The key idea here is to multiply the L.H.S. matrices and equate to the R.H.S, thereafter we solve the simultaneous equations so formed in \(x\) and \(y\).y matrix multiplication,
\begin{aligned}
\left(\begin{array}{c}
3(x)+(-4)(y) \\
5(x)+7 y
\end{array}\right) &=\left(\begin{array}{c}
10 \\
3
\end{array}\right) \\
\left(\begin{array}{l}
3 x-4 y \\
5 x+7 y
\end{array}\right) &=\left(\begin{array}{c}
10 \\
3
\end{array}\right)
\end{aligned}
by matrix equality, we equate the entries of the matrix of the L.H.S and that of the R.H.S
i.e. \(3 x-4 y=10\)
\(5 x+7 y=3\)
Using elimination method;
first to obtain \(y\), we eliminate \(x\) by multiplying equation
(1) by 5 and (2) by 3 to become
\begin{aligned}
15 x-20 y=50 \\
-15 x+21 y=9 \\
-41 y=41 \\
y=& \frac{-41}{41}=-1
\end{aligned}
and solving for \(x\) we substitute \(y=-1\) into (1)
\begin{array}{l}
3 x-4(-1)=10 \\
3 x+4=10 \\
3 x=10-4=6 \\
3 x=6 ; x=2 \\
\therefore x=2, y=-1
\end{array}