Why { $z-x-y=0$ , $z-2x=0$ , $2x+y-3z=0$ } cannot be solved this way? I was recently solving a system of linear equations, 3 equations and 3 unknowns. I first solved via Row Reduction of the matrix and got a valid answer, but my friend attempted to solve the system using informal algebra methods and got the wrong answer. I know his answer is wrong, but I am struggling to explain what mathematical rule he broke.
Here is the system:
$z-x-y=0$
$z-2x=0$
$2x+y-3z=0$
Combining the first and third equation, one gets $x=-2y$. Plugging this back into equation one, one gets $z=-y$. Setting $x=1$, one gets the vector $<1,-1/2,1/2>$. This vector is valid for equations 1 and 3, but not for equation 2.
Now I know that this is not the proper technique for solving a system of three variables and that equation 2 was not used so how should one expect it to be satisfied. I know that this solution is wrong, but I am unsure how to explain what is wrong about it other than saying "that's not the way it's done." I personally made this mistake when first learning linear algebra and "that's not the way it's done" is all my teacher could say. If anyone has a better explanation for what exactly is wrong about this, I would greatly appreciate. Also, since equation 2 is not being used, it is 2 equations, 3 unknowns so there should be 2 free variables, not one (again I think this will occur if elimination is "done correctly").
 A: You wrote, correctly, that the second equation was not used at all. And the solutions of the system which consists of the other two equations are the triplets $(x,y,z)$ such that $x=-2y$ and that $z=-y$ indeed. Finally, there is not reason at all to pick $x=1$.
A: Let me present an extreme version of your friend's argument:

By the first equation, we have
$$  z = x + y $$
Then, simply choose $ x = y = 1 $ and conclude $ z = 2 $.
Thus, $(1,1,2)$ is a solution to the system!

Or to a ridiculous degree:

By ignoring all equations, we can choose $x = y = z = 1$ and so $(1,1,1)$ is a solution!

I hope this illuminates the issue with your friend's solution. Each equation potentially restricts the possible solutions to the system, and so each equation not used potentially un-restricts those solutions to ones which are not actually valid.
One more example to demonstrate this fact:

Consider the system of two equations for one variable,
$$ \begin{align*} x &= 1 \\ x &= 2 \end{align*} $$
By limiting our scope to only one equation, we may conclude either that $1$ is a solution or that $2$ is a solution, but in reality there are (obviously) no solutions, and so both are wrong.

A: Your friend used just two equations, which (since they were linearly independent) is enough to limit the solution space to one dimension.
The correct conclusion is not that
$\langle x,y,z\rangle = \langle 1,-\frac12,\frac12\rangle$
is the solution, but that the solution has the form
$$\langle x,y,z\rangle = \left\langle t,-\frac12t,\frac12t\right\rangle$$
for some real number $t.$
This is true. The correct solution does have that form.
Now compare this with your solution and see if you can tell what value(s) of $t$ give a solution to all three equations.
A: We can begin by substituting $(2)$ into $(1)$ or $(3)$. The problem is, using $(2)$, we find a contradiction between $(4)$ and  $(6)$
\begin{align}
z-x-y=0  \implies -x-y+z&=0    \tag{1}\\
z-2x=0   \implies z&=2x  \tag{2}\\
2x+y-3z=0\implies  2(x)+y-3z&=0  \tag{3}
\end{align}
Substitute $(2)$ into $(1)$  then
$(4)$ into $(3)$
\begin{align} 
-x-y+(2x)=0\implies x&=y  \tag{4}\\
2(x)+y-3z=0\implies 3y-3z&= 0\tag{5}\\
\implies y=z\implies y&=2x \tag{6}
\end{align}
or we can combine $(1)$ and $(3)$ as you did.
Sometimes rearranged things are earier to "see"
\begin{align}
z-x-y=0  \implies -x-y+z=0    \tag{1}\\
z-2x=0   \implies -2x+0y+z=0  \tag{2}\\
2x+y-3z=0\implies  2(x)+y-3z=0  \tag{3}
\end{align}
Add $(1)$ to $(3)$, substitute $(4)$ into $(3)$ then  $(5)$ into $(6)$, etc.
\begin{align}
x-2z=0\implies x&=(2z)       \tag{4}   \\
\implies z&=\frac{x}{2}  \tag{5}\\
2(2z)+y-3z=0\implies y&=(-z) \tag{6}\\
\implies y&=\frac{-x}{2} \tag{7}
\end{align}
Setting $x=1\quad 
\bigg(x,\dfrac{-x}{2},\dfrac{x}{2}\bigg)
=\bigg(1,\dfrac{-1}{2},\dfrac{1}{2}\bigg) $
A: It might be useful to see what you actually did solve: the set of three equations:
$$z-x-y=0$$
$$x=1$$
$$2x+y-3z=0$$
I've removed the equation that wasn't used, and replaced it with an equation that was used. <1,−1/2,1/2> is a solution to this particular problem. For this particular problem, I wouldn't start by combining 1 and 3. Instead, I'd first substitute 2 into 1 and then 2 into 3.
It's still a similar sort of problem: the intersection of three planes. But if should be fairly clear that by choosing different planes, you can get different solutions.
