Answer
$(-3,2), (3,2), (-2,-3),(2,-3)$
Work Step by Step
We are given the system:
$\begin{cases}
x^2-y=7\\
x^2+y^2=13
\end{cases}$
We will use the addition method. Multiply Equation 1 by -1 and add it to Equation 2 to eliminate $x$ and determine $y$:
$\begin{cases}
-x^2+y=-7\\
x^2+y^2=13
\end{cases}$
$-x^2+y+x^2+y^2=-7+13$
$y^2+y-6=0$
$y^2-2y+3y-6=0$
$y(y-2)+3(y-2)=0$
$(y-2)(y+3)=0$
$y-2=0\Rightarrow y_1=2$
$y+3=0\Rightarrow y_2=-3$
Substitute each of the values of $y$ in Equation 1 to determine $x$:
$x^2-y=7$
$y_1=2\Rightarrow x^2-2=7\Rightarrow x^2=9\Rightarrow x_1=-3,x_2=3$
$y_2=-3\Rightarrow x^2-(-3)=7\Rightarrow x^2=4\Rightarrow x_1=-2,x_2=2$
The system's solutions are:
$(-3,2), (3,2), (-2,-3),(2,-3)$
