How to solve absolute value equations and inequalities

The key idea

Both 5 and $-5$ are 5 steps from zero, so $|5| = |-5| = 5$ . That is why an absolute value equation usually has two answers: whatever is inside the bars can be the positive number or its opposite.

|u| = c \;\Longrightarrow\; u = c \;\text{ or }\; u = -c \qquad (c \ge 0)

The four patterns (u is whatever is inside the bars, c is a positive number)
Pattern	Rewrite as	What the answer looks like
$\|u\| = c$	$u = c$ or $u = -c$	two numbers
$\|u\| < c$	$-c < u < c$	one segment (an "and" inequality)
$\|u\| > c$	$u < -c$ or $u > c$	two rays pointing apart (an "or" inequality)
$\|u\| =$ a negative number	nothing to rewrite	no solution, since distance is never negative

Worked examples

Example 1: the basic split

Problem Solve $|x - 4| = 9$ .

The absolute value is already alone. Split into two cases: the inside is 9, or the inside is $-9$ .
$x - 4 = 9 \quad\text{or}\quad x - 4 = -9$
Add 4 in each case.
$x = 13 \quad\text{or}\quad x = -5$
Check: $|13 - 4| = |9| = 9$ and $|-5 - 4| = |-9| = 9$ . Both work. On a number line, 13 and $-5$ are both 9 steps from 4.

Answer $x = 13$ or $x = -5$

Example 2: isolate first

Problem Solve $3|2x + 1| - 4 = 11$ .

Do not split yet. Add 4 to both sides.
$3|2x + 1| = 15$
Divide by 3. Now the absolute value is alone.
$|2x + 1| = 5$
Split into two cases.
$2x + 1 = 5 \quad\text{or}\quad 2x + 1 = -5$
Solve each one.
$x = 2 \quad\text{or}\quad x = -3$

Answer $x = 2$ or $x = -3$

Example 3: an absolute value inequality

Problem Solve $|2x - 3| \le 7$ .

Less than means the inside is trapped between $-7$ and 7.
$-7 \le 2x - 3 \le 7$
Add 3 to all three parts.
$-4 \le 2x \le 10$
Divide all three parts by 2.
$-2 \le x \le 5$
Graph: closed circles at $-2$ and 5 with the segment between them shaded.

Answer $-2 \le x \le 5$

Example 4 (test-hard): a variable outside the bars

Problem Solve $|x - 3| = 2x$ .

Split into two cases as usual.
$x - 3 = 2x \quad\text{or}\quad x - 3 = -2x$
Solve each case.
$x = -3 \quad\text{or}\quad x = 1$
Check $x = -3$ : the left side is $|-6| = 6$ , but the right side is $2(-3) = -6$ . They do not match, so $-3$ is extraneous. Throw it out.
Check $x = 1$ : the left side is $|-2| = 2$ and the right side is $2(1) = 2$ . It works.

Answer $x = 1$ only.

Common mistakes

Splitting before you isolate. In $3|2x + 1| - 4 = 11$ , the 3 and the $-4$ are outside the bars. Fix: undo them first, so the bars stand alone, then split.
Solving $|x + 2| = -4$ anyway. No distance is negative, so there is no solution. Fix: after isolating, glance at the sign of the number on the other side.
Using "and" for a greater-than inequality. $|u| > c$ means $u$ is far from zero in either direction, so it is $u < -c$ or $u > c$ .
Skipping the check when x is outside the bars. If the right side contains x, it can turn negative and create an extraneous answer, like $-3$ in Example 4.
Thinking $|u|$ just means "drop the minus sign" of a variable. $|x - 4|$ is not $x + 4$ . Fix: treat the whole inside as one quantity.

Quick methods

Practice

5 practice questions

Solve $|x + 6| = 2$ .
1. $x = -4$ or $x = -8$
2. $x = 4$ or $x = -8$
3. $x = -4$ only
4. $x = 4$ or $x = 8$
Show answer
Answer: $x = -4$ or $x = -8$
Split: $x + 6 = 2$ gives $x = -4$ , and $x + 6 = -2$ gives $x = -8$ . Both are 2 steps from $-6$ . Stopping at $-4$ misses the second case.
Solve $2|x - 1| + 3 = 1$ .
1. $x = 0$ or $x = 2$
2. $x = 0$
3. $x = 2$
4. No solution
Show answer
Answer: No solution
Subtract 3: $2|x - 1| = -2$ . Divide by 2: $|x - 1| = -1$ . An absolute value can never be negative, so no x works. The answer $x = 0$ or $x = 2$ comes from ignoring the negative sign.
Solve $|x - 5| < 3$ .
1. $2 < x < 8$
2. $x < 2$ or $x > 8$
3. $-8 < x < -2$
4. $x < 8$
Show answer
Answer: $2 < x < 8$
Less than gives an "and" inequality: $-3 < x - 5 < 3$ . Add 5 to all three parts: $2 < x < 8$ . These are the numbers less than 3 steps from 5. The "or" choice is the answer to $|x - 5| > 3$ .
Solve $|3x + 3| \ge 12$ .
1. $-5 \le x \le 3$
2. $x \le -5$ or $x \ge 3$
3. $x \le -3$ or $x \ge 3$
4. $x \ge 3$
Show answer
Answer: $x \le -5$ or $x \ge 3$
Greater than gives "or": $3x + 3 \ge 12$ gives $x \ge 3$ , and $3x + 3 \le -12$ gives $3x \le -15$ , so $x \le -5$ . The choice with $-3$ just copies the 3 with a minus sign instead of solving the second case.
Which equation has no solution?
1. $|x| = 0$
2. $|x - 2| = 3$
3. $|x| + 4 = 1$
4. $-|x| = -5$
Show answer
Answer: $|x| + 4 = 1$
Subtracting 4 gives $|x| = -3$ , which is impossible. $|x| = 0$ has one solution (0), $|x - 2| = 3$ has two (5 and $-1$ ), and $-|x| = -5$ is the same as $|x| = 5$ , with two solutions.

Frequently asked questions

Why do absolute value equations have two answers?

Absolute value measures distance from zero, and there are two numbers at any positive distance: one on each side. So $|u| = 7$ means $u$ is 7 or $-7$ . If the distance is 0, there is only one answer, and if it is negative, there is none.

Can an absolute value equation have no solution?

Yes. Once the absolute value is alone, if it equals a negative number, like $|x - 1| = -1$ , there is no solution, because a distance cannot be negative. Isolate first, though: $|x| - 4 = -1$ looks negative but becomes $|x| = 3$ , which has two answers.

What is an extraneous solution?

It is an answer your algebra produces that does not actually work in the original equation. With absolute value, this happens when x also appears outside the bars, as in $|x - 3| = 2x$ . Plug each answer back in and drop any that fail.

How do I know whether an absolute value inequality is "and" or "or"?

Look at the direction once the absolute value is alone on the left. Less than ( $<$ or $\le$ ) means close to the center, so it becomes one "and" segment. Greater than ( $>$ or $\ge$ ) means far away, so it becomes two "or" pieces.