Lesson 1.7, For use with pages 51-58

ANSWER

1. 3x + 15 = –42

2. 5x – 8 ≤ 7

ANSWER

Solve the equation or inequality.

–19

x ≤ 3

Lesson 1.7, For use with pages 51-58

3. 2x + 1 < –3 or 2x + 1 > 5

Solve the equation or inequality.

ANSWER x < –2 or x > 2

4. In the next 2 weeks you need to work at least 30 hours. If you can work h hours this week and then twice as many hours next week, how many hours must you work this week?

ANSWER at least 10 h

EXAMPLE 1 Solve a simple absolute value equation

Solve |x – 5| = 7. Graph the solution.

SOLUTION

| x – 5 | = 7

x – 5 = – 7 or x – 5 = 7

x = 5 – 7 or x = 5 + 7

x = –2 or x = 12

Write original equation.

Write equivalent equations.

Solve for x.

Simplify.

EXAMPLE 1

The solutions are –2 and 12. These are the values of x that are 7 units away from 5 on a number line. The graph is shown below.

ANSWER

Solve a simple absolute value equation

EXAMPLE 2 Solve an absolute value equation

| 5x – 10 | = 45

5x – 10 = 45 or 5x –10 = – 45

5x = 55 or 5x = – 35

x = 11 or x = – 7

Write original equation.

Expression can equal 45 or – 45 .

Add 10 to each side.

Divide each side by 5.

Solve |5x – 10 | = 45.

SOLUTION

EXAMPLE 2 Solve an absolute value equation

The solutions are 11 and –7. Check these in the original equation.

ANSWER

Check:

| 5x – 10 | = 45

| 5(11) – 10 | = 54?

|45| = 45?

45 = 45

| 5x – 10 | = 45

| 5(– 7 ) – 10 | = 54?

45 = 45

| – 45| = 45?

EXAMPLE 3

| 2x + 12 | = 4x

2x + 12 = 4x or 2x + 12 = – 4x

12 = 2x or 12 = – 6x

6 = x or –2 = x

Write original equation.

Expression can equal 4x or – 4 x

Add – 2x to each side.

Solve |2x + 12 | = 4x. Check for extraneous solutions.

SOLUTION

Solve for x.

Check for extraneous solutions

EXAMPLE 3

| 2x + 12 | = 4x

| 2(– 2) +12 | = 4(–2)?

|8| = – 8?

8 = –8

Check the apparent solutions to see if either is extraneous.

Check for extraneous solutions

| 2x + 12 | = 4x

| 2(6) +12 | = 4(6)?

|24| = 24?

24 = 24

The solution is 6. Reject – 2 because it is an extraneous solution.

ANSWER

CHECK

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

1. | x | = 5

SOLUTION

| x | = 5

| x | = – 5 or | x | = 5

x = –5 or x = 5

Write original equation.

Write equivalent equations.

Solve for x.

for Examples 1, 2 and 3

GUIDED PRACTICE for Examples 1, 2 and 3

The solutions are –5 and 5. These are the values of x that are 5 units away from 0 on a number line. The graph is shown below.

ANSWER

– 3

– 4

– 2

– 1

0

1 2

3

4

5

6

7

– 5

– 6

– 7

5 5

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

2. |x – 3| = 10

SOLUTION

| x – 3 | = 10

x – 3 = – 10 or x – 3 = 10

x = 3 – 10 or x = 3 + 10

x = –7 or x = 13

Write original equation.

Write equivalent equations.

Solve for x.

Simplify.

for Examples 1, 2 and 3

GUIDED PRACTICE

The solutions are –7 and 13. These are the values of x that are 10 units away from 3 on a number line. The graph is shown below.

ANSWER

– 3

– 4

– 2

– 1

0

1

2

3

4

5

6

7

– 5

– 6

– 7

8

9

10

11

12

13

10 10

for Examples 1, 2 and 3

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

SOLUTION

| x + 2 | = 7

x + 2 = – 7 or x + 2 = 7

x = – 7 – 2 or x = 7 – 2

x = –9 or x = 5

Write original equation.

