Why are free variables in systems of linear eqs used as parameter In describing the solution of a system of linear equations with many solutions, why do we use a free variable as a parameter to describe the other variables in the solution? Why do we not we use a leading variable? Since by the commutative property of addition we can swap between the free and leading variables, e.g. x + y + z = x + z + y; the solution set will essentially be the same (albeit having different orders).
Definitions: 


*

*A free variable is a parameter that is not a leading variable.

*A leading variable is the first variable that has a non-zero coefficient in reduced form.

*These definitions are most easily understood with respect to the Echelon form of a system of linear equations expressed as a Matrix. See this link for details.


For example:
Let $S$ be the solution set of the system
$$\begin{align*}
x+y+z &= 3\\
y-z &= 4
\end{align*}$$
Using the free variable $z$ as the parameter
$$S = \{(-2z-1, z+4, z)\mid z\in\mathbb{R}\}.$$ 
Using the leading variable $y$ as the parameter
$$S = \{(-2y+7, y, y-4)\mid y\in\mathbb{R}\}.$$ 
 A: Suppose you and I (and a few hundred other people) independently solve a system of linear equations, and we want to compare our answers. If we have taken Gaussian elimination to reduced row-echelon form, and have expressed our answers with the free variables as parameters, we can immediately compare our answers; if there is any difference at all, then (at least) one of us is wrong. If I used the free variables, and you used the leading variables, then we have some calculating to do before we can see whether we got the same answer. 
So, it's really a matter of selecting a standard form so everyone can agree on an answer without having to do a lot more calculation to check. 
EDIT: Lots of things that come later depend on solving systems of equations (e.g., finding a basis for the kernel of a linear transformation, finding a basis for an eigenspace of a matrix), and it continues to be useful to have a standard form. 
A: When you have more variables than equations, the solution of the system of linear equations is not just a point, but a line, or a plane etc.
Suppose you have $n$ variables and $m$ equations, $m<n$. You may "fix" extra $n-m$ variables and consider those just as numbers. By doing this, you essentially reduce the number of variables to $m$, equal to the number of equations, and express all other variables in terms of constants and fixed variables (parameters). And it does not matter which variables you fix and use as parameters (as long as each equation still contains a non-fixed variable): you are going to get the same set of solutions, just parametrized in different ways.
Consider your example. You have 3 variables, which means that you have the whole 3D space of possible solutions: $$(x,y,z)$$. Each linear equation defines a plane in this space: $ax+by+cz=d$ is an equation for a plane. If you have two such equations, then the set of possible solutions is either a line, or the empty set (if two planes are parallel, e.g. $x+y+z=1$ and $x+y+z=2$). In your example, it is a line. You have already given two different ways to describe (parametrize) this line. Another way would be to fix $x$ and solve for $y$ and $z$: $\{(x,\frac{7-x}{2},-\frac{x+1}{2})|x\in\mathbb{R}\}$.
In fact, you do not have to fix some variables. Instead, you may add $n-m$ equations (of course, parametrized by new external variables). For example, let as add a new equation: $y+z=2t$. Now we can solve the system of 3 equations with 3 variables ($x,y,z$) as usually and obtain the following answer: $\{(3-2t,t+2,t-2)|t\in\mathbb{R}\}$. You can easily check that this set is exactly as any of the 3 previous ones: to get the first solution from this one just let $t=z+2$, the second one -- $t=y-2$, and the third one $t=\frac{3-x}{2}$.
