Solve the following simultaneous equations:

$\dfrac{3}{x + y} + \dfrac{2}{x - y} = 3,\;\dfrac{2}{x + y} + \dfrac{3}{x - y} = \dfrac{11}{3}$

Given,

Equations:

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

$\dfrac{2}{x + y} + \dfrac{3}{x - y} = \dfrac{11}{3}$ ……….(2)

Multiplying equation (1) by 2, we get:

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

Multiplying equation (2) by 3, we get:

$\Rightarrow 3\Big(\dfrac{2}{x + y} + \dfrac{3}{x - y}\Big) = \dfrac{11}{3} \times 3 \\[1em] \Rightarrow \Big(\dfrac{6}{x + y} + \dfrac{9}{x - y}\Big) = 11 \text{ ………(4) }$

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

$\Rightarrow \Big(\dfrac{6}{x + y} + \dfrac{9}{x - y}\Big) - \Big(\dfrac{6}{x + y} + \dfrac{4}{x - y}\Big) = 11 - 6 \\[1em] \Rightarrow \Big(\dfrac{6}{x + y} + \dfrac{9}{x - y} - \dfrac{6}{x + y} - \dfrac{4}{x - y}\Big) = 5 \\[1em] \Rightarrow \Big(\dfrac{9}{x - y} - \dfrac{4}{x - y}\Big) = 5 \\[1em] \Rightarrow \Big(\dfrac{5}{x - y}\Big) = 5 \\[1em] \Rightarrow 5 = 5(x - y) \\[1em] \Rightarrow x - y = 1 \text{ ………(5) }$

Substituting value of x - y from equation (5) in equation (1), we get:

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

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

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

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

⇒ x + y = 3

⇒ 2 + y = 3

⇒ y = 3 - 2

⇒ y = 1.

Hence, x = 2, y = 1.