Solution of two variables system

Question

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 ?

Answer 1

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.