What are the solution pairs of the following to simultaneous quadratic equation.

Question

$$ x^2 + y = 12 $$ and $$ y^2 + x = 12 $$

I tried by eliminating y but got stuck in biquadratic equation.

Answer 1

HINT

We have

$$x^2+y=y^2+x\iff x^2-y^2-x+y=0\iff (x-y)(x+y)-(x-y)=0$$

$$\iff (x-y)(x+y-1)=0$$

then we have two cases

• $x=y \implies x^2+x-12=0$
• $y=1-x \implies x^2+(1-x)-12=0$

from here you can solve by quadratic equation for $x$ and then find the corresponding values for $y$.

Answer 2

We have $$x^2-y^2+y-x=0$$ or $$(x-y)(x+y-1)=0.$$ Can you end it now?

Answer 3

with $$x=12-y^2$$ we get $$(12-y^2)^2+y-12=0$$ and this is $$(y-3) (y+4) \left(y^2-y-11\right)=0$$