Question 8044
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.