Absolute Value — Definition and Properties

Most absolute value problems come down to one move: turn $|\,\text{something}\,|$ into two ordinary cases. The reason that move is allowed is worth understanding before you use it, because it also tells you when there is no solution at all.

The Formula and Its Symbols

For a real number $x$,

Here $|x|$ is read "the absolute value of $x$," and it means the distance from $x$ to $0$ on the number line. That distance is never negative, which is why $|5| = 5$ and $|-5| = 5$: the two numbers sit on opposite sides of $0$ but the same distance from it. Absolute value does not "make everything positive" — it leaves nonnegative numbers alone and flips the sign of negative ones.

The properties you will use most:

1. $|x| \ge 0$ for every real $x$.
2. $|x| = 0$ only when $x = 0$.
3. $|-x| = |x|$.
4. $|ab| = |a||b|$ for real $a$ and $b$.
5. If $b \ne 0$, then $\left|\frac{a}{b}\right| = \frac{|a|}{|b|}$.

The condition $b \ne 0$ matters in the last property because division by $0$ is undefined.

Why the Two-Case Method Works

The two-case rule is not an arbitrary trick — it falls straight out of the distance meaning. An expression like $|a-b|$ is the distance between $a$ and $b$, so $|x-3| = 5$ literally asks: which numbers sit a distance of $5$ from $3$? On a number line, exactly two points satisfy that — one $5$ units to the right, one $5$ units to the left. That geometric fact is what guarantees two cases (and what tells you there are none if the required distance is negative). For example,

so the distance between $2$ and $7$ is $5$. Holding onto "distance from $0$" is the most reliable way to read any absolute value expression.

Worked Example: Solve $|x - 3| = 5$

This asks for numbers whose distance from $3$ is $5$. A point $5$ units to the right of $3$:

A point $5$ units to the left of $3$:

So the solutions are

The general pattern: if $|u| = k$ and $k > 0$, solve both $u = k$ and $u = -k$.

Practice It Yourself

Solve $|x + 4| = 9$, reading it as "distance from $-4$ is $9$." Write the two matching cases:

so $x = 5$ or $x = -13$. Check your answers by substituting back: $|5 + 4| = |9| = 9$ and $|-13 + 4| = |-9| = 9$. Both work.

Calculation Traps to Watch

• Thinking $|x|$ can be negative. For real numbers it is always at least $0$; an equation like $|u| = k$ with $k < 0$ has no real solution.
• Solving only one case. Stopping at $x = 8$ in the worked example misses the second point that is also $5$ units from $3$.
• Confusing $|-x|$ with $-|x|$. They are different: $|-x| = |x|$ is zero or positive, while $-|x|$ is zero or negative.

A Quick Pre-Check for $|u| = k$

Before solving, look at the right side. If $k < 0$, stop — no real solution, because a distance cannot be negative. If $k = 0$, the inside must be $0$. If $k > 0$, expect two cases unless both lead to the same value.

Absolute value shows up wherever size matters more than direction: distance on a number line, error or deviation from a target, equations and inequalities, and formulas where only magnitude should remain. It returns later in coordinate geometry, calculus, and complex numbers, though the exact meaning shifts with the setting — but the two-case method, grounded in distance, stays the workhorse.