Absolute Value Equation Calculator

Absolute Value Equation Calculator

Solve real equations of the form a|x + b| = c and identify whether there are two solutions, one solution, no real solution, or all real numbers.

Last updated: August 2026 | By Summa Calculator

Absolute Value Equation Solver
1 |x + 0| = 5
Solutions
x₁ = 5
x₂ = -5

What This Solver Handles

This calculator solves equations with one absolute-value expression in the specific form a|x + b| = c.

The expression |x + b| is the distance from x to -b. After accounting for the coefficient a, the equation asks which points lie a fixed distance from that center.

It does not solve arbitrary typed equations, equations with absolute-value expressions on both sides, nonlinear expressions inside the bars, or systems of equations.

Why the Number of Solutions Changes

When a ≠ 0, divide by a:

|x + b| = c / a
c / a > 0

Two solutions: x = -b ± c/a.

c / a = 0

One solution: x = -b.

c / a < 0

No real solution, because absolute value cannot be negative.

A negative coefficient a is valid. What matters is the sign of the isolated value c/a, not the sign of c by itself.

Worked Example: 2|x - 3| = 10

2|x - 3| = 10
|x - 3| = 5
x - 3 = 5 or x - 3 = -5
x = 8 or x = -2

Check: 2|8 - 3| = 10 and 2|-2 - 3| = 10, so both values satisfy the original equation.

The Special Case a = 0

If a = 0, the left side is zero for every real x, regardless of b. The equation becomes 0 = c.

Therefore, a = 0, c = 0 gives all real numbers, while a = 0, c ≠ 0 gives no real solution.

Common Questions

Why are there usually two answers?

A positive distance from the center -b can be reached on either side, giving x = -b + c/a and x = -b - c/a after the absolute value is isolated.

Can a negative coefficient be used?

Yes. Divide by a first. If c/a is non-negative, real solutions may exist; if c/a is negative, there is no real solution.

Does the solver accept decimals?

Yes. It accepts finite decimals, zero, negative values, and scientific notation for a, b, and c.

Should I check the answers?

Yes. Substituting each reported x into the original equation is a direct way to confirm the equality.

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