Solution:

The given system of equation is:
$\begin{array}{c}
3 \mathrm{x}-5 \mathrm{y}-19=0 \\
-7 \mathrm{x}+3 \mathrm{y}+1=0
\end{array}$
When multiplying equation(i) by 3 and equation(ii) by 5 , we obtain:
$9 x-15 y=57 \quad \ldots \ldots$ (iii)
$-35 x+15 y=-5 \ldots \ldots$ (iv)
Subtracting equation(iii) from equation(iv) we obtain:
$\begin{array}{l}
-26 x=(57-5)=52 \\
\Rightarrow x=-2
\end{array}$
When substituting the value of $\mathrm{x}=-2$ in equation(i), we obtain:
$\begin{array}{l}
-6-5 y-19=0 \\
\Rightarrow 5 y=(-6-19)=-25 \\
\Rightarrow y=-5
\end{array}$
As a result, the solution is $x=-2$ and $y=-5$.