Solve (a) |3x + 1| < 8 and (b) |x - 2| ≥ 5.

Step 1. (a). $-9 < 3x < 7$, so $-3 < x < \tfrac{7}{3}$, where $\tfrac{7}{3} \approx 2.33$: the interval $\left(-3, \tfrac{7}{3}\right)$.

Step 2. (b). $x \geq 7$ or $x \leq -3$: $(-\infty, -3] \cup [7, \infty)$.