side.

Factor the perfect square tri-nomial on the left side.



x2 + ba

x + b2

4a2 = -c # 4aa # 4a

+ b2

4a2

b2

4a2x2 + ba

x + b2

4a2 = - ca

+ b2

4a2

12a b

ab = b

2a and a b

2ab 2

= b2

4a2

ba

ca

x2 + ba

x = - ca

x2 x2 + ba

x + ca

= 0

ax2 + bx + c = 0

x2a Z 0ax2 + bx + c = 0

ax2 + bx + c = 0

A

Solving Quadratic Equations by the Quadratic Formula SECTION 10.2 645

Objectives

Solve quadratic equations by usingthe quadratic formula.

Determine the number and type ofsolutions of a quadratic equationby using the discriminant.

Solve geometric problems modeledby quadratic equations.

C

B

A


QUADRATIC FORMULA

A quadratic equation written in the form has thesolutions

x = -b ; 2b2 - 4ac2a

ax2 + bx + c = 0


HELPFUL HINT

To replace a, b, and c correctly in the quadratic formula, write thequadratic equation in standard form .ax2 + bx + c = 0

Answers

1. , 2.


e 3 + 1112

, 3 - 111

2fe - 5

2, -2 f

Practice Problem 1Solve: 2x2 + 9x + 10 = 0

Practice Problem 2Solve: 2x2 - 6x - 1 = 0

Example 1 Solve:

Solution: This equation is in standard form with , ,and . We substitute these values into the quadraticformula.

Quadratic formula

Let , , and .


Example 2 Solve:

Solution: First we write the equation in standard form by subtract-ing 3 from both sides.

Now , , and . We substitute thesevalues into the quadratic formula.



e 2 + 1102

, 2 - 110

2f

=2 A2 ; 110 B

2 # 2= 2 ; 110

2

= 4 ; 1404

= 4 ; 21104

= 4 ; 116 + 244

=- A-4 B ; 2 A-4 B 2 - 4 A2 B A-3 B

2 A2 B

x = -b ; 2b2 - 4ac2a

c = -3b = -4a = 2

2x2 - 4x - 3 = 0

2x2 - 4x = 3

e - 13

, -5 f

x = -16 + 146

= - 13

or x = -16 - 14

6= - 30

6= -5

= -16 ; 11966

= -16 ; 146

= -16 ; 1256 - 606

c = 5b = 16a = 3 =-16 ; 2162 - 4 A3 B A5 B

2 A3 B

x = -b ; 2b2 - 4ac2a

c = 5b = 16a = 3

3x2 + 16x + 5 = 0

✓ CONCEPT CHECK

For the quadratic equation ,which substitution is correct?a. , , and b. , , and c. , , and d. , , and c = -7b = 1a = 1

c = 7b = 0a = 0c = 7b = 0a = 1c = -7b = 0a = 1

x2 = 7


Answers

3. ,

4. ,

✓ Concept Check answers may vary

-1 + 3i178

fe -1 - 3i178

e 3 + 1332

, 3 - 133

2f

Practice Problem 3

Solve: 16

x2 - 12

x - 1 = 0

Example 3 Solve:

Solution: We could use the quadratic formula with , ,

and . Instead, let’s find a simpler, equivalent stan-

dard form equation whose coefficients are not fractions.First we multiply both sides of the equation by 4 to

clear the fractions.

Simplify.

Now we can substitute , , and intothe quadratic formula and simplify.


Example 4 Solve:

Solution: The equation in standard form is .Thus, , , and in the quadratic formula.



e -1 + i1356

, -1 - i135

6f

= -1 ; 1-356

= -1 ; i1356

p = -1 ; 212 - 4 A3 B A3 B2 A3 B = -1 ; 11 - 36

6

c = 3b = 1a = 33p2 + p + 3 = 0

p = -3p2 - 3

E2 + 12, 2 - 12F

= 4 ; 182

= 4 ; 2122

=2 A2 ; 12 B

2= 2 ; 12

= 4 ; 116 - 82

m =- A-4 B ; 2 A-4 B 2 - 4 A1 B A2 B

2 A1 B

c = 2b = -4a = 1

m2 - 4m + 2 = 0

4 a 14

m2 - m + 12b = 4 # 0

c = 12

b = -1a = 14

14

m2 - m + 12

= 0

Practice Problem 4Solve: x = -4x2 - 4

✓ CONCEPT CHECK

What is the first step in solvingusing the quadratic

formula?-3x2 = 5x - 4

HELPFUL HINT

To simplify the expression in the preceding example, note

that 2 is factored out of both terms of the numerator before simplifying.

4 ; 21104

=2 A2 ; 110 B

2 # 2= 2 ; 110

2

4 ; 21104

USING THE DISCRIMINANT

In the quadratic formula , the radicand is

called the discriminant because when we know its value, we can discrimi-nate among the possible number and type of solutions of a quadratic equa-tion. Possible values of the discriminant and their meanings are summa-rized next.

Example 5 Use the discriminant to determine the number and typeof solutions of .

Solution: In , , , and . Thus,

Since , this quadratic equation has one realsolution.


Solution: In this equation, , , and . Then. Since is neg-

ative, this quadratic equation has two complex but notreal solutions.


Solution: In this equation, , , and . Then

Since is positive, this quadratic equation hastwo real solutions.


The quadratic formula is useful in solving problems that are modeled byquadratic equations.

C

b2 - 4ac

b2 - 4ac = A-7 B 2 - 4 A2 B A-4 B = 81

c = -4b = -7a = 2

2x2 - 7x - 4 = 0

b2 - 4ac02 - 4 A3 B A2 B = -24b2 - 4ac =c = 2b = 0a = 3

3x2 + 2 = 0

b2 - 4ac = 0

b2 - 4ac = 22 - 4 A1 B A1 B = 0

c = 1b = 2a = 1x2 + 2x + 1 = 0

x2 + 2x + 1

b2 - 4acx = -b ; 2b2 - 4ac2a

B


DISCRIMINANT

The following table corresponds the discriminant of a qua-dratic equation of the form with the number andtype of solutions of the equation.

b2 � 4ac Number and Type of Solutions

Positive Two real solutionsZero One real solutionNegative Two complex but not real solutions

ax2 + bx + c = 0b2 - 4ac

Answers5. 1 real solution, 6. 2 complex but not realsolutions, 7. 2 real solutions

Practice Problem 5Use the discriminant to determine the number and type of solutions of

.x2 + 4x + 4 = 0


.5x2 + 7 = 0


.3x2 - 2x - 2 = 0


Answer8. 9 feet

Example 8 Calculating Distance SavedAt a local university, students often leave the sidewalkand cut across the lawn to save walking distance. Giventhe diagram below of a favorite place to cut across thelawn, approximate how many feet of walking distance astudent saves by cutting across the lawn instead of walk-ing on the sidewalk.

Solution: 1. UNDERSTAND. Read and reread the problem. Youmay want to review the Pythagorean theorem.

2. TRANSLATE. By the Pythagorean theorem, we haveIn words:

Translate:3. SOLVE. Use the quadratic formula to solve.

Square and 50.

Set the equation equal to 0.

Divide by 2.

Here, , , . By the quadraticformula,

Simplify.

Check: 4. INTERPRET. Check your calculations in the qua-dratic formula. The length of a side of a triangle can’tbe negative, so we reject . Since

feet, the walking distance alongthe sidewalk is

feet.

State: A person saves or 18 feet of walking distance bycutting across the lawn.

68 - 50

x + Ax + 20 B L 24 + A24 + 20 B = 68

-10 + 5146 L 24-10 - 5146

= -10 ; 5146

= -20 ; 1100 # 462

= -20 ; 101462

= -20 ; 1400 + 42002

= -20 ; 146002

x =-20 ; 2202 - 4 A1 B A-1050 B

2 # 1

c = -1050b = 20a = 1

x2 + 20x - 1050 = 0

2x2 + 40x - 2100 = 0

Ax + 20 Bx2 + x2 + 40x + 400 = 2500

x2 + Ax + 20 B 2 = 502

A leg B 2 + A leg B 2 = Ahypotenuse B 2

x + 2050 feet

x

Practice Problem 8Given the diagram below, approxi-mate to the nearest foot how many feetof walking distance a person saves by cutting across the lawn instead of walk-ing on the sidewalk.

x + 820 feet

x


Example 9 An object is thrown upward from the top of a 200-footcliff with a velocity of 12 feet per second. The height h ofthe object after t seconds is

How long after the object is thrown will it strike theground? Round to the nearest tenth of a second.

Solution: 1. UNDERSTAND. Read and reread the problem.2. TRANSLATE. Since we want to know when the

object strikes the ground, we want to know when theheight , or

3. SOLVE. First, divide both sides of the equation by .


Here, , , and . By the quadraticformula,

Check: 4. INTERPRET. Check your calculations in the qua-dratic formula. Since the time won’t be negative,

we reject the proposed solution .

State: The time it takes for the object to strike the

ground is exactly seconds 3.9 seconds.L3 + 18098

3 - 18098

= 3 ; 18098

= 3 ; 19 + 8008

t =- A-3 B ; 2 A-3 B 2 - 4 A4 B A-50 B

2 # 4

c = -50b = -3a = 4

-40 = 4t2 - 3t - 50

-4

0 = -16t2 + 12t + 200

h = 0

200 feet

h = -16t2 + 12t + 200

Practice Problem 9Use the equation given in Example 9to find how long after the object isthrown it will be 100 feet from theground. Round to the nearest tenth ofa second.

Answer9. 1.7 sec

651


MENTAL MATH

Identify the values of a, b, and c in each quadratic equation.

1. 2. 3.

4. 5.

EXERCISE SET 10.2

Use the quadratic formula to solve each equation. See Examples 1 through 4.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. 3y2 + 6y + 5 = 010y2 + 10y + 3 = 0

12

y2 = y + 12

13

y2 - y - 16

= 018

x2 + x = 52

25

y2 + 15

y = 35

16

x2 + x + 13

= 012

x2 - x - 1 = 0

x2 - 13 = 5x3m2 - 7m = 311n2 - 9n = 1

8m2 - 2m = 7y2 + 5y + 3 = 0x2 + 7x + 4 = 0

y2 + 10y + 25 = 0x2 - 6x + 9 = 05x2 - 3 = 14x

2y = 5y2 - 3p2 + 11p - 12 = 0m2 + 5m - 6 = 0A

6x2 - x = 0x2 + 9 = 0

7x2 - 4 = 02x2 - 5x - 7 = 0x2 + 3x + 1 = 0

MENTAL MATH ANSWERS

1.

2.

3.

4.

5.

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

10.3 SOLVING EQUATIONS BY USING QUADRATIC METHODS

SOLVING EQUATIONS THAT ARE QUADRATIC IN FORM

In this section, we discuss various types of equations that can be solved inpart by using the methods for solving quadratic equations.

Once each equation is simplified, you may want to use these steps whendeciding what method to use to solve the quadratic equation.

The first example is a radical equation that becomes a quadratic equa-tion once we square both sides.

Example 1 Solve:

Solution: Recall that to solve a radical equation, first get the radi-cal alone on one side of the equation. Then square bothsides.

Add to both sides.

Square both sides.

Set the equation equal to 0.

Check:

True False


Example 2 Solve:

Solution: In this equation, x cannot be either 2 or 0, because thesevalues cause denominators to equal zero. To solve for x,we first multiply both sides of the equation by to clear the fractions. By the distributive property, thismeans that we multiply each term by .x Ax - 2 B

x Ax - 2 B

3xx - 2

- x + 1x

= 6x Ax - 2 B

E9F -4 = 0 0 = 0

4 - 2 - 6 � 0 9 - 3 - 6 � 04 - 14 - 6 � 09 - 19 - 6 � 0x - 1x - 6 = 0x - 1x - 6 = 0Let x = 4Let x = 9

x = 9 x = 4

x - 9 = 0 or x - 4 = 0

Ax - 9 B Ax - 4 B = 0

x2 - 13x + 36 = 0

x2 - 12x + 36 = x

1x x - 6 = 1x

x - 1x - 6 = 0

A

Solving Equations by Using Quadratic Methods SECTION 10.3 653

Objectives

Solve various equations that arequadratic in form.

Solve problems that lead to qua-dratic equations.

B

A

SOLVING A QUADRATIC EQUATION

Step 1. If the equation is in the form , use the squareroot property and solve. If not, go to Step 2.

Step 2. Write the equation in standard form: .

Step 3. Try to solve the equation by the factoring method. If not possi-ble, go to Step 4.

Step 4. Solve the equation by the quadratic formula.

ax2 + bx + c = 0

Aax + b B 2 = c

Practice Problem 1Solve: x - 1x - 1 - 3 = 0

Answers

1. , 2. e 1 + 1132

, 1 - 113

2fE5F

Practice Problem 2

Solve: 2x

x - 1- x + 2

x= 5

x Ax - 1 B



Answers3. , 4.

✓ Concept Check: a. true, b. false

E-1, -6FE3, -3, 2i, -2iF

Practice Problem 3Solve: x4 - 5x2 - 36 = 0

✓ CONCEPT CHECK

a. True or False? The maximum num-ber of solutions that a quadraticequation can have is 2.

b. True or False? The maximum num-ber of solutions that an equation inquadratic form can have is 2.

Practice Problem 4Solve: Ax + 4 B 2 - Ax + 4 B - 6 = 0

Simplify.

Multiply.

Simplify.

This equation cannot be factored using integers, so wesolve by the quadratic formula.

Neither proposed solution will make the denominators 0.


Example 3 Solve:

Solution: First we factor the trinomial.

Factor.

Factor further.



Example 4 Solve:

Solution: Notice that the quantity is repeated in this equa-tion. Sometimes it is helpful to substitute a variable (inthis case other than x) for the repeated quantity. We willlet . Then

becomes

Factor.

To solve, we use the zero factor property.


Solve. y = 4 y = -1

y - 4 = 0 or y + 1 = 0

Ay - 4 B Ay + 1 B = 0

Ax - 3 B 2 - 3 Ax - 3 B - 4 = 0

y2 - 3y - 4 = 0 Let x � 3 � y.

y = x - 3

Ax - 3 BAx - 3 B 2 - 3 Ax - 3 B - 4 = 0

E2, -2, i, - iF p = ;1-1 = ; i

p = 2 p = -2 p2 = -1

Set each factor equal to0 and solve.

p - 2 = 0 or p + 2 = 0 or p2 + 1 = 0

Ap - 2 B Ap + 2 B Ap2 + 1 B = 0

Ap2 - 4 B Ap2 + 1 B = 0

p4 - 3p2 - 4 = 0

p4 - 3p2 - 4 = 0

e -1 + 1334

, -1 - 133

4f

= -1 ; 1334

= -1 ; 11 + 324

Let a = 2, b = 1, and c = -4 inthe quadratic formula.

Simplify.

x = -1 ; 212 - 4 A2 B A-4 B2 # 2

2x2 + x - 4 = 0

3x2 - x2 + x + 2 = 6

3x2 - Ax2 - x - 2 B = 6

3x2 - Ax - 2 B Ax + 1 B = 6

x Ax - 2 B a 3xx - 2

b - x Ax - 2 B a x + 1xb = x Ax - 2 B c 6

x Ax - 2 B d

To find values of x, we substitute back. That is, we substi-tute for y.

Both 2 and 7 check. The solution is .

Example 5 Solve:

Solution: The key to solving this equation is recognizing that. We replace with m so that

becomes

Now we solve by factoring.

Factor.


Since , we have

Both 8 and 27 check. The solution set is .

SOLVING PROBLEMS THAT LEAD TO QUADRATIC EQUATIONS

The next example is a work problem. This problem is modeled by a ratio-nal equation that simplifies to a quadratic equation.

Example 6 Finding Work TimeTogether, an experienced typist and an apprentice typistcan process a document in 6 hours. Alone, the experi-enced typist can process the document 2 hours faster thanthe apprentice typist can. Find the time in which eachperson can process the document alone.

B

E8, 27F x = 33 = 27 or x = 23 = 8

x1>3 = 3 or x1>3 = 2

m = x1>3 m = 3 m = 2

m - 3 = 0 or m - 2 = 0

Am - 3 B Am - 2 B = 0

m2 - 5m + 6 = 0

m2 - 5m + 6 = 0

Ax1>3 B 2 - 5x1>3 + 6 = 0

x1>3x2>3 = Ax1>3 B 2x2>3 - 5x1>3 + 6 = 0

E2, 7F

x = 7 x = 2

x - 3 = 4 or x - 3 = -1

x - 3


Answers5. , 6. Doug,

hours; Karen, hours9 + 1101

2L 9.5

11 + 11012

L 10.5E8, 125F

HELPFUL HINT

When using substi-tution, don’t forgetto substitute backto the original vari-able.

HELPFUL HINT

Example 3 can be solved using substitution also. Think ofas

Then let , and

T b solve and substitute back. The solution set will

be the same.x2 - 3x - 4 = 0

x = p2Ap2 B 2 - 3p2 - 4 = 0

p4 - 3p2 - 4 = 0

Practice Problem 5Solve: x2>3 - 7x1>3 + 10 = 0

Practice Problem 6Together, Karen and Doug Lewis canclean a strip of beach in 5 hours. Alone,Karen can clean the strip of beach onehour faster than Doug. Find the timethat each person can clean the strip ofbeach alone. Give an exact answer anda one decimal place approximation.

Solution: 1. UNDERSTAND. Read and reread the problem. Thekey idea here is the relationship between the time(hours) it takes to complete the job and the part of thejob completed in one unit of time (hour). For exam-ple, because they can complete the job together in 6hours, the part of the job they can complete in 1 hour

is . Let

the time in hours it takes the apprentice typ-ist to complete the job alone

the time in hours it takes the experiencedtypist to complete the job alone

We can summarize in a chart the information dis-cussed.

2. TRANSLATE.

In words:

T T T T T

Translate:

3. SOLVE.

Now we can substitute , , and into the quadratic formula and simplify.

x =- A-14 B ; 2 A-14 B 2 - 4 A1 B A12 B

2 A1 B = 14 ; 11482

c = 12b = -14a = 1

0 = x2 - 14x + 12

6x - 12 + 6x = x2 - 2x

6 Ax - 2 B + 6x = x Ax - 2 B6x Ax - 2 B #

1x

+ 6x Ax - 2 B # 1

x - 2= 6x Ax - 2 B #

16

Multiply bothsides by the LCD,6x(x - 2).

Use the distribu-tive property.

6x Ax - 2 B a 1x

+ 1x - 2

b = 6x Ax - 2 B # 16

1x

+ 1x - 2

= 16

16

=1x - 2

+1x

part of jobcompletedtogether in

1 hour

isequal

to

part of jobcompleted byexperienced

typist in1 hour

addedto

part of jobcompleted by

apprenticetypist in1 hour

x - 2 =

x =

16


Part of JobTotal Hours to CompletedComplete Job in 1 Hour

Apprentice Typist x

Experienced Typist

Together 616

1x - 2

x - 2

1x


Answer7. 50 mph to the beach; 60 mph returning

Using a calculator or a square root table, we see thatrounded to one decimal place. Thus,

4. INTERPRET.

Check: If the apprentice typist completes the job alone in 0.9hours, the experienced typist completes the job alone in

hours. Since this is not possi-ble, we reject the solution of 0.9. The approximate solu-tion is thus 13.1 hours.

State: The apprentice typist can complete the job alone inapproximately 13.1 hours, and the experienced typist cancomplete the job alone in approximately

hours.

Example 7 Beach and Fargo are about 400 miles apart. Asalesperson travels from Fargo to Beach one day at acertain speed. She returns to Fargo the next day and

drives 10 mph faster. Her total travel time was hours.

Find her speed to Beach and the return speed to Fargo.

Solution: 1. UNDERSTAND. Read and reread the problem. Let

the speed to Beach, so

the return speed to Fargo.

Then organize the given information in a table.

2. TRANSLATE.

In words:

Translate:443

=400x + 10

+400x

14 23

hours=

returntime toFargo

+time toBeach

x + 10 = x =

x + 10 mph

x mph

400 miles

400 miles

Beach Fargo

1423

x - 2 = 13.1 - 2 = 11.1

x - 2 = 0.9 - 2 = -1.1

x L 14 + 12.22

= 13.1 or x L 14 - 12.22

= 0.9

x L 14 ; 12.22

1148 L 12.2

Practice Problem 7A family drives 500 miles to the beachfor a vacation. The return trip wasmade at a speed that was 10 mph

faster. The total traveling time was

hours. Find the speed to the beach andthe return speed.

1813

distance rate time

400 x

400400

x + 10x + 10

400x

#=

To Beach

Return to Fargo


3. SOLVE.


Set each factor equal to 0. Factor.

4. INTERPRET.

Check: The speed is not negative, so it's not . The number50 does check.

State: The speed to Beach was 50 mph and her return speed toFargo was 60 mph.

-5511

x = - 6011

or -5511

x = 50

11x + 60 = 0 or x - 50 = 0

0 = A11x + 60 B Ax - 50 BSet equation equal to 0.

0 = 11x2 - 490x - 3000

300x + 3000 + 300x = 11x2 + 110x

3 Ax + 10 B # 100 + 3x # 100 = x Ax + 10 B # 11Use the distributive property.

3x Ax + 10 B # 100x

+ 3x Ax + 10 B # 100

x + 10= 3x Ax + 10 B #

113

Multiply both sides by theLCD, 3x(x + 10).

3x Ax + 10 B a 100x

+ 100x + 10

b = 3x Ax + 10 B # 113

100x

+ 100x + 10

= 113

400x

+ 400x + 10

= 443

659

Focus On HistoryTHE EVOLUTION OF SOLVING QUADRATIC EQUATIONS

The ancient Babylonians (circa 2000 B.C.) are sometimes creditedwith being the first to solve quadratic equations. This is only par-tially true because the Babylonians had no concept of an equation.However, what they did develop was a method for completing thesquare to apply to problems that today would be solved with aquadratic equation. The Babylonians only recognized positivesolutions to such problems and did not acknowledge the existenceof negative solutions at all.

Babylonian mathematical knowledge influenced much of theancient world, most notably Hindu Indians. The Hindus were thefirst culture to denote debts in everyday business affairs with nega-tive numbers. With this level of comfort with negative numbers,the Indian mathematician Brahmagupta (598–665 A.D.) extendedthe Babylonian methods and was the first to recognize negativesolutions to quadratic equations. Later Hindu mathematiciansnoted that every positive number has two square roots: a positivesquare root and a negative square root. Hindus allowed irrationalsolutions (quite an innovation in the ancient world!) to quadraticequations and were the first to realize that quadratic equationscould have 0, 1, or 2 real number solutions. They did not, however,acknowledge complex numbers and, therefore, could not solveequations with solutions requiring the square root of a negativenumber.

Complex numbers were finally developed by European mathe-maticians during the 17th and 18th centuries. Up until that time,what we would today consider complex solutions to quadraticequations were routinely ignored by mathematicians.

10.4 NONLINEAR INEQUALITIES IN ONE VARIABLE

SOLVING POLYNOMIAL INEQUALITIES

Just as we can solve linear inequalities in one variable, so we can also solvequadratic inequalities in one variable. A quadratic inequality is an inequal-ity that can be written so that one side is a quadratic expression and theother side is 0. Here are examples of quadratic inequalities in one variable.Each is written in standard form.

A solution of a quadratic inequality in one variable is a value of the vari-able that makes the inequality a true statement.

The value of an expression such as will sometimes be pos-itive, sometimes negative, and sometimes 0, depending on the value substi-tuted for x. To solve the inequality , we are looking forall values of x that make the expression less than 0, or nega-tive. To understand how we find these values, we’ll study the graph of thequadratic function .

Notice that the x-values for which y or is positive are sep-arated from the x values for which y or is negative by the val-ues for which y or is 0, the x-intercepts. Thus, the solution setof consists of all real numbers from to 5, or in inter-val notation, .

It is not necessary to graph to solve the relatedinequality . Instead, we can draw a number line repre-senting the x-axis and keep the following in mind: A region on the numberline for which the value of is positive is separated from aregion on the number line for which the value of is negativeby a value for which the expression is 0.

Let’s find these values for which the expression is 0 by solving therelated equation:

Factor.


Solve. x = 5 x = -2

x - 5 = 0 or x + 2 = 0

Ax - 5 B Ax + 2 B = 0

x2 - 3x - 10 = 0

x2 - 3x - 10x2 - 3x - 10

x2 - 3x - 10 6 0y = x2 - 3x - 10

A-2, 5 B -2x2 - 3x - 10 6 0x2 - 3x - 10

x2 - 3x - 10x2 - 3x - 10

12108642

2−2

−6−8

−10−12

−4−6−8−10−12 4 6 8 10 12

x-valuescorresponding to negative

y-values

y = x2 − 3x − 10

x

y

y = x2 - 3x - 10

x2 - 3x - 10x2 - 3x - 10 6 0

x2 - 3x - 10

x2 - 3x + 11 ≥ 02x2 + 9x - 2 6 03x2 + 2x - 6 7 0x2 - 10x + 7 ≤ 0

A

Nonlinear Inequalities in One Variable SECTION 10.4 661

Objectives

Solve polynomial inequalities ofdegree 2 or greater.

Solve inequalities that containrational expressions with variablesin the denominator.

B

A


These two numbers and 5, divide the number line into three regions.We will call the regions A, B, and C. These regions are important because,if the value of is negative when a number from a region issubstituted for x, then is negative when any number in thatregion is substituted for x. The same is true if the value of ispositive for a particular value of x in a region.

To see whether the inequality is true or false in eachregion, we choose a test point from each region and substitute its value forx in the inequality . If the resulting inequality is true, theregion containing the test point is a solution region.

The values in region B satisfy the inequality. The numbers and 5 arenot included in the solution set since the inequality symbol is . The solu-tion set is , and its graph is shown.

Example 1 Solve:

Solution: First we solve the related equation .

The two numbers and 3 separate the number line intothree regions, A, B, and C.

Now we substitute the value of a test point from eachregion. If the test value satisfies the inequality, everyvalue in the region containing the test value is a solution.

The points in regions A and C satisfy the inequality. Thenumbers and 3 are not included in the solution sincethe inequality symbol is . The solution set is

, and its graph is shown.

The steps in the margin may be used to solve a polynomial inequality.


) )−3 3

A B C

T F T

A- q, -3 B ´ A3, q B 7-3

−3 3

A B C

-3

x = -3 x = 3

x + 3 = 0 or x - 3 = 0

Ax + 3 B Ax - 3 B = 0

Ax + 3 B Ax - 3 B = 0

Ax + 3 B Ax - 3 B 7 0

)(−2 5

A B C

F T F

A-2, 5 B 6-2

−2 5

A B C

x2 - 3x - 10 6 0

x2 - 3x - 10 6 0

x2 - 3x - 10x2 - 3x - 10

x2 - 3x - 10

-2


Answer1.

✓ Concept Check: The solutions found in Step 2have a value of zero in the original inequality.

A- q, -4 B ´ A2, q B

SOLVING A POLYNOMIAL

INEQUALITY

Step 1. Write the inequality instandard form and thensolve the related equation.

Step 2. Separate the number lineinto regions with the solu-tions from Step 1.

Step 3. For each region, choose atest point and determinewhether its value satisfiesthe original inequality.

Step 4. The solution set includesthe regions whose testpoint value is a solution. Ifthe inequality symbol is

or , the values fromStep 1 are solutions; if or , they are not.7

6≥≤

Practice Problem 1Solve: Ax - 2 B Ax + 4 B 7 0

✓ CONCEPT CHECK

When choosing a test point in Step 4,why would the solutions from Step 2not make good choices for testpoints?

Test Point

Region Value Result

A False.

B 0 True.

C 6 False. A1 B A8 B 6 0

A-5 B A2 B 6 0

A-8 B A-1 B 6 0-3

Ax - 5 B Ax + 2 B 6 0

Test Point

Region Value Result

A True.

B 0 False.

C 4 True. A7 B A1 B 7 0

A3 B A-3 B 7 0

A-1 B A-7 B 7 0-4

Ax + 3 B Ax - 3 B 7 0

Example 2 Solve:

Solution: First we solve the related equation .

The numbers 0 and 4 separate the number line into threeregions, A, B and C.

Check a test value in each region in the original inequal-ity. Values in region B satisfy the inequality. The num-bers 0 and 4 are included in the solution since theinequality symbol is . The solution set is , and itsgraph is shown.

Example 3 Solve:

Solution: First we solve . By inspec-tion, we see that the solutions are , and 5. They sep-arate the number line into four regions, A, B, C, and D.Next we check test points from each region.

The solution set is , and its graph isshown. We include the numbers , 1 and 5 because theinequality symbol is .

SOLVING RATIONAL INEQUALITIES

Inequalities containing rational expressions with variables in the denomi-nator are solved by using a similar procedure.

Example 4 Solve:

Solution: First we find all values that make the denominator equalto 0. To do this, we solve , or .

Next, we solve the related equation .x + 2x - 3

= 0

x = 3x - 3 = 0

x + 2x - 3

≤ 0

B

] [ ]1−2T F T F5

A B DC

≤-2

A- q, -2 D ´ C1, 5 D

1−2 5

A B DC

-2, 1Ax + 2 B Ax - 1 B Ax - 5 B = 0

Ax + 2 B Ax - 1 B Ax - 5 B ≤ 0

[ ]0 4

A B C

F T F

C0, 4 D≤

0 4

A B C

x = 0 or x = 4

x Ax - 4 B = 0

x2 - 4x = 0

x2 - 4x = 0

x2 - 4x ≤ 0

Nonlinear Inequalities in One Variable SECTION 10.4 663

Answers2. , 3. , 4. A-5, 3 DA- q, -5 D ´ C-1, 2 DC0, 6 D

Practice Problem 2Solve: x2 - 6x ≤ 0

Practice Problem 3Solve: Ax - 2 B Ax + 1 B Ax + 5 B ≤ 0

Test Point

Region Value Result

A True.B 0 False.C 2 True.D 6 False. A8 B A5 B A1 B ≤ 0

A4 B A1 B A-3 B ≤ 0 A2 B A-1 B A-5 B ≤ 0A-1 B A-4 B A-8 B ≤ 0-3

Ax - 5 B ≤ 0Ax + 2 B Ax - 1 B

Practice Problem 4

Solve: x - 3x + 5

≤ 0

Multiply both sides by the LCD, .

Now we place these numbers on a number line and pro-ceed as before, checking test point values in the originalinequality.

Choose -3 from region A. Choose 0 from region B. Choose 4 from region C.

True. False.

False.

The solution set is . This interval includes because satisfies the original inequality. This intervaldoes not include 3, because 3 would make the de-nominator 0.

The steps in the margin may be used to solve a rational inequality withvariables in the denominator.

Example 5 Solve:

Solution: First we find values for x that make the denominatorequal to 0.

Next we solve .

Multiply both sides bythe LCD, .

Simplify.

We use these two solutions to divide a number line intothree regions and choose test points. Only a test pointvalue from region B satisfies the original inequality. The

solution set is , and its graph is shown.

)(− 7

2−1

A B C

F T F

a - 72

, -1 b

− 72

−1

A B C -7

2= x

7 = -2x

5 = -2x - 2

x + 1Ax + 1 B #

5x + 1

= Ax + 1 B # -2

5x + 1

= -2

x = -1x + 1 = 0

5x + 1

6 -2

)[−2 3

A B C

F T F

-2-2C-2, 3 B

16

≤ 0

6 ≤ 0 - 23

≤ 0 -1-6

≤ 0

4 + 24 - 3

≤ 00 + 20 - 3

≤ 0-3 + 2-3 - 3

≤ 0

x + 2x - 3

≤ 0x + 2x - 3

≤ 0x + 2x - 3

≤ 0

x = -2

x + 2 = 0

x - 3x + 2x - 3

= 0

−2 3

A B C


Answer

5. A- q, 2 B ´ a 72

, q b

Practice Problem 5

Solve: 3

x - 26 2

SOLVING A RATIONAL INEQUALITY

Step 1. Solve for values that makeall denominators 0.

