Answer
$\{(-5,3)\}$.
Work Step by Step
The given system of equations is
$x+4y=7$
$3x+5y=0$
The augmented matrix is
$\Rightarrow \left[\begin{array}{cc|c}
1 & 4 & 7\\
3 & 5 & 0
\end{array}\right]$
Perform $R_2\rightarrow R_2-3 R_1$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 4 & 7\\
3-3(1) & 5-3(4) & 0-3(7)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 4 & 7\\
0 & -7 & -21
\end{array}\right]$
Perform $R_2\rightarrow (\frac{1}{-7}) R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 4 & 7\\
(\frac{1}{-7})0 &(\frac{1}{-7})( -7) & (\frac{1}{-7})(-21)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 4 & 7\\
0 &1 & 3
\end{array}\right]$
Perform $R_1\rightarrow R_1-4R_2$.
$\Rightarrow \left[\begin{array}{cc|c}
1-4(0) & 4-4(1) & 7-4(3)\\
0 &1 & 3
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{cc|c}
1 & 0 & -5\\
0 &1 & 3
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=-5$
and
$\Rightarrow y=3$.
The solution set is $\{(x,y)\}=\{(-5,3)\}$.
