Copyright © 2013 Pearson Education, Inc. Section 2.2 Linear Equations.

Copyright © 2013 Pearson Education, Inc.

Section 2.2

Linear Equations

If an equation is linear, writing it in the form ax + b = 0 should not require any properties or processes other than the following:

• using the distributive property to clear parentheses,

• combining like terms,

• applying the addition property of equality.

Page 100

Example

Determine whether the equation is linear. If the equation is linear, give values for a and b.a. 9x + 7 = 0 b. 5x3 + 9 = 0 c. Solutiona. The equation is linear because it is in the form ax + b = 0 with a = 9 and b = 7.b. This equation is not linear because it cannot be written in the form ax + b = 0. The variable has an exponent other than 1.c. This equation is not linear because it cannot be written in the form ax + b = 0. The variable appears in the denominator of a fraction.

57 0x

Page 100

Example

Complete the table for the given values of x. Then solve the equation 4x – 6 = −2.

Solution

x −3 −2 −1 0 1 2 3

4x − 6 −18

From the table 4x – 6 = −2 when x = 1. Thus the solution to 4x – 6 = −2 is 1.

x −3 −2 −1 0 1 2 3

4x − 6 −18 −14 −10 −6 −2 2 6

Page 101

Slide 5

STEP 1: Use the distributive property to clear any parentheses on each side of the equation. Combine any like terms on each side.

STEP 2: Use the addition property of equality to get all of the terms containing the variable on one side of the equation and all other terms on the other side of the equation. Combine any like terms on each side.

STEP 3: Use the multiplication property of equality to isolate the variable by multiplying each side of the equation by the reciprocal of the number in front of the variable (or divide each side by that number).

STEP 4: Check the solution by substituting it in the given equation.

Solving a Linear Equation Symbolically Page 102

Example

Solve each linear equation.a. 12x − 15 = 0 b. 3x + 19 = 5x + 5Solutiona. 12x – 15 = 0 b. 3x + 19 = 5x + 5

12x − 15 + 15 = 0 + 15

12x = 15 12 15

12 12

x

5

4x

3x − 3x + 19 = 5x − 3x + 5

19 = 2x + 5

19 − 5 = 2x + 5 − 5

14 = 2x 14

2 2

2x

7 x

Page 103

Example

Solve the linear equation. Check your solution. x + 5 (x – 1) = 11

Solution

6x − 5 = 11

16

6x

6 16

6 6

x

x + 5 (x – 1) = 11 x + 5x – 5 = 11

6x − 5 + 5 = 11 + 5 6x = 16

8

3

Page 104

Example (cont)

Check: x + 5 (x – 1) = 11

8 85 1 11

3 3

8 8 35 11

3 3 3

8 55 11

3 3

8 2511

3 3

3311

3

11 11

The answer checks, the

solution is 8.3

Page 104

Example

Solve the linear equation. Solution

1 15

2 6x x

1 156 6

2 6x x

3 30x x

2 30x

15x

1 15

2 6x x

The solution is 15.

Page 105

Example

Solve the linear equation. Solution

5.3 0.8 7x

5.3 0.8 7x

5.3 0.10 18 7 0x 53 8 70x

53 8 753 0 53x 8 17x 8 17

8 8

x

17

8x

The solution is17.

8

Page 106

Equations with No Solutions or Infinitely Many Solutions

An equation that is always true is called an identity and an equation that is always false is called a contradiction.

Slide 11

Page 107

Example

Determine whether the equation has no solutions, one solution, or infinitely many solutions.a.10 – 8x = 2(5 – 4x)b.7x = 9x + 2(12 – x) c.6x = 4(x + 5)

Solution

10 – 8x = 2(5 – 4x)

10 – 8x = 10 – 8x

– 8x = – 8x

0 = 0

Because the equation 0 = 0 is always true, it is an identity and there are infinitely many solutions.

Page 107

Example (cont)

b. 7x = 9x + 2(12 – x)

7x = 9x + 24 – 2x

7x = 7x + 24

0 = 24

Because the equation 0 = 24 is always false, it is a contradiction and there are no solutions.

c. 6x = 4(x + 5)

6x = 4x + 20

2x = 20

x = 10

Thus there is one solution.

Page 107

Slide 14

Page 107

End of chapter problems

• do 42, 44, 46, 56 on page 109

• do 69, 70 on page 109

DONE

Objectives

• Basic Concepts• Solving Linear Equations• Applying the Distributive Property• Clearing Fractions and Decimals• Equations with No Solutions or Infinitely Many Solutions