algebra connections parent guide can be successful in mathematics as long as they are willing to...

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Algebra Connections Parent Guide

Managing Editors Leslie Dietiker Phillip and Sala Burton Academic High School San Francisco, CA

Evra Baldinger Phillip and Sala Burton Academic High School San Francisco, CA

Contributing Editors to the Parent Guide Bev Brockhoff Glen Edwards Middle School Lincoln, CA

Elizabeth Coyner Christian Brothers High School Sacramento, CA

Brian Hoey Christian Brothers High School Sacramento, CA

Pat King Oliver Wendell Holmes Junior High School Davis, CA

Bob Petersen Rosemont High School Sacramento, CA

Technical Manager Rebecca Harlow Berkeley, CA

Program Directors Leslie Dietiker Judy Kysh, Ph.D. Phillip and Sala Burton Academic High School Departments of Mathematics and Education San Francisco, CA San Francisco State University Brian Hoey Tom Sallee, Ph.D. Christian Brothers High School Department of Mathematics Sacramento, CA University of California, Davis

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Copyright © 2006 by CPM Educational Program. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher. Requests for permission should be made in writing to: CPM Educational Program, 1233 Noonan Drive, Sacramento, CA 95822. Email: [email protected]. 1 2 3 4 5 6 7 8 9 10 09 08 07 06 05 Version 3.0

Printed in the United States of America ISBN 1-931287-49-X

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Introduction to the Parent Guide

Welcome to the Algebra Connections Parent Guide. The purpose of this guide is to assist you should your child need help with homework or the ideas in the course. We believe all students can be successful in mathematics as long as they are willing to work and ask for help when they need it. We encourage you to contact your child’s teacher if your student has additional questions that this guide or other resources do not answer. Assistance with most homework problems is available at www.hotmath.com.

This guide was written to address the major topics in each chapter of the textbook. Each section of the Parent Guide begins with a title bar and the section(s) of the book that it addresses. In many cases the explanation box at the beginning of the section refers you to one or more Math Notes boxes in the student text for additional information about the fundamentals of the idea. Detailed examples follow a summary of the concept or skill and include complete solutions. The examples are similar to the work your child has done in class. Additional problems, with answers, are provided for your child to practice.

There will be some topics that your child understands quickly and some concepts that may take longer to master. The big ideas of the course take time to learn. This means that students are not necessarily expected to master a concept when it is first introduced. When a topic is first introduced in the textbook, there will be several problems to do for practice. Subsequent lessons and homework assignments will continue to practice the concept or skill over weeks and months so that mastery will develop over time.

Practice and discussion are required to understand mathematics. When your child comes to you with a question about a homework problem, often you may simply need to ask your child to read the problem and then ask her/him what the problem is asking. Reading the problem aloud is often more effective than reading it silently. When you are working problems together, have your child talk about the problems. Then have your child practice on his/her own.

Below is a list of additional questions to use when working with your child. These questions do not refer to any particular concept or topic. Some questions may or may not be appropriate for some problems.

• What have you tried? What steps did you take?

• What didn't work? Why didn't it work?

• What have you been doing in class or during this chapter that might be related to this problem?

• What does this word/phrase tell you?

• What do you know about this part of the problem?

• Explain what you know right now.

• What do you need to know to solve the problem?

• How did the members of your study team explain this problem in class?

• What important examples or ideas were highlighted by your teacher?

• Can you draw a diagram or sketch to help you?

• Which words are most important? Why?

• What is your guess/estimate/prediction?

• Is there a simpler, similar problem we can do first?

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• How did you organize your information? Do you have a record of your work?

• Have you tried Guess and Check, making a list, looking for a pattern, etc.?

If your student has made a start at the problem, try these questions. • What do you think comes next? Why?

• What is still left to be done?

• Is that the only possible answer?

• Is that answer reasonable?

• How could you check your work and your answer?

• How could your method work for other problems?

If you do not seem to be making any progress, you might try these questions. • Let's look at your notebook, class notes, and Learning Log. Do you have them?

• Were you listening to your team members and teacher in class? What did they say?

• Did you use the class time working on the assignment? Show me what you did.

• Were the other members of your team having difficulty with this as well? Can you call your study partner or someone from your study team?

This is certainly not a complete list; you will probably come up with some of your own questions as you work through the problems with your child. Ask any question at all, even if it seems too simple to you.

To be successful in mathematics, students need to develop the ability to reason mathematically. To do so, students need to think about what they already know and then connect this knowledge to the new ideas they are learning. Many students are not used to the idea that what they learned yesterday or last week will be connected to today’s lesson. Too often students do not have to do much thinking in school because they are usually just told what to do. When students understand that connecting prior learning to new ideas is a normal part of their education, they will be more successful in this mathematics course (and any other course, for that matter). The student’s responsibilities for learning mathematics include the following:

• Actively contributing in whole class and study team work and discussions.

• Completing (or at least attempting) all assigned problems and turning in assignments in a timely manner.

• Checking and correcting problems on assignments (usually with their study partner or study team) based on answers and solutions provided in class.

• Studying the tutorial solutions to homework problems available at www.hotmath.com.

• Asking for help when needed from his or her study partner, study team, and/or teacher.

• Attempting to provide help when asked by other students.

• Taking notes and using his/her Learning Log when recommended by the teacher or the text.

• Keeping a well-organized notebook.

• Not distracting other students from the opportunity to learn.

Assisting your child to understand and accept these responsibilities will help him or her to be successful in this course, develop mathematical reasoning, and form habits that will help her/him become a life-long learner.

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Table of Contents Chapter 1 Problem Solving…………………………………………… 1 Interpreting Graphs 1 Solving Problems with Guess and Check 4 Chapter 2 Variables and Proportions……………………………… 6 Variables and Combining Like Terms 6 Writing and Simplifying Algebraic Expressions 8 Solving Equations 10 Proportions 12 Chapter 3 Graphs and Equations…………………………………… 14 Graphs, Tables, and Rules 14 Examining and Using Equations 19 Chapter 4 Multiple Representations………………………………… 22 Multiple Representations 22 Solving Linear Systems: The Equal Values Method 27 Chapter 5 Multiplication and Proportions…………………………. 30 Using Rectangles to Multiply 30 Solving Equations with Multiplication 32 Solving Multi-Variable Equations 34 Setting Up Proportions 35 Solving Proportions 37 Chapter 6 Systems of Equations……………………………………… 39 Writing Equations 39 Solving Systems by Substitution 43 Elimination Method 46

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Chapter 7 Linear Relationships……………………………………… 50 Slope—A Measure of Steepness 50 Writing an Equation Given the Slope and a Point on the Line 53 Writing the Equation of a Line Given Two Points 56 Chapter 8 Quadratics…………………………………………………. 58 Factoring Quadratics 58 Using the Zero Product Property 62 Using the Quadratic Formula 65 Chapter 9 Inequalities………………………………………………… 70 Solving Inequalities 70 Graphing Inequalities 72 Systems of Inequalities 75 Chapter 10 Simplifying and Solving………………………………… 78 Simplifying Expressions 78 Multiplication and Division of Rational Expressions 80 Solving by Rewriting: Fraction Busters 82 Multiple Methods for Solving Equations 84 Solving Absolute Value and Quadratic Inequalities 86 Solving by Completing the Square 88 Laws of Exponents 90 Chapter 11 Functions and Relations………………………………… 93 Relations and Functions 93 Transformations of a Function 97 Intercepts and Intersections 99 Chapter 12 Algebraic Extensions……………………………………. 102 Factoring Shortcuts 102 Addition and Subtraction of Rational Expressions 104 Work and Mixture Problems 106

Chapter 1: Problem Solving 1

INTERPRETING GRAPHS 1.1.1 and 1.1.2

Every point on a graph represents two pieces of information. The two pieces of information correspond to the labels on each of the two axes. In general, a point on a graph is named or interpreted (described) by first considering the horizontal axis, then the vertical axis. For information about using the (x, y) coordinate plane, see the Math Notes box on page 10 in the student text.

Example 1 Example 2 Explain all that you can about points A and B on the graph below. By reading the axes we see that the graph shows us the relationship between the number of years a person has been in school and the number of books that person has read. The farther to the right a point is, the more years a person has spent in school. The higher a point is vertically, the more books the person has read. Point A represents someone who has not been in school very many years and, consequently, has not read many books. Point B represents someone who has been in school for several years and has read a fair number of books. Another way to interpret the graph is to note that point B represents a person who has been in school about three times longer than the person represented by point A and who has read about three times as many books.

Explain all that you can about points A and B on the graph below. This graph shows the relationship between the temperature of a cup of coffee and the time since the coffee was poured. The farther to the right a point is, the more time has passed since the coffee was poured. The higher a point is vertically, the greater the temperature. Point A represents a cup of coffee which was poured recently and is still hot. Point B represents a cup of coffee that was poured awhile ago and is cooler in temperature.

Years in School

Num

ber

of b

ooks

rea

d

B

A

Time since poured in cup

Tem

pera

ture

of

coff

ee

B

A

Algebra Connections Parent Guide 2

AB

C

A

B

C AB

C

A

B C

A

B

C

CAB

Example 3

Find the coordinates of points A and B in the graph at right. To find the coordinates of point A, start at the center of the graph where the x- and y-axes cross (this is the point (0, 0), called the origin). Go to the right along the x-axis until you reach the point below point A. This number, in this case 1.5, is the x-coordinate of point A. Then go up or down from there to the point. The number along the vertical y-axis that matches the point, in this case 0.5, is the y-coordinate of point A. The coordinates are written (1.5, 0.5), called an ordered pair, because the x-coordinate is always written first. An x-coordinate is negative if it is to the left of the y-axis; a y-coordinate is negative if it is below the x-axis.

1 2–1–2

1

2

–1

–2

x

y

A

B

C

D

E

Problems

Explain what points A, B, and/or C represent on each of these graphs. 1. 2. 3. 4. 5. 6. 7.

A

Use the graph in Example 3 above to find the coordinates of each of the following points. 8. Point C 9. Point D 10. Point E

Age of a car

Val

ue o

f a

car

Age

Hei

ght

Years as a scout Num

ber o

f bad

ges e

arne

d

Months (Jan.-Dec.) T

empe

ratu

re

Jon’s Account

Emma’s Account

Time

Am

ount

of

mon

ey

Time since kicked

Hei

ght o

f fo

otba

ll

CompUs

AOH

Time spent on the internet

Cos

t


Answers 1. Point A shows that a newer car has greater value. Point C is older than point B and point C

has less value than points A or B. 2. Point A is older than point B but taller. Point C is the oldest, taller than point B but shorter

than point A. 3. Point C has been a scout the longest and has the least number of badges. Point B has the

most badges. Point A has been a scout the shortest time and has more badges than point C. 4. The temperature at point A is lowest and it is earliest in the year. At points B and C, later

in the year, the temperatures are about the same and warmer than point A. 5. Point A is the point in time when Jon’s and Emma’s accounts have the same amount of

money in them. 6. Point A is when the football is kicked, point B is the highest point, and point C is when it

hits the ground. 7. Point A is AOH’s base price, point B is CompUs’s base price, and point C is where the cost

of using the internet is the same. 8. C(2, -1.5) 9. D(-1, 1.5) 10. E(0, -1)


SOLVING PROBLEMS WITH GUESS AND CHECK 1.2.1 and 1.2.3

Guess and Check is a problem solving strategy students can use to solve many types of problems, especially complex word problems. When using Guess and Check, students begin by choosing a value for one part of the problem (the guess), then use the information in the problem to develop mathematical relationships based on the guess. The result of the guess is compared to the expected outcome (answer) of the problem. This information is used to refine the guess to find a value closer to the answer. It is usually helpful to organize the information in a table so that patterns and relationships are easier to see. See the Math Notes box on page 28 for vocabulary words that are commonly used in word problems.

Example The product of two numbers is 126. One of the numbers is 5 more than the other. Find the two numbers. Start by guessing a value for the first number, such as 5. Use the information in the problem to find the value of the second number. The second number is 10 because 5 more than 5 is 10. Calculate the product of the two numbers, compare the result to the expected product, and record the work in a table.

Guess First Number

Second number

Product

Check. Is the product 126?

5 5+ 5 = 10 5(10) = 50 No, too low

Because the product is 50, well below the expected 126, the guess of 5 is too low and the next guess should be higher. Make as many guesses as it takes to get the correct answer.

Guess First Number

Second number

Product

Check. Is the product 126?

5 5+ 5 = 10

5(10) = 50 No, too low

10 5 + 10 = 15 10(15) = 150 No, too high

8 5+ 8 = 13

8(13) = 104 No, too low

9 5 + 9 = 14 9(14) = 126 Yes. The two numbers with a product of 126 are 9 and 14.


Problems Use Guess and Check to solve each of these problems. State the solution in a complete sentence. 1. The product of two numbers is 450. The smaller number is seven less than the larger

number. What are the numbers? 2. The area of a rectangle is 810 square centimeters. The length is 3 centimeters greater than

the width. Find the length and width of the rectangle. 3. The perimeter of a triangle is 47 inches. The first side is twice the length of the second

side. The third side is seven more than the second side. What is the length of each side? 4. The perimeter of a triangle is 53 inches. The second side is 3 inches longer than the first side.

The third side is 1.5 times the length of the second side. What is the length of each side? 5. Yanavi drew a rectangle on graph paper. One side of the rectangle is two units longer

than the other. The area of the rectangle is 399 squares. What is the length of each side? 6. Aimee cut a 126 centimeter piece of string into two pieces so that one piece is two times as

long as the other. How long is each piece? 7. Maria Elena has 84 feet of fence to put around her rectangular flower bed. How long and how

wide will the flower bed have to be so that she has 437 square feet of area to plant flowers? 8. Three friends, Riccardo, Giacomo, and Chiara, mowed lawns to earn money to buy a

canoe. They all mowed the same number of lawns. Chiara also received a $7 tip. Together they earned $574 including the tip. How much money did each friend earn?

9. The sum of the ages of Peggy and Elizabeth is 98 years. Elizabeth is 14 years younger than

Peggy. How old is each person? 10. Hannah and Mary each earn $10 per hour to paint. Together they earned $460. Mary

painted for 14 more hours than Hannah. How long did each girl paint?

Answers

1. The numbers are18 and 25. 2. The width is 27 cm and the length is 30 cm.

3. The side lengths are 10, 17, and 20 in. 4. The side lengths are 13, 16, and 24 inches.

5. The side lengths are 19 and 21 units. 6. The lengths of the pieces are 42 and 84 cm.

7. The dimensions are19 and 23 feet. 8. Riccardo and Giacomo each earned $189; Chiara earned $196.

9. Elizabeth is 42 and Peggy is 56. 10. Hannah painted 16 hours, Mary 30 hours.

Algebra Connections Parent Guide6

VARIABLES AND COMBINING LIKE TERMS 2.1.1 and 2.1.2

Algebraic expressions can be represented by the perimeters and areas of algebra tiles(rectangles and squares) and combinations of algebratiles. The dimensions of each tile are shown along itssides and the area is shown on the tile itself in thefigures at left. When using the tiles, perimeter is thedistance around the exterior of a figure. The area ofa figure is the sum of the areas of the individualpieces (length times width of each piece).

Combining terms that have the same area to write a simpler expression is calledcombining like terms. When working without tiles or pictures of the tiles, simply add orsubtract the coefficients of the like terms. The Math Notes boxes on pages 49 and 57explain the standard order of operations and their use with algebra tiles.

Using Algebra Tiles

Example 1 Example 2 Example 3

Perimeter= 4x + 4y + 2 Perimeter= 6y + 8 Perimeter= 8x + 6Area= x2 + xy + y2 + x Area= 2y2 + 3y + 2 Area= 3x2 + 7x + 2

Combining Like Terms

Example 4 3x2 + 7x + 2 + x2 + 5 + 2x = 4x2 + 9x + 7

Example 5 x2 + xy + y2 + 3xy + y2 + 7 = x2 + 4xy + 2y2 + 7

Example 6: 3x2 − 4x + 3+ −x2 + 3x − 7 = 2x2 − x − 4

1xx yy

1

1 11

xy

y

x

y

y

y2

x

x

x2

y

y

y2

xy

y

x

y

x

xx2 x

x 1

1

y2

y

y

y

y2

y

y11

11

11y y y

1 1 1

1 1 1

y-2xx

xx

xx 1

1

xx2x2x2

Chapter 2: Variables and Proportions 7

ProblemsDetermine the area and the perimeter of each figure.

1. 2. 3.

4. 5. 6.

7. 8. 9.

Simplify each expression by combining like terms.

10. 2x2 + x + 3+ 4x2 + 3x + 5 11. y2 + 2y + x2 + 3y2 + x2

12. x2 − 3x + 2 + x2 + 4x − 7 13. y2 + 2y − 3− 4y2 − 2y + 3

14. 4xy + 3x + 2y − 7 + 6xy + 2x + 7 15. x2 − y2 + 2x + 3y + x2 + y2 + 3y

16. (4x2 + 4x −1) + (x2 − x + 7) 17. (y2 + 3xy + x2 ) + (2y2 + 4xy − x2 )

18. (7x2 − 6x − 9) − (9x2 + 3x − 4) 19. (3x2 − 8x − 4) − (5x2 + x +1)

Answers 1. P = 4x + 4

A = x2 + 2x 2. P = 4y + 6

A = y2 + 3 3. P = 4x + 8

A = 4x + 4

4. P = 4y + 4A = y2 + 2y +1

5. P = 4y + 4A = x2 + 2y

6. P = 2x + 2y + 6A = 2x + 2y + 2

7. P = 8x + 6A = 4x2 + 6x + 2

8. P = 6xA = x2 + x +1

9. P = 6x + 4yA = 2x2 + xy + y2

10. 6x2 + 4x + 8 11. 4y2 + 2y + 2x2 12. 2x2 + x − 5

13. −3y2 14. 10xy + 5x + 2y 15. 2x2 + 2x + 6y

16. 5x2 + 3x + 6 17. 3y2 + 7xy 18. −2x2 − 9x − 5

19. −2x2 − 9x − 5

x2 x x1 11 1xx

xx1 1 1y2

y2

1y

y x2yy

xx

yy

11

x2

x2 x2

x2xx

x x

xx 1 1

x2 x1

y2

x2x2 xy


WRITING AND SIMPLIFYING 2.1.3 through 2.1.7ALGEBRAIC EXPRESSIONS

An expression mat is an organizing tool that is used to represent algebraic expressions.Pairs of expression mats can be modified to make an equation mat (see next section). Theupper half of an expression mat is the positive region and the lower half is the negativeregion. Positive algebra tiles are shaded and negative tiles are blank. A matching pair oftiles with one tile shaded and the other one blank represents zero (0).

Tiles may be removed from or moved on an expression mat in one of three ways:(1) removing the same number of opposite tiles in the same region; (2) flipping a tilefrom one region to another. Such moves create “opposites” of the original tile, so ashaded tile becomes unshaded and an unshaded tile becomes shaded; and (3) removing anequal number of identical tiles from both the “+” and “-” regions. See the Math Notesbox on page 60.

Examples

Example 1 3x + 2 − (2x − 3)

3x − 4 can berepresentedvarious ways.

+

_

xxx

+

_

xxx

+

_x

x

x= +1= –1

The expressionmats at rightall representzero.

+

_

xx

xx

+

_

xx

xxx

x +

_

= +1= –1

Expressions canbe simplified bymoving tiles tothe top (changethe sign) andlooking forzeros.

