To solve an inequality containing an absolute value, treat the "<",
"
≤
", ">", or "
≥
" sign as an "=" sign, and solve the equation
as in Absolute Value Equations. The resulting values of
*x*
are called boundary points or critical points.

Plot the boundary points on the number line, using closed circles if the original inequality contained a ≤ or ≥ sign, and open circles if the original inequality contained a < or > sign. If you are unsure which type of circle to use, test each critical point in the original inequality; if it satisfies the inequality, use a closed circle.

If there are 2 boundary points, the number line will be divided into 3 regions. Pick a point in each region--not a critical point--and test this value in the original inequality. If it satisfies the inequality, draw a dark line over the entire region; if one point in a region satisfies the inequality, all the points in that region will satisfy the inequality. Make sure that each region is tested, because the solution set may consist of multiple regions.

*Example 1*: Solve and graph:
| *x* + 1| < 3
.

Solve
| *x* + 1| = 3
:

- Inverse operations: None to reverse.
- Separate:
*x*+ 1 = 3 or*x*+ 1 = - 3 . - Solve:
*x*= 2 or*x*= - 4 . - Check: | 2 + 1| = 3 ? Yes. | - 4 + 1| = 3 ? Yes.

Critical Points of
| *x* + 1| < 3

Left:Graph the inequality:x= - 5 : | - 5 + 1| < 3 ? No.

Middle:x= 0 : | 0 + 1| < 3 ? Yes.

Right:x= 3 : | 3 + 1| < 3 ? No.

Graph of
| *x* + 1| < 3

*Example 2*: Solve and graph:
4| 2*x* - 1|≥20
.

Solve
4| 2*x* - 1| = 20
:

- Inverse operations:
| 2
*x*- 1| = 5 . - Separate:
2
*x*- 1 = 5 or 2*x*- 1 = - 5 . - Solve:
*x*= 3 or*x*= - 2 . - Check: 4| 2(3) - 1| = 20 ? Yes. 4| 2(- 2) - 1| = 20 ? Yes.

Critical Points of
4| 2*x* - 1|≥20

Left:Graph the inequality:x= - 3 : 4| 2(- 3) - 1|≥20 ? Yes.

Middle:x= 0 : 4| 2(0) - 1|≥20 ? No.

Right:x= 4 : 4| 2(4) - 1|≥20 ? Yes.

Graph of
4| 2*x* - 1|≥20