Consider the following linear model.
\begin{align} \min&\quad z = c^{t}x\\ \mathrm{s.t.}&\quad Ax = b\\ &\quad e^{t} x = 1 \\ &\quad x_{i}\geqslant0 \quad \forall i \in \{1,...,n-1\} \end{align}
where $b, c \in \mathbb{R}^{n}$ and $A_{ij} = 1$ if $ i = j$ or $j = n$ and $A_{ij} = 0$ otherwise. Finally, $e^{t}_{i} = 1$ for all $i \in \{1,...,n\}$ and $x_n$ is a free variable.
Use the $e^{t} x = 1$ restriction to get rid of $x_n$. Can you do the same if $x_n$ is not a free variable?
I've tried solving it for $n = 3$ for simplicity, so we have $ A =\left(\begin{array}{cccc} 1 & 0 & 1 \\ 0& 1 & 1\\ 0 & 0 & 1\\ \end{array}\right) $
And we get the following restrictions: \begin{align} x_1 + x_3 = b_1 \\ x_2 + x_3 = b_2 \\ x_3 = b_3 \\ x_1 + x_2 + x_3 = 1 \\ \end{align}
Which leads to
\begin{align} x_1 = b_1 - b_3 \\ x_2 = b_2 - b_3 \\ x_1 + x_2 = 1 - b_3 \end{align}
Now if I understand this right we've gotten rid of $x_3$ since it's forced to be equal to $b_3$ and we can replace that in all the equations.
I think this is probably wrong though since $x_3$ being a free variable is not involved in any point. Furthermore, if we let $x_3 \geq 0$ I can find no contradiction and it feels like I'm just moving symbols around.
Any help would be greatly appreciated.
If the constraint $x_3=b_3$ holds, then we can replace it everywhere (after all we need to satisfy this condition in the feasible set also). Furthermore, if $x_3\geq 0$ and $b_3<0$, we will have a contradiction, else not. So there is no question of moving symbols around, the optimization makes sense.