I've been using this site to learn the basics of the simplex algortihm: https://www.zweigmedia.com/RealWorld/tutorialsf4/framesSimplex2.html
Now I'm trying to understand how simplex works when the linear program contains both $\ge$ and $\le$ constraints, and how simplex handles conflicting constraints.
As a very simple example, consider this linear program, obviously this has no solution:
Maximize $p = x$ where $x \ge 10$ and $x \le 5$
My starting tableau looks like this (using $s$ and $t$ as slack and surplus variables):
$$ \begin{array}{c|ccc|cc} &x&s&t&p&A\\ \hline *s&1&-1&0&0&10\\ t&1&0&1&0&5\\ \hline p&-1&0&0&1&0 \end{array} $$
The pivot column is the column with the the largest positive number in the first starred row.
In this case the that's the first column. The row is the row with the lowest positive ratio, $\frac{10}{1} > \frac{5}{1}$ so that's the second row. Pivoting results in the following tableau:
$$ \begin{array}{c|ccc|cc} &x&s&t&p&A\\ \hline s&0&-1&-1&0&5\\ x&1&0&1&0&5\\ \hline p&0&0&1&1&5 \end{array} $$
How can I tell from the tableau that this is not a valid solution? Obviosly I can substitute $x = 5$ in the original constraints and see that they don't match, but is there a more direct way?
We know that $s \geq 0$ and $t \geq 0$, hence $-s-t \leq 0$.
but the first row of the tableu says $-s-t =5 \leq 0$ which is a contradiction.
Also, note that all basic variable should be nonnegative, here $s=-5$.