Answer
$(0,-3),\left(\dfrac{12}{5},\dfrac{9}{5}\right)$
Work Step by Step
We are given the system:
$\begin{cases}
2x-y=3\\
x^2+y^2=9
\end{cases}$
We will use the substitution method. Solve Equation 1 for $y$ and substitute the expression of $y$ in Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
y=2x-3\\
x^2+(2x-3)^2=9
\end{cases}$
$x^2+(2x-3)^2=9$
$x^2+4x^2-12x+9=9$
$5x^2-12x=0$
$x(5x-12)=0$
$x=0\Rightarrow x_1=0$
$5x-12=0\Rightarrow x_2=\dfrac{12}{5}$
Substitute each of the values of $x$ in Equation 1 to determine $y$:
$y=2x-3$
$x_1=0\Rightarrow y=2(0)-3\Rightarrow y_1=-3$
$x_2=\dfrac{12}{5}\Rightarrow y=2\left(\dfrac{12}{5}\right)-3\Rightarrow y_2=\dfrac{9}{5}$
The system's solutions are:
$(0,-3),\left(\dfrac{12}{5},\dfrac{9}{5}\right)$
Check the solutions:
$x_1=0; y_1=-3$
$2(0)-(-3)\stackrel{?}{=}3$
$3=3\checkmark$
$(0)^2+(-3)^2\stackrel{?}{=}9$
$0+9\stackrel{?}{=}9$
$9=9\checkmark$
$x_1=\dfrac{12}{5}; y_1=\dfrac{9}{5}$
$2\left(\dfrac{12}{5}\right)-\dfrac{9}{5}\stackrel{?}{=}3$
$\dfrac{24}{5}-\dfrac{9}{5}\stackrel{?}{=}3$
$3=3\checkmark$
$\left(\dfrac{12}{5}\right)^2+\left(\dfrac{9}{5}\right)^2\stackrel{?}{=}9$
$\dfrac{144}{25}=\dfrac{81}{25}\stackrel{?}{=}9$
$9=9\checkmark$
Therefore the solutions satisfy the given equations.