Linear Programming - Simplex Method Let,
$\begin{array}{lcl}Min: z  =10x_{1}+5x_{2}  \\ \\ 20x_{1}+50x_{2} \geq 200 \\ 50x_{1}+10x_{2} \geq 150 \\30x_{1}+30x_{2} \geq 210 \\x_{1},x_{2} \geq 0\end{array}$
a linear programming problem.
In standart form one have,
$\begin{array}{lcl}Min: z  =10x_{1}+5x_{2}  \\ \\ 20x_{1}+50x_{2}+s_{1} = 200 \\ 50x_{1}+10x_{2}+s_{2} = 150 \\30x_{1}+30x_{2}+s_{3} = 210 \\x_{1},x_{2},s_{1},s_{2},s_{3} \geq 0\end{array}$
The previous form is associate to the  basic feasible solution, $\left( 0,0,200,150,210\right)$.
Because the goal is to minimize $z$, let $x_{2}$ increase. Because the slack variables have to be positive, one have:
$\begin{array}
\\200-50x_{2}\geq0
\\150-10x_{2}\geq0
\\210-30x_{2}\geq0
\end{array}$
Solving the previous system, one get that the maximum value of $x_{2}$ is $4$. When $x_{2}=4$ one get $s_{1}=0$.
The new basic feasible solution is $\left( 0,4,0,110,90\right)$. And, by making $x_{2}=4-\frac{2}{5}x_{1}-\frac{1}{50}s_{1}$, the standart form associated is,
$\begin{array}{lcl}Min: z  =20 +8x_{1}-\frac{1}{10}s_{1}  \\ \\ \frac{2}{5}x_{1}+x_{2}+\frac{1}{50}s_{1} = 4 \\ 46x_{1}-\frac{1}{5}s_{1}+s_{2} = 110 \\18x_{1}-\frac{3}{5}s_{1}+s_{3} = 90 \\x_{1},x_{2},s_{1},s_{2},s_{3} \geq 0\end{array}$
Now, and because the goal still is minimize $z$, let $s_{1}$ increase. But by the first restrition, if $s_{1}$ increase (and because $x_{1}=0$) $x_{2}$ will decrease and one would get back to the first basic feasible solution.
What is wrong with my thoughts? Thanks

It should be:
$\begin{array}{lcl}Min: z  =10x_{1}+5x_{2}  \\ \\ 20x_{1}+50x_{2}-s_{1} = 200 \\ 50x_{1}+10x_{2}-s_{2} = 150 \\30x_{1}+30x_{2}-s_{3} = 210 \\x_{1},x_{2},s_{1},s_{2},s_{3} \geq 0\end{array}$
