An absolute value equation splits into two cases. |x − 3| = 5 becomes x − 3 = 5 and x − 3 = −5, giving x = 8 and x = −2. Two answers, because two numbers sit five units from 3. If the right-hand side is negative there are none at all — a distance is never negative.
Two cases, always
The bars come off twice: once as written, once with the other side negated.
It has to be twice because two numbers share every positive absolute value, one on each side of zero. Solving only the first case finds one of them and silently loses the other, which is the commonest way to get half marks on this topic.
Isolate the bars first. 2|x − 1| + 3 = 11 has to become |x − 1| = 4 before it splits. Splitting while the bars are still tangled in other terms produces two wrong equations rather than two right ones.
When the right-hand side is exactly zero the two cases collapse into one, because zero is the only number whose absolute value is zero. One solution, not two — and this page says so rather than printing the same root twice.
And versus or
This is the distinction the page is really for, and it is worth stating as a rule and then as a reason.
Less-than is an AND. |x − 3| < 5 means the distance from 3 is under 5, so x is caught between two bounds: −2 < x < 8. One interval, and both conditions hold at once.
Greater-than is an OR. |x − 3| > 5 means the distance is more than 5, so x is out beyond one bound or the other: x < −2 or x > 8. Two rays, and no number satisfies both.
The error to avoid is writing the second as a chain — 8 < x < −2. It looks like the first answer and it is not an answer at all: it asks for a number simultaneously above 8 and below −2, and describes the empty set.
A mnemonic that survives: less-thand and great-or. Silly, and it works.
When the right-hand side misbehaves
Three cases where the answer is decided before any algebra, and each is a different sentence.
Negative, with an equals or a less-than: nothing. No absolute value is negative, so none can equal −5 or be less than it. The solution set is empty, and no case-splitting is needed to find that out.
Negative, with a greater-than: everything. Every absolute value is at least zero, so every one of them exceeds any negative number. Every real number is a solution.
Zero: an equation has exactly one solution; |u| < 0 has none; |u| ≥ 0 has every real number; and |u| > 0 has every real number except the single point where u is zero.
These are the ones that get missed, because the instinct is to start splitting cases immediately. Checking the right-hand side first takes a second and settles four of the possibilities outright.
Reading it as a distance
|x − 3| is “the distance from x to 3”. Once you read it that way the answers stop needing algebra at all.
|x − 3| = 5 asks which numbers are exactly 5 from 3. Walk 5 either way from 3: −2 and 8. Done.
|x − 3| < 5 asks which are closer than 5 to 3 — the stretch between them. |x − 3| > 5 asks which are further — everything outside.
This reading is also why the graph of y = |x − 3| has its corner at 3, and why a tolerance is naturally written with absolute value bars: |measured − target| ≤ tolerance says exactly what a tolerance means.
Where it gets used
Tolerance and specification. A part is in spec when |measured − nominal| is within tolerance. That is an absolute value inequality, and it is the less-than case.
Error bounds. “Accurate to within 0.5” is |estimate − true| ≤ 0.5, which is how numerical methods state their guarantees.
Quality control. Flagging outliers uses the greater-than case: a reading is suspect when its distance from the mean exceeds a threshold.
Analysis. The epsilon-delta definition of a limit is two nested absolute value inequalities, which is why this topic is a prerequisite for calculus rather than a detour.
Sources and methodology
The case split and the interval conventions are curriculum standards; these are the references.
Method. Roots are computed as exact rationals, so |3x − 1| = 5 returns −4/3 rather than −1.3333333 — which matters because a fraction is the answer a marker wants. The three degenerate right-hand sides are branched on rather than derived: a negative value is refused before any algebra, because an absolute value is a distance and no distance is negative, and a zero collapses the two cases of an equation into one. The suite verifies every solution by substituting it back into the original statement, and checks the inequalities by testing a point inside each interval and a point beyond each bound. That engine is verified on every change against 140 hand-written assertions, including that every returned root satisfies the original equation by substitution across 693 generated cases, and that a midpoint satisfies each less-than while a point beyond the upper bound satisfies each greater-than. The count and the per-case breakdown are published on the formula verification page.
Read the guide
To convert a solution set between interval, set-builder and inequality notation, the Interval Notation Calculator covers all three.