Step 2. Solve the related equation.

Step 3. Separate the number lineinto regions with the solu-tions from Steps 1 and 2.

Step 4. For each region, choose atest point and determinewhether its value satisfiesthe original inequality.

Step 5. The solution set includesthe regions whose testpoint value is a solution.Check whether to includevalues from Step 2. Be surenot to include values thatmake any denominator 0.

10.5 QUADRATIC FUNCTIONS AND THEIR GRAPHS

GRAPHING f(x) � x2 � kWe graphed the quadratic function in Section 7.5. In that sec-tion, we discovered that the graph of a quadratic function is a parabolaopening upward or downward. Now, as we continue our study, we will dis-cover more details about quadratic functions and their graphs.

First, let’s recall the definition of a quadratic function.

Notice that equations of the form , where , alsodefine quadratic functions since y is a function of x or .

Recall that if , the parabola opens upward and if , theparabola opens downward. Also, the vertex of a parabola is the lowestpoint if the parabola opens upward and the highest point if the parabolaopens downward. The axis of symmetry is the vertical line that passesthrough the vertex.

Example 1 Graph and on the same set ofaxes.

Solution: First we construct a table of values for amd plot thepoints. Notice that for each x-value, the correspondingvalue of must be 3 more than the correspondingvalue of since and . Inother words, the graph of is the same asthe graph of shifted upward 3 units. The axisof symmetry for both graphs is the y-axis.

f Ax B = x2g Ax B = x2 + 3

g Ax B = x2 + 3f Ax B = x2f Ax Bg Ax B

f Ax B

g Ax B = x2 + 3f Ax B = x2

x

y

Vertex

Axisof

Symmetry

f(x) = ax2 + bx + c,a > 0

x

y

Vertex

Axisof

Symmetry

f(x) = ax2 + bx + c,a < 0

a 6 0a 7 0y = f Ax Ba Z 0y = ax2 + bx + c

f Ax B = x2

A

Quadratic Functions and Their Graphs SECTION 10.5 665

Objectives

Graph quadratic functions of theform .




Graph quadratic functions of theform .f Ax B = a Ax - h B 2 + k

Ef Ax B = ax2

Df Ax B = Ax - h B 2 + k

Cf Ax B = Ax - h B 2Bf Ax B = x2 + k

A

QUADRATIC FUNCTION

A quadratic function is a function that can be written in the form, where a, b, and c are real numbers and .a Z 0f Ax B = ax2 + bx + c

Practice Problem 1Graph and on the same set of axes.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

g Ax B = x2 + 4f Ax B = x2

Answer1.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

g(x) = x2 + 4

f(x) = x2

1 3 4 5

(0, 4)

(0, 0)


In general, we have the following properties.

Examples Graph each function.

2.

The graph of is obtained by shiftingthe graph of upward 2 units.

3.

The graph of is obtained by shiftingthe graph of downward 3 units.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

g(x) = x2 − 3

y = x2

1 3

(0, −3)

4 5

y = x2g Ax B = x2 - 3

g Ax B = x2 - 3

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

F(x) = x2 + 2y = x2

1 3

(0, 2)

4 5

y = x2F Ax B = x2 + 2

F Ax B = x2 + 2

67

54321

1−1

−2−3−4

−2−3−4−5−6 2 3 4

(0, 3)

(0, 0)

g(x) = x2 + 3

f(x) = x2

x

y


GRAPHING THE PARABOLA DEFINED BY f(x) � x2 � k

If k is positive, the graph of is the graph of shifted upward k units.

If k is negative, the graph of is the graph of shifted downward units.

The vertex is , and the axis of symmetry is the y-axis.A0, k B@k @

y = x2f Ax B = x2 + k

y = x2f Ax B = x2 + k

Answers2.

3.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

g(x) = x2 − 2

1 3 4 5

(0, −2)

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

F(x) = x2 + 1

1 3 4 5

(0, 1)

Practice Problems 2–3Graph each function.

2.

3.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

g Ax B = x2 - 2

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

F Ax B = x2 + 1

x

4 71 4

0 0 31 1 42 4 7

cEach y-valueis increased

by 3.

-1-2

g Ax B = x2 + 3f Ax B = x2

GRAPHING f(x) � (x � h)2

Now we will graph functions of the form .

Example 4 Graph and on the same setof axes.

Solution: By plotting points, we see that for each x-value, the cor-responding value of is the same as the value of when the x-value is increased by 2. Thus, the graph of

is the graph of shifted to theright 2 units. The axis of symmetry for the graph of

is also shifted 2 units to the right and isthe line .

In general, we have the following properties.

Examples Graph each function.

5.

The graph of is obtained by shiftingthe graph of to the right 3 units.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

G(x) = (x − 3)2

y = x2

x = 3

1 3

(3, 0)

4 5

y = x2G Ax B = Ax - 3 B 2

G Ax B = Ax - 3 B 2

g(x) = (x − 2)2f(x) = x2

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

x = 2

1 3

(2, 0)

(0, 0)

4 5

x = 2g Ax B = Ax - 2 B 2

f Ax B = x2g Ax B = Ax - 2 B 2f Ax Bg Ax B

g Ax B = Ax - 2 B 2f Ax B = x2

f Ax B = Ax - h B 2B


Answer

4.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

g(x) = (x − 1)2f(x) = x2

1 3 4 5(1, 0)(0, 0)

x x

-2 4 0 4-1 1 1 1

0 0 2 01 1 3 12 4 4 4

cEach x-valueincreased by 2corresponds tosame y-value.

g Ax B = Ax - 2 B 2f Ax B = x2

GRAPHING THE PARABOLA DEFINED BY f(x) � (x � h)2

If h is positive, the graph of is the graph of shifted to the right h units.

If h is negative, the graph of is the graph of shifted to the left units.

The vertex is , and the axis of symmetry is the vertical line .x = hAh, 0 B@h @

y = x2f Ax B = Ax - h B 2y = x2f Ax B = Ax - h B 2

Practice Problem 4

Graph and on the same set of axes.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

Ax - 1B 2g AxB =fAxB = x2

on the same set of axes.

6.

The equation can be written as. The graph of

is obtained by shifting the graph ofto the left 1 unit.

GRAPHING f(x) � (x � h)2 � kAs we will see in graphing functions of the form , itis possible to combine vertical and horizontal shifts.

Example 7 Graph:

Solution: The graph of is the graph ofshifted 3 units to the right and 1 unit up. The ver-

tex is then , and the axis of symmetry is . Afew ordered pair solutions are plotted to aid in graphing.

GRAPHING f(x) � ax2

Next, we discover the change in the shape of the graph when the coefficientof is not 1.x2

D

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

F(x) = (x − 3)2 + 1

x = 3

1 3

(3, 1)

4 5

x = 3A3, 1 By = x2F Ax B = Ax - 3 B 2 + 1

F Ax B = Ax - 3 B 2 + 1

f Ax B = Ax - h B 2 + k

C

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

F(x) = (x + 1)2 y = x2

x = −1

1 3

(−1, 0)

4 5

y = x2

Cx - A-1 B D 2F Ax B =Cx - A-1 B D 2F Ax B =

F Ax B = Ax + 1 B 2F Ax B = Ax + 1 B 2


Answers5.

6.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

F(x) = (x + 2)2

1 3 4 5(−2, 0)

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

G(x) = (x − 4)2

1 3 4 5(4, 0)

GRAPHING THE PARABOLA DEFINED BY f(x) � (x � h)2 � k

The parabola has the same shape as .The vertex is , and the axis of symmetry is the vertical line .x = hAh, k B y = x2

x

1 52 24 25 5

F Ax B = Ax - 3 B 2 + 1

Practice Problems 5–6Graph each function.5.

6.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

F Ax B = Ax + 2 B 2

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

G Ax B = Ax - 4 B 2


Answers7.

8.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

g(x) = 2x2

f(x) = x2

h(x) = 13 x2

1 3 4 5

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

F(x) = (x − 2)2 + 3(2, 3)

1 3 4 5


x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

F Ax B = Ax - 2 B 2 + 3

Example 8 Graph , , and on the

same set of axes.

Solution: Comparing the table of values, we see that for each x-value, the corresponding value of is triple the corre-sponding value of . Similarly, the value of ishalf the value of .

The result is that the graph of is narrowerthan the graph of and the graph of

is wider. The vertex for each graph is ,

and the axis of symmetry is the y-axis.

Example 9 Graph:

Solution: Because , a negative value, this parabola opensdownward. Since and , the parabola isnarrower than the graph of . The vertex is ,and the axis of symmetry is the y-axis. We verify this byplotting a few points.

A0, 0 By = x22 7 1@ -2 @ = 2

a = -2

f Ax B = -2x2

A0, 0 Bh Ax B = 12

x2

f Ax B = x2

g Ax B = 3x2

6789

101112

5432

1−1

−2

−2−3−4 2 3 4 5 x

y

g(x) = 3x2

f(x) = x2

h(x) = x212

f Ax Bh Ax Bf Ax B

g Ax B

h Ax B = 12

x2g Ax B = 3x2f Ax B = x2

Practice Problem 8Graph , , and


x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

h Ax B = 13

x2

g Ax B = 2x2f Ax B = x2

x x

4 121 3

0 0 0 01 1 1 32 4 2 12

-1-1-2-2

g Ax B = 3x2f Ax B = x2

x

2

0 0

1

2 2

12

12

-1

-2

h Ax B = 12

x2

GRAPHING THE PARABOLA DEFINED BY f(x) � ax2

If a is positive, the parabola opens upward, and if a is negative, theparabola opens downard.

If , the graph of the parabola is narrower than the graph of.

If , the graph of the parabola is wider than the graph of .y = x2@a @ 6 1

y = x2@a @ 7 1

GRAPHING f(x) � a (x � h)2 � kNow we will see the shape of the graph of a quadratic function of the form

.

Example 10 Graph: . Find the vertex and the

axis of symmetry.

Solution: The function may be written as

. Thus, this graph is the same

as the graph of shifted 2 units to the left and

5 units up, and it is wider because a is . The vertex is

, and the axis of symmetry is . We plot afew points to verify.

789

101112

54321

1−1

−2

−3−4−5 2 3 4 5 6 7 x

y

x = −2

g(x) = (x + 2)2 + 512

(−2, 5)

x = -2A-2, 5 B12

y = x2

g Ax B = 12Cx - A-2 BD 2 + 5

g Ax B = 12Ax + 2 B 2 + 5

g Ax B = 12Ax + 2 B 2 + 5

f Ax B = a Ax - h B 2 + k

E

321

1−1

−2−3−4−5−6−7−8−9

−10−11

−2 2 3 4 5

(0, 0)

f(x) = −2x2

x

y


x

7

5

0 7

512

-1

-2

512

-3

-4

g Ax B = 12Ax + 2 B 2 + 5

Answers

9.

10.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

(−3, −4)

x = −3

1 3 4 5

f(x) = 2(x + 3)2 − 4

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 2

f(x) = −3x21 3 4 5

(0, 0)


x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f Ax B = -3x2

x

0 012 -8

-2

-2-1-8-2

f Ax B = -2x2

Practice Problem 10Graph: . Findthe vertex and axis of symmetry.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f Ax B = 2 Ax + 3 B 2 - 4


In general, the following holds.

TRY THE CONCEPT CHECK IN THE MARGIN. ✓ CONCEPT CHECK

Which description of the graph ofis cor-

rect?a. The graph opens downward and

has its vertex at .b. The graph opens upward and has

its vertex at .c. The graph opens downward and

has its vertex at .d. The graph is narrower than the

graph of .y = x2

A-3, -4 BA-3, 4 BA-3, 4 B

f Ax B = -0.35 Ax + 3 B 2 - 4

GRAPH OF A QUADRATIC FUNCTION

The graph of a quadratic function written in the formis a parabola with vertex . If , the

parabola opens upward, and if , the parabola opens downward.The axis of symmetry is the line whose equation is .

x

y

(h, k)

f(x) = a(x − h)2 + k

a > 0x = h

x

y

(h, k)

a < 0

x = h

x = ha 6 0

a 7 0Ah, k Bf Ax B = a Ax - h B 2 + k



GRAPHING CALCULATOR EXPLORATIONS

Use a grapher to graph the first function of each pair that follows. Then use its graph to predict the graph ofthe second function. Check your prediction by graphing both on the same set of axes.

1. ; 2. ;

3. ; 4. ;

5. ; 6. ; g Ax B = 1x - 4 - 2G Ax B = 1x - 2F Ax B = @x + 4 @ + 3f Ax B = @x + 4 @

g Ax B = Ax - 3 B 3 + 2h Ax B = x3 + 2f Ax B = @x - 5 @H Ax B = @x @

H Ax B = x3 - 2g Ax B = x3G Ax B = 1x + 1F Ax B = 1x

673

Focus On Business and CareerFINANCIAL RATIOS

A financial ratio is a number found with a rational expression that tells something about acompany’s activities. Such ratios allow a comparison between the financial positions of twocompanies, even if the values of the companies’ financial data are very different. Here aresome common financial ratios:j The current ratio gauges a company’s ability to pay its short-term debts. It is given by the formula

The higher the value of this ratio, the better able the company is to pay off its short-term debts.j The total asset turnover ratio gauges how effectively a company is using all of its resources to generate

sales of its products and services. It is given by the formula

The higher the value of this ratio, the more effective the company is at utilizing its resources for sales gen-eration.

j The gross profit margin ratio gauges how effectively the company is making pricing decisions as well ascontrolling production costs. It is given by the formula

The higher the value of this ratio, the better the company is doing with regards to controlling costs andpricing products.

j The price-to-earnings (P/E) ratio gauges the stock market’s view of a company with respect to risk. It isgiven by the formula

A company with low risk will generally have a high P/E ratio. A high P/E ratio also translates into bettergrowth potential for the company’s earnings.

For all of these ratios, a higher-than-industry-average ratio is generally considered to be a sign of goodfinancial health.

ADDITIONAL DEFINITIONS

j Assets—things of value that are owned by a company. Current assets include cash and assets that can beconverted into cash quickly. Total assets are all things of value, including property and equipment, ownedby the company. Current assets and total assets may be found on a company’s consolidated balance sheetor statement of financial position in an annual report.

j Liabilities—what a company owes to creditors. Current liabilities include any debts expected to come duewithin the next year. Current liabilities may be found on a company’s consolidated balance sheet or state-ment of financial position in an annual report.

j Sales—the total amount of money collected by a company from the sales of its goods or services. Sales(or sometimes noted as “net sales”) may be found on a company’s consolidated statement ofincome/earnings/operations in an annual report.

j Cost of sales—a company’s cost of inventory actually sold to customers. This is also sometimes referredto as “cost of goods/merchandise sold.” Cost of sales may be found on a company’s consolidated state-ment of income/earnings/operations in an annual report.

j Earnings per share—the value of a company’s earnings available for each share of common stock held bystockholders. Earnings per share (or sometimes noted as net income per share) may be found on a com-pany’s consolidated statement of income/earnings/operations in an annual report.

j Stock market price per share—the current price of a company’s share of stock as given on one of themajor stock markets. Current share prices may be found in newspapers or on the World-Wide Web.

GROUP ACTIVITY

Locate annual reports for two companies involved in similar industries. Using the informationand definitions given above, compute these four financial ratios for each company. Then com-pare the companies’ ratios and discuss what the ratios indicate about the two companies.Which company do you think is in better overall financial health? Why?

P>E ratio =stock market price per share

earnings per share

gross profit margin ratio = sales - cost of salessales

total asset turnover ratio = salestotal assets

current ratio = total current assetstotal current liabilities

10.6 FURTHER GRAPHING OF QUADRATIC FUNCTIONS

WRITING QUADRATIC FUNCTIONS IN THE FORM y � a (x � h)2�kWe know that the graph of a quadratic function is a parabola. If a quadraticfunction is written in the form

we can easily find the vertex and graph the parabola. To write a qua-dratic function in this form, we need to complete the square. (See Section10.1 for a review of completing the square.)

Example 1 Graph: . Find the vertex and anyintercepts.

Solution: The graph of this quadratic function is a parabola. Tofind the vertex of the parabola, we complete the squareon the binomial . To simplify our work, we let

.

Let .

Now we add the square of half of to both sides.

and

From this equation, we can see that the vertex of theparabola is , a point in quadrant IV, and the axisof symmetry is the line .

Notice that . Since , the parabola opensupward. This parabola opening upward with vertex

will have two x-intercepts.To find them, we let or .

The two x-intercepts are 6 and . To find the y-intercept, we let .

The y-intercept is . The sketch ofis shown.f Ax B = x2 - 4x - 12

-12

f A0 B = 02 - 4 # 0 - 12 = -12

x = 0-2

6 = x -2 = x

0 = x - 6 or 0 = x + 2

0 = Ax - 6 B Ax + 2 B0 = x2 - 4x - 12

y = 0f Ax BA2, -16 Ba 7 0a = 1

x = 2A2, -16 B

f Ax B = Ax - 2 B 2 - 16

y = Ax - 2 B 2 - 16

y + 16 = Ax - 2 B 2Add 4 to both sides.

Factor the trinomial.


Replace y with f(x).

y + 12 + 4 = x2 - 4x + 4

A-2 B 2 = 412A-4 B = -2

-4

Add 12 to both sides to get thex-variable terms alone.

y + 12 = x2 - 4x

f Ax B = y y = x2 - 4x - 12

f Ax B = yx2 - 4x

f Ax B = x2 - 4x - 12

Ah, k Bf Ax B = a Ax - h B 2 + k

A

Further Graphing of Quadratic Functions SECTION 10.6 675

Objectives

Write quadratic functions in theform .

Derive a formula for finding thevertex of a parabola.

Find the minimum or maximumvalue of a quadratic function.

C

By = a Ax - h B 2 + k

A

Practice Problem 1Graph: . Find thevertex and any intercepts.

x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

f Ax B = x2 - 4x - 5

Answer1. vertex: ; x-intercepts: ; y-inter-cept:

x

y

42

68

10

−4−6−8

−10

−4 −2−6−8−10 42 6 8 10

(2, −9)

(0, −5)

(−1, 0) (5, 0)

-5-1, 5A2, -9 B



Solution: We replace with y and complete the square on x towrite the equation in the form .

Replace with y.

Get the x-variable terms alone.

Next we factor 3 from the terms so that thecoefficient of is 1.

Factor out 3.

The coefficient of x is 1. Then and .

Since we are adding inside the parentheses, we are

really adding , so we must add to the left

side.

Add to both sides.

Replace y with f(x).

Then , , and . This means that the

parabola opens upward with vertex and that

the axis of symmetry is the line .

To find the y-intercept, we let . Then

f A0 B = 3 A0 B 2 + 3 A0 B + 1 = 1

x = 0

x = - 12

a - 12

, 14b

k = 14

h = - 12

a = 3

f Ax B = 3 ax + 12b 2

+ 14

14 y = 3 ax + 1

2b 2

+ 14

Simplify the left side andfactor the right side. y - 1

4= 3 ax + 1

2b 2

y - 1 + 3 a 14b = 3 ax2 + x + 1

4b

3 a 14b3 a 1

4b

14

a 12b 2

= 14

12A1 B = 1

2

y - 1 = 3 Ax2 + x Bx2

3x2 + 3x

y - 1 = 3x2 + 3x

f Ax B y = 3x2 + 3x + 1

y = a Ax - h B 2 + kf Ax B

f Ax B = 3x2 + 3x + 1

(6, 0)(−2, 0)

(0, −12)

(2, −16)

x = 2

f(x) = x2 − 4x − 12

1214

108642

2−2

−6−8

−10−12−14−16

−4−6−8−10−12 4 8 10 12 x

y


HELPFUL HINT

Parabola Opens UpwardVertex in I or II: no x-interceptsVertex in III or IV: 2 x-intercepts

Parabola Opens DownwardVertex in I or II: 2 x-interceptsVertex in III or IV: no x-intercepts.

x

yII

III

I

twox-intercepts

nox-intercept

IV

x

y

II

III

I

no x-intercept

twox-intercepts

IV

Answer

2. vertex: ; y-intercept: 5

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

12( ),− 9

2(0, 5)

a- 12

, 92b

Practice Problem 2Graph: . Findthe vertex and any intercepts.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f Ax B = 2x2 + 2x + 5the vertex and any intercepts.


Practice Problem 3Graph: . Findthe vertex and any intercepts.

x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

f Ax B = -x2 - 2x + 8

Answer3. vertex: ; x-intercepts: ; y-inter-cept: 8

x

y

42

68

10

−4−6−8

−10

−4 −2−6−8−10 42 6 8 10

(−1, 9)

(−4, 0) (2, 0)

(0, 8)

-4, 2A-1, 9 B

This parabola has no x-intercepts since the vertex is inthe second quadrant and it opens upward. We use thevertex, axis of symmetry, and y-intercept to graph theparabola.


Solution: We write in the form by completingthe square. First we replace with y.

Factor from the terms .

The coefficient of x is 2. Then and . We

add 1 to the right side inside the parentheses and addto the left side.


Replace y with .

Since , the parabola opens downward with vertexand axis of symmetry .

To find the y-intercept, we let and solve for y.Then

Thus, 3 is the y-intercept.To find the x-intercepts, we let y or and

solve for x.

Let .

Now we divide both sides by so that the coefficient ofis 1:x2

-1

f Ax B = 0 0 = -x2 - 2x + 3

f Ax B = -x2 - 2x + 3

f Ax B = 0

f A0 B = -02 - 2 A0 B + 3 = 3

x = 0x = -1A-1, 4 Ba = -1

f Ax B f Ax B = -1 Ax + 1 B 2 + 4

y = -1 Ax + 1 B 2 + 4

Simplify the left sideand factor the right side. y - 4 = -1 Ax + 1 B 2

y - 3 - 1 A1 B = -1 Ax2 + 2x + 1 B-1 A1 B

12 = 112A2 B = 1

-x2 - 2x-1y - 3 = -1 Ax2 + 2x BSubtract 3 from both sides toget the x-variable terms alone.y - 3 = -x2 - 2x

y = -x2 - 2x + 3

f Ax B = -x2 - 2x + 3

f Ax Ba Ax - h B 2 + kf Ax B

f Ax B = -x2 - 2x + 3

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

12( ),−

12

14

y = 3x2 + 3x + 1

(0, 1)

x = −

HELPFUL HINT

This can be written as . Notice that

the vertex is .A-1, 4 BA-1 B D 2 + 4-1Cx -

f Ax B =

the vertex and any intercepts.

Divide both sides by -1.

Simplify.

Factor.


Solve.

The x-intercepts are and 1. We use these points tograph the parabola.

DERIVING A FORMULA FOR FINDING THE VERTEX

Recall from Section 7.5 that we introduced a formula for finding the vertexof a parabola. Now that we have practiced completing the square, we willshow that the x-coordinate of the vertex of the graph of or

can be found by the formula . To do so, we com-

plete the square on x and write the equation in the form .First we get the x-variable terms alone by subtracting c from both sides.

Factor a from the terms .

Now we add the square of half of , or , to the right side

inside the parentheses. Because of the factor a, what we really added is

and this must be added to the left side.

subtract from both sides.

Compare this form with or and see that h is

, which means that the x-coordinate of the vertex of the graph of

is .-b2a


-b2a

y = a Ax - h B 2 + kf Ax B

b2

4a y = a ax + b2ab 2

+ c - b2

4a

Simplify the left side andfactor the right side.Add c to both sides and

y - c + b2

4a= a ax + b

2ab 2

y - c + a a b2

4a2 b = a ax2 + ba

x + b2

4a2 ba a b2

4a 2b

a b2ab 2

= b2

4a2ba

ax2 + bxy - c = a ax2 + ba

x by - c = ax2 + bx

y = ax2 + bx + c

y = Ax - h B 2 + k

x = -b2a

y = ax2 + bx + c

f Ax B

B

(0, 3)

(1, 0)

(−1, 4)

(−3, 0)

f(x) = −x2 − 2x + 3

x = −1

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

-3

x = 1 x = -3

x + 3 = 0 or x - 1 = 0

0 = Ax + 3 B Ax - 1 B 0 = x2 + 2x - 3

0

-1= -x2

-1- 2x

-1+ 3

-1


Let’s use this vertex formula in the margin to find the vertex of theparabola we graphed in Example 1.

Example 4 Find the vertex of the graph of .

Solution: In the quadratic function , noticethat , , and . Then

The x-value of the vertex is 2. To find the correspondingor y-value, find . Then

The vertex is . These results agree with our find-ings in Example 1.

FINDING MINIMUM AND MAXIMUM VALUES

The quadratic function whose graph is a parabola that opens upward has aminimum value, and the quadratic function whose graph is a parabola thatopens downward has a maximum value. The or y-value of the vertexis the minimum or maximum value of the function.

Recall from Section 8.2 that the discriminant, , tells us howmany solutions the quadratic equation has. It also tells us how many x-intercepts the graph of a quadratic equation

has.

No x-intercepts One x-intercept Two x-interceptsb2 - 4ac 7 0b2 - 4ac = 0b2 - 4ac 6 0

y = x2 - 2x - 3y = x2 - 2x + 1y = x2 + 2x + 3

54321

1−1−2−3−4−5

−2−3−4−5 2 3 4 5 x

y54321

1−1−2−3−4−5

−2−3−4−5 2 3 4 5 x

y54321

1−1−2−3−4−5

−2−3−4−5 2 3 4 5 x

y

y = ax2 + bx + c

0 = ax2 + bx + cb2 - 4ac

321

1−1

−3−2

−4−5−6−7

−2−3−4−5−6 x

y

Minimumvalue

Vertex (h, k)

67

54321

1−1

−3−2

2 3 4 5 6 7 x

y

Maximumvalue

Vertex (h, k)

f Ax B

C

A2, -16 Bf A2 B = 22 - 4 A2 B - 12 = 4 - 8 - 12 = -16

f A2 Bf Ax B

-b2a

=- A-4 B

2 A1 B = 2

c = -12b = -4a = 1f Ax B = x2 - 4x - 12

f Ax B = x2 - 4x - 12

Practice Problem 4Find the vertex of the graph of

. Compare yourresult with the result of Practice Prob-lem 1.

f Ax B = x2 - 4x - 5


VERTEX FORMULA

The graph of, when

, is a parabola with vertex

a -b2a

, f a -b2ab b


Answer4. A2, -9 B

result with the result of Practice Prob-lem 1.

✓ CONCEPT CHECK

Without making any calculations, tellwhether the graph of

has a maxi-mum value or a minimum value.Explain your reasoning.

f Ax B = 7 - x - 0.3x2


Answers

5. Maximum height: feet in seconds

✓ Concept Check: has a maximum valuesince it opens downward.

f Ax B516

101916

Practice Problem 5An object is thrown upward from thetop of a 100-foot cliff. Its height aboveground after t seconds is given by thefunction .Find the maximum height of the objectand the number of seconds it tookfor the object to reach its maximumheight.

f A t B = -16t2 + 10t + 100


Example 5 Finding Maximum HeightA rock is thrown upward from the ground. Its height infeet above ground after t seconds is given by the function

. Find the maximum height of therock and the number of seconds it took for the rock toreach its maximum height.

Solution: 1. UNDERSTAND. The maximum height of the rock isthe largest value of . Since the function

is a quadratic function, its graphis a parabola. It opens downward since .Thus, the maximum value of is the or y-valueof the vertex of its graph.

2. TRANSLATE. To find the vertex , notice thatfor , , , and .We will use these values and the vertex formula

3. SOLVE.

4. INTERPRET. The graph of is a parabola open-

ing downward with vertex . This means that

the rock’s maximum height is feet, or feet,

which was reached in second.58

614

254

a 58

, 254b

f A t Bf a 5

8b = -16 a 5

8b2

+ 20 a 58b = -16 a 25

64b + 25

2= - 25

4+ 50

4= 25

4

h = -b2a

= -20-32

= 58

a -b2a

, f a -b2ab b

c = 0b = 20a = -16f A t B = -16t 2 + 20tAh, k B

f A t Bf A t B -16 6 0f A t B = -16t2 + 20t

f A t B

f A t B = -16t2 + 20t

and the number of seconds it took for the object to reach its maximumheight.



SECTION 10.1 SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE

SQUARE ROOT PROPERTY

If b is a real number and if , then .a = ;1ba2 = b

Solve:

x = -3 ; 114

x + 3 = ;114

Ax + 3 B 2 = 14

TO SOLVE A QUADRATIC EQUATION IN X BY

COMPLETING THE SQUARE

Step 1. If the coefficient of is not 1, divide bothsides of the equation by the coefficient of .

Step 2. Get the variable terms alone.

Step 3. Complete the square by adding the squareof half of the coefficient of x to both sides.

Step 4. Write the resulting trinomial as the squareof a binomial.

Step 5. Use the square root property.

x2

x2

Solve:

1.

2.

3. and

4.

5. x = 2 ; 110

x - 2 = ;110

Ax - 2 B 2 = 10x2 - 4x + 4 = 6 + 4

A-2 B 2 = 412A-4 B = -2

x2 - 4x = 6

x2 - 4x - 6 = 0

3x2 - 12x - 18 = 0

SECTION 10.2 SOLVING QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA

QUADRATIC FORMULA

A quadratic equation written in the formhas solutions

x = -b ; 2b2 - 4ac2a

ax2 + bx + c = 0

Solve:

, ,

x = 1 ; 1132

x =- A-1 B ; 2 A-1 B 2 - 4 A1 B A-3 B

2 # 1

c = -3b = -1a = 1

x2 - x - 3 = 0

SECTION 10.3 SOLVING EQUATIONS BY USING QUADRATIC METHODS

Substitution is often helpful in solving an equationthat contains a repeated variable expression.

Solve:

Let . Then

Let

x = 1 x = 12

Substituteback. 2x + 1 = 3 2x + 1 = 2

m = 3 or m = 2

Am - 3 B Am - 2 B = 0

m = 2x + 1. m2 - 5m + 6 = 0

m = 2x + 1

A2x + 1 B 2 - 5 A2x + 1 B + 6 = 0


SECTION 10.4 NONLINEAR INEQUALITIES IN ONE VARIABLE

TO SOLVE A POLYNOMIAL INEQUALITY

Step 1. Write the inequality in standard form.Step 2. Solve the related equation.Step 3. Use solutions from Step 2 to separate the

number line into regions.Step 4. Use a test point to determine whether val-

ues in each region satisfy the originalinequality.

Step 5. Write the solution set as the union ofregions whose test point values are solu-tions.

Solve: 1.2.

3.

4. Test PointRegion Value Result

A True.

B 1 False.

C 7 True.

5.

The solution set is .A- q, 0 D ´ C6, q B] ]0 6

72 ≥ 6 A7 B 12 ≥ 6 A1 B

A-2 B 2 ≥ 6 A-2 B-2

x2 ≥ 6x

0 6

A B C

x = 0 or x = 6x Ax - 6 B = 0

x2 - 6x = 0x2 - 6x ≥ 0

x2 ≥ 6x

TO SOLVE A RATIONAL INEQUALITY

Step 1. Solve for values that make all denominators0.

Step 2. Solve the related equation.Step 3. Use solutions from Steps 1 and 2 to sepa-

rate the number line into regions.Step 4. Use a test point to determine whether val-

ues in each region satisfy the originalinequality.

Step 5. Write the solution set as the union ofregions whose test point value is a solution.

Solve:

1. Set denominator equal to 0.

2.

Multiply by .

3.

4. Only a test value from region B satisfies the orig-inal inequality.

5.

