Answer
$x=-1$ or $x=2$
$y=0$ or $y=\pm\sqrt 3$
Work Step by Step
$\begin{cases}
x^2-y^2=1\\
2x^2-y^2=x+3
\end{cases}$
Multiplying the first system of equation by -1 and adding it to the second system of equation...
$\begin{cases}
-x^2+y^2=-1\\
2x^2-y^2-x=3\\
-- -- -- -- \\
x^2-x=2
\end{cases}$
Therefore, $x^2-x-2=0$.
Solving for the trinomial using factoring, find two numbers whose sum is $-1$ and whose multiple is $-2$. the number is $-2$ and $+1$.
$x^2-x-2=x^2+x-2x-2=0$,
$x(x+1)-2(x+1)=(x+1)(x-2)=0$,
Thus, $x=-1$ or $x=2$. Substituting back in, $y=0$ or $y=\pm\sqrt 3$
