convert absolute form into linear programming problem I would like to convert this problem into a Linear Programming Problem : 
$\min |x|+|y|+|z|$
subject to $x+y \leq 1$ 
$2x+z=3$.
The solution to this problem is given chapter and here. But I still do not get it. Could anyone please explain it explicitly the problem solution. Especially the part how the objective value is formed.
Thanks in advance.
 A: Let's call the following model model $A$:
$\min |x|+|y|+|z|$
subject to $x+y \leq 1$ 
$2x+z=3$.
and the following model model $B$:
$\min x_1+x_2+|y|+|z|$
subject to $x_1-x_2+y \leq 1$ 
$2x_1-2x_2+z=3$.
$x_1,x_2 \geq 0$
I claim that the two models are equivalent. 
To see this, suppose the optimal solution from  model $A$ that is  $(x_A, y_A, z_A)$. We can let 
$$(x_{1B},x_{2B},y_B,z_B)=\begin{cases} (x_A,0,y_A,z_A)& x_A\geq0 \\ (0,-x_A,y_A,z_A)& x_A <0\end{cases}$$
and we can see that it is feasible for model $B$ and obtain the same value. Hence the optimal value for model $B$ cannot be worse than model $A$. the question is can it do better?
To verify that this is not the case, suppose we have an optimal solution for model $B$. $(x_{1B},x_{2B},y_B,z_B)$, we want to construct a solution to model $A$ that shares the same value.
I claim that $x_{1B}x_{2B}=0$, i.e., one of them is zero. Otherwise, we can construct another feasible solution $(x_{1B}-\min(x_{1B},x_{2B}),x_{2B}-\min(x_{1B},x_{2B}),y_B,z_B)$ that is feasible and has a smaller objective value, contradicting optimality of $(x_{1B},x_{2B},y_B,z_B)$.
Now, we can verify that 
$$(x_A,y_A,z_A)=(x_{1B}-x_{2B},y_B,z_B)$$
is indeed feasible for model $A$.
If $x_{1B}=0$, then $x_A=x_{1B}-x_{2B}=-x_{2B}$, and $$|x_A|=-(-x_{2B})=x_{2B}=0+x_{2B}=x_{1B}+x_{2B}$$
If $x_{2B}=0$, then $x_A=x_{1B}-x_{2B}=x_{1B}$, and $$|x_A|=x_{1B}=x_{1B}+0=x_{1B}+x_{2B}$$
and hence the two optimization problem is equivalent.
To answer your original problem, repeat the same procedure for $y$ and $z$.
