Solve : $\dfrac{2}{3}x + 5 ≤ \dfrac{1}{2}x + 6; x ∈ W$

$\dfrac{2}{3}x + 5 ≤ \dfrac{1}{2}x + 6$

⇒ $\dfrac{2}{3}x - \dfrac{1}{2}x ≤ 6 - 5$

⇒ $\dfrac{2}{3}x - \dfrac{1}{2}x ≤ 1$

since L.C.M. of denominator 3 and 2 = 6, multiply each term with 6 to get:

⇒ $\dfrac{2 \times 6}{3}x - \dfrac{1 \times 6}{2}x ≤ 1 \times 6$

⇒ 2 $\times$ 2x - 1 $\times$ 3x ≤ 6

⇒ 4x - 3x ≤ 6

⇒ x ≤ 6

∵ x is a whole number.

Hence, solution set = {0, 1, 2, 3, 4, 5, 6}.