Question 8044: Five years ago, a man was seven times as old as his son. Five years hence, father will be three times as old as his son. Find their present ages.
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! Find their present ages...so lets define:
Let x = father's age now
Let y = son's age now.
Five years ago, father was (x-5) and son was (y-5) and father was 7 times older than son, so (x-5) = 7(y-5)
In five years time, father will be (x+5) and son will be (y+5) and father will be 3 times older than son, so (x+5) = 3(y+5)
(x-5) = 7(y-5)
--> x-5 = 7y-35
--> x-7y = -30
(x+5) = 3(y+5)
--> x+5 = 3y+15
--> x-3y = 10
Subtract second equation from first to remove the x-term. We are left with -4y = -40. So y = -40/-4.
--> y = 10
therefore (x-5) = 7(y-5) gives
x-5 = 7(10-5)
x-5 = 7(5)
x-5 = 35
x = 40
Father is now 40
Son is now 10
Check: 5 years ago, 35 and 5...7 times difference
Check: 5 years time, 45 and 15...3 times difference
jon.
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