Converting a linear optimization problem with absolute values

I think there is something wrong with the problem statement: it has ##x_1## and ##x_2## in the objective but ##x_1## and ##x_3## in the constraint.
Mod note: This is now fixed.
As written, the problem does not have a finite solution, since we can have ##x_1 \rightarrow -\infty##, ##x_3 = 6 - x_1 \rightarrow +\infty## and ##x_2 = 0##. If you really intend to have ##x_1 \geq 0## you need to say so.

In any case, you need only one constraint, not three. Basically, it is one of the standard methods for such problems to just replace ##x_2## by ##x_2^+ - x_2^-## (with ##x_2^+, x_2^- \geq 0##) and to write ##x_2^+ + x_2^-## in place of ##|x_2|## in the objective.

 

There are two *different* ways of handling such problems, and you have mixed them up. For problems in which |x| appears only in the objective, and with a coefficient c > 0, we can:
(1) Replace ##x## in the constraints by ##x^+ - x^-## and ##|x|## by ##x^+ + x^-## in the objective; or
(2) keep x unchanged in the constraints, replace |x| in the objective by a new variable z, and add two constraints z ≥ x and z ≥ -x.

 

Stop worrying; your formulation is OK exactly as it stands. When you try to make z small, that will restrict x_2 in the constraints, so everything *really is* connected together. If you don't believe it, try solving both formulations (either by hand or by using an LP computer package).