Lesson Objectives
• Demonstrate an understanding of how to solve a compound inequality with "or"
• Demonstrate an understanding of how to solve a compound inequality with "and"
• Learn how to solve an absolute value inequality
• Learn how to identify an absolute value inequality with an infinite number of solutions
• Learn how to identify an absolute value inequality with no solution

## How to Solve an Absolute Value Inequality

In our last lesson, we learned how to solve absolute value equations. In this lesson, we will take the next step and learn how to solve absolute value inequalities. Before we jump in, let's again think about the absolute value operation. The absolute value of a number is the distance between that number and 0 on the number line. Since distance is non-negative, the absolute value of a number can only be 0 or a positive number. Let's think about the absolute value of 8 using a number line: From our number line, we can see that 8 is 8 units away from zero on the number line. This means the absolute value of 8 is 8.
|8| = 8
Additionally, the opposite of 8, (-8) will also have an absolute value of 8. Opposites will always have the same absolute value. Opposites are defined as numbers that have the same distance away from zero on the number line, but that lie on opposite sides. Let's look at the absolute value of (-8) using a number line: What could we say is the solution for an inequality such as:
|x| > 8
If we think about this carefully, we are saying that we can replace x with any number whose absolute value or distance from zero is greater than 8. This means any number larger than 8 or any number smaller than -8. This means we can show our solution as:
x > 8 or x < -8

### Solving Absolute Value Inequalities of the form |ax + b| > k

• Isolate the absolute value operation on one side of the inequality
• Set up a compound inequality with "or"
• 1st Inequality » Remove the absolute value bars
• 2nd Inequality » Remove the absolute value bars, flip the inequality sign, change the constant on the right into its opposite
Let's look at an example.
Example 1: Solve each inequality, write in interval notation, graph the interval.
-8 + 9|10x - 10| ≥ 82
Step 1) Isolate the absolute value operation.
9|10x - 10| ≥ 90
Divide each side by 9:
|10x - 10| ≥ 10
Step 2) Set up a compound inequality with "or"
Inequality 1) Remove the absolute value bars:
10x - 10 ≥ 10
Inequality 2) Remove the absolute value bars, flip the inequality sign, change the constant on the right into its opposite:
10x - 10 ≤ -10
The solution for our absolute value inequality will be the union of the two solution sets.
10x - 10 ≥ 10 or 10x - 10 ≤ -10
10x - 10 ≥ 10
10x ≥ 20
x ≥ 2
10x - 10 ≤ -10
10x ≤ 0
x ≤ 0
x ≤ 0 or x ≥ 2
Interval Notation:
(-∞, 0] ∪ [2, ∞)
Graphing the Interval: Example 2: Solve each inequality, write in interval notation, graph the interval.
2|9 + 3x| - 2 ≥ 40
Step 1) Isolate the absolute value operation.
2|9 + 3x| ≥ 42
Divide each side by 2:
|9 + 3x| ≥ 21
Step 2) Set up a compound inequality with "or"
Inequality 1) Remove the absolute value bars:
9 + 3x ≥ 21
Inequality 2) Remove the absolute value bars, flip the inequality sign, change the constant on the right into its opposite:
9 + 3x ≤ -21
The solution for our absolute value inequality will be the union of the two solution sets.
9 + 3x ≥ 21 or 9 + 3x ≤ -21
9 + 3x ≥ 21
3x ≥ 12
x ≥ 4
9 + 3x ≤ -21
3x ≤ -30
x ≤ -10
x ≤ -10 or x ≥ 4
Interval Notation:
(-∞, -10] ∪ [4, ∞)
Graphing the Interval: Now that we understand how to solve an absolute value inequality with greater than, let's think about an absolute value inequality with less than. Let's suppose we changed our example inequality into:
|x| < 8
For this inequality, we are saying that we can replace x with any number whose absolute value or distance from zero is less than 8. This means any number between 8 and -8.

