Answer
$(0,0), (-2,2), (2,2)$
Work Step by Step
We are given the system:
$\begin{cases}
x^2-2y=0\\
x^2+(y-2)^2=4
\end{cases}$
We will use the addition method. Multiply Equation 1 by -1 and add it to Equation 2 to eliminate $x$ and determine $y$:
$\begin{cases}
-x^2+2y=0\\
x^2+(y-2)^2=4
\end{cases}$
$-x^2+2y+x^2+(y-2)^2=0+4$
$2y+y^2-4y+4=4$
$y^2-2y=0$
$y(y-2)=0$
$y=0\Rightarrow y_1=0$
$y-2=0\Rightarrow y_2=2$
Substitute each of the values of $y$ in Equation 1 to determine $x$:
$x^2-2y=0$
$y_1=0\Rightarrow x^2-2(0)=0\Rightarrow x^2=0\Rightarrow x_1=0$
$y_2=2\Rightarrow x^2-2(2)=0\Rightarrow x^2=4\Rightarrow x_2=-2,x_3=2$
The system's solutions are:
$(0,0), (-2,2), (2,2)$
The numbers are 8 and 12.
