Modulus equations and inequalities practice questions.

Determine the solution set of |x − 2| < 3.

• −1 < x < 5
• x < 5
• −5 < x < 1
• x < −1 or x > 5

Answer: −1 < x < 5

|x − 2| < 3 means the distance from x to 2 is less than 3, i.e. −3 < x − 2 < 3, so −1 < x < 5. The union of outer intervals would solve |x − 2| > 3 instead.

Determine the solution set of |3x + 1| ≥ 5.

• −2 ≤ x ≤ 4/3
• x ≤ −2 or x ≥ 4/3
• x ≤ −4/3 or x ≥ 2
• x ≥ 4/3

Answer: x ≤ −2 or x ≥ 4/3

|3x + 1| ≥ 5 splits into 3x + 1 ≥ 5, giving x ≥ 4/3, or 3x + 1 ≤ −5, giving x ≤ −2. For a linear expression and a positive right-hand side, a 'greater than' modulus inequality gives two outward rays; the bounded interval −2 ≤ x ≤ 4/3 solves |3x + 1| ≤ 5 instead.

Determine the number of real solutions of |x² − 4| = 2.

• 2
• 3
• 4
• 0

Answer: 4

The equation splits into x² − 4 = 2, giving x = ±√6, and x² − 4 = −2, giving x = ±√2 — four solutions in total. Forgetting the negative case x² − 4 = −2 leads to the wrong answer 2.

Solve the equation |x + 1| = |2x − 4|.

• x = 1 only
• x = 5 only
• x = −1 and x = 5
• x = 1 and x = 5

Answer: x = 1 and x = 5

With moduli on both sides, set x + 1 = 2x − 4, giving x = 5, and x + 1 = −(2x − 4), giving x = 1; both check in the original equation. Equivalently, squaring both sides gives 3x² − 18x + 15 = 0, i.e. x = 1 or x = 5.

Determine the solution set of |x| > |x − 2|.

• x > 2
• x < 1
• x > 1
• 0 < x < 2

Answer: x > 1

Both sides are non-negative, so square: x² > x² − 4x + 4, which simplifies to 4x > 4, i.e. x > 1. Geometrically, x must be closer to 2 than to 0, so x lies to the right of the midpoint 1.

Solve the inequality |2x − 5| ≤ x + 1.

• 4/3 ≤ x ≤ 6
• x ≤ 6
• −6 ≤ x ≤ 4/3
• x ≥ 4/3

Answer: 4/3 ≤ x ≤ 6

|2x − 5| ≤ x + 1 is equivalent to −(x + 1) ≤ 2x − 5 ≤ x + 1; the left part gives x ≥ 4/3 and the right part gives x ≤ 6. Both boundary values give equality, so the solution is the closed interval 4/3 ≤ x ≤ 6.