Absolute Value Equations & Inequalities
Absolute value measures distance from zero — so |x| = 5 has two answers, not one. Learn to split every absolute value equation into cases, isolate the bars first, and read inequalities as distance statements on the number line.
1 Absolute value is distance
The absolute value |x| is the distance from x to 0 on the number line. Distance is never negative — and a distance of 5 happens at two places: 5 and −5.
Every absolute value equation with a positive right side splits into two ordinary equations. A negative right side, like |x| = −2, has no solution — distance cannot be negative.
Inequalities are distance statements too: |x − a| < b means “x is within b of a” (between a−b and a+b), while |x − a| > b means “x is farther than b from a” (two outer rays).
2 Worked examples
Five problems covering every case you will meet.
Solve |x − 3| = 5.
The bars are already alone. Split: x − 3 = 5 or x − 3 = −5.
So x = 8 or x = −2. Check: |8−3| = 5 and |−2−3| = 5. Both work.
Solve 2|x + 1| = 10.
Divide by 2 first: |x + 1| = 5. Then split: x + 1 = 5 or x + 1 = −5.
So x = 4 or x = −6. Splitting before isolating is the classic trap — never skip the divide.
Solve |x − 2| < 3.
“Within 3 of 2”: −3 < x − 2 < 3. Add 2 everywhere.
So −1 < x < 5 — one connected interval.
Solve |x + 4| ≥ 6.
“At least 6 away from −4”: x + 4 ≥ 6 or x + 4 ≤ −6.
So x ≤ −10 or x ≥ 2. On the number line these are two rays pointing outward.
Solve |x − 5| = −2.
An absolute value equals a negative number. Distance can never be negative.
No solution. Write “no solution”, not x = 0.
3 Common mistakes
Three traps, each with the wrong version and the fix.
Wrong: |x − 3| = 5 gives x = 8 only. Right: two cases — x − 3 = 5 or x − 3 = −5, so x = 8 or x = −2.
Wrong: 2|x + 1| = 10 split as 2(x + 1) = 10 or 2(x + 1) = −10. Right: divide by 2 first: |x + 1| = 5, then split.
Wrong: |x − 2| < 3 becomes x < 5 or x > −1. Right: less-than squeezes into one interval: −1 < x < 5. “Or” belongs to greater-than.
4 Quick checks
Try these yourself, then reveal the answer.
5 Key points
Remember
- Isolate the absolute value first, then split into two cases.
- |x − a| = b (b > 0) gives two solutions: x = a + b or x = a − b.
- |x − a| < b is one interval: a − b < x < a + b.
- |x − a| > b is two rays: x < a − b or x > a + b.
- An absolute value equal to a negative number has no solution.