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Absolute Value Equation Calculator

Enter the coefficients of your absolute value equation in the form |ax + b| + c = d and this calculator solves it instantly. It finds both roots, shows the step-by-step split into positive and negative cases, tells you whether there are two solutions, one solution, or no solution, and plots the solutions on a number line. Switch to inequality mode to solve |ax + b| + c < d or > d as well.

Your details

Choose equation to find exact roots; choose an inequality to find the solution interval.
The number multiplying x inside the absolute value bars. For |x + 3| = 5, a = 1.
The constant added to ax inside the absolute value bars. For |2x - 1| = 7, b = -1.
A number added to the absolute value expression on the left side. For |2x+3| + 1 = 8, c = 1. Set to 0 if there is no outside constant.
The value on the right side of the equation or inequality. For |2x+3| = 7, d = 7.
Solution typeSolution found
Two solutions

Whether the equation has two, one, or no real solutions.

Solution x₁x = 2
Solution x₂x = -5
Isolated |ax+b|7
Solution interval-
Solution x₁-5
Solution x₂2

Result: Two solutions.

  • The equation |2x + 3| = 7 splits into two linear equations: 2x + 3 = 7 and 2x + 3 = -7.
  • Both solutions must be verified by substituting back; absolute value equations can produce extraneous roots if d - c is incorrectly negative.
  • The two solutions are symmetric around x = -3/2, which is where |2x + 3| = 0.

Next stepVerify each solution by substituting it back into the original equation. If the result equals d, the solution is valid.

Formula

ax+b+c=d    ax+b=dc    ax+b=(dc) or ax+b=(dc)|ax + b| + c = d \implies |ax + b| = d - c \implies ax + b = (d - c) \text{ or } ax + b = -(d - c)

Worked example

Solve |2x + 3| + 1 = 8. Step 1: subtract 1 from both sides to get |2x + 3| = 7. Step 2: split into 2x + 3 = 7 and 2x + 3 = -7. Step 3: solve each - from the first, 2x = 4 so x = 2; from the second, 2x = -10 so x = -5. Both solutions check out.

What is an absolute value equation?

An absolute value equation contains the absolute value operator |...|, which returns the non-negative magnitude of any real number. The equation |ax + b| = k has two solutions when k is positive, one solution when k is zero, and no real solutions when k is negative, because absolute values are always greater than or equal to zero. Understanding this three-case structure is the foundation of solving every absolute value problem.

How to solve |ax + b| + c = d step by step

The standard method has three steps. First, isolate the absolute value on one side: subtract c from both sides to get |ax + b| = d - c. Let k = d - c. If k < 0, stop - there is no solution. Second, split the equation into two linear cases: Case 1 is ax + b = k and Case 2 is ax + b = -k. Third, solve each linear equation for x by applying standard algebra. Case 1 gives x = (k - b) / a. Case 2 gives x = (-k - b) / a. Always verify both answers by substituting back into the original equation.

Solving absolute value inequalities

Absolute value inequalities follow a similar split strategy but produce interval solutions instead of discrete points. For |ax + b| < k (with k positive), the solution is a bounded interval: -k < ax + b < k, which simplifies to the open interval between the two boundary points. For |ax + b| > k, the solution is two unbounded rays: ax + b < -k or ax + b > k. The boundary points themselves are found by solving the corresponding equation |ax + b| = k, giving x values that separate the solution from the non-solution region.

Common mistakes and how to avoid them

The most frequent errors are: (1) Forgetting to isolate the absolute value first before splitting into cases - if c is non-zero, move it to the right side before splitting. (2) Forgetting the negative case - |ax + b| = k produces two equations, not one. (3) Failing to verify solutions - occasionally an algebraic manipulation produces a value that does not actually satisfy the original equation (an extraneous solution), so always substitute back. (4) Assuming every absolute value equation has two solutions - if k = 0 there is only one, and if k < 0 there are none.

Absolute value equation types and their solutions

ConditionNumber of solutionsForm of solutionsExample
k > 0 2 solutions x = (k - b)/a and x = (-k - b)/a|x - 3| = 5 gives x = 8, x = -2
k = 0 1 solution x = -b/a|x - 3| = 0 gives x = 3
k < 0 No solution None (|...| cannot be negative)|x - 3| = -2 has no solution

Summary of all cases for |ax + b| = k (where k = d - c after isolating the absolute value).

Frequently asked questions

Why does |ax + b| = k give two solutions?

The absolute value of a number equals k whenever the number itself is k or -k. So |ax + b| = k is really two equations in disguise: ax + b = k (the positive case) and ax + b = -k (the negative case). Each linear equation gives one value of x, so the original absolute value equation has two solutions when k is positive.

What does it mean when there is no solution?

If the right-hand side, after isolating the absolute value, is a negative number, the equation has no solution. An absolute value can never be negative (it is always zero or positive), so no value of x can satisfy |ax + b| = negative number. For example, |x + 5| = -3 has no solution.

How do I handle |ax + b| + c = d when c is not zero?

Subtract c from both sides first to isolate the absolute value: |ax + b| = d - c. Then proceed with the standard two-case split using k = d - c. If d - c is negative, there is no solution. This calculator does this isolation step automatically.

How does solving an absolute value inequality differ from an equation?

An equation |ax + b| = k gives discrete point solutions. An inequality gives an interval. For |ax + b| < k the solution is the open interval between the two boundary points (the values where equality holds). For |ax + b| > k the solution is the union of two rays extending outward from those boundary points. The boundary points are found by solving the equation first.

What if the coefficient a is negative?

The calculator handles negative a correctly. For example, |-2x + 6| = 8 gives Case 1: -2x + 6 = 8, so x = -1; and Case 2: -2x + 6 = -8, so x = 7. The formula x = (k - b) / a works for any non-zero a, positive or negative. The only restriction is a cannot be zero, because then there is no variable inside the absolute value.

How do I check whether my solution is correct?

Substitute each solution back into the original equation and verify that both sides are equal. For |2x + 3| = 7 with solution x = 2: |2(2) + 3| = |7| = 7, which matches. For x = -5: |2(-5) + 3| = |-7| = 7, which also matches. Both solutions are confirmed.

Sources

Written by Dr. Rajiv Menon, PhD Applied Mathematician · Bengaluru, India

Applied mathematician bridging algebraic theory and computational tools for students, engineers, and everyday problem-solvers.

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