# Prove solution does not exist for inequalities system

I have an inequalities sytem like the following:

Example

> x+y+z <= A
> x+y <= B
> x+z > C
> y+z > D
> x >= E

Let A,B,C,D,E be any constant value.

Notice that the coeficient of the variables are always 1. So an equation matrix will always look like this:

(1   1   1 | <= A)
(1   1   0 | <= B)
(1   0   1 | <  C)
(0   1   1 | >  D)
(1   0   0 | >= E)

Each row represents a plane in 3D space.

The solution to this sytem has a geometrical 3D representation which is a volumne space. So if the volume is greater than 0, there is a solution!

Is there a way to tell if all this equations have at least one feasible solution? or in other words, is there a "volume" contoured by this planes??

• it does have a solution for example $x=10$ and $y=40$. Perhaps you got one of the inequalities switched. – Anurag A May 29 '14 at 1:58
• en.wikipedia.org/wiki/Fourier%E2%80%93Motzkin_elimination – Batman May 29 '14 at 2:33
• This is just an example, i want to find an algorythm that works with any linear inequalities system – Herno May 29 '14 at 2:33