Solving system of inequalities

Hi everyone,

I have the following set of equations:

x+y+z=1
x+y+0.5z >= 0.9,
x+y+0.5z <= 0.9,
x+y+1.5z >= 1,
x+y+1.5z <= 1,

Of course, I know how to solve it if all the >= and <= where the equal sign. How can I deal with such equations?

[Abdusalam Alkhwaji]

rearrange the first equation,so that x = 1-y-z,then reorganize your other 4-inequality equations. that means you will solve only the 4 eqns which do non have x. then you can evaluate x from the above eqn

Answer 1

[Roger Stafford]

As they stand, your particular five conditions are equivalent to the three equations:

 x+y+z=1 
 x+y+0.5z = 0.9, 
 x+y+1.5z = 1,

However there can be no solution to these equations, as can be seen if you subtract twice the first equation from the sum of the second two equations, which would result in the impossible equation

 0 = (0.9+1)-2*1

[Simo]

Actually, these are the exact equations I want to solve:

x+y+z=1
x+y+0.94z>=0.9
x+y+0.94z<=1
x+y+0.54z>=0.6
x+y+0.54z<=0.75
x+0.664y+0.3130z>=0.4
x+0.664y+0.3130z<=0.55
x+0.26y+0.228z>=0.2
x+0.26y+0.228z<=0.35
0.7360x+0.176y+0.09z>=0.12
0.7360x+0.176y+0.09z<=0.22 
0.401x+0.05y+0.031z>=0.05
0.401x+0.05y+0.031z<=0.1

They can be solved, and the solution should be 0.05, 0.3 and 0.65 for x, y and z respectively.