Absolute Value Equation Calculator (solve |ax + b| = c or |ax + b| = cx + d)

Understanding the absolute value equation solver — concept and principle

This calculator solves equations of the form |ax + b| = c or, more generally, |ax + b| = cx + d. It splits the equation into its two cases, solves each, and rejects any candidate that would make the right side negative. Use it to check algebra homework step by step.

How it is calculated

|A| = B (with B ≥ 0) means A = B or A = −B.

Case 1: ax + b = cx + d, so x = (d − b) ÷ (a − c)

Case 2: ax + b = −(cx + d), so x = −(b + d) ÷ (a + c)

• A candidate is kept only if the right side cx + d is 0 or more at that x; otherwise it is extraneous and listed as rejected.
• With c = 0 the right side is the constant d: no solution if d < 0, one solution if d = 0, two if d > 0.
• If a case reduces to an identity, every x on one side of −b/a works.

The principle

An absolute value is a distance and cannot be negative. So |A| equals a non-negative number B exactly when A is B units from zero, on either side: A = B or A = −B. Each case is an ordinary linear equation. When the right side contains x, solving the cases can produce a value at which cx + d is negative; such a value cannot satisfy the original equation, which is why every candidate must be checked.

Worked example

Default inputs: a = 2, b = −3, c = 0, d = 7, which is |2x − 3| = 7.

1. The right side is the constant 7, which is positive, so two solutions are expected.
2. Case 1: 2x − 3 = 7, so 2x = 10 and x = 5.
3. Case 2: 2x − 3 = −7, so 2x = −4 and x = −2.
4. Check: |2·5 − 3| = |7| = 7 and |2·(−2) − 3| = |−7| = 7. None rejected.
5. Solutions: x = −2; x = 5, number of solutions 2.

How to read the result

On a graph, the solutions are where the V-shaped graph of y = |ax + b| meets the line y = cx + d. A V and a line can meet twice, once, or not at all, and can share a whole ray when the line lies along one arm of the V. If the calculator reports a rejected value, that is where the line would meet the reflected, upside-down part of the V that does not exist.

Common mistakes

• Solving only the positive case and missing the second solution.
• Forgetting to check candidates when x appears on the right side.
• Writing |2x − 3| = −7 as having solutions; an absolute value cannot equal a negative number.
• Distributing the minus sign only to the first term in case 2.

Frequently asked questions

How many solutions can an absolute value equation have?

For |ax + b| = d: none if d < 0, one if d = 0, two if d > 0 (with a ≠ 0).

What is an extraneous solution?

A value produced by the algebra that does not satisfy the original equation.

How do I solve |x − 4| = 2x?

Enter a = 1, b = −4, c = 2, d = 0. Case 1 gives x = −4, which is rejected because 2x would be negative; case 2 gives x = 4/3.

Sources: OpenStax, College Algebra 2e (2021), §2.6 Other Types of Equations