# SOLVING ABSOLUTE VALUE INEQUALITIES

The general form of an absolute value inequality is

|ax + b|  k

or

|ax + b| ≥ k

Method 1 : (Less Than or Equal to)

Solve the absolute value inequality given below

|x + 2| ≤ 3

Solution :

We can solve the absolute value inequality |x + 2| ≤ 3 as shown below. Let us graph the solution of the first branch x ≤ 1. Let us graph the solution of the second branch x ≥ -5. If we combine the above two graphs, we will get a graph as given below. From the above graph, the solution for |x + 2| ≤ 3 is

-5 ≤ x ≤ 1

Method 2 : (Greater Than or Equal to)

Solve the absolute value inequality given below

|x - 3| ≥ 1

Solution :

We can solve the absolute value inequality |x - 3| ≥ 1 as shown below. Let us graph the solution of the first branch  x ≥ 4 Let us graph the solution of the second branch  x ≤ 2 If we combine the above two graphs, we will get a graph as given below. From the above graph, the solution for |x - 3| ≥ 1 is

(-∞, 2] U [3, +∞)

## Solved Examples

Solve the following absolute value inequalities :

Example 1 :

|2x + 1|  5

Solution :

Solve :

2x + 1 ≤ 5  or  2x + 1 ≥ -5

2x ≤ 4  or  2x ≥ -6

x ≤ 2  or  x ≥ -3

Hence, the solution is

-3 ≤ x ≤ 2

Example 2 :

|3x + 5|  7

Solution :

Solve :

3x + 5 ≥ 7  or  3x + 5 ≤ -7

3x ≥ 2  or  3x ≤ -12

≥ 2/3  or  x ≤ -4

Hence the solution is

(-∞, -4] U [2/3, +∞)

Example 3 :

|x - 1| + 2  5

Solution :

Solve :

|x - 1| + 2  5

Subtract 2 from each side.

|x - 1| ≤ 3

x - 1 ≤ 3  or  x - 1 ≥ -3

≤ 4  or  x ≥ -2

Hence the solution is

-2 ≤ x ≤ 4

Example 4 :

|2x - 3| - 5 ≥ 7

Solution :

Solve :

|2x - 3| - 5 ≥ 7

|2x - 3| ≥ 12

2x - 3 ≥ 12  or  2x - 3 ≤ -12

2x ≥ 15  or  2x ≤ -9

≥ 15/2  or  x ≤ -9/2

Hence the solution is

(-∞, -9/2] U [15/2, +∞)

Example 5 :

2|x + 1|  6

Solution :

Solve :

2|x + 1| ≤ 6

Divide each side by 2.

|x + 1| ≤ 3

x + 1 ≤ 3  or  x + 1 ≥ -3

≤ 2  or  x ≥ -4

Hence the solution is

-4 ≤ x ≤ 2

Example 6 :

5|x - 3|  15

Solution :

Solve :

5|x - 3|  15

Divide each side by 5.

|x - 3| ≥ 3

x - 3 ≥ 3  or  x - 3 ≤ -3

≥ 6  or  x ≤ 0

Hence the solution is

(-∞, 0] U [6, +∞)

Example 7 :

2|x + 3| + 5 ≤ 13

Solution :

Solve :

2|x + 3| + 5  13

Subtract 5 from each side.

2|x + 3| ≤ 8

Divide each side by 2.

|x + 3| ≤ 4

x + 3 ≤ 4  or  x + 3 ≥ -4

≤ 1  or  x ≥ -7

Hence the solution is

-7 ≤ x ≤ 1

Example 8 :

5|x +7| - 2 ≥ 18

Solution :

Solve :

5|x +7| - 2 ≥ 18

5|x +7| ≥ 20

Divide each side by 5.

|x +7| ≥ 4

x + 7 ≥ 4  or  x + 7 ≤ -4

≥ -3  or  x ≤ -11

Hence the solution is

(-∞, -11] U [-3, +∞)

Example 9 :

|x + 3| < 13

Solution :

Solve :

|x + 3| < 13

x + 3 < 13  or  x + 3 > -13

x < 10  or  x > -16

Hence the solution is

-16 < x < 10

Example 10 :

|x +7| > 18

Solution :

Solve :

|x +7| > 18

x + 7 > 18  or  x + 7 < -18

x > 11  or  x < -25

Hence the solution is

(-∞, -25) U (11, +∞)

Kindly mail your feedback to v4formath@gmail.com

## Recent Articles 1. ### Representing a Decimal Number

Apr 01, 23 11:43 AM

Representing a Decimal Number

2. ### Comparing Irrational Numbers Worksheet

Mar 31, 23 10:41 AM

Comparing Irrational Numbers Worksheet