I'm starting to solve linear program exercises with the Simplex method.
$$max \ (3x_1-x_2)$$ $$\begin{cases} x_1-x_2 \le 3\\ 2x_1\le x_2\\ x_1+x_2\ge 12\\ x_2 \le 10\\ x_1,x_2 \ge 0 \end{cases}$$
I transform it to the standard form...
$$\begin{cases} x_1-x_2 + x_3 = 3\\ \color{red}{2x_1 + 0 + 0 + x_4 = x_2}\\ x_1+x_2 + 0 + 0 - x_5 = 12\\ 0 + x_2 + 0 + 0 + 0 + x_6 = 10\\ x_1,x_2 \ge 0 \end{cases}$$
Look at the second row. The righthand of the equation is $x_2$.
It is my understanding that the righthand of the constraints must be non-negative so this program can be in standard form. Since it is $x_2$ in the second row, this isn't necessarily the case.
I imagined that I should simply move the $x_2$ to the left side. However I am hesitant do that, because I guess that affects the matrix and hence the result. What exactly should one do in this scenario?
In the original problem you simply rewrite the second inequality as
$$2x_1-x_2\le 0$$