Solve :

$\dfrac{3}{x+y} + \dfrac{2}{x-y} = 2$ and

$\dfrac{9}{x+y} - \dfrac{4}{x-y} = 1$

Let $\dfrac{1}{x + y}$ = u and $\dfrac{1}{x - y}$ = v.

The given equations are :

3u + 2v = 2

9u - 4v = 1

Multiply the first equation by 3 and subtract the second equation from it.

⇒ (3u + 2v = 2) x 3

$\begin{matrix} & 9u & + & 6v & = & 6 \ & 9u & - & 4v & = & 1 \ & - & + & & & - \ \hline & & & 10v & = & 6 - 1\ \Rightarrow & & & 10v & = & 5 \end{matrix}$

⇒ v = $\dfrac{5}{10}$

⇒ v = $\dfrac{1}{2}$

Substituting the value of v in first equation, we get:

⇒ 3u + 2 $\times \dfrac{1}{2}$ = 2

⇒ 3u + 1 = 2

⇒ 3u = 2 - 1

⇒ 3u = 1

⇒ u = $\dfrac{1}{3}$

So, x + y = $\dfrac{1}{u}$ = 3 and x - y = $\dfrac{1}{v}$ = 2

⇒ x + y = 3

And, x - y = 2

Adding both the equations, we get:

$\begin{matrix} & x & + & y & = & 3 \ & x & - & y & = & 2 \ \hline & 2x & & & = & 3 + 2\ \Rightarrow & 2x & & & = & 5 \end{matrix}$

⇒ x = $\dfrac{5}{2}$

Substituting the value of x in first equation, we get:

⇒ $\dfrac{5}{2}$ + y = 3

⇒ y = 3 - $\dfrac{5}{2}$

⇒ y = $\dfrac{6 - 5}{2}$

⇒ y = $\dfrac{1}{2}$

Hence, the value of x = $\dfrac{5}{2}$ and y = $\dfrac{1}{2}$.