Answer: d
|4x + 1| ≤ 5
⇒ - 5 ≤ 4x + 1 ≤ 5
⇔ -6 ≤ 4x ≤ 4
⇔ -1.5 ≤ x ≤ 1
⇒ x ∈ [-1.5; 1]
Step-by-step explanation:
