About the Absolute Value Equation Solver
This absolute value equation solver handles equations and inequalities of the form k·|ax + b| + m ? c, where ? is =, <, ≤, > or ≥. It first isolates the absolute value, then splits the problem into its two cases — the expression inside equals (or is bounded by) the positive value and the negative value — and solves each one, so you see exactly the working a teacher expects.
It is designed for algebra students and anyone checking homework on absolute value. Equations give zero, one or two solutions; "less than" inequalities give a single interval between two values (an AND, or compound, inequality); "greater than" inequalities give two rays heading outward (an OR inequality). Results are shown both as inequalities and in interval notation.
The tool also catches the special cases that trip people up: an absolute value can never equal a negative number, |x| < 0 has no solution, and |x| ≥ −2 is true for every real number. If k is negative, dividing by it flips the inequality sign, and the solver does this for you.
How to use the absolute value equation solver
- 1Write your problem as k|ax + b| + m ? c (k = 1 and m = 0 for a plain |ax + b|).
- 2Enter k, a, b, m and c, and pick the relation (=, <, ≤, >, ≥).
- 3Read the isolated form and the two cases.
- 4Copy the solution as inequalities or in interval notation.
- 5Use the graph: solutions are where the V-shaped line meets or lies below/above the horizontal line.
Formula and method
First isolate the absolute value by subtracting m and dividing by k; if k is negative the inequality sign reverses. With u = ax + b and R the isolated right side, the absolute value |u| is the distance of u from 0. So |u| = R means u is R units from zero in either direction, giving the two cases u = R and u = −R.
For |u| < R the distance must be less than R, which is the single compound inequality −R < u < R. For |u| > R the distance must exceed R, giving u < −R or u > R. Each case is then solved as an ordinary linear equation or inequality in x (dividing by a negative a flips the sign). If R is negative, equations and "less than" inequalities have no solution and "greater than" inequalities hold for all x.
- k, m
- Multiplier and constant outside the absolute value
- a, b
- Coefficient and constant inside: u = ax + b
- c
- Right-hand side
- R
- Isolated right side (c − m)/k
Worked examples
|2x − 3| = 7
Case 1: 2x − 3 = 7 gives x = 5. Case 2: 2x − 3 = −7 gives x = −2. Both check: |10 − 3| = 7 and |−4 − 3| = 7.
|x + 4| < 3
A "less than" inequality becomes −3 < x + 4 < 3. Subtracting 4 throughout gives −7 < x < −1 — every number within 3 units of −4.
3|x − 2| + 1 ≥ 10
Subtract 1 and divide by 3 to get |x − 2| ≥ 3. Then x − 2 ≤ −3 or x − 2 ≥ 3, so x ≤ −1 or x ≥ 5.
Negative right side: |5x + 1| = −4
An absolute value is a distance and can never be negative, so there is no x with |5x + 1| = −4. No need to solve the two cases.
Frequently asked questions
How do you solve an absolute value equation?+
Isolate the absolute value, then write two equations: the inside equals the positive right side, and the inside equals the negative right side. Solve both and check each answer in the original equation, especially when x also appears outside the bars.
Why does an absolute value equation have two solutions?+
Absolute value measures distance from zero, and two numbers are a given distance from zero: R and −R. So |u| = 5 means u = 5 or u = −5, which usually leads to two values of x.
When does an absolute value equation have no solution?+
When, after isolating the absolute value, the other side is negative — for example |x − 1| = −3. A distance cannot be negative, so there is no solution. Similarly |u| < 0 has no solution.
What is the difference between AND and OR absolute value inequalities?+
|u| < R (or ≤) is an AND inequality: u must be between −R and R, giving one interval. |u| > R (or ≥) is an OR inequality: u is below −R or above R, giving two separate rays. A memory aid is "less thAND, greatOR".
Do I flip the inequality sign with absolute values?+
You flip it when multiplying or dividing by a negative number, as with any inequality — for example dividing both sides of −2|x| > −6 by −2 gives |x| < 3. For the negative case of |u| > R you also write u < −R, which reverses the direction.