copyright © 2005 pearson education, inc. solving linear equations 1.4

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Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

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Page 1: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Copyright © 2005 Pearson Education, Inc.

Solving Linear Equations

1.4

Page 2: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-2 Copyright © 2005 Pearson Education, Inc.

Definitions

Algebra: a generalized form of arithmetic. Variables: used to represent numbers Algebraic expression: a collection of variables,

numbers, parentheses, and operation symbols. Examples:

24 2, 4, 4(3 5), , 8 2

3 5

xx x y y y

x

Page 3: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-3 Copyright © 2005 Pearson Education, Inc.

Order of Operations

1. First, perform all operations within parentheses or other grouping symbols (according to the following order).

2. Next, perform all exponential operations (that is, raising to powers or finding roots).

3. Next, perform all multiplication and divisions from left to right.

4. Finally, perform all additions and subtractions from left to right.

Page 4: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-4 Copyright © 2005 Pearson Education, Inc.

Example: Evaluating an Expression

Evaluate the expression x2 + 4x + 5 for x = 3. Solution:

x2 + 4x + 5

= 32 + 4(3) + 5

= 9 + 12 + 5

= 26

Page 5: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-5 Copyright © 2005 Pearson Education, Inc.

Example: Substituting for Two Variables

Evaluate when x = 3 and y = 4. Solution:

2 24 3 5x xy y

2 2

2 2

4(3) 3(3)(4) 5(4 )

4(9) 36 5(16)

36 36 80

0 80

8

4 3 5

0

x xy y

Page 6: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Copyright © 2005 Pearson Education, Inc.

Linear Equations in

One Variable

Page 7: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-7 Copyright © 2005 Pearson Education, Inc.

Definitions

Like terms are terms that have the same variables with the same exponents on the variables.

Unlike terms have different variables or different

exponents on the variables.

2 22 , 7 5 , 8x x x x

3 22 , 7 5 , 6x x x

Page 8: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-8 Copyright © 2005 Pearson Education, Inc.

Properties of the Real NumbersEquality Principles

Associative property of multiplication

(ab)c = a(bc)

Associative property of addition

(a + b) + c = a + (b + c)

Commutative property of multiplication

ab = ba

Commutative property of addition

a + b = b + a

Distributive propertya(b + c) = ab + ac

Page 9: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-9 Copyright © 2005 Pearson Education, Inc.

Example: Combine Like Terms

8x + 4x

= (8 + 4)x

= 12x

5y 6y

= (5 6)y

= y

x + 15 5x + 9

= (1 5)x + (15+9)

= 4x + 24

3x + 2 + 6y 4 + 7x

= (3 + 7)x + 6y + (2 4)

= 10x + 6y 2

Page 10: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-10 Copyright © 2005 Pearson Education, Inc.

Solving Equations

Addition Property of Equality

If a = b, then a + c = b + c for all real numbers a, b, and c.

Find the solution to the equation

x 9 = 24.

x 9 + 9 = 24 + 9

x = 33

Check: x 9 = 24

33 9 = 24 ? 24 = 24 true

Page 11: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-11 Copyright © 2005 Pearson Education, Inc.

Solving Equations continued

Subtraction Property of Equality

If a = b, then a c = b c for all real numbers a, b, and c.

Find the solution to the equation

x + 12 = 31.

x + 12 12 = 31 12

x = 19

Check: x + 12 = 31

19 + 12 = 31 ? 31 = 31 true

Page 12: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-12 Copyright © 2005 Pearson Education, Inc.

Solving Equations continued

Multiplication Property of Equality

If a = b, then a • c = b • c for all real numbers a, b, and c, where c 0.

Find the solution to the equation

9.

7

x

1

97

(7 7 9)7

7

x

x

1 7

x63

63x

Page 13: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-13 Copyright © 2005 Pearson Education, Inc.

Solving Equations continued

Division Property of Equality

If a = b, then for all real numbers a, b, and c, c 0.

Find the solution to the equation 4x = 48.

a b

c c

4

4 48

4 48

124

x

x

x

Page 14: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-14 Copyright © 2005 Pearson Education, Inc.

General Procedure for Solving Linear Equations If the equation contains fractions, multiply both sides of

the equation by the lowest common denominator (or least common multiple). This step will eliminate all fractions from the equation.

Use the distributive property to remove parentheses when necessary.

Combine like terms on the same side of the equal sign when possible.

Page 15: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-15 Copyright © 2005 Pearson Education, Inc.

General Procedure for Solving Linear Equations continued Use the addition or subtraction property to

collect all terms with a variable on one side of the equal sign and all constants on the other side of the equal sign. It may be necessary to use the addition or subtraction property more than once. This process will eventually result in an equation of the form ax = b, where a and b are real numbers.

Page 16: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-16 Copyright © 2005 Pearson Education, Inc.

General Procedure for Solving Linear Equations continued Solve for the variable using the division or

multiplication property. This will result in an answer in the form x = c, where c is a real number.

Page 17: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-17 Copyright © 2005 Pearson Education, Inc.

