Applying simplex, I found out the optimal solution to be $z^*=16$ and $x^{*} = (8,0,0,0,12)$ where $x_4,x_5$ are slack variables.
Now, suppose we add constraint $x_2+2x_3 = 3$
So, If I were to add this constraint to the optimal tableau, we see that this does ${\bf not}$ affect the values of $z_j-c_j$ and so optimality is not changed. This suggests that we have to consider $x_2+2x_3 \leq 3 $ and $x_2+2x_3 \geq 3$ which is same as $-x_2-2x_3 \leq -3$ and so after adding slack variables $x_6,x_7$,then we see that the feasibility is changed (primal feasibility). In this situation, do we run the dual simplex?
or the way to handle this situation is different than the way I thought?
$(8,0,0)$ doesn't satisfy $x_2+2x_3=3.$ The value on the LHS is less than the $RHS$.
$$\max 2x_1+x_2-x_3-Mx_6$$ subject to $$x_1+2x_2+x_3+x_4=8$$
$$-x_1+x_2-2x_3 + x_5=4$$
$$x_2+2x_3+x_6=3$$
$$x \ge 0$$
At $(x_1, x_2, x_3, x_4, x_5, x_6)=(8,0,0,0,12,3)$, we have $x_1, x_5, x_6$ being basic variables and we can run primal simplex.
If the previous basis matrix is $B$, now, our basis matrix is $\begin{bmatrix} B & 0 \\ 0 & 1\end{bmatrix}$ with its inverse being $\begin{bmatrix} B^{-1} & 0 \\ 0 & 1\end{bmatrix}.$ The main entries of the simplex tables are $$\begin{bmatrix} B^{-1} & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix} A & 0\\ a_3^T & 1\end{bmatrix}=\begin{bmatrix} B^{-1}A & 0\\ a_3^T & 1\end{bmatrix}$$
$$\begin{bmatrix} B^{-1} & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix} b \\ b_3\end{bmatrix}=\begin{bmatrix} B^{-1}b \\ b_3 \end{bmatrix}$$