Invalid subtraction when solving system of equations?

Question

I'm trying to solve these two equations: $$\begin{cases} 1-4x(x^2+y^2)=0 \\ 1-4y(x^2+y^2)=0 \end{cases}$$ and I tried to do it by subtracting the first equation from the second, yielding $(4x-4y)(x^2+y^2)=0$. Clearly this is satisfied when $x=y$, which gives $(x,y)=(\frac{1}{2},\frac{1}{2})$, or when $x^2+y^2=0$, which gives $(x,y)=(0,0)$.

But when I plug them back in, it's obvious that $(x,y)=(0,0)$ is not a solution. Where did I go wrong?

Answer 1

You didn't go wrong. Subtracting the equations will preserve any existing solution, but may add others, which aren't solutions to the original equations (as here). So you need to feed the possible answers back into the original equations, as you have done to check that they are valid.

Answer 2

$$\begin{cases} 1-4x(x^2+y^2)=0 \\ 1-4y(x^2+y^2)=0 \end{cases}$$

Let $S$ the set of solutions.

You can summarise the logic steps:

1. If $(x,y)$ is a solution: $$4x(x^2+y^2)=4y(x^2+y^2) \implies x^2+y^2=0 \text{ or }x=y \\ \implies (x,y)=(0,0) \text{ or } \\ x=y=\frac{1}{4(x^2+y^2)} =\frac{1}{8x^2}\implies x=y=\frac 1{8^{1/2}}=\frac 12 $$ You have proved $S\subset\{(1/2,1/2),(0,0)\}$.
2. If $x=(1/2,1/2)$ the equation is true. If $x=(0,0)$ the equation is false. You have proved $S=\{(1/2,1/2)\}$.

Answer 3

First $x=y$ yields $8x^3=1$

$x=\dfrac{1}{2},\dfrac{-1+\sqrt3i}{4},\dfrac{-1-\sqrt3i}{4}$

$\therefore(x,y)=\left(\dfrac{1}{2},\dfrac{1}{2}\right),\left(\dfrac{-1+\sqrt3i}{4},\dfrac{-1+\sqrt3i}{4}\right),\left(\dfrac{-1-\sqrt3i}{4},\dfrac{-1-\sqrt3i}{4}\right)$

But $x^2+y^2=0$ yields $1=0$ which is not useful.

Note than $x^2+y^2=0$ should not immediately implies $x=y=0$ , which should have no concept abut the false solution $(x,y)=(0,0)$ .