+

_

xxx

xx

+

_

xxxxx

+

_

x

= +1= –1

3x + 2 − (2x − 3) 3x + 2 − 2x + 3 x + 5


Example 2 1− (2y − 3) + y − 2

1− (2y − 3) + y − 2 1− 2y + 3+ y − 2 −y + 2ProblemsSimplify each expression.

1. 2. 3.

4. 5. 6.

7. 3+ 5x − 4 − 7x 8. −x − 4x − 7 9. −(−x + 3)10. 4x − (x + 2) 11. 5x − (−3x + 2) 12. x − 5 − (2 − x)13. 1− 2y − 2y 14. −3x + 5 + 5x −1 15. 3− (y + 5)16. −(x + y) + 4x + 2y 17. 3x − 7 − (3x − 7) 18. −(x + 2y + 3) − 3x + y

Answers 1. 0 2. 2x + 2 3. 2y + 2 4. −5x + 2 5. 2y −1 6. −y + 5 7. −2x −1 8. −5x − 7 9. x − 310. 3x − 2 11. 8x − 2 12. 2x − 713. −4y +1 14. 2x + 4 15. −y − 216. 3x + y 17. 0 18. −4x − y − 3

+

_

yy

y +

_

yy

y +

_

y= +1= –1

+

_

+

_

xxx

x

+

_yy

+

_

xx x

x

x

+

_

y

y

+

_

y

y

y


SOLVING EQUATIONS 2.1.8 and 2.1.9

Combining two expression mats into an equation mat creates a concrete model forsolving equations. Practicing solving equations using the model will help studentstransition to solving equations abstractly with better accuracy and understanding.

In general, and as shown in the first example below, the negative in front of theparenthesis causes everything inside to “flip” from the top to the bottom or the bottom tothe top of an expression mat, that is, all terms in the expression change signs. Aftersimplifying the parentheses, simplify each expression mat. Next, isolate the variables onone side of the equation mat and the non-variables on the other side by removingmatching tiles from both sides. Then determine the value of the variable. Students shouldbe able to explain their steps. See the Math Notes box on page 69.

Procedure and Example

Solve x + 2 − (−2x) = x + 5 − (x − 3) .

First build the equation on the equation mat.

Second, simplify each side using legalmoves on each expression mat, that is, oneach side of the equation mat.

Isolate x-terms on one side and non-x-termson the other by removing matching tilesfrom both sides of the equation mat.

Finally, since both sides of the equation areequal, determine the value of x.

+

_

+

_xx x

xx

+

_

+

_

x

xxx

x

x + 2 + 2x = x + 5 − x + 3+

_

+

_

xx

x

3x + 2 = 8+

_

+

_

xxx

3x = 6x = 2


Once students understand how to solve equations using an equation mat, they may usethe visual experience of moving tiles to solve equations with variables and numbers. Theprocedures for moving variables and numbers in the solving process follow the samerules. Note: When the process of solving an equation ends with different numbers oneach side of the equal sign (for example, 2 = 4), there is no solution to the problem. Whenthe result is the same expression or number on each side of the equation (for example,x + 2 = x + 2 ) it means that all numbers are solutions. See Section 3.2.2-3.2.4 in thisguide.

Example 1 Solve 3x + 3x −1 = 4x + 9

3x + 3x −1 = 4x + 96x −1 = 4x + 92x = 10x = 5

Example 2 Solve −2x +1− (−3x + 3) = −4 + (−x − 2)

−2x +1− (−3x + 3) = −4 + (−x − 2)−2x +1+ 3x − 3 = −4 − x − 2

x − 2 = −x − 62x = −4x = −2

ProblemsSolve each equation. 1. 2x − 3 = −x + 3 2. 1+ 3x − x = x − 4 + 2x 3. 4 − 3x = 2x − 6 4. 3+ 3x − (x − 2) = 3x + 4 5. −(x + 3) = 2x − 6 6. −4 + 3x −1 = 2x +1+ 2x 7. −x + 3 = 10 8. 5x − 3+ 2x = x + 7 + 6x 9. 4y − 8 − 2y = 4 10. 9 − (1− 3y) = 4 + y − (3− y)11. 2x − 7 = −x −1 12. −2 − 3x = x − 2 − 4x13. −3x + 7 = x −1 14. 1+ 2x − 4 = −3− (−x)15. 2x −1−1 = x − 3− (−5 + x) 16. −4x − 3 = x −1− 5x17. 10 = x + 6 + 2x 18. −(x − 2) = x − 5 − 3x19. 6 − x − 3 = 4x − 8 20. 0.5x − (−x + 3) = x − 5

Answers 1. 2 2. 5 3. 2 4. 1 5. 1 6. −6 7. −7 8. no answer 9. 6 10. −711. 2 12. all numbers 13. 2 14. 0 15. 216. no answer 17. 18. −7 19. 20. −4

problemsimplifyadd 1, subtract 4x on each sidedivide

problemremove parenthesis (flip)simplifyadd x, add 2 to each sidedivide

113 2 15

Solution

Solution


PROPORTIONS 2.2.1 through 2.2.3

Students solve proportional reasoning (ratio) problems in a variety of ways. They mayfind the number or cost per one unit and then multiply by the number of units. They mayalso organize work in a table. Later in the course students will use ratio equations orproportions. See Section 5.2.1-5.2.2 for this topic.

Example 1

Mrs. Salem’s student assistant can correct 18 homework papers in 10 minutes. At this rate howmany papers can she correct during a 55 minute class?

18÷10 = 1.8 papers per minute1.8 x 55 = 99 papers

Example 2

Toby uses 7 tubes of toothpaste every 10 months. How many tubes will he use in 24 months?How long will 35 tubes last him?

about 17 tubes 50 months

÷10

x 24

÷10

x 24

Tubes Months 7 10

0.7 1

16.8 24

7 tubes 35 tubes = so = 10 months 50 months

x 5

x 5

or


Problems

Solve and explain your reasoning

1. Alice knows six cups of rice will make enough Spanish rice to feed 15 people. How muchrice is needed to feed 75 people?

2. Elaine can plant 16 flowers in 10 minutes. How many can she plant in 25 minutes? 3. Ivanna needs to buy 360 cherries for a large salad. She can buy nine cherries for $0.57.

How much will 360 cherries cost? 4. A plane travels 3400 miles in eight hours. How far would it travel in six hours at the same

rate? 5. Leslie can write a 500-word essay in a hour. If she writes an essay in 20 minutes,

approximately how many words should the essay contain? 6. About eight out of every 100 people in the state have red hair. If a typical classroom in

the state has 25 students, how many would you expect to have red hair? If the typicalmiddle school has 650 students, how many would you expect to have red hair?

7. When Carlos rides his bike to school, it takes 15 minutes to go 8 blocks. If he rides at thesame speed, how long should it take him to travel 30 blocks?

8. Simba the cat is on a diet. Ten pounds of special low-fat food costs $22.50. How muchwould 30 pounds cost? How much would 36 pounds cost?

9. Elizabeth came to bat 110 times in 20 games. How many times should she expect to batin 62 games?

10. Ly can deliver 32 newspapers in 25 minutes on his bike. Next week he needs to deliver80 newspapers in the same neighborhood. How long should it take him if he works at thesame rate as he did for 32 newspapers?

Answers

1. 30 cups 2. 40 flowers 3. $22.80 4. 2550 miles 5. ≈167 words

6. 2 & 52 people 7. 56.25 min. 8. $67.50; $81.00 9. 341 at-bats 10. 62.5 min.


GRAPHS, TABLES, AND RULES 3.1.2 through 3.1.6

Three ways to write relationships for data are tables, words (descriptions), and rules. The pattern in tables between input (x) and output (y) values usually establishes the rule for a relationship. If you know the rule, it may be used to generate sets of input and output values. A description of a relationship may be translated into a table of values or a general rule (equation) that describes the relationship between the input values and output values. Each of these three forms of relationships may be used to create a graph to visually represent the relationship. See the Math Notes boxes on pages 102, 106, 110, 119, and 161.

Example 1 Complete the table by determining the relationship between the input (x) values and output (y) values, write the rule for the relationship, then graph the data. input (x) 4 -3 5 0 3 -2 x output (y) 8 4 10 0 -2 -4

Begin by examining the four pairs of input values: 4 and 8, 5 and 10, 0 and 0, -2 and -4. Determine what arithmetic operation(s) are applied to the input value of each pair to get the second value. The operation(s) applied to the first value must be the same in all four cases to produce each given output value. In this example, the second value in each pair is twice the first value. Since the pattern works for all four points, make the conjecture that the rule is y(output) = 2x(input) . This makes the missing values -3 and -6, 2 and 4, -1 and -2, 3 and 6. The rule is y = 2x . Finally, graph each pair of data on an xy-coordinate system, as shown at right.

2 4–2–4

2

4

6

8

10

–2

–4

–6

x

y

Chapter 3: Graphs and Equations 15

Example 2 Complete the table by determining the relationship between the input (x) and output (y) values, then write the rule for the relationship. input (x) 2 -1 4 -3 0 -2 1 x output (y) 3 -3 7 -7 -1 -5 1

Use the same approach as Example 1. In this table, the relationship is more complicated than simply multiplying the input value or adding (or subtracting) a number. Use a Guess and Check approach to try different patterns. For example, the first pair of values could be found by the rule x +1 , that is, 2 +1 = 3 . However, that rule fails when you check it for -1 and -3: !1+1 " !3 . From this guess you know that the rule must be some combination of multiplying the input value and then adding or subtracting to that product. The next guess could be to double x. Try it for the first two or three input values and see how close each result is to the known output values: for 2 and 3, 2(2) = 4 ; for -1 and -3, 2(!1) = !2 ; and for 4 and 7, 2(4) = 8 . Notice that each result is one more than the actual output value. If you subtract 1 from each product, the result is the expected output value. Make the conjecture that the rule is y(output) = 2x(input) !1 and test it for the other input values: for -3 and -7, 2(!3) !1 = !7 ; for 0 and -1, 2(0) !1 = !1; for -2 and -5, 2(!2) !1 = !5 ; and for 1 and 1, 2(1) !1 = 1 . So the rule is y = 2x !1 . Example 3 Complete the table below for y = !2x +1 , then graph each of the points in the table. input (x) -4 -3 -2 -1 0 1 2 3 4 x output (y)

Replace x with each input value, multiply by -2, then add 1. The results are ordered pairs: (-4, 9), (-3, 7), (-2, 5), (-1, 3), (0, 1), (1, -1), (2, -3), (3, -5), and (4, -7). Plot these points on the graph (see Chapter 1 if you need help with the fundamentals of graphing).

2 4–2–4

2

4

6

8

10

–2

–4

–6

–8

x

y


2 4–2

2

4

6

8

10

–2

y

x

Example 4 Complete the table below for y = x2 ! 2x +1, then graph the pairs of points and connect them with a smooth curve. input (x) -2 -1 0 1 2 3 4 x output (y) x

2! 2x +1

Replace x in the equation with each input value. Square the value, multiply the value by -2, then add both of these results and 1 to get the output (y) value for each input (x) value. The results are ordered pairs: (-2, 9), (-1, 4), (0, 1), (1, 0), (2, 1), (3, 4), and (4, 9). Example 5 Make an x → y table for the graph at right, then write a rule for the table.

Working left to right on the graph, read the coordinates of each point and record them in the table. Guess and Check by multiplying the input value, then adding or subtracting numbers to get the output value. For example you could start by multiplying the input value by 2: 2(!4) = !8 , 3(!4) = !12 , 2(!3) = !6 , etc. The results are not close to the correct output value. The product is also the opposite sign (+ -) of what you want. Your next choice could be to multiply by -2: !2(!4) = 8 , !2(!3) = 6 , !2(!2) = 4 . Each result is three more than the expected output value, so make the conjecture that the rule is y = !2x ! 3 . Test it for the remaining points: !2(!1) ! 3 = !1 , !2(0) ! 3 = !3 , and !2(1) ! 3 = !5 . The rule is y = !2x ! 3 .

input (x) -4 -3 -2 -1 0 1 x output (y) 5 3 1 -1 -3 -5 !2x ! 3

2–2–4

2

4

6

–2

–4

–6

x

y


Problems

Complete the table. Then write a rule relating x and y. 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

Complete a table for each rule, then graph and connect the points. For each rule, start with a table like the one below.

Input (x) -3 -2 -1 0 1 2 3 Output (y)

13. y = 3x ! 2 14. y = 1

2x +1 15. y = !x + 2 16. y = x2 ! 6

10 14

5 9

20

-4

-7 -3

3

x

Input(x) Output(y)

22 19

5 2

11

-8

-9 -12

12

x

Input(x) Output(y)

10 -30

-3 9

4 -12

-1

-5

15

x

Input(x) Output(y)

10 25

0 -5

15

-8

-7

6 13

x

Input(x) Output(y)

-4 0

0 2

12

-4

-16 -6

6

x

Input(x) Output(y)

4 7

-6 -3

6

2

-12 -9

1

x

Input(x) Output(y)

3 3

0 -3

13

-12

16 29

9

x

Input(x) Output(y)

2 -5

0 1

-4

-20

-13 40

-17

x

Input(x) Output(y)

9 3

-6 -2

4

0

3 1

-3

x

Input(x) Output(y)

-3 -7

-2 -4

-1 -1

0 2

1 5

2 8

x

Input(x) Output(y)

-6 1

-4 2

-2 3

0 4

2 5

4 6

x

Input(x) Output(y)

-2 5

-1 2

0 1

1 2

2 5

3 10

x

Input(x) Output(y)


2–2

2

4

6

8

–2

–4

–6

–8

–10

–12

x

y

2–2

2

–2

y

x

Answers

1. 24, -8, 7; y = x + 4

2. 8, -5, 9; y = x ! 3 3. 40, -1, -26; y = 3x ! 5

4. 1

3, 15, -45; y = !3x

5. 8, -12, 8; y = !1

2x + 2 6. 3, 5, -2; y = x + 3

7. 8, -27, 15; y = 2x ! 3

8. 13, 7, 6; y = !3x +1 9. 12, 0, -1; y = x3

10. y = 3x + 2

11. y = 1

2x + 4 12. y = x2 +1

13.

14. input (x) -3 -2 -1 0 1 2 3 output (y) -0.5 0 0.5 1 1.5 2 2.5

15. input (x) -3 -2 -1 0 1 2 3 output (y) 5 4 3 2 1 0 -1

16. input (x) -3 -2 -1 0 1 2 3 output (y) 3 -2 -5 -6 -5 -2 3

input (x) -3 -2 -1 0 1 2 3 output (y) -11 -8 -5 -2 1 4 7

2 4–2–4

2

4

–2

x

y

2 4–2–4

2

4

–2

–4

–6

x

y


EXAMINING AND USING EQUATIONS 3.2.2 through 3.2.4

Part of Chapter 2 taught you how to simplify algebraic expressions and how to solve single variable equations. The second part of Chapter 3 examines solutions to single variable equations to consider cases where there is no solution and cases where there is more than one solution. In addition, students use their equation solving skills to solve application problems. See the Math Notes box on page 127.

Example 1 One Solution Solve 10 ! 2x = 6x + 3! 5 .

Simplify the right side of the equation.

Add !6x and !10 to both sides of the equation.

Divide both sides by !8 .

10 ! 2x = 6x + 3! 5

10 ! 2x = 6x ! 2

!8x = !12

x =!12!8,!so!x = 3

2!or !1.5

Example 2 Multiple Solutions Solve 2x + 2 + x = 4x + 2 ! x .

Simplify both sides of the equation.


2x + 2 + x = 4x + 2 ! x

3x + 2 = 3x + 2

0 = 0

Notice that both sides of the equation are the same after simplifying them, that is, 3x + 2 = 3x + 2 . When you continue to solve this equation, the result is 0 = 0 . Since both of these statements are always true, any number substituted for x in the original equation will make it true. Thus, the solution to equations with this result is all numbers. Example 3 No Solution Solve 4x + 7 = 3x + 6 + x .

Simplify the right side of the equation.


4x + 7 = 3x + 6 + x

4x + 7 = 4x + 6

0 = !1

Since 0 = !1 is never true, solutions that lead to a false statement have no solution, that is, there is no value for x that makes the equation true.


Example 4 Tamica started collecting CDs several weeks ago. Her sister gave her five CDs to start her collection. Use the data table to predict how many CDs she will have: a) after nine weeks; b) at the end of one year (52 weeks); and c) at the end of any week. Weeks collecting CDs (x) 0 1 2 3 4 5 6 7 8 9 Number of CDs owned (y) 5 8 11 14 17 20

For part (a): Each week she buys three more CDs, so in four weeks she will add 12 CDs to the 20 that she already has for a total of 32 CDs. For parts (b) and (c): While part (b) could be solved for 52 weeks in the same way as part (a), finding the rule for the data will make this and any other question about the table quick and easy to answer. The input value represents the number of weeks Tamica has been collecting CDs. Each week this number increases by three. Express this relationship as 3x . Since she was given five CDs before she purchased her first three CDs, her collection will always have five more CDs than she has purchased herself. So the number of CDs she has at the end of any week can be expressed as 3x + 5 . Written as a rule, the answer to part (c) is y = 3x + 5 and the answer to part (b) for x = 52 weeks is 3(52) + 5 = 161 CDs. Problems Solve each equation. Look for situations that have no solution or more than one solution. 1. 3x + 5 = x !1 2. 4x + 2 ! x +1 = 2x ! 3+ x + 6

3. 3x ! 7 + 2x + 4 = 4x ! 6 + x 4. 3+ 4x ! 7 + 2x = x + 4 + x

5. 4 + x ! 3x + 2 = 4x + 2 ! 3x +1 6. 2x ! 3+ 2 + 4x = 4x + 3! (4 ! 2x)

7. 4x + 3! 2x = !x ! (!3x + 3) 8. 4x + 3! 2x = !x ! (3x + 3) 9. Amber lives near a new part of Denver where a home building company starts construction

on 24 new homes each month. The developer currently has 144 homes either completed or under construction. In how many months will the builder have more than 1,000 home sites (that is, completed homes and homes under construction) in the area? Make an x→y table, find the first few values, then write a rule and use it to answer the question.

10. Jabari measures the height of a new house plant in his room once a week. When he planted

it, it was 15 centimeters tall. After measuring it for five weeks he noticed that it seems to grow about 1.5 centimeters per week. If the plant continues to grow at this rate, in how many weeks will the plant be 100 centimeters tall? Make an x→y table, find the first few values, then write a rule and use it to answer the question.


11. Adriana estimates that there are 280 plums on a tree in her yard. If she picks six plums each

day, how many plums will be left on the tree after 10 days? After 30 days? Make an x→y table for the first few values, then write a rule and use it to answer the questions.

Answers 1. -3 2. all numbers 3. no solution 4. 2

5. 1 6. all numbers 7. no solution 8. -1 9. Month (x) 0 1 2 3 4 x Number houses (y) 144 168 192 216 240 24x +144

The builder will have more than 1,000 home sites in the area in 36 months. (Exact answer: 35.6 months) 10. Number of weeks (x) 0 1 2 3 4 5 x Height of plant (y) 15 16.5 18 19.5 21 22.5 1.5x +15

The plant will be 100 centimeters tall in 57 weeks. (Exact answer: 56.6 weeks) 11. Days (x) 0 1 2 3 4 x Number of plums on tree (y) 280 274 268 262 256 280 ! 6x

There will be 220 plums on the tree in 10 days. There will be 100 plums on the tree in 30 days.


