Absolute Value Solver
Solve absolute value equations and inequalities, with the split into cases shown.
for |2x − 3|, distance from zero
- 1Split into two equations.2x − 3 = 7 or 2x − 3 = −7, because both are that far from zero.
Absolute value is distance from zero, so |x| = 7 has two answers, 7 and −7. An equation splits into two cases: |2x − 3| = 7 gives 2x − 3 = 7 or 2x − 3 = −7, so x = 5 or x = −2. Inequalities split the same way.
Distance is the whole idea
|x| asks how far x sits from zero, ignoring direction. That single definition explains every rule that follows. Two numbers share each positive distance — one on each side of zero — which is why an absolute value equation normally has two solutions. Zero has only itself at distance zero, and no number has a negative distance, which is why |x| = −3 has no solution at all.
More generally |ax + b| measures how far the expression ax + b lands from zero. Solving is therefore a two-part job: work out which distances satisfy the relation, then translate each one back into a statement about x.
Why less-than and greater-than behave differently
A small distance traps you near zero, so |x| < 4 means x sits between −4 and 4 — a single interval. A large distance pushes you away from zero in either direction, so |x| > 4 means x < −4 or x > 4 — two intervals with a gap in the middle. Students often write the greater-than case as a double inequality; that is the single most common error on this topic, because no number is both less than −4 and greater than 4.
the same split drives the inequality cases
- 1 Isolate the absolute value first. For 2|x − 1| + 3 < 9, subtract 3 and divide by 2 to get |x − 1| < 3.
- 2 Check the sign of the right-hand side. A negative bound means no solution for = and <, but every real number for >, since a distance is never negative.
- 3 Split into cases. For |2x − 3| = 7 write 2x − 3 = 7 and 2x − 3 = −7; for a less-than, write the double inequality −3 < x − 1 < 3 instead.
- 4 Solve each case separately. 2x − 3 = 7 gives x = 5, and 2x − 3 = −7 gives x = −2.
- 5 Combine and check. The answer is x = 5 or x = −2; substituting back gives |7| = 7 and |−7| = 7, both correct.
The four cases at a glance
Assumes c > 0. The negative and zero bounds are the special cases the tool flags for you.
| Form | Becomes | Shape of the answer |
|---|---|---|
| |u| = c | u = c or u = −c | Two separate values |
| |u| < c | −c < u < c | One interval around the centre |
| |u| ≤ c | −c ≤ u ≤ c | One closed interval |
| |u| > c | u < −c or u > c | Two intervals with a gap |
| |u| ≥ c | u ≤ −c or u ≥ c | Two closed intervals |
| |u| = negative | Impossible | No solution |
Isolate before you split
The bars have to stand alone before the case split is valid. In 2|x − 1| + 3 < 9 the 3 and the 2 belong to the outside, so subtract and divide first; splitting while they are still attached produces an answer that does not check. And if the multiplier outside is negative, dividing by it flips the relation — −2|x| + 8 ≥ 2 becomes |x| ≤ 3, not |x| ≥ 3.
Two boundary cases are worth knowing by sight. |u| ≤ 0 can only happen when u = 0, so the interval collapses to a single point. And |u| > 0 holds everywhere except at that same point, which is why the answer reads as all real numbers with one value removed.