Write equivalent equations.

Solve for x.

Simplify.

3. |x + 2| = 7

for Examples 1, 2 and 3

GUIDED PRACTICE

The solutions are –9 and 5. These are the values of x that are 7 units away from – 2 on a number line.

ANSWER

for Examples 1, 2 and 3

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

4. |3x – 2| = 13

SOLUTION

3x – 2 = 13 or 3x – 2 = – 13

Write original equation.

Solve for x.

Simplify.

|3x – 2| = 13

Write equivalent equations.

x = or x = 5 3–3

2

for Examples 1, 2 and 3

x = –13 + 2

3or x =

13 + 23

GUIDED PRACTICE

The solutions are – and 5.

ANSWER

33

2

for Examples 1, 2 and 3

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

5. |2x + 5| = 3x

| 2x + 5 | = 3x

2x + 5 = – 3x or 2x + 5 = 3x

x = 1 or x = 5

Write original equation.

Write Equivalent equations.

Simplify

SOLUTION

for Examples 1, 2 and 3

2x + 3x = 5 or 2x – 3x = –5

GUIDED PRACTICE

The solution of is 5. Reject 1 because it is an extraneous solution.

ANSWER

for Examples 1, 2 and 3

| 2x + 12 | = 4x

| 2(1) +12 | = 4(1)?

|14| = 4 ? 14 = –8

Check the apparent solutions to see if either is extraneous.

| 2x + 5 | = 3x

| 2(5) +5 | = 3(5) ?

|15| = 15?

15 = 15

CHECK

GUIDED PRACTICE

Solve the equation. Check for extraneous solutions.

6. |4x – 1| = 2x + 9

SOLUTION

4x – 1 = – (2x + 9) or 4x – 1 = 2x + 9

Write original equation.

Solve For xx = or x = 5 31

1

|4x – 1| = 2x + 9

Write equivalent equations.

for Examples 1, 2 and 3

4x + 2x = – 9 + 1 or 4x – 2x = 9 + 1 Rewrite equation.

GUIDED PRACTICE

ANSWER

The solutions are – and 5. 3

11

for Examples 1, 2 and 3

| 4x – 1 | = 2x + 9

| 4(5) – 1 | = 2(5) + 9?

|19| = 19 ?

19 = 19

Check the apparent solutions to see if either is extraneous.

| 4x – 1 | = 2x + 9

|4( -1 ) – 1 | = 2 ( - 1 )+ 931

3

1?

CHECK

| | = ? 3

– 19

319

= 3

– 19

319

EXAMPLE 4 Solve an inequality of the form |ax + b| > c

Solve |4x + 5| > 13. Then graph the solution.

SOLUTION

First Inequality Second Inequality

4x + 5 < – 13 4x + 5 > 13

4x < – 18 4x > 8

x < – 92

x > 2

Write inequalities.

Subtract 5 from each side.

Divide each side by 4.

The absolute value inequality is equivalent to

4x +5 < –13 or 4x + 5 > 13.

EXAMPLE 4

ANSWER

Solve an inequality of the form |ax + b| > c

The solutions are all real numbers less than or greater than 2. The graph is shown below.– 9

2

EXAMPLE 5 Solve an inequality of the form |ax + b| ≤ c

A professional baseball should weigh 5.125 ounces, with a tolerance of 0.125 ounce. Write and solve an absolute value inequality that describes the acceptable weights for a baseball.

Baseball

SOLUTION

Write a verbal model. Then write an inequality.STEP 1

EXAMPLE 5 Solve an inequality of the form |ax + b| ≤ c

STEP 2 Solve the inequality.

Write inequality.

Write equivalent compound inequality.

