Quadratic Simultaneous equation

Question

$$4x(x^2+y^2-1)=0$$ $$4y(x^2+y^2+1)=0$$ Solving Simultaneous equation $(x,y)=(0,0) ,(1,0)$ or $, (-1,0)$

I try to use substitution but couldnt find $x$ and $y$ properly, and (0,0) always be one of the solutions?

Answer 1

From the second equation, since $x^2+y^2 +1 > 0$, we have $y=0$

Substitute $y=0$ into the first equation, we end up with

$$4x(x^2-1)=0$$

Now we can solve for $x$.

Answer 2

• If $x=0$ then the second equation gives $4y(y^2+1)=0 \implies y=0\,$, thus the solution $\boxed{(0,0)}$.

• Else if $x \ne0, y=0$ the first equation gives $4x(x^2-1)=0 \implies x=\pm1\,$, thus $\boxed{(\pm1, 0\,)}\,$.

• Otherwise if $x,y \ne 0\,$, dividing the first equation by $4x \ne 0$, the second one by $4y \ne 0$ and
  adding the two gives ...  subtracting the two (@LordSharktheUnknown's comment) gives $2=0\,$, obviously impossible, so there are no further solutions in this case.