2 4–2

2

4

6

8

10

12

–2

x

y

MULTIPLE REPRESENTATIONS 4.1.1 through 4.1.7

The first part of Chapter 4 ties together several ways to represent the same relationship. The basis for any relationship is a consistent pattern that connects input and output values. This course uses tile patterns to help visualize algebraic relationships. (Note: In this course we consider tile patterns to be continuous relationships and graph them with a continuous line or curve.) These relationships may also be displayed on a graph, in a table, or as an equation. In each situation, all four representations show the same relationship. Students learn how to use each format to display relationships as well as how to switch from one representation to another. We use the diagram at right to show the connections between the various ways to display a relationship and call it the “representations web.” See the Math Notes box on page 161.

Example 1

At this point in the course we use the notion of growth to help understand linear relationships. For example, a simple tile pattern may start with two tiles and grow by three tiles in each successive figure as shown below.

fig. 0 fig. 1 fig. 2 fig. 3 fig. 4

The picture of the tile figures may also be described by an equation in y = mx + b form, where x and y are variables and m represents the growth rate and b represents the starting value of the pattern. In this example, y = 3x + 2 , where 2 represents the number of tiles in the original figure (usually called “figure 0”) and 3 is the “growth factor” that describes the rate at which each successive figure adds tiles to the previous figure. This relationship may also be displayed in a table, called an “x→y table,” as shown below. The rule is written in the last column of the table. Figure number (x) 0 1 2 3 4 x Number of tiles (y) 2 5 8 11 14 3x + 2

Finally, the relationship may be displayed on an xy-coordinate graph by plotting the points in the table as shown at right. The highlighted points on the graph represent the tile pattern. The line represents all of the points described by the equation y = 3x + 2 .

Graph Rule

Pattern

Table

Chapter 4: Multiple Representations 23

2 4

5

10

15

20

y

x

Example 2 Draw figures 0, 4, and 5 for the tile pattern below. Use the pattern to predict the number of tiles in figure 100, describe the figure, write a rule that will give the number of tiles in any figure, record the data for the first six tiles (figures 0 through 5) in a table, and graph the data.

fig. 0 fig. 1 fig. 2 fig. 3 fig. 4 fig. 5 Each figure adds four tiles: two tiles to the top row and two tiles to the lower portion of the figure. Figure 0 has two tiles, so the rule is y = 4x + 2 and figure 100 has 4(100) + 2 = 402 tiles. There are 202 tiles in the top row and 200 tiles in the lower portion of figure 100. The table is:

Figure number (x) 0 1 2 3 4 5 x Number of tiles (y) 2 6 10 14 18 22 4x + 2

The graph is shown at right.

Example 3 Use the table below to determine the rule in y = mx + b form that describes the pattern.

input (x) -2 -1 0 1 2 3 4 5 x output (y) -8 -5 -2 1 4 7 10 13

The constant difference between the output values is the growth rate, that is, the value of m. The output value paired with the input value x = 0 is the starting value, that is, the value of b. So this table can be described by the rule: y = 3x ! 2 . Note: If there is no constant difference between the output values for consecutive integer input values, then the rule for the pattern is not in the form y = mx + b .

+3 +3 +3 +3 +3 +3 +3


2

2

4

–2

–4

–6

x

yExample 4 Use the graph at right to create an x→y table, then write a rule for the pattern it represents. First transfer the coordinates of the points into an x→y table.

input (x) 0 1 2 3 x output (y) 5 1 -3 -7

Using the method described in Example 3, that is, noting that the growth rate between the output values is -4 and the value of y at x = 0 is 5, the rule is: y = !4x + 5 . Problems 1. Based on the tile pattern below, draw figures 0, 4, and 5. Then find a rule that will give the

number of tiles in any figure and use it to find the number of tiles in figure 100. Finally, display the data for the first six figures (numbers 0-5) in a table and on a graph.

fig. 0 fig. 1 fig. 2 fig. 3 fig. 4 fig. 5 2. Based on the tile pattern below, draw figures 0, 4, and 5. Then find a rule that will give the

number of tiles in any figure and use it to find the number of tiles in figure 100. Finally, display the data for the first six figures (numbers 0-5) in a table and on a graph.

fig. 0 fig. 1 fig. 2 fig. 3 fig. 4 fig. 5


Use the patterns in the tables and graphs to write rules for each relationship. 3.

input (x) -3 -2 -1 0 1 2 3 4 5 output (y) -11 -8 -5 -2 1 4 7 10 13

4. input (x) -3 -2 -1 0 1 2 3 4 5 output (y) 10 8 6 4 2 0 -2 -4 -6

5. 6.

2 4–2–4

2

4

–2

–4

–6

y

x

2 4–2–4

2

4

6

–2

x

y


2 4

5

10

15

20

x

y

Answers 1. fig. 0 fig. 4 fig. 5 The rule is y = 2x + 5 . Figure 100 will have 205 tiles. It will have a base of three tiles, with 102 tiles extending up from the right tile in the base and 100 tiles extending to the right of the top tile in the vertical extension above the base.

2. fig. 0 fig. 4 fig. 5 The rule is y = 4x +1 . Figure 100 will have 401 tiles in the shape of an “X” with 100 tiles on each “branch” of the X, all connected to a single square in the middle.

Figure number (x) 0 1 2 3 4 5 Number of tiles (y) 1 5 9 13 17 21

3. y = 3x ! 2 4. y = !2x + 4 5. y = 2x ! 3 6. y = !x

Figure number (x) 0 1 2 3 4 5 Number of tiles (y) 5 7 9 11 13 15

2 4

5

10

15

x

y


SOLVING LINEAR SYSTEMS: 4.2.1 through 4.2.3 THE EQUAL VALUES METHOD

Two lines on an xy-coordinate grid intersect at a point unless they are parallel or the equations are different forms of the same line. The point of intersection is the only pair of (x, y) values that will make both equations true. One way to find the point of intersection is to graph the two lines. However, graphing is both time-consuming and, in many cases, not exact, because the result is only a close approximation of the coordinates. When two equations are written in the y = mx + b form, we can take advantage of the fact that both y values are the same (equal) at the point of intersection. For example, if two lines are described by the equations y = !2x + 5 and y = x !1 , and we know that both y values are equal, then the other two sides of the equations must also be equal to each other. We say that both right sides of these equations have “equal values” at the point of intersection and write!2x + 5 = x !1 . We can solve this equation in the usual way and find that x = 2 . Now we know the x-coordinate of the point of intersection. Since this value will be the same in both of the original equations at the point of intersection, we can substitute x = 2 in either equation to solve for y: y = !2(2) + 5 so y = 1 or y = 2 !1and y = 1. So the two lines in this example intersect at (2, 1).

Example 1 Find the point of intersection for y = 5x +1 and y = !3x !15 . Substitute the equal parts of the equations. Solve for x.

5x +1 = !3x !15

8x = !16

x = !2

Replace x with !2 in either original equation y = 5(-2) + 1 or y = -3(-2) - 15

and solve for y. y = 5(!2) +1

y = !10 +1

y = !9

or y = !3(!2) !15

y = 6 !15

y = !9

The two lines intersect at (-2, -9).


Example 2 Highland has a population of 12,200 that has been increasing at a rate of 300 people per year. Lowville has a population of 21,000 which is declining by 250 people per year. Assuming that the rates do not change, in how many years will the populations be equal? The first step in the solution is to write an equation in y = mx + b form that describes the population conditions in each city. In this example, let x equal the number of years from now and y be the population at any particular time. Then the equation to represent Highland's population x years from now is y = 300x +12200 . Similarly, the equation representing Lowville's population x years from now is y = !250x + 21000 . As usual, the rate of change is the value of m and the starting population is the value of b. Now that we know that y = 300x +12200 and y = !250x + 21000 , the next step is to use the “Equal Values” method to write one equation using x , then solve for x.

300x +12200 = !250x + 21000

550x = 8800

x = 16

Use the value of x to find y. y = 300(16) +12200

= 17000

The solution is (16, 17000). This means that 16 years from now, both cities will have the same population of 17,000 people. Problems

Use the Equal Values method to find the point of intersection (x, y) for each pair of linear equations. 1. y = x ! 6

y = 12 ! x

2. y = 3x ! 5

y = x + 3

3. x = 7 + 3y

x = 4y + 5

4. x = !3y +10

x = !6y ! 2

5. y = x + 7

y = 4x ! 5

6. y = 7 ! 3x

y = 2x ! 8

7. Jacques will wash the windows of a house for $15.00 plus $1.00 per window. Ray will

wash them for $5.00 plus $2.00 per window. Let x be the number of windows and y be the total charge for washing them. Write an equation that represents how much each person charges to wash windows. Solve the system of equations and explain what the solution means and when it would be most economical to use each window washer.


8. Cross Country Movering (CCM) charges $2000 plus $0.90 per pound to move a house full of furniture from Maryland to California. GlobalCity (GC) charges $3500 plus $0.40 per pound for the same move. Write two equations that represent each company's charges. What do your variables represent? Solve the system of equations, then decide who you would hire and why.

9. Misha and Noraa want to buy season passes for a ski lift but neither of them has the $225

needed to purchase a pass. Noraa decides to get a job that pays $6.25 per hour. She has nothing saved right now but she can work four hours each week. Misha already has $80 and plans to save $15 of her weekly allowance. Who will be able to purchase a pass first?

10. Ginny is raising pumpkins to enter a contest to see who can grow the heaviest pumpkin. Her

best pumpkin weighs 22 pounds and is growing 2.5 pounds per week. Martha planted her pumpkins late. Her best pumpkin weighs 10 pounds but she expects it to grow 4 pounds per week. Assuming that their pumpkins grow at these rates, in how many weeks will their pumpkins weigh the same? How much will they weigh? If the contest ends in seven weeks, who will have the heavier pumpkin at that time?

11. Larry and his sister, Betty, are saving money to buy their own laptop computers. Larry has

$215 and can save $35 each week. Betty has $380 and can save $20 each week. When will Larry and Betty have the same amount of money?

Answers 1. (9, 3) 2. (4, 7) 3. (13, 2)

4. (22, -4) 5. (4, 11) 6. (3, -2) 7. Let x = number of windows, y = cost. Jacques: y = 1x +15 ; Roy: y = 2x + 5 . The

solution is (10, 25), which means that the cost to wash 10 windows is $25. For fewer than 10 windows use Roy; for more than 10 windows, use Jacques.

8. Let x = pounds, y = amount charged. CCM: y = 0.90x + 2000 ; GC: y = 0.40x + 3500 .

The solution is (3000, 4700). For fewer than 3000 pounds, use CCM.

9. Let x = weeks, y = total savings. Misha: y = 15x + 80 ; Noraa: y = 25x . The solution is (8, 200). Both of them will have $200 in 8 weeks, so Noraa will have $225 in 9 weeks and be able to purchase the lift pass first. An alternative solution is to write both equations, then substitute 225 for y in each equation and solve for x. In this case, Noraa can buy a ticket in 9 weeks, Misha in 9.67 weeks.

10. Let x = weeks and y = weight of the pumpkin. Ginny: y = 2.5x + 22 ; Martha: y = 4x +10 .

The solution is (8, 42), so their pumpkins will weigh 42 pounds in 8 weeks. Ginny would win (39.5 pounds to 38 pounds for Martha).

11. Let x = weeks, y = total money saved. Larry: y = 35x + 215 ; Betty: y = 20x + 380 . The

solution is (11, 600). They will both have $600 in 11 weeks.


USING RECTANGLES TO MULTIPLY 5.1.1 through 5.1.3

Two ways to find the area of a rectangle are: as a product of the (height) ! (base) or as the sum of the areas of individual pieces of the rectangle. For a given rectangle these two areas must be the same, so Area as a product = Area as a sum. Algebra tiles and, later, generic rectangles, provide area models to help multiply expressions in a visual, concrete manner. See the Math Notes box on page 218.

Example 1 Using Tiles The algebra tile pieces x2 + 6x + 8 are arranged into a rectangle inside the cornerpiece as shown at right. The area of the rectangle can be written as the product of its base and height or as the sum of its parts. area as a product area as a sum

Example 2 Using Generic Rectangles A generic rectangle allows us to organize the problem in the same way as the first example without needing to draw the individual tiles. It does not have to be drawn accurately or to scale. Multiply x ! 3( )(2x +1) . ! ! x ! 3( ) 2x +1( ) = 2x2 ! 5x ! 3 area as a product area as a sum

x2

x

x

x

x

x

x

x

cornerpiece

height

base

x

(x + 4)

base

123(x + 2)

height

123!!!!!!!=!!!!!!!x2 + 6x + 8

area

1 24 34

-3

x

2x +1

-3

x

2x +1

!3

x 2x2

!6x

Chapter 5: Multiplication and Proportions 31

Problems Write a statement showing: area as a product equals area as a sum. 1.

2. 3.

4.

5. 6.

Multiply.

7. 3x + 2( ) 2x + 7( ) 8. 2x !1( ) 3x +1( ) 9. 2x( ) x !1( )

10. 2y !1( ) 4y + 7( ) 11. y ! 4( ) y + 4( ) 12. y( ) x !1( )

13. 3x !1( ) x + 2( ) 14. 2y ! 5( ) y + 4( ) 15. 3y( ) x ! y( )

16. 3x ! 5( ) 3x + 5( ) 17. 4x +1( )2 18. x + y( ) x + 2( )

19. 2y ! 3( )2 20. x !1( ) x + y +1( ) 21. x + 2( ) x + y ! 2( )

Answers 1. x +1( ) x + 3( ) = x2 + 4x + 3 2. x + 2( ) 2x +1( ) = 2x2 + 5x + 2

3. x + 2( ) 2x + 3( ) = 2x2 + 7x + 6 4. x ! 5( ) x + 3( ) = x2 ! 2x !15

5. 6 3y ! 2x( ) = 18y !12x 6. x + 4( ) 3y ! 2( ) = 3xy ! 2x +12y ! 8

7. 6x2+ 25x +14 8. 6x

2! x !1 9. 2x

2! 2x

10. 8y2+10y ! 7 11. y

2!16 12. xy ! y

13. 3x2+ 5x ! 2 14. 2y

2+ 3y ! 20 15. 3xy ! 3y

2

16. 9x2! 25 17. 16x

2+ 8x +1 18. x

2+ 2x + xy + 2y

19. 4y2!12y + 9 20. x

2+ xy ! y !1 21. x

2+ xy + 2y ! 4

6

3y !2x

x –5

x +3

3y

x + 4

–2 !5x x2 3x

!15 18y !12x

!2x

3xy

!8

12y


SOLVING EQUATIONS WITH MULTIPLICATION 5.1.4

To solve an equation with multiplication, first use the Distributive Property or a generic rectangle to rewrite the equation without parentheses, then solve in the usual way. See the Math Notes box on page 198.

Example 1 Solve 6 x + 2( ) = 3 5x +1( ) Use the Distributive Property Subtract 6x Subtract 3 Divide by 9

6x +12 = 15x + 3

12 = 9x + 3

9 = 9x

1 = x Example 2 Solve x 2x ! 4( ) = 2x +1( ) x + 5( ) Rewrite using generic rectangles Subtract 2x2 Subtract 11x Divide by !15

2x2 ! 4x = 2x2 +11x + 5

!4x = 11x + 5

!15x = 5

x =5

!15= !

1

3


Problems Solve each equation.

1. 3 c + 4( ) = 5c +14

2. x ! 4 = 5 x + 2( )

3. 7(x + 7) = 49 ! x

4. 8(x ! 2) = 2 2 ! x( )

5. 5x ! 4 x ! 3( ) = 8

6. 4y ! 2 6 ! y( ) = 6

7. 2x + 2(2x ! 4) = 244

8. x 2x ! 4( ) = 2x +1( ) x ! 2( )

9. x !1( ) x + 7( ) = x +1( ) x ! 3( )

10. x + 3( ) x + 4( ) = x +1( ) x + 2( )

11. 2x ! 5 x + 4( ) = !2 x + 3( )

12. x + 2( ) x + 3( ) = x2 + 5x + 6

13. x ! 3( ) x + 5( ) = x2 ! 7x !15

14. x + 2( ) x ! 2( ) = x + 3( ) x ! 3( )

15. 1

2x x + 2( ) = 1

2x + 2( ) x ! 3( )

Answers

1. !1

2. !3.5 3. 0

4. 2

5. !4 6. 3

7. 42

8. 2 9. 0.5

10. !2.5

11. !14 12. all numbers

13. 0 14. no solution 15. !12


SOLVING MULTI-VARIABLE EQUATIONS 5.1.5

Solving equations with more than one variable uses the same process as solving an equation with one variable. The only difference is that instead of the answer always being a number, it may be an expression that includes numbers and variables. The usual steps may include: removing parentheses, simplifying by combining like terms, removing the same thing from both sides of the equation, moving the desired variables to one side of the equation and the rest of the variables to the other side, and possibly division or multiplication.

Example 1 Solve for y

Subtract 3x

Divide by !2

Simplify

3x ! 2y = 6

!2y = !3x + 6

y = !3x+6!2

y = 3

2x ! 3

Example 2 Solve for y

Subtract 7

Distribute the 2

Subtract 2x

Divide by 2

Simplify

7 + 2(x + y) = 11

2(x + y) = 4

2x + 2y = 4

2y = !2x + 4

y = !2x+42

y = !x + 2

Example 3 Solve for x y = 3x ! 4 Add 4 y + 4 = 3x

Divide by 3 y + 4

3= x

Example 4 Solve for t I = prt

Divide by pr I

pr= t

Problems

Solve each equation for the specified variable.

1. y in 5x + 3y = 15

2. x in 5x + 3y = 15 3. w in 2l + 2w = P

4. m in 4n = 3m !1

5. a in 2a + b = c 6. a in b ! 2a = c

7. p in 6 ! 2(q ! 3p) = 4 p 8. x in y = 1

4x +1 9. r in 4(r ! 3s) = r ! 5s

Answers (Other equivalent forms are possible.)

1. y = !5

3x + 5 2. x = !

3

5y + 3 3. w = !l +

P

2

4. m =4n+1

3 5. a =

c!b

2 6. a =

c!b

!2!or !

b!c

2

7. p = q ! 3 8. x = 4y ! 4 9. r =7s

3


SETTING UP PROPORTIONS 5.2.1 and 5.2.2

A convenient way to set up proportions is to arrange the information in a table. Once the information is placed into a labeled table, the proportion can be written directly from the table. See the Math Notes box on page 211.

Example 1 A tree casts a 43-foot shadow. At the same time, a 4 1

2 foot tall boy casts a 10-foot

shadow. Determine the height of the tree. Eliminate the denominators. Divide by 10.

x

4.5=43

10

45x

4.5

!"#

$%&= 45

43

10

!"#

$%&

10x = 193.5

x = 19.35

Example 2 Kim noticed that 100 vitamins cost $1.89. At this rate, how much should 350 vitamins cost? Eliminate the denominators. Divide by 100.

100350

=1.89

x

350x100

350

!"#

$%&= 350x

1.89

x

!"#

$%&

100x = 661.5

x = $6.62

height (ft) shadow (ft) x 43

4.5 10

Vitamins (#) Cost ($) 100 1.89 350 x


Problems Solve each problem by writing a proportion and solving it. 1. Joe came to bat 464 times in 131 games. At this rate, how many times should he expect to

bat in a full season of 162 games? 2. Mario’s car needs 12 gallons to go 320 miles. At the same rate, how far can he travel with

10 gallons of gas? 3. If 50 empty soda cans weigh 3 1

2 pounds, how much would 70 empty soda cans weigh?

