Answer
$x=-1$ or $x=\frac{1}{2}$.
$y=\pm\sqrt {2}$ or $y=\pm\sqrt {\frac{7}{2}}$.
Work Step by Step
$\begin{cases}
x-y^2+3=0\\
2x^2+y^2-4=0
\end{cases}$
Adding the system of equations,
$\begin{cases}
x-y^2+3=0\\
2x^2+y^2-4=0\\
-- -- -- --\\
2x^2+x-1=0
\end{cases}$
Solving for the trinomial by factoring,
$2x^2+x-1=0$, find two numbers whose sum is $+1$, and whose multiple is $-2$.
$2x^2+2x-x-1=2x(x+1)-1(x+1)=0$
$(2x-1)(x+1)=0$
Thus, $x=-1$ or $x=\frac{1}{2}$.
Substituting it back into the equation, $y=\pm\sqrt {2}$ or $y=\pm\sqrt {\frac{7}{2}}$.
