2.5Absolute Value Equations and Inequalities
Absolute value measures distance from zero. Equations ask "where is the distance exactly equal to a value?" — giving two solutions. Inequalities ask "where is the distance less than or greater than a value?" — giving a compound inequality you already know how to solve.
Why This Matters
Absolute value equations and inequalities appear in engineering tolerances, quality control, measurement error, and SAT/ACT problems. Equations find exact boundary values; inequalities describe acceptable ranges — together they model real-world precision requirements.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Essential Question
How does thinking about absolute value as distance from zero explain why equations produce two solutions and why inequalities reduce to compound inequalities?
Part 1 — Absolute Value Equations
Lesson Overview — Equations
An absolute value equation has the form |expression| = k. Because absolute value measures distance from zero, |x| = k means x is exactly k units from 0 — so there are two cases: the expression equals k (positive case) or the expression equals −k (negative case). The Two-Case Method: set the expression inside equal to k and also equal to −k, then solve each equation separately. Always check both solutions in the original equation, because extraneous solutions can arise when the absolute value expression is not fully isolated. Three special cases to know: if k > 0, there are two solutions; if k = 0, there is exactly one solution (the expression equals zero); if k < 0, there is no solution (absolute value is never negative).
The Two-Case Method — At a Glance
k > 0 → Two Solutions
|expression| = k
- Case 1: expression = k
- Case 2: expression = −k
- Solve each; check both.
k = 0 → One Solution
|expression| = 0
- Only one case: expression = 0
- One solution.
k < 0 → No Solution
|expression| = negative
- Absolute value ≥ 0 always.
- Cannot equal a negative.
- Solution: ∅ (empty set).
Worked Examples — Equations
Solve: |x| = 7
The absolute value equals a positive number, so there are two cases.
Case 1: x = 7
Case 2: x = −7
Check Case 1: |7| = 7 ✓
Check Case 2: |−7| = 7 ✓
Solve: |x − 3| = 5
The absolute value is already isolated. Apply the Two-Case Method.
Case 1: x − 3 = 5 → x = 8
Case 2: x − 3 = −5 → x = −2
Check Case 1: |8 − 3| = |5| = 5 ✓
Check Case 2: |−2 − 3| = |−5| = 5 ✓
Solve: |2x + 4| = 10
The absolute value is isolated. Apply the Two-Case Method.
Case 1: 2x + 4 = 10 → 2x = 6 → x = 3
Case 2: 2x + 4 = −10 → 2x = −14 → x = −7
Check Case 1: |2(3) + 4| = |10| = 10 ✓
Check Case 2: |2(−7) + 4| = |−10| = 10 ✓
Solve: |3x − 6| = 0 (Special Case)
The absolute value equals zero — only one case.
3x − 6 = 0 → 3x = 6 → x = 2
Check: |3(2) − 6| = |0| = 0 ✓
Exactly one solution.
Solve: |x + 2| = −4 (Special Case)
The right side is negative.
Absolute value is always ≥ 0 — it can never equal a negative number.
No solution is possible.
Guided Practice — Equations
Guided Practice Video: Absolute Value Equations and Inequalities
Watch the guided practice walkthrough for absolute value equations and inequalities, then complete the problems below.
Video by Sang Real Math
Watch on YouTube ↗Answers are in the Answer Key section.
Solve: |x| = 9
Hint: Two cases: x = 9 or x = −9. Check both: |9| = 9 ✓, |−9| = 9 ✓.
Solve: |x + 5| = 3
Hint: Case 1: x + 5 = 3 → x = −2. Case 2: x + 5 = −3 → x = −8. Check both.
Solve: |4x − 8| = 12
Hint: Case 1: 4x − 8 = 12 → 4x = 20 → x = 5. Case 2: 4x − 8 = −12 → 4x = −4 → x = −1.
Solve: |2x + 6| = 0
Hint: Only one case: 2x + 6 = 0 → x = −3. One solution.
Solve: |x − 1| = −7
Hint: The right side is negative. Absolute value can never be negative. No solution (∅).
Part 2 — Absolute Value Inequalities
Lesson Overview — Inequalities
An absolute value inequality compares the distance of an expression from zero to a fixed number. Because absolute value measures distance, every absolute value inequality rewrites as a compound inequality. The direction of the inequality symbol tells you which type: the Less-Than Rule says |x| < a is equivalent to −a < x < a (an AND compound inequality — a bounded segment). The Greater-Than Rule says |x| > a is equivalent to x < −a OR x > a (an OR compound inequality — two outward rays). These rules also apply to ≤ and ≥. For expressions like |ax + b| < c, isolate the absolute value first if needed, then apply the rule and solve the resulting compound inequality. The special cases |expression| ≤ 0 (only solution is when the expression equals zero) and |expression| > 0 (all reals except where the expression is zero) are worth memorizing.
