Answer
$\{(-2,1)\}$.
Work Step by Step
The given system of equations is
$3x-2y=-8$
$x+6y=4$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cc|c}
3 & -2 & -8\\
1 & 6 & 4
\end{array}\right]$
Perform $R_1\leftrightarrow R_2$.
Swap row one and row two.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 6 & 4 \\
3 & -2 & -8
\end{array}\right]$
Perform $R_2\rightarrow R_2-3\times R_1$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 6 & 4 \\
3-3(1) & -2-3(6) & -8-3(4)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 6 & 4 \\
0 & -20 & -20
\end{array}\right]$
Perform $R_2\rightarrow R_2/(-20)$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 6 & 4 \\
0/(-20) & -20/(-20) & -20/(-20)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 6 & 4 \\
0 & 1 & 1
\end{array}\right]$
Perform $R_1\rightarrow R_1-6 R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1-6(0) & 6-6(1) & 4-6(1) \\
0 & 1 & 1
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 &0 & -2 \\
0 & 1 & 1
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=-2$
and
$\Rightarrow y=1$.
The solution set is $\{(x,y)\}=\{(-2,1)\}$.
