Why are free variables free? I'm having trouble understanding this: The free variables are called free because they can take on any value; none of the equations relates any of them to each other. Why can we set free variables to an arbitrary s or t?  
 A: Consider this system of equations:
$$
\begin{cases}
x+y+z = 2 \\
x-y-z = 0
\end{cases}
$$
If we convert it to a matrix, we get
$$
\begin{bmatrix}
  1 & 1  & 1  & 2 \\
  1 & -1 & -1 & 0
\end{bmatrix}
$$
which reduces to
$$
\begin{bmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & \boxed{1} & 1
\end{bmatrix}
$$
which indicates that our original system can be rewritten
$$
\begin{cases}
x     &= 1 \\
y + z &= 1
\end{cases}
$$
with $z$ being "free".
Now recall what it means for a set of values $(x,y,z)$ to be a solution to our system: It means if we plug in those values for $x$, $y$, and $z$ all the equations are satisfied. Hence $(1,1,0)$ is a solution, whereas $(2,1,1)$ is not (because the second equation becomes false).
When we simplified our system down to
$$
\begin{cases}
x     &= 1 \\
y + z &= 1
\end{cases}
$$
we learn that any solution to the system must have $x=1$, but the only other restriction is that $y+z=1$. Any pair of $y$ and $z$ that add to $1$ will work as a solution (alongside $x=1$). How can we find such $(y,z)$ pairs? We can let one of the variables be "free" to be any value, and then solve for the other variable. In this case, we choose $z$ to be free. So if I choose $z=1$, then I can solve for $y=0$. If $z=2$, then $y=-1$. In general, if $z=t$, then $y = 1-t$.
So then we can characterize any solution to our system by saying it must have this form:
$$ (x,y,z) = (1,\ 1-t,\ t) $$
where $t$ is free to be anything, and its value determines $y$. Using free variables allows us to write an explicit formula that can compute all the solutions of our system.
