Answer
$\{(4,-2)\}$.
Work Step by Step
The given system of equations is
$3x-5y=22$
$4x-2y=20$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cc|c}
3 & -5 & 22\\
4 & -2 & 20
\end{array}\right]$
Perform $R_1\rightarrow \frac{R_1}{3}$.
$\Rightarrow \left[\begin{array}{cc|c}
3/3 & -5/3 & 22/3\\
4 & -2 & 20
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -5/3 & 22/3\\
4 & -2 & 20
\end{array}\right]$
Perform $R_2\rightarrow R_2-4\times R_1$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -5/3 & 22/3\\
4-4\times 1 & -2-4\times (-5/3) & 20-4\times (22/3)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -5/3 & 22/3\\
0 & 14/3 & -28/3
\end{array}\right]$
Perform $R_2\rightarrow \frac{3}{14}\times R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -5/3 & 22/3\\
(3/14)0 & (3/14)(14/3) & (3/14)(-28/3)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & -5/3 & 22/3\\
0 & 1 & -2
\end{array}\right]$
Perform $R_1\rightarrow R_1+(5/3)\times R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1+(5/3)\times 0 & -5/3+(5/3)\times1 & (22/3)+(5/3)\times (-2)\\
0 & 1 &-2
\end{array}\right]$
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 $\{(x,y)\}=\{(4,-2)\}$.
