Solve :

$\dfrac{2xy}{x + y} = \dfrac{3}{2}$

$\dfrac{xy}{2x - y} = -\dfrac{3}{10}$

x + y ≠ 0 and 2x - y ≠ 0

Simplifying first equation :

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

Simplifying second equation :

$\Rightarrow \dfrac{xy}{2x - y} = -\dfrac{3}{10} \\[1em] \Rightarrow \dfrac{2x - y}{xy} = -\dfrac{10}{3} \\[1em] \Rightarrow \dfrac{2x}{xy} - \dfrac{y}{xy} = -\dfrac{10}{3} \\[1em] \Rightarrow \dfrac{2}{y} - \dfrac{1}{x} = -\dfrac{10}{3} \text{ ……….(2)}$

Adding equations (1) and (2), we get :

$\Rightarrow \dfrac{1}{y} + \dfrac{1}{x} + \Big(\dfrac{2}{y} - \dfrac{1}{x}\Big) = \dfrac{4}{3} + \Big(-\dfrac{10}{3}\Big) \\[1em] \Rightarrow \dfrac{1}{y} + \dfrac{2}{y} + \dfrac{1}{x} - \dfrac{1}{x} = \dfrac{4 - 10}{3} \\[1em] \Rightarrow \dfrac{3}{y} = \dfrac{-6}{3} \\[1em] \Rightarrow y = \dfrac{3 \times 3}{-6} \\[1em] \Rightarrow y = -\dfrac{9}{6} = -\dfrac{3}{2}.$

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

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

Hence, $x = \dfrac{1}{2} \text{ and } y = -\dfrac{3}{2}$.