Solve the following simultaneous equations:

$\dfrac{22}{x + y} + \dfrac{15}{x - y} = 5,\dfrac{55}{x + y} + \dfrac{40}{x - y} = 13$

Given,

Equations:

$\dfrac{22}{x + y} + \dfrac{15}{x - y} = 5$ ……….(1)

$\dfrac{55}{x + y} + \dfrac{40}{x - y} = 13$ ……….(2)

Multiplying equation (1) by 5, we get:

$\Rightarrow 5\Big(\dfrac{22}{x + y} + \dfrac{15}{x - y}\Big) = 5 \times 5 \\[1em] \Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{75}{x - y}\Big) = 25 \text{ ……….(3) }$

Multiplying equation (2) by 2, we get:

$\Rightarrow 2\Big(\dfrac{55}{x + y} + \dfrac{40}{x - y}\Big) = 13 \times 2 \\[1em] \Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{80}{x - y}\Big) = 26 \text{ ……….(4) }$

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

$\Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{80}{x - y}\Big) - \Big(\dfrac{110}{x + y} + \dfrac{75}{x - y}\Big) = 26 - 25 \\[1em] \Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{80}{x - y} - \dfrac{110}{x + y} - \dfrac{75}{x - y}\Big) = 1 \\[1em] \Rightarrow \Big(\dfrac{80 - 75}{x - y}\Big) = 1 \\[1em] \Rightarrow \Big(\dfrac{5}{x - y}\Big) = 1 \\[1em] \Rightarrow 5 = (x - y) \\[1em] \Rightarrow x - y = 5 \text{ ………(5) }$

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

$\Rightarrow \dfrac{22}{x + y} + \dfrac{15}{x - y} = 5 \\[1em] \Rightarrow \dfrac{22}{x + y} + \dfrac{15}{5} = 5 \\[1em] \Rightarrow \dfrac{22}{x + y} + 3 = 5 \\[1em] \Rightarrow \dfrac{22}{x + y} = 5 - 3 \\[1em] \Rightarrow \dfrac{22}{x + y} = 2 \\[1em] \Rightarrow x + y = \dfrac{22}{2} \\[1em] \Rightarrow x + y = 11 \text{ …….(6) }$

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

$\Rightarrow x + y + x - y = 11 + 5 \\[1em] \Rightarrow 2x = 16 \\[1em] \Rightarrow x = \dfrac{16}{2} = 8.$

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

⇒ x + y = 11

⇒ 8 + y = 11

⇒ y = 11 - 8

⇒ y = 3

Hence, x = 8, y = 3.