Simplex Method with negative R.H.S Question:
Maximize $2x_1 - 6x_2$
Subject to 
\begin{align*}
-x_1 - x_2 - x_3 &\leq -2 \\
2 \, x_1 - x_2 + x_3 &\leq 1 
\end{align*}
My Process:
I create an auxiliary problem: 
Maximize $-x_0$ 
Subject to
$-x_1 - x_2 - x_3 -x_0 \leq -2 \\
2x_1 - x_2 + x_3 -x_0 \leq 1 $
Introduce slack variables:
$w_1 = -2 + x_1 + x_2 + x_3 + x_0 \\
w_2 = 1 - 2x_1 + x_2 - x_3 + x_0   \\
P = -x_0  \\
$
$x_0$ enters basis, $w_1$ leaves. 
$x_0 = 2 - x_1 - x_2 - x_3 + w_1 \\
w_2 = 3 - 3x_1 - 2x_3 + w_1      \\
P = -2 + x_1 + x_2 + x_3 - w_1   
$
$x_1$ enters basis, $x_0$ leaves.
$x_1 = 2 - x_2 - x_3 + w_1 - x_0  \\
w_2 = -3 + 3x_2 + x_3 - 2w_1 + 3x_0 \\
P = 0 - x_0 $
I'm not sure what to do here. Since $w_2 = -3$ but $P = 0 - x_0$ meaning the solution to the auxiliary problem is optimal, but not feasible. Have I made an arithmetic error? Or am I doing this wrong. 
I would like to solve this using the auxiliary problem method as discussed in my text. I'm not supposed to use a tableau. 

I figured out my error. I forget to apply one Minimum Ratio Test and should have applied a pivot to $x_0$ instead of $w_2$. Since choosing $w_2$ to leave the basis provides a tighter bound.
for $x_0$ to be $\geq 0$
$x_0 \leq 2 - x_1 \implies x_1 \leq 2$
for $w_2$ to be $\geq 0$
$w_2 \leq 3 - 3x_1 \implies x_1 \leq 1$ 
