\(\frac{22}{x+y}\)+\(\frac{15}{x-y}\) = 5
\(\frac{55}{x+y}\)+\(\frac{45}{x-y}\) = 14
Multiplying eq1 by 3 and subtracting from eq2
⇒ - 11/(x + y) = - 1
⇒ x + y = 11 ---------- (3)
Multiplying eq1 by 5 and eq2 by 2 and subtracting
⇒ 15/(x – y) = 3
⇒ x – y = 5 ------ (4)
(3) + (4)
⇒ 2x = 16
⇒ x = 8
Thus,
y = 3