Description for Correct answer:
Given, \( \Large \frac{3}{x+y}+\frac{2}{x-y}=2 \) ...(i)

and \( \Large \frac{9}{x+y}-\frac{4}{x-y}=1 \) ...(ii)

Let x + y = a and x - y = b

On multiplying Eq. (i) by 3 and subtracting from Eq. (ii), we get

\( \Large \frac{9}{a}-\frac{4}{b}=1 \)

\( \Large \frac{9}{a}+\frac{6}{b}=6 \)

\( \Large \frac{-10}{b}=5 \)

Therefore, b = 2

Now, on putting the value of b in Eq. (i), we get

\( \Large \frac{3}{a}+\frac{2}{2}=2 \)

\( \Large \frac{3}{a}=2-1 \)

= a = 3 = x+y = 3 ...(iii)

and x-y = 2 ...(iv)

On subtracting Eq. (iv) from Eq. (iii), we get

2y = 1

Therefore, \( \Large y=\frac{1}{2} \)

Now, putting \( \Large y=\frac{1}{2} \) in Eq. (iii), we get

\( \Large x+\frac{1}{2}=3 \)

= x = \( \Large 3-\frac{1}{2}=\frac{6-1}{2}=\frac{5}{2} \)

Therefore, \( \Large \frac{x}{y}= \frac{\frac{5}{2}}{\frac{1}{2}}=\frac{5}{2} \times \frac{2}{1}=5 \)