Add 5.125 to each expression.

|w – 5.125| ≤ 0.125

– 0.125 ≤ w – 5.125 ≤ 0.125

5 ≤ w ≤ 5.25

So, a baseball should weigh between 5 ounces and 5.25 ounces, inclusive. The graph is shown below.

ANSWER

EXAMPLE 6

The thickness of the mats used in the rings, parallel bars, and vault events must be between 7.5 inches and 8.25 inches, inclusive. Write an absolute value inequality describing the acceptable mat thicknesses.

Gymnastics

SOLUTION

STEP 1 Calculate the mean of the extreme mat thicknesses.

Write a range as an absolute value inequality

EXAMPLE 6

Mean of extremes = = 7.875 7.5 + 8.25

2

Find the tolerance by subtracting the mean from the upper extreme.

STEP 2

Tolerance = 8.25 – 7.875

Write a range as an absolute value inequality

= 0.375

EXAMPLE 6

STEP 3 Write a verbal model. Then write an inequality.

A mat is acceptable if its thickness t satisfies |t – 7.875| ≤ 0.375.

ANSWER

Write a range as an absolute value inequality

GUIDED PRACTICE for Examples 5 and 6

Solve the inequality. Then graph the solution.

10. |x + 2| < 6

The absolute value inequality is equivalent to x + 2 < 6 or x + 2 > – 6

First Inequality Second Inequality

x + 2 < 6 x + 2 > – 6

x < 4 x > – 8

Write inequalities.

Subtract 2 from each side.

SOLUTION

GUIDED PRACTICE for Examples 5 and 6

ANSWER

The solutions are all real numbers less than – 8 or greater than 4. The graph is shown below.

GUIDED PRACTICE for Examples 5 and 6

Solve the inequality. Then graph the solution.

11. |2x + 1| ≤ 9

The absolute value inequality is equivalent to 2x + 1 < 9 or 2x + 1 > – 9

First Inequality Second Inequality

2x < 8 2x > – 10

Write inequalities.

Subtract 1 from each side.

SOLUTION

2x + 1 > – 92x + 1 < 9

x < 4 x > – 5Divide each side 2

GUIDED PRACTICE for Examples 5 and 6

ANSWER

The solutions are all real numbers less than – 5 or greater than 4. The graph is shown below.

GUIDED PRACTICE for Examples 5 and 6

12. |7 – x| ≤ 4

Solve the inequality. Then graph the solution.

The absolute value inequality is equivalent to 7 – x < 4 or 7 – x > – 4

First Inequality Second Inequality

– x < – 3 – x > – 11

Write inequalities.

Subtract 7 from each side.

SOLUTION

x < 3 x > 11Divide each side (– ) sign

7 – x > – 47 – x < 4

GUIDED PRACTICE for Examples 5 and 6

ANSWER

The solutions are all real numbers less than 3 or greater than 11. The graph is shown below.

GUIDED PRACTICE for Examples 5 and 6

13. Gymnastics: For Example 6, write an absolute value inequality describing the unacceptable mat thicknesses.

SOLUTION

STEP 1 Calculate the mean of the extreme mat thicknesses.

Mean of extremes = = 7.875 7.5+ 8.25

2

Find the tolerance by subtracting the mean from the upper extreme.

STEP 2

Tolerance = 8.25 – 7.875 = 0.375

GUIDED PRACTICE for Examples 5 and 6

STEP 3

A mat is unacceptable if its thickness t satisfies |t – 7.875| > 0.375.

ANSWER

GUIDED PRACTICE for Example 4

Solve the inequality. Then graph the solution.

7. |x + 4| ≥ 6

x < – 10 or x > 2 The graph is shown below.

ANSWER

GUIDED PRACTICE for Example 4

Solve the inequality. Then graph the solution.

8. |2x –7|>1

ANSWER

x < 3 or x > 4 The graph is shown below.

GUIDED PRACTICE for Example 4

Solve the inequality. Then graph the solution.

9. |3x + 5| ≥ 10

ANSWER

x < – 5 or x > 123

The graph is shown below.