The solution set is .A-2, 1 B)(1−2

1−2

A B C

-2 = x 4 = -2x 6 = -2x + 2

Ax - 1 B 6 = -2 Ax - 1 B6

x - 1= -2

x = 1x - 1 = 0

6x - 1

6 -2


SECTION 10.5 QUADRATIC FUNCTIONS AND THEIR GRAPHS

GRAPH OF A QUADRATIC FUNCTION

The graph of a quadratic function written in the formis a parabola with vertex

. If , the parabola opens upward; if, the parabola opens downward. The axis of

symmetry is the line whose equation is .

(h, k)

a > 0x = h

x

y

(h, k)

a < 0

x = hf(x) = a(x − h)2 + k

x

y

x = ha 6 0

a 7 0Ah, k Bf Ax B = a Ax - h B 2 + k

Graph: The graph is a parabola with vertex and axis of

symmetry . Since is positive, the graphopens upward.

g(x) = 3(x − 1)2 + 4

(1, 4)

x = 1

x

y

a = 3x = 1A1, 4 B

g Ax B = 3 Ax - 1 B 2 + 4

SECTION 10.6 FURTHER GRAPHING OF QUADRATIC FUNCTIONS

The graph of , , is aparabola with vertex

.a -b2a

, f a -b2ab b

a Z 0f Ax B = ax2 + bx + c Graph: . Find the vertex and x- and y-intercepts.

The vertex is .

The x-intercept points are and .

The y-intercept point is .

(−2, 0)

(0, −8)(1, −9)

(4, 0)

x

y

A0, -8 Bf A0 B = 02 - 2 # 0 - 8 = -8

A-2, 0 BA4, 0 Bx = 4 or x = -2

0 = Ax - 4 B Ax + 2 B0 = x2 - 2x - 8

A1, -9 B f A1 B = 12 - 2 A1 B - 8 = -9

-b2a

=- A-2 B2 # 1

= 1

f Ax B = x2 - 2x - 8

685

11In this chapter, we discuss two closely related functions: exponentialand logarithmic functions. These functions are vital in applications ineconomics, finance, engineering, the sciences, education, and otherfields. Models of tumor growth and learning curves are two examples ofthe uses of exponential and logarithmic functions.

11.1 The Algebra of Functions11.2 Inverse Functions11.3 Exponential Functions11.4 Logarithmic Functions11.5 Properties of Logarithms11.6 Common Logarithms, Natural

Logarithms, and Change ofBase

11.7 Exponential and LogarithmicEquations and ProblemSolving

An earthquake is a series of vibrations in the crust of theearth. The size, or magnitude, of an earthquake is measured onthe Richter scale. The magnitude of an earthquake can rangebroadly, from barely detectable (2.5 or less on the Richter scale)to massively destructive (7.0 or greater on the Richter scale).According to the United States Geological Survey, earthquakesare an everyday occurrence. In 1997, there were a total of20,824 earthquakes around the world, or an average of 57earthquakes per day. However, most of these were minortremors with magnitudes of 3.9 or less, and many could only bedetected by seismographs. Only 0.09% of the earthquakesoccurring during 1997 could be classified as major earthquakes(7.0 or greater on the Richter scale). Even so, earthquakes wereresponsible for 2907 deaths that year.

Exponential and Logarithmic Functions C H A P T E R

11.1 THE ALGEBRA OF FUNCTIONS

ADDING, SUBTRACTING, MULTIPLYING, AND DIVIDING FUNCTIONS

As we have seen in earlier chapters, it is possible to add, subtract, multiply,and divide functions. Although we have not stated it as such, the sums,differences, products, and quotients of functions are themselves functions. For example, if and , their product,

, is a new function. We can use thenotation to denote this new function. Finding the sum, differ-ence, product, and quotient of functions to generate new functions is calledthe algebra of functions.

Example 1 If and , find:

a.b.c.

d.

Solution: Use the algebra of functions and replace by and by . Then we simplify.

a.

b.

c.

d. , where

There is an interesting but not surprising relationship between thegraphs of functions and the graph of their sum, difference, product, andquotient. For example, the graph of can be found by adding thegraph of to the graph of . We add two graphs by adding corre-sponding y-values.

g Ax Bf Ax B Af + g B Ax B

x Z 32

a fgb Ax B =

f Ax Bg Ax B = x - 1

2x - 3

= 2x2 - 5x + 3 = Ax - 1 B A2x - 3 B

Af # g B Ax B = f Ax B # g Ax B = -x + 2 = x - 1 - 2x + 3 = Ax - 1 B - A2x - 3 B

Af - g B Ax B = f Ax B - g Ax B = 3x - 4 = Ax - 1 B + A2x - 3 B

Af + g B Ax B = f Ax B + g Ax B2x - 3g Ax B x - 1f Ax B

a fgb Ax B

Af # g B Ax BAf - g B Ax BAf + g B Ax B

g Ax B = 2x - 3f Ax B = x - 1

Af # g B Ax Bf Ax B # g Ax B = 3x Ax + 1 B = 3x2 + 3x

g Ax B = x + 1f Ax B = 3x

A

The Algebra of Functions SECTION 11.1 687

Objectives

Add, subtract, multiply, and dividefunctions.

Compose functions.B

A

ALGEBRA OF FUNCTIONS

Let f and g be functions. New functions from f and g are defined asfollows:

SumDifferenceProduct

Quotient a fgb Ax B =

f Ax Bg Ax B

Af # g B Ax B = f Ax B # g Ax BAf - g B Ax B = f Ax B - g Ax BAf + g B Ax B = f Ax B + g Ax B

Practice Problem 1If and ,find:

a.

b.

c.

d. a fgb Ax B

Af # g B Ax BAf - g B Ax BAf + g B Ax B

g Ax B = 3x - 1f Ax B = x + 3

Answers1. a. , b. , c. ,

d. where x Z 13

x + 33x - 1

3x2 + 8x - 3-2x + 44x + 2


COMPOSITION OF FUNCTIONS

Another way to combine functions is called function composition. Tounderstand this new way of combining functions, study the tables below.They show degrees Fahrenheit converted to equivalent degrees Celsius,and then degrees Celsius converted to equivalent degrees Kelvin. (TheKelvin scale is a temperature scale devised by Lord Kelvin in 1848.)

Suppose that we want a table that shows a direct conversion fromdegrees Fahrenheit to kelvins. In other words, suppose that a table isneeded that shows kelvins as a function of degrees Fahrenheit. This caneasily be done because in the tables, the output of the first table is the sameas the input of the second table. The new table is as follows.

Since the output of the first table is used as the input of the second table,we write the new function as . The new function is formed fromthe composition of the other two functions. The mathematical symbol forthis composition is . Thus, .

It is possible to find an equation for the composition of the two functionsand . In other words, we can find a function that converts

degrees Fahrenheit directly to kelvins. The function

converts degrees Fahrenheit to degrees Celsius, and the functionconverts degrees Celsius to kelvins. Thus,

In general, the notation f(g(x)) means “f composed with g” and can be written as ( )(x). Also , or , means “g composedwith f.”

Ag ø f B Ax Bg Af Ax B Bf ø g

AK ø C B Ax B = K AC Ax B B = K a 59Ax - 32 B b = 5

9Ax - 32 B + 273.15

K AC B = C + 273.15

C Ax B = 59Ax - 32 B

K Ax BC Ax BAK ø C B Ax B = K AC Ax B BAK ø C B Ax B

K AC Ax B B

B

(f + g)(0) = 5

(f + g)(6) = 10

(f + g)(x)

g(x)

f(x)g(6) = 6

f(6) = 4g(3) = 5g(0) = 4

f(0) = 1 f(3) = 212

(f + g)(3) = 712

x

y

21

3456789

−1 21 3 4 5 6 7 8 9

688 CHAPTER 11 Exponential and Logarithmic Functions

Degrees Fahrenheit (Input) 32 68 149 212

Degrees Celsius (Output) 0 20 65 100-25-35C Ax B =

-13-31x =

Degrees Celsius (Input) 0 20 65 100

Kelvins (Output) 238.15 248.15 273.15 293.15 338.15 373.15K AC B =

-25-35C =


Kelvins (Output) 238.15 248.15 273.15 293.15 338.15 373.15K AC Ax B B =

-13-31x =

The Algebra of Functions SECTION 11.1 689

Answers2. a. 49; 19, b. ; ,3. a. , b. , 4. a. ,b. Ag ø f B Ax B

Ah ø g B Ax B1x + 11x + 12x2 + 14x2 + 4x + 1

COMPOSITE FUNCTIONS

The composition of functions f and g is

Af ø g B Ax B = f Ag Ax B B

HELPFUL HINT

does not mean the same as .

while Af # g B Ax B = f Ax B # g Ax BAf ø g B Ax B = f Ag Ax B BAf # g B Ax BA f ø g B Ax B

Example 2 If and , find each composition.

a. and b. and

Solution:

a.Since , then .

Since , then .

b.Replace with .

Square .

Replace with .

Example 3 If and , find each composition.

a.b.

Solution: a.b.

Example 4 If , , and , write eachfunction as a composition with f, g, or h.

a.b. G Ax B = 5x - 2

F Ax B = 1x - 2

h Ax B = 1xg Ax B = x - 2f Ax B = 5x

Ag ø f B Ax B = g Af Ax B B = g A @x @ B = @x @ - 2Af ø g B Ax B = f Ag Ax B B = f Ax - 2 B = @x - 2 @Ag ø f B Ax BAf ø g B Ax B

g Ax B = x - 2f Ax B = @x @g Ax2 B = x2 + 3 = x2 + 3

x2f Ax B = g Ax2 BAg ø f B Ax B = g Af Ax B B

Ax + 3 B = x2 + 6x + 9f Ax + 3 B = Ax + 3 B 2 = Ax + 3 B 2

x + 3g Ax B = f Ax + 3 BAf ø g B Ax B = f Ag Ax B B

= 4 + 3 = 7f A2 B = 22 = 4f Ax B = x2 = g A4 BAg ø f B A2 B = g Af A2 B B

= 52 = 25g A2 B = 2 + 3 = 5g Ax B = x + 3 = f A5 B

Af ø g B A2 B = f Ag A2 B B

Ag ø f B Ax BAf ø g B Ax BAg ø f B A2 BAf ø g B A2 B

g Ax B = x + 3f Ax B = x2 Practice Problem 2If and , findeach composition.

a. and

b. and Ag ø f B Ax BAf ø g B Ax BAg ø f B A3 BAf ø g B A3 B

g Ax B = 2x + 1f Ax B = x2

HELPFUL HINT

In Examples 2 and 3, notice that . In gen-eral, may or may not equal .Af ø g B Ax BAg ø f B Ax B

Ag ø f B Ax B Z Af ø g B Ax B

Practice Problem 3If and , findeach composition.

a.

b. Ag ø f B Ax BAf ø g B Ax B

g Ax B = x + 1f Ax B = 1x

Practice Problem 4

If , , and, write each function as a

composition of f, g, or h.

a.

b. G Ax B = 2x + 5

F Ax B = @x + 5 @

h Ax B = @x @g Ax B = x + 5f Ax B = 2x


GRAPHING CALCULATOR EXPLORATIONSIf and , then

.

To visualize this addition of functions with a grapher, graph

, ,

Use a TABLE feature to verify that for a given x value,. For example, verify that when , ,

and .Y3 = 2 + 4 = 6Y2 = 4Y1 = 2x = 0Y1 + Y2 = Y3

10

10

−10

−10

y = x2 + 413

y = x + 212

y = x2 + x + 613

12

Y3 = 13

x2 + 12

x + 6Y2 = 13

x2 + 4Y1 = 12

x + 2

= 13

x2 + 12

x + 6

= a 12

x + 2 b + a 13

x2 + 4 bAf + g B Ax B = f Ax B + g Ax B

g Ax B = 13

x2 + 4f Ax B = 12

x + 2

Solution: a. Notice the order in which the function F operates onan input value x. First, 2 is subtracted from x, and thenthe square root of that result is taken. This means that

. To check, we find :

b. Notice the order in which the function G operates onan input value x. First, x is multiplied by 5, and then 2is subtracted from the result. This means that

. To check, we find :

Ag ø f B Ax B = g Af Ax B B = g A5x B = 5x - 2

g ø fG = g ø f

Ah ø g B Ax B = h Ag Ax B B = h Ax - 2 B = 1x - 2

h ø gF = h ø g

11.2 INVERSE FUNCTIONS

In the next section, we begin a study of two new functions: exponential andlogarithmic functions. As we learn more about these functions, we will dis-cover that they share a special relation to each other; they are inverses ofeach other.

Before we study these functions, we need to learn about inverses. Webegin by defining one-to-one functions.

DETERMINING WHETHER A FUNCTION IS ONE-TO-ONE

Study the following table.

Recall that since each degrees Fahrenheit (input) corresponds to exactlyone degrees Celsius (output), this table of inputs and outputs does describea function. Also notice that each output corresponds to a different input.This type of function is given a special name—a one-to-one function.

Does the set , , , describe a one-to-onefunction? It is a function since each x-value corresponds to a unique y-value. For this particular function f, each y-value corresponds to a uniquex-value. Thus, this function is also a one-to-one function.

Examples Determine whether each function described is one-to-one.

1. , , ,

The function f is one-to-one since each y-value corre-sponds to only one x-value.

2. , , ,

The function g is not one-to-one because the y-value9 in and corresponds to two different x-values.

3. , , ,

The function h is one-to-one since each y-value corre-sponds to only one x-value.

4.

A-5, -5 B FA10, 10 BA2, 2 Bh = E A1, 1 BA-3, 9 BA3, 9 B

A0, 0 B FA-3, 9 BA-4, 2 Bg = E A3, 9 B

A7, 3 B FA-1, 0 BA5, 4 BfE A6, 2 B

ONE-TO-ONE FUNCTION

For a one-to-one function, each x-value (input) corresponds to only oney-value (output) and each y-value (output) corresponds to only one x-value (input).

A7, 6 B FA-3, 5 BA2, 2 Bf = E A0, 1 B

A

Inverse Functions SECTION 11.2 691

Objectives

Determine whether a function is aone-to-one function.

Use the horizontal line test todecide whether a function is a one-to-one function.

Find the inverse of a function.

Find the equation of the inverse ofa function.

Graph functions and their inverses.E

DC

B

A


Degrees Celsius (Output) 0 20 65 100-25-35

-13-31

Mineral (Input) Talc Gypsum Diamond Topaz Stibnite

Hardness on the Mohs Scale (Output) 1 2 10 8 2

Answers1. one-to-one, 2. not one-to-one, 3. one-to-one, 4. not one-to-one, 5. not one-to-one

Practice Problems 1–5Determine whether each functiondescribed is one-to-one.

1. , , ,

2. , , ,

3. , , ,

4.

5.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

A5, 6 B FA3, 4 BA1, 2 Bh = E A0, 0 BA-6, 2 B F

A2, 14 BA6, 3 Bg = E A-3, 2 BA4, -2 B F

A5, 0 BA-1, 1 Bf = E A7, 3 B

State Colo- Missis- New(Input) rado sippi Nevada Mexico Utah

Number of Colleges andUniversities (Output) 9 44 13 44 21

Source: The Chronicle of Higher Education, Vol. XLV, No. 1,

August 28, 1998.



This table does not describe a one-to-one function sincethe output 2 corresponds to two different inputs, gypsumand stibnite.

5.

This graph does not describe a one-to-one function since the y-value corresponds tothree different x-values, ,

, and 3, as shown to theright.

USING THE HORIZONTAL LINE TEST

Recall that we recognize the graph of a function when it passes the verticalline test. Since every x-value of the function corresponds to exactly one y-value, each vertical line intersects the function’s graph at most once. Thegraph shown next, for instance, is the graph of a function.

Is this function a one-to-one function? The answer is no. To see why not,notice that the y-value of the ordered pair , for example, is the sameas the y-value of the ordered pair . This function is therefore not one-to-one.

To test whether a graph is the graph of a one-to-one function, we canapply the vertical line test to see if it is a function, and then apply a similarhorizontal line test to see if it is a one-to-one function.

67

54321

1−1

−2

−2−3−4−5−6−7 2 3 4 5 6 7

(−3, 3) (3, 3)

x

y

A3, 3 BA-3, 3 B

B

-1-2

-1

x

y

(3, −1)(−2, −1)

(−1, −1) 21

345

−2−3−4−5

−2 −1−3−4 21 3 4

x

y

21

345

−2−3−4−5

−2 −1−3−4 21 3 4

HORIZONTAL LINE TEST

If every horizontal line intersectsthe graph of a function at mostonce, then the function is a one-to-one function.


Practice Problem 6Use the vertical and horizontal linetests to determine whether each graphis the graph of a one-to-one function.

a.

b.

c.

d.

e.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

Answers6. a. not one-to-one, b. not one-to-one,c. one-to-one, d. one-to-one, e. not one-to-one

Example 6 Use the vertical and horizontal line tests to determinewhether each graph is the graph of a one-to-one function.

a. b.

c. d.

e.

Solution: Graphs a, b, c, and d all pass the vertical line test, so onlythese graphs are graphs of functions. But, of these, only band c pass the horizontal line test, so only b and c aregraphs of one-to-one functions.

FINDING THE INVERSE OF A FUNCTION

One-to-one functions are special in that their graphs pass the vertical andhorizontal line tests. They are special, too, in another sense: We can find its

C

HELPFUL HINT

All linear equations are one-to-one functions except those whosegraphs are horizontal or vertical lines. A vertical line does not pass thevertical line test and hence is not the graph of a function. A horizontalline is the graph of a function but does not pass the horizontal line testand hence is not the graph of a one-to-one function.

34

21

1−1

−2−3−4

−2−3−4 2 x

y

321

1−1

−2

−2−3−4 2 3 4 x

y

4321

1−1

−2−3

−2−3−4 2 3 4 x

y

45

321

1−1

−2−3−4−5

−2−3−4 2 3 x

y

54321

1−1

−2

−2−3−4 2 3 4 x

y

inverse function for any one-to-one function by switching the coordinatesof the ordered pairs of the function, or the inputs and the outputs. Forexample, the inverse of the one-to-one function

is the function

Notice that the ordered pair of the function, for example,becomes the ordered pair of its inverse.

Also, the inverse of the one-to-one function , ,is , , . For a function f, we use the notation

, read “f inverse,” to denote its inverse function. Notice that since thecoordinates of each ordered pair have been switched, the domain (set ofinputs) of f is the range (set of outputs) of , and the range of f is thedomain of . See the definition of inverse function in the margin.

Example 7 Find the inverse of the one-to-one function:

, , ,

Solution: , , , c c c c————————————————


FINDING THE EQUATION OF THE INVERSE OF A FUNCTION

If a one-to-one function f is defined as a set of ordered pairs, we can findby interchanging the x- and y-coordinates of the ordered pairs. If a one-

to-one function f is given in the form of an equation, we can find the equa-tion of by using a similar procedure.

Example 8 Find an equation of the inverse of .

Solution:

Step 1. Replace with y.

Step 2. Interchange x and y.x = y + 3

f Ax By = x + 3

f Ax B = x + 3

f Ax B = x + 3

f-1

f-1

D

Switch coordinatesof each ordered pair.

A4, 4 B FA-6, 3 BA7, -2 Bf-1 = E A1, 0 BA4, 4 B FA3, -6 BA-2, 7 Bf = E A0, 1 B

f-1f-1

f -1A1, 9 B FA10, 5 BE A-3, 2 BA9, 1 B F A5, 10 Bf = E A2, -3 BA-35, -31 BA-31, -35 B


Answers7. , , , ,8.

✓ Concept Check: belongs to f -1A5, 1 Bf-1 Ax B = x + 6

A-8, -7 B FA0, 0 BA13, -1 Bf -1 = EA-4, 2B


Degrees Celsius (Output) 0 20 65 100-25-35

-13-31

Degrees Celsius (Input) 0 20 65 100

Degrees Fahrenheit (Output) 32 68 149 212-13-31

-25-35

INVERSE FUNCTION

The inverse of a one-to-one func-tion f is the one-to-one function

that consists of the set of allordered pairs where belongs to f.

Ax, y BAy, x Bf-1

HELPFUL HINT

The symbol is the single sym-bol used to denote the inverse ofthe function f. It is read as “finverse.” This symbol does not

mean .1f

f-1

Practice Problem 7

Find the inverse of the one-to-one function: , , ,A-7, -8 B F

A0, 0 BA-1, 13Bf = EA2, -4B

✓ CONCEPT CHECK

Suppose that is a one-to-onefunction. If the ordered pair belongs to f, name one point that weknow must belong to the inverse func-tion .f -1

A1, 5 Bf Ax B

FINDING AN EQUATION OF THE INVERSE OF A ONE-TO-ONE

FUNCTION f (x)


Step 2. Interchange x with y.

Step 3. Solve the equation for y.

Step 4. Replace y with the notation .f -1 Ax B

f Ax B

Practice Problem 8Find the equation of the inverse of

.f Ax B = x - 6


Answers

9.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f(x) = 2x + 3

f−1(x) = x − 32

f-1 Ax B = x - 32

Step 3. Solve for y.

Step 4. Replace y with .

The inverse of is . Noticethat, for example,

Example 9 Find the equation of the inverse of .Graph f and on the same set of axes.

Solution:


Step 2. Interchange x and y.



Now we graph and on the same set of axes.

Both and are linear

functions, so each graph is a line.

x x

1 10 0

0 0

GRAPHING INVERSE FUNCTIONS

Notice that the graphs of f and in Example 9 are mirror images of eachother, and the “mirror” is the dashed line . This is true for every func-tion and its inverse. For this reason, we say that the graphs of f and aresymmetric about the line .y = x

f-1y = x

f-1

E

f(x) = 3x − 5

f−1(x) =

y = x

, 0( )53

0,

(−2, 1)(−5, 0)

(1, −2)

(0, −5)

( )53 x + 5

3

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

53

53

-5-5-2-2

y = f-1 Ax By = f Ax Bf -1 Ax B = x + 5

3f Ax B = 3x - 5

f-1 Ax B = x + 53

f Ax B = 3x - 5

f -1 Ax Bf Ax Bf-1 Ax Bf-1 Ax B = x + 5

3

y = x + 53

3y = x + 5

x = 3y - 5

f Ax B y = 3x - 5

f Ax B = 3x - 5

f-1f Ax B = 3x - 5

The coordinates areswitched, as expected.

f A1 B = 1 + 3 = 4 and f-1 A4 B = 4 - 3 = 1

Ordered pair: (1, 4) Ordered pair: (4, 1)

f-1 Ax B = x - 3f Ax B = x + 3

f-1 Ax Bf-1 Ax B = x - 3

x - 3 = y

Practice Problem 9Find an equation of the inverse of

. Graph f and onthe same set of axes.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f -1f Ax B = 2x + 3

Practice Problem 10Graph the inverse of each function.a.

b. y

x

y

y = x

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f

x

f

y

y = x

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5


Answers10. a.

b.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

y = xf

f −1

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

y = x

f

f −1

To see why this happens, study the graph of a few ordered pairs and theirswitched coordinates.

Example 10 Graph the inverse of each function.

Solution: The function is graphed in blue and the inverse isgraphed in red.

a. b.

x

f

f−1

y

y = x

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5x

f

f−1

y

y = x

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

y = x

(2, 1)

(1, 2)(−4, 3)

(−5, −3)

(3, −4)(−3, −5)

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

GRAPHING CALCULATOR EXPLORATIONSA grapher can be used to visualize functions and theirinverses. Recall that the graph of a function f and its inverse

are mirror images of each other across the line . To seethis for the function , use a square window andgraph

the given function:

its inverse:

and the line:

y = 3x + 2 y = x

10

15−15

−10

y = x − 23

Y3 = x

Y2 = x - 23

Y1 = 3x + 2

f Ax B = 3x + 2y = xf-1

11.3 EXPONENTIAL FUNCTIONS

In earlier chapters, we gave meaning to exponential expressions such as , where x is a rational number. For example,

Three factors; each factor is 2

Three factors; each factor is

When x is an irrational number (for example, ), what meaning can wegive to ?

It is beyond the scope of this book to give precise meaning to if x isirrational. We can confirm your intuition and say that is a real number,and since , then . We can also use a calculatorand approximate : . In fact, as long as the base b is posi-tive, is a real number for all real numbers x. Finally, the rules of expo-nents apply whether x is rational or irrational, as long as b is positive.

In this section, we are interested in functions of the form ,where . A function of this form is called an exponential function.

GRAPHING EXPONENTIAL FUNCTIONS

Now let’s practice graphing exponential functions.

Example 1 Graph the exponential functions defined by and on the same set of axes.

Solution: To graph these functions, we find some ordered pairsolutions, plot the points, and connect them with asmooth curve.

g Ax B = 3xf Ax B = 2x

A

b 7 0f Ax B = bx

bx213 L 3.321997213

21 6 213 6 221 6 13 6 2213

2x213

13

1223> 2 = A21> 2 B 3 = 12 # 12 # 12

23 = 2 # 2 # 2

2x

Exponential Functions SECTION 11.3 697

Objectives

Graph exponential functions.

Solve equations of the form.

Solve problems modeled by expo-nential equations.

Cbx = by

BA

EXPONENTIAL FUNCTION

A function of the form

is called an exponential function if , b is not 1, and x is a realnumber.

b 7 0

f Ax B = bx

f Ax B = 2x

x 0 1 2 3

1 2 4 814

12

f Ax B-2-1

g Ax B = 3x

x 0 1 2 3

1 3 9 2719

13

g Ax B-2-1

Practice Problem 1Graph the exponential function

.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f Ax B = 4x

Answer1.

f(x) = 4x

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5


A number of things should be noted about the two graphs of exponen-tial functions in Example 1. First, the graphs show that and

are one-to-one functions since each graph passes the verticaland horizontal line tests. The y-intercept of each graph is 1, but neithergraph has an x-intercept. From the graph, we can also see that the domainof each function is all real numbers and that the range is . We canalso see that as x-values are increasing, y-values are increasing also.

Example 2 Graph the exponential functions and


Solution: As before, we find some ordered pair solutions, plot thepoints, and connect them with a smooth curve.

Each function in Example 2 again is a one-to-one function. The y-inter-cept of both is 1. The domain is the set of all real numbers, and the range is

.Notice the difference between the graphs of Example 1 and the graphs

of Example 2. An exponential function is always increasing if the base is

A0, q B

56

y = x

7

4321

1−1−2−3−4 2 3 x

y

( )21

y = x( )31

y = a 13b x

y = a 12b x

A0, q B

g Ax B = 3xf Ax B = 2x

56789

4321

1

g(x) = 3x

f(x) = 2x

−1−2 2 3 4 x

y


y = a 12b x

x 0 1 2 3

y 1 2 418

14

12

-2-1

y = a 13b x

x 0 1 2 3

y 1 3 9127

19

13

-2-1

Answer2.

f(x) = )( x15

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5


.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f Ax B = a 15b x

greater than 1. When the base is between 0 and 1, the graph is alwaysdecreasing. The following figures summarize these characteristics of expo-nential functions.

Example 3 Graph the exponential function .

Solution: As before, we find and plot a few ordered pair solutions.Then we connect the points with a smooth curve.


SOLVING EQUATIONS OF THE FORM bx � by

We have seen that an exponential function is a one-to-one function.Another way of stating this fact is a property that we can use to solve expo-nential equations.

Example 4 Solve:

Solution: We write 16 as a power of 2 and then use the uniquenessof to solve.

2x = 24

2x = 16

bx

2x = 16

y = bx

B

x

y

42

68

10

−4−6−8

−10

−4 −2−6−8−10 42 6 8 10

f(x) = 3x + 2

f Ax B = 3x + 2

56

4321

1−1−2−3 2 3

(1, b)

f(x) = bx,for 0 < b < 1

(0, 1)x

y

56

4321

1−1−2−3−4 2 3

(1, b)

f(x) = bx,for b > 1

(0, 1)

x

y


Answers3.

✓ Concept Check: b and c are exponential

f(x) = 2x−1

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

y = 3x + 2x 0

y 9 3 119

13

-4-3-2-1


.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f Ax B = 2x - 1

✓ CONCEPT CHECK

Which functions are exponential func-tions?

a. b.

c. d. w Ax B = A2x B 2h Ax B = 5x - 2

g Ax B = a 23b x

f Ax B = x3

UNIQUENESS OF bx

Let and . Then is equivalent to .x = ybx = byb Z 1b 7 0


Answers

4. , 5. , 6. E-2Fe 32fE3F


Since the bases are the same and are nonnegative, by theuniqueness of we then have that the exponents areequal. Thus,

To check, we replace x with 4 in the original equation.The solution set is .

Example 5 Solve:

Solution: Since both 9 and 27 are powers of 3, we can use theuniqueness of .

Write 9 and 27 as powers of 3.

Use the uniqueness of .


To check, we replace x with in the original equation.


Example 6 Solve:

Solution: We write both 4 and 8 as powers of 2, and then use theuniqueness of .



Check to see that the solution set is .

There is one major problem with the preceding technique. Often the twosides of an equation, for example, cannot easily be written as pow-ers of a common base. We explore how to solve such an equation with thehelp of logarithms later.

SOLVING PROBLEMS MODELED BY EXPONENTIAL EQUATIONS

The bar graph on the next page shows the increase in the number of cellu-lar phone users. Notice that the graph of the exponential function

approximates the heights of the bars. This is just oneexample of how the world abounds with patterns that can be modeled byexponential functions. To make these applications realistic, we use num-bers that warrant a calculator. Another application of an exponential func-tion has to do with interest rates on loans.

y = 3.638 A1.443 Bx

C

4 = 3x

E6F 6 = x

bx 2x + 6 = 3x

22x + 6 = 23x

A22 Bx + 3 = A23 Bx 4x + 3 = 8x

bx

4x + 3 = 8x

e 32f

32

x = 32

bx 2x = 3

32x = 33

A32 Bx = 33

9x = 27

bx

9x = 27

E4Fx = 4

bx

Practice Problem 6Solve: 9x - 1 = 27x



Answer7. approximately 5.08%

Practice Problem 7As a result of the Chernobyl nuclearaccident, radioactive debris was car-ried through the atmosphere. One im-mediate concern was the impact thatdebris had on the milk supply. The per-cent y of radioactive material in rawmilk after t days is estimated by

. Estimate the ex-pected percent of radioactive materialin the milk after 30 days.

y = 100 A2.7 B-0.1t

The exponential function defined by models the pat-

tern relating the dollars A accrued (or owed) after P dollars are invested (or loaned) at an annual rate of interest r compounded n times each yearfor t years. This function is known as the compound interest formula.

Example 7 Using the Compound Interest FormulaFind the amount owed at the end of 5 years if $1600 isloaned at a rate of 9% compounded monthly.

Solution: We use the formula , with the follow-ing values:

(the amount of the loan)

(the annual rate of interest)

(the number of times interest is compounded each year)

(the duration of the loan, in years)

Compound interest formula

Substitute known values.

To approximate A, use the or key on your cal-culator.

Thus, the amount A owed is approximately $2505.09.

When interest is compounded continuously, the formula is used,where r is the annual interest rate and interest is compounded continuouslyfor t years. The number e used in the formula is an irrational numberapproximately equal to 2.7183.

A = Pert

2505.0896

^yx

= 1600 A1.0075 B 60

= 1600 a1 + 0.0912b 12A5B

A = P a1 + rnb nt

t = 5

n = 12

r = 9% = 0.09

P = $1600

A = P a1 + rnb nt

A = P a1 + rnb nt

50

45

40

35

30

25

20

15

10

5

0

Num

ber

of S

ubsc

ribe

rs(i

n m

illio

ns)

1991 1994 1996199519931992Year

1989 1990

Cellular Phone Users

y = 3.638(1.443)x

where x = 0 correspondsto the year 1989, and so on.

Example 8 Finding the Amount Owed on a LoanFind the amount owed at the end of 5 years if $1600 isloaned at a rate of 9% compounded continuously.