4. There is a $34 tax on an $800 motor scooter. How much tax would there be on a $1000

motor scooter? 5. A dozen donuts cost $3.50. How much should three donuts cost? 6. In 35 minutes, Suki’s car goes 25 miles. If she continues at the same rate, how long will it

take her to drive 90 miles? 7. In a city of three million people, 3,472 were surveyed. Of those surveyed, 28 of them

watched the last Lonely Alien movie. If the survey represents the city’s TV viewing habits, about how many people in the city watched the movie?

8. Julio runs 3

10 mile in 1 1

2 minutes. At that rate how long will it take him to run a mile?

9. It is now 7:51 p.m. The movie that you have waited three weeks to see starts at 8:15 p.m.

While standing in line for the movie, you count 146 people ahead of you. Nine people buy their tickets in 70 seconds. Will you be able to buy your ticket before the movie starts?

10. A biscuit recipe uses 1

2 teaspoon of baking powder for 3

4 cup of flour. How much baking

powder is needed for three cups of flour? 11. The germination rate for zinnia seeds is 78%. This means that 78 out of every 100 seeds

will sprout and grow. If Jim wants 60 plants for his yard, how many seeds should he plant? 12. L.J.’s car has a gas tank which holds 19 gallons of fuel. If the gas tank was full and he

used eight gallons to drive 200 miles, does the car have enough gas to go another 250 miles?

Answers

1. about 574 times 2. about 267 miles 3. 4.9 pounds 4. $42.50 5. $0.88 6. 126 minutes 7. about 24,194 people 8. 5 minutes 9. yes, in about 19 min. 10. 2 teaspoons 11. about 77 seeds 12. yes, 275 miles


SOLVING PROPORTIONS 5.2.2

To solve proportions (equal ratio equations) remove the denominators by multiplying both fractions (ratios) by the common denominator. If you cannot determine the common denominator, multiply the denominators and then multiply both fractions (ratios) by the result. Then solve in the usual manner. Another approach is to begin by multiplying to remove the denominator of the fraction (ratio) with the variable. In the first example below, multiply by 12; in the second example, multiply by 3. When there is a variable in both fractions, multiply by the product of both denominators.

Example 1 Solve for x x

12=7

10

Multiply by 60 60 x

12

!"#

$%&= 60

7

10

!"#

$%&

Simplify 5x = 42

Divide by 5 x =42

5

x = 8.4

Example 2 Solve for p 11

5=p + 2

3

Multiply by 15 15 11

5

!"#

$%&= 15

p + 2

3

!"#

$%&

Simplify 33 = 5 p + 2( ) Distribute the 5 33 = 5p +10 Subtract 10 23 = 5p

Divide by 5 p =23

5

p = 4.6

Example 3 Solve for x

Multiply by 12 Simplify Subtract 4x and 9

x + 3

4=x !13

12x + 3

4

"#$

%&'= 12

x !13

"#$

%&'

3(x + 3) = 4(x !1)

3x + 9 = 4x ! 4

!x = !13

x = 13


Problems Solve each proportion.

1. x

20=3

5

2. x ! 4

12=2

3 3. 100

40=5

2x

4. 6x

5=x ! 3

10

5. 8x

7=13

2 6. 3! x

4=x ! 7

10

7. 6x

11=35

4

8. 6 + x

2=4 ! x

8 9. 9 ! x

6=24

2

10. 3x

10=24

9

11. 4x

5=x ! 2

7 12. 6x

7=42

3

13. 4x

7=81

15 14. x

100=5

20 15. x

100=7

35

Answers

1. 12

2. 12 3. 1

4. ! "0.27

5. ≈5.69 6. ≈ 4.14

7. ≈ 16.04

8. !4 9. !63

10. ≈ 8.89

11. ! "0.43 12. 16.3

13. 9.45 14. 25 15. 20

Chapter 6: Systems of Equations 39

WRITING EQUATIONS 6.1.1 through 6.1.3

An equation is a mathematical sentence that conveys information to the reader. It uses variables and operation symbols (like +, - , /, =) to represent relationships for a situation. Technology has made it easier to solve equations, but to use these algebraic operating systems, you need to be able to write an equation. An equation may contain one or more variables. Solving a one variable equation (refer to sections 3.2, 5.1 and 5.2 in the textbook) involves different procedures than solving two variable equations. Students are asked to practice writing both types of equations in Chapter 6. See the Math Notes box on page 234.

Example 1 Just as there are variations in the vocabulary used in English sentences, there may be alternative ways to write an equation. The sum of the lengths of two adjacent sides of a rectangle is 15. Using one variable: If x = the length, then the width = 15! x , so x + (15 ! x) = 15 Using two variables: l + w = 15 , where l is the length and w is the width. Example 2 Equations are easier to use if the variable is defined for the reader. a) A collection of quarters, dimes and

nickels is worth $6.00. Let q = quarters, d = dimes, and n = nickels.

0.25q + 0.10d + 0.05n = 6.00

b) The cost of cell phone calls priced at 5¢ per minute. Letc = cost, m = minutes.

c = 0.05m


Example 3 Drawing a diagram with labels or creating a Guess & Check table can help identify relationships between quantities. The perimeter of a rectangle is 60 cm. The width is 4 times the length. Write an equation and solve for the length. Guess & Check Table:

Length (guess) Width Perimeter Check? ? 10 5

4(10) 4(5)

2(10) + 2(40) 2(5) + 2(20)

100 50

too big too small

l 4(l) 2l + 2(4l) = 60 Drawing a diagram: Using one variable: 2l + 2(4l) = 60

Using two variables and two equations: w = 4l2l + 2w = 60

Example 4 A “Let” statement describes what the variable represents. Mike spent $11.19 on a bag containing red and blue candies. The bag weighed 11 pounds. The red candy costs $1.29 a pound and the blue candy costs $0.79 a pound. Write a system of equations for this problem. Let x = the weight of the red candy and y = the weight of the blue candy.

x + y = 111.29x + 0.79y = 11.19

The first equation represents the weight of the bag. The second equation represents the cost of the candy.

4l

l or

w

l


Problems Write an equation using one variable or a system of equations. Do NOT solve your equations. 1. A rectangle is three times as long as it is wide. Its perimeter is 36. Find each side length. 2. A rectangle is twice as long as it is wide. Its area is 72. Find each side length. 3. The sum of two consecutive odd integers is 76. Find the numbers. 4. Nancy started the year with $425 in the bank and is saving $25 a week. Seamus started with

$875 and is spending $15 a week. When will they both have the same amount of money in the bank?

5. Oliver earns $50 a day and $7.50 for each package he processes at Company A. His

paycheck on his first day was $140. How many packages did he process? 6. Dustin has a collection of quarters and pennies. The total value is $4.65. There are 33

coins. How many quarters and pennies does he have? 7. A one pound mixture of raisins and peanuts cost $7.50. The raisins cost $3.25 a pound and

the peanuts cost $5.75 a pound. How much of each ingredient is in the mixture? 8. An adult ticket at an amusement park costs $24.95 and a child’s ticket costs $15.95. A

group of 10 people paid $186.50 to enter the park. How many were adults? 9. Katy weighs 105 pounds and is gaining 2 pounds a month. James weighs 175 pounds and is

losing 3 pounds a month. When will they weigh the same? 10. Holmes Junior High has x students. Harper Middle School has 125 fewer students than

Holmes. When the two schools are merged there will be 809 students. How many students attend each school?


Answers (Other equivalent forms are possible.)

One Variable Equation System of Equations Let Statement

1.

2w + 2(3w) = 36

l = 3w2w + 2l = 36

Let l = length, w = width

2. w(2w) = 72

l = 2wlw = 72

Let l = length, w = width

3. m + (m + 2) = 76 m + n = 76n = m + 2

Let m = the first odd integer and n = the next consecutive odd integer

4. 425 + 25x = 875 !15x y = 425 + 25xy = 875 !15x

Let x = the number of weeks and y = the total money in the bank

5. 50 + 7.5p = 140

6. 0.25q + 0.01(33! q) = 4.65

q + p = 330.25q + 0.01p = 4.65

Let q = number of quarters, p = number of pennies

7. 3.25r + 5.75(1! r) = 7.5

r + p = 13.25r + 5.75p = 7.5(1)

Let r = weight of raisins and p = weight of peanuts

8. 24.95a +15.95(10 ! a) = 186.5 a + c = 1024.95a +15.95c = 186.5

Let a = number of adult tickets and c = number of child’s tickets

9. 105 + 2m = 175 ! 3m w = 105 + 2mw = 175 ! 3m

Let m = the number of months and w = the weight of each person

10. x + (x !125) = 809 x + y = 809y = x !125

Let x = Holmes students and y = Harper students


SOLVING SYSTEMS BY SUBSTITUTION 6.2.1

A system of equations has two or more equations with two or more variables. In Section 4.2, students were introduced to solving a system by looking at the intersection point of the graphs. They also learned the algebraic “equal values” solution method. Graphing and the Equal Values method are convenient when both equations are in y-form. The substitution method is used to change a two-variable system of equations to a one-variable equation. This method is useful when one of the variables is isolated in one of the equations in the system. Two substitutions must be made to find both the x- and y-values for a complete solution. See the Math Notes boxes on pages 248, 252, and 258.

The equation x + y = 9 has infinite possibilities for solutions: (10, -1) (2, 7) (0, 9)…, but if

y = 4 then there is only one possible value for x. That value is easily seen when we replace (substitute for) y with 4 in the original equation: x + 4 = 9 , so x = 5when y = 4 . Substitution and this observation are the basis for the following method to solve systems of equations. Example 1 When solving a system of equations, we replace a variable with an algebraic expression.

Solve: y = 2x

x + y = 9

x + y = 9 First replace y with 2x so there is only one variable in the

second equation.

x + 2x = 9

3x = 9x = 3

Solve as usual.

y = 2(3) Use x = 3 to substitute in either original equation to find y = 6 the y value.

The solution is (3, 6) and makes both equations true. When graphing, this is the intersection point of the two lines.


Example 2 Substitute the variable that is convenient. Look for the variable that is by itself on one side of the equation.

Solve: 4x + y = 8x = 5 ! y

4(5 ! y) + y = 820 ! 4y + y = 8

20 ! 3y = 8!3y = !12y = 4

x = 5 ! 4x = 1

Since x is alone, substitute 5 ! y in the first equation for x, then solve as usual. Then substitute y = 4 in the original equation to find x.

The solution is (1, 4). Check by substituting this point into each equation to confirm that you are correct. Example 3 Not all systems have a solution. If the substitution results in an equation that is not true, then there is NO SOLUTION. This graph would have two parallel lines, so there is no point of intersection.

Solve: y = 7 ! 3x3x + y = 10

3x + (7 ! 3x) = 103x ! 3x + 7 = 10

7 " 10

Substitute 7 ! 3x in the second equation for y. No Solution.

There may also be an INFINITE NUMBER OF SOLUTIONS. This graph would appear as a single line for the two equations.

Solve: y = 4 ! 2x!4x ! 2y = !8

!4x ! 2(4 ! 2x) = !8!!!!!4x ! 8 + 4x = !8

!8 = !8

Substitute 4 ! 2x in the second equation for y. Always true for any value of x or y.


Example 4 Sometimes you need to solve one of the equations for x or y to use this method. Also be careful when a negative sign appears in the equation. Check your final solution to be sure it works in both of the original equations.

Solve:

y ! 2x = !73x ! 4y = 8

3x ! 4(2x ! 7) = 83x ! 8x + 28 = 8

!5x + 28 = 8!5x = !20

x = 4

Solve the first equation for y to get y = 2x ! 7 , then substitute 2x ! 7 in the second equation for y.

Then substitute again to find y.

y = 2 4( ) ! 7y = 1

Check: 1! 2(4) = !7 and 3(4) ! 4(1) = 8 .

The solution is (4, 1). This would be the intersection point of the two lines. Problems 1. y = !3x

4x + y = 2

2. y = 7x ! 52x + y = 13

3. x = !5y ! 4x ! 4y = 23

4. x + y = 10y = x ! 4

5. y = 5 ! x

4x + 2y = 10

6. 3x + 5y = 23y = x + 3

7. y = !x ! 22x + 3y = !9

8. y = 2x ! 3

!2x + y = 1

9. x = 12 y +

12

2x + y = !1

10. a = 2b + 4

b ! 2a = 16

11. y = 3! 2x4x + 2y = 6

12. y = x +1

x ! y = 1

Answers 1. (2, -6) 2. (2, 9) 3. (11, -3)

4. (7, 3) 5. (0, 5) 6. (1, 4)

7. (3, -5) 8. No solution 9. (0, -1)

10. (-12, -8) 11. Infinite solutions 12. No solution


ELIMINATION METHOD 6.2.3 and 6.2.4

In previous work with systems of equations, sections 6.1 and 6.2, one of the variables was usually alone on one side of one of the equations. In that situation, you can substitute to rewrite a two-variable equation as a one-variable equation, then solve the resulting equation. Systems of linear equations in standard form (that is, ax + by = c ) have both variables on the same side of both equations. To solve this type of system, we can eliminate one of the variables by focusing on the coefficients. (The coefficient is the number in front of the variable.) This method is called the elimination method and uses addition or subtraction, sometimes combined with multiplication or division, to rewrite the system as a one variable equation. After solving for one variable, we then substitute to find the other. A complete solution includes both an x and a y value. See the Math Notes boxes on pages 264 and 287 as well as those on pages 252 and 258.

Example 1

Solve: x ! y = 22x + y = 1

Notice that y and – y are opposites.

Adding the two equations together will eliminate the y-term because y + !y = 0 . This is sometimes called the addition method of elimination.

Add vertically. x ! y = 2

+2x + y = 1

Solve for x. 3x + 0 = 3

x = 1

Substitute the value of x in either of the original equations to find the y value.

2(1) + y = 12 + y = 1

y = !1 The point of intersection is (1, -1).


Example 2

Solve: 3x + 6y = 243x + y = !1

Notice that 3x ! 3x = 0 .

If the coefficients are the same, subtract the second equation to eliminate the variable, then substitute the value of y into one of the original equations to find x.

3x + 6y = 24!(3x + y = !1)

3x + 6y = 24!3x ! y = +10 + 5y = 25

y = 5

3x + 6(5) = 243x + 30 = 24

3x = !6x = !2

The solution is (-2, 5). It makes both equations true and names the point where the two lines intersect on a graph. Example 3 If neither of the pairs of coefficients are the same, multiply one or both of the equations to make one pair of coefficients the same or opposites. Then add or subtract as needed to eliminate a variable term.

Solve: x + 3y = 74x ! 7y = !10

Multiplying the first equation by 4 or -4 will allow you to eliminate the x. Be sure to multiply everything on the equation’s left side and right side to maintain equivalence.

One possible process: multiply the first equation by 4. 4(x + 3y = 7)! 4x +12y = 28 4x +12y = 28

!(4x ! 7y = !10)19y = 38y = 2

4x ! 7(2) = !104x !14 = !10

4x = 4x = 1

Use this equation in the system with the original second equation. Subtract and solve for y, then substitute in either original equation to find x. Solution is (1, 2).

If you prefer the addition method, multiply the first equation by – 4:

!4(x + 3y = 7)"!4x !12y = !28 . Then add and solve.

then


Example 4 If one multiplication will not eliminate a variable, multiply both equations by different numbers to get coefficients that are the same or opposites.

Solve: 8x ! 7y = 53x ! 5y = 9

Two ways to eliminate x:

3(8x ! 7y = 5)" 24x ! 21y = 158(3x ! 5y = 9)" 24x ! 40y = 72

or 3(8x ! 7y = 5)" 24x ! 21y = 15

!8(3x ! 5y = 9)"!24x + 40y = !72

Two ways to eliminate y:

5(8x ! 7y = 5)" 40x ! 35y = 257(3x ! 5y = 9)" 21x ! 35y = 63

or !5(8x ! 7y = 5)"!40x + 35y = !257(3x ! 5y = 9)" 21x ! 35y = 63

The examples above are all equivalent systems and have the same solution of (-2, -3). Example 5 The special cases of no solution and infinite solutions can also occur. See the Math Notes box on page 258.

Solve: 4x + 2y = 62x + y = 3

Dividing the first equation by 2 produces 2x + y = 3 .

The two equations are identical, so when graphed there would be one line with infinite solutions because the same ordered pairs are true for both equations.

Solve: 2x + y = 34x + 2y = 8

Multiplying the first equation by 2 produces 4x + 2y = 6 .

There are no numbers that could make both equations true at the same time, so the lines are parallel and there is no solution, that is, no point of intersection.


SUMMARY OF METHODS TO SOLVE SYSTEMS

Method When to Use Example Equal values Both equations in y-form.

y = x ! 2y = !2x +1

Substitution One variable is alone on one side of one equation.

y = !3x !13x + 6y = 24

Add to eliminate Equations in standard form with opposite coefficients.

x + 2y = 213x ! 2y = 7

Subtract to eliminate Equations in standard form with same coefficients.

x + 2y = 33x + 2y = 7

Multiply to eliminate When nothing else works. In this case you could multiply the first equation by 3 and the second equation by 2, then subtract to eliminate x.

2x ! 5y = 33x + 2y = 7

Problems 1. 2x + y = 6

!2x + y = 2

2. !4x + 5y = 0!6x + 5y = !10

3. 2x ! 3y = !9

x + y = !2

4. y ! x = 42y + x = 8

5. 2x ! y = 412 x + y = 1

6. !4x + 6y = !20

2x ! 3y = 10

7. 6x ! 2y = !164x + y = 1

8. 6x ! y = 4

6x + 3y = !16

9. 2x ! 2y = 52x ! 3y = 3

10. y ! 2x = 6y ! 2x = !4

11. 4x ! 4y = 14

2x ! 4y = 8

12. 3x + 2y = 125x ! 3y = !37

Answers 1. (1, 4)

2. (5, 4) 3. (!3 , 1)

4. (0, 4)

5. (2, 0) 6. Infinite

7. (!1, 5)

8. ( !16 , !5 ) 9. (4.5, 2)

10. No solution 11. (3, !12 ) 12. (!2 , 9)


2 4–2–4

2

4

–2

–4

x

y

2 4–2–4

2

4

–2

–4

x

y

2 4–2–4

2

4

–2

–4

x

y

SLOPE—A MEASURE OF STEEPNESS 7.1.3 through 7.1.5

Students have used the equation y = mx + b throughout this course to graph lines and describe patterns. When the equation is written in y-form, the m is the coefficient of x and is the slope of the line. It indicates the direction of the line and its steepness. The constant, b, is the y-intercept, written (0, b), and indicates where the line crosses the y-axis. See the Math Notes boxes on pages 283, 291, and 298.

