Quick answer

If |expression| = k and k ≥ 0, solve expression = k and expression = −k. If k < 0, there is no real solution.

Formula

  • |x| = a → x = a or x = −a (when a ≥ 0)
  • |ax + b| = c → ax + b = c or ax + b = −c

Introduction

Absolute value equations appear in algebra, physics, and data problems where magnitude is fixed but sign is unknown. The bars hide whether the inside is positive or negative.

The reliable method is isolate, split, solve each branch, and check. Skipping the check invites extraneous roots when you square both sides.

The Absolute Value Calculator evaluates |x| for checks, while this page teaches the two-branch method. Compare with absolute value inequalities when the symbol is < or > instead of =.

Why two cases?

Both 5 and −5 are five units from zero, so |x| = 5 has two solutions unless the right side is zero.

Inside more complex expressions, the same idea applies: whatever is between the bars can be positive or negative while the outer magnitude stays fixed.

If the right side is negative, like |x| = −4, there is no real solution because a magnitude cannot equal a negative number.

Graphical reasoning appears in the absolute value graph article: a horizontal line y = k crosses the V-shape in zero, one, or two points.

Core templates

  • |x| = a → x = ±a when a ≥ 0
  • |x| = a → no solution when a &lt; 0
  • |x| = 0 → x = 0
  • |ax + b| = c → ax + b = c or ax + b = −c (when c ≥ 0)

Write both equations before you solve either one. Mixing algebra across branches causes lost solutions or extra ones.

When |ax + b| = 0, you get a single linear equation ax + b = 0 instead of two distinct cases.

After linear equations, study interval answers where solutions are ranges, not isolated points.

Solving procedure

  1. Isolate the absolute value. Get |something| alone on one side when possible. Clear fractions or parentheses first.
  2. Split into two equations. Set the inside equal to k and to −k when k ≥ 0.
  3. Solve each linear equation. Simplify separately; do not mix the branches on the same line.
  4. Check every candidate. Substitute back into the original equation. Discard values that fail.

Worked equations

Solve |2x − 1| = 7. Then 2x − 1 = 7 gives x = 4, and 2x − 1 = −7 gives x = −3. Both satisfy the original equation.

Solve |x| = 0. Only x = 0 works.

Solve |x + 2| = −1. No real solution because |x + 2| ≥ 0 cannot equal −1.