How did they get the standard form of this LP?

Question

I'm confused to what method they are using to get the standard form of this LP? Why is there an "e" variable? I need help with this problem.

Answer 1

That's a surplus variable, the purpose is to convert the first inequality into an equality.

Notice that if $a \ge b$, it is equivalent to there exists an $e \ge 0$ such that $a-e = b$.

To see this, if $a \ge b$, we have $a-b \ge 0$, we then let $e=a-b$.

Conversely, if $e=a-b$ where $e \ge 0$, then we have $a-b \ge 0$.