Answer
The answer is
$ x=3 $ and $x=-3 $.
Work Step by Step
The given system of linear equations is
$2x+y=3 $
$x-3y=12 $
The augmented matrix is
$\left[\begin{array}{cc|c}
2& 1 &3 \\
1&-3 &12 \\
\end{array}\right]$
Perform $R_2\rightarrow R_2-\frac{1}{2}\cdot R_1$.
Multiply row one by $-\frac{1}{2}$ and subtract from row two.
$\left[\begin{array}{cc|c}
2& 1 &3 \\
1-\frac{1}{2}\cdot 2&-3-\frac{1}{2}\cdot1 &12-\frac{1}{2}\cdot3 \\
\end{array}\right]$
Simplify.
$\left[\begin{array}{cc|c}
2& 1 &3 \\
1-1&-3-\frac{1}{2} &12-\frac{3}{2} \\
\end{array}\right]$
$\left[\begin{array}{cc|c}
2& 1 &3 \\
0&\frac{-6-1}{2} &\frac{24-3}{2} \\
\end{array}\right]$
$\left[\begin{array}{cc|c}
2& 1 &3 \\
0&\frac{-7}{2} &\frac{21}{2} \\
\end{array}\right]$
Perform $R_2\rightarrow \frac{2}{-7}\cdot R_2$.
Multiply row two by $\frac{2}{-7}$.
$\left[\begin{array}{cc|c}
2& 1 &3 \\
\frac{2}{-7} \cdot 0&\frac{2}{-7}\cdot \frac{-7}{2} &\frac{2}{-7}\cdot \frac{21}{2} \\
\end{array}\right]$
Simplify.
$\left[\begin{array}{cc|c}
2& 1 &3 \\
0&1 &-3 \\
\end{array}\right]$
Perform Perform $R_1 \rightarrow R_1- R_2$.
Subtract row two from row one.
$\left[\begin{array}{cc|c}
2-0& 1-1 &3+3 \\
0&1 &-3 \\
\end{array}\right]$
Simplify.
$\left[\begin{array}{cc|c}
2& 0 &6 \\
0&1 &-3 \\
\end{array}\right]$
Use back substitution to solve the linear system.
$ 2x=6 $
$x=3 $
and $y=-3 $.
