Why do absolute value equations often have two solutions?

When |A| = 5, the quantity A could be 5 or it could be -5—both have an absolute value of 5. Splitting into these two cases and solving each usually yields two different values of x. However, if c = 0, there is only one solution.

What does it mean if c is negative in the equation |ax + b| = c?

It means the equation has no solution. Absolute value is never negative, so there's no value of ax + b that can have an absolute value equal to a negative number. Simply write 'no solution' without further work.

How do I know if my solution is correct?

Substitute both solutions back into the original equation. For example, if you solved |3x - 2| = 10 and got x = 4, check: |3(4) - 2| = |12 - 2| = 10 ✓. Always verify both solutions this way.

Can an absolute value equation have exactly one solution?

Yes, when c = 0. For instance, |x - 5| = 0 is true only when x - 5 = 0, so x = 5 is the unique solution. When c > 0, you typically get two solutions, but the special case c = 0 always gives exactly one.

Absolute Value Equations

Solve equations with absolute value by splitting into two cases where the expression inside equals the positive value or its negative.

|ax + b| = c \iff ax + b = \pm c

Solve a problem with Absolute Value Equations

Type the problem. The solver will use Absolute Value Equations where Absolute Value Equations is the right tool, and tell you when it is not.

What each symbol means

What Absolute Value Equations takes

Symbol	Meaning
$a$	The coefficient (multiplier) of x; determines how quickly the expression inside the absolute value changes as x changes.
$b$	The constant term added to ax; shifts the expression left or right horizontally when graphed.
$c$	The value that the absolute value expression equals; must be non-negative ( $c \geq 0$ ), otherwise no solution exists.
$x$	The unknown variable being solved for; usually produces two different values because the absolute value removes sign information.

When to use it

Use this when you need to solve an equation where a variable expression is inside absolute value bars, like $|2x - 3| = 7$ .

Level

Usually taught in: Algebra I

Worked examples

1. Solve a simple absolute value equation with small integers

Problem

Solve

|2x - 3| = 7

for x.

$2x - 3 = 7 \text{ or } 2x - 3 = -7$
Apply the definition of absolute value: if $|A| = 7$ , then $A = 7$ or $A = -7$ .
$2x = 10 \text{ or } 2x = -4$
Add 3 to both sides of each equation.
$x = 5 \text{ or } x = -2$
Divide each by 2 to isolate x.

Answer: $x = 5 \text{ or } x = -2$

We split the absolute value equation into two cases. Solving the first case $2x - 3 = 7$ gives $x = 5$ , and solving the second case $2x - 3 = -7$ gives $x = -2$ . Both values satisfy the original equation because $|2(5) - 3| = 7$ and $|2(-2) - 3| = 7$ .

2. Solve an equation with a subtraction and negative coefficient

Problem

Solve

|4 - 3x| = 10

for x.

$4 - 3x = 10 \text{ or } 4 - 3x = -10$
Rewrite using the absolute value rule: if $|A| = 10$ , then $A = 10$ or $A = -10$ .
$-3x = 6 \text{ or } -3x = -14$
Subtract 4 from both sides in each case.
$x = -2 \text{ or } x = \frac{14}{3}$
Divide by $-3$ in each case; $6 \div (-3) = -2$ and $-14 \div (-3) = \frac{14}{3}$ .

Answer: $x = -2 \text{ or } x = \frac{14}{3}$

This example shows that solutions can be negative or fractional. Verify by substituting: $|4 - 3(-2)| = |4 + 6| = 10$ ✓ and $|4 - 3 \cdot \frac{14}{3}| = |4 - 14| = 10$ ✓. Always divide carefully when the coefficient of x is negative.

3. Solve a word problem with a video game scoring system

Problem

In a video game, a player earns a score bonus

B = |6L - 30|

where L is the character level. If the bonus equals 18 points, what character levels give this bonus?

$|6L - 30| = 18$
Set up the equation using the given formula.
$6L - 30 = 18 \text{ or } 6L - 30 = -18$
Split into two cases using the absolute value rule.
$6L = 48 \text{ or } 6L = 12$
Add 30 to both sides in each case.
$L = 8 \text{ or } L = 2$
Divide by 6 to find the character levels.

Answer: $L = 8 \text{ or } L = 2$

Both level 2 and level 8 characters earn a bonus of 18 points. This demonstrates that absolute value equations model situations where different inputs produce the same output. In this context, both solutions are valid.

Common mistakes

Where Absolute Value Equations usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\|2x - 3\| = 7$ and solving only $2x - 3 = 7$ to get $x = 5$ , stopping there.	Forgetting the second case misses half the solutions; absolute value equations always require checking both the positive and negative cases.	Always write both cases: $2x - 3 = 7$ and $2x - 3 = -7$ , solve each separately, and report both solutions: $x = 5$ or $x = -2$ .
Attempting to solve $\|3x + 2\| = -5$ by splitting into $3x + 2 = -5$ or $3x + 2 = 5$ .	Absolute value is always non-negative, so $\|A\|$ can never equal a negative number. Attempting to solve wastes time and generates invalid solutions.	Recognize immediately: if $c < 0$ , the equation has no solution. Write 'no solution' and stop.
Solving $\|x + 1\| = 4$ to get $x = 3$ or $x = -5$ , then not checking these answers.	Arithmetic errors during splitting or solving can go undetected, allowing you to report incorrect answers with confidence.	Substitute both solutions back: if $x = 3$ , then $\|3 + 1\| = 4$ ✓; if $x = -5$ , then $\|-5 + 1\| = \|-4\| = 4$ ✓. Always verify.

Tips and when to use something else

Always split the absolute value into two separate cases: if $|A| = c$ , write both $A = c$ and $A = -c$ , then solve each as a linear equation.
Before you solve, check whether $c$ is negative. If so, stop and write 'no solution'—absolute value can never be negative.
Substitute both of your solutions back into the original equation to verify you didn't make an arithmetic mistake.
When an equation has other terms with the absolute value, like $3|x - 1| + 5 = 20$ , isolate the absolute value expression first using inverse operations and Combining Like Terms before splitting into cases.

Related formulas

Absolute ValueAbsolute value measures the distance a number is from zero without regard to direction, so it is always positive or zero.Piecewise FunctionsA piecewise function uses different formulas for different input ranges, letting you define relationships that change behavior at specific boundaries.Combining Like TermsCombining like terms merges coefficients of identical variables to simplify expressions and solve equations more efficiently.

Frequently asked questions

Why do absolute value equations often have two solutions?: When $|A| = 5$ , the quantity $A$ could be 5 or it could be $-5$ —both have an absolute value of 5. Splitting into these two cases and solving each usually yields two different values of x. However, if $c = 0$ , there is only one solution.
What does it mean if c is negative in the equation $|ax + b| = c$ ?: It means the equation has no solution. Absolute value is never negative, so there's no value of $ax + b$ that can have an absolute value equal to a negative number. Simply write 'no solution' without further work.
How do I know if my solution is correct?: Substitute both solutions back into the original equation. For example, if you solved $|3x - 2| = 10$ and got $x = 4$ , check: $|3(4) - 2| = |12 - 2| = 10$ ✓. Always verify both solutions this way.
Can an absolute value equation have exactly one solution?: Yes, when $c = 0$ . For instance, $|x - 5| = 0$ is true only when $x - 5 = 0$ , so $x = 5$ is the unique solution. When $c > 0$ , you typically get two solutions, but the special case $c = 0$ always gives exactly one.

Need a different method?

The full solver is not scoped to one formula — type any problem and it will pick the method.

Open the math solver

Reviewed 2026-09-18