### Solving Inequalities Containing Absolute Value

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:

1. Inverse operations: None to reverse.
2. Separate: x + 1 = 3 or x + 1 = - 3.
3. Solve: x = 2 or x = - 4.
4. Check: | 2 + 1| = 3 ? Yes. | - 4 + 1| = 3 ? Yes.
Thus, the critical points are x = 2 and x = - 4. Graph these as open circles: Test regions:
Left: x = - 5: | - 5 + 1| < 3 ? No.
Middle: x = 0: | 0 + 1| < 3 ? Yes.
Right: x = 3: | 3 + 1| < 3 ? No.
Graph the inequality:

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

Solve 4| 2x - 1| = 20:

1. Inverse operations: | 2x - 1| = 5.
2. Separate: 2x - 1 = 5 or 2x - 1 = - 5.
3. Solve: x = 3 or x = - 2.
4. Check: 4| 2(3) - 1| = 20 ? Yes. 4| 2(- 2) - 1| = 20 ? Yes.
Thus, the critical points are x = 3 and x = - 2. Graph these as closed circles: Test regions:
Left: 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 the inequality: