Answer
$(x,y)=(12,-5)$.
Work Step by Step
The given system of equations is
$3x+2y=26$
$x+y=7$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cc|c}
3 & 2 & 26\\
1 & 1 & 7
\end{array}\right]$
Perform $R_1\rightarrow \frac{R_1}{3}$.
$\Rightarrow \left[\begin{array}{cc|c}
3/3 & 2/3 & 26/3\\
1 & 1 & 7
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 2/3 & 26/3\\
1 & 1 & 7
\end{array}\right]$
Perform $R_2\rightarrow R_2- R_1$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 2/3 & 26/3\\
1-1 & 1-(2/3) & 7-(26/3)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 2/3 & 26/3\\
0 & 1/3 & -5/3
\end{array}\right]$
Perform $R_2\rightarrow 3\times R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 2/3 & 26/3\\
3\times 0 & 3\times(1/3) &3\times( -5/3)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 2/3 & 26/3\\
0 & 1 &-5
\end{array}\right]$
Perform $R_1\rightarrow R_1-(2/3)\times R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1 -(2/3)0& 2/3-(2/3)1 & 26/3-(2/3)(-5)\\
0 & 1 &-5
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 0 & 26/3+10/3\\
0 & 1 &-5
\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc|c}
1 & 0 & 36/3\\
0 & 1 &-5
\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc|c}
1 & 0 & 12\\
0 & 1 &-5
\end{array}\right]$
Write each row of the matrix as an equation.
$\Rightarrow x=12$
and
$\Rightarrow y=-5$.
The solution set is $(x,y)=(12,-5)$.
