Why doesn't adding the equations in this system of equations find the solution? I have two equations: $x^2 - y = 0$ and $y^2 - x = 0$. Adding them gives $x^2 - x + y^2 - y = 0$, and completing the square results in $(x - 1/2)^2 + (y-1/2)^2 = 1/2$. This suggests that there is an infinite number of solutions on the circle centered at (1/2, 1/2) with a radius of 1/4. However, only the solutions 0 and 1 actually work in the original equations. 
I understand that the original equations result in $y = x^2$, and, substituting this into the other equation, $x(x^3 - 1) = 0$, which provides the solutions 0 and 1, but why don't all points on the circle defined above work as solutions in these equations since, after all, the circle was derived from those equations? 
 A: Your logical steps are 
$(x^2 - y = 0) \wedge (y^2 - x = 0) \implies (x^2-x+y^2-y=0)$. The implication doesn't go the ohter way around.
The core of it is that if $A=0$ and $B=0$, then $A+B=0$. But if $A+B=0$ then we can't say that $A=0$ and $B=0$, because it's possible for example that $A=5$ and $B=-5$.
Sometimes we gloss over $x$ and $y$ as formal symbols, but in situations like these I always like to think of $x$ and $y$ as actual concrete numbers. Imagine that you have two concrete numbers $x$ and $y$, and if you compute $x^2-y$ and $y^2-x$ you get $0$ each time. Then for sure if you compute $(x^2-y)+(y^2-x)$ you will get $0$ also. But now imagine you have two concrete numbers (maybe different, maybe not; but I will still call them $x, y$) such that when you compute $(x^2-y)+(y^2-x)$ you get zero. But this does not mean for sure that if you compute $x^2-y$ and $y^2-x$ you will get zero for them (though if you get $0$ for one, you will for sure get $0$ for the other). You can find such examples by simply taking a point on the circle you describe which does not satisfy the two smaller equations. 
A: The statement “$A=0$ and $B=0$” isn't the same thing as “$A+B=0$”.
(But it is the same thing as “$A=0$ and $A+B=0$”.)
