Solve :

$\dfrac{3}{2x} + \dfrac{2}{3y} = -\dfrac{1}{3}$

$\dfrac{3}{4x} + \dfrac{1}{2y} = -\dfrac{1}{8}$

Given, equations :

$\dfrac{3}{2x} + \dfrac{2}{3y} = -\dfrac{1}{3}$ …….(1)

$\dfrac{3}{4x} + \dfrac{1}{2y} = -\dfrac{1}{8}$ …….(2)

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

$\Rightarrow \dfrac{1}{2} \times \Big(\dfrac{3}{2x} + \dfrac{2}{3y}\Big) = \dfrac{1}{2} \times -\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{3}{2x} + \dfrac{1}{2} \times \dfrac{2}{3y} = -\dfrac{1}{6} \\[1em] \Rightarrow \dfrac{3}{4x} + \dfrac{1}{3y} = -\dfrac{1}{6} \text{ ……..(3)}$

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

$\Rightarrow \Big(\dfrac{3}{4x} + \dfrac{1}{2y}\Big) - \Big(\dfrac{3}{4x} + \dfrac{1}{3y}\Big) = -\dfrac{1}{8} - \Big(-\dfrac{1}{6}\Big) \\[1em] \Rightarrow \dfrac{3}{4x} - \dfrac{3}{4x} + \dfrac{1}{2y} - \dfrac{1}{3y} = -\dfrac{1}{8} + \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{3 - 2}{6y} = \dfrac{-3 + 4}{24} \\[1em] \Rightarrow \dfrac{1}{6y} = \dfrac{1}{24} \\[1em] \Rightarrow y = \dfrac{24}{6} = 4.$

Substituting value of y in equation (1), we get :

$\Rightarrow \dfrac{3}{2x} + \dfrac{2}{3 \times 4} = -\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{3}{2x} + \dfrac{1}{6} = -\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{3}{2x} = -\dfrac{1}{3} - \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{3}{2x} = \dfrac{-2 - 1}{6} \\[1em] \Rightarrow \dfrac{3}{2x} = \dfrac{-3}{6} \\[1em] \Rightarrow x = \dfrac{3 \times 6}{2 \times -3}\\[1em] \Rightarrow x = \dfrac{18}{-6} = -3.$

Substituting value of x in equation (1), we get :

$\Rightarrow \dfrac{3}{2 \times -3} + \dfrac{2}{3y} = -\dfrac{1}{3} \\[1em] \Rightarrow -\dfrac{1}{2} + \dfrac{2}{3y} = -\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{2}{3y} = -\dfrac{1}{3} + \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{2}{3y} = \dfrac{-2 + 3}{6} \\[1em] \Rightarrow \dfrac{2}{3y} = \dfrac{1}{6} \\[1em] \Rightarrow y = \dfrac{2 \times 6}{3} \\[1em] \Rightarrow y = \dfrac{12}{3} = 4.$

Hence, x = -3 and y = 4.