Intermediate Algebra (11th edition)

NOW TRY

Solving a Linear Inequality with Fractions

Solve , and graph the solution set.

To clear fractions, multiply each side by the least common denominator, 6.

Multiply by 6, the LCD.

Distributive property

- 41 x- 32 - 36315 - x 2 Multiply.

6 c-

2

3

1 x- 3 2d- 6 a

1

2

b 66 c

1

2

15 - x2d

6 c-

2

3

1 x- 32 -

1

2

d 66 c

1

2

15 - x2d

-

2

3

1 x- 32 -

1

2

6

1

2

15 - x 2

-^23 1 x- 32 -^12612 15 - x 2

EXAMPLE 5

96 CHAPTER 2 Linear Equations, Inequalities, and Applications

NOW TRY
EXERCISE 5
Solve and graph the solution
set.

3

4

1 x- 22 +

1

2

7

1

5

1 x- 82

NOW TRY ANSWERS

5. A-^1211 , qB

Check that the solution set is 1 - 6, q 2 .See the graph in FIGURE 14.

–8 –7 –6 –5 –4 –3 –2 –1 0

FIGURE 14 NOW TRY

OBJECTIVE 3 Solve linear inequalities with three parts.For some applica-

tions, it is necessary to work with a three-part inequalitysuch as

where is between3 and 8.

Solving a Three-Part Inequality

Solve , and graph the solution set.

To solve this inequality, we subtract 2 from eachof the three parts of the inequality.

1 6 x 66

3 - 26 x+ 2 - 268 - 2

3 6 x+ 2 68

36 x+ 268

EXAMPLE 6

x+ 2

36 x+ 26 8,

–2 –1 0 1

12

• 11

Be careful

here.

Step 1 Distributive property

Step 2 Add 3x.

Subtract 9.

Step 3 Multiply by

Change to

x7- 6

6 7.

- 11 - x 27 - 1162 - 1.

- x 66

- x+ 9 - 9615 - 9

- x+ 9615

- 4 x+ 9 + 3 x 615 - 3 x+ 3 x

- 4 x+ 9615 - 3 x

- 4 x+ 12 - 3615 - 3 x

01234567

FIGURE 15

6. 1 1, 5 2

–1012345

Subtract 2 from

all three parts.

Thus, xmust be between 1 and 6 so that will be between 3 and 8. The solution

set, 1 1, 6 2 ,is graphed in FIGURE 15.

x+ 2

NOW TRY
EXERCISE 6
Solve and graph the solution
set.

• 16 x- 263

Reverse the inequality

symbol when multiplying

by a negativenumber.