= 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 $5y_1+y_2=43/5\ge5$, 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:
$$
\boxed{x_*=(0,1/5,2/5)^T,\quad y_*=(8/5,3/5)^T,\quad c^Tx_*=b^Ty_*=14/5.}
$$
<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.
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