 Solution

Try the dual intersection $3y_1+2y_2=6$, $y_1+4y_2=4$. It gives $y=(8/5,3/5)$; the remaining constraint is $5y_1+y_2=43/5\ge 5$, so it is dual feasible. Its strictly positive slack in the first column suggests $x_1=0$, while positivity of both $y$ entries suggests $Ax=b$. Solving $3x_2+x_3=1$, $2x_2+4x_3=2$ gives $x=(0,1/5,2/5)$, which is primal feasible.

Their objective values agree:Weak duality certifies that both are optimal. The strict first-column dual slack and the two active primal equations also show uniqueness of the primal optimum; positivity of $x_2,x_3$ forces the two dual equations, so the dual optimum is unique as well.