Solve :

(i) $\dfrac{x}{2} + 5 \le \dfrac{x}{3} + 6$, where x is a positive odd integer

(ii) $\dfrac{2x + 3}{3} \ge \dfrac{3x - 1}{4}$, where x is a positive even integer

(i) Solving,

$\Rightarrow \dfrac{x}{2} + 5 \le \dfrac{x}{3} + 6 \\[1em] \Rightarrow \dfrac{x}{2} - \dfrac{x}{3} \le 6 - 5 \\[1em] \Rightarrow \dfrac{3x - 2x}{6} \le 1 \\[1em] \Rightarrow \dfrac{x}{6} \le 1 \\[1em] \Rightarrow x \le 6.$

Since, x is a positive odd integer

∴ Solution set = {1, 3, 5}.

(ii) Solving,

$\Rightarrow \dfrac{2x + 3}{3} \ge \dfrac{3x - 1}{4} \\[1em] \Rightarrow \dfrac{2x + 3}{3} - \dfrac{3x - 1}{4} \ge 0 \\[1em] \Rightarrow \dfrac{4(2x + 3) - 3(3x - 1)}{12} \ge 0 \\[1em] \Rightarrow 8x + 12 - 9x + 3 \ge 0 \\[1em] \Rightarrow -x + 15 \ge 0 \\[1em] \Rightarrow x \le 15.$

Since, x is a positive even integer

∴ Solution set = {2, 4, 6, 8, 10, 12, 14}.