Answer
The answer is
$ x=-1 $ and $ y=-2 $
Work Step by Step
The given system of linear equations is
$ 3x-5y=7 $
$x-y=1$
The augmented matrix is
$\left[\begin{array}{cc|c}
3 & -5 & 7\\
1 & -1 & 1
\end{array}\right]$
Perform $R_2\rightarrow R_2-\frac{1}{3}\cdot R_1$.
$\left[\begin{array}{cc|c}
3 & -5 & 7\\
1-\frac{1}{3}\cdot 3 & -1-\frac{1}{3}\cdot (-5) & 1-\frac{1}{3}\cdot 7
\end{array}\right]$
Simplify.
$\left[\begin{array}{cc|c}
3 & -5 & 7\\
1-1 & -1+\frac{5}{3} & 1-\frac{7}{3}
\end{array}\right]$
$\left[\begin{array}{cc|c}
3 & -5 & 7\\
0 & \frac{-3+5}{3} & \frac{3-7}{3}
\end{array}\right]$
$\left[\begin{array}{cc|c}
3 & -5 & 7\\
0 & \frac{2}{3} & \frac{-4}{3}
\end{array}\right]$
Perform $R_1\rightarrow R_1+\frac{15}{2}\cdot R_2$.
$\left[\begin{array}{cc|c}
3 & -5+5 & 7-10\\
0 & \frac{2}{3} & \frac{-4}{3}
\end{array}\right]$
Simplify.
$\left[\begin{array}{cc|c}
3 & 0 & -3\\
0 & \frac{2}{3} & \frac{-4}{3}
\end{array}\right]$
Use back substitution to solve the linear system.
$3x=-3$
$x=-1 $
and
$\frac{2}{3}\cdot y=\frac{-4}{3} $
$ y=-2 $
