Solve the following simultaneous linear equations:

$\dfrac{30}{x - y} + \dfrac{44}{x + y} = 10$

$\dfrac{40}{x - y} + \dfrac{55}{x + y} = 13$

Substituting $\dfrac{1}{x - y} = a \text{ and } \dfrac{1}{x + y} = b$ in above equations we get,

30a + 44b = 10 …….(i)

40a + 55b = 13 …….(ii)

Multiplying (i) by 4 and (ii) by 3 we get,

120a + 176b = 40 …….(iii)

120a + 165b = 39 …….(iv)

Subtracting eq. (iii) from (iv) we get,

⇒ 120a + 165b - (120a + 176b) = 39 - 40

⇒ 165b - 176b = -1

⇒ -11b = -1

⇒ b = $\dfrac{1}{11}$.

$\therefore \dfrac{1}{x + y} = \dfrac{1}{11} \\[1em] \Rightarrow x + y = 11 ………(v)$

Substituting value of b in (iv),

⇒ 120a + $165 \times \dfrac{1}{11}$ = 39

⇒ 120a + 15 = 39

⇒ 120a = 24

⇒ a = $\dfrac{24}{120} = \dfrac{1}{5}$.

$\therefore \dfrac{1}{x - y} = \dfrac{1}{5} \\[1em] \Rightarrow x - y = 5 ……..(vi)$

Adding eq. (v) and (vi) we get,

⇒ x + y + (x - y) = 11 + 5

⇒ 2x = 16

⇒ x = 8.

Substituting value of x in (v)

⇒ x + y = 11

⇒ 8 + y = 11

⇒ y = 11 - 8 = 3.

Hence, x = 8 and y = 3.