Answer
The required value is $x=\frac{32}{7},y=\frac{-20}{7}$
Work Step by Step
We know that the system can be rewritten as:
$\begin{align}
& 5x-6y=40 \\
& 3x+2y=8
\end{align}$
By adding three times the second equation to the first equation:
$\begin{align}
& 5x-6y+3\left( 3x+2y \right)=40+3\left( 8 \right) \\
& 5x-6y+9x+6y=40+24 \\
& 14x=64 \\
& x=\frac{32}{7}
\end{align}$
Substitute the value of $x=\frac{32}{7}$ in the first equation and solve for y:
$\begin{align}
& 5\left( \frac{32}{7} \right)-6y=40 \\
& -6y=40-\frac{160}{7} \\
& \frac{-6y}{-6}=\frac{1}{-6}\left( \frac{120}{7} \right) \\
& y=\frac{-20}{7}
\end{align}$
Thus, the value of $x=\frac{32}{7}$ and $y=\frac{-20}{7}$.