### Solving Absolute Value Inequalities of the form |ax + b| < k

• Isolate the absolute value operation on one side of the inequality
• Set up a compound inequality with "and"
• 1st Inequality » Remove the absolute value bars
• 2nd Inequality » Remove the absolute value bars, flip the inequality sign, change the constant on the right into its opposite
• We can write this as a three-part inequality: -k < ax + b < k
Let's look at an example.
Example 3: Solve each inequality, write in interval notation, graph the interval.
8|-2x - 1| + 10 ≤ 50
Step 1) Isolate the absolute value operation on one side of the inequality.
Subtract 10 away from each side:
8|-2x - 1| ≤ 40
Divide each side of the inequality by 8:
|-2x - 1| ≤ 5
Step 2) Set up a compound inequality with "and".
We can set this up as a three-part inequality:
-5 ≤ -2x - 1 ≤ 5
-4 ≤ -2x ≤ 6
Divide each part by -2, remember to flip the direction of the inequality symbols:
2 ≥ x ≥ -3
Rewrite in the proper order:
-3 ≤ x ≤ 2
Interval Notation:
[-3,2]
Graphing the Interval: Example 4: Solve each inequality, write in interval notation, graph the interval.
-3|5 - 2x| - 4 > -25
Step 1) Isolate the absolute value operation on one side of the inequality.
Add 4 to each side of the inequality:
-3|5 - 2x| > -21
Divide each side by -3, remember to flip the direction of the inequality symbol
|5 - 2x| < 7
Step 2) Set up a compound inequality with "and".
We can set this up as a three-part inequality:
-7 < 5 - 2x < 7
Subtract 5 from each part:
-12 < -2x < 2
Divide each part by -2, remember to flip the direction of the inequality symbols:
6 > x > -1
Rewrite in the proper order:
-1 < x < 6
Interval Notation:
(-1,6)
Graphing the Interval: ### Solving Absolute Value Inequalities with an Infinite Number of Solutions

In some cases, we will see an absolute value inequality that has an infinite number of solutions. This occurs when the absolute value operation is said to be larger than a negative value. Since the absolute value of a number is always non-negative, the result of the operation will always be larger than a negative value. This type of problem results in an answer of "all real numbers". Let's look at an example.
Example 5: Solve each inequality, write in interval notation, graph the interval.
-9|1 - 9x| + 3 < 75
Let's begin by isolating the absolute value operation:
-9|1 - 9x| < 72
|1 - 9x| > -8
Now let's stop for a moment and think about this problem. We know that the absolute value of any number is always non-negative. This means no matter what we plug in for x, we will get a true statement. The absolute value of (1 - 9x) will always be larger than -8 since it can only be 0 or any larger number. When this occurs, we will stop and say "all real numbers".
Interval notation:
(-∞, ∞)
Graphing the Interval: ### Solving Absolute Value Inequalities with No Solution

Additionally, we have a case where we don't have a solution. This will occur when the absolute value operation is said to be less than 0. Since the absolute value of a number is always non-negative, this is not possible and we won't have a solution. Let's look at an example.
Example 6: Solve each inequality, write in interval notation, graph the interval.
6 - 5|-10x + 1| > 51
Let's begin by isolating the absolute value operation:
-5|-10x + 1| > 45
|-10x + 1| < -9
No matter what we replace x with, the absolute value of (-10x + 1) can never be less than -9. For this problem, we will say there is "no solution" or we can write the symbol for the empty set "∅" to show that our solution set contains no elements.

#### Skills Check:

Example #1

Solve each inequality. $$6|{-}2x - 1| - 2 ≥ 52$$

A
$$x ≤ -\frac{8}{5}\hspace{.1em}\text{or}\hspace{.1em}x ≥ 0$$
B
$$x ≤ -\frac{9}{5}\hspace{.1em}\text{or}\hspace{.1em}x ≥ 4$$
C
$$x ≤ -5 \hspace{.1em}\text{or}\hspace{.1em}x ≥ 4$$
D
$$-5 ≤ x ≤ 4$$
E
$$-2 ≤ x ≤ \frac{11}{5}$$

Example #2

Solve each inequality. $$|8x - 5| + 9 ≤ 52$$

A
$$x ≤ -3 \hspace{.1em}\text{or}\hspace{.1em}x ≥ 9$$
B
$$-3 ≤ x ≤ 10$$
C
$$-1 ≤ x ≤ 3$$
D
$$-\frac{19}{4}≤ x ≤ 6$$
E
$$x ≤ -\frac{19}{4}\hspace{.1em}\text{or}\hspace{.1em}x ≥ 6$$

Example #3

Solve each inequality. $$-\frac{1}{2}\left|\frac{1}{2}x + \frac{21}{8}\right| - \frac{3}{2}≤ -\frac{63}{16}$$

A
$$\frac{1}{2}≤ x ≤ 1$$
B
$$\text{No Solution}$$
C
$$-15 ≤ x ≤ \frac{9}{2}$$
D
$$x ≤ -\frac{3}{2}\hspace{.1em}\text{or}\hspace{.1em}x ≥ 8$$
E
$$x ≤ -15 \hspace{.1em}\text{or}\hspace{.1em}x ≥ \frac{9}{2}$$         