Solve the following pair of equations:

$\dfrac{2}{3(2x - y)} + \dfrac{1}{2(x + 2y)} = \dfrac{5}{12}, \dfrac{1}{2x - y} - \dfrac{2}{x + 2y} = \dfrac{1}{6}$

Substituting $\dfrac{1}{2x - y}$ = a and $\dfrac{1}{x + 2y}$ = b in above equations, we get :

⇒ $\dfrac{2}{3}a + \dfrac{1}{2}b = \dfrac{5}{12}$ ……….(1)

⇒ a - 2b = $\dfrac{1}{6}$ ………(2)

Multiplying equation (1) by $\dfrac{3}{2}$ we get :

$\Rightarrow \dfrac{3}{2}\Big(\dfrac{2}{3}a + \dfrac{1}{2}b\Big) = \dfrac{3}{2} \times \dfrac{5}{12} \\[1em] \Rightarrow a + \dfrac{3}{4}b = \dfrac{5}{8} \text{ ………..(3)}$

Subtracting equation (2) from (3), we get :

$\Rightarrow a + \dfrac{3}{4}b - (a - 2b) = \dfrac{5}{8} - \dfrac{1}{6} \\[1em] \Rightarrow a - a + \dfrac{3}{4}b + 2b = \dfrac{15 - 4}{24} \\[1em] \Rightarrow \dfrac{3b + 8b}{4} = \dfrac{11}{24} \\[1em] \Rightarrow \dfrac{11b}{4} = \dfrac{11}{24} \\[1em] \Rightarrow b = \dfrac{11}{24} \times \dfrac{4}{11} = \dfrac{1}{6}. \\[1em] \therefore \dfrac{1}{x + 2y} = \dfrac{1}{6} \\[1em] \Rightarrow x + 2y = 6 \\[1em] \Rightarrow x = 6 - 2y \text{ …………(4)}$

Substituting b = $\dfrac{1}{6}$ in equation (2), we get :

$\Rightarrow a - 2 \times \dfrac{1}{6} = \dfrac{1}{6} \\[1em] \Rightarrow a - \dfrac{1}{3} = \dfrac{1}{6} \\[1em] \Rightarrow a = \dfrac{1}{6} + \dfrac{1}{3} \\[1em] \Rightarrow a = \dfrac{1 + 2}{6} \\[1em] \Rightarrow a = \dfrac{3}{6} \\[1em] \Rightarrow a = \dfrac{1}{2} \\[1em] \therefore \dfrac{1}{2x - y} = \dfrac{1}{2}\\[1em] \Rightarrow 2x - y = 2 \text{ ……(5)}$

Substituting value of x from equation (4) in (5), we get :

⇒ 2(6 - 2y) - y = 2

⇒ 12 - 4y - y = 2

⇒ 12 - 5y = 2

⇒ -5y = 2 - 12

⇒ -5y = -10

⇒ y = $\dfrac{-10}{-5}$ = 2.

Substituting y = 2 in equation (4), we get :

⇒ x = 6 - 2y

⇒ x = 6 - 2(2) = 6 - 4 = 2.

Hence, x = 2 and y = 2.