Simplex method in prime how to use the constrains maximise s.t. constraints.

[Gil]

There is no alternate solution, but just the one you found, ie solution1.

But to your vector solution1, you could create a more general solution always including your initial solution :

general solution =
solution1 (always to be included) + mu*[vector d],
with mu a free real.

Your solution1 is equivalent to
general solutions +mu=0 * [vector d].

Note further that Z-Max/Min(x) = always 0
for x = mu× (vector d), but not including solution 1.

Finding d is not (the normal) part of simplex algorithm.

As said, nice to get, as in my program HP50G EMU48 posted in that forum, the following general solution:
[ '2/3' 0 '4/3' 0 0 ]
:
+µ[>0]
*
d,
with d : [ 1 1 0 0 0 ].

Try for instance with µ=10
x1: 2/3+10×1
x2: 0+ 10×1
x3: 4/3+10×0
x4: 0+10×0
x5: 0+10×0