Solve: (2x + 1)/3 + 15 < 17; x ∈ W
(2x + 1)/3 + 15 < 17; x ∈ W
⇒ (2x + 1)/3 ≤ 17 – 15 = 2
⇒ 2x + 1 ≤ 6 ⇒ 2x ≤ 5
⇒ x ≤ (5/2) = 2. ½
But x ∈ W
∴ x = 0, 1, 2
∴ Solution set is = {0, 1, 2}