Solution: We use the formula , where

(the amount of the loan)

(the rate of interest)

(the 5-year duration of the loan)


Now we can use a calculator to approximate the solution.

The total amount of money owed is approximately$2509.30.

A L 2509.30

= 1600e0.45

= 1600e0.09A5BA = Pert

t = 5

r = 9% = 0.09

P = $1600

A = Pert


GRAPHING CALCULATOR EXPLORATIONSWe can use a graphing calculator and its TRACE feature to solve PracticeProblem 7 graphically.

To estimate the expected percent of radioactive material in the milk after 30 days, enter . The graph does not appear on a standardviewing window, so we need to determine an appropriate viewing window. Becauseit doesn’t make sense to look at radioactivity before the Chernobyl nuclear acci-dent, we use . We are interested in finding the percent of radioactivematerial in the milk when , so we choose to leave enough spaceto see the graph at . Because the values of y are percents, it seems appro-priate that . (We also use and .) Now we graph the function.

We can use the TRACE feature to obtain an approximation of the expectedpercent of radioactive material in the milk when . (A TABLE feature mayalso be used to approximate the percent.) To obtain a better approximation, let’suse the ZOOM feature several times to zoom in near .

The percent of radioactive material in the milk 30 days after the Chernobylaccident was 5.08%, accurate to two decimal places.

Use a grapher to find each percent. Approximate your solutions so that they are accurate to two decimal places.

1. Estimate the expected percent of radioactive material in the milk 2 days after the Chernobyl nuclear acci-dent.




x = 30

x = 30

X = 30.001039 Y = 5.0800021

1

Yscl = 10Xscl = 10 ≤ y ≤ 100x = 30

Xmax = 35x = 30Xmin = 0

Y1 = 100 A2.7 B-0.1x

350

100

Practice Problem 8Find the amount owed at the end of 3years if $1200 is loaned at a rate of 8%compounded continuously.

Answer8. $1525.50

703


EXERCISE SET 11.3

Graph each exponential function. See Examples 1 through 3.

1. 2. 3. 4.

5. 6. 7. 8.

9. 10.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = -3xy = -2x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = a 13b x

+ 2y = a 12b x

- 2y = a 15b x

y = a 14b x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = 3x - 1y = 1 + 2xy = 5xy = 4x

A

11.4 LOGARITHMIC FUNCTIONS

USING LOGARITHMIC NOTATION

Since the exponential function is a one-to-one function, it has aninverse. We can create a table of values for by switching the coordinatesin the accompanying table of values for .

x x

0 1 1 01 2 2 12 4 4 23 8 8 3

The graphs of and its inverse are shown in the margin. Notice that thegraphs of f and are symmetric about the line , as expected.

Now we would like to be able to write an equation for . To do so, wefollow the steps for finding the equation of an inverse:

Step 1. Replace by y.



At this point, we are stuck. To solve this equation for y, a new notation, thelogarithmic notation, is needed.

The symbol means “the power to which b is raised to produce aresult of x.” In other words,

We say that is “the logarithm of x to the base b” or “the log of x tothe base b.” See the logarithmic definition in the margin.

Before returning to the function and solving it for y in terms ofx, let’s practice using the new notation .

It is important to be able to write exponential equations with logarithmicnotation, and vice versa. The following table shows examples of both forms.

logbxx = 2y

logbx

logb x = y means by = x

logbx

x = 2y

y = 2xf Ax Bf Ax B = 2x

f -1y = xf-1

f Ax B

-112

12

-1

-214

14

-2

-318

18

-3

y = f -1 Ax By = f Ax Bf Ax B = 2x

f-1f Ax B = 2x

A

Logarithmic Functions SECTION 11.4 705

Objectives

Write exponential equations withlogarithmic notation and write log-arithmic equations with exponen-tial notation.

Solve logarithmic equations byusing exponential notation.

Identify and graph logarithmicfunctions.

C

B

A

LOGARITHMIC DEFINITION

If , and , then

for every and every realnumber y.

x 7 0

y = logb x means x = by

b Z 1b 7 0


34

21

1−1

−2−3−4

−2−3−4 2 3 4 5

f(x) = 2x

f −1(x)

x

y

Logarithmic Equation Corresponding Exponential Equation

81>3 = 2 log8 2 = 13

4-2 = 116

log4 116

= -2

23 = 8 log2 8 = 3 60 = 1 log6 1 = 0 32 = 9 log3 9 = 2

HELPFUL HINT

Notice that a logarithm is an expo-nent. In other words, log3 9 is thepower that we raise 3 to in order toget 9.

Examples Write as an exponential equation.

1. means

2. means

3. means

Examples Write as a logarithmic equation.

4. means

5. means

6. means

Example 7 Find the value of each logarithmic expression.

a.

b.

c.

Solution: a. because .

b. because .

c. because .

SOLVING LOGARITHMIC EQUATIONS

The ability to interchange the logarithmic and exponential forms of a state-ment is often the key to solving logarithmic equations.

Example 8 Solve:

Solution:

Write as an exponential equation.


Example 9 Solve:

Solution:


Even though , the base b of a logarithm mustbe positive. The solution set is .

Example 10 Solve:

Solution:


Write 1 as .


The solution set is .E0Fbx x = 0

30 3x = 30

3x = 1

log3 1 = x

log3 1 = x

E5FA-5 B 2 = 25

x = 5

x2 = 25

logx 25 = 2

logx 25 = 2

E125F 125 = x

53 = x

log5 x = 3

log5 x = 3

B

91>2 = 19 = 3log9 3 = 12

10-1 = 110

log10 110

= -1

42 = 16log4 16 = 2

log9 3

log10 110

log4 16

log5 315 = 1

351>3 = 315

log6 136

= -26-2 = 136

log9 729 = 393 = 729

21> 2 = 12log2 12 = 12

6-1 = 16

log6 16

= -1

52 = 25log5 25 = 2


Answers

1. , 2. , 3. ,

4. , 5. ,

6. , 7. a. 3, b. , c.

8. , 9. , 10. E0FE3FE16F12

-2log7 317 = 1

3

log2 18

= -3log3 81 = 4

31>2 = 138-1 = 18

72 = 49

PROPERTIES OF LOGARITHMS

If b is a real number, and, then

1.2.3. blogbx = x

logb bx = x

logb 1 = 0b Z 1

b 7 0

Practice Problems 1–3Write as an exponential equation.

1. 2.

3. log3 13 = 12

log8 18

= -1log7 49 = 2

Practice Problems 4–6Write as a logarithmic equation.

4. 5.

6. 71>3 = 317

2-3 = 18

34 = 81

Practice Problem 7Find the value of each logarithmicexpression.

a. b.

c. log25 5

log3 19

log2 8

Practice Problem 8Solve: log2 x = 4

Practice Problem 9Solve: logx 9 = 2

Practice Problem 10Solve: log2 1 = x

Practice Problem 11Simplify.

a. b.

c. d. 3log3 107log7 5

log11 11-4log6 63

In Example 10, we illustrated an important property of logarithms. Thatis, is always 0. This property as well as two important others are givenin the margin on page 706.

To see that , change the logarithmic form to exponentialform. Then, means . In exponential form, the statementis true, so in logarithmic form, the statement is also true.

Example 11 Simplify.

a. b.c. d.

Solution: a. From property 2, .b. From property 2, .c. From property 3, .d. From property 3, .

GRAPHING LOGARITHMIC FUNCTIONS

Let us now return to the function and write an equation for itsinverse, . Recall our earlier work.

Step 1. Replace by y.


Having gained proficiency with the notation , we can now completethe steps for writing the inverse equation.



Thus, defines a function that is the inverse function of thefunction . The function or is called a logarith-mic function.

TRY THE CONCEPT CHECK IN THE MARGIN.We can explore logarithmic functions by graphing them.

Example 12 Graph the logarithmic function .

Solution: First we write the equation with exponential notation as. Then we find some ordered pair solutions that

satisfy this equation. Finally, we plot the points and con-nect them with a smooth curve. The domain of this func-tion is , and the range is all real numbers.

Since is solved for x, we choose y-values andcompute corresponding x-values.

If ,

If ,

If ,

If , x = 2-1 = 12

y = -1

x = 22 = 4y = 2

x = 21 = 2y = 1321

4

112

−1

−2−3

−2−3 2 3 4 5

(2, 1)(4, 2)

(1, 0)

, −1(

y = log2 x

x

y

6

)

x = 20 = 1y = 0

x = 2yA0, q B

2y = x

y = log2 x

y = log2 xf-1 Ax Bf Ax B = 2x

f-1 Ax B = log2 x

f-1 Ax B = log2 xf-1 Ax B y = log2 x

logb x

x = 2y

y = 2x f Ax B f Ax B = 2x

f-1 Ax B f Ax B = 2x

C2log26 = 65log53 = 3log7 7

-1 = -1log3 3

2 = 2

2log265log53log7 7

-1log3 32

bx = bxlogb bx = x

logb bx = x

logb 1

Logarithmic Functions SECTION 11.4 707

Answers

11. a. 3, b. , c. 5, d. 10

✓ Concept Check: ; answers may varyA9, 2 B-4

LOGARITHMIC FUNCTION

If x is a positive real number, b is aconstant positive real number, andb is not 1, then a logarithmic func-tion is a function that can bedefined by

The domain of f is the set of posi-tive real numbers, and the range off is the set of real numbers.

f Ax B = logb x

✓ CONCEPT CHECK

Let and .These two functions are inverses ofeach other. Since is an orderedpair solution of , what orderedpair do we know to be a solution of

? Explain why.f Ax Bg Ax BA2, 9 B

g Ax B = 3xf Ax B = log3 x

y

1 02 14 2

-112

x = 2y

Example 13 Graph the logarithmic function .

Solution: We can replace with y, and write the result withexponential notation.

Replace with y.

Write in exponential form.

Now we can find ordered pair solutions that satisfy

, plot these points, and connect them with a

smooth curve.

If ,

If ,

If ,

If ,

The domain of this function is , and the range isthe set of all real numbers.

The following figures summarize characteristics of logarithmic functions.

21

43

5

1−1

−2−3

−2 2 3 4 5

(b, 1)

(1, 0)

f(x) = logb x,if 0 < b < 1

x

y

6 721

43

67

5

1−1

−2−3

−2 2 3 4 5

(b, 1)

(1, 0)

f(x) = logb x, if b > 1

x

y

6 7

A0, q B

21

1

13

−1

−2−3

−2 2 3 4 5(3, −1)

(9, −2)

(1, 0)

, 1( )

f(x) = log1/3x

x

y

6 7 8 9

x = a 13b -2

= 9y = -2

x = a 13b -1

= 3y = -1

x = a 13b 1

= 13

y = 1

x = a 13b 0

= 1y = 0

a 13b y

= x

a 13b y

= x

f Ax B y = log1>3 xf Ax B = log1>3 x

f Ax Bf Ax B = log1>3 x

Practice Problem 13Graph the logarithmic function

.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

f Ax B = log1>2 x


Answers12.

y = log4x

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

y

1 0

1

39 -2

-1

13

x = a 13b y

Answer13.

f(x) = log1/2x

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

Practice Problem 12Graph the logarithmic function

.

x

y

21

345

−2−3−4−5

−2 −1−3−4−5 21 3 4 5

y = log4 x

EXERCISE SET 11.4

Write each as an exponential equation. See Examples 1 through 3.

1. 2. 3.

4. 5.

Write each as a logarithmic equation. See Examples 4 through 6.

6. 7. 8.

9. 10.

Find the value of each logarithmic expression. See Example 7.

11. 12. 13.

14. 15.

Solve. See Examples 8 through 10.

16. 17. 18.

19. 20. logx 49 = 2log2 x = 3

log3 x = 4log2 8 = xlog3 9 = x

B

log25 5log2 132

log3 19

log3 9log2 8

e3 = x104 = 1000

102 = 10053 = 12524 = 16

log10 1000 = 3log5 125

= -2

log3 127

= -3log2 32 = 5log6 36 = 2

A

709


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Name _________________________________________________________________________

710

Graph each logarithmic function. See Examples 12 and 13.

21. 22. 23.

24. 25.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

f Ax B = log5 xf Ax B = log1>2 x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

f Ax B = log1>4 xy = log2 xy = log3 xC

11.5 PROPERTIES OF LOGARITHMS

In the previous section we explored some basic properties of logarithms.We now introduce and explore additional properties. Because a logarithmis an exponent, logarithmic properties are just restatements of exponentialproperties.

USING THE PRODUCT PROPERTY

The first of these properties is called the product property of logarithmsbecause it deals with the logarithm of a product.

To prove this, we let and . Now we write each log-arithm with exponential notation.

is equivalent tois equivalent to

When we multiply the left sides and the right sides of the exponential equa-tions, we have that

If we write the equation in equivalent logarithmic form, wehave

But since and , we can write

Let and .

In other words, the logarithm of a product is the sum of the logarithmsof the factors. This property is sometimes used to simplify logarithmicexpressions.

Example 1 Write as a single logarithm:

Solution: Use the product property.


Solution:

USING THE QUOTIENT PROPERTY

The second property is the quotient property of logarithms.

B

log2 Ax + 2 B + log2 x = log2 C Ax + 2 B # x D = log2 Ax2 + 2x B

log2 Ax + 2 B + log2 x

= log11 30

log11 10 + log11 3 = log11 A10 # 3 Blog11 10 + log11 3

N = logb yM = logb xlogb xy = logb x + logb y

N = logb yM = logb x

logb xy = M + N

xy = bM + N

xy = AbM B AbN B = bM + N

bN = ylogb y = NbM = xlogb x = M

logb y = Nlogb x = M

A

Properties of Logarithms SECTION 11.5 711

Objectives

Use the product property of loga-rithms.

Use the quotient property of loga-rithms.

Use the power property of loga-rithms

Use the properties of logarithmstogether.

D

C

B

A

PRODUCT PROPERTY OF LOGARITHMS

If x, y, and b are positive real numbers and , then

logb xy = logb x + logb y

b Z 1

Practice Problem 1Write as a single logarithm:log2 7 + log2 5

Answers1. , 2. log3 Ax2 - 9x Blog2 35

Practice Problem 2Write as a single logarithm:log3 x + log3 Ax - 9 B


The proof of the quotient property of logarithms is similar to the proof ofthe product property. Notice that the quotient property says that the loga-rithm of a quotient is the difference of the logarithms of the dividend anddivisor.



Solution: Use the quotient property.


Solution:

Use the quotient property.

USING THE POWER PROPERTY

The third and final property we introduce is the power property of loga-rithms.

Examples Use the power property to rewrite each expression.

5.

6.

USING MORE THAN ONE PROPERTY

Many times we must use more than one property of logarithms to simplifya logarithmic expression.

Examples Write as a single logarithm.

7.

= log5 72 = log5 A9 # 8 B = log5 9 + log5 8

Use the powerproperty.

Use the productproperty.

2 log5 3 + 3 log5 2 = log5 32 + log5 2

3

D

log4 12 = log4 21>2 = 1

2 log4 2

log5 x3 = 3 log5 x

C

log3 Ax2 + 5B - log3 Ax2 + 1B = log3 x2 + 5x2 + 1

log3 Ax2 + 5B - log3 Ax2 + 1B = log10 9

log10 27 - log10 3 = log10 273

log10 27 - log10 3


QUOTIENT PROPERTY OF LOGARITHMS

If x, y, and b are positive real numbers and , then

logb xy

= logb x - logb y

b Z 1

Answers3. , 4. , 5. ,

6. , 7. , 8, 8.


log2 A2x - 1 B 5

xlog4 200

13

log7 4

5 log3 xlog3 x3 + 4x2 + 2

log7 5

POWER PROPERTY OF LOGARITHMS

If x and b are positive real numbers, , and r is a real number,then

logb xr = r logb x

b Z 1

✓ CONCEPT CHECK

Which of the following is the correct

way to rewrite ?

a.b.

c.

d. log5 14

log5 7log5 2

log5 A7 - 2 Blog5 7 - log5 2

log5 72

Practice Problem 3Write as a single logarithm:log7 40 - log7 8

Practice Problem 4Write as a single logarithm:log3 Ax3 + 4 B - log3 Ax2 + 2 B


5. 6. log7 314log3 x

5

Practice Problems 7–8Write as a single logarithm.

7.

8. 5 log2 A2x - 1 B - log2 x

3 log4 2 + 2 log4 5

Use the power property to rewriteeach expression.

8.

Examples Write each expression as sums or differences oflogarithms.

9. Use the quotient property.

Use the product property.

10. Use the quotient property.

Use the power property.


Examples If and , use the properties oflogarithms to evaluate each expression.

11. Write 6 as .


Substitute given values.

Simplify.

12. Write 9 as .


Substitute the given value.

Simplify.

13. Write as .


Substitute the given value.

Simplify. = 0.215

= 12

A0.43 B = 1

2 logb 2

21>212logb 12 = logb 21>2

= 1.36 = 2 A0.68 B = 2 logb 3

32logb 9 = logb 32

= 1.11 = 0.43 + 0.68 = logb 2 + logb 3

2 # 3logb 6 = logb A2 # 3 Blogb 3 = 0.68logb 2 = 0.43

= 5 log2 x - 2 log2 y

log2 x5

y2 = log2 Ax5 B - log2 Ay2 B = log3 5 + log3 7 - log3 4

log3 5 # 7

4= log3 A5 # 7 B - log3 4

= log9 x3

x + 1


Use the quotientproperty.

3 log9 x - log9 Ax + 1 B = log9 x3 - log9 Ax + 1 B

Properties of Logarithms SECTION 11.5 713

Answers9. ,10. , 11. 2.07, 12. 2.42,13. 0.28

✓ Concept Check: The properties do not giveany way to simplify the logarithm of a sum;answers may vary

64 log3 x - 3 log3 y

log7 6 + log7 2 - log7 5

HELPFUL HINT

Notice that we are not able to simplify further a logarithmic expres-sion such as . None of the basic properties gives a wayto write the logarithm of a difference in some equivalent form.

log5 A2x - 1 B

Practice Problems 9–10Write each expression as sums or dif-ferences of logarithms.

9.

10. log3 x4

y3

log7 6 # 2

5

✓ CONCEPT CHECK

What is wrong with the following?

Use a numerical example to demon-strate that the result is incorrect.

= 2 log10 x + log10 5log10 Ax2 + 5 B = log10 x

2 + log10 5

Practice Problems 11–13If and , usethe properties of logarithms to evalu-ate each expression.

11.

12.

13. logb 314

logb 49

logb 28

logb 7 = 1.21logb 4 = 0.86


Focus On the Real WorldSOUND INTENSITY

The decibel (dB) measures sound intensity, or the relative loudness or strength of a sound.One decibel is the smallest difference in sound levels that is detectable by humans. The decibelis a logarithmic unit. This means that for approximately every 3-decibel increase in soundintensity, the relative loudness of the sound is doubled. For example, a 35 dB sound is twice asloud as a 32 dB sound.

In the modern world, noise pollution has increasingly become a concern. Sustained exposureto high sound intensities can lead to hearing loss. Regular exposure to 90 dB sounds can even-tually lead to loss of hearing. Sounds of 130 dB and more can cause permanent loss of hearinginstantaneously.

The relative loudness of a sound D indecibels is given by the equation

where I is the intensity of a sound givenin watts per square centimeter. Somesound intensities of common noises arelisted in the table in order of increasingsound intensity.

GROUP ACTIVITY

1. Work together to create a table of therelative loudness (in decibels) of thesounds listed in the table.

2. Research the loudness of other common noises. Add these sounds and their decibel levels toyour table. Be sure to list the sounds in order of increasing sound intensity.

D = 10 log10 I

10-16

SOME SOUND INTENSITIES OF COMMON NOISES

IntensityNoise (watts/cm2)

Whispering

Rustling leaves

Normal conversation

Background noise in a quiet residence

Typewriter

Air conditioning

Freight train at 50 feet

Vacuum cleaner

Nearby thunder

Air hammer

Jet plane at takeoff

Threshold of pain 10-4

10-6

10-6.5

10-7

10-8

10-8.5

10-10

10-11

10-12.2

10-13

10-14.2

10-15

11.6 COMMON LOGARITHMS, NATURAL LOGARITHMS, AND CHANGE OF BASE

In this section we look closely at two particular logarithmic bases. Thesetwo logarithmic bases are used so frequently that logarithms to their basesare given special names. Common logarithms are logarithms to base 10.Natural logarithms are logarithms to base e, which we introduced in Section11.3. The work in this section is based on the use of the calculator which hasboth the common “log” and the natural “log” keys.

APPROXIMATING COMMON LOGARITHMS

Logarithms to base 10—common logarithms—are used frequently becauseour number system is a base 10 decimal system. The notation log x meansthe same as .

Example 1 Use a calculator to approximate log 7 to four decimalplaces.

Solution: Press the following sequence of keys:

or

To four decimal places,

EVALUATING COMMON LOGARITHMS OF POWERS OF 10To evaluate the common log of a power of 10, a calculator is not needed.According to the property of logarithms,

It follows that if b is replaced with 10, we have

Examples Find the exact value of each logarithm.

2.

3.

4.

5.

As we will soon see, equations containing common logs are useful mod-els of many natural phenomena.

log 110 = log 101>2 = 12

log 1000 = log 103 = 3

log 110

= log 10-1 = -1

log 10 = log 101 = 1

HELPFUL HINT

The base of this logarithm is understood to be 10.

log 10x = x

logb bx = x

B

log 7 L 0.8451

ENTER7LOGLOG7

COMMON LOGARITHMS

log x means log10 x

log10 x

A

LNLOG

Common Logarithms, Natural Logarithms, and Change of Base SECTION 11.6 715

Objectives

Identify common logarithms andapproximate them with a calculator.

Evaluate common logarithms ofpowers of 10.

Identify natural logarithms and ap-proximate them with a calculator.

Evaluate natural logarithms ofpowers of e.

Use the change of base formula.E

D

C

B

A

Practice Problem 1 Use a calculator to aproximate log 21to four decimal places.

Answers

1. 1.3222, 2. 2, 3. , 4. 4, 5. 13

-2

Practice Problems 2–5Find the exact value of each logarithm.

2. log 100

3.

4. log 10,000

5. log 3110

log 1

100


Example 6 Solve: . Give an exact solution, and thenapproximate the solution to four decimal places.

Solution: Remember that the base of a common log is understoodto be 10.

Write with exponential notation.

The exact solution is . To four decimal places,.

APPROXIMATING NATURAL LOGARITHMS

Natural logarithms are also frequently used, especially to describe naturalevents; hence the label “natural logarithm.” Natural logarithms are loga-rithms to the base e, which is a constant approximately equal to 2.7183. Thenotation is usually abbreviated to ln x. (The abbreviation ln is read“el en.”)

Example 7 Use a calculator to approximate ln 8 to four decimalplaces.

Solution: Press the following sequence of keys:

or

To four decimal places,

EVALUATING NATURAL LOGARITHMS OF POWERS OF eAs a result of the property , we know that , or

.

Examples Find the exact value of each natural logarithm.

8.

9.

Example 10 Solve: . Give an exact solution, and thenapproximate the solution to four decimal places.

Solution: Remember that the base of a natural logarithm is under-stood to be e.

Write with exponential notation.

Solve for x.

The exact solution is . To four decimal places,

x L 49.4711.

e5

3

e5

3= x

e5 = 3x

ln 3x = 5

ln 3x = 5

ln 51e = ln e1>5 = 15

ln e3 = 3

ln ex = xloge e

x = xlogb bx = x

D

ln 8 L 2.0794

ENTER8lnln8

NATURAL LOGARITHMS

ln x means loge x

loge x

C

x L 15.8489101.2

101.2 = x

log x = 1.2

log x = 1.2


Answers6. , 7. 2.3979,

8. 5, 9. , 10. x = e10

7; x L 3146.6380

12

x = 102.9; x L 794.3282

HELPFUL HINT

The under-stood baseis 10.

HELPFUL HINT

The under-stood baseis e.

Practice Problem 6Solve: . Give an exact solu-tion, and then approximate the solu-tion to four decimal places.

log x = 2.9

Practice Problem 7Use a calculator to approximate ln 11to four decimal places.

Practice Problems 8–9Find the exact value of each naturallogarithm.

8. 9. ln 1eln e5

Practice Problem 10Solve: . Give an exact solu-tion, and then approximate the solu-tion to four decimal places.

ln 7x = 10

USING THE CHANGE OF BASE FORMULA

Calculators are handy tools for approximating natural and common loga-rithms. Unfortunately, some calculators cannot be used to approximatelogarithms to bases other than e or 10—at least not directly. In such cases,we use the change of base formula.

Example 11 Approximate to four decimal places.

Solution: We use the change of base property to write as aquotient of logarithms to base 10.

Use the change of base property.

Approximate the logarithms by calculator.

Simplify by calculator.

To four decimal places, .


log5 3 L 0.6826

L 0.6826063

L 0.47712130.69897

log5 3 =log 3log 5

log5 3

log5 3

E

Common Logarithms, Natural Logarithms, and Change of Base SECTION 11.6 717

Answer11. 0.8271

✓ Concept Check: f Ax B =log xlog 5

CHANGE OF BASE

If a, b, and c are positive real numbers and neither b nor c is 1, then

logb a =logc alogc b

Practice Problem 11Approximate to four decimalplaces.

log7 5

✓ CONCEPT CHECK

If a graphing calculator cannotdirectly evaluate logarithms to base 5,describe how you could use the graph-ing calculator to graph the function

.f Ax B = log5 x


Focus On HistoryTHE INVENTION OF LOGARITHMS

Logarithms were the invention of John Napier (1550–1617), a Scottish land owner and theolo-gian. Napier was also fascinated by mathematics and made it his hobby. Over a period of 20years in his spare time, he developed his theory of logarithms, which were explained in hisLatin text Mirifici Logarithmorum Canonis Descriptio (A Description of an Admirable Tableof Logarithms), published in 1614. He hoped that his discovery would help to simplify themany time-consuming calculations required in astronomy. In fact, Napier’s logarithms revolu-tionized astronomy and many other advanced mathematical fields by replacing “the multiplica-tions, divisions, square and cubical extractions of great numbers, which besides the tediousexpense of time are for the most part subject to many slippery errors” with related numbersthat can be easily added and subtracted instead. His discovery was a great time-saving device.Some historians suggest that the use of logarithms to simplify calculations enabled Germanastronomer Johannes Kepler to develop his three laws of planetary motion, which in turnhelped English physicist Sir Isaac Newton develop his theory of gravitation. Two hundredyears after Napier’s discovery, the French mathematician Pierre de Laplace wrote that loga-rithms, “by shortening the labors, doubled the life of the astronomer.”

Napier’s original logarithm tables had several flaws: They did not actually use a particularlogarithmic base per se and log 1 was not defined to be equal to 0. An English mathematician,Henry Briggs, read Napier’s Latin text soon after it was published and was very impressed byhis ideas. Briggs wrote to Napier, asking to meet in person to discuss his wonderful discoveryand to offer several improvements. The two mathematicians met in the summer of 1615. Briggssuggested redefining logarithms to base 10 and defining log . Napier had also thought ofusing base 10 but hadn’t been well enough to start a new set of tables. He asked Briggs toundertake the construction of a new set of base 10 tables. And so it was that the first table ofcommon logarithms was constructed by Briggs over the next two years. Napier died in 1617before Briggs was able to complete his new tables.

CRITICAL THINKING

Locate a table of common logarithms and describe how to use it. Give several examples. Explainwhy a table of common logarithms would have been invaluable to many calculations before theinvention of the hand-held calculator.

1 = 0

719


EXERCISE SET 11.6

Use a calculator to approximate each logarithm to four decimal places. See Exam-ples 1 and 7.

1. log 8 2. log 6 3. log 2.31 4. log 4.86

5. ln 2 6. ln 3 7. ln 0.0716 8. ln 0.0032

9. log 12.6 10. log 25.9

Find the exact value of each logarithm. See Examples 2 through 5, 8, and 9.

11. log 100 12. log 10,000 13. 14.

15. 16. 17. 18.

19. 20. ln e5log 103

ln 51eln 41eln e4ln e2

log 110

log 1

1000

DB

CA

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

11.7 EXPONENTIAL AND LOGARITHMIC EQUATIONS

AND PROBLEM SOLVING

SOLVING EXPONENTIAL EQUATIONS

In Section 11.3 we solved exponential equations such as by writing16 as a power of 2 and using the uniqueness of :

Write 16 as .


To solve an equation such as , we use the fact that isa one-to-one function. Another way of stating this fact is as a property ofequality.

Example 1 Solve: . Give an exact answer and a four-decimal-place approximation.

Solution: We use the logarithm property of equality and take thelogarithm of both sides. For this example, we use thecommon logarithm.

Take the common log of both sides.

Use the power property of logarithms.

Divide both sides by log 3.

The exact solution is . To approximate to four deci-

mal places, we have

The solution set is , or approximately .

SOLVING LOGARITHMIC EQUATIONS

By applying the appropriate properties of logarithms, we can solve a broadvariety of logarithmic equations.

Example 2 Solve:

Solution: Notice that must be positive, so x must be greaterthan 2. With this in mind, we first write the equation withexponential notation.

x - 2

log4 Ax - 2 B = 2

B

E1.7712Fe log 7log 3

f

log 7log 3

L 0.8450980.4771213

L 1.7712

log 7log 3

x =log 7log 3

x log 3 = log 7

log 3x = log 7

3x = 7

3x = 7

f Ax B = logb x3x = 7

bx x = 4

242x = 24

2x = 16

bx2x = 16

A

Exponential and Logarithmic Equations and Problem Solving SECTION 11.7 721

Objectives

Solve exponential equations.

Solve logarithmic equations.

Solve problems that can be mod-eled by exponential and logarith-mic equations.

CBA

LOGARITHM PROPERTY OF EQUALITY

Let a, b, and c be real numbers such that and are realnumbers and b is not 1. Then

is equivalent to a = clogb a = logb c

logb clogb a

Practice Problem 1Solve: . Give an exact answerand a four-decimal-place aproxima-tion.

2x = 5

Answers

1. , 2. E4Fe log 5log 2

f ; E2.3219F


Practice Problem 2Solve: log3 Ax + 5 B = 2


Practice Problem 3Solve: log6 x + log6 Ax + 1 B = 1

Practice Problem 4Solve: log Ax - 1 B - log x = 1


To check, we replace x with 18 in the original equation.

Let .

True.


Example 3 Solve:

Solution: Notice that must be positive, so x must be greaterthan 1. We use the product property on the left side ofthe equation.


Next we write the equation with exponential notationand solve for x.


Factor.


Recall that cannot be a solution because x must begreater than 1. If we forgot this, we would still reject after checking. To see this, we replace x with in theoriginal equation.

Let .

Because the logarithm of a negative number is undefined,is rejected. Check to see that the solution set is .

Example 4 Solve:

Solution: We use the quotient property of logarithms on the leftside of the equation.

Use the quotient property.

100 = x + 2x

Write using exponentialnotation. 102 = x + 2

x

log x + 2

x= 2

log Ax + 2 B - log x = 2

log Ax + 2 B - log x = 2

E2F-1

x = -1log2 A-1 B + log2 A-1 - 1 B � 1

log2 x + log2 Ax - 1 B = 1

-1-1

-1

2 = x -1 = x

0 = x - 2 or 0 = x + 1

0 = Ax - 2 B Ax + 1 B 0 = x2 - x - 2

21 = x2 - x

log2 Ax2 - x B = 1

log2 Cx Ax - 1 B D = 1

log2 x + log2 Ax - 1 B = 1

x - 1

log2 x + log2 Ax - 1 B = 1

E18F 42 = 16

log4 16 � 2

x = 18log4 A18 - 2 B � 2

log4 Ax - 2 B = 2

18 = x

16 = x - 2

42 = x - 2

log4 Ax - 2 B = 2

Answers3. , 4. ¤E2F

Exponential and Logarithmic Equations and Problem Solving SECTION 11.7 723

Answers5. approximately 16,601 lemmings, 6.years

1134

Practice Problem 5Use the equation in Example 5 to estimate the lemming population in 8months.

