Equations and inequalities with absolute values

Definition and meaning

The absolute value of a number is its distance from zero:

$$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$$

So $|5| = 5$, $|-5| = 5$ and $|0| = 0$. An absolute value is never negative. The second line looks odd at first: if $x = -5$, then $-x = 5$, which is positive.

More useful still: $|x - a|$ is the distance between $x$ and $a$ on the number line. Then $|x - 2| = 3$ asks for the points at distance 3 from 2, which are $-1$ and $5$, and $|x + 4| < 1$ asks for the points within 1 of $-4$, because $x + 4 = x - (-4)$.

Useful properties: $|ab| = |a| \cdot |b|$, $\left|\frac{a}{b}\right| = \frac{|a|}{|b|}$, $|{-a}| = |a|$ and $\sqrt{x^2} = |x|$. But $|a + b|$ is not $|a| + |b|$ in general: $|3 + (-5)| = 2$, while $|3| + |-5| = 8$.

Equations

Form 1: $|A| = k$, with $k$ a number. If $k > 0$, then $A = k$ or $A = -k$: two equations. If $k = 0$, then $A = 0$. If $k < 0$, there is no solution, since an absolute value cannot be negative.

Form 2: $|A| = |B|$. Two numbers with the same absolute value are equal or opposite, so $A = B$ or $A = -B$.

Form 3: $|A| = B$, where $B$ contains the unknown. Now the right side might be negative for some $x$, and those $x$ cannot be solutions. Either:

• solve $A = B$ and $A = -B$, then keep only the solutions with $B \ge 0$; or
• split by the sign of $A$: where $A \ge 0$ solve $A = B$; where $A < 0$ solve $-A = B$; and keep each solution only if it lies in its own case.

Both routes give the same answer. The check is not optional: this form is built to produce false solutions.

Inequalities

For $k > 0$:

$$|A| < k \iff -k < A < k \qquad\qquad |A| > k \iff A < -k \ \text{ or } \ A > k$$

The first says "within $k$ of zero", one interval; the second says "farther than $k$ from zero", two separate pieces. The same holds with $\le$ and $\ge$, endpoints included.

If $k \le 0$, think instead of solving: $|A| < -2$ has no solution; $|A| > -2$ holds for every $x$ where $A$ is defined; $|A| \le 0$ holds only where $A = 0$.

More than one absolute value: split at critical points

For an expression such as $|x - 1| + |x + 2|$, each absolute value changes formula where its inside is zero, here at $x = 1$ and $x = -2$. These critical points cut the line into intervals. In each interval every absolute value has a fixed sign, so you can drop the bars, solve an ordinary equation and keep the solution only if it lies in that interval.

Interval	$x - 1$	$x + 2$	Expression
$x < -2$	negative	negative	$-(x - 1) - (x + 2) = -2x - 1$
$-2 \le x < 1$	negative	non-negative	$-(x - 1) + (x + 2) = 3$
$x \ge 1$	non-negative	non-negative	$(x - 1) + (x + 2) = 2x + 1$

The middle row has a distance meaning too: for any $x$ between $-2$ and $1$, the distances to $-2$ and to $1$ add up to the gap between them, which is 3.

The graph in one sentence

The graph of $y = |x - a|$ is a V with its corner at $(a, 0)$. The functions unit draws it in detail; for now, the V shape explains why $|A| = k$ (a horizontal line meeting the V) has two, one or no solutions.