Set of solutions with no solution, but one found by manipulation Consider the set of equations:
$$
\begin{cases}
x^2 &= -4y-10\\y^2 &= 6z-6\\z^2 &= 2x+2\\
\end{cases}$$
With $x,y,z$ being real numbers.
By adding the three equations, after simple manipulations, we easily obtain
$$
(x-1)^2+(y+2)^2+(z-3)^2=0
$$
Which yields
$$
\begin{cases}
x&=1\\y&=-2\\z&=3\\
\end{cases}
$$
However, plugging that in we find that the solution is invalid. How can one reason about this? In other words, why is it invalid? What causes it to be invalid?

 A: A solution to three equations will be a solution to the sum of the equations. But the solution to the sum need not be a solution to all three.
Consider $x^2 = 6z-6; y^2 = 2x + 2; z^2 = -4y-10$.  Those are different equations and will have different solutions but if you add them you get the same sums.
Even worse we could have $x^2 = -4y+10; y^2 =6z-10; z^2 = 2x - 14$ and the sum will be the same but obvious can't have the same solution.
You are basically losing information when you sum them as you are no longer requiring that $x^2$ must equal $-4y -10$ (etc) but can be some other combination.
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An even simpler example could be $2x + 3y + z = 17; x-y+2z =7; x+z = 6$.  Three equations, three unknowns.  One solution $x=2;y=3;z=4$
But if we add them:  $4x + 2y +4z = 30$.  One equation, three unknowns.  Infinite number of solutions of which $x=2;y=3;z=4$ is just one of them.
A: What you found is a necessary but not sufficient condition.  This is what happens in general when you take linear combinations of your initial equations.
If you back substitute, you'll end up with a degree 8 polynomial with 8 complex roots.  The manipulations you did aren't wrong - there are indeed 0 real roots.
Consider the simpler set of equations:
$$x^2=9 \\
y^2=16 \\
z^2=-25$$
By inspection, the 8 roots are $(\pm3,\pm4,\pm5i)$.
If you add the equations, you get
$$x^2+y^2+z^2=0$$
Well, $(0,0,0)$ is clearly a solution of the new equation, but not the original set.  OTOH, if you plug in the 8 true roots into this new derived equation, you will find that it's still true.  So, as I said, necessary but not sufficient.
A: This is how I think about it. When you use your initial set of three equations, which I'll call (A), to obtain your final equation, which I'll call (B), what you're saying is the following: "If $x,y,z$ satisfy (A), then they satisfy (B)".
Importantly, this reasoning only gives you a one-sided implication (A implies B), and not necessarily an equivalence (A if and only if B). That is, the converse ("If $x,y,z$ satisfy (B), then they satisfy (A)") may not be true. Indeed, you've found an example where this converse is false: you found $x,y,z$ which satisfy (B) but not (A).
Here's a simpler example. Start with the equality $x=2$, then square both side to get $x^2=4$. What this tells us is: "If $x=2$, then $x^2=4$". But the converse is not true: the equation $x^2=4$ is also satisfied by $x=-2$.
