Absolute Value Equations and Inequalities

10 Solved Problems with Step-by-Step Solutions

Solving absolute value equations and inequalities is a core skill in algebra. Because the absolute value $|x|$ represents the distance from zero, most problems require evaluating two distinct cases.

Need a refresher? Check our absolute value tutorial before diving into the problems below.

Core Rules & Equivalences

Use these fundamental equivalences to eliminate absolute value symbols (where $k > 0$):

Problem Type Algebraic Equivalence
Equations \( |x| = k \iff x = k \text{ or } x = -k \)
Inequalities (<) \( |x| \lt k \iff -k \lt x \lt k \)
Inequalities (>) \( |x| \gt k \iff x \gt k \text{ or } x \lt -k \)

Part 1: Absolute Value Equations

Question 1: Solve \( 2|-2x - 2| - 3 = 13 \)

1. Isolate the absolute value: Add 3, then divide by 2:

\[ | - 2 x - 2| = 8 \]

2. Apply the rule \( x = k \) or \( x = -k \):

  • \( -2x - 2 = 8 \implies -2x = 10 \implies x = -5 \)
  • \( -2x - 2 = -8 \implies -2x = -6 \implies x = 3 \)

Solution Set: \( \{-5, 3\} \)

Question 2: Solve \( |x - 1| = 2x + 1 \)

Case A: \( x - 1 \ge 0 \implies x - 1 = 2x + 1 \implies x = -2 \). (Does not satisfy \( x \ge 1 \))

Case B: \( x - 1 \lt 0 \implies -(x - 1) = 2x + 1 \implies x = 0 \). (Satisfies \( x \lt 1 \))

Verification: At \( x=0 \), \( |0-1| = 2(0)+1 \implies 1 = 1 \). (True)

Solution Set: \( \{0\} \)

Question 3: Solve \( |2x - 1| = x^2 \)

Case 1: \( 2x - 1 = x^2 \implies x^2 - 2x + 1 = 0 \implies (x-1)^2 = 0 \implies x = 1 \)

Case 2: \( 2x - 1 = -x^2 \implies x^2 + 2x - 1 = 0 \)

\[ x = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = -1 \pm \sqrt{2} \]

Solution Set: \( \{1, -1-\sqrt{2}, -1+\sqrt{2}\} \)

Question 4: Solve \( -2|x + 1| - 2 = 4 \)
\[ -2|x + 1| = 6 \implies |x + 1| = -3 \]

Conclusion: No solution, as absolute value cannot be negative.

Question 5: Solve \( |x - 1| = |x^2 - 2x + 1| \)

Identity: \( x^2 - 2x + 1 = (x - 1)^2 \). Let \( u = |x - 1| \):

\[ u = u^2 \implies u(u - 1) = 0 \]

Solution Set: \( \{0, 1, 2\} \)

Part 2: Absolute Value Inequalities

Question 6: Solve \( |x + 2| \lt 3 \)
\[ -3 \lt x + 2 \lt 3 \implies -5 \lt x \lt 1 \]

Interval: \( (-5, 1) \)

Question 7: Solve \( -3|-2x + 4| \le 4 \)
\[ |-2x + 4| \ge -\frac{4}{3} \]

True for all real numbers. Solution: \( (-\infty, \infty) \)

Question 8: Solve \( |-2x - 4| \gt 9 \)
  • \( -2x - 4 \gt 9 \implies x \lt -6.5 \)
  • \( -2x - 4 \lt -9 \implies x \gt 2.5 \)

Interval: \( (-\infty, -6.5) \cup (2.5, \infty) \)

Question 9: Solve \( |x + 1| \le |x^2 + 2x + 1| \)

Solution Set: \( (-\infty , -2] \cup \{-1\} \cup [0 , \infty) \)

Sign analysis graph
Question 10: Solve \( x + 2 \lt |x^2 - 4| \)

Case 1 (|x| ≥ 2): \( (-\infty, -2) \cup (3, \infty) \)

Case 2 (|x| < 2): \( (-2, 1) \)

Combined: \( (-\infty, -2) \cup (-2, 1) \cup (3, \infty) \)

Graphical check

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