Solve :

$\dfrac{20}{x + y} + \dfrac{3}{x - y} = 7$

$\dfrac{8}{x - y} - \dfrac{15}{x + y} = 5$

Given, equations :

$\dfrac{20}{x + y} + \dfrac{3}{x - y} = 7$ …….(1)

$\dfrac{8}{x - y} - \dfrac{15}{x + y} = 5$ ………….(2)

Multiplying equation (1) by 8, we get :

$\Rightarrow 8\Big(\dfrac{20}{x + y} + \dfrac{3}{x - y}\Big) = 8 \times 7 \\[1em] \Rightarrow \dfrac{160}{x + y} + \dfrac{24}{x - y} = 56 \text{ ……….(3)}$

Multiplying equation (2) by 3, we get :

$\Rightarrow 3\Big(\dfrac{8}{x - y} - \dfrac{15}{x + y}\Big) = 3 \times 5 \\[1em] \Rightarrow \dfrac{24}{x - y} - \dfrac{45}{x + y} = 15 \text{ ………(4)}$

Subtracting equation (4) from (3), we get :

$\Rightarrow \dfrac{160}{x + y} + \dfrac{24}{x - y} - \Big(\dfrac{24}{x - y} - \dfrac{45}{x + y}\Big) = 56 - 15 \\[1em] \Rightarrow \dfrac{160}{x + y} + \dfrac{45}{x + y} + \dfrac{24}{x - y} - \dfrac{24}{x - y} = 41 \\[1em] \Rightarrow \dfrac{205}{x + y} = 41 \\[1em] \Rightarrow x + y = \dfrac{205}{41} \\[1em] \Rightarrow x + y = 5 \text{ ……..(5)}$

Substituting value of x + y from equation 5 in equation 1, we get :

$\Rightarrow \dfrac{20}{x + y} + \dfrac{3}{x - y} = 7 \\[1em] \Rightarrow \dfrac{20}{5} + \dfrac{3}{x - y} = 7 \\[1em] \Rightarrow 4 + \dfrac{3}{x - y} = 7 \\[1em] \Rightarrow \dfrac{3}{x - y} = 7 - 4 \\[1em] \Rightarrow \dfrac{3}{x - y} = 3 \\[1em] \Rightarrow x - y = \dfrac{3}{3} \\[1em] \Rightarrow x - y = 1 \text{ ……..(6)}$

Adding equation (5) and (6), we get :

⇒ (x + y) + (x - y) = 5 + 1

⇒ 2x = 6

⇒ x = $\dfrac{6}{2}$

⇒ x = 3.

Substituting value of x in equation (6), we get :

⇒ 3 - y = 1

⇒ y = 3 - 1 = 2.

Hence, x = 3 and y = 2.