Solve : $\dfrac{2x + 1}{3}$ + 15 ≤ 17; x ∈ W.

$\dfrac{2x + 1}{3}$ + 15 ≤ 17

Multiply each term with 3 to get:

⇒ $\dfrac{3(2x + 1)}{3}$ + 15 $\times$ 3 ≤ 17 $\times$ 3

⇒ 2x + 1 + 45 ≤ 51

⇒ 2x + 46 ≤ 51

⇒ 2x ≤ 51 - 46

⇒ 2x ≤ 5

⇒ x ≤ $\dfrac{5}{2}$

⇒ x ≤ 2.5

∵ As x is a whole number.

Hence, solution set = {0, 1, 2}.