SOLUTION: A man 38 years old has a son who is 14 years old. How many years ago was the father seven times as old as his son?

Question 168965: A man 38 years old has a son who is 14 years old. How many years ago was the father seven times as old as his son?
Found 2 solutions by nerdybill, ankor@dixie-net.com:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
A man 38 years old has a son who is 14 years old. How many years ago was the father seven times as old as his son?
.
Let x = number of years ago when father was 7 times older than his son
.
38-x = 7(14-x)
38-x = 98-7x
38+6x = 98
6x = 60
x = 10 years

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A man 38 years old has a son who is 14 years old. How many years ago was the
father seven times as old as his son?
:
Let x = no. of year for this to be true
:
38 - x = 7(14 - x)
:
38 - x = 98 - 7x
:
-x + 7x = 98 - 38
:
6x = 60
:
x = 10 yrs ago
;
:
Check solution
38 - 10 = 7(14 - 10)

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