Example 1 If m is positive, the line goes upward from left to right. If m is negative, the line goes downward from left to right. If m = 0 then the line is horizontal. y = x

y = !x y = 0x + 3 or

y = 3 Example 2 When m = 1 , as in

y = x , the line goes upward by one unit each time it goes over one unit to the right. Steeper lines have a larger m value, that is, m > 1. Flatter lines have an m value between 0 and 1, usually in the form of a fraction.

y = 3x y = 1

3x

2 4–2–4

2

4

–2

–4

x

y

2 4–2–4

2

4

–2

–4

x

y

Chapter 7: Linear Relationships 51

Example 3 Slope is written as a ratio. If the line is drawn on a set of axes, a slope triangle can be drawn between any two convenient points (that is, where grid lines cross and coordinates are integers), as shown in the graph at right. Count the vertical distance and the horizontal distance on the dashed sides of the slope triangle. Write the distances in a ratio:

m =vertical !!

horizontal !!. Another form used is:

!y

!x . The symbol

! means change. The order in the fraction is important: the numerator (top of the fraction) must be the vertical distance and the denominator (bottom of the fraction) must be the horizontal distance. Parallel lines have the same steepness and direction, so they have the same slope, as shown in the graph at right. If !y = 0 , then the line is horizontal and has a slope of zero, that is, m = 0 . If !x = 0 , then the line is vertical and its slope is undefined, so we say that it has no slope. Example 4 When the vertical and horizontal distances are not easy to determine, you can find the slope by drawing a generic slope triangle and using it to find the lengths of the vertical (!y ) and horizontal (!x ) segments. The figure at right shows how to find the slope of the line that passes through the points (-21, 9) and (19, -15). First graph the points on unscaled axes by approximating where they are located, then draw a slope triangle. Next find the distance along the vertical side by noting that it is 9 units from point B to the x-axis then 15 units from the x-axis to point C, so !y is 24. Then find the distance from point A to the y-axis (21) and the distance from the y-axis to point B (19). !x is 40. This slope is negative because the line goes downward from left to right, so the slope is ! 24

40= !

3

5.

3–3

3

–3

y

x

y = 2x ! 3,!y = 2x +1

!y = 2 !x = 3

3–3

3

–3

y

x

AB(-21,9)

(19,-15)(19,-15) C

!x =40

!y =24


Problems Is the slope of these lines negative, positive or zero? 1.

y

x

y

x

2.

y

x

y

x

3.

y

x

y

x

Identify the slope in these equations. State whether the graph of the line is steeper or flatter than y = x or y = !x , whether it goes up or down from left to right, or if it is horizontal or vertical. 4. y = 3x + 2 5. y = !

1

2x + 4 6. y = 1

3x ! 4

7. 4x ! 3 = y 8. y = !2 +1

2x 9. 3+ 2y = 8x

10. y = 2 11. x = 5 12. 6x + 3y = 8 Without graphing, find the slope of each line based on the given information. 13. !y = 27!!x = "8 14. !x = 15!!y = 3 15. !y = 7!!x = 0

16. Horizontal ! = 6

Vertical ! = 0 17. Between (5, 28) and

(64, 12) 18. Between (-3, 2) and

(5, -7) Answers 1. zero 2. negative 3. positive

4. Slope = 3, steeper, up 5. Slope = - 12

, flatter, down 6. Slope = 13

, flatter, up

7. Slope = 4, steeper, up 8. Slope = 12

, flatter, up 9. Slope = 4, steeper, up

10. horizontal 11. vertical 12. Slope = -2, steeper, down

13. !27

8 14. 3

15=1

5 15. undefined

16. 0 17. !16

59 18. !

9

8


WRITING AN EQUATION GIVEN THE SLOPE 7.3.1 and 7.3.2 AND A POINT ON THE LINE

In earlier work students used substitution in equations like y = 2x + 3 to find x and y pairs that make the equation true. Students recorded those pairs in a table, then used them as coordinates to graph a line. Every point (x, y) on the line makes the equation true. Later in the course, students used the patterns they saw in the tables and graphs to recognize and write equations in the form of y = mx + b . The “b” represents the y-intercept of the line, the “m” represents the slope, while x and y represent the coordinates of any point on the line. Each line has a unique value for m and a unique value for b, but there are infinite (x, y) values for each linear equation. The slope of the line is the same between any two points on that line. We can use this information to write equations without creating tables or graphs. See the Math Notes box on page 308.

Example 1 What is the equation of the line with a slope of 2 that passes through the point (10, 17)? Write the general equation of a line. Substitute the values we know: m, x, and y. Solve for b. Write the complete equation using the values of m (2) and b (-3).

y = mx + b

17 = 2(10) + b

17 = 20 + b

!3 = b

y = 2x ! 3


Example 2 This algebraic method can help us write equations of parallel and perpendicular lines. Parallel lines never intersect or meet. They have the same slope, m, but different y-intercepts, b. What is the equation of the line parallel to y = 3x ! 4 that goes through the point (2, 8)? Write the general equation of a line. y = mx + b Substitute the values we know: m, x, and y. Since the lines are parallel, the slopes are equal. 8 = 3(2) + b

Solve for b. 8 = 6 + b

2 = b

Write the complete equation. y = 3x + 2

Example 3 Perpendicular lines meet at a right angle, that is, 90° . They have opposite, reciprocal slopes and may have different y-intercepts. If one line is steep and goes upward, then the line perpendicular to it must be shallow and go downward. Reciprocals are pairs of numbers that when multiplied equal one. For example, 3 and 1

3 are

reciprocals as are 52

and 25

. Opposites have different signs: positive or negative. Examples of

opposite reciprocal slopes are ! 15

and 5; ! 23

and 32

; and !2 and 12

. Write the equation of the line that is perpendicular to y = 2x + 4 and goes through the point (6, 3). Write the general equation of a line. y = mx + b Substitute the values we know: m, x, and y. Since the line is perpendicular, the new slope is ! 1

2. 3 = !

12(6) + b

Solve for b. 3 = !3+ b

6 = b

Write the complete equation. y = !

1

2x + 6


Problems Write the equation of the line with the given slope that passes through the given point. 1. slope = 5, (3, 13)

2. slope = ! 5

3, (3, -1) 3. slope = -4 , (-2, 9)

4. slope = 32

, (6, 8)

5. slope = 3, (-7, -23) 6. slope = 2, ( 52

, -2)

Write the equation of the line parallel to the given line that goes through the given point. 7. y = 3

5x + 2 (0, 0)

8. y = 4x !1 (-2, -6) 9. y = !2x + 5 (-4, -2)

10. y = 4x + 5 (-6, -28)

11. y = 1

3x !1 (6, 9) 12. y = 3x + 8 (0, 1

2)

Write the equation of the line perpendicular to the given line that goes through the given point. 13. y = 3

5x + 2 (0, 0)

14. y = 4x !1 (-2, 6) 15. y = !2x + 5 (-4, -2)

16. y = 1

3x !1 (6, -2)

17. y = 4x + 5 (20, -28) 18. y = 3x + 1

2 (0, 1

2)

Answers 1. y = 5x ! 2 2. y = !

5

3x + 4 3. y = !4x +1

4. y = 3

2x !1 5. y = 3x ! 2 6. y = 2x ! 7

7. y = 3

5x 8. y = 4x + 2 9. y = !2x !10

10. y = 4x ! 4 11. y = 1

3x + 7 12. y = 3x + 1

2

13. y = !5

3x 14. y = !

1

4x + 5 1

2 15. y = 1

2x

16. y = !3x +16 17. y = !1

4x ! 23 18. y = !

1

3x + 1

2


WRITING THE EQUATION OF A LINE GIVEN TWO POINTS 7.3.3

Students now have all the tools they need to find the equation of a line passing through two given points. Recall that the equation of a line requires a slope and a y-intercept in y = mx + b . Students can write the equation of a line from two points by creating a slope

triangle and calculating !y!x

as explained in sections 7.1.3 through 7.1.5. See the Math Notes box on page 314.

Example 1 Write the equation of the line that passes through the points (1, 9) and (-2, -3). Draw a generic slope triangle. Calculate the slope using the given two points. Write the general equation of a line. y = mx + b Substitute m and one of the points for x and y, in this case (1, 9). 9 = 4(1) + b Solve for b. 5 = b Write the complete equation. y = 4x + 5 Example 2 Write the equation of the line that passes through the points (8, 3) and (4, 6). Draw a generic slope triangle. Calculate the slope using the given two points. Write the general equation of a line. y = mx + b Substitute m and one of the points for x and y, in this case (8, 3). 3 = !

34(8) + b

Solve for b. 3 = !6 + b

9 = b

Write the complete equation. y = !3

4x + 9

(1, 9)

(-2, -3)

12

3

!y

!x=12

3,

m = 4

(4, 6)

(8, 3)

4

3 !y

!x=3

4,

m = "3

4


Problems Write the equation of the line containing each pair of points. 1. (1, 1) and (0, 4)

2. (5, 4) and (1, 1) 3. (1, 3) and (-5, -15)

4. (-2, 3) and (3, 5)

5. (2, -1) and (3, -3) 6. (4, 5) and (-2, -4)

7. (1, -4) and (-2, 5) 8. (-3, -2) and (5, -2) 9. (-4, 1) and (5, -2) Answers

1. y = !3x + 4

2. y = 3

4x + 1

4

3. y = 3x

4. y = 2

5x + 3 4

5

5. y = !2x + 3 6. y = 3

2x !1

7. y = !3x !1 8. y = !2 9. y = !1

3x ! 1

3


FACTORING QUADRATICS 8.1.1 and 8.1.2

Chapter 8 introduces students to quadratic equations. These equations can be written in the form of

y = ax2+ bx + c and, when graphed, produce a curve called a parabola.

There are multiple methods that can be used to solve quadratic equations. One of them requires students to factor. Students have used algebra tiles to build rectangles of quadratic expressions. Later, they used the formula for the area of a rectangle, (base)(height) = area , to create generic rectangles and write equations expressing the area as both a sum and a product. In the figure below, the length and width of the rectangle are factors since they multiply together to produce the quadratic x2 + 6x + 8 . Notice the 4x and 2x are located diagonally from each other. They are like terms and can be combined and written as 6x . The dimensions of the rectangle are (x + 2) and (x + 4) . The area is x2 + 6x + 8 .

The factors of x2 + 6x + 8 are (x + 2) (x + 4) . The ax2 term (1x2 ) and the constant (8) are always diagonal to one another in a generic rectangle. In this example, the diagonal product is 8x2 and the c term is the product: 2(4) = 8 . The two x-terms are the other diagonal and can be combined into a sum when they are like terms. The b term of the quadratic is the sum of the factors: 2x + 4x = 6x . Their diagonal product is (2x)(4x) = 8x

2 , the same as the other diagonal. (See the Math Notes box on page 332 for why the products of the two diagonals are always equal.) To factor, students need to think about sums and products simultaneously. Students can Guess & Check or use a “diamond problem” from Chapter 1 to help organize their sums and products. See the Math Notes box on page 338.

4x 8

x2 2x

4

+ x

x + 2

Chapter 8: Quadratics 59

Example 1 Factor x2

+ 7x +12 . Sketch a generic rectangle with 4 sections. Write the x2 and the 12 along one diagonal.

Find two terms whose product is 12x2 and whose sum is 7x (3x and 4x). Students are familiar with this situation as a “diamond problem” from Chapter 1. Write these terms as the other diagonal.

Find the base and height of the rectangle by using the partial areas. Write the complete equation. (x + 3)(x + 4) = x

2+ 7x +12

Example 2 The same process works with negative values. Factor x2

+ 7x ! 30 . Sketch a generic rectangle with 4 sections. Write the x2 and the -30 along one diagonal. Find two terms whose product is -30x2 and whose sum is 7x. Write these terms as the other diagonal.

Find the base and height of the rectangle. Pay particular attention to the signs so the sum is 7x, not -7x.

Write the complete equation. (x ! 3)(x +10) = x

2+ 7x ! 30

3x 12

x2 4x

3 + x

x + 4

3x 12

x2 4x

12

x2

–3x !30

x2 10x

–3x !30

x2 10x

3 – x

x + 10


Example 3 Factor x2

!15x + 56 . Sketch a generic rectangle with 4 sections. Write the x2 and the 56 along one diagonal. Find two terms whose product is 56x2 and whose sum is -15x. Write these terms as the other diagonal. Find the base and height of the rectangle. Mentally multiply the factors to be sure they are correct. Pay close attention to the signs of the factors. Write the complete equation. (x ! 7)(x ! 8) = x2 !15x + 56

Example 4 Factor 2x

2+ 7x + 6 .

Sketch a generic rectangle with 4 sections. Write the 2x2 and the 6 along one diagonal. Find two terms whose product is 12x2 and whose sum is 7x. Write these terms as the other diagonal.

Find the base and height of the rectangle. Write the complete equation. (2x + 3)(x + 2) = 2x2 + 7x + 6

–8x 56

x2 –7x

–8x 56

x2 –7x

8 – x

x – 7

3x 6

2x2 4x

3x 6

2x2 4x

3 + 2x

x + 2


Example 5 Factor 12x2 !19x + 5 . Sketch a generic rectangle with 4 sections. Write the 12x2 and the 5 along one diagonal. Find two terms whose product is 60x2 and whose sum is -19x. Write these terms as the other diagonal.

Find the base and height of the rectangle. Check the signs of the factors. Write the complete equation. (3x !1)(4x ! 5) = 12x2 !19x + 5 Problems 1. x

2+ 5x + 6 2. 2x

2+ 5x + 3 3. 3x

2+ 4x +1 4. x

2+10x + 25

5. x2+15x + 44 6. x

2+ 7x + 6 7. x

2+11x + 24 8. x

2+ 4x ! 32

9. 4x2+12x + 9 10. 12x

2+11x ! 5 11. x

2+ x ! 72 12. 3x

2! 20x ! 7

13. x2!11x + 28 14 2x

2+11x ! 6 15. 2x

2+ 5x ! 3 16. x

2! 3x !10

17. 4x2!12x + 9 18. 3x

2+ 2x ! 5 19. 6x

2! x ! 2 20. 9x

2!18x + 8

Answers 1. (x + 2)(x + 3) 2. (x +1)(2x + 3) 3. (3x +1)(x +1) 4. (x + 5)(x + 5)

5. (x +11)(x + 4) 6. (x + 6)(x +1) 7. (x + 8)(x + 3) 8. (x + 8)(x ! 4)

9. (2x + 3)(2x + 3) 10. (3x !1)(4x + 5) 11. (x ! 8)(x + 9) 12. (x ! 7)(3x +1)

13. (x ! 4)(x ! 7) 14. (x + 6)(2x !1) 15. (x + 3)(2x !1) 16. (x ! 5)(x + 2)

17. (2x ! 3)(2x ! 3) 18. (3x + 5)(x !1) 19. (2x +1)(3x ! 2)

20. (3x ! 4)(3x ! 2)

–15x 5

12x2 – 4x

–15x 5

12x2 – 4x

5 – 4x

3x – 1


USING THE ZERO PRODUCT PROPERTY 8.2.3 and 8.2.4

A parabola is a symmetrical curve. Its highest or lowest point is called the vertex. It is formed using the equation

y = ax2+ bx + c . Students have been graphing parabolas by

substituting values for x and solving for y. This can be a tedious process if the range of x values is unknown. One possible shortcut is to find the x-intercepts first, then find the vertex and other convenient points. To find the x-intercepts, substitute 0 for y and solve the quadratic equation, 0 = ax

2+ bx + c . Students will learn multiple methods to solve

quadratic equations in this course. The method described below utilizes two ideas:

(1) When the product of two or more numbers is zero, then one of the numbers must be zero.

(2) Some quadratics can be factored into the product of two binomials, where coefficients and constants are integers.

This procedure is called the Zero Product Method or Solving by Factoring. See the Math Notes boxes on pages 349 and 361.

Example 1 Find the x-intercepts of y = x2 + 6x + 8 . The x-intercepts are located on the graph where y = 0 , so write the quadratic expression equal to zero, then solve for x. x2 + 6x + 8 = 0 Factor the quadratic. (x + 4)(x + 2) = 0 Set each factor equal to 0. (x + 4) = 0 or (x + 2) = 0 Solve each equation for x. x = !4 or x = !2 The x-intercepts are (-4, 0) and (-2, 0). You can check your answers by substituting them into the original equation. (!4)2 + 6(!4) + 8"16 ! 24 + 8" 0

(!2)2 + 6(!2) + 8" 4 !12 + 8" 0


Example 2 Solve 2x

2+ 7x !15 = 0.

Factor the quadratic. (2x ! 3)(x + 5) = 0 Set each factor equal to 0. (2x ! 3) = 0 or (x + 5) = 0 Solve for each x. 2x = 3

x =3

2 or x = !5

Example 3 If the quadratic does not equal 0, rewrite it algebraically so that it does, then use the zero product property. Solve 2 = 6x2 ! x . Set the equation equal to 0.

Factor the quadratic.

Solve each equation for x.

2 = 6x2 ! x

0 = 6x2 ! x ! 2

0 = (2x +1)(3x ! 2)

(2x +1) = 0!!!!!!!or !!!!!!!(3x ! 2) = 0

2x = !1!!!!!or !!!!!!!!!!!!!!!3x = 2

x = !12!!!!!!!!!!!!!!!!!!!!!!!!!x =

23

Example 4 Solve 9x

2! 6x +1= 0 .

Factor the quadratic. Solve each equation for x. Notice the factors are the same so there will be only one x value. For this parabola, the x-intercept is the vertex.

9x2 ! 6x +1 = 0

(3x !1)(3x !1) = 0

(3x !1) = 0

3x = 1

x =13


Problems Solve for x.

1. x2! x !12 = 0 2. 3x

2! 7x ! 6 = 0 3. x

2+ x ! 20 = 0

4. 3x2+11x +10 = 0 5. x

2+ 5x = !4 6. 6x ! 9 = x2

7. 6x2 + 5x ! 4 = 0 8. x2! 6x + 8 = 0 9. 6x2 ! x !15 = 0

10. 4x2 +12x + 9 = 0 11. x2!12x = 28 12. 2x

2+ 8x + 6 = 0

13. 2 + 9x = 5x2 14. 2x2 ! 5x = 3 15. x2= 45 ! 4x

Answers

1. 4 or !3 2. !2

3 or 3 3. !5 or 4

4. !5

3 or !2 5. !4 or !1 6. 3

7. !4

3or 1

2 8. 4 or 2 9. !

3

2or 5

3

10. !3

2 11. 14 or !2 12. !1 or !3

13. !1

5 or 2 14. !

1

2or 3 15. 5 or -9


USING THE QUADRATIC FORMULA 8.3.1

When a quadratic equation is not factorable, another method is needed to solve for x. The Quadratic Formula can be used to calculate the roots of the equation, that is, the x-intercepts of the quadratic. The Quadratic Formula can be used with any quadratic equation, factorable or not. There may be two, one or no solutions, depending on whether the parabola intersects the x-axis twice, once, or not at all.

For any quadratic ax2 + bx + c = 0 , x = !b ± b2 ! 4ac

2a. The ± symbol is read as “plus

or minus.” It is shorthand notation that tells you to calculate the formula twice, once with + and again with – to get both x-values. To use the formula, the quadratic equation must be written in standard form: ax2 + bx + c = 0 . This is necessary to correctly identify the values of a, b, and c. Once the equation is in standard form and equal to 0, a is the coefficient of the x2 term, b is the coefficient of the x term and c is the constant. See the Math Notes boxes on pages 358 and 361 and problem 12-52 on page 510.

Example 1 Solve 2x

2! 5x ! 3 = 0 .

Identify a, b, and c. Watch your signs carefully. a = 2,!b = !5,!c = !3

Write the quadratic formula. x =!b ± b2 ! 4ac

2a

Substitute a, b, and c in the formula and do the initial calculations. x =

!(!5) ± (!5)2 ! 4(2)(!3)

2(2)

x =5 ± 25 ! (!24)

4

Simplify the . x =5 ± 49

4

Calculate both values of x. x =5+7

4=12

4= 3 or x = 5!7

4=

!2

4= !

