I transformed the linear programming problem, where $x_1$ and $x_2$ also has to be non-negative: $$ \begin{array}{rrrrrrrl} (P) & \text{min. } & \xi & = & x_1 & + & 3x_2 &\\ & \text{st. } & -x_1 & & &\geq & -2.5\\ & & & & x_2 &\geq & 1.5\\ & & x_1 & + & 2x_2 &\geq & 5 \\ \end{array} $$ to standard form: $$ \begin{array}{rrrrrrrl} (P) & \text{-max. } & -\xi & = & -x_1 & - & 3x_2 &\\ & \text{st. } & x_1 & & &\leq & 2.5\\ & & & - & x_2 &\leq & -1.5\\ & & -x_1 & - & 2x_2 &\leq & -5 \\ \end{array} $$
I know this problem easily can be solved with dual simplex, however, would like to solve it with primal simplex, if it is possible. I can add the parameter $\mu=5+\epsilon$, when making the simplex table, so the primal simplex becomes feasible: \begin{array}{r|rrrrr|r} & x_1 & x_2 & w_1 & w_2 & w_3 & \\ \hline & -1 + \mu & -3 + \mu & 0 & 0 & 0 & 0 \\ \hline w_1 & 1 & 0 & 1 & 0 & 0 & 2.5 + \mu \\ w_2 & 0 & -1 & 0 & 1 & 0 & -1.5 + \mu \\ w_3 & -1 & -2 & 0 & 0 & 1 & -5 + \mu \end{array}
However, my minimum ratio test becomes: $$ \text{min}\left\{\frac{7.5+\epsilon}{0},-3.5-\epsilon,\frac{-\epsilon}{2}\right\} $$ It should then be the one, that are minimum of these, however the minimum shall be positive, and not zero, however, none of these fulfill this, when having chosen the second column, $x_2$, as my pivot column as $-3+\mu<-1+\mu$.
Any help on what to do? I don't understand what I'm doing wrong unfortunately.