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Introduce nonnegative slack variables $s_1,s_3$, a nonnegative surplus variable $s_2$, and an artificial variable $a$. Phase I of the two-phase simplex method minimizes $w=a$. The initial simplex dictionary isWith the nonbasic variables zero this is feasible for the artificial problem. Enter $x_1$; the minimum ratios are $9/4,12/6,3/3$, so $s_3$ leaves. Solving its row for $x_1$ and substituting givesNow enter $x_3$. The ratios from decreasing basic variables are $5/(4/3)=15/4$ and $6/3=2$, so $a$ leaves. The resulting dictionary isSetting the nonbasic variables zero gives $a=0$, so Phase I has reached its minimum and found feasibility. Remove $a$ and reinstate the true objective $z$. Substitution of the basic variables givesEnter $x_2$. The decreasing rows give ratios $(5/3)/(2/3)=5/2$ for $x_1$ and $(7/3)/(7/3)=1$ for $s_1$, so $s_1$ leaves. Phase II then hasEvery nonbasic variable is nonnegative and every reduced cost is positive. Thus the simplex optimality criterion provesAll three original constraints are active. Positivity of all three reduced costs also proves uniqueness: equality in the objective lower bound forces $s_1=s_2=s_3=0$, which fixes the displayed solution.