Can the Lever Rule be Modified for a Three-Component Problem?

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vin300
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Is it possible or not to solve the typical kind of problem given more than two components; take this self made problem: A quantity measuring 12500 units be divided into three parts such that 15% first, 20% second and 30% third amount to 25% original, that is 3125. What would be each divided value? I'm only looking for a solution as simple as two-component problem, nevertheless may try until some level of difficulty.
 
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vin300 said:
Is it possible or not to solve the typical kind of problem given more than two components; take this self made problem: A quantity measuring 12500 units be divided into three parts such that 15% first, 20% second and 30% third amount to 25% original, that is 3125. What would be each divided value? I'm only looking for a solution as simple as two-component problem, nevertheless may try until some level of difficulty.
If I understand this correctly, you will have two linear equations with three variables, and the condition that the three variables are positive. This kind of problems normally can have an infinite number of solutions.

In your example, you have
a+b+c=12500
15a+20b+30c=312500
a>0, b>0, c>0

You can solve the two equations for b and c in terms of a, and then get bounds for a by requiring that a>0, b>0, c>0.
The solutions are (if I didn't make a computation error):
b=6250-3a/2
c=6250+a/2
Requiring b>0 gives 0<a<4166.66
 
Yes I did reckon in a few hours time that this kind will produce infinite results, and also explains why it remained unseen. So sorry for the waste of time.