Lesson 91: Solving Absolute-Value Inequalities

New Concept

An absolute-value inequality is an inequality with at least one absolute-value expression.

What’s next

Next, you'll translate these inequalities into compound statements and solve for the range of possible solutions on a number line.

Property

For an inequality in the form $|K| < a$, where $K$ represents a variable expression and $a > 0$, the solution is $-a < K < a$. This is an "AND" compound inequality, meaning $K > -a$ AND $K < a$.

Explanation

Think of this as a 'distance limit.' Your value must be so close to zero that its distance is less than a certain number. This traps your answer between two boundaries, creating a single, cozy interval. You must be greater than the negative boundary AND less than the positive one simultaneously.

Examples

$|x - 5| \le 3$ is solved as $-3 \le x - 5 \le 3$, which simplifies to the solution $2 \le x \le 8$.
$|x| + 7.4 \le 9.8$ first simplifies to $|x| \le 2.4$, which means the solution is $-2.4 \le x \le 2.4$.

Expressions inside the absolute value shift the solution's center. Let's see this key idea in action.

Example Problem

Solve the inequality $|x - 2| \le 5$ and graph the solution.

Property

For an inequality in the form $|K| > a$, where $K$ represents a variable expression and $a > 0$, the solution is $K < -a$ OR $K > a$.

Explanation

This rule is for when your value is far from zero. Your distance must be greater than a certain number, which means you can either be far out in the positive numbers OR far out in the negative numbers. Since you can't be in both places at once, the solution splits into two separate regions.

Examples

$|x + 7| > 3$ is solved as $x + 7 < -3$ OR $x + 7 > 3$, giving the solution $x < -10$ OR $x > -4$.
$-2|x| < -6$ simplifies to $|x| > 3$ (remember to flip the sign!), so the solution is $x < -3$ OR $x > 3$.

Property

If an inequality simplifies to $|K| \le (\text{negative number})$, there is no solution ($\{\}$ or $\emptyset$). If it simplifies to $|K| \ge (\text{negative number})$, the solution is all real numbers ($\mathbb{R}$).

Explanation

This is a logic check! Absolute value represents a distance, and distance cannot be negative. Asking for a distance to be less than or equal to -2, like in $|x| \le -2$, is impossible. On the flip side, asking for a distance to be greater than -5 is always true for any real number.

Examples

$|x| + 6 \le 4$ simplifies to $|x| \le -2$. Since absolute value cannot be negative, there is no solution, $\emptyset$.
$|x| + 6 > 1$ simplifies to $|x| > -5$. Since absolute value is always non-negative, this is true for all real numbers, $\mathbb{R}$.