Which variable enters the basis from $\hat c$ in the revised Simplex method?

Question

I had an issue understanding wich matrix enters the bases in a linear programing exercise done during a lecture : the variable that entered the basis is one which has negative coefficient in $\hat c$ in the case of a $\max$ problem.

$$ \begin{cases} \begin{aligned} \max \ & 2x_1&+ 3x_2\\ &3x_1&+5x_2&\le 15\\ &4x_1&+x_2&\le 8\\ &x_1&+x_2&\ge 1\\ \forall i, x_i \end{aligned} \end{cases} $$

was transformed with slack variables into :

$$ \begin{cases} \begin{aligned} \max \ & 2x_1&+ 3x_2\\ &3x_1&+5x_2&+x_3&&&= 15\\ &4x_1&+x_2&&+x_4&&= 8\\ &x_1&+x_2&&&+x_5&= 1\\ \forall i, x_i \end{aligned} \end{cases} $$

I think there is a mistake in the last line, shouldn't it be : $-x_1-x_2+x_5= 1$? Because it leads me to an absurdity about which variable enters the matrix when doing revised simplex method:

\begin{align*} \hat{c} &= c -\Pi A\\ &=\begin{pmatrix} 2 & 3 & 0 & 0 & 0 \end{pmatrix}-\begin{pmatrix} 0 & 0 & 2 \end{pmatrix}\begin{pmatrix} 3& 5 & 1 & 0 & 0\\ 4 & 1 & 0 & 1 & 0\\ 1 & 1 & 0 & 0 & 1 \end{pmatrix}\\ \hat{c}&=\begin{pmatrix} 0 & 1 & 0 & 0 & -2 \end{pmatrix} \end{align*}

Then it was said that $x_5$ enters the matrix. Yet, its coefficient is negative in $\hat c$.

Answer 1

The last line of the formulation in standard form is wrong. It should read $x_1+x_2-x_5 = 1$ ($x_5$ is subtracted as an excess variable).

Indeed $x_5$ enters the basis since the coefficient is negative and it is a $\max$ problem. No contradiction there.