Solve :

x + y = 7xy

2x - 3y = -xy

Given, equations : x + y = 7xy and 2x - 3y = -xy

Dividing both the sides of first equation by xy, we get :

$\Rightarrow \dfrac{x + y}{xy} = \dfrac{7xy}{xy} \\[1em] \Rightarrow \dfrac{x}{xy} + \dfrac{y}{xy} = 7 \\[1em] \Rightarrow \dfrac{1}{y} + \dfrac{1}{x} = 7 \\[1em]$

Multiplying both sides of the above equation by 3, we get :

$\Rightarrow 3\Big(\dfrac{1}{y} + \dfrac{1}{x}\Big) = 3 \times 7 \\[1em] \Rightarrow \dfrac{3}{y} + \dfrac{3}{x} = 21\text{ …….(1)}$

Dividing both the sides of second equation by xy, we get :

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

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

$\Rightarrow \Big(\dfrac{3}{y} + \dfrac{3}{x}\Big) + \Big(\dfrac{2}{y} - \dfrac{3}{x}\Big) = 21 + (-1) \\[1em] \Rightarrow \dfrac{5}{y} = 20 \\[1em] \Rightarrow y = \dfrac{5}{20} = \dfrac{1}{4}.$

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

$\Rightarrow \dfrac{3}{\dfrac{1}{4}} + \dfrac{3}{x} = 21 \\[1em] \Rightarrow 12 + \dfrac{3}{x} = 21 \\[1em] \Rightarrow \dfrac{3}{x} = 21 - 12 \\[1em] \Rightarrow \dfrac{3}{x} = 9 \\[1em] \Rightarrow x = \dfrac{3}{9} = \dfrac{1}{3}.$

Hence, $x = \dfrac{1}{3} \text{ and } y = \dfrac{1}{4}$.