The Two Rules — At a Glance
Less-Than Rule → AND
- |x| < a means x is within a distance of a from 0.
- Rewrites as: −a < x < a
- Also works for ≤: |x| ≤ a → −a ≤ x ≤ a
- Graph: a segment between −a and a.
- Interval: (−a, a)
Graph: −4 < x < 4
Open at −4 and 4 → (−4, 4)
Greater-Than Rule → OR
- |x| > a means x is more than a distance of a from 0.
- Rewrites as: x < −a OR x > a
- Also works for ≥: |x| ≥ a → x ≤ −a OR x ≥ a
- Graph: two rays pointing outward.
- Interval: (−∞, −a) ∪ (a, ∞)
Graph: x < −3 OR x > 3
Open at −3 and 3 → (−∞, −3) ∪ (3, ∞)
Worked Examples — Inequalities
Solve, graph, and write in interval notation: |x| < 4
Apply the Less-Than Rule: |x| < a → −a < x < a.
|x| < 4 → −4 < x < 4
Graph: open circle at −4, open circle at 4, shade the segment between them.
Interval notation: (−4, 4)
Check: Test x = 0: |0| = 0 < 4 ✓ — in solution.
Check: Test x = 5: |5| = 5 < 4 ✗ — not in solution (correct).
Graph: −4 < x < 4
Open at −4 and 4 · segment → (−4, 4)
Solve, graph, and write in interval notation: |x| > 3
Apply the Greater-Than Rule: |x| > a → x < −a OR x > a.
|x| > 3 → x < −3 OR x > 3
Graph: open circle at −3, shade left; open circle at 3, shade right.
Interval notation: (−∞, −3) ∪ (3, ∞)
Check: Test x = −5: |−5| = 5 > 3 ✓ — in solution.
Check: Test x = 1: |1| = 1 > 3 ✗ — not in solution (correct).
Graph: x < −3 OR x > 3
Open at −3 and 3 · two rays → (−∞, −3) ∪ (3, ∞)
Solve, graph, and write in interval notation: |2x − 1| ≤ 5
Apply the Less-Than Rule (≤ version): |2x − 1| ≤ 5 → −5 ≤ 2x − 1 ≤ 5
Add 1 to all three parts: −5 + 1 ≤ 2x ≤ 5 + 1 → −4 ≤ 2x ≤ 6
Divide all three parts by 2 (positive, no flip): −2 ≤ x ≤ 3
Graph: closed circle at −2, closed circle at 3, shade between.
Interval notation: [−2, 3]
Check: Test x = 0: |2(0) − 1| = 1 ≤ 5 ✓.
Check: Test x = 4: |2(4) − 1| = 7 ≤ 5 ✗ — not in solution (correct).
Graph: −2 ≤ x ≤ 3
Closed at −2 and 3 · segment → [−2, 3]
Solve, graph, and write in interval notation: |3x + 6| > 9
Apply the Greater-Than Rule: |3x + 6| > 9 → 3x + 6 < −9 OR 3x + 6 > 9
Left branch: 3x + 6 < −9 → 3x < −15 → x < −5
Right branch: 3x + 6 > 9 → 3x > 3 → x > 1
Combine with OR: x < −5 OR x > 1
Graph: open circle at −5, shade left; open circle at 1, shade right.
Interval notation: (−∞, −5) ∪ (1, ∞)
Check: Test x = −6: |3(−6) + 6| = |−12| = 12 > 9 ✓.
Check: Test x = 0: |3(0) + 6| = 6 > 9 ✗ — not in solution (correct).
Graph: x < −5 OR x > 1
Open at −5 and 1 · two rays → (−∞, −5) ∪ (1, ∞)
Solve: |x − 4| ≤ 0 (Special Case)
Absolute value is always ≥ 0 for any real number.
The only way |x − 4| ≤ 0 is if |x − 4| = 0 exactly.
|x − 4| = 0 → x − 4 = 0 → x = 4
There is exactly one solution: x = 4.
Graph: a single closed dot at 4 (not a segment or ray).
Interval notation: {4} (a single-element set)
Check: |4 − 4| = |0| = 0 ≤ 0 ✓.
Guided Practice — Inequalities
Answers are in the Answer Key section.
Solve and graph: |x| ≤ 6
Hint: Less-Than Rule (≤ version): |x| ≤ 6 → −6 ≤ x ≤ 6. Closed circles at −6 and 6, shade between. Interval: [−6, 6].
Graph your answer (|x| ≤ 6):
Solve and graph: |x| ≥ 2
Hint: Greater-Than Rule (≥ version): |x| ≥ 2 → x ≤ −2 OR x ≥ 2. Closed circles at −2 and 2, shade outward. Interval: (−∞, −2] ∪ [2, ∞).
