Answer
$\{(0,\pm\sqrt 5)\}$
Work Step by Step
1. Given $\begin{cases} 2x^2+y^2=5 \\ 3x^2+2y^2=10 \end{cases}$ use elimination, multiply -2 to the first equation, we have $\begin{cases} -4x^2-2y^2=-10 \\ 3x^2+2y^2=10 \end{cases}$
2. Add up the two equations to get $-x^2=0$, thus $x=0$
3. Use the first equation to get $2(0)^2+y^2=5$, thus $y=\pm\sqrt 5$
4. Thus the solution set is $\{(0,\pm\sqrt 5)\}$
