How this instrument works
|x − a| = b asks a single question: which numbers sit exactly b units from the point a on the number line? Distance can run in either direction, so the bars split into two ordinary linear equations, x − a = b and x − a = −b, and solving each gives the pair x = a + b and x = a − b. Nothing about absolute value is being solved directly; the bars are removed by naming both directions a number can travel from a, and each direction is then handled as plain algebra.
Picture a on a ruler and mark two ticks, one b units to its right and one b units to its left — those two ticks are the entire solution set, and no third point works, because a distance is a single non-negative number, not a range. When b is positive the two ticks are distinct. Set b to zero and both formulas collapse onto a itself, giving one solution instead of two. Ask for a negative distance, b < 0, and the equation has no solution at all, since no real quantity travels a negative number of units from anything.
The word absolute here traces back to Gauss, who wrote of a complex number's magnitudo absoluta in 1801 to mean its distance from zero in the plane; restricted to the real line, that is exactly what these bars measure. A close relative, |x − a| ≤ b, asks the same distance question but accepts everything no farther than b from a — a filled interval of infinitely many points rather than this equation's two isolated ones, and that split-versus-filled contrast is the one worth keeping straight between the two.
- Enter the reference point in the a field — the number inside the bars that both solutions sit around, as in |x − a| = b.
- Enter the required distance in the b field — how far each solution sits from a; it must be zero or positive for a real answer to exist.
- Read Solution 1 (x = a + b) for the point b units above a.
- Read Solution 2 (x = a − b) for the point b units below a, and check that a negative b produces no usable pair.
Worked example — solving |x − 3| = 5
Set a = 3 and b = 5 into |x − a| = b, meaning: which numbers are exactly 5 units from 3? Splitting the bars gives two linear equations, x − 3 = 5 and x − 3 = −5, and solving each in turn returns the pair of answers — Solution 1 reads x1 = a + b = 3 + 5 = 8, and Solution 2 reads x2 = a − b = 3 − 5 = −2.
Check both answers against the original equation: |8 − 3| = |5| = 5, and |−2 − 3| = |−5| = 5 — both land exactly 5 units from 3, one on each side, confirming the pair is complete and that no third solution is hiding anywhere else on the line.
Questions
How many solutions does an absolute value equation have?
Two, in general — one for each direction from a, unless b = 0 (a single solution, x = a) or b is negative (no real solution, since a distance cannot be negative). For |x − a| = b with b > 0, the pair x = a + b and x = a − b is always the complete answer, never more and never fewer.
What is the most common mistake when solving |x − a| = b?
Solving only the branch x − a = b and stopping there, missing x − a = −b entirely. The bars represent two directions away from a, and dropping the negative case throws away half the solution set; both linear equations must be solved and both results checked back into the original equation.
Why does |x − a| = b split into two separate equations?
Because absolute value strips the sign of whatever sits inside the bars, so x − a must equal either +b or −b before the bars were applied, and both possibilities are equally consistent with |x − a| = b. Algebra can't distinguish them without splitting into the two linear cases and solving each on its own.
What happens if b is negative?
There is no solution. |x − a| measures a distance, and a distance is never negative, so asking for it to equal a negative number describes a situation no real x can satisfy. A calculator built on this formula should report no valid pair for such an input rather than force an answer.
How is an absolute value equation different from an absolute value inequality?
An equation like |x − a| = b names exactly two points, a + b and a − b. Change the equals sign to ≤ and the same expression describes every point in between as well — a solid interval rather than two isolated ones — which is why the two problems, built from the same bars, call for different solving strategies entirely.
Can I solve |x − a| = b by squaring both sides instead?
Yes — squaring gives (x − a)^2 = b^2, a quadratic that factors into the same two roots, a + b and a − b, since squaring removes the sign ambiguity the bars represent. It works, but it is more arithmetic than a linear expression needs, and squaring can introduce extraneous roots in messier equations, so the direct case split is usually faster and safer.