The absolute value of a number is its distance measured on the number line from 0. Absolute value inequalities are inequality statements containing absolute value expressions.

The simplest form of absolute value inequalities are of one of the following forms:

1. | x | < k  or | x | $\leq$ k, which can be equivalently represented by correspondingly as -k < x < k and -k $\leq$ x $\leq$ k
2. | x | > k  or | x | $\geq$ k, which are written correspondingly in equivalent form as x < -k  or x > k and x $\leq$ -k or x $\geq$ k.

Absolute value inequalities are solved algebraically, by taking the expression inside the absolute value notation as both positive and negative. The inequalities are also solved by writing the inequality in expanded form. The solution of an absolute value inequality is often graphically on a number line. The graphical representation of absolute value inequalities are shown below.

## Absolute Inequalities Solver

The key to solving an absolute value inequality lies in writing it as an equivalent compound inequality. The equivalent form for the general four types of inequalities and the condition that makes the inequality a true statement are given below:

 Inequality Equivalent Compound inequality Condition for the inequality to be true | ax + b | < k -k < ax + b < k k > 0 | ax + b | $\leq$ k -k $\leq$ ax + b $\leq$ k k $\geq$ 0 | ax + b | > k ax + b < -k or ax + b > k k > 0 | ax + b | $\geq$ k ax + b $\leq$ -k or ax + b $\geq$ k k $\geq$ 0

## Solve Absolute Value Inequalities

Given below are some of the examples on solving absolute value inequalities.

### Solved Examples

Question 1: Solve | 2y - 3 | < 11
Solution:
- 11 < 2y - 3 < 11                    Equivalent compound inequality
- 8 < 2y < 14                          Add 3 to each expression in the inequality
- 4 < y < 7                             Divide each expression by 2 to get the final solution.
Hence, the solution to the given inequality is all real numbers greater than - 4 but less than 7.
The solution can be written in interval form as (-4, 7).

Question 2: Solve | 3x + 5 | $\geq$ 17
Solution:
The equivalent form for the above inequality is the disjoint compound inequality.
3x + 5 $\leq$ - 17    or    3x + 5 $\geq$ 17
3x $\leq$ - 22          or    3x  $\geq$ 12                    Subtract 5 from each inequality

x $\leq$ $\frac{-22}{3}$   or    x $\geq$ 4                   Divide each inequality by 3.

Hence, the solution for the inequality is all real numbers which are either less than or equal to $\frac{-22}{3}$ or greater than or equal to 4.

The solution is written in interval form as $(-\infty,$$\frac{-22}{3}$$ ]$ U $[4,\infty )$

## Graphing Absolute Value Inequalities

The equivalent compound inequality is used for the graphing purpose. The graphs of the original inequalities which are solved algebraically earlier and their solutions are shown below:
The original inequality | 2y - 3 | < 11 is graphed below.

The solution we got algebraically for the above inequality - 4 < y < 7 is graphed below.

The original form of other solved inequality | 3x + 5 | $\geq$ 17 is graphed below.

The solution graph x $\leq$ $\frac{-22}{3}$   or    x $\geq$ 4 is shown as follows:

The solution of an absolute value inequality is better understood when graphed on the number line.

## Properties of Absolute Value Inequalities

An important property of absolute value inequalities is expressed in the form of triangle inequalities.
1. | a + b | $\leq$ | a | + | b |
2. | a - b | $\geq$ | a | - | b |

We can verify these properties substituting some values for a and b.
Let a = 7 and b = 3
| a + b | = | 7 + 3 | = | 3 | + | 7 | = | a | + | b |
| a - b | = | 7 - 3 | = 4 =  | 7 | - | 3 |

Let a = -5 and b = 8
| a + b | = | -5 + 8 | = 3 < 13 = | -5 | + | 8 |
| a - b | = | -5 - 8 | = 13 > -3 = | -5 | - | 8 |.

Reciprocal Properties:

| x | > k  <=> $\frac{1}{|x|}$ < $\frac{1}{k}$, where k > 0.
| x | < k  <=> $\frac{1}{|x|}$ > $\frac{1}{k}$, where k > 0.

## Absolute  Value Inequalities Examples

Given below are some of the examples on absolute value inequalities.

### Solved Examples

Question 1: Solve $\frac{1}{|3x+1|}$ $\geq$ 4
Solution:
|3x + 1| $\leq$ $\frac{1}{4}$             (Reciprocal property of absolute value inequalities)

-$\frac{1}{4}$ $\leq$ 3x + 1 $\leq$ $\frac{1}{4}$  (Equivalent Compound inequality)

-$\frac{1}{4}$ - 1 $\leq$ 3x $\leq$ $\frac{1}{4}$ - 1

-$\frac{5}{4}$ $\leq$ 3x $\leq$ -$\frac{3}{4}$

-$\frac{5}{12}$ $\leq$ x $\leq$ -$\frac{1}{4}$    (Inequality solved.)

The solution in interval notation is [ -$\frac{5}{12}$, -$\frac{1}{4}$ ]

Question 2: Solve | $\frac{x+1}{2}$ - $\frac{2x-1}{3}$ | < $\frac{1}{6}$
Solution:
6 x | $\frac{x+1}{2}$ - $\frac{2x-1}{3}$ | < $\frac{1}{6}$ x 6            (Multiply the inequality by 6)

| 3(x + 1) - 2(2x - 1) | < 1
| 3x + 3 - 4x + 2 | < 1
| -x + 5 | < 1
-1 < -x + 5 < 1                           (Equivalent compound inequality)
- 6 < -x < - 4
4 < x < 6                                   (Inequality solved.)
The solution in interval notation is ( 4, 6 ).