Solve : $\dfrac{2}{3}$ x + 8 < 12 ; x ∈ W.

$\dfrac{2}{3}$ x + 8 < 12

⇒ $\dfrac{2}{3}$ x < 12 - 8

⇒ $\dfrac{2}{3}$ x < 4

⇒ x < $\dfrac{4 \times 3}{2}$

⇒ x < $\dfrac{12}{2}$

⇒ x < 6

∵ x is a whole number.

Hence, solution set = {0, 1, 2, 3, 4, 5}.