Solution:

The given eq. are:
$\begin{array}{l}
\frac{5}{x}-\frac{3}{y}=1 \ldots \ldots(i) \\
\frac{3}{2 x}+\frac{2}{3 y}=5 \ldots \ldots (ii)
\end{array}$
Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we obtain:
$5 u-3 v=1 \ldots \ldots$ (iii)
$\Rightarrow \frac{3}{2} u+\frac{2}{3} v=5$
$\Rightarrow \frac{9 u+4 v}{6}=5$
$\Rightarrow 9 u+4 v=30 \ldots \ldots$ (iv)
On multiplying equation(iii) by 4 and equation(iv) by 3, we obtain:
$20 \mathrm{u}-12 \mathrm{v}=4 \ldots \ldots \ldots(\mathrm{v})$
$27 \mathrm{u}+12 \mathrm{v}=90 \ldots \ldots$.. (vi)
On adding equation(iv) and equation(v), we obtain:
$\begin{array}{l}
47 \mathrm{u}=94 \Rightarrow \mathrm{u}=2 \\
\Rightarrow \frac{1}{x}=2 \Rightarrow \mathrm{x}=\frac{1}{2}
\end{array}$
On substituting $\mathrm{x}=\frac{1}{2}$ in equation(i), we obtain:
$\begin{array}{l}
\frac{5}{1 / 2}-\frac{3}{y}=1 \\
\Rightarrow 10-\frac{3}{y}=1 \Rightarrow \frac{3}{y}=(10-1)=9 \\
y=\frac{3}{9}=\frac{1}{3}
\end{array}$
As a result, the required solution is $\mathrm{x}=\frac{1}{2}$ and $\mathrm{y}=\frac{1}{3}$.