 Solution

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.