Answer
$\{(4,-2)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
2x& +y &=&6\\
3x& -2y &=&16
\end{matrix}\right.$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cc|c}
2& 1 & 6\\
3 & -2 & 16
\end{array}\right]$
Perform $R_1\rightarrow R_1/2$.
$\Rightarrow \left[\begin{array}{cc|c}
2/2& 1/2 & 6/2\\
3 & -2 & 16
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1& 1/2 & 3\\
3 & -2 & 16
\end{array}\right]$
Perform $R_2\rightarrow R_2-3R_1$.
$\Rightarrow \left[\begin{array}{cc|c}
1& 1/2 & 3\\
3-3(1) & -2-3(1/2) & 16-3(3)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1& 1/2 & 3\\
0 & -7/2 & 7
\end{array}\right]$
Perform $R_2\rightarrow R_2(-2/7)$.
$\Rightarrow \left[\begin{array}{cc|c}
1& 1/2 & 3\\
0(-2/7) & -7/2(-2/7) & 7(-2/7)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1& 1/2 & 3\\
0 & 1 & -2
\end{array}\right]$
Perform $R_1\rightarrow R_1-(1/2) R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1-(1/2)(0)& 1/2-(1/2)(1) & 3-(1/2)(-2)\\
0 & 1 & -2
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1& 0 & 4\\
0 & 1 & -2
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=4$
and
$\Rightarrow y=-2$.
The solution set is $\{(x,y)\}=\{(4,-2)\}$.
