$$ x^2 + y = 12 $$ and $$ y^2 + x = 12 $$
I tried by eliminating y but got stuck in biquadratic equation.
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$$x^2+y=y^2+x\iff x^2-y^2-x+y=0\iff (x-y)(x+y)-(x-y)=0$$
then we have two cases
from here you can solve by quadratic equation for $x$ and then find the corresponding values for $y$.