Answer
$\{(3,1)\}$.
Work Step by Step
The given system of equations is
$3x-y=8$
$x-5y=-2$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cc|c}
3 & -1 & 8\\
1 & -5 & -2
\end{array}\right]$
Perform $R_1\leftrightarrow R_2$.
Swap the second and the first row.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -5 & -2\\
3 & -1 & 8
\end{array}\right]$
Perform $R_2\rightarrow R_2-3\times R_1$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -5 & -2\\
3-3(1) & -1-3(-5) & 8-3(-2)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -5 & -2\\
0 & 14 & 14
\end{array}\right]$
Perform $R_2\rightarrow \frac{R_2}{14}$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -5 & -2\\
0/14 & 14/14 & 14/14
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -5 & -2\\
0 & 1 & 1
\end{array}\right]$
Perform $R_1\rightarrow R_1+5R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1+5(0) & -5+5(1) & -2+5(1)\\
0 & 1 & 1
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 0 & 3\\
0 & 1 & 1
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=3$
and
$\Rightarrow y=1$.
The solution set is $\{(x,y)\}=\{(3,1)\}$.
