Solve system of linear equations, using matrix method.

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

$A=\left[\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$

Now, $|A|=15-14=1 \neq 0$.

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

Now,

$A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)$

$\therefore A^{-1}=\left[\begin{array}{rr}3 & -2 \\ -7 & 5\end{array}\right]$

$\therefore A^{-1}=\left[\begin{array}{rr}3 & -2 \\ -7 & 5\end{array}\right]$

$\therefore X=A^{-1} B=\left[\begin{array}{rr}3 & -2 \\ -7 & 5\end{array}\right]\left[\begin{array}{l}4 \\ 5\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}12-10 \\ -28+25\end{array}\right]=\left[\begin{array}{c}2 \\ -3\end{array}\right]$

Hence, $x=2$ and $y=-3$.