Practice Problem 6How long does it take an investment of$1000 to double if it is invested at 6%compounded quarterly?

Multiply both sides by x.

Subtract x from both sides.


Check to see that the solution set is .

SOLVING PROBLEMS MODELED BY EXPONENTIAL

AND LOGARITHMIC EQUATIONS

Logarithmic and exponential functions are used in a variety of scientific,technical, and business settings. A few examples follow.

Example 5 Estimating Population SizeThe population size y of a community of lemmings variesaccording to the relationship . In this formula,t is time in months, and is the initial population at time0. Estimate the population after 6 months if there wereoriginally 5000 lemmings.

Solution: We substitute 5000 for and 6 for t.

Let and .

Multiply.

Using a calculator, we find that . In6 months the population will be approximately12,300 lemmings.

Example 6 Doubling an InvestmentHow long does it take an investment of $2000 to doubleif it is invested at 5% interest compounded quarterly?

The necessary formula is , where A is

the accrued amount, P is the principal invested, r is theannual rate of interest, n is the number of compoundingperiods per year, and t is the number of years.

Solution: We are given that and . Com-pounding quarterly means 4 times a year, so . Theinvestment is to double, so A must be $4000. We substi-tute these values and solve for t.


Simplify .


Take the logarithm of both sides.

Use the power property. log 2 = 4t A log 1.0125 B log 2 = log 1.01254t

2 = A1.0125 B 4t

1 + 0.054

4000 = 2000 A1.0125 B 4t

4000 = 2000 a1 + 0.054b 4t

A = P a1 + rnb nt

n = 4r = 5% = 0.05P = $2000

A = P a1 + rnb nt

y L 12,298.016

= 5000e0.9

y0 = 5000t = 6 = 5000e0.15A6By = y0e

0.15t

y0

y0

y = y0e0.15t

C

e 299f

x = 299

99x = 2

100x = x + 2


GRAPHING CALCULATOR EXPLORATIONSUse a grapher to find how long it takes an investment of $1500 to triple if it is invested at 8% interestcompounded monthly.

First, let , , and (for 12 months) in the formula

Notice that when the investment has tripled, the accrued amount A is $4500. Thus,

Determine an appropriate viewing window and enter and graph the equations

and

The point of intersection of the two curves is the solution. The x-coordinate tells how long it takes for theinvestment to triple.

Use a TRACE feature or an INTERSECT feature to approximate the coordinates of the point of intersec-tion of the two curves. It takes approximately 13.78 years, or 13 years and 10 months, for the investment to triplein value to $4500.

Use this graphical solution method to solve each problem. Round each answer to the nearest hundredth.

1. Find how long it takes an investment of $5000 to grow to $6000 if it is invested at 5% interest compoundedquarterly.

2. Find how long it takes an investment of $1000 to double if it is invested at 4.5% interest compounded daily.(Use 365 days in a year.)

3. Find how long it takes an investment of $10,000 to quadruple if it is invested at 6% interest compoundedmonthly.

4. Find how long it takes $500 to grow to $800 if it is invested at 4% interest compounded semiannually.

Y2 = 4500

5000

0 20

y = 4500

0.08121 +y = 1500 ( ) 12x

Y1 = 1500 a1 + 0.0812b 12x

4500 = 1500 a1 + 0.0812b 12t

A = P a1 + rnb nt

n = 12r = 0.08P = $1500

Divide both sides by 4 log 1.0125.

Approximate by calculator.

It takes 14 years for the money to double in value.

13.949408 L t

log 2

4 log 1.0125= t



SECTION 11.1 THE ALGEBRA OF FUNCTIONS

SECTION 11.2 INVERSE FUNCTIONS

ALGEBRA OF FUNCTIONS

SumDifferenceProduct

Quotient afgb Ax B =

f Ax Bg Ax B

Af # g B Ax B = f Ax B # g Ax B Af - g B Ax B = f Ax B - g Ax B Af + g B Ax B = f Ax B + g Ax B

If and ,

a fgb Ax B =

f Ax Bg Ax B = 7x

x2 + 1

= 7x3 + 7x2

Af # g B Ax B = f Ax B # g Ax B = 7x Ax2 + 1 B= 7x - x2 - 1

Af - g B Ax B = f Ax B - g Ax B = 7x - Ax2 + 1 B Af + g B Ax B = f Ax B + g Ax B = 7x + x2 + 1

g Ax B = x2 + 1f Ax B = 7x

COMPOSITE FUNCTIONS

The notation means “f composed with g.”

Ag ø f B Ax B = g Af Ax B BAf ø g B Ax B = f Ag Ax B B

Af ø g B Ax BIf and , find

.

= x2 - 10x + 26

= Ax - 5 B 2 + 1

= f Ax - 5 BAf ø g B Ax B = f Ag Ax B B

Af ø g B Ax Bg Ax B = x - 5f Ax B = x2 + 1

If f is a function, then f is a one-to-one function onlyif each y-value (output) corresponds to only one x-value (input).

HORIZONTAL LINE TEST

If every horizontal line intersects the graph of a func-tion at most once, then the function is a one-to-onefunction.

Determine whether each graph is a one-to-one func-tion.

Graphs A and C pass the vertical line test, so onlythese are graphs of functions. Of graphs A and C,only graph A passes the horizontal line test, so onlygraph A is the graph of a one-to-one function.

x

yB

x

yC

x

yA



The inverse of a one-to-one function f is the one-to-one function that is the set of all ordered pairs

such that belongs to f.

TO FIND THE INVERSE OF A ONE-TO-ONE

FUNCTION f(x)

Step 1. Replace with y.Step 2. Interchange x and y.Step 3. Solve for y.Step 4. Replace y with .f -1 Ax B

f Ax B

Aa, b BAb, a B f-1Find the inverse of .

Replace with y.

Interchange x and y.

Solve for y.

Replace y with .

The inverse of is .f -1 Ax B = x - 72

f Ax B = 2x + 7

f -1 Ax Bf-1 Ax B = x - 72

y = x - 72

2y = x - 7

x = 2y + 7

f Ax B y = 2x + 7

f Ax B = 2x + 7

SECTION 11.3 EXPONENTIAL FUNCTIONS

A function of the form is an exponentialfunction, where , , and x is a real num-ber.

b Z 1b 7 0f Ax B = bx Graph the exponential function .

x y

0 11 42 16

Solve: 2x + 5 = 8

x

y

14

-1

116

-2

y = 4x

UNIQUENESS OF

If and , then is equivalent to.

COMPOUND INTEREST FORMULA

where r is the annual interest rate for P dollars com-pounded n times a year for t years.

CONTINUOUSLY COMPOUNDED INTEREST FORMULA

where r is the annual interest rate for P dollarsinvested for t years.

A = Pert

A = P a1 + rnb n t

x = ybx = byb Z 1b 7 0

bx

Write 8 as .



Find the amount in an account at the end of 3 yearsif $1000 is invested at an interest rate of 4% com-pounded continuously.

Here, years, , and .

L $1127.50

= 1000e0.04A3BA = Pert

r = 0.04P = $1000t = 3

x = -2

bx x + 5 = 3

23 2x + 5 = 23

SECTION 11.4 LOGARITHMIC FUNCTIONS

LOGARITHMIC DEFINITION

If and , then

means

for any positive number x and real number y.

PROPERTIES OF LOGARITHMS

If b is a real number, and , thenb Z 1b 7 0

x = byy = logb x

b Z 1b 7 0

LOGARITHMIC CORRESPONDING

FORM EXPONENTIAL STATEMENT

91>2 = 3 log9 3 = 12

52 = 25log5 25 = 2

, , blogb x = xlogb bx = xlogb 1 = 0 , , 3log36 = 6log7 7

2 = 2log5 1 = 0



SECTION 11.5 PROPERTIES OF LOGARITHMS

LOGARITHMIC FUNCTION

If and , then a logarithmic function is afunction that can be defined as

The domain of f is the set of positive real numbers,and the range of f is the set of real numbers.

f Ax B = logb x

b Z 1b 7 0

Graph: Write . Plot the ordered pair

solutions listed in the table, and connect them witha smooth curve.

x y

3 11 0

x

y

-219

-113

y = log3 x as 3y = xy = log3 x

Let x, y, and b be positive numbers and .

PRODUCT PROPERTY

QUOTIENT PROPERTY

POWER PROPERTY

logb xr = r logb x

logb xy

= logb x - logb y

logb xy = logb x + logb y

b Z 1 Write as a single logarithm:

Power property

Product property

Quotient property = log5 36x

y + 2

= log5 36 # x - log5 Ay + 2 B = log5 6

2 + log5 x - log5 Ay + 2 B2 log5 6 + log5 x - log5 Ay + 2 B

SECTION 11.6 COMMON LOGARITHMS, NATURAL LOGARITHMS, AND CHANGE OF BASE

COMMON LOGARITHMS

means log10 xlog x ln 7 = loge 7 L 1.94591

log 5 = log10 5 L 0.69897

NATURAL LOGARITHMS

ln x means

CHANGE OF BASE

logb a = logc alogc b

loge x

log8 12 = log 12log 8

SECTION 11.7 EXPONENTIAL AND LOGARITHMIC EQUATIONS AND PROBLEM SOLVING

LOGARITHM PROPERTY OF EQUALITY

Let and be real numbers and .Then

is equivalent to a = clogb a = logb c

b Z 1logb clogb a

Solve:

Log property of equality

Power property

Divide both sides by log 2.

Use a calculator. x L 2.3219

x =log 5log 2

x log 2 = log 5

log 2x = log 5

2x = 5

729

AThe word geometry is formed from the Greek words geo, meaning earth,and metron, meaning measure. Geometry literally means to measure theearth. In this chapter we learn about various geometric figures and theirproperties, such as perimeter, area, and volume. Knowledge of geometrycan help us solve practical problems in real-life situations. For instance,knowing certain measures of a circular swimming pool allows us to cal-culate how much water it can hold.

A.1 Lines and Angles

A.2 Plane Figures and Solids

A.3 Perimeter

A.4 Area

A.5 Volume

A.6 Square Roots and thePythagorean Theorem

A.7 Congruent and SimilarTriangles

Just outside of Cairo, Egypt, is a famous plateau called Giza.This is the home of the Great Pyramids, the only surviving entryon the list of the Seven Wonders of the Ancient World. Thereare three pyramids at Giza. The largest, and oldest, was built asthe tomb of the Pharaoh Khufu around 2550 B.C. This pyramidis made from over 2,300,000 blocks of stone weighing a total of6.5 million tons. It took about 30 years to build this monument.Prior to the 20th century, Khufu’s pyramid was the tallest build-ing in the world. Khufu’s son Khafre is responsible for the sec-ond-largest pyramid at Giza during his rule as pharaoh between2520 to 2494 B.C. The smallest of the pyramids at Giza is cred-ited to Menkaure, believed to be the son of Khafre and grand-son of Khufu.

Geometry A P P E N D I X

A.1 LINES AND ANGLES

IDENTIFY LINES, LINE SEGMENTS, RAYS, AND ANGLES

Let’s begin with a review of two important concepts—plane and space.A plane is a flat surface that extends indefinitely. Surfaces like a plane

are a classroom floor or a blackboard or whiteboard.

Space extends in all directions indefinitely. Examples of objects in spaceare houses, grains of salt, bushes, your Basic College Mathematics text-book, and you.

The most basic concept of geometry is the idea of a point in space. Apoint has no length, no width, and no height, but it does have location. Wewill represent a point by a dot, and we will label points with letters.

A line is a set of points extending indefinitely in two directions. A linehas no width or height, but it does have length. We can name a line by anytwo of its points. A line segment is a piece of a line with two endpoints.

A ray is a part of a line with one endpoint. A ray extends indefinitely inone direction. An angle is made up of two rays that share the same end-point. The common endpoint is called the vertex.

The angle in the figure above can be named

orc c

The vertex is the middle point.

Rays BA and BC are sides of the angle.

jxjBjCBAjABC

B

C

A

Vertex

xA

B

Ray AB or AB

A B

Line segment AB or AB

A B

Line AB or AB

P

Point P

Plane

A

Lines and Angles SECTION A.1 731

Objectives

Identify lines, line segments, rays,and angles.

Classify angles as acute, right, ob-tuse, or straight.

Identify complementary and sup-plementary angles.

Find measures of angles.D

C

B

A

732 APPENDIX A Geometry

Answers1. a. line segment; line segment RS or ,b. ray; ray AB or -AS

B, c. line; line EF or dE

SF,

d. angle; , or or ,2. , jSTRjRTS

jVjHVTjTVH

RS

Practice Problem 1Identify each figure as a line, a ray, aline segment, or an angle. Then namethe figure using the given points.

a.

b.

c.

d. H

T

V

E

F

B

A

R

S

Example 1 Identify each figure as a line, a ray, a line segment, or anangle. Then name the figure using the given points.

a. b.

c. d.

Solution: Figure (a) extends indefinitely in two directions. It is lineCD or

dCD

S.

Figure (b) has two endpoints. It is line segment EF or .

Figure (c) has two rays with a common endpoint. It is, , or .

Figure (d) is part of a line with one endpoint. It is ray PTor -PTS.

Example 2 List other ways to name .

Solution: Two other ways to name are and . We may not use the vertexalone to name this angle because threedifferent angles have T as their vertex.

CLASSIFYING ANGLES

An angle can be measured in degrees. The symbol for degrees is a small,raised circle, °. There are 360° in a full revolution, or full circle.

of a revolution measures . An angle that measures

180° is called a straight angle.

180°

R S T

∠RST is a straight angle.

A360° B = 180°12

12

360°

B

jRTQjQTRjy

Q

T

S

Ryz

jy

jNjONMjMNO

EF

P

T

N

O

M

EF

C

D

Practice Problem 2Use the figure for Example 2 to listother ways to name .jz


Answers3. a. acute, b. straight, c. obtuse, d. right

of a revolution measures . An angle that measures 90°

is called a right angle. The symbol is used to denote a right angle.

An angle whose measure is between 0° and 90° is called an acute angle.

An angle whose measure is between 90° and 180° is called an obtuseangle.

Example 3 Classify each angle as acute, right, obtuse, or straight.

a. b.

c. d.

Solution: a. is a right angle, denoted by .b. is a straight angle.c. is an acute angle. It measures between 0° and 90°.d. is an obtuse angle. It measures between 90°

and 180°.

IDENTIFYING COMPLEMENTARY AND SUPPLEMENTARY ANGLES

Two angles that have a sum of 90° are called complementary angles. We saythat each angle is the complement of the other.

and are complementary angles becauseT T

60° + 30° = 90°

jSjR

C

jQjTjSjR

QT

S

R

120°95°

Obtuse angles

40°

62°

Acute angles

∠ABC is a right angle.

A

B C

A360° B = 90°14

14

Practice Problem 3Classify each angle as acute, right, ob-tuse, or straight.

a.

b.

c. d.

P

O

N

M


Answers4. 54°, 5. 92°, 6. 39°

✓ Concept Check: False; the complement of a

48° angle is 42°, the supplement of a 48° angle

is 132°.

Practice Problem 4Find the complement of a 36° angle.

Two angles that have a sum of 180° are called supplementary angles. Wesay that each angle is the supplement of the other.

and are supplementary angles becauseT T

Example 4 Find the complement of a 48° angle.

Solution: The complement of an angle that measures 48° is anangle that measures .

Example 5 Find the supplement of a 107° angle.

Solution: The supplement of an angle that measures 107° is anangle that measures .


FINDING MEASURES OF ANGLES

Measures of angles can be added or subtracted to find measures of relatedangles.

Example 6 Find the measure of .

Solution:

Two lines in a plane can be either parallel or intersecting. Parallel linesnever meet. Intersecting lines meet at a point. The symbol is used to indi-cate “is parallel to.” For example, in the figure .p ‘ q

‘

jx = 87° - 52° = 35°

52°

87°

T

x

QR

S

jx

D

180° - 107° = 73°

90° - 48° = 42°

125°55°

M N

Supplementary angles125° + 55° = 180°

125° + 55° = 180°

jNjM

60° 30°R S

Complementary angles60° + 30° = 90°

Practice Problem 5Find the supplement of an 88° angle.

✓ CONCEPT CHECK

True or false? The supplement of a48° angle is 42°. Explain.

Practice Problem 6Find the measure of .

97°

136°

D

y

B

A

C

jy


Some intersecting lines are perpendicular. Two lines are perpendicularif they form right angles when they intersect. The symbol is used todenote “is perpendicular to.” For example, in the figure .

When two lines intersect, four angles are formed. Two of these anglesthat are opposite each other are called vertical angles. Vertical angles havethe same measure. Two angles that share a common side are called adja-cent angles. Adjacent angles formed by intersecting lines are supplemen-tary. That is, they have a sum of 180°.

Example 7 Find the measure of , and if the measure of is 42°.

Solution: Since and are vertical angles, they have the same measure, so measures 42°.

Since and are adjacent angles, their measures have asum of 180°. So measures .

Since and are vertical angles, they have the samemeasure. So measures 138°.jz

jzjy

180° - 42° = 138°jyjyjt

jxjxjt

yt

zx

jtjzjx, jy

ab

cd

Vertical angles:∠ a and ∠ c∠ d and ∠ b

Adjacent angles:∠ a and ∠ b∠ b and ∠ c∠ c and ∠ d∠ d and ∠ a

n

m

Perpendicular lines

n›m›

Intersecting lines

p

q

Parallel lines

Practice Problem 7Find the measure of , , and .

112°b

ac

jcjbja

Answer7. ; ; jc = 68°jb = 68°ja = 112°


Answer8. ; ; jz = 140°jy = 40°jx = 40°

A line that intersects two or more lines at different points is called atransversal. Line l is a transversal that intersects lines m and n. The eightangles formed have special names. Some of these names are:

Corresponding Angles: and , and , and ,

and Alternate Interior Angles:

and , and

When two lines cut by a transversal are parallel, the following are true:

Example 8 Given that and that the measure of is 100°, findthe measures of , , and .

Solution:

The measure of . and are vertical angles.

The measure of . and jz are corresponding angles.

The measure of . jz and j y are supplementary angles.jy = 180° - 100° = 80°jxjz = 100°

jwjxjx = 100°

m

n

xw

yz

jzjyjxjwm ‘ n

PARALLEL LINES CUT BY A TRANSVERSAL

If two parallel lines are cut by a transversal, then the measures of cor-responding angles are equal and alternate interior angles are equal.

a bm

nfe

c d

g h

l

jejdjfjc

jhjdjfjbjgjcje

ja

Practice Problem 8Given that and that the measureof , find the measures of ,

, and .

m

n

wx

yz

jzjyjxjw = 40°

m ‘ n

A.2 PLANE FIGURES AND SOLIDS

In order to prepare for the sections ahead in this chapter, we first reviewplane figures and solids.

IDENTIFYING PLANE FIGURES

Recall from Section A.1 that a plane is a flat surface that extends indefi-nitely.

A plane figure is a figure that lies on a plane. Plane figures, like planes,have length and width but no thickness or depth.

A polygon is a closed plane figure that basically consists of three or moreline segments that meet at their endpoints.

A regular polygon is one whose sides are all the same length and whoseangles are the same measure.

A polygon is named according to the number of its sides.Some triangles and quadrilaterals are given special names, so let’s study

these special polygons further. We will begin with triangles. The sum of themeasures of the angles of a triangle is 180°.


Solution: Since the sum of the measures of the three angles is 180°,we have

measure of

To check, see that .

We can classify triangles according to the lengths of their sides. (We willuse tick marks to denote sides and angles in a figure that are equal.)

95° + 35° + 50° = 180°

ja = 180° - 95° - 35° = 50°

95°35°

a

ja

F H KJG I

A B C D E

Plane

A

Plane Figures and Solids SECTION A.2 737

Objectives

Identify plane figures.

Identify solid figures.BA

Number Figureof Sides Name Examples

3 Triangle A, F4 Quadrilateral B, E, G5 Pentagon H6 Hexagon I7 Heptagon C8 Octagon J9 Nonagon K

10 Decagon D


25°

110°x

jx

Answer1. 45°


Answer2. 65°

One other important type of triangle is a right triangle. A right triangleis a triangle with a right angle. The side opposite the right angle is called thehypotenuse and the other two sides are called legs.


Solution: We know that the measure of the right angle, , is 90°.Since the sum of the measures of the angles is 180°, wehave

measure of

Now we review some special quadrilaterals. A parallelogram is a specialquadrilateral with opposite sides parallel and equal in length.

A rectangle is a special parallelogram that has four right angles.

A square is a special rectangle that has all four sides equal in length.

jb = 180° - 90° - 30° = 60°

30°

b

jb

hypotenuseleg

leg

Equilateral triangle Isoceles triangle Scalene triangle

All three sides arethe same length. Also,all three angles havethe same measure.

Two sides are the same length. Also,the angles oppositethe equal sideshave equal measure.

No sides are thesame length. Noangles are the same measure.


25°

y

jy

HELPFUL HINT

From the previous example, canyou see that in a right triangle,the sum of the other two acuteangles is 90°? This is because

c c cright sum of sum ofangle’s other angles’measure two measures

angles’measures

90° + 90° = 180°

Plane Figures and Solids SECTION A.2 739

A rhombus is a special parallelogram that has all four sides equal inlength.

A trapezoid is a quadrilateral with exactly one pair of opposite sides par-allel.

TRY THE CONCEPT CHECK IN THE MARGIN.In addition to triangles and quadrilaterals, circles are common plane fig-

ures. A circle is a plane figure that consists of all points that are the samefixed distance from a point c. The point c is called the center of the circle.The radius of a circle is the distance from the center of the circle to anypoint of the circle. The diameter of a circle is the distance across the circlepassing through the center. The diameter is twice the radius and the radiusis half the diameter.

= =

T T T T Td = 2 r r =

Example 3 Find the diameter of the circle.

Solution: The diameter is twice the radius.

The diameter is 10 centimeters.

IDENTIFYING SOLID FIGURES

Recall from Section A.1 that space extends in all directions indefinitely.A solid is a figure that lies in space. Solids have length, width, and height

or depth.A rectangular solid is a solid that consists of six sides, or faces, all of

which are rectangles.

height

width

length

B

d = 2 # 5 cm = 10 cm

d = 2 # r

5 cm

d2

#

diameter2

radiusradius#2diameter

centerradius

diameter

parallelsides

✓ CONCEPT CHECK

True or false? All quadrilaterals areparallelograms. Explain.

Answers3. 8 in.

✓ Concept Check: false

Practice Problem 3Find the radius of the circle.

16 in.


Answer4. 14 mi

A cube is a rectangular solid whose six sides are squares.

A pyramid is shown below.

A sphere consists of all points in space that are the same distance froma point c. The point c is called the center of the sphere. The radius of asphere is the distance from the center to any point of the sphere. The diam-eter of a sphere is the distance across the sphere passing through the center.

The radius and diameter of a sphere are related in the same way as theradius and diameter of a circle.

or

Example 4 Find the radius of the sphere.

Solution: The radius is half the diameter.

The radius is 18 feet.

The cylinders we will study have bases that are in the shape of circles andare perpendicular to their height.

The cones we will study have bases that are circles and are perpendicu-lar to their height.

height

circular base

height

circular base

r = 36 feet2

= 18 feet

r = d2

36 ft

r = d2

d = 2 # r

radius

center

diameter

height

squarebase

Practice Problem 4Find the diameter of the sphere.

7 mi

A.3 PERIMETER

USING FORMULAS TO FIND PERIMETERS

Recall from Section 1.2 that the perimeter of a polygon is the distancearound the polygon. This means that the perimeter of a polygon is the sumof the lengths of its sides.

Example 1 Find the perimeter of the rectangle below.

Solution:

Notice that the perimeter of the rectangle in Example 1 can be writtenas .

c clength width

In general, we can say that the perimeter of a rectangle is always

As we have just seen, the perimeter of some special figures such as rec-tangles form patterns. These patterns are given as formulas. The formulafor the perimeter of a rectangle is shown next.

Example 2 Find the perimeter of a rectangle with a length of11 inches and a width of 3 inches.

Solution: We will use the formula for perimeter and replace theletters by their known lengths.

The perimeter is 28 inches.

= 28 in.

= 22 in. + 6 in.

Replace l with 11 in. and w with3 in.

= 2 # 11 in. + 2 # 3 in.

P = 2 # l + 2 # w

11 in.

3 in.

2 # length + 2 # width

2 # A9 inches B + 2 # A5 inches B

= 28 inches

perimeter = 9 inches + 9 inches + 5 inches + 5 inches

9 in.

5 in.

A

Perimeter SECTION A.3 741

Objectives

Use formulas to find the perimeterof special geometric figures.

Use formulas to find the circumfer-ence of a circle.

B

A

Practice Problem 1Find the perimeter of the rectangularlot shown below.

60 ft

80 ft

Answers1. 280 ft, 2. 64 cm

PERIMETER OF A RECTANGLE

In symbols, this can be written as

length

length

widthwidth

P = 2 # l + 2 # w

perimeter = 2 # length + 2 # width

Practice Problem 2Find the perimeter of a rectangle witha length of 22 cm and a width of 10 cm.

Recall that a square is a special rectangle with all four sides the samelength. The formula for the perimeter of a square is shown next.

Example 3 Finding Perimeter of a Table TopFind the perimeter of a square table top if each side is5 feet long.

Solution: The formula for the perimeter of a square is .We will use this formula and replace s by 5 feet.

The perimeter of the square table top is 20 feet.

The formula for the perimeter of a triangle with sides of length a, b, andc is given next.

= 20 feet

= 4 # 5 feet

P = 4 # s

P = 4 # s

5 ft

5 ft


PERIMETER OF A SQUARE

In symbols,

side

side

side side

P = 4 # s

= 4 # side

Perimeter = side + side + side + side

Answer3. 200 yd

PERIMETER OF A TRIANGLE

In symbols,

Side b

Side c

Side a

P = a + b + c

Perimeter = side a + side b + side c

Practice Problem 3How much fencing is needed to en-close a square field 50 yd on a side?

50 yd


Answers4. 21 cm, 5. 17 km, 6. 92 m

Practice Problem 4Find the perimeter of a triangle if thesides are 5 cm, 9 cm, and 7 cm inlength.

Practice Problem 5Find the perimeter of the trapezoidshown.

3 km3 km

7 km

4 km

Example 4 Find the perimeter of a triangle if the sides are 3 inches,7 inches, and 6 inches.

Solution: The formula is , where a, b, and c are thelengths of the sides. Thus,

The perimeter of the triangle is 16 inches.

Recall that to find the perimeter of other polygons, we find the sum ofthe lengths of their sides.

Example 5 Find the perimeter of the trapezoid shown.

Solution: To find the perimeter, we find the sum of the lengths ofits sides.

The perimeter is 12 centimeters.

Example 6 Finding the Perimeter of a RoomFind the perimeter of the room shown below.

Solution: To find the perimeter of the room, we first need to findthe lengths of all sides of the room.

Now that we know the measures of all sides of the room,we can add the measures to find the perimeter.

9 feet

This sidemust measure

15 ft − 9 ft = 6 ft

15 feet

7 feet10 feet

This sidemust measure

10 ft − 7 ft = 3 ft

9 ft

7 ft10 ft

15 ft

perimeter = 3 cm + 2 cm + 5 cm + 2 cm = 12 cm

3 cm

6 cm

2 cm1 cm

= 16 in.

= 3 in. + 7 in. + 6 in.

P = a + b + c

P = a + b + c

6 in.

3 in.

7 in.

Practice Problem 6Find the perimeter of the room shown.

15 m

26 m

7 m

20 m

The perimeter of the room is 50 feet.

Example 7 Calculating the Cost of BaseboardA rectangular room measures 10 feet by 12 feet. Find thecost to install new baseboard around the room if the costof the baseboard is $0.66 per foot.

Solution: First we find the perimeter of the room.

The cost for the baseboard is

The cost of the baseboard is $29.04.

USING FORMULAS TO FIND CIRCUMFERENCES

Recall from Section 4.4 that the distance around a circle is given a specialname called the circumference. This distance depends on the radius or thediameter of the circle.

The formulas for circumference are shown next.

B

cost = 0.66 # 44 = 29.04

= 44 feet

= 24 feet + 20 feet

Replace l with 12 feet and wwith 10 feet. = 2 # 12 feet + 2 # 10 feet

P = 2 # l + 2 # w

= 50 ft

perimeter = 10 ft + 9 ft + 3 ft + 6 ft + 7 ft + 15 ft

15 ft

9 ft

6 ft

7 ft

3 ft

10 ft


Answer7. $684

Practice Problem 7

CIRCUMFERENCE OF A CIRCLE

In symbols,

where .p L 3.14 or p L 227

C = 2 # p # r or C = p # d

Circumference = 2 # p # radius or Circumference = p # diameter

centerradius

diameter

A rectangular lot measures 60 ft by120 ft. Find the cost to install fencingaround the lot if the cost of fencing is$1.90 per foot.

Practice Problem 8An irrigation device waters a circularregion with a diameter of 20 yd. Whatis the circumference of the wateredregion?

20 yd


✓ CONCEPT CHECK

The distance around which figure isgreater: a square with side length 5 in.or a circle with radius 3 in.?

Answers8. 62.8 yd

✓ Concept Check: a square with length 5 in.

To better understand circumference and (pi), try the following experi-ment. Take any can and measure its circumference and its diameter.

The can in the figure above has a circumference of 23.5 centimeters and adiameter of 7.5 centimeters. Now divide the circumference by the diameter.

Try this with other sizes of cylinders and circles—you should always geta number close to 3.1. The exact ratio of circumference to diameter is p.

(Recall that or .)

Example 8 Mary Catherine Dooley plans to install a border of newtiling around the circumference of her circular spa. If herspa has a diameter of 14 feet, find its circumference.

Solution: Because we are given the diameter, we use the formula.

Replace d with 14 feet.

The circumference of the spa is exactly 14 p feet. By re-placing p with the approximation 3.14, we find that the cir-cumference is approximately feet.


14 feet # 3.14 = 43.96

= 14 p ft

= p # 14 ft

C = p # d

C = p # d

p L 227

p L 3.14

circumferencediameter

= 23.5 cm7.5 cm

L 3.1

1 2 3 4 5 6 7

1211

cm

232221201924 25 26 27 28 29 30 31

23

p


Focus On HistoryTHE PYTHAGOREAN THEOREM

Pythagoras was a Greek teacher who lived from about 580 B.C. to501 B.C. He founded his school at Croton in southern Italy. Theschool was a tightly knit community of eager young students withthe motto “All is number.” Pythagoras himself taught by lecturingon the subjects of arithmetic, music, geometry, and astronomy—all based on mathematics. In fact, Pythagoras and his studentsheld numbers to be sacred and tried to find mathematical order inall parts of life and nature. Pythagoras’ followers, known asPythagoreans, were sworn to not reveal any of the Master’s teach-ings to outsiders. It is for this reason that very little of his life orteachings are known today—Pythagoreans were forbidden fromrecording any of the Master’s teachings in written form, andPythagoras left none of his own writings.

Traditionally, the so-called Pythagorean theorem is attributedto Pythagoras himself. However, some scholars question whetherPythagoras ever gave any rigorous proof of the theorem at all!There is ample evidence that several ancient cultures had knowl-edge of this important theorem and used its results in practicalmatters well before the time of Pythagoras.

j A Babylonian clay tablet (#322 in the Plimpton collection atColumbia University) dating from 1900 B.C. gives numerical evi-dence that the Babylonians were well aware of what we todaycall the Pythagorean theorem.

j The ancient Chinese knew of a very simple geometric proof ofthe Pythagorean theorem, which can be seen in the ArithmeticClassics of the Gnomon and the Circular Paths of Heaven, datingfrom about 600 B.C.

j The ancient Egyptians used the Pythagorean theorem to formright angles when they needed to measure a plot of land. Theyput equally spaced knots in pieces of rope so they could form atriangle that had a right angle by stretching the rope out on theground. They used equally spaced knots to ensure that thelengths of the measuring ropes were in the ratio 3:4:5.