1

2

The solution is x = 3 or x = !

1

2. The x-intercepts are (3, 0) and (! 1

2, 0).


Example 2 Solve 3x2 + 5x +1 = 0 . Identify a, b, and c. a = 3,!b = 5,!c = 1


2a


!(5) ± (5)2 ! 4(3)(1)

2(3)

x =!5 ± 25 !12

6

Simplify the . x =!5 ± 13

6

The solution is x = !5 + 13

6or x = !5 ! 13

6 . These are the exact values of x.

The x values are the x-intercepts of the parabola. To graph these points, use a calculator to find the decimal approximation of each one. The x-intercepts are ( ! "0.23 , 0) and ( ! "1.43 , 0).

Example 3 Solve 25x2 ! 20x + 4 = 0 . Identify a, b, and c. a = 25,!b = !20,!c = 4


2a


!(!20) ± (!20)2 ! 4(25)(4)

2(25)

x =20 ± 400 ! 400

50

Simplify the . x =20 ± 0

50

This quadratic has only one solution: x = 2

5.


Example 4 Solve x2 + 4x +10 = 0 . Identify a, b, and c. a = 1,!b = 4,!c = 10


2a


!(4) ± (4)2 ! 4(1)(10)

2(1)

x =!4 ± 16 ! 40

2

Simplify the . x =!4 ± !24

2

It is impossible to take the square root of a negative number; therefore this quadratic has no real solution. The graph would still be a parabola, but there would be no x-intercepts. This parabola is above the x-axis. Example 5 Solve (3x +1)(x + 2) = 1 . Rewrite in standard form. (3x +1)(x + 2) = 1

3x2 + 7x + 2 = 1

3x2 + 7x +1 = 0

Identify a, b, and c. a = 3,!b = 7,!c = 1


2a


!(7) ± (7)2 ! 4(3)(1)

2(3)

x =!7 ± 49 !12

6

Simplify . x =!7 ± 37

6

The x-intercepts are (! "0.15,!0) and (! "2.18,!0) .


Example 6 Solve 3x

2+ 6x +1= 0 .

Identify a, b, and c. a = 3,!b = 6,!c = 1


2a


!(6) ± (6)2 ! 4(3)(1)

2(3)

x =!6 ± 36 !12

6

Simplify . x =!6 ± 24

6

The x-intercepts are ( ! "1.82 , 0) and ( ! "0.18 , 0). Math Note 8.3.3 describes another form of this expression that can be written by simplifying the square root. The result is equivalent to the exact values above. Factor the 24 , then simplify by taking the square root of 4. 24 = 4 6 = 2 6

Simplify the fraction by dividing every term by 2.

x =!6 ± 2 6

6

x =!3± 6

3


Problems Solve each of these using the Quadratic Formula. 1. x

2! x ! 2 = 0 2. x

2! x ! 3 = 0 3. !3x

2+ 2x +1 = 0

4. !2 ! 2x2= 4x 5. 7x = 10 ! 2x

2 6. !6x2! x + 6 = 0

7. 6 ! 4x + 3x2= 8 8. 4x

2+ x !1 = 0 9. x

2! 5x + 3 = 0

10. 0 = 10x2! 2x + 3 11. x(!3x + 5) = 7x !10 12. (5x + 5)(x ! 5) = 7x

Answers 1. x = 2 or x = !1 2.

x =1± 13

2

! 2.30!or !"1.30

3. x = !1

3 or x = 1

4. x = !1 5. x =

!7± 129

4

! 1.09!or !"4.59

6. x =

1± 145

!12

! "1.09!or !0.92

7. x =

4± 40

6=2± 10

3

! 1.72!or !"0.39

8. x =

!1± 17

8

! 0.39!or !"0.64

9. x =

5± 13

2

! 4.30!or !0.70

10. No solution

11. x =

2± 124

!6=1± 31

!3

! "2.19!or !1.52

12. x =

27± 1229

10

! 6.21!or !"0.81


SOLVING INEQUALITIES 9.1.1 and 9.1.2

To solve an inequality in one variable, first change it to an equation and solve. Place the solution, called a “boundary point,” on a number line. This point separates the number line into two regions. The boundary point is included in (! or ! shown by a solid dot) or excluded from (> or < shown by an open dot) the solution depending on the inequality sign. Next, choose a number from each region and check if it is true or false in the original inequality. If it is true, then every number in that region is a solution to the inequality. If it is false, then no number in that region is a solution to the inequality. See the Math Notes box on page 386.

Example 1 Solve 3x ! x + 2( ) " 0 Change to an equation and solve. Place the solution (boundary point) on the number line. Because x = 1 is also a solution to the inequality (! ), we use a solid dot. Test a number from each side of the boundary point in the original inequality. The solution is: x ! 1.

3x ! x + 2( ) = 0

3x ! x ! 2 = 0

2x = 2

x = 1

Test x = 0 Test x = 3 3 !0 " 0 + 2( ) # 0

"2 # 0

3 ! 3" 3+ 2( ) # 0

4 # 0

False True

Example 2

Solve: !x + 6 > x + 2 Change to an equation and solve. Place the solution (boundary point) on the number line. Because the original problem is a strict inequality (>), x = 2 is not a solution, so we use an open dot. Test a number from each side of the boundary point in the original inequality. The solution is: x < 2 .

!x + 6 = x + 2

!2x = !4

x = 2

Test x = 0 Test x = 4 !0 + 6 > 0 + 2

6 > 2 !4 + 6 > 4 + 2

2 > 6

True False

1 2 3 4 0 -1 -2 -3 -4 x

1 2 3 40-1-2-3-4 x

1 2 3 4 0 -1 -2 -3 -4 k

1 2 3 4 0 -1 -2 -3 -4 k

Chapter 9: Inequalities 71

Problems Solve each inequality.

1. 4x !1 " 7

2. 2 x ! 5( ) " 8 3. 3! 2x < x + 6

4. 1

2x > 5 5. 3 x + 4( ) > 12

6. 2x ! 7 " 5 ! 4x

7. 3x + 2 < 11

8. 4 x ! 6( ) " 20 9. 1

4x < 2

10. 12 ! 3x > 2x +1

11. x!5

7" !3 12. 3(5 ! x) " 7x !1

13. 3y ! 2y + 2( ) " 7 14. m+2

5<2m

3 15. m!2

3"2m+1

7

Answers

1. x ! 2

2. x ! 9 3. x > !1

4. x > 10

5. x > 0 6. x ! 2

7. x < 3

8. x ! 11 9. x < 8

10. x <11

5 11. x ! "16

12. x ! 1.6

13. y ! 9 14. m >6

7 15. m ! 17


GRAPHING INEQUALITIES 9.2.1 and 9.2.2

To graph an inequality in two variables, first graph the corresponding equation. This graph is know as the boundary line (or curve), since all points that make the inequality true lie on one side or the other of the line. Before you graph the equation, decide whether the line or curve is part of the solution or not, that is, whether it is solid or dashed. If the inequality symbol is either ≤ or ≥, then the boundary line is part of the inequality and it must be solid. If the symbol is either < or >, then the boundary line is dashed. Next, decide which side of the boundary line must be shaded to show the part of the graph that represent all values that make the inequality true. Choose a point not on the boundary line. Substitute this point into the original inequality. If the inequality is true for the test point, then shade the graph on this side of the boundary line. If the inequality is false for the test point, then shade the opposite side of the line. See the Math Notes box on page 393. CAUTION: If you need to rearrange the inequality in order to graph it, such as putting it in slope-intercept form, always use the original inequality to test a point, NOT the rearranged form.

Example 1 Graph the inequality y > 3x ! 2. First, graph the line y = 3x ! 2, but draw it dashed since > means the boundary line is not part of the solution. Next, test the point (-2, 4) that lies to the left of the boundary line. 4 > 3(!2) ! 2 , so 4 > !8 . Since the inequality is true for this test point, shade the region left of the boundary line.


Example 2 Graph the inequality y ! x2 " 2x " 8. First, graph the parabola y = x

2! 2x ! 8 and draw it solid,

since ≤ means the boundary curve is part of the solution. Next, test the point (2, 2) above the boundary curve.

2 ! 22" 2 #2 " 8 , so 2 ! "8

Since the inequality is false for this test point above the curve, shade below the boundary curve. Problems Graph each of the following inequalities on a separate set of axes.

1. y ! 3x +1

2. y ! "2x + 3 3. y > 4x ! 2

4. y < !3x ! 5

5. y ! 3 6. x > 1

7. y > 2

3x + 8 8. y < !

3

5x ! 7 9. 3x + 2y ! 7

10. !4x + 2y < 3

11. y ! x2" 3 12. y ! x

2+ 2x

13. y < 4 ! x2 14. y ! x + 2 15. y ! " x + 3

Answers

1.

2.

3.


4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.


SYSTEMS OF INEQUALITIES 9.3.1 and 9.3.2

To graph systems of inequalities, follow the same procedure outlined in the previous section but do it twice—once for each inequality. The solution to the system of inequalities is the overlap of the shading from the individual inequalities.

Example 1 Graph the system y ! 1

2x + 2 and y > !

2

3x +1.

Graph the lines y = 1

2x + 2 and

y = !2

3x +1 . The first is solid

and the second is dashed. Test the point (-4, 5) in the first inequality.

5 !

1

2"4( ) + 2 , so 5 ! 0

This inequality is false, so shade on the opposite side of the boundary line from (-4, 5), that is, below the line.

5 > !2

3!4( ) +1 , so 5 > 11

3

Test the same point in the second inequality. This inequality is true, so shade on the same side of the boundary line as (-4, 5), that is, above the line.

The solution is the overlap of the two shaded regions shown by the darkest shading in the second graph above right.


Example 2 Graph the system y ! "x + 5 and y ! x2 "1 . Graph the line y = !x + 5 and the parabola y = x2 !1 with a solid line and curve.

2 ! " "1( ) + 5 , so 2 ! 6

Test the point (-1, 2) in the first inequality. This inequality is true, so shade on the same side of the boundary line as (-1, 2), that is, below the line.

2 ! "1( )2"1 , so 2 ! 0

Test the same point in the second inequality. This inequality is also true, so shade on the same side of the boundary curve as (-1, 2), that is, inside the curve.

The solution is the overlap of the two shaded regions shown by the darkest shading in the second graph above right. Problems

Graph each of the following pairs of inequalities on the same set of axes.

1. y > 3x ! 4 !and !y " !2x + 5 2. y ! "3x " 6!and !y > 4x " 4

3. y < !3

5x + 4 !and !y < 1

3x + 3 4. y < !

3

7x !1!and !y > 4

5x +1

5. y < 3!and !y > 1

2x + 2 6. x ! 3!and !y < 3

4x " 4

7. y ! 2x +1!and !y " x2# 4 8. y < !x + 5!and !y " x

2+1

9. y < !x + 6!and !y " x ! 2 10. y < !x2+ 5!and !y " x !1

Answers

1.

2.


3.

4.

5.

6.

7.

8.

9.

10.


SIMPLIFYING EXPRESSIONS 10.1.1

To simplify rational expressions, find factors in the numerator and denominator that are the same and then write them as fractions equal to 1. For example,

6

6= 1 x

2

x2= 1 (x + 2)

(x + 2)= 1 (3x ! 2)

(3x ! 2)= 1

Notice that the last two examples involved binomial sums and differences. Only when sums or differences are exactly the same does the fraction equal 1. The rational expressions below cannot be simplified:

6 + 5( )

6 x

3+ y

x3

x

x + 2 3x ! 2

2

As shown in the examples below, most problems that involve simplifying rational expressions will require that you factor the numerator and denominator. Note that in all cases we assume the denominator does not equal zero, so in example 4 below the simplification is only valid provided x ! "6 or 2. See the Math Notes boxes on pages 410 and 413. One other special situation is shown in the following examples:

!2

2= !1 !x

x= !1 !x ! 2

x + 2"

! x + 2( )

x + 2"!1 5 ! x

x ! 5"

! x ! 5( )

x ! 5"!1

Again assume the denominator does not equal zero.

Example 1

12

54=2 !2 ! 3

2 ! 3 ! 3 ! 3=2

9 since 2

2 = 33

= 1

Example 2

6x3y2

15x2y4=2 ! 3 ! x

2! x ! y

2

5 ! 3 ! x2! y

2! y

2=2x

5y2

Example 3 12 x !1( )

3

x + 2( )

3 x !1( )2

x + 2( )2=4 " 3 x !1( )

2

x !1( ) x + 2( )

3 x !1( )2

x + 2( ) x + 2( )

=4(x !1)

(x + 2) since 3

3, x !1( )

2

x !1( )2

, and x + 2x + 2

= 1

Example 4

x2! 6x + 8

x2+ 4x !12

=x ! 4( ) x ! 2( )

x + 6( ) x ! 2( )

=x ! 4( )

x + 6( ) since x ! 2( )

x ! 2( )= 1

Chapter 10: Simplifying and Solving 79

Problems Simplify each of the following expressions completely. Assume the denominator does not equal zero.

1. 2 x + 3( )

4 x ! 2( ) 2. 2 x ! 3( )

6 x + 2( ) 3. 2 x + 3( ) x ! 2( )

6 x ! 2( ) x + 2( )

4. 4 x ! 3( ) x ! 5( )

6 x ! 3( ) x + 2( ) 5. 3 x ! 3( ) 4 ! x( )

15 x + 3( ) x ! 4( ) 6. 15 x !1( ) 7 ! x( )

25 x +1( ) x ! 7( )

7. 24 y ! 4( ) y ! 6( )

16 y + 6( ) 6 ! y( ) 8. 36 y + 4( ) y !16( )

32 y +16( ) 16 ! y( ) 9. x + 3( )

2

x ! 2( )4

x + 3( )4

x ! 2( )3

10. 5 ! x( )2

x ! 2( )2

x + 5( )4

x ! 2( )3

11. 5 ! x( )4

3x !1( )2

x ! 5( )4

3x ! 2( )3

12. 12 x ! 7( ) x + 2( )4

20 x ! 7( )2

x + 2( )5

13. x2+ 5x + 6

x2+ x ! 6

14. 2x2+ x ! 3

x2+ 4x ! 5

15. x2+ 4x

2x + 8

16. 24 3x ! 7( ) x +1( )6

20 3x ! 7( )3

x +1( )5

17. x2!1

x +1( ) x ! 2( ) 18. x

2! 4

x +1( )2

x ! 2( )

19. x2! 4

x2+ x ! 6

20. x2!16

x3+ 9x

2+ 20x

21. 2x2! x !10

3x2+ 7x + 2

Answers

1. x + 3( )

2 x ! 2( ) 2. x ! 3( )

3 x + 2( ) 3. x + 3( )

3 x + 2( )

4. 2 x ! 5( )

3 x + 2( ) 5. !

x ! 3( )

5 x + 3( ) 6. !

3 x !1( )

5 x +1( )

7. !3 y ! 4( )

2 y + 6( ) 8. !

9 y + 4( )

8 y +16( ) 9. x ! 2( )

x + 3( )2

10. 5 ! x( )2

x + 5( )4

x ! 2( ) 11. 3x !1( )

2

3x ! 2( )3

12. 3

5 x ! 7( ) x + 2( )

13. x + 2

x ! 2 14. 2x + 3

x + 5 15. x

2

16. 6 x +1( )

5 3x ! 7( )2

17. x !1

x ! 2 18. x + 2

x +1( )2

19. x + 2

x + 3 20. x ! 4

x(x + 5) 21. 2x ! 5

3x +1


MULTIPLICATION AND DIVISION 10.1.2 OF RATIONAL EXPRESSIONS

Multiplication or division of rational expressions follows the same procedure used with numerical fractions. However, it is often necessary to factor the polynomials in order to simplify them. When dividing rational expressions, change the problem to multiplication by inverting (flipping) the second expression (or any expression that follows a division sign) and completing the process as you do for multiplication. As in the previous section, remember that simplification assumes that the denominator is not equal to zero. See the Math Notes box on page 413.

Example 1 Multiply x

2+ 6x

x + 6( )2!x2+ 7x + 6

x2"1

and simplify the result.

After factoring, the expression becomes: After multiplying, reorder the factors:

Since (x + 6)x + 6( )

= 1 and (x +1)x +1( )

= 1 , simplify:

x(x + 6)

x + 6( ) x + 6( )!(x + 6)(x +1)

(x +1)(x "1)

(x + 6)

x + 6( )!(x + 6)

(x + 6)!

x

(x "1)!(x +1)

x +1( )

1 !1 !x

x "1!1!!=>!!

x

x "1 for x ! 6,!"1,!or !1 .

Example 2 Divide x

2! 4x ! 5

x2! 4x + 4

÷x2! 2x !15

x2+ 4x !12

and simplify the result.

First change to a multiplication expression by inverting (flipping) the second fraction: After factoring, the expression is: Reorder the factors (if you need to):

Since x ! 5( )

x ! 5( )= 1 and x ! 2( )

x ! 2( )= 1 , simplify:

x2! 4x ! 5

x2! 4x + 4

"x2+ 4x !12

x2! 2x !15

x ! 5( ) x +1( )

x ! 2( ) x ! 2( )"x + 6( ) x ! 2( )

x ! 5( ) x + 3( )

x ! 5( )

x ! 5( )"x ! 2( )

x ! 2( )"x +1( )

x ! 2( )"x + 6( )

x + 3( )

x +1( )

x ! 2( )"x + 6( )

x + 3( )

Thus, x2! 4x ! 5

x2! 4x + 4

÷x2! 2x !15

x2+ 4x !12

= x +1( )

x ! 2( )"x + 6( )

x + 3( ) or x

2+ 7x + 6

x2+ x ! 6

for x ! "3,!2,!or !5 .


Problems Multiply or divide each pair of rational expressions. Simplify the result. Assume the denominator is not equal to zero.

1. x2+ 5x + 6

x2! 4x

"4x

x + 2 2. x

2! 2x

x2! 4x + 4

÷4x

2

x ! 2

3. x2!16

(x ! 4)2"x2! 3x !18

x2! 2x ! 24

4. x2! x ! 6

x2+ 3x !10

"x2+ 2x !15

x2! 6x + 9

5. x2! x ! 6

x2! x ! 20

"x2+ 6x + 8

x2! x ! 6

6. x2! x ! 30

x2+13x + 40

"x2+11x + 24

x2! 9x +18

7. 15 ! 5xx2! x ! 6

÷5x

x2+ 6x + 8

8. 17x +119x2+ 5x !14

÷9x !1

x2! 3x + 2

9. 2x2! 5x ! 3

3x2!10x + 3

"9x

2!1

4x2+ 4x +1

10. x2!1

x2! 6x ! 7

÷x3+ x

2! 2x

x ! 7

11. 3x ! 21x2! 49

÷3x

x2+ 7x

12. x2! y

2

x + y"1

x ! y

13. y2! y

w2! y

2÷y2! 2y +1

1! y 14. y

2! y !12

y + 2÷

y ! 4

y2! 4y !12

15. x2+ 7x +10

x + 2÷x2+ 2x !15

x + 2

Answers

1. 4 x + 3( )

x ! 4 2. 1

4x 3. x + 3

x ! 4

4. x + 2x ! 2

5. x + 2x ! 5

6. x + 3x ! 3

7. !x ! 4x

8. 17 x !1( )

9x !1 9. 3x +1

2x +1

10. 1

x x + 2( ) 11. 1 12. 1

13. !y

w2! y

2 14. y + 3( ) y ! 6( ) 15. x + 2

x ! 3


SOLVING BY REWRITING: FRACTION BUSTERS 10.1.3 and 10.1.4

Equations with fractions and/or decimals can be converted into equivalent equations without fractions and/or decimals and then solved in the usual manner. Equations can also be made simpler by dividing a common numerical factor out of each term. Fractions can be eliminated from an equation by multiplying BOTH sides (and all terms) of an equation by the common denominator. If you cannot determine a common denominator, then multiply the entire equation by the product of all of the denominators. We call the term used to eliminate the denominators a fraction buster. Also remember to check your answers. See the Math Notes boxes on pages 416 and 418.

