Absolute Value Equation and Inequality Calculator
Solve |ax + b| = c and absolute value inequalities (<, ≤, >, ≥) step by step, with both cases, interval notation and the distance form |x − h| = d.
How it works
|u| = c (c > 0) → u = c or u = −c · |u| < c → −c < u < c · |u| > c → u < −c or u > c · |u| = negative → no solution
The absolute value |u| is the distance of u from 0, so |u| = c has two solutions when c is positive (u = c and u = −c), one when c is 0 and none when c is negative. A "less than" inequality becomes one compound (and) inequality, −c < u < c, whose solution is a single interval; a "greater than" inequality splits into two (or) pieces, u < −c or u > c, giving two rays. Dividing by |a| gives the distance form |x − h| = d: the solutions are the points at distance d from h.
Worked example
Solve |2x − 3| = 7. Split into 2x − 3 = 7 or 2x − 3 = −7. The first gives x = 5, the second x = −2, so the solution is x = −2 or x = 5. In distance form |x − 3/2| = 7/2: both answers lie 3.5 units from 1.5.
Sources: Absolute value equations — Math is Fun · Absolute value — Wikipedia
Frequently asked questions
Why does an absolute value equation usually have two answers?
Two numbers have the same distance from zero: c and −c. So the expression inside the bars can equal either one, which gives two linear equations to solve.
What if the right side is negative?
An absolute value can never be negative, so |ax + b| = −4 or |ax + b| < −4 has no solution, while |ax + b| > −4 is true for every x.
When is the answer one interval and when is it two?
“Less than” (< or ≤) keeps x close to the centre, giving one interval between the two boundary points. “Greater than” (> or ≥) pushes x away from it, giving two rays joined by ∪.
What if there is a number added outside the bars?
Isolate the absolute value first. For |2x − 3| + 4 = 11 subtract 4 to get |2x − 3| = 7, then enter a = 2, b = −3, c = 7.
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