Answer: $($ a) $x=2, y=3$

Solution:
The given system is
$\begin{array}{l}
\frac{2 x}{3}-\frac{y}{2}=-\frac{1}{6}\dots \dots(i) \\
\frac{x}{2}+\frac{2 y}{3}=3\dots \dots(ii)
\end{array}$
Multiplying equation(i) and equation(ii) by 6 , we get
$\begin{array}{l}
4 x-3 y=-1\dots \dots(iii) \\
3 x+4 y=18\dots \dots(iv)
\end{array}$
Multiplying equation(iii) by 4 and equation(iv) by 3 and adding, we obtain
$\begin{array}{l}
16 x+9 x=-4+54 \\
\Rightarrow x=\frac{50}{25}=2
\end{array}$
Now, putting $\mathrm{x}=2$ in equation(iv), we have
$3 \times 2+4 y=18 \Rightarrow y=\frac{18-6}{4}=3$
Therefore, $x=2$ and $y=3$.