Example 1

Solve: 0.12x + 7.5 = 0.2x + 3

Multiply to remove the decimals.

100 ! (0.12x + 7.5 = 0.2x + 3)

12x + 750 = 20x + 300

Solve in the usual manner.

!8x = !450

x = 56.25

Example 2

Solve: 25x2 +125x +150 = 0

Divide each term by 25 (a common factor).

x2+ 5x + 6 = 0

Solve in the usual manner.

x + 2( ) x + 3( ) = 0 x = !2 or x = !3

Example 3

Solve: x2+x

6= 7

Multiply both sides of the equation by 6, the common denominator, to remove the fractions.

6 ! x

2+ x

6( ) = 6(7) Distribute and solve as usual.

6 !x

2+ 6 !

x

6= 6 ! 7

3x + x = 42

4x = 42

x =42

4=21

2= 10.5

Example 4

Solve: 52x

+1

6= 8

Multiply both sides of the equation by 6x , the common denominator, to remove the fractions.

6x ! 52x

+ 16( ) = 6x(8)

Distribute and solve as usual.

6x !5

2x+ 6x !

1

6= 6x !8

15 + x = 48x

15 = 47x

x =15

47" 0.40


Problems Rewrite each equation in a simpler form and then solve the new equation.

1. x3+x

2= 5 2. 3000x + 2000 = !1000 3. 0.02y !1.5 = 17

4. x2+x

3!x

4= 12 5. 50x2 ! 200 = 0 6. x

9+2x

5= 3

7. 3x10

+x

10=15

10 8. 3

2x+5

x=13

6 9. x2 ! 2.5x +1 = 0

10. 23x

!1

x=1

36 11. 0.002x = 5 12. 10 + 5

x+3

3x= 11

13. 0.3(x + 7) = 0.2(x ! 2) 14. x + x

2+3x

5= 21 15. 32 ! 3x " 32 !1 = 32 !8

16. 5 + 2x+5

4x=73

12 17. 17

2x +1=17

5 18. 2 + 6

x+6

3x= 3

19. 2.5x2 + 3x + 0.5 = 0 20. x

x ! 2=

7

x ! 2

Answers

1. 6 2. !1 3. 925

4. 1447

! 20.57 5. ±2 6. 13523

! 5.87

7. 154= 3.75 8. 3 9. 1

2 or 2

10. -12 11. 2500 12. 6

13. -25 14. 10 15. 3

16. 3 17. 2 18. 8

19. ! 15

or !1 20. 7 Note: x cannot be 2.


MULTIPLE METHODS FOR 10.2.1 through 10.2.3 SOLVING EQUATIONS

Equations may be solved in a variety of ways. These sections in the textbook use three methods that allow students to find the solutions to sometimes complex equations in more efficient ways. The methods are called rewriting, looking inside, and undoing. Example 4 is a reminder that when solving an equation it is necessary to find all of the solutions. Problems with squares may have two solutions, one solution, or no solution.

x2= 144

x = ±12

x ! 3( )2= 0

x = 3 2x +1( )

2= !7

See the Math Notes boxes on pages 424 and 430.

Example 1 Rewriting

Solve: x

2+x

5= 3

This problem can be rewritten without fractions using fraction busters.

10 ! x

2+ x

5( ) = 10(3)5x + 2x = 30

7x = 30

x = 307" 4.29

Example 2 Undoing

Solve: x +1 = 6

This problem has a square root and a sum, so subtracting and then squaring both sides of the equation can undo it.

x +1!1 = 6 !1

x = 5

x( )2

= 52

x = 25

Example 3 Looking Inside

Solve: x +12

= !6

Looking inside the fraction can solve this problem. Since something

2= !6 , the numerator

must be !12 . x +1 = !12

x = !13

Example 4 Looking Inside

Solve 2x + 3 = 11

Looking inside the absolute value can solve this problem. Since 11 = !11 = 11 ,

2x + 3 = 11 or 2x + 3 = !11 . Solving both equations yields two answers:

2x = 8 or 2x = !14 x = 4 or x = !7

no real number squared is negative


Problems Solve each equation. Find all.

1. 3x +1

2= 5 2. 4(x !1) + 3 = 15 3. 2x + 5 = 10

4. 10 ! (x + 7) = 5 5. 3(2x ! 7) = !21 6. 8 +y

2

!"#

$%&= 10

7. x ! 3 = 7 8. x +1( )2

= 81 9. x

2!x

5= 3

10. x ! 2 = 5 11. 2 x ! 3 = 8 12. 4(x !1) = 16

13. 2x +1 = 9 14. y + 7

3= 10 15. m

3!2m

5=1

5

16. x2+ 5 = 4 17. 3y !1

3= 10 18. 2 3x !1( ) + 7 = !13

19. y !1( )2

= 9 20. 20 ! 3x( ) = 10

Answers

1. 3 2. 4 3. 95

2= 47.5

4. !2 5. 0 6. 4

7. 100 8. 8 or !10 9. 10

10. 7 or !3 11. 19 12. 5

13. 4 or !5 14. 23 15. !3

16. no solution 17. 31

3= 10.3 18. !3

19. 4 or !2 20. 10

3


SOLVING ABSOLUTE VALUE 10.2.4 and 10.2.5 AND QUADRATIC INEQUALITIES

There are several methods for solving absolute value and quadratic inequalities, but one method that works for all kinds of inequalities is to change the inequality to an equation, solve it, then put the solution(s) on a number line. The solution(s), called “boundary point(s),” divide the number line into regions. Check any point in each region in the original inequality. If that point is true, then all the points in that region are solutions. If that point is false, then none of the points in that region are solutions. The boundary points are included in (! or !" ) or excluded from (>!or !< ) the solution depending on the inequality sign.

Example 1 Solve: x ! 3 " 5 Change to an equation and solve.

x ! 3 = 5

x ! 3 = 5 or x ! 3 = !5

x = 8 or x = !2 (the boundary points)

1 2 3 4 5 6 7 8–1–2–3 0x

Choosing x = !3,!0,!9 to test in the original inequality, x = !3 is false, x = 0 is true, and x = 9 is false. false true false

1 2 3 4 5 6 7 8–1–2–3 0x

The solution is all numbers greater than or equal to !2 and less than or equal to 8, written as !2 " x " 8 .

Example 2 Solve: 2y +1 > 9 Change to an equation and solve.

2y +1 = 9

2y +1 = 9 or 2y +1 = !9

y = 4 or y = !5 (the boundary points)

1 2 3 4 5 6–1–2–3–4–5–6 0

Choosing y = !6,!0,!5 to test in the original equation, y = !6 is true, y = 0 is false, and y = 5 is true.

true false true

1 2 3 4 5 6–1–2–3–4–5–6 0

The solution is all numbers less than !5 or numbers greater than 4, written as y < !5 or y > 4 .


Example 3 Solve: x2 ! 3x !18 < 0

Change to an equation and solve. x2! 3x !18 = 0

x ! 6( ) x + 3( ) = 0 x = 6 or x = !3 (the boundary points)

1 2 3 4 5 6 7–1–2–3–4 0x

Choosing x = !4,!0,!7 to test in the original equation, x = !4 is false, x = 0 is true, and x = 7 is false.

false true false

1 2 3 4 5 6 7–1–2–3–4 0x

The solution is all numbers greater than 3 and less than 6, written as !3 < x < 6 .

Example 4 Solve: m2! 3 " 1

Change to an equation and solve. m2! 3 = 1

m2= 4

m = ±2 (the boundary points)

1 2 3–1–2–3 0x

Choosing m = !3,!0,!3 to test in the original equation, m = !3 is true, m = 0 is false, and m = 3 is true.

true false true

1 2 3–1–2–3 0x

The solution is all numbers less than or equal to !2 or all numbers greater than or equal to 2, written as m ! "2 or m ! 2 .

Problems

Solve each inequality.

1. x + 4 ! 7 2. x2+ 6x + 8 < 0 3. y

2! 5y > 0

4. x ! 5 " 8 5. x ! 5 " 8 6. y2! 5y < 0

7. 4r ! 2 > 8 8. x2! 3x ! 4 < 0 9. 3x ! 12

10. !x2! 9x !14 < 0 11. 1! 3x " 13 12. y

2! 16

13. 2x ! 3 > 15 14. 3x2+ 7x ! 6 " 0 15. 5x > !15

16. !2 x ! 3 + 6 < !4 17. x2+ 4x ! 8 < 4 18. y

2+ 6y + 9 > 0

19. 4 ! d " 7 20. x 7x ! 26( ) " 8

Answers

1. x ! 3 or x ! "11 2. !4 < x < !2 3. y < 0 or y > 5

4. !13 " x " 13 5. !3 " x " 13 6. 0 < y < 5

7. r < !3

2 or r > 5

2 8. !1 < x < 4 9. !4 " x " 4

10. x < !7 or x > !2 11. !4 < x <14

3 12. !4 " y " 4

13. x < !6 or x > 9 14. x ! "3 or x ! 2

3 15. all numbers

16. x > 8 or x < !2 17. !6 < x < 2 18. all numbers except!3

19. !3 " d " 11 20. !2

7" x " 4


SOLVING BY COMPLETING THE SQUARE 10.3.1 and 10.3.2

As shown below, taking the square root of both sides of an equation, keeping in mind that there are usually two roots, can solve equations of the form x ± a( )

2

= positive!number .

x2= 25

x = ±5

x ! 3( )2= 25

x ! 3 = ±5 x ! 3( )

2= 5

x ! 3 = ± 5

x = 3 ± 5,!x = 8,!!2 x = 3 ± 5

The process of converting an equation of the form x2 + bx + c = 0 into x ± a( )

2

= positive!number is called completing the square. This process is illustrated in the following examples. See the Math Notes boxes on pages 438 and 444.

Example 1 Solve x2 + 4x + 4 = 25 The left side already is a perfect square. Factor and take the square root of each side. Simplify.

(x + 2)2 = 25 x + 2 = ±5 x = !2 ± 5,!x = 3,!!7

Example 2 Solve x2 !10x + 22 = 0 Isolate the constant. We need to make x2 !10x into a perfect square. Taking half the x coefficient and squaring it will accomplish this. The 25 that is put into the parentheses to complete the square on the left side of the equation must be balanced by adding 25 to the other side of the equation. Factor and simplify.

x2!10x = !22

(x

2!10x + ?) = !22 ? = !10

2

"

#

$

%

2

= 25

(x

2!10x + 25) = !22 + 25

x ! 5( )

2

= 3

x ! 5( ) = ± 3

x = 5 ± 3


Example 3 Solve x2 + 5x + 2 = 0 Isolate the 2. We need to make x2 + 5x into a perfect square. Again, taking half of the x coefficient and squaring the result will accomplish the task. The 25

4that was added to the parenthesis must

be balanced by adding 254

to the right side of the equation (and simplifying). Factor and simplify.

x2+ 5x = !2

(x2 + 5x + ?) = !2 ? = 5

2( )2

= 25

4

(x2+ 5x +

25

4) = !2 +

25

4

(x +5

2)2= !

8

4+25

4=17

4

(x +5

2) = ±

17

4= ±

17

2

x = !5

2±17

2or !5 ± 17

2! "0.44,!"4.56

Problems

Solve each quadratic equation by completing the square.

1. x + 2( )2

= 3

2. y2! 6y + 9 = 25 3. x

2+ 2x ! 3 = 0

4. x2+ 8x = 5

5. x

2+ 6x + 2 = 0 6. x

2+10x ! 75 = 0

7. x2! 6x + 2 = 0

8. y

2+ 5y = 14 9. x

2+ 6x +1 = !10

10. x2! 2x ! 3 = 0

11. x

2+ 4x = 3 12. x

2+ 3x = 3

13. x2! 3x !13.75 = 0 14. x

2+ x ! 3 = 0 15. x

2!10x +12 = 6

Answers

1. !2 ± 3

2. 8 or !2 3. 1 or !3

4. !4 ± 21

5. !3 ± 7 6. 5 or !15

7. 3 ± 7

8. -7 or 2 9. no real solution

10. !1 or 3 11. !2 ± 7 12. !3± 21

2

13. !2.5 or 5.5 14. 1± 13

2 15. 5 ± 19


LAWS OF EXPONENTS 10.4.1 through 10.4.3

The basic laws of exponents are listed here. In all expressions with fractions we assume the denominator does not equal zero. See the Math Notes box on page 452. (1) x

a! x

b= x

a+b examples: x3 ! x4 = x7 ;27 !24 = 211

(2) xa

xb= x

a!b examples: x10

x4= x

6 ; 24

27= 2

!3

(3) x

a( )b

= xab examples: x4( )

3

= x12 ; 2x3( )

5

= 25! x

15= 32x

15

(4) x0= 1 examples: 20 = 1 ; !3( )

0

= 1 ; 1

4

!"#

$%&0

= 1

(5) x!n

=1

xn

examples: x!3 = 1

x3

; y!4 = 1

y4

; 4!2=1

42=1

16

(6) 1

x!n

= xn examples: 1

x!5

= x5 ; 1

x!2

= x2 ; 1

3!2

= 32= 9

(7) x

m

n = xmn examples: x

2

3 = x23 ; y

1

2 = y

Example 1 Simplify: 2xy3( ) 5x2y4( ) Reorder: 2 !5 ! x ! x

2! y

3! y

4 Using (1): 10x

3y7

Example 2 Simplify; 14x

2y12

7x5y7

Separate: 14

7

!"#

$%&'x2

x5

!"#

$%&'y12

y7

!"#

$%&

Using (2) and (5): 2x!3y5=2y

5

x3

Example 3 Simplify: 3x

2y4( )3

Cube each factor 33! x

2( )3

! y4( )3

Using (3): 27x

6y12


3( )!2

Using (5): 1

2x3( )2

Using (3): 1

22! x

3( )2=1

4x6



7y3

15x!2y3

Separate: 10

15

!"#

$%&'x7

x(2

!"#

$%&'y3

y3

!"#

$%&

Using (2): 2

3x9y0

Using (4): 2

3x9!1 =

2

3x9=2x

9

3

Example 6

Simplify: x

1

6!"#

$%&

2

Using (3): x

2

6 = x

1

3 Using (7): x

13= x

3

Problems Simplify each expression. Final answers should contain no parenthesis, negative or fractional exponents.

1. y5! y

7 2. b4!b

3!b

2 3. 86!8

"2

4. y5( )2

5. 3a( )4 6. m

8

m3

7. 12m8

6m!3

8. x3y2( )3

9.

y4( )2

y3( )2

10. 15x2y5

3x4y5 11. 4c4( ) ac3( ) 3a5c( ) 12. 7x3y5( )

2

13. 4xy2( ) 2y( )

3 14. 4

x2

!"#

$%&3

15. 2a

7( ) 3a2( )6a

3

16. 5m3n

m5

!"#

$%&

3

17. 3a2x3( )2

2ax4( )3

18. x3y

y4

!"#

$%&

4

19. 6x8y2

12x3y7

!"#

$%&

2

20. 2x

5y3( )3

4xy4( )2

8x7y12

21. x

!3

22. 2x!3 23. 2x( )

!3 24. 2x3( )0

25. 5

1

2 26. 2x

3

!"#

$%&'2

27. x

3

4

28. 8

1

3 29. 25

3

2 30. x

3

2

x


Answers

1. y12 2. b

9 3. 84

4. y10 5. 81a

4 6. m5

7. 2m11 8. x

9y6 9. y

2

10. 5

x2

11. 12a6c8 12. 49x

6y10

13. 32xy5 14. 64

x6

15. a6

16. 125n3

m6

17. 72a7x18 18. x

12

y12

19. x10

4y10

20. 16x10y5 21. 1

x3

22. 2

x3

23. 1

8x3

24. 1

25. 5 26. 9

4x2

27. x34

28. 83

= 2 29. 253 = 125 30. x

3

2!1

= x

1

2 = x

Chapter 11: Functions and Relations 93

RELATIONS AND FUNCTIONS 11.1.2 through 11.1.4

Relations and Functions establish a correspondence between the input values (usually x) and the output values (usually y) according to the particular relation or function. Often this correspondence can be described by an equation and a graph. Relations are functions if, for each input value, there is only one output value. Relations and functions work like a machine, as shown in the diagram in Example 1. A relation or function is given a name that is usually a letter, such as f or g. The notation f (x) represents the output when x is processed by the machine. (Note: f (x) is read, “f of x.”) When x is put into the machine, f (x) , the value of a relation or function for a specific x-value, comes out. See the Math Notes box on page 473.

Example 1 Numbers are put into the relation or function machine (in this case, the function f (x) = 2x +1) one at a time, and then the relation or function performs the operation(s) on each input to determine each output. For example, when x = 3 is put into the function f (x) = 2x +1, the machine multiplies 3 by 2 and adds 1 to get the output, which is 7. The notation f (3) = 7 shows that the function named f connects the input (3) with the output (7). This also means the point (3, 7) lies on the graph of the function. Example 2 a. If f (x) = x ! 2 then f (11) = ? f (11) = 11! 2 = 9 = 3 b. If g(x) = 3! x2 then g(5) = ? g(5) = 3! (5)

2= 3! 25 = !22

c. If f (x) = x + 3

2x ! 5 then f (2) = ? f (2) =

2 + 3

2 !2 " 5=5

"1= "5

x = 3

f (x) = 2x +1

inputs

outputs f (3) = 7


g(x)

Example 3 A relation in which each input has only one output is called a function. Examining the graphs below, g(x) is a function while f (x) is not.

f (x)

Each input (x) has only one output (y). g(!2) = 1,!g(0) = 3,!g(4) = !1 and so on.

Each input greater than !3has two y-values associated with it. f (1) = 2 and f (1) = !2

Example 4 The set of all possible inputs of a relation is called the domain, while the set of all possible outputs of a relation is called the range. For g(x) in Example 3 the domain is “all numbers from !2 through 4” and the range is “all numbers from !1 through 3.” For f (x) in Example 3, since the graph starts at !3 and continues forever to the right, the domain is “all numbers greater than or equal to !3 .” The graph of f (x) extends in both the positive and negative y directions forever, so the range is “all numbers.” Example 5 For the graph at right, since the x-values extend forever in both directions the domain is “all numbers.” The y-values start at 1 and go higher so the range is “all numbers greater or equal to 1.”


Problems Determine the outputs for the following relations and the given inputs. 1.

2. 3.

4. f (x) = (5 ! x)2

f (8) = ? 5. g(x) = x

2! 5

g(!3) = ? 6. f (x) =

2x + 7

x2! 9

f (3) = ?

7. h(x) = 5 ! x h(9) = ?

8. h(x) = 5 ! x h(9) = ?

9. f (x) = !x2

f (4) = ? Determine if each relation is a function. Then state its domain and range.

10.

4–4

4

–4

x

y

x

11.

4–4

4

–4

x

y

x

12.

4–4

4

–4

x

y

x

13.

x

y

4

4

-4

-4

14.

