Absolute Value Inequalities

Absolute value as distance

The absolute value $|x - a|$ is the distance between $x$ and $a$ on the number line. It does not matter which side of $a$ you are on. This one idea turns every absolute value inequality into ordinary inequalities.

Drag the slider to move the graph of $y = |x - a|$ and compare it with the horizontal line $y = 3$. The inequality $|x - a| < 3$ is true where the V is below the line.

Isolate the absolute value first

The translations above only work when the absolute value is alone on one side. If something else is added, multiplied, or subtracted, undo it first with inverse operations.

1. Isolate the absolute value expression $|u|$.

2. Rewrite as "and" ($|u| < k$) or "or" ($|u| > k$).

3. Solve the resulting linear inequalities.

Two facts help. First, $|-u| = |u|$, so $|-x + 6| = |x - 6|$. Second, if you multiply or divide both sides by a negative number, the inequality sign flips.

When there is no solution, or every number works

Since $|u| \ge 0$ always, some inequalities can be settled just by thinking about signs.

Tolerances and margins of error

The most common real-life use of absolute value inequalities is a tolerance: a measurement must be close to a target value, with a maximum allowed error.

Systems of absolute value inequalities

A system means all conditions must hold at once. Solve each inequality separately, then keep only the numbers that are in both solution sets: the intersection. Sketching both solutions on stacked number lines makes the overlap easy to see.

Intersections also answer "is this guaranteed?" questions. If a machine's temperature is only known to satisfy one inequality, and a safety rule demands another, check whether every allowed value also satisfies the rule. If the allowed interval sticks out past the rule, it is not guaranteed.

Quadratic and rational expressions inside the bars

When the expression inside the bars is quadratic or rational, the method does not change: translate to a compound inequality, then solve each piece with the tools for that type.

Rational expressions. If the variable is in a denominator, first note which $x$ make the denominator $0$ and exclude them. Since $|x - 3| > 0$ for every other $x$, you can multiply both sides by it without flipping the sign.

Parameters. A parameter such as $a$ is treated like a number. Just keep track of its sign: if $a > 0$, then $\sqrt{a^2} = a$.

A variable expression on the right side

Sometimes the "number" $k$ is really an expression such as $x + 1$. The same translations still work:

When a piece is a quadratic inequality, solve it by factoring and testing intervals, exactly as in the quadratic inequalities section. For $|x^2 - 4x| > 3x$ you get $x^2 - 4x > 3x$ or $x^2 - 4x < -3x$. Solve each one and take the union.

Two absolute values

There are three good strategies. Pick the one that matches the problem.

Strategy 1: think of distances

$|x - 2| < |x + 4|$ says "$x$ is closer to $2$ than to $-4$." The points equally far from both are at their midpoint, $\dfrac{2 + (-4)}{2} = -1$. Everything to the right of the midpoint is closer to $2$.

So the solution of $|x - 2| < |x + 4|$ is $x > -1$, that is, $(-1, \infty)$.

Strategy 2: square both sides

Both $|A|$ and $|B|$ are never negative, so $|A| < |B| \iff A^2 < B^2$ (and the same for $\le$, $\ge$, $>$). Squaring removes the bars because $|A|^2 = A^2$.

This also handles fractions like $\left|\dfrac{2x - 1}{x + 2}\right| \ge 1$. Since $x \ne -2$, this is the same as $|2x - 1| \ge |x + 2|$. Square both sides, solve, and finally remove $x = -2$ from the answer if it is in it.

Strategy 3: split at the critical points

For sums such as $|x - 1| + |x - 3|$ there is no shortcut by squaring. Instead cut the number line at the values where each expression inside the bars is $0$. On each piece the signs are fixed, so you can drop the bars.

The distance picture agrees: the total of the distances from $x$ to $-1$ and $2$ is at least $3$, and it reaches $5$ when $x$ is $1$ unit beyond each end.

Challenge problems: parameters, speed, and piecewise expressions

Containing an interval

$|x - k| \le r$ has the solution set $[k - r, k + r]$. For this interval to lie inside $[p, q]$, both endpoints must be in $[p, q]$: $k - r \ge p$ and $k + r \le q$. Solve both for $k$ and intersect.

Speed and distance traveled

Speed is always nonnegative, which is why it is often an absolute value. On a speed-versus-time graph, the distance traveled is the area between the graph and the time axis.

A piecewise expression inside the bars

If an expression like $d(x) = |x + 1| - |x - 3|$ sits inside another absolute value, first rewrite $d(x)$ without bars on each piece (split at the critical points, as in Strategy 3). On $x < -1$ it is $-4$; on $-1 \le x \le 3$ it is $2x - 2$; on $x > 3$ it is $4$. Then solve $|d(x)| \le 2$ on each piece, and keep only solutions inside that piece. Here that gives the single piece $[-1, 3]$, where $|2x - 2| \le 2$ means $0 \le x \le 2$. The solution is $[0, 2]$.

Summary

Back to the thermostat: $|T - 72| \le 3$ gives $[69, 75]$, so the room stays between $69^\circ$F and $75^\circ$F.