Dual LP from Primal LP, how? $$\min x +y + z$$
so that
$$ x + y =2$$
and
$$y + z = 3$$
where, $x,y,z >0$. How to create the dual?
[Something like this?]
$$ \max 2s + 3t$$
so that
$$ ...+... =1$$
and 
$$ ...+...=1$$ 
 A: The underlying idea of duality is very simple: there are no non-trivial linear consequences of linear inequalities and identities. The trivial consequence is just a linear combination of the given conditions with some coefficients, which are the unknowns for the dual problem.
So, you have 5 conditions: $x\ge0$, $y\ge 0$, $z\ge 0$, $x+y=2$ and $y+z=3$. You want to use them to estimate $x+y+z$ from below. You can add them up with coefficients $a,b,c,d,e$ to get $ax+by+cz+d(x+y)+e(y+z)\ge 2d+3e$ where $a,b,c$ must be $\ge 0$ not to revert the signs of the inequalities but $d,e$ can be arbitrary because the identities survive multiplication by any constants. Now, if you want to draw the conclusion about $x+y+z$ from here, you should have exactly it on the left, so we must have $a+d=1$, $b+d+e=1$, $c+e=1$. This, together with the already mentioned $a,b,c\ge 0$ gives you $6$ restrictions. Now, you want to get as large lower bound as possible and you get $2d+3e$, so the objective is $2d+3e\to\max$. That's all there is to it. 
A: Primal
$$\min c'\bar{x}$$
$$A\bar{x} \leq \bar{b}$$ 
$$\bar{x}\geq0$$
Dual
$$\max \bar{b}' \bar{y}$$
$$A^{T} \bar{y} \leq \bar{c}$$
$$\bar{y}\geq0$$
Now work out the things.
