Solve the following systems of equations:

The given equations are:

$\frac{2}{3 x+2 y}+\frac{3}{3 x-2 y}=\frac{17}{5}$

$\frac{5}{3 x+2 y}+\frac{1}{3 x-2 y}=2$

Let $\frac{1}{3 x+2 y}=u$ and $\frac{1}{3 x-2 y}=v$ then equations are

$2 u+3 v=\frac{17}{5}$ $\ldots(i)$

$5 u+v=2 \ldots(i i)$

Multiply equation (ii) by 3 and subtract (ii) from (i), we get

$2 u+3 v=\frac{17}{5}$

Put the value of $u$ in equation $(i)$, we get

$2 \times \frac{1}{5}+3 v=\frac{17}{5}$

$\Rightarrow 3 v=3$

$\Rightarrow v=1$

Then 

$\frac{1}{3 x+2 y}=\frac{1}{5}$$\ldots(i i i)$

$\Rightarrow 3 x+2 y=5$

$\frac{1}{3 x-2 y}=1$$\ldots(i v)$

$\Rightarrow 3 x-2 y=1$

Add both equations, we get

Put the value of $x$ in equation (iii) we get

$3 \times 1+2 y=5$

$\Rightarrow 2 y=2$

$\Rightarrow y=1$

Hence the value of $x=1$ and $y=1$.