Answer
The required solution is $\left\{ \left( -5,0 \right),\left( 4,3 \right) \right\}$
Work Step by Step
The given equations are,
$ x-3y=-5$ (I)
And
${{x}^{2}}+{{y}^{2}}-25=0$ (II)
From equation (I)
$ x=3y-5$ (III)
Substitute $ x=3y-5$ in equation (II)
$\begin{align}
& {{\left( 3y-5 \right)}^{2}}+{{y}^{2}}-25=0 \\
& 9{{y}^{2}}-30x+25+{{y}^{2}}-25=0 \\
& 10\left( {{y}^{2}}-3y \right)=0 \\
& y\left( y-3 \right)=0
\end{align}$
Now,
$ y=0,3$
Substitute $ y=0$ in equation (III):
$\begin{align}
& x=3\left( 0 \right)-5 \\
& x=-5 \\
\end{align}$
Substitute $ y=3$ in equation (III):
$\begin{align}
& y=3\left( 3 \right)-5 \\
& y=4 \\
\end{align}$
Thus, the solution is $\left\{ \left( -5,0 \right),\left( 4,3 \right) \right\}$.
