Question 963109
First equation has {{{-y^2}}} and second equation has {{{y^2}}}.  Just sum the corresponding MEMBERS...


{{{x+2x^2-1=0}}}


You can work and finish from that?
GOOD.  Understood:


{{{system(x-y^2+9=0,2x^2+y^2-10=0)}}}


{{{x-y^2+9+2x^2+y^2-10=0+0}}}


{{{x+2x^2+9-10-y^2+y^2=0}}}


{{{x+2x^2-1+0=0}}}


{{{2x^2+x-1=0}}}

{{{(2x-1)(x+1)=0}}}


{{{x=1/2}}}  OR  {{{x=-1}}}
Which is part of the solution...