What is awry with this proof? Let $x=5$, $y=7$, $z=6$
$x+y = 2z$
Rearranging, $x-2z = -y$
and $x = -y+2z$
Multiply both sides respectively. $x^2-2xz = y^2-2yz$
$$x^2-2xz+z^2 = y^2-2yz+z^2$$
$$(x-z)^2 = (y-z)^2$$
$$x-z = y-z$$
Hence $x=y$, or $5 = 7$
Well, the conclusion is clearly false, but what went wrong? I think it may be the step in which one square roots both sides because it's taking out one solution?
 A: What you have is 
$$(x-z)^2=(y-z)^2$$
Note that $y-z$ is positive, but $x-z$ is negative. Thus you have to consider the absolut value when taking the square root.
This means
$$|x-z| = y-z$$
But not 
$$x-z = y-z$$
A: If $(x-z)^2 = (y-z)^2$ then $x-y = \pm(y-z)$.
A: Hint $\ $ When debugging proofs on abstract objects, the error may become simpler to spot after specializing to more concrete objects. The symbols $\rm\:x,y,z\:$ denote abstract numbers, so let's specialize them to their concrete number values: $\rm\:x = 5,\: y=7,\: z = 6,\:$ yielding this "proof"
$$\begin{eqnarray}
 5 + 7 &=&\: 2\cdot 6 \\
 5- 2\cdot 6 &=&\: -7 \\
\cdots\ &=&\ \cdots  \\
(5-6)^2\! &=&\: (7-6)^2  \\
 \color{#c00}{5-6}\ \ \:&=&\:\ \ \color{#c00}{7-6}\: 
\end{eqnarray}$$ 
Now we can spot which inference is incorrect by determining the first $\rm\color{#c00}{false\ equation}$ above. If equation number $\rm\: n\!+\!1\:$ is false then the inference from equation $\rm\:n\:$ to $\rm\:n\!+\!1\:$ is incorrect. Doing so we find that last equation being false, which reveals the culprit inference $\,(-1)^2 = 1^2\color{#c00}{\Rightarrow\, -1 = 1}$ 
Analogous methods prove helpful generally: when studying abstract objects and something is not clear, look at concrete specializations to gain further insight on the general case.
A: You have $x=−y+2z$ and $x−2z=−y$  then you multiply both sides.
I think you are computing $-xy$ in two ways: for the first $-xy=y^2-2zy$
for the second $-xy=-2zx+x^2$ so you have $y^2-2zy=x^2-2zx$.  Adding $z^2$ to both sides
gives $(y-z)^2=(x-z)^2$. But now taking square roots gives $|y-z|=|x-z|$. 
If you try to remove the absolute value signs then you get $y-z=x-z$ if both are positive or negative but $y-z=-(x-z)$ if these have opposite signs. In the first case this leads to $y=x$, and in the second case $y=-x+2z$. For your example, the second case applies
$7=-5+2\cdot6$. 