Graph your answer (|x| ≥ 2):
Solve and graph: |x + 3| < 4
Hint: Less-Than Rule: −4 < x + 3 < 4. Subtract 3 from all parts: −7 < x < 1. Open circles at −7 and 1, shade between. Interval: (−7, 1).
Graph your answer (|x + 3| < 4):
Solve and graph: |2x − 4| ≥ 6
Hint: Greater-Than Rule: 2x − 4 ≤ −6 OR 2x − 4 ≥ 6. Left: 2x ≤ −2 → x ≤ −1. Right: 2x ≥ 10 → x ≥ 5. Interval: (−∞, −1] ∪ [5, ∞).
Graph your answer (|2x − 4| ≥ 6):
Solve: |x + 1| < 0
Hint: Absolute value is always ≥ 0. It can never be strictly less than 0. Solution: no solution (∅).
Key Vocabulary
Absolute Value Equation
An equation that contains an absolute value expression, such as |x| = 5 or |2x − 3| = 7. Solved using the Two-Case Method.
Example: |x − 3| = 5 → x = 8 or x = −2
Absolute Value Inequality
An inequality that contains an absolute value expression, such as |x| < 4 or |2x − 1| ≥ 3.
Example: |x − 5| < 2 means x is within 2 units of 5
Less-Than Rule (AND)
|x| < a → −a < x < a. The solution is a bounded segment.
|x| < 4 → (−4, 4)
Greater-Than Rule (OR)
|x| > a → x < −a OR x > a. The solution is two outward rays.
|x| > 3 → (−∞, −3) ∪ (3, ∞)
Tolerance
In manufacturing, the acceptable range of variation from a target value — modeled by an absolute value inequality.
Example: |m − 50| ≤ 0.5 means the mass m must be within 0.5 g of 50 g
Extraneous Solution
A value that satisfies a transformed equation but not the original. Always check solutions in the original absolute value equation.
Example: Check both cases in |2x + 1| = 5 by substituting back.
Interval Notation — Visual Reference
Interactive Practice — 7 Questions
How many solutions does |x| = 5 have?
Solve |2x − 4| = 10.
|x| < 5 is equivalent to which compound inequality?
|x| > 3 is equivalent to which compound inequality?
Solve |2x − 1| ≤ 5.
Solve |3x + 6| > 9.
What is the solution of |x − 4| ≤ 0?
Independent Practice
Answers are in the Answer Key section.
Equations
Independent Practice
Solve: |x| = 11
Solve: |x − 6| = 4
Solve: |3x + 9| = 15
Solve: |x + 2| = 0
Solve: |2x − 5| = −1
ChallengeInequalities
Independent Practice
Solve and graph: |x| < 7
Solve and graph: |x − 2| ≤ 4
Solve and graph: |x + 5| > 3
Solve and graph: |4x − 8| ≥ 12
A machine part must be within 0.02 cm of 5.00 cm. Write and solve an absolute value inequality for the acceptable length L.
ChallengeInequality Problem 1 graph:
Inequality Problem 2 graph:
Inequality Problem 3 graph:
Inequality Problem 4 graph:
Common Mistakes
Forgetting the negative case in an absolute value equation — writing only x = k and missing x = −k.
Always write both cases: expression = k AND expression = −k. Solve each separately and check both.
Applying the Two-Case Method to an inequality — writing x = k or x = −k when the problem has < or >.
Equations use the Two-Case Method. Inequalities use the Less-Than Rule (AND) or Greater-Than Rule (OR).
Applying the Greater-Than Rule to a less-than inequality — writing x < −a OR x > a when the symbol is <.
Less-than (< or ≤) always gives an AND (segment): −a < x < a. Greater-than (> or ≥) always gives an OR (two rays): x < −a OR x > a.
Forgetting to flip the left inequality sign when writing the AND form — writing a < x < a instead of −a < x < a.
The left boundary is always the negative of a. |x| < 4 → −4 < x < 4, not 4 < x < 4.
Saying |x + 1| < 0 has the solution x = −1.
Absolute value is always ≥ 0, so it can never be strictly negative. The solution is no solution (∅).
Math Tips
Two-Case Method for equations: |expression| = k always splits into two equations — expression = k and expression = −k. Solve both, check both.
Think distance: |x − a| < r means x is within r units of a. |x − a| = r means x is exactly r units from a (two points).
Less → AND, Greater → OR: "Less than" gives a bounded segment (AND); "greater than" gives two outward rays (OR).
Special cases: |expression| = 0 → one solution. |expression| < 0 → no solution. |expression| ≤ 0 → only the zero of the expression. |expression| = negative → no solution.
SAT/ACT tip: absolute value problems often appear as tolerance or error problems. Translate the words into |x − target| ≤ tolerance first.