SOLUTION: Three brothers, A,B,C, bought a farm for $5,000. A could pay for it alone by borrowing 1/3 of B's money and 1/2 of C's money. B could pay for it by borrowing 1/2 of A's and 1/4 of

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: Three brothers, A,B,C, bought a farm for $5,000. A could pay for it alone by borrowing 1/3 of B's money and 1/2 of C's money. B could pay for it by borrowing 1/2 of A's and 1/4 of       Log On


   

Question 451883: Three brothers, A,B,C, bought a farm for $5,000. A could pay for it alone by borrowing 1/3 of B's money and 1/2 of C's money. B could pay for it by borrowing 1/2 of A's and 1/4 of C's money. C could pay for it by borrowing 1/4 of A's and 1/6 of B's money. How much money does each have?

Answer by pedjajov(51) About Me  (Show Source):
You can put this solution on YOUR website!
Brothers has a, b and c dollars. From the text we have

a%2Bb%2F3%2Bc%2F2=5000, multiply by 6
a%2F2%2Bb%2Bc%2F4=5000, multiply by 4
a%2F4%2Bb%2F6%2Bc=5000, multiply by 6

6a%2B2b%2B+3c=30000 -> a=(30000-2b-3c)/6, substitute in the second
2a%2B4b%2B++c=20000
3a%2B2b%2B12c=60000

6a%2B+2b%2B+3c=30000
2%2830000-2b-3c%29%2F6%2B4b%2Bc=20000, multiply by 6
2%2830000-2b-3c%29%2B24b%2B6c=120000
60000-4b-6c%2B24b%2B6c=120000
20b=60000
b=3000

Going back to a we have ->
a=%2830000-2%2A3000+-+3c%29%2F6
a=%2824000-3c%29%2F6, let use it in third equation together with value for b

3%2824000-3c%29%2F6%2B2%2A3000%2B12c=60000, multiply by 6
3%2824000-3c%29%2B36000%2B72C=360000
72000-9C%2B36000%2B72C=360000
63C=252000
C=4000

One more time going back to a we have
a=%2824000-3%2A4000%29%2F6
a=12000%2F6
a=2000

So brothers have
A=$2000
B=$3000
C=$4000