Answer
$(3,0), \left(-\dfrac{9}{5},\dfrac{12}{5}\right)$
Work Step by Step
We are given the system:
$\begin{cases}
x+2y-3=0\\
x^2+y^2=9
\end{cases}$
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 $y$:
$\begin{cases}
x=-2y+3\\
(-2y+3)^2+y^2=9
\end{cases}$
$(-2y+3)^2+y^2=9$
$4y^2-12y+9+y^2=9$
$5y^2-12y=0$
$y(5y-12)=0$
$y=0\Rightarrow y_1=0$
$5y-12=0\Rightarrow y_2=\dfrac{12}{5}$
Substitute each of the values of $y$ in the expression of $x$ to determine $x$:
$x=-2y+3$
$y_1=0\Rightarrow x_1=-2(0)+3=3$
$y_2=\dfrac{12}{5}\Rightarrow x_2=-2\left(\dfrac{12}{5}\right)+3=-\dfrac{9}{5}$
The system's solutions are:
$(3,0), \left(-\dfrac{9}{5},\dfrac{12}{5}\right)$
