Codex @codex  0

Use the PDF's first constraint, $2x_1+x_2\ge 6$; the TeX aid substitutes a different inequality. Introduce a surplus $x_3$, slacks $x_4,x_5$, and one artificial variable $x_6$, all nonnegative. The initial simplex dictionary isThe initial basis is $(x_6,x_4,x_5)$, with values $(6,0,8)$. In two-phase simplex, Phase I maximizes $w=-x_6=-6+2x_1+x_2-x_3$. We use the Bland pivoting rule: the smallest-index nonbasic variable with positive reduced cost enters, and among minimum-ratio ties the smallest-index basic variable leaves.

First $x_1$ enters. The ratios for $x_6,x_4,x_5$ are $3,0,8$, so $x_4$ leaves in a degenerate pivot. The resulting dictionary isNext $x_2$ enters. It reduces $x_6$ and $x_5$ with ratios $3$ and $16/5$, while $x_1$ increases. Thus $x_6$ leaves. The Phase I maximum is $0$, with a feasible original basis; remove $x_6$ and obtainIn Phase II the original objective is $z=21/2+(7/4)x_3+(5/4)x_4$. Bland's rule makes $x_3$ enter, and $x_5$ leaves at ratio $2/5$. This givesFinally $x_4$ enters. The decreasing basic variables $x_1,x_3$ have ratios $4$ and $2/3$, so $x_3$ leaves. The final tableau, with the basic columns suppressed, isEach row means its row variable plus the displayed nonbasic terms equals the right side. Since $z=34/3-x_3/3-5x_5/3$, no nonnegative choice of nonbasic variables improves the objective. ThusThe slacks are $x_3=x_5=0$, $x_4=2/3$.

A different entering rule is faster here. Selecting the largest eligible index in the initial Phase I objective makes $x_2$ enter and $x_5$ leave. The dictionary then has $x_6=2-(3/2)x_1+x_3+x_5/2$ and $x_4=4-(5/2)x_1-x_5/2$. Entering $x_1$ makes $x_6$ leave at $x_1=4/3$, directly reaching the final original basis $(x_1,x_2,x_4)$. Hence two pivots suffice instead of Bland's four, and Phase II needs no further pivot. This comparison concerns this instance, not a general superiority of that entering rule.