# Nice unanswered question by Maggie

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"100 bushels of corn are divided among 100 men, women, and children. Men receive 3 bushels each, women 2 bushels and children 1/2 bushel each. How can the bushels be distributed? is there more than one solution? if so, find the other solutions."

let the number of men be x
the number of women be y
and the number of children be (100-x-y)

then 3x + 2y + (1/2)(100-x-y) = 100
6x + 4y + 100 - x - y = 200
5x + 3y = 100

this is a linear relations where one obvious solution is (20,0)
the 'slope' is -5/3, that is,
for every decrease of 3 in the x's we can increase y by 5 and we have another solution.
so other solutions are
(17,5)
(14,10)
(11,15) etc.
of course we have to check if the number of children is still valid.
e.g. for (11,15)
we would have 11 men, 15 women and 100-11-15 or 74 children
check: 11(3) + 15(2) + (1/2)(74) = 100 YEAH!

keep my pattern going for the rest of the results.

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