CRITICAL THINKING

In the Egyptian use of the Pythagorean theorem, if three pieces ofrope were used and one piece had a total of 10 knots equally spacedand another piece had a total of 13 knots with the same equal spac-ing, how many knots with the same equal spacing would a third pieceof rope need so that the three pieces form a right triangle with the firsttwo pieces as the legs of the triangle? Draw a figure and explain youranswer. (Assume that each piece of rope was knotted at both ends.)

747


EXERCISE SET A.3

Find the perimeter of each figure. See Examples 1 through 6.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

20 m

13 m

10 m 4 m

10 m

9 m

15 ft

7 ft

10 ft 8 ft

8 ft

2 yd

3 yd

Parallelogram

35 cm

25 cmParallelogram

8 units

4 units10 units

9 in.

5 in. 7 in.

46 mi

Square

9 cm

Square

4 m

10 mRectangle17 ft

15 ft Rectangle

A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Name _________________________________________________________________________

748

Find the perimeter of each figure. See Example 6.

11. 12.

13. 14.

15. 16.

22 km

12 km

5 km6 km

12 mi

34 mi10 mi

8 mi

16 cm

2 cm

11 cm4 cm

3 cm

9 cm

4 ft3 ft

6 ft

5 ft

15 ft

13 in.

30 in.

6 in.

13 in.

17 m

20 m

28 m20 m

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Find the circumference of each circle. Give the exact circumference and then an approx-imation. Use . See Example 8.

17. 18.

19. 20.

50 ft

8 mi

6 in.

17 cm

p L 3.14B

A.4 AREA

FINDING AREA

Recall that area measures the amount of surface of a region. Thus far, weknow how to find the area of a rectangle and a square. These formulas, aswell as formulas for finding the areas of other common geometic figures,are given next.

A

Area SECTION A.4 749

Objective

Find the area of geometric figures.A

AREA FORMULAS OF COMMON GEOMETRIC FIGURES

Geometric Figure Area Formula

Area of a rectangle:

Area of a square:

Area of a triangle:

Area of a parallelogram:

Area of a trapezoid:

A = 12

# Ab + B B # h

area = 12

# Aone base + other Base B # height

A = b # hArea = base # height

A = 12

# b # h

Area = 12

# base # height

A = s # s = s2 Area = side # side

A = lwArea = length # width

side

side

base

height

base

TRAPEZOID

PARALLELOGRAM

TRIANGLE

SQUARE

RECTANGLE

height

other base or B

one base or b

height

length

width


Answers1. 25 sq. in., 2. 17.64 sq. yd, 3. 396 sq. m

Practice Problem 1Find the area of the triangle.

6 14 in.

8 in.

Practice Problem 2Find the area of the square.

4.2 yd

4.2 yd

Use these formulas for the following examples.

Example 1 Find the area of the triangle.

Solution:

1

1

The area is 56 square centimeters.

Example 2 Find the area of the parallelogram.

Solution:

The area is 5.1 square miles.

Example 3 Find the area of the figure.

4 ft

5 ft

8 ft

12 ft

HELPFUL HINT

When finding the area of figures, be sure all measurements arechanged to the same unit before calculations are made.

= 5.1 square miles

= 3.4 miles # 1.5 miles

A = b # h

3.4 mi

1.5 mi

= 56 square centimeters

= 2 # 7 # 82

square centimeters

= 12

# 14 cm # 8 cm

A = 12

# b # h

8 cm

14 cm

HELPFUL HINT

Area is always measured in square units.

Practice Problem 3Find the area of the figure.

18 m

18 m

24 m

12 m

Solution: Split the figure into two rectangles. To find the area ofthe figure, we find the sum of the areas of the two rec-tangles.

Notice that the length of Rectangle 2 is or 8 feet.

To better understand the formula for area of a circle, try the following.Cut a circle into many pieces as shown.

The circumference of a circle is . This means that the circumfer-ence of half a circle is half of , or .p # r2 # p # r

2 # p # r

π ⋅ r

π ⋅ r

= 72 square feet

= 32 square feet + 40 square feet

Area of the Figure = Area of Rectangle 1 + Area of Rectangle 2

= 40 square feet

= 8 feet # 5 feet

Area of Rectangle 2 = l # w

12 feet - 4 feet

= 32 square feet

= 8 feet # 4 feet

Area of Rectangle 1 = l # w

4 ft

Rectangle 2

12 − 4 or 8 ft

Rec

tang

le 1

5 ft

8 ft

12 ft

Area SECTION A.4 751

HELPFUL HINT

The figure in Example 3 can also be split into two rectangles asshown.

1Rectangle

4 feet

Rectangle 2 5 feet

8 feet

12 feet

Then unfold the two halves of the circle and place them together asshown.

The figure on the right is almost a parallelogram with a base of anda height of r. The area is

T T

This is the formula for area of a circle.

Example 4 Find the area of a circle with a radius of 3 feet. Find theexact area and an approximation. Use 3.14 as anapproximation for .

Solution: We let feet and use the formula

To approximate this area, we substitute 3.14 for .

The exact area of the circle is square feet, which isapproximately 28.26 square feet.


9p

= 28.26 square feet

9 # p square feet L 9 # 3.14 square feet

p

= 9 # p square feet = p # A3 feet B 2

A = p # r 2

r = 3

3 ft

p

= p # r2

= Ap # r B # r

height#baseA =

p # r

π ⋅ r

r

π ⋅ r

π ⋅ r


AREA FORMULA OF A CIRCLE

Circle

Area of a circle:

(A fraction approximation for is .)

(A decimal approximation for is 3.14.)p

227

p

A = p # r 2Area = p # Aradius B 2

radius

Answers4.

✓ Concept Check: a square 10 in. long on eachside

49p sq. cm L 153.86 sq. cm

Practice Problem 4Find the area of the given circle. Findthe exact area and an approximation.Use 3.14 as an approximation for .

7 cm

p

✓ CONCEPT CHECK

Use estimation to decide which figurewould have a larger area: a circle ofdiameter 10 in. or a square 10 in. longon each side.

753


EXERCISE SET A.4

Find the area of the geometric figure. If the figure is a circle, give an exact area and thenuse the given approximation for to approximate the area. See Examples 1 through 4.

1. 2. 3.

4. 5. 6.

7. Use 3.14 for . 8. Use for . 9.

10. 11.

12. 13. 4 yd

4 yd

7 yd

Trapezoid6 in.

Trapezoid

8 in.5 in. 12

5 m

9 m

4 m

Trapezoid4.25 cm

3 cm

Parallelogram

r = 2 cmd = 3 in.

5.25 ft

Parallelogram

7 ft

p227

p

5 ft 7 ft

6 yd

5 yd

4 ft

5 ft

12

6 yd

3 yd

12

7 ft

2.75 ft Rectangle

3.5 m

Rectangle2 m

pA

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

Name _________________________________________________________________________

754

14. 15.

16. 17.

18. 19.

20. 21.

22. 25 cm

15cm

5cm

12cm

5 mi

10mi

17 mi

3 mi

10 km

5 km

6 km

4 km

7 cm

3 cm

cm12 cm

12cm11

24 m

6 m

Parallelogram

5 in.

in.4 12 Parallelogram

3 cm

cm414Parallelogram

7 ft

ft514

Parallelogram

10 ft

5 ft

3 ft Trapezoid

14.

15.

16.

17.

18.

19.

20.

21.

22.

755

Name _________________________________________________________________________

23. 24.

25. Use for .

r = 6 in.

p227

4in.

5 in.

5cm

3 cm

23.

24. .

25.

A.5 VOLUME

FINDING VOLUME

Volume is a measure of the space of a region. The volume of a box or can,for example, is the amount of space inside. Volume can be used to describethe amount of juice in a pitcher or the amount of concrete needed to poura foundation for a house.

The volume of a solid is the number of cubic units in the solid. A cubiccentimeter and a cubic inch are illustrated.

Formulas for finding the volumes of some common solids are given next.

Actual size

1 cubic centimeter 1 cubic inch

1 centimeter

1 centimeter

1 inch

1 inch

1 inch1 centimeter

Actual size

A

Volume SECTION A.5 757

Ojective

Find the volume of solids.A

VOLUME FORMULAS OF COMMON SOLIDS

Solid Volume Formulas

Volume of a rectangular solid:

Volume of a cube:

Volume of a sphere:

Volume of a circular cylinder:

V = p # r2 # h

Volume = p # Aradius B 2 # height

V = 43

# p # r3

Volume = 43

# p # Aradius B 3

V = s3Volume = side # side # side

V = l # w # hVolume = length # width # height

length

Rectangular Solid

widthheight

sideside

side

CUBE

SPHERE

radius

CIRCULAR CYLINDER

radius

height


VOLUME FORMULAS OF COMMON SOLIDS

Solid Volume Formulas

Volume of a cone:

Volume of a square-based pyramid:

V = 13

# s2 # h

Volume = 13

# Aside B 2 # height

V = 13

# p # r2 # h

= 13

# p # Aradius B 2 # height

Volume

CONE

SQUARE-BASED PYRAMID

radius

height

height

side

Practice Problem 1

✓ CONCEPT CHECK

Juan is calculating the volume of thefollowing rectangular solid. Find theerror in his calculation.

= 27 cubic cm

= 14 + 8 + 5

Volume = l + w + h

5 cm

8 cm14 cm

Practice Problem 2Draw a diagram and approximate the

volume of a ball of radius cm. Use

for .p227

12

Example 1 Find the volume of a rectangular box that is 12 incheslong, 6 inches wide, and 3 inches high.

Solution:

The volume of the rectangular box is 216 cubic inches.


Example 2 Approximate the volume of a ball of radius 3 inches.

Use the approximation for .

3 in.

p227

V = 12 inches # 6 inches # 3 inches = 216 cubic inches

V = l # w # h

12 in.

3 in.

6 in.

HELPFUL HINT

Volume is always measured in cubic units.

Answers

1. 40 cu. ft, 2. cu. cm

✓ Concept Check

= 560 cu. cm = 14 # 8 # 5

Volume = l # w # h

1121

Draw a diagram and find the volumeof a rectangular box that is 5 ft long,2 ft wide, and 4 ft deep.

Volume SECTION A.5 759

Practice Problem 3Approximate the volume of a cylinderof radius 5 in. and height 7 in. Use 3.14for .p

Answer3. 549.5 cu. in.

Solution:

1

1

or

The volume is approximately .

Example 3 Approximate the volume of a can that has a -inch

radius and a height of 6 inches. Use for .

Solution: Using the formula for a circular cylinder, we have

T———

or approximately

The volume is approximately 231 cubic inches.

= 231 cubic inches

L 227

# 494

# 6 cubic inches

= p # a 72

inches b 2

# 6 inches

312

= 72

V = p # r2 # h

6 in.

312

in.

p227

312

11317

cubic inches

11317

cubic inches = 7927

= 4 # 22 # 3 # 93 # 7

cubic inches

= 43

# 227

# 27 cubic inches

L 43

# 227

A3 inches B 3V = 4

3 # p # r3


Focus On the Real WorldTHE COST OF ROAD SIGNS

There are nearly 4 million miles of streets and roads in the United States. With streets, roads,and highways comes the need for traffic control, guidance, warning, and regulation. Road signsperform many of these tasks. Just in our routine travels, we see a wide variety of road signsevery day. Think how many road signs must exist on the 4 million miles of roads in the UnitedStates. Have you ever wondered how much signs like these cost?

The cost of a road sign generally depends on the type of sign. Costs for several types of signsand sign posts are listed in the table. Examples of various types of signs are shown below.

The cost of a sign is based on its area. For diamond, square, or rectangular signs, the area isfound by multiplying the length (in feet) times the width (in feet). Then the area is multipliedby the cost per square foot. For signs with irregular shapes, costs are generally figured as if thesign were a rectangle, multiplying the height and width at the tallest and widest parts of thesign.

GROUP ACTIVITY

Locate four different kinds of road signs on or near your campus. Measure the dimensions of eachsign, including the height of the post on which it is mounted. Using the cost data given in the table,find the minimum and maximum cost of each sign, including its post. Summarize your results in atable, and include a sketch of each sign.

20SPEEDLIMIT

75

95INTERSTATE

R R

STOP

NOPARKING

ANYTIME

8855

RESTAREA

MARICOPA

COUNTY

93KingmanEXIT 1 MILE

U-channel

Regulatory Warning Marker Large Guide Posts

Square tube

Steel breakaway posts

3rd StRIGHT LANE

ALBANY

ROCHESTER

BUFFALO

32

248

315

ROAD SIGN COSTS

Type of Sign Cost

Regulatory, warning, marker $15–$18 per square foot

Large guide $20–$25 per square foot

Type of Post Cost

U-channel $125–$200 each

Square tube $10–$15 per foot

Steel breakaway posts $15–$25 per foot

761


EXERCISE SET A.5

Find the volume of each solid. See Examples 1 through 3. Use for .

1. 2.

3. 4.

5. 6.

7. 8. in.134

9 in.

10 in.

10 ft

6 ft

2 yd

3 yd

8 cm

4 cm

4 cm

8 cm

8 cm

8 cm

3 mi

4 in.

3 in.

6 in.

p227A

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

Name _________________________________________________________________________

762

9. 10.

1 ft

9 cm

5 cm

Solve.

11. Find the volume of a cube with edges

of inches.

131 in.

113

12. A water storage tank is in the shape ofa cone with the pointed end down. Ifthe radius is 14 ft and the depth of the tank is 15 ft, approximate the vol-ume of the tank in cubic feet. Use

for .

15 ft

14 ft

p227

13. Find the volume of a rectangular box2 ft by 1.4 ft by 3 ft.

14. Find the volume of a box in the shapeof a cube that is 5 ft on each side.

15. Find the volume of a pyramid with asquare base 5 in. on a side and aheight of 1.3 in.

9.

10.

11.

12.

13.

14.

15.

A.6 SQUARE ROOTS AND THE PYTHAGOREAN THEOREM

FINDING SQUARE ROOTS

The square of a number is the number times itself. For example,

The square of 5 is 25 because or .The square of 3 is 9 because or .The square of 10 is or .

The reverse process of squaring is finding a square root. For example,

A square root of 9 is 3 because .A square root of 25 is 5 because .A square root of 100 is 10 because .

We use the symbol , called a radical sign, to name square roots. Forexample,

because .because .

Example 1 Find each square root.

a. b. c. d.

Solution: a. because .b. because .c. because .d. because .

Example 2 Find:

Solution: because .

Example 3 Find:

Solution: because .

APPROXIMATING SQUARE ROOTS

Thus far, we have found square roots of perfect squares. Numbers like

, 36, , and 1 are called perfect squares because their square root is a

whole number or a fraction. A square root such as cannot be written asa whole number or a fraction since 5 is not a perfect square.

Although cannot be written as a whole number or a fraction, it canbe approximated by estimating, by using a table, or by using a calculator.

15

15

425

14

B

25

# 25

= 425A 4

25= 2

5

A 425

16

# 16

= 136A 1

36= 1

6

A 136

92 = 81181 = 912 = 111 = 1

62 = 36136 = 672 = 49149 = 7

18111136149

SQUARE ROOT OF A NUMBER

A square root of a number a is a number b whose square is a. We usethe radical sign 1–to name square roots.

52 = 25125 = 532 = 919 = 3

1

102 = 10052 = 25

32 = 9

10 # 10 = 1001023 # 3 = 9325 # 5 = 2552

A

Square Roots and the Pythagorean Theorem SECTION A.6 763

Objectives

Find the square root of a number.

Approximate square roots.

Use the Pythagorean theorem.CBA

Practice Problem 1Find each square root.

a.

b.

c.

d. 10

1121

164

1100

Answers1. a. 10, b. 8, c. 11, d. 0, 2. , 3. 3

412

Practice Problem 2

Find: A14

Practice Problem 3

Find: A 916


HELPFUL HINT

is approximately 6.557. This means that if we multiply 6.557 by6.557, the product is close to 43.

6.557 * 6.557 = 42.994249

143

Answers4. 3.317, 5. 5.385

PYTHAGOREAN THEOREM

In any right triangle,

leg

leg

hypotenuse

A leg B 2 + Aother leg B 2 = Ahypotenuse B 2

FINDING AN UNKNOWN LENGTH OF A RIGHT TRIANGLE

or

leg = 3Ahypotenuse B 2 - Aother leg B 2

hypotenuse = 3A leg B 2 + Aother leg B 2

Practice Problem 4Use a calculator to approximate thesquare root of 11 to the nearest thou-sandth.

Practice Problem 5Approximate to the nearest thou-sandth.

129

Example 4 Use a calculator to approximate the square root of 43 tothe nearest thousandth.

Solution:

Example 5 Approximate to the nearest thousandth.

Solution:

USING THE PYTHAGOREAN THEOREM

One important application of square roots has to do with right triangles.Recall that a right triangle is a triangle in which one of the angles is a rightangle, or measures 90° (degrees). The hypotenuse of a right triangle is theside opposite the right angle. The legs of a right triangle are the other twosides. These are shown in the following figure. The right angle in the trian-gle is indicated by the small square drawn in that angle.

The following theorem is true for all right triangles.

Using the Pythagorean theorem, we can use one of the following formu-las to find an unknown length of a right triangle.

leg

leg

hypotenuse

C

132 L 5.657

132

143 L 6.557

Example 6 Find the length of the hypotenuse of the given righttriangle.

Solution: Since we are finding the hypotenuse, we use the formula

Putting the known values into the formula, we have

Legs are 6 feet and 8 feet.

The hypotenuse is 10 feet long.

Example 7 Approximate the length of the hypotenuse of the givenright triangle. Round the length to the nearest wholeunit.

Solution:

The hypotenuse is exactly meters, which is approx-imately 20 meters.

Example 8 Find the length of the leg in the given right triangle. Give the exact length and a two-decimal-placeapproximation.

7 in.

5 in.

1389

L 20

= 1389

= 1289 + 100

The legs are 10 metersand 17 meters.

From a calculator

= 3A17 B 2 + A10 B 2hypotenuse = 3A leg B 2 + Aother leg B 2

10 m

17 m

= 10

= 1100

= 136 + 64

hypotenuse = 3A6 B 2 + A8 B 2


8 ft

6 fthypotenuse


Answers6. 20 ft, 7. 11 km, 8. 172 ft ≠ 8.49 ft

Practice Problem 6Find the length of the hypotenuse ofthe given right triangle.

16 ft

12 ft

Practice Problem 7

7 km

9 km

Practice Problem 8Find the length of the leg in the givenright triangle. Give the exact lengthand a two-decimal-place approxima-tion.

11 ft

7 ft

Approximate the length of the hy-potenuse of the given right triangle.Round to the nearest whole unit.


Answers9. 113 yd

✓ Concept Check: a. 82 + 152 = 172

✓ CONCEPT CHECK

The following lists are the lengths ofthe sides of two triangles. Which setforms a right triangle?a. 8, 15, 17b. 24, 30, 40

Solution: Notice that the hypotenuse measures 7 inches and thatthe length of one leg measures 5 inches. Since we arelooking for the length of the other leg, we use the for-mula


The length of the leg is exactly inches, which isapproximately 4.90 inches.


Example 9 Finding the Diagonal Length of a City BlockA standard city block is a square that measures 300 feeton a side. Find the length of the diagonal of a city blockrounded to the nearest whole foot.

Solution: The diagonal is the hypotenuse of a right triangle, so weuse the formula


The length of the diagonal is approximately 424 feet.

L 424

= 1180,000

= 190,000 + 90,000

The legs are both 300feet.

From a calculator

hypotenuse = 3A300 B 2 + A300 B 2


300 ft

300 ft

c

124

L 4.90

= 124

= 149 - 25

The hypotenuse is 7 inches and theother leg is 5 inches.

From a calculator

leg = 3A7 B 2 - A5 B 2

leg = 3Ahypotenuse B 2 - Aother leg B 2

Practice Problem 9A football field is a rectangle measur-ing 100 yards by 53 yards. Draw a dia-gram and find the length of the diago-nal of a football field to the nearestyard.


CALCULATOR EXPLORATIONSFINDING SQUARE ROOTS

To simplify or approximate square roots using a calculator, locate

the key marked .To simplify , for example, press the keys

or

The display should read . Then

To approximate , press the keys

or

The display should read . This is an approximation for. A three-decimal-place approximation is:

Is this answer reasonable? Since 10 is between perfect squares 9and 16, is between and . Our answer is rea-sonable since 3.162 is between 3 and 4.

Simplify.

1. 2.

Approximate each square root. Round each answer to the nearestthousandth.

3. 4.

5. 6. 156197

119115

167611024

116 = 419 = 3110

110 L 3.162

110

3.16227766

101–1–10

110

164 = 8

8

641–1–64

164

1–


Focus On Business and CareerPRODUCT PACKAGING

Suppose you have just developed a new product that you wouldlike to market. You will need to think about who would like to buyit, where and how it should be sold, for how much it will sell, howto package it, and other pressing concerns. Although all of theseitems are important to think through, many package designersbelieve that the packaging in which a product is sold is at least asimportant as the product itself.

Product packaging contains the product, keeping it from leakingout, keeping it fresh if perishable, providing protective cushioningagainst breakage, and keeping all the pieces together as a bundle.Product packaging also provides a way to give information aboutthe product: what it is, how to use it, for whom it is designed, howit is beneficial or advantageous, who to contact if more informa-tion is needed, what other products are necessary to use with theproduct, etc. Product packaging must be pleasing and eyecatchingto the product’s audience. It must be capable of selling its contentswithout further assistance.

CRITICAL THINKING

1. How can a knowledge of geometry be helpful in the packagingdesign process?

2. Design two different packages for the same product that haveroughly the same volume. Does one package “look” larger thanthe other? How could this be useful to a package designer?

769


EXERCISE SET A.6

Find the unknown length of each right triangle. If necessary, approximate the length tothe nearest thousandth. See Examples 6 through 8.

1. 2.

3. 4.

Sketch each right triangle and find the length of the side not given. If necessary, approximatethe length to the nearest thousandth. See Examples 6 through 8.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20. hypotenuse = 7, leg = 6hypotenuse = 2, leg = 1

leg = 110, leg = 132leg = 30, leg = 30

leg = 30, leg = 15leg = 35, leg = 28

leg = 45, hypotenuse = 117leg = 5, hypotenuse = 13

leg = 27, leg = 36leg = 2, leg = 16

leg = 32, leg = 19leg = 10, leg = 14

leg = 48, hypotenuse = 53leg = 6, hypotenuse = 10

leg = 9, leg = 12leg = 3, leg = 4

9 yd3 yd

10cm

12cm

24 ft?

25 ft

12 in.

?5 in.

C

ANSWERS

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

A.7 CONGRUENT AND SIMILAR TRIANGLES

DECIDING WHETHER TRIANGLES ARE CONGRUENT

Two triangles are congruent if they have the same shape and the same size.In congruent triangles, the measures of corresponding angles are equal andthe lengths of corresponding sides are equal. The following triangles arecongruent.

Since these triangles are congruent, the measures of corresponding anglesare equal.

Angles with Equal Measure: and , and , and

Also, the lengths of corresponding sides are equal.

Equal Corresponding Sides: and , and , and

Any one of the following may be used to determine whether two trian-gles are congruent.

FDCAEFBCDEAB

jFjCjEjBjDjA

D

6 in.

5 in.3 in.

E F

A

6 in.

5 in.3 in.

B C

A

Congruent and Similar Triangles SECTION A.7 771

Objectives

Decide whether two triangles arecongruent.

Find the ratio of correspondingsides in similar triangles.

Find unknown lengths of sides insimilar triangles.

C

B

A

CONGRUENT TRIANGLES

Angle-Side-Angle (ASA)

If the measures of two angles of a triangle equal the measures of twoangles of another triangle and the lengths of the sides between eachpair of angles are equal, the triangles are congruent.

For example, these two triangles are congruent by Angle-Side-Angle.

Side-Side-Side (SSS)

If the lengths of the three sides of a triangle equal the lengths of thecorresponding sides of another triangle, the triangles are congruent.

For example, these two triangles are congruent by Side-Side-Side.

11 m10 m

12 m

D

FE

11 m10 m

12 m

A

CB

17 cm

D

FE

45°

35°

17 cm

A

B C

45°

35°


CONGRUENT TRIANGLES, CONTINUED

Side-Angle-Side (SAS)

If the lengths of two sides of a triangle equal the lengths of correspon-ing sides of another triangle, and the measures of the angles betweeneach pair of sides are equal, the triangles are congruent.

For example, these two triangles are congruent by Side-Angle-Side.

50°23 ft

14 ft

E

FD

50°23 ft

14 ft

B

CA

Practice Problem 1Determine whether triangle MNO iscongruent to triangle RTS.

3 mi

90°

4 mi

S R

Q

3 mi

90°

4 miM

O

N

Example 1 Determine whether triangle ABC is congruent to triangleDEF.

Solution: Since the lengths of all three sides of triangle ABC equalthe lengths of all three sides of triangle DEF, the trian-gles are congruent.

In Example 1, notice that once we know that the two triangles are con-gruent, we know that all three corresponding angles are congruent.

FINDING THE RATIOS OF CORRESPONDING SIDES IN SIMILAR TRIANGLES

Two triangles are similar if they have the same shape but not necessarilythe same size. In similar triangles, the measures of corresponding angles areequal and corresponding sides are in proportion. The following trianglesare similar.

Since these triangles are similar, the measures of corresponding anglesare equal.

1410

12

D

E F

75

6

A

B C

B

12 in.

8 in.

13 in.

D

EF

12 in. 8 in.

13 in.

B

A C

Answer1. congruent

Congruent and Similar Triangles SECTION A.7 773

Practice Problem 2Find the ratio of corresponding sidesfor the similar triangles QRS andXYZ.

13 m ZY

X

9 m SR

Q

Practice Problem 3

510

n4

Angles with Equal Measure: and , and , and

Also, the lengths of corresponding sides are in proportion.

Sides in Proportion: , ,

The ratio of corresponding sides is .

Example 2 Find the ratio of corresponding sides for the similartriangles ABC and DEF.

Solution: We are given the lengths of two corresponding sides.Their ratio is

FINDING UNKNOWN LENGTHS OF SIDES IN SIMILAR TRIANGLES

Because the ratios of lengths of corresponding sides are equal, we can useproportions to find unknown lengths in similar triangles.

Example 3 Given that the triangles are similar, find the missinglength n.

Solution: Since the triangles are similar, corresponding sides are inproportion. Thus, the ratio of 2 to 3 is the same as theratio of 10 to n, or

To find the unknown length n, we set cross productsequal.


The missing length is 15 units.

n = 15

n = 302

2 # n = 30

n10

32 =

23

= 10n

210 3

n

C

12 feet17 feet

= 1217

17 ft

D

E F12 ft

A

B C

12

CAFD

= 714

= 12

BCEF

= 612

= 12

ABDE

= 510

= 12

jFjCjEjBjDjA

Answers

2. , 3. n = 89

13

Given that the triangles are similar,find the missing length n.


✓ CONCEPT CHECK

The following two triangles are simi-lar. Which vertices of the first triangleappear to correspond to which ver-tices of the second triangle.

A

B O

N

M

C

Practice Problem 4Tammy Shultz, a firefighter, needs toestimate the height of a burning build-ing. She estimates the length of hershadow to be 8 feet long and the lengthof the building’s shadow to be 60 feetlong. Find the height of the building ifshe is 5 feet tall.

5 ft

8 ft

60 ft

n

Answers4. 37.5 ft

✓ Concept Check: A corresponds to O; B corre-sponds to N; C corresponds to M

TRY THE CONCEPT CHECK IN THE MARGIN.Many applications involve a diagram containing similar triangles. Sur-

veyors, astronomers, and many other professionals use ratios of similar tri-angles continually in their work.

Example 4 Finding the Height of a TreeMel Rose is a 6-foot-tall park ranger who needs to knowthe height of a particular tree. He notices that when theshadow of the tree is 69 feet long, his own shadow is 9feet long. Find the height of the tree.

Solution: 1. UNDERSTAND. Read and reread the problem.Notice that the triangle formed by the sun’s rays, Mel,and his shadow is similar to the triangle formed by thesun’s rays, the tree, and its shadow.

2. TRANSLATE. Write a proportion from the similartriangles formed.

or (ratio in lowest terms)

3. SOLVE for n.


4. INTERPRET. Check to see that replacing n with 46 inthe proportion makes the proportion true. State yourconclusion: The height of the tree is 46 feet.

46 = n

1383

= n

138 = n # 3

6 # 23 = n # 3

233

n6 =

6n

= 323

length of Mel’s shadowlength of tree’s shadow

dd

6n

= 969

SS

Mel’s heightheight of tree

6 ft

9 ft 69 ft

n

Appendix A Highlights 775

APPENDIX A HIGHLIGHTSDEFINITIONS AND CONCEPTS EXAMPLES

SECTION A.1 LINES AND ANGLES

A line is a set of points extending indefinitely in twodirections. A line has no width or height, but it doeshave length. We can name a line by any two of itspoints.

Line AB or dAB

SA B

A line segment is a piece of a line with two end-points.

Line segment AB or AB

A B

A ray is a part of a line with one endpoint. A rayextends indefinitely in one direction.

Ray AB or AB! A B

An angle is made up of two rays that share the sameendpoint. The common endpoint is called thevertex.

A

C

B

Vertex

An angle that measures 180° is called a straightangle.

is a straight angle.jRST

R S

180°

T

An angle that measures 90° is called a right angle.The symbol is used to denote a right angle.

is a right angle.jABC

B C

A

An angle whose measure is between 0° and 90° iscalled an acute angle.

An angle whose measure is between 90° and 180° iscalled an obtuse angle.

45°

120° 95°

Acute angles

Obtuse angles

62°

Two angles that have a sum of 90° are called com-plementary angles. We say that each angle is thecomplement of the other.

R SComplementary angles

60°

60° + 30° = 90°

30°


Two angles that have a sum of 180° are called sup-plementary angles. We say that each angle is thesupplement of the other

M NSupplementary angles

125°55°

125° + 55° = 180°

When two lines intersect, four angles are formed.Two of these angles that are opposite each otherare called vertical angles. Vertical angles have thesame measure.

Two of these angles that share a common side arecalled adjacent angles. Adjacent angles formed inintersecting lines are supplementary.

Vertical angles:and and

Adjacent angles:and and and and jajd

jdjc

jcjb

jbja

jbjd

jcja

acd b

A line that intersects two or more lines at differentpoints is called a transversal. Line l is a transversalthat intersects lines m and n. The eight anglesformed have special names. Some of these namesare:

Corresponding Angles: and , and , and , and

Alternate Interior Angles: and , andje

jdjfjc

jhjdjfjbjgjcjeja

m

n

l

c da b

e fg h

PARALLEL LINES CUT BY A TRANSVERSAL

If two parallel lines are cut by a transversal, then themeasures of corresponding angles are equal and themeasures of alternate interior angles are equal.

SECTION A.2 PLANE FIGURES AND SOLIDS

The sum of the measures of the angles of a triangle is180°.

Find the measure of .

The measure of jx = 180° - 85° - 45° = 50°

85°

45°x

jx

SECTION A.1 (CONTINUED)

A right triangle is a triangle with a right angle. Theside opposite the right angle is called the hypote-nuse and the other two sides are called legs.

hypotenuse

leg

leg


SECTION A.2 (CONTINUED)

For a circle or a sphere:

=R T b

=

b

r = d2

diameter2

radius

d = 2 # r

radius#2diameter

Find the diameter of the circle.

