SOLUTION: A man's age is three times the sum of the ages of his two sons. one of whom is twice old as the other. In four years the sum of the son's ages will be half of their father's age.

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Question 1085690: A man's age is three times the sum of the ages of his two sons.
one of whom is twice old as the other. In four years the sum of
the son's ages will be half of their father's age. find their ages?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

>>...two sons, one of whom is twice old as the other.<< 

So:
 x = younger son's age now
2x = older son's age now.

>>...A man's age is three times the sum of the ages 
of his two sons...<<

So: 
man's age now =s 3(x + 2x), which =s 3(3x) which =s 9x  

>>...In four years...<< 

Younger sons' age in 4 years will be x+4
Older son's age in 4 years will be 2x+4
Man's age in 4 years will be 9x+4 

>>...the sum of the son's ages (in four years)...<<

So, sum of ages =s x+4 + 2x+4 =s 3x+8 <--sum of sons ages 
in 4 years.  

>>...will be half of their father's age. 

So, 

3x+8 = 1%2F2(9x+4)

Solve that for x, which will be the younger son's age.
Multiply by 2 to get the older son's age.
Add them and multiply by 3 to get the man's age.

Edwin