 Solution

For any feasible primal vector, twice the second constraint plus one third of the third givesSince $x_1,x_4\ge 0$,The vector $(0,11/6,1/3,0)$ is feasible and attains this bound. It is optimal, by this direct inequality proof of weak duality, without assuming the simplex method or a duality theorem. Equality forces $x_1=x_4=0$ and both positively weighted constraints tight, so it also proves uniqueness.