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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?

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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.

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  • $\begingroup$ Note also that if you multiply the first question by $y$ and the second by $x$ before subtracting, you immediately get $y-x=0$ so you kill the more complicated term. Always worth noticing a short cut. $\endgroup$ – Mark Bennet Apr 4 '14 at 16:01
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    $\begingroup$ I've learned to be careful with division, since you may end up dividing by zero, but this is the first time subtraction has caused any problems for me. So do I always have to double-check my solutions when I use any elementary operations to solve a system of equations? $\endgroup$ – Andrea Apr 4 '14 at 16:27
  • $\begingroup$ Here in the US, we don’t make it clear to students what they’re doing when they “solve” an equation. That’s really the uniqueness part of the story: here, if there is a solution, it’s either $(1/2,1/2))$ or $(0,0)$. But the existence part is what we call “checking”: to see whether the possibilities you found really do satisfy the equation(s). Here, OP has done the checking and found that one of the apparent possibilities is in fact not a solution. In teaching, we haven’t been stressing strongly enough that checking is an essential part of the process. $\endgroup$ – Lubin Apr 4 '14 at 16:34
  • $\begingroup$ @Lubin That's really strange, considering I can't recall running into this problem before. Maybe it's just an uncommon occurrence. So what exactly is it that generates these "false" solutions? That is, are there any circumstances under which I can be certain that no incorrect solutions are produced by elementary operations (subtraction etc.)? $\endgroup$ – Andrea Apr 4 '14 at 17:22
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    $\begingroup$ Subtraction the first from the second equation does not change the solution set. However, forgetting the first equation will change the solution set. Clearly, when you treat the case $x=y$, the first equation is still present in your calculations. The error happens when you seem to forget this in the case $x=y=0$ (which by the way is already excluded as a solution by the first case). So it is not a question of plugging things back in, but of satisfying the full transformed system. $\endgroup$ – Dr. Lutz Lehmann Apr 4 '14 at 18:31
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$$\begin{cases} 1-4x(x^2+y^2)=0 \\ 1-4y(x^2+y^2)=0 \end{cases}$$

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 $S=\{(1/2,1/2)\}$.
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  • $\begingroup$ So to clarify, if I use elementary operations to simplify a system of equations, I still always need to verify that they are indeed solutions to the original system? $\endgroup$ – Andrea Apr 4 '14 at 16:29
  • $\begingroup$ @Andreas exacly. $\endgroup$ – mookid Apr 4 '14 at 16:54
  • $\begingroup$ @Andreas: In general yes, in this case no, since $(0,0)$ does not satisfy all equations of the transformed system. $\endgroup$ – Dr. Lutz Lehmann Apr 4 '14 at 18:33
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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 $(x,y)=(0,0)$ .

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