<s>[INST] Solution of two variables system this might be a very simple question but it has its importance for me. In my multivariable calculus class, I have to find all $(x,y)$ in $R^2$ verifying the system 
I will call $(S)$ : 
\begin{cases}
3(x-y)^2+8x-3=0 \\
-3(x-y)^2+3=0
\end{cases}
What I did : 
This system is equivalent to the following : 
\begin{cases}
3(x-y)^2+8x-3=0 \\
3(-(x-y)^2+1)=0
\end{cases} 
Equivalent to : 
$$\begin{cases}
3(x-y)^2+8x-3=0 \\
3(1-x+y)(1+x-y)=0
\end{cases}$$
By the second last equation, we get 
$$(S) \Leftrightarrow \begin{cases}
3(x-y)^2+8x-3=0 \quad (1)\\
(x-y)=1 \text{ or } (x-y)=-1\quad (2)
\end{cases}$$
Then, we substitute in $(1)$. 
$$(x-y)=1 \text{ or }(x-y)=-1 \Rightarrow x=0$$
So we found two points $(0,-1)$ and $(0,1)$ verifying $(S)$. My question is : how can we know that there are no other points ? Is it a question of factorisation ? 
 [/INST] Note that the first equation gives $3(x-y)^2 = 3-8x$, and the second gives $3(x-y)^2 = 3$. So we equate to get $3-8x = 3 \implies x = 0$. Now $3(x-y)^2 = 3$ becomes $(-y)^2 = 1$, so $y$ is $1$ or $-1$, and it is straightforward to check that both $(x,y) = (0,1), (0,-1)$ satisfy both equations.
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