Answer
$(-2,0), (2,0), (0,-4)$
Work Step by Step
We are given the system:
$\begin{cases}
y=x^2-4\\
16x^2+4y^2=64
\end{cases}$
Rewrite the system:
$\begin{cases}
x^2-y=4\\
16x^2+4y^2=64
\end{cases}$
We will use the addition method. Multiply Equation 1 by -16 and add it to Equation 2 to eliminate $x$ and determine $y$:
$\begin{cases}
-16(x^2-y)=-16(4)\\
16x^2+4y^2=64
\end{cases}$
$-16(x^2-y)+16x^2+4y^2=-16(4)+64$
$-16x^2+16y+16x^2+4y^2=0$
$4y^2+16y=0$
$4y(y+4)=0$
$y=0\Rightarrow y_1=0$
$y+4=0\Rightarrow y_2=-4$
Substitute each of the values of $y$ in Equation 1 to determine $x$:
$x^2-y=4$
$y_1=0\Rightarrow x^2-0=4\Rightarrow x^2=0\Rightarrow x=\pm 2\Rightarrow x_1=-2,x_2=2$
$y_1=-4\Rightarrow x^2-(-4)=4\Rightarrow x^2=0\Rightarrow x_3=0$
The system's solutions are:
$(-2,0), (2,0), (0,-4)$