= 2 # 6 feet = 12 feet

d = 2 # r

6 ft

SECTION A.3 PERIMETER

PERIMETER FORMULAS

Rectangle:

Square:

Triangle:

Circumference of a Circle:

or

where or p L 227

p L 3.14

C = p # dC = 2 # p # r

P = a + b + c

P = 4 # s

P = 2 # l + 2 # w

Find the perimeter of the rectangle.

The perimeter is 86 meters.

= 86 m

= 56 m + 30 m

= 2 # 28 m + 2 # 15 m

P = 2 # l + 2 # w

28 m

15 m

SECTION A.4 AREA

AREA FORMULAS

Rectangle:

Square:

Triangle:

Parallelogram:

Trapezoid:

Circle:

A = p # r2

A = 12

# Ab + B B # h

A = b # h

A = 12

# b # h

A = s2

A = l # w

Find the area of the square.

square centimeters

The area of the square is 64 square centimeters.

= 64

= A8 cm B 2A = s2

8 cm

R


SECTION A.5 VOLUME

VOLUME FORMULAS

Rectangular Solid:

Cube:

Sphere:

Right Circular Cylinder:

Cone:

Square-Based Pyramid:

V = 13

# s2 # h

V = 13

# p # r2 # h

V = p # r2 # h

V = 43

# p # r3

V = s3

V = l # w # h

Find the volume of the sphere. Use for .

cubic inches

or cubic inches331121

= 70421

= 4 # 22 # 83 # 7

L 43

# 227

# A2 inches B 3V = 4

3 # p # r3

4 in.

p227

SECTION A.6 SQUARE ROOTS AND THE PYTHAGOREAN THEOREM

SQUARE ROOT OF A NUMBER

A square root of a number a is a number b whosesquare is a. We use the radical sign to namesquare roots.

1

, 1100 = 10, 11 = 119 = 3

PYTHAGOREAN THEOREM

hypotenuse

leg

leg

A leg B 2 + Aother leg B 2 = Ahypotenuse B 2Find the hypotenuse of the given triangle.

inches

inches L 8.5

= 173

= 19 + 64

The legs are 3 and 8 inches. = 3A3 B 2 + A8 B 2


hypotenuse3 in.

8 in.

TO FIND AN UNKNOWN LENGTH OF A RIGHT

TRIANGLE

leg = 2 Ahypotenuse B 2 - Aother leg B 2hypotenuse = 3A leg B 2 + Aother leg B 2


SECTION A.7 CONGRUENT AND SIMILAR TRIANGLES

Congruent triangles have the same shape and thesame size. Corresponding angles are equal and cor-responding sides are equal.

6

53

E

Congruent Triangles

F

D

6

53

B C

A

Similar triangles have exactly the same shape but notnecessarily the same size. Corresponding angles areequal and the ratios of the lengths of correspondingsides are equal.

, ,

CAFD

= 412

= 13

BCEF

= 618

= 13

ABDE

= 39

= 13

34

618

129

B

A

E

D

FC

ANSWERS TO SELECTED EXERCISES

Chapter R PREALGEBRA REVIEW

EXERCISE SET R.2 1. 2. 3. 4. 5. 6. 7. 1 8. 5 9. 7 10. 11. 12.

13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36.

EXERCISE SET R.3 1. 2. 3. 4. 5. 8.71 6. 34.0734 7. 6.5 8. 5.5 9. 15.22 10. 24.69

11. 56.431 12. 76.9931 13. 598.23 14. 823.75 15. 0.12 16. 0.63 17. 67.5 18. 891 19. 43.274 20. 44.16521. 84.97593 22. 115.03226 23. 0.094 24. 5.85 25. 70 26. 40 27. 5.8 28. 3.6 29. 840 30. 96031. 0.6 32. 0.6 33. 0.23 34. 0.45 35. 0.594 36. 63.452 37. 98,207.2 38. 68,936.5 39. 12.35 40. 42.98841. 42. 0.4375 43. 0.625 44. 45. 46. 0.36 47. 0.031 48. 0.02249. 1.35 50. 4.17 51. 0.8149 52. 0.61 53. 0.823; 0.816 54. 87.6% 55. 52.1% 56. 50% 57. 10%58. 6.98 years 59. a. 14.8 pounds b. 89.2 pounds c. 211.2 pounds 60. 64%

Chapter 1 REAL NUMBERS AND INTRODUCTION TO ALGEBRA

EXERCISE SET 1.2 1. 9 2. 4 3. 1 4. 5. 1 6. 10 7. 11 8. 19 9. 8 10. 14 11. 45 12. 50

EXERCISE SET 1.3 1. 9 2. �3 3. �14 4. �20 5. 1 6. 2 7. �12 8. �5 9. �5 10. �11 11. �12

12. 2 13. �4 14. 4 15. 7 16. 2 17. �2 18. �3 19. 0 20. 0

EXERCISE SET 1.4 1. �10 2. �20 3. �5 4. �3 5. 19 6. 17 7. 8. 9. 2 10. 28 11. �11

12. �12 13. 11 14. 9 15. 5 16. 12 17. 37 18. 48 19. �6.4 20. �2.9

EXERCISE SET 1.5 1. �24 2. �40 3. �2 4. �28 5. 50 6. 66 7. �12 8. �16 9. 42 10. 54

11. �18 12. �15 13. 14. 15. 16. 17. �7 18. �15 19. 0.14 20. 0.15

EXERCISE SET 1.6 1. �9 2. �2 3. 4 4. 3 5. �4 6. �12 7. 0 8. 0 9. �5 10. �3 11. undefined

12. undefined 13. 3 14. 5 15. �15 16. �7 17. 18. 19. 20. 21. �1

EXERCISE SET 1.7 1. 4x+4y 2. 7a+7b 3. 9x � 54 4. 11y � 44 5. 6x+10 6. 35+40y 7. 28x � 218. 24x � 3 9. 18+3x 10. 2x+10 11. �2y+2z 12. �3z+3y 13. �21y � 35 14. �10r � 5515. 5x+20m+10 16. 24y+8z � 48 17. �4+8m � 4n 18. �16 � 8p � 20 19. �5x � 2 20. �9r � 521. �r+3+7p

EXERCISE SET 1.8 1. approx. 7.8 million 2. approx. 1.6 million 3. 2002 4. approx. 7.4 million5. 1994 6. 1995 7. 1986, 1987, or 1989, 1990 8. 1986 or 1987 9. approx. 52%10. approx. 49% 11. 1994 12. answers may vary

Chapter 2 EQUATIONS, INEQUALITIES, AND PROBLEM SOLVING

EXERCISE SET 2.1 1. �7 2. 17 3. like 4. unlike 5. 15y 6. 5x 7. 13w 8. �4c 9. �7b � 910. 3g � 2 11. �m � 6 12. �3a � 2 13. �8 14. �11 15. 5y+20 16. 7r+21 17. 7d � 1118. 10x � 3 19. 4y+11 20. �4x � 9

EXERCISE SET 2.2 1. x � 3 2. x � 11 3. x � �2 4. y � 10 5. x � �14 6. z � �16 7. r � 0.5 8. t � 2.4

9. f � 10. c �5

24512

1180

2027

-85

-187

1560

=14

2436

=23

124

310

-18

16

310

0.16 L 0.170.54 L 0.550.3 L 0.333

723100

186100

910

610

4344

1718

138

75

16

566

25

57

5235

6521

1112

2321

15

13

411

572

380

2147

16

18

1835

3061

83

15

45

1415

35

59

37

34

12

2025

1620

A1

ANSWERS

MENTAL MATH 1. a � 9 2. c � 6 3. b � 2 4. t � 2 5. x � �5

EXERCISE SET 2.3 1. x � �4 2. x � 7 3. x � 0 4. x � 0 5. x � 12 6. y � �8 7. x � �12 8. n � �209. d � 3 10. v � 2 11. x � 10 12. x � 9 13. x � �20 14. x � 28 15. a � 0

EXERCISE SET 2.6 1. h � 3 2. r � 65 3. h � 3 4. V � 336 5. h � 20 6. h � 12 7. c � 12 8. A � 27

9. r � 2.5 10. A � 63.585 11. T � 3 12. P � 3,200,000 13. h � 15 14. V � 113.04 15. h � 16. r �

17. W � 18. n � 19. y � 7 � 3x 20. y � 13+x

EXERCISE SET 2.7 1. 11.2 2. 880 3. 55% 4. 20% 5. 180 6. 360 7. 4.6 8. 120.4 9. 50

10. 125 11. 30% 12. 120% 13. x � 4 14. x � 15. x � 16. x � 17. x � 18. a �

19. x � 7 20. y � 40

MENTAL MATH 1. 2. 3. 4. 5. 6.

7. 8.

EXERCISE SET 2.8 1. (–q, –1] 2. (–3, q] 3. (–q, 11] 4. (4, q) 5. (–13, q) 6. 7. (–q, 7]

8. 9. [0, q) 10. 11. (–q, –5] 12. [–1, q)

Chapter 3 GRAPHING EQUATIONS AND INEQUALITIES

MENTAL MATH 1. answers may vary; Ex. (5, 5), (7, 3) 2. answers may vary; Ex. (0, 6), (6, 0)

EXERCISE SET 3.1

3. a � b 4. answers may vary 5. (0, 0) 6. 7. (3, 2) 8. (�1, 3) 9. (�2, �2) 10. (0, �1) 11. (2, �1)

12. (2, 0) 13. (0, �3) 14. (�2, 3) 15. (1, 3) 16. (1, �1) 17. (�3, �1) 18. (�2, 0)

EXERCISE SET 3.21. (6, 0); (4, �2); (5, �1) 2. (0, �4); (6, 2); (�1, �5) 3. (1, �4); (0, 0); (�1, 4)

(−1, 4)

(1, −4)

(0, 0)x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(0, −4)(−1, −5)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

(6, 2)

(4, −2)(5, −1)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

(6, 0)

a312

, 0 b

a 143

, q ba 12

, q bc 83

, q bEx @x ≥ 8FEx @x ≤ 2F

Ex @x 6 4FEx @x 7 4FEx @x ≤ 7FEx @x ≥ 10FEx @x 7 7FEx @x 6 6F

152

214

34

509

163

T

mr

V

LH

C

2pf

5g

A2

ANSW

ERS

1. (1, 5) is in quadrant I, is in quadrant II, (�5, �2) is in

1. quadrant III, (2, �4) is in quadrant IV, (�3, 0) lies on the x-axis,(0, 1) lies on the y-axis

(−3, 0)

(−5, −2)

(2, −4)

(0, −1)

(1, 5)−1, 41

2)(

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

a -1, 412b 2. (2, 4) is in quadrant I, (�2, 1) is in quadrant II, (�3, �3) is in

2. quadrant III, (5, �4) is in quadrant IV, lies on the 2. x-axis, (0, 2) lies on the y-axis

(5, −4)

(−2, 1)

(−3, −3)

(0, 2)

(2, 4)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

, 0)( 343

a334

, 0 b

4. (1, �5); (0, 0); (�1, 5) 5. (0, 0); (6, 2); (�3, �1) 6. (0, 0); (�4, �2); (2, 1)

7. (0, 3); (1, �1); (2, �5) 8. (0, 2); (1, �3); (2, 8) 9.

10. 11. 12.

13. 14. 15.

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(2, −8)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10(1, −3)

(0, 2)

(1, −1)

(2, −5)

(0, 3)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(−4, −2)

(2, 1)(0, 0)x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5(−3, −1)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

(0, 0)(6, 2)

(−1, 5)

(1, −5)

(0, 0)x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

A3

ANSWERS

16. 17. 18.

19. 20. 21.

22. 23. 24.

25.

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

A4

ANSW

ERS

EXERCISE SET 3.31. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

(0, 10)

(−5, 0)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

(0, 6)

(−2, 0)x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

(0, 0)x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

(0, 0)x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(0, 2)

(3, 0)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(4, 0)

(0, −2)

(0, 4)(−8, 0)

x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10

(0, 3)(−6, 0)x

y

2−2

2468

10

−4−6−8

−10

−4−6−8−10 4 6 8 10x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(0, 0)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(0, 0)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(0, 4)

(−4, 0)

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(3, 0)

(0, −3)

A5

ANSWERS

13. 14. 15.

16. 17. 18.

MENTAL MATH 1. upward 2. downward 3. horizontal 4. vertical

EXERCISE SET 3.4 1. m � 2. m � 3. m � 4. m � 5. m � 6. m � �5 7. m � �1 8. m �

9. m � 10. m � �5 11. m � 12. m � 5 13. m � �2 14. m � �2 15. m � 5 16. m �

Chapter 4 EXPONENTS AND POLYNOMIALS

MENTAL MATH 1. base: 3; exponent: 2 2. base: 5; exponent: 4 3. base: �3; exponent: 6 4. base: 3; exponent: 75. base: 4; exponent: 2

EXERCISE SET 4.1 1. 49 2. �9 3. �5 4. 9 5. �16 6. 4 7. �8 8. 135 9. 4 10. 15011. x7 12. y3 13. x36 14. y35 15. p7q7 16. x2 17. y 18. 4 19. �125 20. p6q5

MENTAL MATH 1. 2. 3. y6 4. x3 5. 4y3 6. 16y7

EXERCISE SET 4.2 1. 2. 3. 4. 5. x4 6. y3 7. p4 8. y6 9. 7.8 � 104 10. 9.3 � 109

11. 1.67 � 10�6 12. 1.7 � 10�7 13. 6.35 � 10�3 14. 1.94 � 10�3

EXERCISE SET 4.3 1. 1; –3x; 5 2. 2x3; –1; 4 3. �5; 3.2; 1; �5 4. 9.7; �3; 1; 5. (a) 6; (b) 5 6. (a) �10; (b) �12

7. (a) �2; (b) 4 8. (a) �4; (b) �3 9. (a) �15; (b) �16 10. (a) �6; (b) �1 11. 23x2 12. 14x3 13. 12x2 � y14. 3k3+11 15. 7s 16. 9ab-11a 17. �10xy+y 18. 4x2-7xy+3y2 19. 3a2-16ab+4b2 20. –3xy2+4

EXERCISE SET 4.4 1. 12x+12 2. 6x2+16 3. –3x2+10 4. 4x2-5 5. –3x2+4 6. –x+14 7. 5x2+2y2

8. –2x+9 9. y+8 10. 2x2+7x-16 11. 5x-9 12. 4x-3 13. 6y+13 14. 11y+7 15. –2x2+8x-116. –2a-b+1 17. 10x-4+5y 18. 3x2+5 19. 9a2-4b2+22 20. 6x2-2xy+19y2

MENTAL MATH 1. x8 2. x8 3. y5 4. y10 5. x14

EXERCISE SET 4.5 1. 24x3 2. 18x3 3. –12.4x12 4. –15.6x8 5. x4 6. x2+7x+12 7. x2+11x+188. x3 � 5x2+13x � 14 9. x3+8x2+7x � 24 10. x4 � 5x3 � 3x2 � 11x+20 11. a4 � a3 � 6a2+7a+1412. 10a3 � 27a2 � 26a � 12 13. –3b3 � 14b2 � 13b+6 14. 49x2y2 � 14xy2+y2 15. x4 � 8x2+16

EXERCISE SET 4.6 1. x2+7x+12 2. x2+6x+5 3. x2+5x-50 4. y2-8y-48 5. 5x2+4x-126. 6y2-31y+35 7. 4y2-25y+6 8. 2x2-31x+99 9. 6x2+13x-5 10. 6x2-10x-4 11. a2 � 4912. b2 � 9 13. x2 � 36 14. x2 � 64 15. 9x2 � 1 16. 16x2 � 25 17. x4 � 25 18. a4 � 36 19. 4y4 � 1 20. 9x4 � 1

-14

1343x3

7x3

136

164

3x3

5x2

23

-

23

-

14

52

87

-

74

52

52

-

43

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(0, 0)x

y

1−1

12345

−2−3−4−5

−2−3−4−5 2 3 4 5

(0, 0)

A6

ANSW

ERS

MENTAL MATH 1. a2 2. y 3. a2 4. p5 5. k3 6. k2

EXERCISE SET 4.7 1. 4x3-2x2+3x+1 2. 12x3+3x 3. 15x-9x4 4. 5p2+6p 5. 2m- 6. 2x+17. ab-b2 8. m¤n-n3

Chapter 5 FACTORING POLYNOMIALS

EXERCISE SET 5.1 1. 3(a+2) 2. 6(3a+2) 3. 15(2x-1) 4. 7(6x-1) 5. x2(x+5) 6. y4(y-6)7. 2y(3y3-1) 8. 5x2(1+2x4) 9. 2x(16y-9x) 10. 5x(2y-3x) 11. a2b2(a5b4-a+b3-1)12. x3y3(x6y3+y2-x+1) 13. 5xy(x2-3x+2) 14. 7xy(2x2+x-1) 15. 4(2x5+4x4-5x3+3)

16. 3(3y6-9y4+6y2+2) 17. x(x3+2x2-4x+1) 18. y(2y6-4y4+3y-2) 19. (x+2)(y+3)

20. (y+4)(z+3)

EXERCISE SET 5.2 1. (x+6)(x+1) 2. (x+4)(x+2) 3. (x-9)(x-1) 4. (x-3)(x-3) 5. (x-6)(x+3)6. (x-6)(x+5) 7. (x+10)(x-7) 8. (x+8)(x-4) 9. prime 10. prime 11. 2(z+8)(z+2)12. 3(x+7)(x+3) 13. 2x(x-5)(x-4) 14. x(x-8)(x+7) 15. (x-4y)(x ± y) 16. (x-11y)(x ± 7y)17. (x ± 12)(x+3) 18. (x ± 4)(x+15) 19. (x-2)(x+1) 20. (x-7)(x+2)

EXERCISE SET 5.3 1. x+4 2. y+5 3. 10x-1 4. 3y-2 5. 4x-3 6. (2x+3)(x ± 5) 7. (3x+2)(x ± 2)8. (y-1)(8y-9) 9. (7x-2)(3x-5) 10. (2x+1)(x-5) 11. (9r-8)(4r+3) 12. (4r-1)(5r+8)13. (3x-7)(x � 9) 14. (5x � 1)(2x � 3) 15. (2x � 5)(x � 1) 16. x(3x+2)(4x+1) 17. a(4a+1)(2a+3)18. 3(7x+5)(x � 3) 19. 2(3x � 5)(2x+1) 20. (3x+4)(4x � 3)

EXERCISE SET 5.4 1. (x+3)(x � 2) 2. (x+3)(x � 5) 3. (x � 4)(x � 7) 4. (x � 6)(x � 2) 5. (y+8)(y � 2)6. (z+10)(z � 7) 7. (3x+4)(x+4) 8. (2x+5)(x+7) 9. (8x � 5)(x � 3) 10. (4x � 9)(x � 8)

MENTAL MATH 1. 12 2. 52 3. (3x)2 4. (4y)2

EXERCISE SET 5.5 1. yes 2. yes 3. no 4. no 5. (x+11)2 6. (x+9)2 7. (x-8)2 8. (x-6)2

9. (4a-3)2 10. (5x+2)2 11. (x2+2)2 12. (m2+5)2 13. (x � 2)(x+2) 14. (x+6)(x � 6) 15. (9 � p)(9+p)16. (10 � t)(10+t) 17. (2r � 1)(2r+1) 18. (3t � 1)(3t+1) 19. (n2+4)(n+2)(n-2) 20. xy(x � 2y)(x+2y)

21. (4-ab)(4+ab) 22. 23.

EXERCISE SET 5.6 1. x=2, x=�1 2. x=�3, x=�2 3. x=6, x=7 4. x=�4, x=10 5. x=�9, x=�176. x � 9, x � 4 7. x � �9, x � 7 8. x � �4, x � 2 9. x � 3, x � 2 10. x � 0, x � 7 11. x � 0, x � 312. x � 0, x � �20 13. x � 0, x � �15 14. x � 4, x � �4 15. x � 3, x � �3 16. x � 8, x � �4 17. x � 8, x � �318. x � , x � �2 19. x � , x � 3 20. x � , x � �9

EXERCISE SET 5.7 1. width � x; length � x+4 2. width: x; length: 2x 3. x and x+2 if x is an odd integer4. x and x+2 if x is an even integer 5. base � x; height � 4x+1 6. height � x; base � 5x � 37. 11 units 8. length � 12 in.; width � 7 in. 9. 15 cm, 13 cm, 70 cm, 22 cm 10. 14 ft, 19 ft, 52 ft

Chapter 6 RATIONAL EXPRESSIONS

EXERCISE SET 6.1 1. 2. 3. 4. 5. can’t simplify 6. 7. 1 8. 1

9. �1 10. �1 11. �5 12. 13. 14. 15. 5x+1 16. 6x � 1 17. 18.

19. x+2 20. 4x

MENTAL MATH 1. 2. 3. 4. 5.

EXERCISE SET 6.2 1. 2. 12 3. x4 4. 5. 6. 7. 8. 9. 2x 10. 1 11. x4 12.

13. 14. 15. x(x+4) 16. 17. 4 18. 19. 20.

MENTAL MATH 1. 1 2. 3. 4. 5. 6.

EXERCISE SET 6.3 1. 2. 3. 4. 7p 5. 4 6. 8 7. 8. 1 9. 3 10. x � 2

11. 12. 13. 14.

EXERCISE SET 6.4 1. 2. 3. 4. 5. 6. 7.35x - 6

4x 1x - 2 23x - 7

1x - 2 2 1x + 2 26x + 5

2x2

20c - 8dx

5d

75a + 6b2

5b

7321a

5x

1x - 1

1x - 6

3y + 5

1a + 5

y + 103 + y

3m

n

x + 77

a + 913

17y

519

5y

87x

9611

32

9x

43x

856

149b2

12y6

9y

213

1x

x2

10-

b2

614

214y

95

4x5

11z3

5y2

7x2

3x

4y

2x

3y

1x - 7

1x - 2

1x - 3

1x - 9

7x

34

1x - 5

1x + 2

13x + 2

14 1x + 2 2

83- 1

473

ay -14b ay +

14bax -

12b ax +

12b

15

13

27m2

7

A7

ANSWERS

8. 9. 10. 0 11. 2 12. 3 13. 3x3 � 4 14. 15. 16.

17. 18. 19. 20.

MENTAL MATH 1. x � 10 2. x � 32 3. z � 36 4. y � 56

EXERCISE SET 6.5 1. x � 30 2. x � 55 3. x � 0 4. x � 0 5. x � �2 6. x � �1 7. x � �5, x � 28. y � �1, y � 7 9. a � 5 10. b � 10 11. x � 3 12. y � 1 13. y � 5 14. x � �3 15. x � 9 16. no solution

EXERCISE SET 6.6 1. 2 2. 4 3. �3 4. �3 or �2 5. 5 6. 3 7. 2 8. 4 9. 2 hr 10. 2 hr 11. 1 min12. 6 min 13. $108.00 14. 2 days 15. 3 hr 16. 6 hr 17. 20 hr 18. first pump: 28 min; second pump: 84 min19. 6 mph 20. time traveled by jet: 3 hr; time traveled by car: 4 hr

EXERCISE SET 6.7 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

12. 13. 14. 15. 16. 2x-1 17. 18. 19. 20.

Chapter 7 GRAPHS AND FUNCTIONS

MENTAL MATH 1. m=–4, b=12 2. 3. m=5, b=0 4. m=–1, b=0 5.

EXERCISE SET 7.11. 2. 3.

4. 5. 6.

7. 8. 9.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(3, 0)x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

(0, 7)

(−5, 2)

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(0, 0)x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(−2, −4)

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(1, 3)

x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

m =12

, b = 6m =23

, b = -72

3 + 40y

3 - 40y

x + y

x - y

569

16

1y - 1

6 + 4y2

6y - 5y2

2x 1x - 5 27x2 + 10

x + 4x - 4

m - n

m + n

3763

2716

15

43

-16

-4x

15-

2722

23

-310

23

23

29

12

29

6 - x

x

2 + m

m

5 1y + 2 23y 1y + 5 2

9b - 45b 1b - 1 2

2 1x + 3 21x - 2 2 2

x + 21x + 3 2 2

5x + 45x2

6-

2x - 3

5 + 10y - y3

y2 12y + 1 2

A8

ANSW

ERS

10. 11.

12. y=–x+1 13. 14. 15.

MENTAL MATH 1. m=–2; (1, 4) 2. m=–3; (4, 6) 3. 4. 5. m=5; (3, –2)

EXERCISE SET 7.2 1. y=3x-1 2. y=4x-19 3. y=–2x-1 4. y=–4x+4 5.

6. y=3x-6 7. y=2x-6 8. y=–2x+1 9. y=–2x+10 10. 11. x=2 12. y=–4

13. y=1 14. x=4 15. x=0 16. y=4x-4 17. y=3x+2 18. 19.

20.

EXERCISE SET 7.4 1. 57 2. 79 3. 499 4. 13 5. 1 6. –1 7. 9 8. 47 9. –16 10. –6 11.

12. 1.6 13. 0.4 14. –8 15. 16. 0.7 17. 6 18. 2.5 19. –8 20. –4.5

EXERCISE SET 7.5 1. a. domain, (–q, q); range, (–q, 5] b. x-intercept points, (–2, 0), (6, 0); y-intercept point, (0, 5) c. (0, 5)d. There is no such point. 2. a. domain, (–q, q); range, [1, q) b. no x-intercept points; y-intercept point, (0, 4)c. There is no such point. d. (–3, 1) 3. a. domain, (–q, q); range, [–4, q) b. x-intercept points, (–3, 0), (1, 0); y-intercept point, (0, –3) c. There is no such point. d. (–1, –4) 4. a. domain, (–q, q); range, [1, q) b. no x-intercept points; y-intercept point, (0, 13) c. There is no such point. d. (2, 1) 5. a. domain, (–q, q); range, (–q, q) b. x-intercept points,(–2, 0), (0, 0), (2, 0); y-intercept point, (0, 0) c. There is no such point. d. There is no such point. 6. a. domain, (–q, q); range,(–q, q) b. x-intercept point, (0, 0); y-intercept point, (0, 0) c. There is no such point. d. There is no such point.

7. 8. 9.

10. 11. 12.

(0, 0)x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(0, 0)x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(0, −2)

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(0, 1)

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(0, 0)x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(0, 0)x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

-73

= -213

103

= 313

y = -32

x - 6

y =14

x + 9y =12

x - 6

y = -12

x - 5

y =12

x + 5

m = -23

; 10, 1 2m =14

; 12, 0 2

y = -3x -15

y = 2x +34

y =12

x - 6

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

A9

ANSWERS

13. 14. 15.

16. 17. 18.

19. 20.

Chapter 8 SYSTEMS OF EQUATIONS AND INEQUALITIES

EXERCISE SET 8.2 1. (2, 1) 2. (15, 5) 3. (�3, 9) 4. (�2, 8) 5. (4, 2) 6. (3, 4) 7. (10, 5) 8. (�2, �4)

9. (2, 7) 10. (1, 5) 11. 12. 13. (�2, 4) 14. (7, �4) 15. (�2, �1) 16. no solution

17. no solution 18. (�3, 1) 19. (3, �1) 20. (�4, �2) 21. (3, 5) 22. (1, 4) 23. 24.

25. (�1, �4)

Chapter 9 RATIONAL EXPONENTS, RADICALS, AND COMPLEX NUMBERS

EXERCISE SET 9.1 1. 2, –2 2. 3, –3 3. 10 4. 20 5. 6. 7. 0.01 8. 0.2 9. 2.646 10. 3.317

11. 4 12. 3 13. 14. 15. –1 16. –5 17. –2 18. –3 19. not a real number 20. not a real number

21. –2 22. 23. –2 24. –1 25. 1 26. –3

EXERCISE SET 9.3 1. 2. 3. 2 4. 3 5. 6. 7. 8. 9.

10. 11. 12. 13. 14. 15. 16. 17. 18. 19.3143

42y

3x

42x3

21511

127

2129

167

4227a2b34220x3A 6n

5m

A 14xy

115xy16x315031361110114

13

34

12

35

12

a -95

, 35ba 2

3, -

13b

a -73

, -443ba -

15

, 435b

x

y

21

345

−2−3−4−5

−1−2−3−4−5 21 3 4 5

(0, 4)

x

y

21

345

−2−3−4−5

−1−2−3−4−5 21 3 4 5

(0, 1)

x

y

21

345

−2−3−4−5

−1−2−3−4−5 21 3 4 5

(0, 0) (2, 0)

(1, 3)

x

y

21

345

−2−3−4−5

−1−2−3−4−5 21 3 4 5

(0, 0) (3, 0)

92, − )(3

2

x

y

42

68

10

−4−6−8

−10

−2−4−6−8−10 42 6 8 10

(5, 0) (7, 0)

(0, 35)

(6, −1)

x

y

84

121620

−8−12−16−20

−4−12−20 4 12 20

(−4, 0)

(0, −24)

(6, 0)

(1, −25)

(−5, 0)

(−3, −4)

(−1, 0)

(0, 5)

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

(−4, −9)

(−7, 0) (−1, 0)

(0, 7)

x

y

42

68

10

−4−6−8

−10

−4−2−6−8−10 42 6 8 10

A10

ANSW

ERS

20. 21. 22. 23. 24. 25. 26. 27.

MENTAL MATH 1. 2. 3. 4. 5. 6. 7. 3 8. 4x+1

EXERCISE SET 9.4 1. 2. 3. 4. 5. 6. 7.

8. 9. 10. 11. 12. 13. 14. 15.

16. 17.

Chapter 10 QUADRATIC EQUATIONS AND FUNCTIONS

EXERCISE SET 10.1 1. {–4, 4} 2. {–7, 7} 3. {– , } 4. {– , } 5. {–3 , 3 } 6. {–2 , 2 }7. 8. {– , } 9. {–8, –2} 10. {1, 5} 11. {6-3 , 6+3 } 12. {–4-3 , –4+3 }

13. 14. 15. {–3i, 3i} 16. {–2i, 2i} 17. {– , }

18. {– , } 19. 20. 21. {1-4i, 1+4i} 22. {–1-5i, –2+5i}

23. 24. {–10- , –10+ } 25. {–3-2i , –3+2i }

MENTAL MATH 1. a=1, b=3, c=1 2. a=2, b=–5, c=–7 3. a=7, b=0, c=–4 4. a=1, b=0, c=9

5. a=6, b=–1, c=0

EXERCISE SET 10.2 1. {–6, 1} 2. {–12, 1} 3. 4. 5. {3} 6. {–5}

7. 8. 9. 10.

11. 12. 13. 14.

15. 16. {–10, 2} 17. 18. 19.

20.

Chapter 11 EXPONENTIAL AND LOGARITHMIC FUNCTIONS

EXERCISE SET 11.3

1. 2. 3.

4. 5. 6.

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = )( x15

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = )( x14y = 3x − 1

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = 1 + 2x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = 5x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = 4x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

e -3 - i163

, -3 + i16

3f

e -5 - i1510

, -5 + i15

10f51 - 12, 1 + 126e 3 - 111

2,

3 + 1112

fe -32

, 1 f5-3 - 17, -3 + 17651 - 13, 1 + 136e 5 - 177

2,

5 + 1772

fe 7 - 1856

, 7 + 185

6f

e 9 - 51522

, 9 + 515

22fe 1 - 157

8,

1 + 1578

fe -5 - 1132

, -5 + 113

2fe 7 - 133

2,

-7 + 1332

fe -

15

, 3 fe -35

, 1 f

12121111115-7 - 15, -7 + 1565-2i13, 2i1365-2i12, 2i126110110

1616e -9 - 164

, -9 + 16

4fe 3 - 212

2,

3 + 2122

f1313121212125-110, 1106151512121111111717

14 + 137x17

10

515x

93 314

14

31113

11136

311215

6x31x5b1b-7a 313a

- 312x29121712 - 151510x15x10x12x-213-212

6 31z12 31x131y31x817613

15x

2y

y1z

6x1y

10

3132x2

312x

3y4 313

42a3

3

418x2

3134

A11

ANSWERS

7. 8. 9.

