SOLUTION: Two years ago, a man was eight times as old as his son. In three years time, he will be four and a half times his son's age then. Find their present ages.

Question 498187: Two years ago, a man was eight times as old as his son. In three years time, he will be four and a half times his son's age then. Find their present ages.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=son's present age
And let y father's present age
x-2=son's age 2 yrs ago; x+3=son's age 3 yrs hence
y-2=fathers age 2 yrs ago; y+3=father's age 3 yrs hence
Now we are told the following:
y-2=8(x-2) simplify
y-2=8x-16
y-8x=-14-------------------eq1
and
y+3=4.5(x+3)simplify
y+3=4.5x+13.5
y-4.5x=10.5----------------eq2
subtract eq1 from eq2
3.5x=24.5
x=7 yrs old-----------------present age of son
from eq1
y-56=-14
y=42---------------present age of father
CK
two years ago:
42-2=8*5
40=40
3 years from present:
42+3=(4.5)*10
45=45
Hope this helps-----ptaylor

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