Absolute Value Equations and Inequalities
Absolute Value Equations
Absolute value measures distance from zero on the number line, and it is always non-negative. This geometric interpretation means that the equation |x| = 5 has two solutions — positive 5 and negative 5 — because both are exactly 5 units from zero.
Absolute value equations and inequalities require you to consider two cases: what happens when the expression inside the bars is positive, and what happens when it is negative. This lesson teaches you the systematic case-splitting approach.
You will also learn the special rules for absolute value inequalities: |x| < a describes a single interval around zero, while |x| > a describes two rays extending outward.
To solve $|ax + b| = c$ where $c \ge 0$:
$$ax + b = c \quad\text{or}\quad ax + b = -c$$
If $c < 0$, there is no solution.
$|2x - 5| = 9$
Case 1: $2x - 5 = 9 \;\Rightarrow\; x = 7$
Case 2: $2x - 5 = -9 \;\Rightarrow\; x = -2$
Solution: $x = 7$ or $x = -2$.
$|x + 3| = -4$
No solution — absolute value cannot be negative.
$3|x - 1| + 2 = 14$
- Isolate: $3|x - 1| = 12 \;\Rightarrow\; |x - 1| = 4$
- $x - 1 = 4 \;\Rightarrow\; x = 5$
- $x - 1 = -4 \;\Rightarrow\; x = -3$
Absolute Value Inequalities
- $|X| < c$ → $-c < X < c$ (AND/compound)
- $|X| > c$ → $X < -c$ or $X > c$ (OR)
$|3x + 1| \le 10$
$-10 \le 3x + 1 \le 10 \;\Rightarrow\; -11 \le 3x \le 9 \;\Rightarrow\; -\dfrac{11}{3} \le x \le 3$
$|x - 4| > 2$
$x - 4 > 2 \;\Rightarrow\; x > 6$ or $x - 4 < -2 \;\Rightarrow\; x < 2$.
Practice Problems
Show Answer Key
1. $x = 10$ or $x = -4$
2. $x = 2$ or $x = -3$
3. No solution
4. $|4x - 3| = 9$; $x = 3$ or $x = -\dfrac{3}{2}$
5. $-8 < x < -2$
6. $x \ge 4$ or $x \le -3$
7. $x = 4$ or $x = -4$
8. $x = 2$ only
9. $|x + 1| > 4$; $x > 3$ or $x < -5$
10. $x = -3$ or $x = 13$
11. All $x \ne -4$
12. $4 \le x \le 8$