4–4

4

–4

x

y

x

15.

x

y

4

4

-4

-4

x = 2

f (x) = ?

f (x) = !2x + 4

x = !6

f (x) = x ! 2

f (x) = ?

f (x) = x +1

f (x) = ?

x = 9


Answers

1. 0

2. 8 3. 4

4. 9

5. 4 6. not possible

7. 2

8. not possible 9. -16

10. yes, each input has one output; domain is all numbers, range is!1 " y " 4

11. no, for example x=0 has two outputs; domain is x ! "3 , range is all numbers

12. yes; domain all numbers, range is !2 " y " 2

13. no; -1 has two outputs; domain is -4,-3, -1, 0, 1, 2, 3, 4, range is -4, -3, -2, -1, 0, 1, 2

14. yes; domain is all numbers, range is y ! "2

15. no, many inputs have two outputs; domain is!2 " x " 4 range is !2 " y " 4


TRANSFORMATIONS OF A FUNCTION 11.1.6

The simplest equation of one shape (e.g., line, parabola, absolute value) is called a parent equation. Changing the parent equation by addition or multiplication moves and changes the size and orientation of the parent graph but does not change the basic shape. These changes are called transformations. The first set of examples shows how the parent graph of a parabola can be moved on an xy-coordinate system. In later courses, you will learn how to make it wider or narrower. Transformations of other functions are done in a similar manner.

Examples y = x

2 the parent graph y = (x ! 3)2 right 3 units y = (x + 2)

2 left 2 units

y = x2! 2 down 2 units y = (x + 3)

2! 2

left 3 and down 2 y = !(x ! 2)

2+ 5

right 2, up 5, and flipped

Problems Predict how each parabola is different from the parent graph.

1. y = (x + 5)2 2. y = x

2+ 5 3. y = !x

2 4. y = x

2! 4 5. y = x

2+ 5 6. y = !(x + 2)

2 7. y = (x ! 5)

2! 3 8. y = !(x + 2)

2+1 9. y = (x ! 3)

2! 5


The parent graphs for absolute value and square root are shown at right. They are transformed exactly the same way as parabolas. Predict how the graph of each equation below is different from the parent graph.

y = x

y = x

10. y = x ! 4

11. y = x + 5 12. y = x + 3

13. y = x + 2

14. y = ! x + 5 15. y = ! x

16. y = ! x + 3

17. y = ! x ! 5 18. y = ! x ! 2 + 5

19. y = x + 3

20. y = x ! 7 21. y = ! x + 3 + 7

Answers

1. left 5

2. up 5 3. flipped

4. down 4

5. up 5 6. left 2, flipped

7. right 5, down 3

8. left 2, up 1, flipped 9. right 3, down 5

10. right 4

11. up 5 12. up 3

13. left 2

14. up 5, flipped 15. flipped

16. left 3, flipped

17. right 5, flipped 18. right 2, up 5, flipped

19. up 3 20. right 7 21. left 3, up 7, flipped


INTERCEPTS AND INTERSECTIONS 11.2.1 and 11.2.2

Intercepts and Intersections both represent a point at which two paths cross, but an intercept specifies where a curve or line crosses an axis, whereas an intersection refers to any point where the graphs of the two equations cross. When the graph is present, the points may be estimated from the graph. For more accuracy, intercepts and intersections may be found using algebra in most cases.

Example 1 Use the graphs of the parabola y = x2 ! 3x !10 and the line y = !2x + 2 at right. The graph shows that the parabola has two x-intercepts (-2, 0) and (5, 0) and one y-intercept (0, -10). The line has x-intercept (1, 0) and y-intercept (0, 2). The parabola and the line cross each other twice yielding two points of intersection: (-3, 8) and (4, -6). Recall that x-intercepts are always of the form (x, 0) so they may be found by making y = 0 and solving for x. Y-intercepts are always of the form (0, y) so they may be found by making x = 0 and solving for y. To find the points of intersection, solve the system of equations. See the algebraic solutions below.

Intercepts To find the x-intercepts of the parabola, make y = 0 and solve for x.

0 = x2! 3x !10

Factor and use the zero-product property.

0 = (x ! 5)(x + 2) x = 5 or x = !2 so (5, 0) and (-2, 0).

To find the y-intercepts of the parabola, make x = 0 and solve for y.

y = 02! 3 "0 !10 = !10 so (0, -10).

To find the x- and y-intercepts of the line follow the same procedure.

If x = 0 , then y = !2 "0 + 2 = 2 so (0, 2). If y = 0 , then 0 = !2x + 2 , x = 1 so (1, 0).

Intersection To find the point(s) of intersection of the system of equations use the Equal Values Method or Substitution.

x2! 3x !10 = !2x + 2

Make one side equal to zero, factor, and use the zero product property to solve for x. (The quadratic formula is also possible.)

x2! x !12 = 0

(x ! 4)(x + 3) = 0

x = 4 or x = !3

Using x = 4 in either equation yields y = !6 so (4, -6).

Using x = !3 in either equation yields y = 8 so (-3, 8).


Example 2 Given the lines y = 1

3x + 8 and y = 1

2x ! 3 determine the intercepts and intersection without

graphing. Using the same methods as above: Intercepts

For the line y = 1

3x + 8 :

If x = 0 , then y = 1

3!0 + 8 = 8 so (0, 8).

If y = 0 , 0 = 1

3x + 8 , x = !24 so (-24, 0).

For the line y = 1

2x ! 3 :

If x = 0 , then y = 1

2!0 " 3 = "3 , so (0, -3).

If y = 0 , then 0 = 1

2x ! 3 , x = 6 so (6, 0).

Intersection

1

3x + 8 =

1

2x ! 3

6 !

1

3x + 8 = 1

2x " 3( )

2x + 48 = 3x !18

66 = x

Substituting x = 66 into either original equation yields y = 30 . The point of intersection is (66, 30).

Problems For each equation, determine the x- and y-intercepts. 1. y = 3x ! 2 2. y = 1

4x ! 2 3. 3x + 2y = 12

4. !x + 3y = 15 5. 2y = 15 ! 3x 6. y = x2 + 5x + 6 7. y = x2 ! 3x !10 8. y = 4x2 !11x ! 3 9. y = x2 + 3x ! 2 Determine the point(s) of intersection

10. y = 5x +1

y = !3x !15

11. y = x + 7

y = 4x ! 5 12.

x = 7 + 3y

x = 4y + 5

13. x = !

1

2y + 4

8x + 3y = 31

14. 4x ! 3y = !10

x = 1

4y !1

15. 2x ! y = 6

4x ! y = 12

16. y = x2! 3x ! 8

y = 2 17. y = x

2! 7

y = 8 + 2x 18. y = x

2+ 2x + 8

y = !4x !1


Answers 1. (0.6,!0),!(0,!!2) 2. (8, 0), (0, -2)

3. (4, 0), (0, 6)

4. (-15, 0), (0, 5)

5. (5, 0), (0, 7.5) 6. (-2, 0), (-3, 0), (0, 6)

7. (5, 0), (-2, 0) (0, -10)

8. (!0.25,!0),!(3,!0) (0, -3)

9. (0.56,!0),!(!3.56,!0) (0, -2)

10. (-2, -9)

11. (4, 11) 12. (13, 2)

13. (3.5, 1)

14. (-0.25, 3) 15. (3, 0)

16. (5, 2), (-2, 2) 17. (5, 18), (-3, 2) 18. (-3, 11)


FACTORING SHORTCUTS 12.1.1

Although most factoring problems can be done with generic rectangles, there are two special factoring patterns that, if recognized, can be done by sight. The two patterns are known as Difference of Squares and Perfect Square Trinomials. The general patterns are as follows: Difference of Squares: a2x2 b2 = ax + b( ) ax b( )

Perfect Square Trinomial: a2x2 + 2abx + b2 = ax + b( )2

Examples

Difference of Squares

x2 49 = (x + 7)(x 7)

4x2 25 = (2x 5)(2x + 5)

x2 36 = (x + 6)(x 6)

9x2 1 = (3x 1)(3x +1)

Perfect Square Trinomials

x2 10x + 25 = (x 5)2

9x2 +12x + 4 = (3x + 2)2

x2 6x + 9 = (x 3)2

4x2 + 20x + 25 = (2x + 5)2

Sometimes removing a common factor reveals one of the special patterns.

Example 1 8x2 50y2 2(4x2 25y2 ) 2(2x + 5y)(2x 5y)

Example 2 12x2 +12x + 3 3(4x2 + 4x +1) 3(2x +1)2

Chapter 12: Algebraic Extensions 103

Problems Factor each Difference of Squares. 1. x2 16

2. x2 25 3. 64m2 25

4. 4 p2 9q2

5. 9x2y2 49 6. x4 25

7. 64 y2 8. 144 25p2 9. 9x4 4y2 Factor each Perfect Square Trinomial. 10. x2 + 4x + 4

11. y2 + 8y +16 12. m2 10m + 25

13. x2 4x +16

14. a2 + 8ab +16b2 15. 36x2 +12x +1

16. 25x2 30xy + 9y2 17. 9x2y2 6xy +1 18. 49x2 +1+14x Factor completely. 19. 9x2 16

20. 9x2 + 24x +16 21. 9x2 36

22. 2x2 + 8xy + 8y2

23. x2y +10xy + 25y 24. 8x2 72

25. 4x3 9x 26. 4x2 8x + 4 27. 2x2 + 8 Answers

1. x + 4( ) x 4( ) 2. x + 5( ) x 5( ) 3. 8m + 5( ) 8m 5( )

4. 2p + 3q( ) 2p 3q( ) 5. 3xy + 7( ) 3xy 7( ) 6. x2 + 5( ) x2 5( )

7. 8 + y( ) 8 y( ) 8. 12 + 5p( ) 12 5p( ) 9. 3x2 + 2y( ) 3x2 2y( )

10. x + 2( )2 11. y + 4( )

2 12. m 5( )

2

13. not possible 14. a + 4b( )2 15. 6x +1( )

2

16. 5x 3y( )2 17. 3xy 1( )

2 18. 7x +1( )

2

19. 3x + 4( ) 3x 4( ) 20. 3x + 4( )2

21. 9 x + 2( ) x 2( )

22. 2 x + 2y( )2 23. y x + 5( )

2 24. 8 x + 3( ) x 3( )

25. x 2x + 3( ) 2x 3( ) 26. 4 x 1( )2 27. 2 x2 + 4( )


ADDITION AND SUBTRACTION OF 12.1.2 and 12.1.3 RATIONAL EXPRESSIONS

Addition and Subtraction of Rational Expressions uses the same process as simple numerical fractions. First, find a common denominator (if necessary). Second, convert the original fractions to equivalent ones with the common denominator. Third, add (or subtract) the new numerators over the common denominator. Finally, factor the numerator and denominator and reduce (if possible). See the Math Notes box on page 503. Note that these steps are only valid provided that the denominator is not zero.

Example 1 The least common multiple of 2 n + 2( ) and n n + 2( )

is 2n n + 2( ) .

32(n+2) +

3n(n+2)

To get a common denominator in the first fraction,

multiply the fraction by nn , a form of the number 1.

Multiply the second fraction by 22 .

=3

2(n+2)nn +

3n(n+2)

22

Multiply the numerator and denominator of each term. It may be necessary to distribute the numerator.

=3n

2n(n+2) +6

2n(n+2)

Add, factor, and simplify the result. (Note: n 0 or 2 ) =

3n+62n(n+2)

3(n+2)2n(n+2)

32n

Example 2 2 x

x + 4+3x + 6

x + 4

2 x + 3x + 6

x + 4

2x + 8

x + 4

2 x + 4( )

x + 42

Example 3 3

x 1

2

x 2

3

x 1

x 2

x 2

2

x 2

x 1

x 1

3x 6 2x + 2

x 1( ) x 2( )

x 4

x 1( ) x 2( )


Problems Add or subtract each expression and simplify the result. In each case assume the denominator does not equal zero.

1. x

x + 2( ) x + 3( )+

2

x + 2( ) x + 3( ) 2.

x

x2 + 6x + 8+

4

x2 + 6x + 8

3. b2

b2 + 2b 3+

9

b2 + 2b 3 4.

2a

a2 + 2a +1+

2

a2 + 2a +1

5. x +10

x + 2+x 6

x + 2 6.

a + 2b

a + b+2a + b

a + b

7. 3x 4

3x + 3

2x 5

3x + 3 8.

3x

4x 12

9

4x 12

9. 6a

5a2 + a

a 1

5a2 + a 10.

x2 + 3x 5

10

x2 2x +10

10

11. 6

x(x + 3)+

2x

x(x + 3) 12.

5

x 7+

3

4(x 7)

13. 5x + 6

x25

x 14.

2

x + 4

x 4

x2 16

15. 10a

a2 + 6a

3

3a +18 16.

3x

2x2 8x+

2

x 4

17. 5x + 9

x2 2x 3+

6

x2 7x +12 18

x + 4

x2 3x 28

x 5

x2 + 2x 35

19. 3x +1

x2 16

3x + 5

x2 + 8x +16 20.

7x 1

x2 2x 3

6x

x2 x 2

Answers

1. 1

x + 3 2.

1

x + 2 3.

b 3

b 1 4.

2

a +1

5. 2 6. 3 7. 1

3 8.

3

4

9. 1

a 10.

x 3

2 11.

2

x 12.

23

4(x 7)=

23

4x 28

13. 6

x2 14.

1

x + 4 15.

9

a + 6 16.

7

2(x 4)=

7

2x 8

17. 5(x + 2)

(x 4) x +1( )=

5x +10

x2 3x 4 18.

14

(x 7) x + 7( )=

14

x2 49

19. 4(5x + 6)

(x 4) x + 4( )2 20.

x + 2

(x 3) x 2( )=

x + 2

x2 5x + 6


WORK AND MIXTURE PROBLEMS 12.2.1 and 12.2.2

Work problems are solved using the concept that if a job can be completed in r units of

time, then its rate (or fraction of the job completed) is 1

r.

Mixture problems are solved using the concept that the product (value or percentage) x (quantity) must be consistent throughout the equation.

Examples of Work Problems John can completely wash and dry the dishes in 20 minutes. His brother can do it in 30 minutes. How long will it take them working together?

With two inflowing pipes open, a water tank can be filled in 5 hours. If the larger pipe can fill the tank alone in 7 hours, how long would the smaller pipe take to fill the tank?

Solution: Let t = the time to complete the

task so 1

t is their rate together. John’s rate is

1

20 and his brother’s rate is

1

30. Since they

are working together we add the two rates together to get the combined rate.

The equation is 1

20+1

30=1

t. Solve using

fraction busters:

60t 120 +

130( ) = 60t 1

t( )3t + 2t = 60

5t = 60

t = 12 minutes

Solution: Let t = the time for the smaller

pipe so 1

t is its rate. The combined rate is

1

5

and the larger pipe’s rate is 1

7. Since they are

working together we add the two rates together to get the combined rate.

The equation is17+1t=15

. Solve using

fraction busters:

35t 17 +

1t( ) = 35t 1

5( )5t + 35 = 7t

35 = 2t

t = 17.5 hours


Examples of Mixture Problems Alicia has 10 liters of an 80% acid solution. How many liters of water should she add to form a 30% acid solution?

A store has candy worth $0.90 a pound and candy worth $1.20 a pound. If the owners want 60 pounds of candy worth $1.00 a pound, how many pounds of each candy should they use?

Solution: Use %( ) liters( ) = %( ) liters( )

Let x = liters of water added so x + 10 is new liters. The equation is

0.8( ) 10( ) = 0.3( ) x +10( ) .

Multiply by 10 to clear decimals and solve.

8( ) 10( ) = 3( ) x +10( )

80 = 3x + 30

50 = 3x

x = 16 23 liters

Solution: Use $( ) lbs.( ) = $( ) lbs.( )

Let x = pounds of $0.90 candy so 60 x = pounds of $1.20 candy. The equation is

0.90(x) +1.20(60 x) = 1.00(60) .

Multiply by 100 to clear decimals and solve.

90(x) +120(60 x) = 100(60)

90x + 7200 120x = 6000

30x = 1200

x = 40

40 lbs. of $0.90 candy and

20 lbs. of $1.20 candy

Note: The second example above could also have been solved using two equations, where x = $0.90 candy and y = $1.20 candy:

x + y = 60 and $0.90x + $1.20y = $60.00

Problems Solve each problem. 1. Susan can paint her living room in 2 hours. Her friend Jaime estimates it would take him

3 hours to paint the same room. If they work together, how long will it take them to paint Susan’s living room?

2. Professor Minh can complete a set of experiments in 4 hours. Her assistant can do it in

6 hours. How long will it take them to complete the experiments working together? 3. With one hose a swimming pool can be filled in 12 hours. Another hose can fill it in

16 hours. How long will it take to fill the pool using both hoses?


4. Together, two machines can harvest a tomato crop in 6 hours. The larger machine can do it

alone in 10 hours. How long does it take the smaller machine to harvest the crop working alone?

5. Steven can look up 20 words in a dictionary in an hour. His teammate Mary Lou can look

up 30 words per hour. Working together, how long will it take them to look up 100 words? 6. A water tank is filled by one pump in 6 hours and is emptied by another pump in 12 hours.

If both pumps are operating, how long will it take to fill the tank? 7. Two crews can service the space shuttle in 12 days. The faster crew can service the shuttle in

20 days alone. How long would the slower crew need to service the shuttle working alone? 8. Janelle and her assistant Ryan can carpet a house in 8 hours. If Janelle could complete the

job alone in 12 hours, how long would it take Ryan to carpet the house working alone? 9. Able can harvest a strawberry crop in 4 days. Barney can do it in 5 days. Charlie would take

6 days. If they all work together, how long will it take them to complete the harvest? 10. How much coffee costing $6 a pound should be mixed with 3 pounds of coffee costing

$4 a pound to create a mixture costing $4.75 a pound? 11. Sam’s favorite recipe for fruit punch requires 12% apple juice. How much pure apple juice

should he add to 2 gallons of punch that already contains 8% apple juice to meet his standards?

12. Jane has 60 liters of 70% acid solution. How many liters of water must be added to form a

solution that is 40% acid? 13. How many pound of nuts worth $1.05 a pound must be mixed with nuts worth $0.85 a pound

to get a mixture of 200 pounds of nuts worth $0.90 a pound? 14. A coffee shop mixes Kona coffee worth $8 per pound with Brazilian coffee worth $5 per

pound. If 30 pounds of the mixture is to be sold for $7 per pound, how many pounds of each coffee should be used?

15. How much tea costing $8 per pound should be mixed with 2 pounds of tea costing

$5 per pound to get a mixture costing $6 per pound? 16. How many liters of water must evaporate from 50 liters of an 8% salt solution to make a

25% salt solution? 17. How many gallons of pure lemon juice should be mixed with 4 gallons of 25% lemon juice

to achieve a mixture which contains 40% lemon juice? 18. Brian has 20 ounces of a 15% alcohol solution. How many ounces of a 50% alcohol solution

must he add to make a 25% alcohol solution?


Answers

1. 1.2 hours

2. 2.4 hours

3. 6 67 6.86 hours

4. 15 hours

5. 2 hours 6. 12 hours

7. 30 days

8. 24 hours

9. 6037 1.62 days

10. 1.8 pounds

11. 111 gallon 12. 45 liters

13. 50 pounds 14. 20 Kona, 10 Brazilian

15. 1 pound

16. 34 liters

17. 1 gallon 18. 8 ounces

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