A good start by Jia and Jeremy from Raffles Institution, Singapore

To balance a bar balance, equal weights are required on each side of the pivot.

Taking a as the distance of the 3 weight from the pivot, and b as the distance of the 2 weight from the pivot. And using each column from the pivot as a distance of one, the weight on the left of the pivot has a moment of 40.

So ßa+2b must equal (that is, balance) \40
So 3a+2b must equal (that is, balance) 40


Matching pairs of a & b could be:
a=5 and b=12.5

a=4 and b=14

And some good algebra from Joan in Edinburgh

If we want a less than b it would be good to find where a=b, that means that the two weights are together at the same place.

If a=b then a and b have to be 8
(from 5a = 40 or 5b = 40 whichever you prefer)

But we want a less than b, so a can?t be more than 8 and the 3 weight can go as near to the pivot as we wish. b will have to get bigger to keep the balance. If the 3 weight does go to zero b will have to be 20. 20 is the furthest out from the pivot that the 2 weight goes if it keeps balanced with the 4 weight on the other side.