Answer
$(-5,0), (4,3)$
Work Step by Step
We are given the system:
$\begin{cases}
x-3y=-5\\
x^2+y^2-25=0
\end{cases}$
We will use the substitution method. Solve Equation 1 for $x$ and substitute the expression of $x$ in Equation 2 to eliminate $x$ and determine $y$:
$\begin{cases}
x=3y-5\\
(3y-5)^2+y^2-25=0
\end{cases}$
$9y^2-30y+25+y^2-25=0$
$10y^2-30y=0$
$10y(y-3)=0$
$y=0\Rightarrow y_1=0$
$y-3=0\Rightarrow y_2=3$
Substitute each of the values of $y$ in the expression of $x$ to determine $x$:
$x=3y-5$
$y_1=0\Rightarrow x_1=3(0)-5=-5$
$y_2=3\Rightarrow x_2=3(3)-5=4$
The system's solutions are:
$(-5,0), (4,3)$
