Solution:

The given equations are
$\begin{array}{l}
\frac{2}{3 x+2 y}+\frac{3}{3 x-2 y}=\frac{17}{5}\dots \dots(i) \\
\frac{5}{3 x+2 y}+\frac{1}{3 x-2 y}=2\dots \dots(ii)
\end{array}$
Substituting $\frac{1}{3 x+2 y}=u$ and $\frac{1}{3 x-2 y}=\mathrm{v}$, in equation(i) and equation(ii), we obtain:
$\begin{array}{l}
2 \mathrm{u}+3 \mathrm{v}=\frac{17}{5}\dots \dots(iii) \\
5 \mathrm{u}+\mathrm{v}=2\dots \dots(iv)
\end{array}$
Multiplying equation(iv) by 3 and subtracting from equation(iii), we obtain:
$\begin{array}{l}
2 \mathrm{u}-15 \mathrm{u}=\frac{17}{5}-6 \\
\Rightarrow-13 \mathrm{u}=\frac{-13}{5} \Rightarrow \mathrm{u}=\frac{1}{5} \\
\Rightarrow 3 \mathrm{x}+2 \mathrm{y}=5 \quad\left(\because \frac{1}{3 x+2 y}=u\right)\dots \dots(v)
\end{array}$
Now, substituting $\mathrm{u}=\frac{1}{5}$ in equation(iv), we obtain
$\begin{array}{l}
1+\mathrm{v}=2 \Rightarrow \mathrm{v}=1 \\
\Rightarrow 3 \mathrm{x}-2 \mathrm{y}=1 \quad\left(\because \frac{1}{3 x-2 y}=v\right)\dots \dots(vi)
\end{array}$
Adding equation(v) and equation(vi), we obtain:
$\Rightarrow 6 \mathrm{x}=6 \Rightarrow \mathrm{x}=1$
Substituting $\mathrm{x}=1$ in $(\mathrm{v})$, we obtain:
$3+2 y=5 \Rightarrow y=1$
As a result, $\mathrm{x}=1$ and $\mathrm{y}=1$.