Mathematics · Algebra

Absolute Value Equations and Inequalities

Reference entry · last updated September 20, 2026

Absolute value measures distance on the real number line. The absolute value \(|x|\) is the distance from \(x\) to \(0\), and \(|x - a|\) is the distance from \(x\) to \(a\). That distance interpretation turns every absolute value equation and inequality into a statement about a position or a range of positions.

1. First principles

Absolute value is defined piecewise [1, 2]:

\( |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases} \)

Two properties drive every result below. First, \(|x| \ge 0\) for all real \(x\), with equality only at zero. Second, \(|x| = |{-x}|\), so an absolute value equation has two candidates unless the expression inside is itself zero.

2. Absolute value equations

For \( a > 0 \), the equation \(|X| = a\) means \(X\) is at distance \(a\) from zero in either direction:

\( |X| = a \;\iff\; X = a \;\text{ or }\; X = -a \)

For \(a = 0\), the only solution is \(X = 0\). For \(a < 0\), there is no solution, because an absolute value cannot be negative. To solve, isolate the absolute value, apply the two-case rule, and solve each case separately.

3. Absolute value inequalities

For \( a > 0 \), a less-than inequality bounds the expression on both sides:

\( |X| < a \;\iff\; -a < X < a \)

A greater-than inequality separates into two rays:

\( |X| > a \;\iff\; X < -a \;\text{ or }\; X > a \)

The same rules hold with \( \le \) and \( \ge \). Less-than produces an intersection, an "and" statement. Greater-than produces a union, an "or" statement. This is the connection to compound inequalities.

4. Distance form

The inequality \(|x - a| < d\) says the distance from \(x\) to \(a\) is less than \(d\), which is the interval \( (a - d, a + d) \). The equation \(|x - a| = d\) says \(x\) is exactly \(d\) units from \(a\), giving \(x = a \pm d\). Centering the absolute value at a point converts every problem into an interval around that point.

5. Worked examples

\( |2x - 1| = 7 \) gives \( 2x - 1 = 7 \) or \( 2x - 1 = -7 \), so \(x = 4\) or \(x = -3\).
\( |x - 3| \le 5 \) gives \( -5 \le x - 3 \le 5 \), so \( -2 \le x \le 8 \), the interval \([-2, 8]\).
\( |x + 1| > 4 \) gives \( x + 1 < -4 \) or \( x + 1 > 4 \), so \( x < -5 \) or \( x > 3 \), the union \( (-\infty, -5) \cup (3, \infty) \).

6. Edge cases

\( |X| = 0 \) has the single solution \(X = 0\).
\( |X| < 0 \) has no solution; \( |X| \le 0 \) has only \(X = 0\).
\( |X| > 0 \) holds for every \(X \neq 0\); \( |X| \ge 0 \) holds for every real \(X\).
When solving, isolate the absolute value first. If the isolated absolute value equals a negative number, stop: there is no solution.

References

^OpenStax, College Algebra 2e, Rice University (Absolute value equations and inequalities).
^James Stewart, Lothar Redlin, and Saleem Watson, Precalculus: Mathematics for Calculus, 7th ed., Cengage Learning, 2015.