Use an augmented matrix to solve each system of equations.

$\begin{array}{l}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{10}\\ \mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{5}\end{array}$

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 matrix is:

$\left[\begin{array}{ccc}2& -4& 10\\ 1& 2& 5\end{array}\right]$

First exchange row 1 with row 2 so as to make the first element of row 1 a 1.

$\left[\begin{array}{ccc}2& -4& 10\\ 1& 2& 5\end{array}\right]\stackrel{R_1\leftrightarrow R_2}{\to }\left[\begin{array}{ccc}1& 2& 5\\ 2& -4& 10\end{array}\right]$

To make first element of row to a 0, multiply row 1 by 2 and subtract the resultant from row 1.

$\left[\begin{array}{ccc}1& 2& 5\\ 2& -4& 10\end{array}\right]\stackrel{R_2-2R_1}{\to }\left[\begin{array}{ccc}1& -2& 5\\ 0& -8& 0\end{array}\right]$

To make the second element in the second row a 1, divide the second row by $-8$.

$\left[\begin{array}{ccc}1& -2& 5\\ 0& -8& 0\end{array}\right]\stackrel{-\frac{1}{8}{R}_2}{\to }\left[\begin{array}{ccc}1& -2& 5\\ 0& 1& 0\end{array}\right]$

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

$\left[\begin{array}{ccc}1& -2& 5\\ 0& 1& 0\end{array}\right]\stackrel{R_1\to R_1+2R_2}{\to }\left[\begin{array}{ccc}1& 0& 5\\ 0& 1& 0\end{array}\right]$

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 $(5,0)$.