PreCalculus

8.2. Solving inequalities with absolute values

For an inequality of the form |a| < b, the distance from the origin of the expression a is less than b. Then both (-b < a) and (a < b) must be true. This statement can be written as a double inequality:

-b < a < b

The value of a can be represented on the real line as:

                    a
-----------(=================)----------
          -b                 b

Equivalently, the inequality is true when a is in the interval (-b, b).

For, an inequality of the form |a| > b, the distance from the origin of the expression a is greater than b. Then, either (-b > a) or (a > b) must be true. The inequality is equivalent to two inequalities:

a < -b    or     b < a

The value of a can be represented on the real line as:

     a                            a
===========)-----------------(==========
          -b                 b

Equivalently, the inequality is true when a is in the range (-∞, -b) or (b, +∞).

Example, solve: |x + 2| < 4

-4 < x + 2 < 4
-6 < x < 2
The inequality is true in the interval (-6, 2).

Example, solve: |x + 4| > 2

x + 4 < -2     2 < x + 4
x < -6        -2 < x
The inequality is true in the intervals (-∞, -6), (-2, +∞).

Combined statements of inequality are possible. For example, the statement:

1 < |x + 1| < 3

is equivalent to two statements:

|x + 1| < 3
|x + 1| > 1

To solve this, the values of x for which both statements are true must be found. We may proceed as follows:

|x + 1| < 3
-3 < x + 1 < 3
-4 < x < 2

|x + 1| > 1
x + 1 < -1   or   1 < x + 1
x < -2       or   0 < x

The solution intervals can be shown graphically as:

|x + 1| < 3      (-----------------------)   
|x + 1| > 1   -----------)       (-----------
both             (-------)       (-------)   
              ---+---+---+---+---+---+---+---
                -4      -2       0       2

The combined inequality is true when the variable x satisfies both parts. That is, the statement is true in the intervals: (-4, -2), (0, 2).