Solution:

The given equations may be written as:
$2 x+y-35=0\dots \dots(i)$
$3 \mathrm{x}+4 \mathrm{y}-65=0 \quad \ldots \ldots(ii)$
Here $a_{1}=2, b_{1}=1, c_{1}=-35, a_{2}=3, b_{2}=4$ and $c_{2}=-65$
On cross multiplication, we obtain:

$\begin{array}{l}
{ }^{1} \sum_{-65}^{x} \sum_{4}^{-35} \sum^{1}{ }^{2} \sum_{4}^{1} \\
\therefore \frac{x}{[1 \times(-65)-4 \times(-35)]}=\frac{y}{[(-35) \times 3-(-65) \times 2]}=\frac{1}{[2 \times 4-3 \times 1]} \\
\Rightarrow \frac{x}{(-65+140)}=\frac{y}{(-105+130)}=\frac{1}{(8-3)} \\
\Rightarrow \frac{x}{75}=\frac{y}{25}=\frac{1}{5} \\
\Rightarrow x=\frac{75}{5}=15, y=\frac{25}{5}=5
\end{array}$
As a result, $x=15$ and $y=5$ is the required solution.