# Absolute Value Inequalities

Absolute value inequality is a type of inequality that contains an absolute or mod (modulus) value sign with a variable inside it. The value of the variable represents its distance from the origin, which can be plotted on a number line.

## Types

Now, depending on whether the expression inside the modulus is positive or negative, there can be two possible cases:

1. With < or â‰¤ symbol
2. With > or â‰¥ symbol

## Rules for Solving

### Inequalities with < or â‰¤

Suppose we consider the absolute value inequality |x| < 6. Here, the distance from the origin to the variable â€˜xâ€™ is less than 6 units.

To solve |x| < 6, we follow the below steps.

Step 1: Splitting the Inequality into Two Cases

The inequality is already simplified. So, |x| < 6 â‡’ -x < 6 and x < 6

Step 2: Determining the Range of the Variable

Using the inversion property in the first inequality to find the value of the variable â€˜x,â€™

-x < 6 â‡’ x > -6

Thus, |x| < 6 â‡’ x > -6 and x < 6, which is written in the set notation as {x | -6 < x < 6} and in the interval notation as (-6, 6)

Step 3: Graphing the Solution

When it is plotted on the number line, we get:

In general, for any real numbers x and y, if |x| < y, then -y < x < y. In this case, the solution always shows an intersection between the two.

Solve and Graph the solution set of the inequality |2x – 1| + 5 â‰¤ 14

Solution:

Given, |2x – 1| + 5 â‰¤ 14
Step 1: Simplifying
Using the subtraction property, we get
|2x – 1| + 5 – 5 â‰¤ 14 – 5
â‡’ |2x – 1| â‰¤ 9
Step 2: Splitting the Inequality into Two Cases
By splitting the above absolute value inequality into two, we get
2x – 1 â‰¥ -9 and 2x – 1 â‰¤ 9
Step 3: Determining the Range of the Variable
Now, 2x – 1 â‰¥ -9 and 2x – 1 â‰¤ 9
â‡’ 2x – 1 + 1 â‰¥ -9 + 1 (by addition property) and 2x – 1 + 1 â‰¤ 9 + 1 (by addition property)
â‡’ 2x â‰¥ -8 and 2x â‰¤ 10
â‡’ x â‰¥ -4 (by division property) and x â‰¤ 5 (by division property)
â‡’ x â‰¥ -4 and x â‰¤ 5
Step 4: Graphing the Solution
Here, we observe that the solution is [-4, 5], and the graph is formed with the â€˜closedâ€™ circles for weak inequalities.

Solve the inequality |6x| – 3 < 9

Solution:

Given, |6x| – 3 < 9
Step 1: SImplifying
|6x| – 3 + 3 < 9 + 3
â‡’ |6x| < 12
Step 2: Splitting the Inequality into Two Cases
By splitting the above absolute value inequality into two separate inequalities, we get
6x > -12 and 6x < 12
Step 3: Determining the Range of the Variable
Now, 6x > -12 and 6x < 12
â‡’ x > -2 and x < 2 (by division property)
Thus, the solution set is {x | -2 < x < 2} or (-2, 2)

### Inequalities With > or â‰¥

Now, let us solve the inequality |x| > 6.

|x| > 6 means the distance from the origin to the variable â€˜xâ€™ is greater than 6 units.

To solve |x| > 6, the steps are:

Step 1: Splitting the Inequality into Two Cases

The inequality is already simplified. So, |x| > 6 â‡’ -x > 6 or x > 6

Step 2: Determining the Range of the Variable

Using the inversion property in the first inequality, we get -x > 6 â‡’ x < -6

Thus, |x| > 6 â‡’ x < -6 or x > 6.

The solution set is written in the set builder notation as {x | x < -6 or x > 6} and in the interval notation as (-âˆž, -6) âˆª (6, âˆž)

Step 3: Graphing the Solution

When it is plotted on the number line, we get

In general, for any real numbers x and y, if |x| > y, then x < -y or x > y.

Solve and graph the solution set of the inequality 5|4 – 3x| > 10

Solution:

Given, 5|4 – 3x| > 10
Step 1: Simplifying
Using the division property, we get
|4 – 2x| > 2
Step 2: Splitting the Inequality into Two Cases
By splitting the above absolute value inequality into two separate inequalities, we get
4 – 2x < -12 or 4 – 2x > 12
Step 3: Determining the Range of the Variable
Now, 4 – 2x < -12 or 4 – 2x > 12
â‡’ 4 – 2x – 4 < -12 – 4 or 4 – 2x – 4 > 12 – 4 (by subtraction property)
â‡’ -2x < -16 or -2x > 8
â‡’ -x < -8 or -x > 4 (by division property)
â‡’ x > 8 or x < -4 (by inversion property)
Thus, the solution set is {x | x < -4 or x > 8} or (-âˆž, -4) âˆª (8, âˆž)
Step 4: Graphing the Solution

Solve and graph the solution set of the inequality: |2x + 5| â‰¥ 9

Solution:

Here, |2x + 5| â‰¥ 9
Step 1: Splitting the Inequality into Two Cases
By splitting the above inequality into two separate inequalities, we get
2x + 5 â‰¤ -9 or 2x + 5 â‰¥ 9
Step 2: Determining the Range of the Variable
Now, 2x + 5 â‰¤ -9 or 2x + 5 â‰¥ 9
â‡’ 2x +5 – 5 â‰¤ -9 – 5 (by subtraction property) or 2x + 5 – 5 â‰¥ 9 – 5 (by subtraction property)
â‡’ 2x â‰¤ -14 or 2x â‰¥ 4
â‡’ x â‰¤ -7 (by division property) or x â‰¥ 2 (by division property)
â‡’ x â‰¤ -7 or x â‰¥ 2
Step 3: Graphing the Solution
Since weak inequalities are used here, circles are â€˜closed,â€™ and the solution in the interval notation is (-âˆž, -7] âˆª [2, âˆž).Â