How do I prove that $x,y$ and $z$ lie on the same line if they are related to the same parameter? Related to this question
How exactly does that work? I would be grateful for a proof but barring that just the name of the theorem, so I can look it up myself.
My work/thoughts so far:

If we start with a system like the one below:
$$y+az=b$$
$$x+fz=g$$
I guess I could solve both sides for z and get something like:
$$z=\frac{b-y}{a}$$
$$z=\frac{g-x}{f}$$
Which I guess implies that:
$$\frac{b-y}{a}=\frac{g-x}{f} $$
$$y=\frac{a(g-x)}{f}+b $$
Which is really just another way of writing $y=kx+m$, so for every value of z, the relationship between x and y can be written as a line. Since there is an infinite number of z, that means we can vizualize all the possible lines as an infinite plane.
If we could define a vectpr z we could restrict the possible solutions to the values in the plane that satisfiy both $y=\frac{a(g-x)}{f}+b $ and the equation for z. It just so happens we have two expressions for z, $z=\frac{b-y}{a}$ and $z=\frac{g-x}{f}$.
I dont' know, where do I go from here?
 A: I’m awfully sorry, but your question is imprecise, especially since I don’t see mention of any parameter anywhere in the title or body. So my attempt at an answer may not satisfy you at all. Nonetheless:
It seems to me that $x$, $y$, and $z$ are to be the coordinates of points in $3$-space $\Bbb R^3$. And that $a$, $b$, $f$, and $g$ are parameters which define the problem. You pretty clearly are of the same opinion.
Furthermore, you have correctly solved for $z$ in the two cases, $z=-\frac1ay+\frac ba$, $z=-\frac1fx+\frac gf\,$. These say, separately, that a choice of $x$ and $y$ gives you $z$. But not necessarily the same $z$ in the two cases.
For instance, say $x=1$ and $y=2$. Then you get, in the first case, the point $(1,2,-\frac2a+\frac ba$) and in the second case $(1,2,\frac1f+\frac gf)$. As you see, the $z$-values need not be the same.
There’s where you went wrong, when you said, “Which I guess implies that…”
You want to think of the situation geometrically, where the first equation defines a particular plane in space (depending on the values of $a$ and $b$), and the second equation defines some other plane in space. (They will not be parallel planes in this case, because in the two cases, the vectors perpendicular are $(0,1,a)$ and $(1,0,f)$, definitely not parallel.) Since the planes are not parallel, they have a line in common. (If your geometric intuition is still in its infancy, think of the east wall of your room and the south wall, which intersect in a vertical line, namely the SE seam between the two walls.)
If you want a parametric representation of this line, the work is done for you. You have the three parametric formulas, in terms of a parameter $t$:
\begin{align}
x&=g-ft\\
y&=b-at\\
z&=t
\end{align}
