Example 1 from web page 

      max    z = x1  +4x2  
such that        x1  +2x2  <= 5  
               -2x1  -x2   >= -5  
                 x1, x2    >= 0 .  
lets say 
             1  -1   -4     =  0
                -1   -2   5 => 0 
                -2   -1   5 => 0

or           x1  x2   z   s1  s2   COST 
             1   4    -1          = 0
             1   2        1       = 5
             2   1            1   = 5   
is now assumed the proper setup 
1.00 4.00 -1.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 
1.00 2.00  0.00 1.00 0.00 5.00 0.00 1.00 0.00 0.00 0.00 0.00 
2.00 1.00  0.00 0.00 1.00 5.00 0.00 0.00 1.00 0.00 0.00 0.00 
0.00 0.00  0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 
0.00 0.00  0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 
0.00 0.00  0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 
for solution
1.00 0.00 0.00 0.00 0.00 1.67  0.00 -0.33  0.67  0.33 -0.67 -1.67 
0.00 1.00 0.00 0.00 0.00 1.67  0.00  0.67 -0.33 -0.67  0.33 -1.67 
0.00 0.00 1.00 0.00 0.00 8.33 -1.00  2.33 -0.67 -2.33  0.67 -8.33 
0.00 0.00 0.00 1.00 0.00 0.00  0.00  0.00  0.00  1.00  0.00  0.00 
0.00 0.00 0.00 0.00 1.00 0.00  0.00  0.00  0.00  0.00  1.00  0.00 
0.00 0.00 0.00 0.00 0.00 0.00  0.00  0.00  0.00  0.00  0.00  1.00 

x1 = 1.67, x2 = 1.67, max z = 8.33

so lets check
      max    z = x1  +4x2  =  1.67 + 4 * 1.67 = 5 * 1.67 = 8.35
such that        x1  +2x2  <= 5  or  3 * 1.67 =  5.01 <= 5  ok?
               -2x1  -x2   >= -5 or -3 * 1.67 = -5.01 => -5 ok?
                 x1, x2    >= 0  ok
This does not look as good as a trial by error, with a little
help from these results, that x1 = 1 and x2 =2 might do a better
job. z = 1+8 = 9, 1+4 = 5, -2-2 = -4 > -5
Looks like the solution. Changing some internal code from integer
to floating point still did not change the results.
Further use and validation is required.