10.

EXERCISE SET 11.4 1. 6¤=36 2. 2fi=32 3. 4. 5. 10‹=1000 6. log™ 16=4

7. log∞ 125=3 8. log¡º 100=2 9. log¡º 1000=4 10. log x=3 11. 3 12. 2 13. –2 14. –5 15.

16. {2} 17. {3} 18. {81} 19. {8} 20. {7}

21. 22. 23.

24. 25.

EXERCISE SET 11.6 1. 0.9031 2. 0.7782 3. 0.3636 4. 0.6866 5. 0.6931 6. 1.0986 7. –2.6367 8. –5.7446

9. 1.1004 10. 1.4133 11. 2 12. 4 13. –3 14. –1 15. 2 16. 4 17. 18. 19. 3 20. 515

14

f(x) = log5x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

f(x) = log1/2x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

f(x) = log1/4x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = log2x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = log3x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

12e

5-2 =1

253-3 =

127

y = − 3x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = − 2x

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = + 2)( x13

x

y

21

345

−2−3−4−5

−2−1−3−4−5 21 3 4 5

y = − 2)( x12

A12

ANSW

ERS

Appendix A GEOMETRY

EXERCISE SET A.3 1. 64 ft 2. 28 m 3. 36 cm 4. 184 mi 5. 21 in. 6. 22 units 7. 120 cm 8. 10 yd 9. 48 ft10. 66 m 11. 96 m 12. 86 in. 13. 66 ft 14. 58 cm 15. 128 mi 16. 90 km 17. 17∏ cm; 53.38 cm18. 12∏ in. 37.68 in 19. 16∏ mi; 50.24 mi 20. 50∏ ft; 157 ft

EXERCISE SET A.4 1. 7 sq. m 2. 19.25 sq. ft 3. sq. yd 4. 11 sq. ft 5. 15 sq. yd 6. 17.5 sq. ft

7. 2.25∏ sq. in.≠7.065 sq. in. 8. 4∏ sq. cm≠ sq. cm 9. 36.75 sq. ft 10. 12.75 sq. cm 11. 28 sq. m 12. sq. in.

13. 22 sq. yd 14. 22.5 sq. ft 15. sq. ft 16. sq. cm 17. sq. in. 18. 24 sq. m 19. 25 sq. cm

20. 82 sq. km 21. 86 sq. mi 22. 360 sq. cm 23. 24 sq. cm 24. 45 sq. in. 25. 36∏ sq. in.≠113.1 sq. in.

EXERCISE SET A.5 1. 72 cu. in. 2. 113 cu. mi 3. 512 cu. cm 4. 128 cu. cm 5. cu. yd 6. cu. ft

7. cu. in. 8. cu. in. 9. 75 cu. cm 10. 18 cu. ft 11. cu. in. 12. 3080 cu. ft 13. 8.4 cu. ft

14. 125 cu. ft 15. cu. in.

EXERCISE SET A.6 1. 13 in. 2. 34.655 ft 3. 6.633 cm 4. 8.485 yd 5. 5 6. 15 7. 8 8. 22.4729. 17.205 10. 37.216 11. 16.125 12. 45 13. 12 14. 108 15. 44.822 16. 33.541 17. 42.426 18. 171.82519. 1.732 20. 3.606

10

56

21027

28

78

5231721

47137

1247

17

2212

1234

36

34

40

12

12

47

14

9

34

A13

ANSWERS

Index

AAcute angle, 733, 775Addition:

associative property of, 199–200commutative property of, 199of complex numbers, 624–25, 631of decimals, 148of fractions, 142–43of functions, 687–88identity for, 201of polynomials, 355, 377of radical expressions, 607–8of real numbers, 177–79solving problems that involve, 179

Addition method, solving systems oflinear equations by, 547–52

Addition property of equality, 225–29,276

using, 235–37Addition property of inequality, 268–70Additive inverses, 179–80, 202, 214Adjacent angles, 735, 776Algebra:

absolute value, 168algebraic expressions, evaluating, 171common sets of numbers, identifying,

165–67equality and inequality symbols,

163–64exponents, 168–71mathematical statements, 163

translating into sentences, 164–65order of operations, 168–71sets of numbers, 163solutions of equations, 172symbols, 163–68

translating words into, 172–73Algebra of functions, 687–90Algebraic expressions:

defined, 171evaluating, 171, 184–85writing, 222, 229, 236–37

America’s Cup sailing competition, 379Angles, 731–32, 775

acute, 733, 775adjacent, 735, 776classifying, 733–34complementary, 733–34degrees, 732finding measures of, 734–36intersecting lines, 734obtuse, 733, 775parallel lines, 734perpendicular lines, 735

right, 733, 775side, 731straight, 732, 775supplementary, 733–34, 776transversal, 736, 776vertical, 735, 776

Angle-side-angle (ASA), 771Area, 749–52, 777Assets, 673Associative property of addition, 199–200Associative property of multiplication,

199–200Axis of symmetry, 516

BBar graphs,

reading, 205–7Base, 169

of exponential expression, 331Best fit equation, 498Binomials, 348, 376

squaring, 364–65Boundary, 320–21Brace symbols, 163Break-even point, 544Briggs, Henry, 718Bureau of Labor Statistics, 559Business terms, 374

CCalculator, 174

checking solutions of equations on,246

equations, checking, 246and exponential expressions, 174graphing, 294, 316, 407, 490, 497, 511,

522, 622, 642, 672, 690, 724entering negative numbers on, 196

order of operations, 174scientific, entering negative numbers

on, 196scientific notation, 344and square roots, 767

Cardinal numbers, See Noun determinersCareers, 454Carry, Max, 559Cartesian coordinate system, 528Center, circle, 739Center of gravity, 529Centers of mass, 529Change of base formula, 717Chinese numerals, 192Circle, 739

circumference of, 744–45

Circle graphs, See Pie chartsCircular cylinder, volume formula of, 757Circumference, 744–45Coefficient, 219, 276, 347, 376Column, matrix, 567Combining like terms, 220–21, 276Common logarithms:

approximating, 715defined, 715of powers of 10, evaluating, 715–16

Common sets of numbers, 167identifying, 165–67

Commutative property of addition, 199Commutative property of multiplication,

199Complementary angles, 733–34, 775

finding, 186Completing the square:

defined, 637–39solving quadratic equations by,

635–36Complex conjugates, 625–26Complex fractions, simplifying, 473–76,

481–82Complex numbers, 623–27, 631

adding, 624–25, 631dividing, 625–26, 631imaginary number, 624imaginary unit, 623multiplying, 625, 631powers of i, finding, 626–27subtracting, 624–25, 631

Complex number system, 623Composite numbers, 135Composition of functions, 688–90Composition with Gray and Light Brown

(Mondrian), 230Compound interest, 640Cone, 740

volume formula of, 758Congruent triangles, 771–72, 779

angle-side-angle (ASA), 771side-angle-side (SAS), 772side-side-side (SSS), 771

Conjugate, 615, 630Consistent system, 561Constant, 347Coordinate plane, See Rectangular

coordinate systemCorner point, 578Cost of sales, 673Cross products, 278Cube, 739

volume formula of, 757

I-1

I-2 INDEX

Cube roots, finding, 592–93, 629Cubic functions, 515

graphing, 519–20Current ratio, 673Cylinder, 740

volume formula of, 757

DData, paired, 283Decibel, 714Decimals, 147–52

adding, 148dividing, 148–49multiplying, 148rounding, 149solving equations containing,

242–44subtracting, 148writing as fractions, 147writing fractions as, 150–51writing as percents, 151–52

“Decreased by”, use of term, 173Degree of a polynomial, 377Degrees, angles, 732Denominators:

rationalizing, 613–14denominators having two terms,

615–17Dependent equations, 536Dependent variable, 506Descartes, Rene, 352, 528, 604Determinants:

defined, 571, 586solving systems of equations using,

571–763 � 3 determinants, 574–752 � 2 determinants, 571

Diagrams, scatter, 283Diameter:

circle, 739sphere, 739

Die Cross (Rudolff), 604Difference, of squares, 365Diffusion, 610Diophantus, 471Discriminant, 648Distributive property of multiplication

over addition, 201Divisibility tests, 136Division:

of complex numbers, 625–26, 631

of decimals, 148–49of fractions, 141–42of functions, 687–88of polynomials, 369–73, 378of rational expressions, 439–40of real numbers, 193–96, 214–15synthetic, 372–73, 378

Domain, 501, 526finding from a graph, 505–6

EEarnings per share, 673Egyptian hieroglyphics numerals, 192Element, set, 163Elementatry row operations, matrix, 567Elements, matrix, 567Elimination method, solving systems of

linear equations by, 547–52Ellipsis in sets, 163Epigram of Diophates, 471Equality:

addition property of, 225–29multiplication property of, 233–37, 276

Equality symbol, 163Equations:

checking solutions of, on calculator,246

defined, 172dependent, 536equivalent, 225, 276with no solution, 244–45simplifying, 228–29

Equations of lines:point-slope form, 493–97, 526slope-intercept form, 487–92

Equivalent equations, 225, 276Equivalent fractions, 139–40, 143Equivalent rational expressions, writing,

448Eratosthenes, 416, 424Euclid, 424Exponential equations:

solving, 721solving problems modeled by, 723–24

Exponential expressions, 169on a calculator, 174simplifying, 340–41

Exponential functions, 697–702, 726graphing, 697–99solving problems modeled by, 700–702

Exponential notation, 169Exponent rules, 340, 599, 629Exponents, 169–71, 331–37, 376

exponential expressions:evaluating, 331–32simplifying, 340–41

exponent rules, summary of, 340negative, 376

simplifying expressions containing,339–40

power of a product rule, 334power of a quotient rule, 334–35power rule, using, 333product rule, using, 332–33quotient rule for, 335–36rational, 597–600, 629scientific notation, 376

writing numbers in, 341–42standard form, converting numbers to,

342–43zero exponent, defining, 335–36

FFactored form, 381Factoring, 381–428

by grouping, 383–84, 399–400defined, 381, 426difference of two squares, 405–6factoring out, 381greatest common factor:

factoring out, 382–83, 390, 396finding, 381–82

numbers, 135perfect square trinomials, 403–5, 427quadratic equations, 411–16

with degree greater than two,414–15

solving quadratic equations by, 411–16trinomials, 387–90trinomials of the form ax2 � bx � c,

393–96, 427by grouping, 399–400, 427

Factorization, 135prime, writing, 135–37

Factors, 135–37, 381Fast-growing careers, 454Financial rations, 673Finite differences, 512First-magnitude stars, 161Fisher, Donald and Doris, 133Flowcharts, 644FOIL method, 363–64Formulas, 277, 741

area:of parallelogram, 749of rectangles, 749of square, 749of trapezoid, 749of triangle, 749

change of base, 717defined, 253mathematicsand problem solving, 253–56, 277quadratic, 645–47solving for a variable, 255–56, 277using to find circumference, 744–45vertex, 518volume:

of cone, 758of cube, 757of cylinder, 757of solids, 757–58of sphere, 757of square-based pyramid, 758

Fractions, 139–43adding, 142–43complex, simplifying, 473–76, 481–82dividing, 141–42equivalent, 139–40, 143fundamental principle of, 139–40lowest terms, 139–40multiplying, 141–42simplifying, 140–41

INDEX I-3

solving equations containing, 242–44

subtracting, 142–43unit, 441–42writing as decimals, 150–51writing decimals as, 147writing ratios as, 260–61

Function notation, 506–7, 527Functions, 501–21, 526

adding, 687–88algebra of, 687–90composition, 688–90cubic, 515

graphing, 519–20defined, 502dividing, 687–88domain, 501, 526

finding from a graph, 505–6exponential, 697–702function notation, 506–7, 527identifying, 502–3inverse, 691–96linear, 507, 515logarithmic, 705–8multiplying, 687–88polynomial:

analyzing graphs of, 515evaluating, 509–10

quadratic, 515graphing, 516–17

quartic, 515range, 501, 526rational, evaluating, 510relations, 501, 503, 526

finding from a graph, 505–6subtracting, 687–88vertical line test, 504–5, 526

Function values, finding, 594Fundamental principle of fractions,

139–40

GGain/loss, calculating, 179Gap, Inc., 133Geometry, 408, 729–79

angles, 731–32, 775acute, 733, 775adjacent, 735, 776classifying, 733–34complementary, 733–34, 775degrees, 732finding measures of, 734–36intersecting lines, 734obtuse, 733, 775parallel lines, 734perpendicular lines, 735right, 733side, 731supplementary, 733–34, 776transversal, 736, 776vertical, 735, 776

area, 749–52, 777congruent triangles, 771–72, 779lines, 731–32, 775line segments, 731–32, 775perimeter, 741–45, 777plane, 731plane figures, 737–39points, 731rays, 731, 775similar triangles, 772–74, 779solid figures, 739–40space, 731square roots, 763–67, 778vertex, 731volume, 757–59, 778

Giza, 729Glaciers, 217Golden rectangle, 230Graham’s law, 610Graph of the equation, 289Graphing:

exponential functions, 697–99, 726

inverse functions, 695–96logarithmic functions, 705–8, 726–27

Graphing calculator, 294, 316, 407, 490,497, 511, 522, 622, 642, 672, 690,724

approximating solutions of systems ofequations using, 537

entering negative numbers on, 196and inverse functions, 696TABLE feature, 690, 702TRACE feature, 497, 522, 702Y�, key, 303, 490

Graphing quadratic functions, 516–17,665–71, 675–80

Graphs, 485–529circle

reading, 207–10misleading, 324of polynomial functions, analyzing,

515, 527reading, 205–10, 216

bar graphs, 205–7line graphs, 207–10

slope-intercept form, 487–89, 526

interpreting, 488–89using to write an equation, 488

See also FunctionsGravity, center of, 529Greatest common factor:

defined, 426factoring out, 382–83, 390finding, 381–82

Great Pyramids, 729Gross profit margin ratio, 673Grouping, factoring by, 383–84Grouping method, 399–400Grouping symbols, 169

HHalf-planes, 320Heron of Alexandria, 628Hieroglyphs, 192Hipparchus, 161Horizontal lines:

graphing, 301–3writing equations of, 494

Horizontal line test, 692–93Hui, Liu, 552Hypotenuse, right triangle, 421, 738, 764,

776

IIdentities, recognizing, 244–45Identity properties, 201–2Imaginary unit, 623Inconsistent system, 536, 561Independent variable, 506Index, 593Inequality:

addition property of, 268–70multiplication property of, 270–72original, 322solution set of, 267symbols, 163–64using properties of, 271–72

Integers, 165Intercepts, 299–306, 326

identifying, 299–300intercept points, 326

finding/plotting, 300–301vertical/horizontal lines, graphing,

301–3Intersecting lines, 734Interval notation, 267–68Inverse functions, 691–96

finding the equation of the inverse,694–95

finding the inverse, 693–94graphing, 695–96horizontal line test, 692–93one-to-one functions, 691–92,

725Inverse properties, 201–2Inverses:

additive, 179–80, 202, 214multiplicative, 193, 202, 215

Irrational numbers, 163, 166–67, 592

JJob offers, assessing, 559

KKepler, Johannes, 718Khafre, 729Khufu, 729

LLa Geometrie (Descartes), 352, 528, 604Laplace, Pierre de, 718

I-4 INDEX

Least common denominator (LCD),finding, 446–48

Least common multiple (LCM), 137–38of a list of numbers, finding, 138

Legs, right triangle, 421, 764, 776Leibniz, Gottfried Wilhelm, 528less/fewer, See fewer/less“Less than”, use of term, 173Liabilities, 673Lightning, 531Like radicals, 607, 630Like terms, 219–20, 276

combining, 220–21, 276Linear equation in one variable, 225, 276Linear equations:

graphing, 290–93identifying special systems of, 535–36systems of:

equations in three variables,561–66

and problem solving, 553–58solving by addition, 547–52solving by graphing, 533–36solving by substitution method,

539–43using Cramer’s rule to solve,

572–74in two variables, 289, 325

Linear functions, 515graphing, 507

Linear inequalities:original inequality, 322and problem solving, 267, 272–74,

278systems of, 577–79in two variables:

boundary, 320–21determining solutions of, 319graphing, 319–23, 327half-planes, 320–21

Linear modeling, 498, 581Line graphs,

reading, 207–10Lines, 731–32, 775

horizontal, 301–3, 494intersecting, 734parallel, 313–15, 494–96, 734perpendicular, 313–15, 494–96, 735vertical, 301–3, 494

Line segments, 731–32, 775midpoint, 508

Logarithmic equations:solving, 706–7, 721–23solving problems modeled by, 723–24

Logarithmic functions, 705–8, 726–27graphing, 707–8

Logarithmic notation, 705Logarithms:

common, 715–16invention of, 718natural, 715–16

properties of, 711–13multiple, using, 712–13product property, 711quotient property, 711–12

Lowest terms, fractions, 139–40

MMathematical model, 580Mathematical statements, 163

translating sentences into, 164–65Math exam, studying for, 274, 338Matrices:

defined, 567, 585solving systems of equations using,

567–70square, 571, 586using to solve a system of three

equations, 569–70Maximum value, finding, 679–80Member, set, 163Metrica (Heron), 628Midpoint, line segment, 508Minimum value, finding, 679–80Minus sign (-), 165Misleading graphs, 324Mondrian, Piet, 230Monomial, 348, 376

dividing by, 369multiplying, 359

by a polynomial, 359Mount Vesuvius, 589Mount Waialeale (Hawaii), 429Multiplication, 169

associative property of, 199–200commutative property of, 199of complex numbers, 625, 631of decimals, 148of fractions, 141–42of functions, 687–88identity for, 201of polynomials, 359–60, 377of radical expressions, 608–9of rational expressions, 437–38of real numbers, 189–90, 214

Multiplication property of equality,233–37, 276

using, 235–37Multiplication property of inequality,

270–72Multiplicative inverses, 193, 202, 215

NNapier, John, 718Natural logarithms, 715

approximating, 716of powers of e, evaluating, 716

Natural numbers, 135, 163Negative exponents, simplifying

expressions containing, 339–40Negative integers, 165Negative net earnings, 374

Negative numbers:entering on a graphing calculator, 196entering on a scientific calculator, 196product of, 189

Negative radical sign, 591Negative square root, 591, 629Net income/loss, 374Newton, Isaac, 718Nine Chapters on the Mathematical Art, 552Notation:

exponential, 169function, 506–7interval, 267–68logarithmic, 705scientific, 341–42, 344, 376set, 267

nth roots, finding, 593Number line, 163, 166Numbers

common sets of, 165–67composite, 135converting to standard form, 342–43factoring, 135finding the absolute value of, 168irrational, 163, 166–67, 592natural, 135, 163prime, 135, 424rational, 166–67real, 167whole, 139, 163writing in scientific notation, 341–42

Number theory, 416Numerators:

rationalizing, 614–15numerators having two terms,

615–17Numerical coefficient,

See Coefficient

OObtuse angle, 733, 775One-to-one functions, 691–92, 725OPEC (Organization of Petroleum

Exporting Countries), 279Opposites, 179–80, 202, 214Ordered pairs of numbers:

defined, 281plotting, 281–82as solution, 533

Ordered triple, 561, 584Order of operations, 169–71

on a calculator, 174Original inequality, 322

PPaired data, 283Parabola, 516–17, 528

vertex, 516, 528finding, 517–19

Parallel lines, 734defined, 314

INDEX I-5

slope of, 313–15writing equations of, 494–96

Parallelogram, 738area formula of, 749

Parentheses,simplifying expressions containing,

221–22Percent equations, solving, 259–60Percents, 147–52

solving problems involving, 260writing as decimals, 151writing decimals as, 152

Perfect squares, 592, 763Perfect square trinomials:

factoring, 403–5, 427recognizing, 403–4writing, 637

Perimeter, 741–45, 777Perpendicular bisector, 508Perpendicular lines, 735

defined, 314slope of, 313–15writing equations of, 494–96

Plane, 731defined, 737

Plane figures:defined, 737identifying, 737–39

Plus sign (�), 165Points, 731Point-slope form, 493–97, 526

of the equation of a line, 493–94using in applications, 496–97writing equations:

of parallel and perpendicular lines,494–96

of vertical and horizontal lines, 494

Polaris, 161Polygon, 737Polynomial functions:

analyzing graphs of, 515, 527evaluating, 509–10, 527

Polynomial inequalities, 661–63Polynomials, 347–56, 376–77

adding, 355, 377in one variable, 356in several variables, 356

binomial, 376binomials, 348, 376

squaring, 364–65coefficient, 347, 376defined, 348, 376degree of a polynomial, 377descending powers, 348dividing, 369–73, 378

by a monomial, 369by a polynomial other than a

monomial, 370–71synthetic division, 372–73, 378

evaluating, 349–50

factoring, 381–428monomial, 348, 376

dividing a polynomial bymultiplying, 359–60, 377

of monomials, 359two polynomials, 359–60vertically, 360

numerical coefficient, 347, 376prime, 389simplifying:

by combining like terms, 350–51polynomials containing several

variables, 351special products, 363–66, 377

defined, 364FOIL method, 363–64multiplying the sum and difference

of two terms, 365–66squaring binomials, 364–65using, 366

subtracting, 355–56, 377in one variable, 356in several variables, 356

term, 347, 376degree of, 348

trinomials, 348, 377, 387–90types of, 348

Pompeii, 589Positive integers, 165Positive square root, 591, 629Power of a product rule, 334Power of a quotient rule, 334–35Power rule:

and radical equations, 617using, 333

Powers of i, finding, 626–27Prealgebra review, 133–58

decimals, 147–52factors, 135–37fractions, 139–43least common multiple (LCM), 137–38percents, 151–52

Price-to-earnings (P/E) ratio, 673Prime factorizations, writing, 135–37Prime numbers, 135, 424Prime polynomial, 389Principal square root, 591Problem solving, 247–57, 277

and formulas, 253–56, 277general strategy for, 247and linear inequalities, 267and quadratic equations, 419–23and radical equations, 617–20and rational equations, 463–68

Product packaging, 768Product property of logarithms, 711Product rule, using, 332–33Product rule for radicals, 601Proportions, 278

defined, 261solving, 261–63

solving problems modeled by, 263–64Pyramids, 739Pythagoras, 746Pythagorean theorem, 421, 620–21, 746

and square roots, 764–66

QQin Dynasty, 552Quadrants, 281Quadratic equations:

defined, 411, 427evolution of solving, 659perfect square trinomials, writing,

637and problem solving, 419–23quadratic functions, graphing, 665–71,

675–80quadratic inequalities in one variable,

661–64polynomial inequalities, 661–63rational inequalities, 663–64

solving by completing the square,635–36, 681

solving by factoring, 411–16solving by the quadratic formula,

645–47, 681discriminant, 648

solving problems leading to, 655–59solving problems modeled by, 640–41,

648–50square root property, using, 635–36standard form, 411zero factor property, 411–13, 427

Quadratic formula, 645–47discriminant, 648

Quadratic functions, 515, 528graphing, 516–17, 665–71, 675–80

Quadratic methods, solving equationsusing, 653–55

Quadrilaterals, 737–38perimeter of, 425

Quartic functions, 515Quotient property of logarithms, 711–12Quotient rule, 335–36

for exponents, 335for radicals, 601–2

RRadical equations, and problem solving,

617–20, 630Radical expressions, 591–94

adding, 607–8, 630cube roots, finding, 592–93, 629denominators, rationalizing, 613–14function values, finding, 594multiplying, 608–9nth roots, finding, 593numerators, rationalizing, 614–15simplifying, 601–4, 629

product rule, 601quotient rule, 601–2

I-6 INDEX

Radical expressions (Cont.)square roots:

approximating, 592finding, 591

subtracting, 607–8, 630using rational exponents to simplify,

600Radical functions, 594, 629Radicals:

quotient rule for, 601–2simplifying, 602–4

Radical sign, 763development of symbol, 604

Radicand, 591Radius:

circle, 739sphere, 739

Range, 501, 526Rational equations:

and problem solving, 463–68, 481finding an unknown number,

463–64finding speeds of vehicles, 465–67finding work rates, 464–65similar triangle problems, 467–68

Rational exponents, 597–600, 629a1/n, defined, 597a�m/n, defined, 598–99am/n, defined, 597–98denominator of, 598exponent rules, 599using to simplify radical expressions,

600Rational expressions, 369, 429–84

with common denominators,adding/subtracting, 445–46, 479

complex fractions, simplifying, 473–76,481–82

converting between units of measure,441–42

defined, 478with different denominators,

adding/subtracting, 451–53, 479–80dividing, 439–40, 479

compared to multiplying, 440equivalent, writing, 448evaluating, 431fundamental principle of, 432, 478with least common denominator,

adding/subtracting, 446–48multiplying, 437–38, 478

compared to dividing, 440rational equations, and problem

solving, 463–68, 481simplifying, 431–34, 478

steps in, 433solving equations for a specified

variable, 460solving equations containing, 457–60,

480undefined, identifying, 431–32

Rational functions, evaluating, 510, 527Rational inequalities, 663–64Rationalizing denominators, 613–14, 630Rationalizing numerators, 614–15, 630Rational numbers, 166–67Ratios, 278

defined, 260and proportions, 261–63writing as fractions, 260–61

Rays, 731, 775Real numbers, 244

adding, 177–79associative property of addition,

199–200associative property of multiplication,

199–200calculator operations, 196commutative property of addition, 199commutative property of

multiplication, 199defined, 167distributive property of multiplication

over addition, 201dividing, 193–96

division involving zero, 194evaluating algebraic expressions,

195solutions of equations, 196

identity properties, 201–2inverse properties, 201–2multiplying, 189–90, 214

evaluating algebraic expressions,190

solutions of equations, 190properties of, 199–203subtracting, 183–85

Reciprocals, 193–94, 202defined, 193

Rectangles, 738area formula of, 749area of, 425perimeter of, 741

Rectangular coordinate system, 281–87,325

ordered pairs of numbers:defined, 281plotting, 281–82

ordered pairs solutions, completing,284–86, 325

scatter diagrams, creating, 283Rectangular solids, 739Regular polygon, 737Relations, 501, 503, 526

finding from a graph, 505–6Revenue, 374, 544Rhombus, 739Right angles, 733, 775Right triangle, 421, 738, 776

finding a unknown length of, 764hypotenuse, 421, 738, 764, 776legs, 421, 764, 776

Road grades, 307Road signs, cost of, 760Rolle, Michel, 604Roman numerals, 192Rounding decimals, 149Row, matrix, 567Row operations, matrix, 567, 586Royal Gorge suspension bridge, 509Rudolff, Christoff, 604

SSals, 673Scatter diagrams, creating, 283Scientific calculator, entering negative

numbers on, 196Scientific notation, 376

entering on a scientific calculator, 344

writing numbers in, 341–42Sequence, 512Set, 163Set notation, 267Seurat, Georges, 230Side:

angle, 731triangle, finding the length of, 467–68

Side-angle-side (SAS), 772Side-side-side (SSS), 771Sieve of Eratosthenes, 416, 424Silicon Graphics Inc., Cray Research

unit, 424Similar triangles, 467–68, 772–74, 779

finding ratios of corresponding sidesin, 772–73

finding unknown lengths of sides in,773–74

side of a triangle, finding the lengthof, 467–68

Simplifying equations, 228–29Simplifying expressions, 219–22

containing parentheses, 221–22Simplifying fractions, 140–41Sirius, 161Sixth-magnitude stars, 161Slope, 309–15, 327

defined, 309of a horizontal line, finding, 312–13of a line given its equation, finding,

311–12of a line given two points, finding,

309–11of a parallel line, 313–15of a perpendicular line, 313–15undefined, 313of a vertical line, finding, 312–13

Slope-intercept form, 487–89, 522interpreting, 488–89using to write an equation, 488

Solid figures, identifying, 739–40Solids, 739

volume formulas of, 757–58

INDEX I-7

Solution:and ordered pairs, 533

Solution set of an inequality, 267Solving an equation, 172Sound intensity, 714Space, 731Space exploration, 368Special products, 365–66

polynomials, 363–66, 377defined, 364FOIL method, 363–64multiplying the sum and difference

of two terms, 365–66squaring binomials, 364–65using, 366

Sphere, 739volume formula of, 757

Square, 738area formula of, 749area of, 425perimeter of, 742

Square-based pyramid, volume formulaof, 758

Square matrix, 571, 586Square root property, using, 635–36Square roots, 763–67, 778

approximating, 592, 763–64defined, 591finding, 591, 763, 767negative, 591, 629positive, 591, 629principal, 591and Pythagorean theorem, 764–66

Square yards, converting to square feet,441

Squaring binomials, 364–65Standard form, 289, 325

converting numbers to, 342–43quadratic equations, 411

Standard window, graphing calculator,294

Stock market price per share, 673Straight angle, 732, 775Substitution method, solving systems of

linear equations by, 539–43“Subtracted from”, use of term, 173Subtraction:

of complex numbers, 624–25, 631

of decimals, 148of fractions, 142–43of functions, 687–88of polynomials, 355–56, 377of radical expressions, 607–8of real numbers, 183–85, 214

solving problems that involve, 185–86

Supplementary angles, 733–34, 776finding, 186

Symbols:algebra, 163–68

translating words into, 172–73grouping, 169and sets of numbers, 163translating words into, 172–73

Symmetry, axis of, 516Synthetic division, 372–73, 378Systems of equations, 531–88

defined, 531, 582and problem solving, 553–58, 583solving, 582

by addition, 547–51, 583by graphing, 533–36by substitution, 539–43, 583linear equations in three variables,

561–66, 583–84using determinants, 571–76using matrices, 567–70

Systems of inequalities, 577–79Systems of linear inequalities,

577–79Systems of three equations, solving

problems modeled by, 565–66

TTerm, 219, 276, 347, 376

constant, 347degree of, 348

TRACE feature, graphing calculator, 497Traite d’Algebre (Rolle), 604Transversal, 736, 776Trapezoid, 739

area formula of, 749Triangles, 737–38

area formula of, 749congruent, 771–72, 779

angle-side-angle (ASA), 771side-angle-side (SAS), 772side-side-side (SSS), 771

finding the dimensions of, 422–23perimeter of, 425, 742–43right, 738, 776similar, 467–68, 772–74, 779

Trinomials, 348, 377, 387–90factoring:

of the form ax2 � bx � c, 393–96,399–400, 426

of the form x2 � bx � c, 387–90,426

negative constant in, 390

perfect square:factoring, 403–5recognizing, 403–4writing, 637

positive constant in, 390sign patterns, 390

UUndefined rational expressions,

identifying, 431–32Undefined slope, 313Unit fractions, 441–42Unit price, 264Unlike terms, 219–20U.S. Bureau of the Census, 329U.S. Bureau of Labor Statistics, 559

VVariable costs, 544Variables, 163

dependent/independent, 506Vertex, 516, 528, 731, 775

finding, 517–19formula, 518, 678–79

Vertical angles, 735, 776Vertical lines:

graphing, 301–3writing equations of, 494

Vertical line test, 504–5, 526Vinculum, 604Volume, 757–59, 778

WWhole numbers, 139, 163Window, graphing calculator, 294

translating into symbols, 172–73

Xx-axis, 281, 325x-coordinate, 282x-intercept, 299, 326x-intercept point, 299

Yy-axis, 281, 325y-coordinate, 282y-intercept, 299, 326y-intercept point, 299

ZZero:

division involving, 194quotients involving, 193

Zero exponent, defining, 335–36Zero factor property, 411–13, 427

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