Solving Absolute Value Equations

An absolute value equation is any equation that contains an expression inside absolute value bars, such as \(|x - 4| = 9\). Recall from Introduction to absolute value that absolute value measures distance from zero on the number line, so it is always zero or positive. That single fact is the key to solving every absolute value equation you will meet.

Because \(|x - 4| = 9\) really means "the distance between \(x\) and \(4\) is \(9\)", there are two numbers that work: one that is \(9\) units above \(4\), and one that is \(9\) units below \(4\). Solving absolute value equations is really about finding both of these possibilities.

For any absolute value equation of the form \(|A| = b\), where \(A\) is an algebraic expression and \(b \ge 0\):

\(A = b\)   or   \(A = -b\)

Both equations are solved separately, giving up to two solutions. If \(b < 0\), the equation has no solution at all, because an absolute value can never equal a negative number.

1. Isolate the absolute value expression completely on one side of the equation.

2. Check the isolated value on the other side. If it is negative, stop, there is no solution.

3. Rewrite the equation as two separate equations: one where the inside expression equals the positive value, and one where it equals the negative value.

4. Solve each equation using ordinary algebra techniques, such as those used to solve quadratic equations by completing the square when the expression is more complex.

5. Substitute both answers back into the original equation to confirm they work.

Solve \(|x - 4| = 9\).

Since the absolute value is already isolated, split into two cases:

\(x - 4 = 9\)   or   \(x - 4 = -9\)

\(x = 13\)   or   \(x = -5\)

Checking: \(|13 - 4| = |9| = 9\) and \(|-5 - 4| = |-9| = 9\). Both solutions work.

Solve \(2|x - 1| - 3 = 5\).

Isolate the absolute value expression first:

\(2|x - 1| = 8\)

\(|x - 1| = 4\)

Now split into two cases:

\(x - 1 = 4\)   or   \(x - 1 = -4\)

\(x = 5\)   or   \(x = -3\)

Solve \(|3x - 2| = -5\).

The right-hand side is already negative, and absolute value can never be negative, so this equation has no solution. There is no need to split into cases at all.

When the other side of an absolute value equation contains a variable, such as \(|x - 2| = 3x + 1\), splitting into cases can produce a solution that does not actually satisfy the original equation. This happens because squaring or rearranging can hide a sign error that only shows up once you substitute the number back in.

Solve \(|x - 2| = 3x + 1\):

\(x - 2 = 3x + 1\)   or   \(x - 2 = -(3x + 1)\)

\(-3 = 2x \Rightarrow x = -\frac{3}{2}\)   or   \(x - 2 = -3x - 1 \Rightarrow 4x = 1 \Rightarrow x = \frac{1}{4}\)

Checking \(x = -\frac{3}{2}\): the right side becomes \(3(-\frac{3}{2}) + 1 = -3.5\), which is negative, but the left side \(|x-2|\) can never be negative, so this solution is extraneous and must be rejected. Checking \(x = \frac{1}{4}\): \(|\frac{1}{4} - 2| = \frac{7}{4}\) and \(3(\frac{1}{4}) + 1 = \frac{7}{4}\), which matches, so \(x = \frac{1}{4}\) is the only solution.

Solving \(|x - 3| = 5\) is the same as asking where the V-shaped graph of \(y = |x - 3|\) crosses the horizontal line \(y = 5\). The graph below shows the two intersection points, which match the algebraic solutions \(x = 8\) and \(x = -2\).

This graphical view is explored further in Absolute value functions, which covers how the shape, vertex, and shifts of these V-shaped graphs behave.

Absolute value equations often appear in word problems involving distance, tolerance, or error margins, since absolute value naturally represents "how far apart" two quantities are. Once you are comfortable solving equations, the next step is usually solving absolute value inequalities, which uses the same two-case idea but produces a range of solutions instead of individual numbers.

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