Solution:

The given equations may be written as:
$\begin{array}{l}
2 x+5 y-1=0\dots \dots(i) \\
2 x+3 y-3=0\dots \dots(ii)
\end{array}$
Here $a_{1}=2, b_{1}=5, c_{1}=-1, a_{2}=2, b_{2}=3$ and $c_{2}=-3$
On cross multiplication, we get:

$\begin{array}{l}
\therefore \frac{x}{[5 \times(-3)-3 \times(-1)]}=\frac{y}{[(-1) \times 2-(-3) \times 2]}=\frac{1}{[2 \times 3-2 \times 5]} \\
\Rightarrow \frac{x}{(-15+3)}=\frac{y}{(-2+6)}=\frac{1}{(6-10)} \\
\Rightarrow \frac{x}{-12}=\frac{y}{4}=\frac{1}{-4} \\
\Rightarrow \mathrm{x}=\frac{-12}{-4}=3, \mathrm{y}=\frac{4}{-4}=-1
\end{array}$
As a result, $x=3$ and $y=-1$ is the required solution.