Answer
$(x,y)=(1,6)$.
Work Step by Step
The given system of equations is
$2x+2y=14$
$-x-2y=-13$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cc|c}
2 & 2 & 14\\
-1 & -2 & -13
\end{array}\right]$
Perform $R_1\rightarrow \frac{R_1}{2}$.
$\Rightarrow \left[\begin{array}{cc|c}
2/2 & 2/2 & 14/2\\
-1 & -2 & -13
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 1 & 7\\
-1 & -2 & -13
\end{array}\right]$
Perform $R_2\rightarrow R_2+R_1$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 1 & 7\\
-1+1 & -2+1 & -13+7
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 1 & 7\\
0 & -1 & -6
\end{array}\right]$
Perform $R_2\rightarrow -1(R_2)$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 1 & 7\\
-1(0) & -1(-1) & -1(-6)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 1 & 7\\
0 & 1 & 6
\end{array}\right]$
Perform $R_1\rightarrow R_1- R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1-0 & 1-1 & 7-6\\
0 & 1 & 6
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 0 & 1\\
0 & 1 &6
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=1$
and
$\Rightarrow y=6$.
The solution set is $\{(x,y)\}=\{(1,6)\}$.
