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Introduce surplus variables $s_1,s_2$, slack $s_3$, and artificial variables $a_1,a_2$, all nonnegative. The two-phase simplex standard-form equations arePhase I maximizes $w=-a_1-a_2$. An initial basic feasible solution has basic variables $(a_1,a_2,s_3)=(1,10,6)$ and all others zero. Its simplex dictionary isLet $x_1$ enter. The simplex ratio test gives $1$ from the $a_1$ row and $10/4$ from the $a_2$ row, so $a_1$ leaves. After this pivot,Choose $x_2$ to enter; the limiting ratios are $6/3=2$ and $6$, so $a_2$ leaves. The resulting dictionary isThus $\boxed{w_{\max }=0}$, attained at $(x_1,x_2,s_3)=(3,2,4)$ with $s_1=s_2=a_1=a_2=0$. Since $w\le 0$ for every feasible Phase I point, this proves feasibility and completes Phase I.

Delete the artificial variables and restore the original objective $z=-2x_1+3x_2$. The resulting Phase II dictionary isLet $s_2$ enter; only $s_3$ decreases, and it reaches zero at $s_2=12$. Pivoting givesAn optimal simplex tableau, with each row recording its equality, isBoth reduced costs in the maximization dictionary are negative. Setting nonbasic $s_1=s_3=0$ therefore givesAs a direct optimality check, $x_1\ge 1+x_2$ implies $z\le -2+x_2\le 4$; the displayed point attains both equalities.

Use the original PDF: the objective is $z=-x_1+3x_2$, and the second constraint has coefficient $2$ on $x_2$. The TeX transcription corrupts both, as well as the nonnegativity condition. Introduce surplus variables $s_1,s_2$, slack variables $s_3,s_4$, and artificial variables $a_1,a_2$, all nonnegative. The two-phase simplex Phase One problem isThe initial basis $(a_1,a_2,s_3,s_4)$ has values $(3,6,2,5)$, and its objective dictionary is $w=9-3x_2+s_1+s_2$. Enter $x_2$: the ratio test makes $s_3$ leave at $x_2=2$. The resulting dictionary includeswith $w=3-3x_1+s_1+s_2+3s_3$. Enter $x_1$ next; $a_1$ leaves at $x_1=1/2$. Nowand $w=3/2-s_1/2+s_2+3s_3/2+3a_1/2$. Enter $s_1$; the ratio test makes $a_2$ leave at $s_1=3$. This produces $x_1=2$, $x_2=4$, $s_1=3$, $s_4=1$ with both artificial variables zero. Since $w\ge 0$ always, the Phase One minimum is zero, proving feasibility and providing an original-variable basis.

Delete the artificial columns. The starting Phase Two dictionary isEnter $s_2$, whose reduced gain is positive. Only $s_4$ decreases, and it leaves at step one. Eliminating $s_2=1-s_3-s_4$ gives the optimal dictionaryEquivalently, with row equations written as the displayed coefficients times the variables equal to the right-hand side, the optimal simplex tableau isThe objective row represents $z+s_3+2s_4=12$. Nonnegative nonbasic variables cannot improve the objective, so reading them as zero provesBoth nonbasic reduced costs are strictly unfavorable, so this optimal point is unique.

Use two-phase simplex to handle the constraint of the opposite inequality direction. Introduce nonnegative slack variables $s_1,s_2$, a surplus variable $s_3$, and an artificial variable $a$. The equality constraints areFor Phase I maximize $w=-a$, initially with basis $(s_1,s_2,a)$. Its initial simplex dictionary isChoose $x_1$ to enter. The simplex ratio test compares $30$, $35/2$ and $40$, so $s_2$ leaves. The new dictionary isSince $w=-45/2+(5/2)x_2-x_3-s_2/2-s_3$, choose $x_2$ to enter. Its limiting ratios are $25/3$ from $s_1$ and $9$ from $a$, so $s_1$ leaves. This givesNow $w=-5/3-(8/3)x_3+s_2/3-(5/3)s_1-s_3$, so let $s_2$ enter. Its limiting ratios are $65$ from $x_1$ and $5$ from $a$; the artificial variable leaves. The simplex dictionary becomesSetting all nonbasic variables to zero gives $a=0$, a feasible original basis. Since $-a\le 0$, Phase I is optimal with value zero; delete $a$.

For Phase II the original objective in this basis isAll nonbasic reduced costs are strictly negative, so no further improving simplex pivot exists. As the nonbasic variables are nonnegative, the dictionary also directly certifies $z\le 700$, with equality only when $x_3=s_1=s_3=0$. ThereforeThe second original constraint has slack five and the other two are tight. As an independent linear programming optimality certificate, multiply the first constraint by $130$ and the third, after reversing it, by $80$. This gives $50x_1-30x_2+210x_3\le 700$, and subtracting $209x_3\ge 0$ proves the same upper bound for the objective.

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.