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# 6.6: Absolute Value Inequalities

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Solve absolute value inequalities.
• Rewrite and solve absolute value inequalities as compound inequalities.
• Solve real-world problems using absolute value inequalities.

## Introduction

Absolute value inequalities are solved in a similar way to absolute value equations. In both cases, you must consider the two options.

1. The expression inside the absolute value is not negative.
2. The expression inside the absolute value is negative.

Then we solve each inequality separately.

## Solve Absolute Value Inequalities

Consider the inequality

$\mid x \mid \leq 3$

Since the absolute value of $x$ represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero is less than or equal to 3. The following graph shows this solution:

Notice that this is also the graph for the compound inequality $-3 \leq x \leq 3$.

Now consider the inequality

$\mid x \mid > 2$

Since the absolute value of $x$ represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero are more than 2. The following graph shows this solution.

Notice that this is also the graph for the compound inequality $x<-2$ or $x > 2$.

Example 1

Solve the following inequalities and show the solution graph.

a) $\mid x \mid <6$

b) $\mid x \mid \geq 2.5$

Solution

a) $\mid x \mid <5$ represents all numbers whose distance from zero is less than 5.

Answer $-5

b) $\mid x \mid \geq 2.5$ represents all numbers whose distance from zero is more than or equal to 2.5.

Answer $x \leq -2.5$ or $x \geq 2.5$

## Rewrite and Solve Absolute Value Inequalities as Compound Inequalities

In the last section you saw that absolute value inequalities are compound inequalities.

Inequalities of the type $\mid x \mid can be rewritten as $-a < x

Inequalities of the type $\mid x \mid can be rewritten as $x <-b$ or $x> b$

To solve an absolute value inequality, we separate the expression into two inequalities and solve each of them individually.

Example 2

Solve the inequality $\mid x-3\mid <7$ and show the solution graph.

Solution

Rewrite as a compound inequality.

Write as two separate inequalities.

$x-3 <7$ and $x-3 <7$

Solve each inequality

$x <10$ and $x >-4$

The solution graph is

Example 3

Solve the inequality $\mid 4x +6 \mid \leq 13$ and show the solution graph.

Solution

Rewrite as a compound inequality.

Write as two separate inequalities

$4x + 5 \leq 13$ and $4x+5 \geq -13$

Solve each inequality:

$4x \leq 8$ and $4x \geq -18$

$x \leq 2$ and $x \geq -\frac{9}{2}$

The solution graph is

Example 4

Solve the inequality $\mid x +12 \mid >2$ and show the solution graph.

Solution

Rewrite as a compound inequality.

Write as two separate inequalities.

$x +12 <-2$ or $x+12>2$

Solve each inequality

$x <-14$ or $x >-10$

The solution graph is

Example 5

Solve the inequality $\mid 8x -15 \mid \geq 9$ and show the solution graph.

Rewrite as a compound inequality.

Write as two separate inequalities.

$8x-15 \leq -9$ or $8x-15 \geq 9$

Solve each inequality

$8x \leq 6$ or $8x \geq 24$

$x \leq \frac{3}{4}$ or $x \geq 3$

The solution graph is

## Solve Real-World Problems Using Absolute Value Inequalities

Absolute value inequalities are useful in problems where we are dealing with a range of values.

Example 6:

The velocity of an object is given by the formula $v = 25t - 80$ where the time is expressed in seconds and the velocity is expressed in feet per seconds. Find the times when the magnitude of the velocity is greater than or equal to 60 feet per second.

Solution

Step 1

We want to find the times when the velocity is greater than or equal to 60 feet per second

Step 2

We are given the formula for the velocity $v = 25t - 80$

Write the absolute value inequality $\mid 25t-80\mid \geq 60$

Step 3

Solve the inequality

$25t-80 \geq 60$ or $25t-80 \leq -60$

$25t\geq 140$ or $25t \leq 20$

$t \geq 5.6$ or $t \leq 0.8$

Answer: The magnitude of the velocity is greater than 60 ft/sec for times less than 0.8 seconds and for times greater than 5.6 seconds.

Step 4 When $t= 0.8$ seconds, $v= 25(0.8) -80 = -60 \ ft/sec$. The magnitude of the velocity is 60 ft/sec. The negative sign in the answer means that the object is moving backwards.

When $t= 5.6$ seconds, $v = 25(0.8) - 80 = -60 \ ft/sec$.

To find where the magnitude of the velocity is greater than 60 ft/sec, check values in each of the following time intervals: $t\leq 0.8$,$0.8 \leq t \leq 5.6$ and $t \geq 5.6$.

Check $t = 0.5$: $v = 25(0.5) - 80 = -67.5 \ ft/sec$

Check $t = 2$: $v = 25(2)- 80 = -30 \ ft/sec$

Check $t = 6$: $v = 25(6)-80 = 70 \ ft/sec$

You can see that the magnitude of the velocity is greater than 60 ft/sec for $t \geq 5.6$ or $t\leq 0.8$.

## Lesson Summary

• Like absolute value equations, inequalities with absolute value split into two inequalities. One where the expression within the absolute value is negative and one where it is positive.
• Inequalities of the type $\mid x \mid can be rewritten as $-a .
• Inequalities of the type $\mid x \mid >b$ can be rewritten as $-x<-b$ or $x>b$.

## Review Questions

Solve the following inequalities and show the solution graph.

1. $|x|\leq6$
2. $|x|>3.5$
3. $|x|<12$
4. $|\frac{x} {5}| \leq 6$
5. $|7x|\geq21$
6. $|x-5|>8$
7. $|x+7|<3$
8. $\big | x-\frac{3} {4} \big | \leq \frac{1} {2}$
9. $|2x-5|\geq13$
10. $|5x+3|<7$
11. $\big | \frac{x} {3}-4 \big | \leq 2$
12. $\big | \frac{2x} {7}+9 \big | > \frac{5} {7}$
13. A three month old baby boy weighs an average of 13 pounds. He is considered healthy if he is 2.5 lbs more or less than the average weight. Find the weight range that is considered healthy for three month old boys.

1. $-6 \leq x \leq 6$
2. $x < -3.5$ or $x > 3.5$
3. $-12 < x < 12$
4. $x < -10$ or $x > 10$
5. $x \leq -3$ or $x \geq 3$
6. $x < -3$ or $x > 13$
7. $-10< x < -4$
8. $\frac{1}{4} \leq x \leq \frac{5}{4}$
9. $x \leq -4$ or $x \geq 9$
10. $-2< x < \frac{4}{5}$
11. $6 \leq x \leq 18$
12. $x < -34$ or $x > -29$
13. A healthy weight is $10.5 \ lb \leq x \leq 15.5 \ lb$.

Feb 22, 2012

Aug 22, 2014