Solve system of linear equations, using matrix method.

The given system of equations can be written in the form of AX = B, where

$A=\left[\begin{array}{ll}4 & -3 \\ 3 & -5\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 7\end{array}\right]$

Now,

$|A|=-20+9=-11 \neq 0$

Thus, A is non-singular. Therefore, its inverse exists.

Now,

$A^{-1}=\frac{1}{|A|}($ adjA $)=-\frac{1}{11}\left[\begin{array}{ll}-5 & 3 \\ -3 & 4\end{array}\right]=\frac{1}{11}\left[\begin{array}{ll}5 & -3 \\ 3 & -4\end{array}\right]$

$\therefore X=A^{-1} B=\frac{1}{11}\left[\begin{array}{ll}5 & -3 \\ 3 & -4\end{array}\right]\left[\begin{array}{l}3 \\ 7\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{ll}5 & -3 \\ 3 & -4\end{array}\right]\left[\begin{array}{l}3 \\ 7\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}15-21 \\ 9-28\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}-6 \\ -19\end{array}\right]=\left[\begin{array}{c}-\frac{6}{11} \\ -\frac{19}{11}\end{array}\right]$

Hence, $x=\frac{-6}{11}$ and $y=\frac{-19}{11}$.