Solve using simplex $$\text{ Max } z=4x_1+3x_2$$ $$\text{s.t } 2x_1+3x_2\leq 6 $$ $$-3x_1+2x_2\leq 3$$ $$2x_1 \leq 5$$ $$2x_1+x_2\leq 4$$ $$ x_1,x_2 \geq 0$$
I have solve it and got:
$So x_1 = 1.5, x_2 = 1 \text{ and } z=9$
One can see that some of the constrains are redundant as we need at most two equation for two unknowns. Can we have two "unused" slack variable in the basis? we just ignore them?
Yes, if the solution in the original variables $x_1$ and $x_2$ is what you're interested in, you can ignore the slack variables in the final tableau. They are necessarily part of the basis since the slack of the second and third restriction is non-zero.