\(\frac{2}{3x+2y}\)+\(\frac{3}{3x-2y}\) = \(\frac{17}{5}\)
\(\frac{5}{3x+2y}\) + \(\frac{1}{3x-2y}\) = 2
Multiplying eq1 by 5 and eq2 by 2 and subtracting
⇒ 13/(3x – 2y) = 13
⇒ 3x – 2y = 1------- (3)
Multiplying eq2 by 3 and subtracting eq1 from it
⇒ 13/(3x +2y) = 13/5
⇒ 3x + 2y = 5 ------ (4)
(3) + (4)
⇒ 6x = 6
⇒ x = 1
Thus,
3 + 2 = 5
⇒ y = 1