What is the difference between |x| a?

If |x| a, you get x > a OR x < -a (using OR, so x is in two separate regions). The AND versus OR distinction is critical: the first gives a bounded interval, and the second gives two unbounded rays.

Why does |x| < a give -a < x < a?

The absolute value |x| represents the distance from x to zero on a number line. If that distance is less than a, then x must be within a units on either side of zero. Graphically, this is all points lying between -a and a.

Can an absolute value be less than zero?

No. Since |x| \geq 0 for all real x, the inequality |x| < 0 is impossible. There is no value of x satisfying this condition, so the solution set is empty (no solution).

How do I solve |ax + b| < c?

First, apply the absolute value rule to get -c < ax + b < c. Then isolate x by subtracting b from all three parts and dividing by a. Remember: if a < 0, reverse all inequality signs when you divide by a. The result is a compound inequality describing all solutions for x.

Absolute Value Inequalities

Q: Why does |x| < a give -a < x < a?

The absolute value |x| represents the distance from x to zero on a number line. If that distance is less than a, then x must be within a units on either side of zero. Graphically, this is all points lying between -a and a.

Q: Can an absolute value be less than zero?

No. Since |x| \geq 0 for all real x, the inequality |x| < 0 is impossible. There is no value of x satisfying this condition, so the solution set is empty (no solution).

Q: How do I solve |ax + b| < c?

First, apply the absolute value rule to get -c < ax + b < c. Then isolate x by subtracting b from all three parts and dividing by a. Remember: if a < 0, reverse all inequality signs when you divide by a. The result is a compound inequality describing all solutions for x.

Absolute value inequalities express ranges of values at a fixed distance from a point. Use them to solve distance, tolerance, or deviation problems.

|x| < a \iff -a < x < a

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What each symbol means

What Absolute Value Inequalities takes

Symbol	Meaning
$a$	The boundary value or threshold in the inequality, which must be positive; it represents the maximum distance from a reference point. If $a$ is negative or zero, the inequality may have no solution or only trivial solutions.
$x$	The variable (or expression inside the absolute value) whose values you are solving for; it can be any real number, and the solution describes the complete set of values satisfying the inequality.

When to use it

When you need to find all values satisfying a condition about distance or magnitude, or when setting tolerance ranges in real-world contexts.

Level

Usually taught in: Algebra II

Worked examples

1. Solve a simple absolute value inequality with addition

Problem

Solve

|x + 1| < 3

for

x

$|x + 1| < 3$
This is our starting inequality. We need to find all values of $x$ where the absolute value of $(x + 1)$ is less than 3.
$-3 < x + 1 < 3$
Apply the absolute value inequality rule: if $|u| < a$ , then $-a < u < a$ . Here $u = x + 1$ and $a = 3$ .
$-3 - 1 < x < 3 - 1$
Subtract 1 from all three parts of the compound inequality to isolate $x$ .
$-4 < x < 2$
Simplify each side to get the final solution: $x$ must be between $-4$ and $2$ .

Answer: $-4 < x < 2$

This problem demonstrates the core technique: apply the rule to convert an absolute value inequality into a compound inequality, then isolate the variable by using the same operations on all three parts. The result is an interval on the number line.

2. Solve an absolute value inequality with a negative coefficient

Problem

Solve

|2 - x| < 3

for

x

$|2 - x| < 3$
We need to find values of $x$ such that the absolute value of $(2 - x)$ is less than 3.
$-3 < 2 - x < 3$
Apply the rule: $|u| < a \iff -a < u < a$ , where $u = 2 - x$ and $a = 3$ .
$-3 - 2 < -x < 3 - 2$
Subtract 2 from all parts to isolate the term containing $x$ .
$-5 < -x < 1$
Simplify: the left side is $-3 - 2 = -5$ , and the right side is $3 - 2 = 1$ .
$-1 < x < 5$
Multiply all parts by $-1$ to solve for $x$ alone. Crucially, reverse all inequality signs when multiplying by a negative number: $-5 < -x < 1$ becomes $5 > x > -1$ , rewritten as $-1 < x < 5$ .

Answer: $-1 < x < 5$

This example requires care because $x$ has a negative coefficient inside the absolute value. Students must remember to reverse inequality signs when multiplying or dividing by a negative number. This is a common mistake.

3. Solve a tolerance problem using absolute value inequality

Problem

A manufacturing plant produces nails that should be 2 inches long. Due to machinery limitations, the actual length can vary by up to 0.1 inches. Write and solve an absolute value inequality to find the range of acceptable nail lengths.

