Solve $\dfrac{2}{3} (x - 1) + 4 < 10$ and represent its solution on a number line. Given the replacement set is {-8, -6, -4, 3, 6, 8, 12}.

$\dfrac{2}{3} (x - 1) + 4 < 10$

⇒ $\dfrac{2}{3}x - \dfrac{2}{3} + 4 < 10$

⇒ $\dfrac{2}{3}x - \dfrac{2}{3} < 10 - 4$

⇒ $\dfrac{2}{3}x - \dfrac{2}{3} < 6$

⇒ $\dfrac{2x - 2}{3} < 6$

By cross multiplying, we get

⇒ 2x - 2 < 6 $\times$ 3

⇒ 2x - 2 < 18

⇒ 2x < 18 + 2

⇒ 2x < 20

⇒ x < $\dfrac{20}{2}$

⇒ x < 10

∵ The replacement set is {-8, -6, -4, 3, 6, 8, 12}.

Hence, solution set = {-8, -6, -4, 3, 6, 8}.