Answer
$\{(2,-5)\}$.
Work Step by Step
The given system of equations is
$5x+7y=-25$
$11x+6y=-8$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cc|c}
5 & 7 & -25\\
11 & 6 & -8
\end{array}\right]$
Perform $R_1\rightarrow \frac{R_1}{5}$.
$\Rightarrow \left[\begin{array}{cc|c}
5/5 & 7/5 & -25/5\\
11 & 6 & -8
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 7/5 & -5\\
11 & 6 & -8
\end{array}\right]$
Perform $R_2\rightarrow R_2-11\times R_1$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 7/5 & -5\\
11-11\times 1 & 6-11\times (7/5) & -8-11\times (-5)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 7/5 & -5\\
0 & -47/5 & 47
\end{array}\right]$
Perform $R_2\rightarrow -\frac{5}{47}\times R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 7/5 & -5\\
-\frac{5}{47}\times 0 & -\frac{5}{47}\times(-47/5) &-\frac{5}{47}\times 47
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 7/5 & -5\\
0 & 1 &-5
\end{array}\right]$
Perform $R_1\rightarrow R_1-(7/5)\times R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1-(7/5)\times 0 & 7/5-(7/5)\times1 & -5-(7/5)\times (-5)\\
0 & 1 &-5
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 0 & 2\\
0 & 1 &-5
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=2$
and
$\Rightarrow y=-5$.
The solution set is $\{(x,y)\}=\{(2,-5)\}$.
