Use an augmented matrix to solve each system of equations.

$\begin{array}{l}x+3y=12\\ 2x-4y=14\end{array}$

To write the equations in augmented matrix place the coefficients of the equations and the constant terms into a matrix separated by a dashed line.

Here, the augmented matrices are:

$\left[\begin{array}{ccc}1& 3& 12\\ 2& -4& 14\end{array}\right]$

To make the first element in the second row a 0, divide the second row by 2 and then subtract row 1 from the resultant row 2.

$\begin{array}{r}\left[\begin{array}{ccc}1& 3& 12\\ 2& -4& 14\end{array}\right]\stackrel{\frac{1}{2}{R}_2}{\to }\left[\begin{array}{ccc}1& 3& 12\\ 1& -2& 7\end{array}\right]\\ \stackrel{R_2-R_1}{\to }\left[\begin{array}{ccc}1& 3& 12\\ 0& -5& -5\end{array}\right]\end{array}$

To make the second element in the second row a 1, divide the second row by $-5$.

$\left[\begin{array}{ccc}1& 3& 12\\ 0& -5& -5\end{array}\right]\stackrel{\frac{1}{-5}{R}_2}{\to }\left[\begin{array}{ccc}1& 3& 12\\ 0& 1& 1\end{array}\right]$

Further row-reduce the augmented matrix, by making the second element in the first row a zero.

$\left[\begin{array}{ccc}1& 3& 12\\ 0& 1& 1\end{array}\right]\stackrel{R_1\to R_1-3R_2}{\to }\left[\begin{array}{ccc}1& 0& 9\\ 0& 1& 1\end{array}\right]$

Here, the first row will give the solution of x, because the coefficient of y is 0 and the coefficient of x is 1. Similarly, the second row will give the solution of y, because the coefficient of x is 0 and the coefficient of y is 1. Therefore, the solution is $\left(9,1\right)$.