Unit 2 · Lesson 6 of 6

Absolute value equations and inequalities

About 25 minutes · 5 steps · one idea: absolute value is distance from 0

You'll be able to: solve equations like \(3|2x - 1| + 4 = 25\) and inequalities like \(|x - 3| \le 5\), and write inequality answers in interval notation.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first; the first one sets up today's lesson.

Solve \(-2x + 5 \ge 11\). Write the answer in interval notation.

Subtract 5: \(-2x \ge 6\).

Divide by \(-2\) and flip the sign: \(x \le -3\).

Answer: \((-\infty, -3]\).

Simplify \(-(x - 3) + 2(x + 1)\).

Distribute: \(-(x - 3) = -x + 3\) and \(2(x + 1) = 2x + 2\).

Combine like terms: \(-x + 2x = x\) and \(3 + 2 = 5\).

Answer: \(x + 5\).

Step 2 of 5The idea

The absolute value \(|A|\) is the distance of \(A\) from 0. So \(|A| = 7\) means \(A\) is 7 steps from 0: either \(7\) or \(-7\).

That one picture gives every rule below. Always get the absolute value alone on one side first.

Equations, after isolating \(|A|\):

  • \(|A| = c\) with \(c\) positive: solve \(A = c\) and \(A = -c\). Two answers.
  • \(|A| = 0\): solve \(A = 0\). One answer.
  • \(|A| = c\) with \(c\) negative: no solution. A distance is never negative.

Inequalities (with \(c\) positive):

  • \(|A| \lt c\) means "closer than \(c\)": \(-c \lt A \lt c\). One piece, between.
  • \(|A| \gt c\) means "farther than \(c\)": \(A \lt -c\) or \(A \gt c\). Two pieces, outside.

The same goes for \(\le\) and \(\ge\), with square brackets.

Why this shows up on the placement test: absolute value questions test whether you remember the second case (the negative one), which is where most points are lost.

Step 3 of 5Worked examples

Example 1: \(3|2x - 1| + 4 = 25\)

  1. Isolate: subtract 4 to get \(3|2x - 1| = 21\). Divide by 3: \(|2x - 1| = 7\).
  2. Split into two equations: \(2x - 1 = 7\) or \(2x - 1 = -7\).
  3. First: \(2x = 8\), so \(x = 4\). Second: \(2x = -6\), so \(x = -3\).
  4. Check: \(|2(4) - 1| = |7| = 7\) and \(|2(-3) - 1| = |-7| = 7\). Both work: \(x = 4\) or \(x = -3\).

Example 2: \(|x - 3| \le 5\), then \(|2x + 1| \gt 7\)

  1. "Less than" means between: \(-5 \le x - 3 \le 5\).
  2. Add 3 to all three parts: \(-2 \le x \le 8\). Interval notation: \([-2, 8]\).
−4−3−2−1012345678910
\(|x - 3| \le 5\): every number within 5 of 3.
  1. "Greater than" means outside: \(2x + 1 \lt -7\) or \(2x + 1 \gt 7\).
  2. Solve each: \(2x \lt -8\), so \(x \lt -4\); or \(2x \gt 6\), so \(x \gt 3\).
  3. Interval notation: \((-\infty, -4) \cup (3, \infty)\).
−6−5−4−3−2−10123456
\(|2x + 1| \gt 7\): two pieces going outward.

Step 4 of 5Practice

For equations, list every solution separated by commas (or type none). For inequalities, use interval notation like (-inf, 2) U (6, inf).

Solve \(|x - 2| = 5\).

Split: \(x - 2 = 5\) or \(x - 2 = -5\).

So \(x = 7\) or \(x = -3\).

Check: \(|5| = 5\) and \(|-5| = 5\). Answer: \(7, -3\).

Solve \(|3x + 1| = 8\).

Split: \(3x + 1 = 8\) or \(3x + 1 = -8\).

First: \(3x = 7\), so \(x = \dfrac{7}{3}\). Second: \(3x = -9\), so \(x = -3\).

Answer: \(\dfrac{7}{3}, -3\).

Solve \(2|x + 4| - 3 = 7\).

Isolate: add 3 to get \(2|x + 4| = 10\), then divide by 2: \(|x + 4| = 5\).

Split: \(x + 4 = 5\) or \(x + 4 = -5\), so \(x = 1\) or \(x = -9\).

Answer: \(1, -9\).

Solve \(|5x - 2| + 6 = 1\).

Isolate: subtract 6 to get \(|5x - 2| = -5\).

An absolute value is a distance, so it can never equal a negative number.

Answer: no solution.

Solve \(|x + 1| \lt 4\).

"Less than" means between: \(-4 \lt x + 1 \lt 4\).

Subtract 1 from all three parts: \(-5 \lt x \lt 3\).

Answer: \((-5, 3)\).

Solve \(|2x - 3| \ge 5\).

"Greater than" means outside: \(2x - 3 \le -5\) or \(2x - 3 \ge 5\).

First: \(2x \le -2\), so \(x \le -1\). Second: \(2x \ge 8\), so \(x \ge 4\).

Answer: \((-\infty, -1] \cup [4, \infty)\).

Solve \(3|x - 2| + 1 \le 13\).

Isolate: subtract 1 to get \(3|x - 2| \le 12\), then divide by 3: \(|x - 2| \le 4\).

Between: \(-4 \le x - 2 \le 4\). Add 2 to all three parts: \(-4 + 2 \le x \le 4 + 2\), that is \(-2 \le x \le 6\).

Answer: \([-2, 6]\).

Which is the solution of \(|x - 4| \gt 2\)?

  1. \((-\infty, -6) \cup (-2, \infty)\)
  2. \((-\infty, 2] \cup [6, \infty)\)
  3. \((-\infty, -2) \cup (2, \infty)\)
  4. \((-\infty, 2) \cup (6, \infty)\)

"Greater than" means outside: \(x - 4 \lt -2\) or \(x - 4 \gt 2\). So \(x \lt 2\) or \(x \gt 6\): \((-\infty, 2) \cup (6, \infty)\). Answer (d).

(a) solves \(|x + 4| \gt 2\) instead, (b) includes 2 and 6 although the sign is strict, and (c) forgets the 4.

Step 5 of 5Wrap-up

You can now isolate an absolute value, split it into two cases, spot when there is no solution, and turn \(|A| \lt c\) into "between" and \(|A| \gt c\) into "outside".

Read or watch next: OpenStax Intermediate Algebra 2e, 2.7 Solve Absolute Value Inequalities (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Algebra (all content): Absolute value equations, functions, & inequalities (short videos). Quick rules: Equations and inequalities cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "How do I remember which absolute value inequality gives an 'or'?" It knows this course and can explain things another way.