The two branches give distinct roots.
Absolute Value Equation Solver
Split the equation. Follow both cases. See and verify every real solution.
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Real coefficientsEQUATION FORM · |ax + b| = c|2x + 3| = 5
Special cases & input notes
Negative c gives no solution. For a = 0, check the constant statement |b| = c. Finite decimal entries are interpreted exactly; at most 60 significant digits per entry and exponents −100 to 100. Input fractions such as 1/3 are not supported, but exact fractional answers are. Plot coordinates are approximate.
The bars create two possible signs.
When c > 0 and a ≠ 0, solve ax + b = c and ax + b = −c, then check both answers.
When c > 0 and a ≠ 0, solve ax + b = c and ax + b = −c, then check both answers.
REAL SOLUTION SET · |2x + 3| = 5x = −4 or x = 1
Two distinct solutions · both verified in the original equation.
Exact solution values
Solution count2 real roots
Case 1 · x₁1
Case 2 · x₂−4
Your roots on the number line
Equal-unit scaleCenter−b/a = −1.5±Distance from centerc/|a| = 2.5
The roots are equally far from the center. The solution set contains just the two marked points.
Case 1 · positive branch
ax + b = c2x + 3 = 5
1Subtract 3 from both sides2x = 5 − 3 = 2
2Divide both sides by 2x₁ = 2 / 2 = 1
CASE 1 SOLUTIONx₁ = 1
Substitute into the original equation|2(1) + 3| = |5| = 5Verified
Case 2 · negative branch
ax + b = −c2x + 3 = −5
1Subtract 3 from both sides2x = −5 − 3 = −8
2Divide both sides by 2x₂ = −8 / 2 = −4
CASE 2 SOLUTIONx₂ = −4
Substitute into the original equation|2(−4) + 3| = |−5| = 5Verified
Substitution checks use the exact roots before any displayed rounding.
Why are there two answers?
A GRAPHICAL CHECK
Two intersections. One equation.
The V-shaped graph y = |2x + 3| meets the horizontal line y = 5 twice: once at x = −4 and once at x = 1.
|ax + b| = c ⟺ ax + b = c or ax + b = −c
For c ≥ 0; solve by division only when a ≠ 0.
For c ≥ 0; solve by division only when a ≠ 0.
Understand every possible outcome
Check c, then aThe branches coincide: x = −b/a.
The equation cannot be satisfied.
The statement is true for every x.
Formula & interpretation
x₁ = (c − b)/a or x₂ = (−c − b)/a
For a ≠ 0 and c ≥ 0, the roots lie around −b/a, at a distance c/|a| from that center.
Do not divide by zero.
If a = 0, check the constant statement |b| = c instead. If c < 0, stop: an absolute value cannot be negative.
If a = 0, check the constant statement |b| = c instead. If c < 0, stop: an absolute value cannot be negative.
Frequently asked questions
Why do we use two equations?
For a nonnegative c, a quantity with magnitude c can be c or −c. When c=0 these are the same value.
Can this equation have one solution?
Yes. If a is nonzero and c=0, its unique solution is x=−b/a.
What if the coefficient a is zero?
The variable disappears. If |b|=c, all real x satisfy the equation. Otherwise there is no solution.
Is every point between the roots a solution?
No. An equality |ax+b|=c gives only its root points when a is nonzero. A filled interval would describe an inequality instead.
Sources & Reference Guides
OpenStaxIntermediate Algebra · Absolute Value EquationsSeparate the two cases and check the resulting roots.OpenStaxPrecalculus · Absolute Value FunctionsDistance, graphs and interpreting intersections.
Independent educational references · No institutional endorsement or expert review claimed.
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Split the equation. Follow both cases. See and verify every real solution.