Example: Solving Equations

Solve 3x 4 = 17.

4

3 4 17

3 4 17

3 21

3 21

4

3 37

x

x

x

x

x

Page 18: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-18 Copyright © 2005 Pearson Education, Inc.

Solve 21 = 6 + 3(x + 2)

21 6 3( 2)

21 6 3 6

21 3 12

1221 3 12

9 3

9 3

3

12

3 3

x

x

x

x

x

x

x

Page 19: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-19 Copyright © 2005 Pearson Education, Inc.

Solve 8x + 3 = 6x + 21

8 3 6 21

8 3 6 21

8 6 18

8 6 18

2 18

2 18

9

3 3

6 6

2 2

x x

x x

x x

x x

x

x

x

x

x

Page 20: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-20 Copyright © 2005 Pearson Education, Inc.

Solve 6(x 2) + 2x + 3 = 4(2x 3) + 2

False, the equation has no solution. The equation is inconsistent.

6( 2) 2 3 4(2 3) 2

6 12 2 3 8 12 2

8 9 8 10

8 8 9 8 8 10

9 10

x x x

x x x

x x

x x x x

Page 21: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-21 Copyright © 2005 Pearson Education, Inc.

Solve 4(x + 1) 6(x + 2) = 2(x + 4)

True, 0 = 0 the solution is all real numbers.

4( 1) 6( 2) 2( 4)

4 4 6 12 2 8

2 8 2 8

2 2 8 2 2 8

8 8

8 8 8 8

0 0

x x x

x x x

x x

x x x x

Page 22: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-22 Copyright © 2005 Pearson Education, Inc.

Proportions

A proportion is a statement of equality between two ratios.

Cross Multiplication If then ad = bc, b 0, d 0.,

a c

b d

Page 23: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-23 Copyright © 2005 Pearson Education, Inc.

To Solve Application Problems Using Proportions Represent the unknown quantity by a variable. Set up the proportion by listing the given ratio on the

left-hand side of the equal sign and the unknown and other given quantity on the right-hand side of the equal sign. When setting up the right-hand side of the proportion, the same respective quantities should occupy the same respective positions on the left and right.

Page 24: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-24 Copyright © 2005 Pearson Education, Inc.

To Solve Application Problems Using Proportions continued For example, an acceptable proportion might be

Once the proportion is properly written, drop the units and use cross multiplication to solve the equation.

Answer the question or questions asked.

miles miles

hour hour

Page 25: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-25 Copyright © 2005 Pearson Education, Inc.

Example

A 50 pound bag of fertilizer will cover an area of 15,000 ft2. How many pounds are needed to cover an area of 226,000 ft2?

754 pounds of fertilizer would be needed.

2 2

50 pounds

15,000 ft 226,000 ft

(50)(226,000) 15,000

11,300,000 15,000

11,300,000 15,000

15,000 15,000

753.33

x

x

x

x

x

Page 26: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-26 Copyright © 2005 Pearson Education, Inc.

Translating Words to Expressions

3x 9The sum of three times a number decreased by 9.

2x + 8Eight more than twice a number

2xTwice a number

x 5Five less than a number

x + 10Ten more than a number

Mathematical ExpressionPhrase

Page 27: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-27 Copyright © 2005 Pearson Education, Inc.

To Solve a Word Problem

Read the problem carefully at least twice to be sure that you understand it.

If possible, draw a sketch to help visualize the problem. Determine which quantity you are being asked to find.

Choose a letter to represent this unknown quantity. Write down exactly what this letter represents.

Write the word problem as an equation. Solve the equation for the unknown quantity. Answer the question or questions asked. Check the solution.

Page 28: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-28 Copyright © 2005 Pearson Education, Inc.

Example

The bill (parts and labor) for the repairs of a car was $496.50. The cost of the parts was $339. The cost of the labor was $45 per hour. How many hours were billed?

Let h = the number of hours billed Cost of parts + labor = total amount

339 + 45h = 496.50

Page 29: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-29 Copyright © 2005 Pearson Education, Inc.

Example continued

The car was worked on for 3.5 hours.

339 45 496.50

339 339 45 496.50 339

45 157.50

45 157.50

45 453.5

h

h

h

h

h

Page 30: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-30 Copyright © 2005 Pearson Education, Inc.

Example

Sandra Cone wants to fence in a rectangular region in her backyard for her lambs. She only has 184 feet of fencing to use for the perimeter of the region. What should the dimensions of the region be if she wants the length to be 8 feet greater than the width?

Page 31: Copyright © 2005 Pearson Education, Inc. Solving Linear Equations 1.4

Slide 6-31 Copyright © 2005 Pearson Education, Inc.

continued, 184 feet of fencing, length 8 feet longer than width Let x = width of region Let x + 8 = length P = 2l + 2w x + 8

x

184 2( ) 2( 8)

184 2 2 16

184 4 16

168 4

42

x x

x x

x

x

x

The width of the region is 42 feet and the length is 50 feet.