Solution:
The given system of equations is
$x-y=3 \dots \dots(i)$
$\frac{x}{3}+\frac{y}{2}=6 \dots \dots(ii)$
From equation(i), write y in terms of $x$ to obtain $y=x-3$
When substituting $\mathrm{y}=\mathrm{x}-3$ in equation(ii), we obtain
$\begin{array}{l}
\frac{x}{3}+\frac{x-3}{2}=6 \\
\Rightarrow 2 x+3(x-3)=36 \\
\Rightarrow 2 x+3 x-9=36 \\
\Rightarrow x=\frac{45}{5}=9
\end{array}$
Substituting $\mathrm{x}=9$ in equation(i), we get
$\begin{array}{l}
9-y=3 \\
\Rightarrow y=9-3=6
\end{array}$
As a result, $x=9$ and $y=6$.