Re: Solving a 3D systems of equations


"Randy Poe" <poespam-trap@xxxxxxxxx> wrote in message
news:1156880936.178409.317800@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Top posting repaired.

payam wrote:
fred botjon wrote:
"payam" <pmrazavi@xxxxxxxxx> wrote in message
news:1156874438.929599.311620@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I am trying to solve the following systems of equations for x, y and
z:

{
xyz=a
x+y+z=w
}

x<y<z (i.e., x is not equal to y or z & y is not equal to z)
'a' and 'w' are given quanitites.

How can I solve the system?

Thanks in advance


you have 2 equations, and 3 unknowns, which means it is underdefined

x = a / (y*z)

a / (y*z) + y + z = w is the best you can do.
Thanks fot the replies, (easpecially to Randy).

Sorry to forget the mentions that x, y and z are natural numbers.

I don't care about the name of the variables, I just want to find the
3 natural numbers that satisfy the system.

Do I still have infinite solutions?


Obviously not. The equation x + y + z = w means that
x is at most w/3.

wrong.

The equation xyz = a means x is at most
cuberoot(a).

wrong.

So there are finitely many possible values for x,
and similarly for y and z.

wrong.


There may actually be only one solution but I don't know if
that is provably true in general.

- Randy



.

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