What if the isolated absolute value equals a negative number?

The equation has no solution. Absolute value represents distance, which cannot be negative. For example, $|x| = -3$ has no real solutions.

What if the isolated absolute value equals zero?

There is exactly one solution. Since positive and negative zero are the same, you only set the inside expression equal to zero and solve.

What if there is an absolute value expression on both sides?

Isolate them such that $|A| = |B|$. Then split into two cases: $A = B$ and $A = -B$. Solve each case normally.

How to solve an absolute value equation

To solve an absolute value equation, isolate the absolute value expression and split the equation into a positive case and a negative case. This method applies to any equation where a variable is constrained within absolute value bars.

The setup

Before applying any algebraic rules, you must isolate the absolute value term on one side of the equals sign. Ensure the other side is a constant or an expression free of absolute values.

The steps

Isolate the absolute value expression so the equation takes the form $|ax + b| = c$ . 2. Check the value of $c$ . If $c < 0$ , there is no solution. 3. Split the equation into two separate equations: $ax + b = c$ and $ax + b = -c$ . 4. Solve both equations independently for $x$ .

Checking the result

Substitute each solved value of $x$ back into the original equation. Both values must satisfy the original equation. This step identifies extraneous solutions, which often appear if the equation involves variables on both sides.

Common errors

A frequent mistake is dropping the absolute value bars without creating the negative case, yielding only half of the solution. Another error is failing to isolate the absolute value term before splitting the equation.

Worked example

Solve for $x$ : $2|x - 4| + 3 = 11$

$2|x - 4| + 3 = 11$ . Subtract 3 from both sides: $2|x - 4| = 8$ . Divide by 2: $|x - 4| = 4$ . Split into two cases. Case 1: $x - 4 = 4$ , which gives $x = 8$ . Case 2: $x - 4 = -4$ , which gives $x = 0$ . The solutions are $x = 8$ and $x = 0$ . Check $x = 8$ : $2|8 - 4| + 3 = 2(4) + 3 = 11$ . Check $x = 0$ : $2|0 - 4| + 3 = 2(4) + 3 = 11$ . Both solutions are valid.

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References: OpenStax College Algebra Section 2.7 · Stewart Precalculus Section 1.7