<s>[INST] Help determining the parameterized solution to a system of linear equation. I am quite new to this area so please bear with me if I am overlooking something glaringly obvious here :)
I am trying to solve the following equation system:
$$\begin{array}{lcl} x + y + z & = & 30 \\ x + 2y & = & 25 \\ 2x + 3y +x& = & 55 \end{array}$$
I am using the matrix notation and write the equation system as (i do not know how to typeset the divisor line between the equality part, sorry about that):
$$\begin{pmatrix} 1 & 1 & 1 & 30 \\ 1 & 2 & 0 & 25\\2 & 3 & 1 & 55\end{pmatrix}$$
This matrix can be written as this: (by multiplying row $2$ with $-1$ and adding it to row $1$, and multiplying the new row $1$ with $-1$ and adding it to row $3$, and finally by multiplying row $2$ with $-1$ and adding it to row $3$).
$$\begin{pmatrix} 1 & 1 & 1 & 30 \\ 1 & 2 & 0 & 25\\2 & 3 & 1 & 55\end{pmatrix}\longleftrightarrow\begin{pmatrix} 0 & -1 & 1 & 5 \\ 1 & 2 & 0 & 25\\2 & 3 & 1 & 55\end{pmatrix} \longleftrightarrow \begin{pmatrix} 0 & -1 & 1 & 5 \\ 1 & 2 & 0 & 25\\2 & 4 & 0 & 50\end{pmatrix} \longleftrightarrow \begin{pmatrix} 0 & -1 & 1 & 5 \\ 1 & 2 & 0 & 25\\0 & 0 & 0 & 0\end{pmatrix}$$
This can be written as:
$$\begin{array}{lcl} -y + z & = & 5 \\ x + 2y & = & 25 \\ 0& = & 0 \end{array}$$
according to my textbook $z$ is a parameter in this equation system. But from the development above, could not $y$ just as well be the parameter?
By solving the equation system differently you could get the final result as:
$$\begin{pmatrix} 1 & 0 & 2 & 35 \\0 & 1 & -1 & -5\\0 & 0 & 0 & 0\end{pmatrix}\longleftrightarrow \begin{array}{lcl} x + 2z & = & 35 \\ y + -z & = & -5 \\ 0& = & 0 \end{array}$$
From this development is is much more clear that $z$ should be the parameter since $x, y$ can be expressed as linear functions of $z$?
I would please like to get some help with how to think about this.
Thank you kindly!
 [/INST] 
From this development is is much more clear that z should be the parameter since x,y can be expressed as linear functions of z?

Precisely. $z$ is a "free" variable (as you can tell by the last row of zeros in both approaches) on which the precise values of $x, y$ depend. Once a value for $z$ is chosen (among infinitely many possible values for $z$, then we can determine $x, y$.
So we can write the solutions as follows: 
\begin{pmatrix} 
35 - 2z \\ 
z - 5 \\
z
\end{pmatrix}
You'll often see the following: Given any assignment $z = \alpha$, then the solution can be expressed:
\begin{pmatrix}
35 - 2\alpha \\
\alpha - 5 \\
\alpha
\end{pmatrix}
I used the latter solution to the system of equations as it is in the preferred reduced row echelon form. But the first approach will yield the same parametrization, which you can see when you write $y$ as a function of $z$, since then $x$ is also a function of $z$. 
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