SOLUTION: A man's age is three times the sum of the ages of his two sons, one of whom is twice as old as the other; in 12 years the sum of the son's ages will be three-fourths of their fathe

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Question 979146: A man's age is three times the sum of the ages of his two sons, one of whom is twice as old as the other; in 12 years the sum of the son's ages will be three-fourths of their father's age. Find their respective ages.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A man's age is three times the sum of the ages of his two sons, one of whom is twice as old as the other;
in 12 years the sum of the son's ages will be three-fourths of their father's age.
Find their respective ages.
:
Let m = the man's age
let s = a son's age
"one of whom is twice as old as the other;" therefore
2s = the other sons age
therefore
3s = the sum of the sons's ages
:
"A man's age is three times the sum of the ages of his two sons,"
m = 3(3s)
m = 9s
:
" in 12 years the sum of the son's ages will be three-fourths of their father's age."
s + 12 + 2s + 12 = 3%2F4(m + 12)
3s + 24 = 3%2F4(m + 12)
multiply both sides by 4
4(3s + 24) = 3(m + 12)
12s + 96 = 3m + 36
replace m with 9s
12s + 96 = 27s + 36
96 - 36 = 27s - 12s
60 = 15s
s = 60/15
s = 4 yrs old is one son and then 8 yrs old is the other son
Dad will be 3(12) = 36 yrs