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Last modified on May 13th, 2024

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.

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

- With
**< or â‰¤**symbol - With
**> or â‰¥**symbol

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**

Using the addition property,

|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)

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

Last modified on May 13th, 2024