$|L - 2| \leq 0.1$
Let $L$ be the actual length in inches. The absolute value $|L - 2|$ represents how far the actual length is from the target of 2 inches. The inequality states this difference must be at most 0.1 inches.
$-0.1 \leq L - 2 \leq 0.1$
Apply the absolute value rule to convert to a compound inequality.
$-0.1 + 2 \leq L \leq 0.1 + 2$
Add 2 to all parts to isolate $L$ .
$1.9 \leq L \leq 2.1$
Simplify: $-0.1 + 2 = 1.9$ and $0.1 + 2 = 2.1$ .

Answer: $1.9 \leq L \leq 2.1 \text{ inches}$

This word problem shows how absolute value inequalities model real-world tolerance ranges. Any nail with length between 1.9 and 2.1 inches is acceptable. This type of problem appears frequently in engineering and manufacturing contexts.

Common mistakes

Where Absolute Value Inequalities usually goes wrong

The mistake	Why it is wrong	The fix
Solving $\|2x - 1\| < 5$ as just $2x - 1 < 5$ , which gives $x < 3$ .	Applying only one half of the absolute value rule (the part about $u < a$ ) while ignoring the other half (that $-a < u$ ). This shortcut misses the complete picture.	Use the complete rule: $-5 < 2x - 1 < 5$ . Then solve: $-4 < 2x < 6$ , and dividing by 2 gives $-2 < x < 3$ .
Writing the solution to $\|x\| < 5$ as $x > -5$ OR $x < 5$ (listing two separate conditions with OR).	Forgetting that a compound inequality requires both conditions to hold at the same time. Using OR instead of AND creates a solution set that includes nearly the entire number line, which is incorrect.	The correct solution is the compound inequality $-5 < x < 5$ , which means $x$ must satisfy $x > -5$ AND $x < 5$ simultaneously.
When solving $\|x + 3\| < 2$ , writing $-2 < x + 3 < 2$ but then jumping directly to $-2 < x < 2$ without subtracting 3 from all parts.	Skipping the algebraic step to isolate $x$ leads to an entirely wrong solution; the student fails to apply the same operation to all three parts of the compound inequality.	After obtaining $-2 < x + 3 < 2$ , subtract 3 from all three parts: $-2 - 3 < x < 2 - 3$ , giving $-5 < x < -1$ .

Tips and when to use something else

Always apply the complete rule: if $|u| < a$ , then $-a < u < a$ (not just $u < a$ alone).
Substitute $u$ for complex expressions inside the absolute value first (e.g., let $u = 3x - 2$ ), solve $|u| < a$ to get the compound inequality, then substitute back and solve for $x$ .
If $a$ is negative (e.g., $|x| < -3$ ), recognize immediately that there is no solution, since absolute value is always non-negative. This differs from Linear Inequalities where solutions almost always exist.
If the inequality uses $>$ instead of $<$ (e.g., $|x| > a$ ), the solution structure is completely different: use OR instead of AND, giving $x > a$ OR $x < -a$ . Refer to Compound Inequalities for the correct method.

Related formulas

Compound InequalitiesCompound inequalities describe a range where a variable falls between two values; use them when bounds constrain something from both above and below.Linear InequalitiesSolve linear inequalities by isolating the variable, flipping the inequality sign when multiplying or dividing by a negative number.Absolute ValueAbsolute value measures the distance a number is from zero without regard to direction, so it is always positive or zero.

Frequently asked questions

What is the difference between $|x| < a$ and $|x| > a$ ?: If $|x| < a$ , you get the compound inequality $-a < x < a$ (using AND, so $x$ is in one interval). If $|x| > a$ , you get $x > a$ OR $x < -a$ (using OR, so $x$ is in two separate regions). The AND versus OR distinction is critical: the first gives a bounded interval, and the second gives two unbounded rays.
Why does $|x| < a$ give $-a < x < a$ ?: The absolute value $|x|$ represents the distance from $x$ to zero on a number line. If that distance is less than $a$ , then $x$ must be within $a$ units on either side of zero. Graphically, this is all points lying between $-a$ and $a$ .
Can an absolute value be less than zero?: No. Since $|x| \geq 0$ for all real $x$ , the inequality $|x| < 0$ is impossible. There is no value of $x$ satisfying this condition, so the solution set is empty (no solution).
How do I solve $|ax + b| < c$ ?: First, apply the absolute value rule to get $-c < ax + b < c$ . Then isolate $x$ by subtracting $b$ from all three parts and dividing by $a$ . Remember: if $a < 0$ , reverse all inequality signs when you divide by $a$ . The result is a compound inequality describing all solutions for $x$ .

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Reviewed 2026-09-18