Question 184785: Hi,
The problem is a common one: The ages of a certain father and son are the same with the digits reversed. Nine years ago, the father was twice as old as the son. How old are they now?
I know what the answer is - that is trivial. I am after an ALGEBRAIC solution, if their is one.
Thank You,
Michael L. Giancola
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The ages of a certain father and son are the same with the digits reversed.
:
here is some algebra with a little "try & see"
;
Let 10x + y = Father's age
then
10y + x = son's age
:
"Nine years ago, the father was twice as old as the son."
(10x+y) - 9 = 2((10y+x)-9)
10x + y - 9 = 20y + 2x - 18
10x - 2x = 20y - y - 18 + 9
8x = 19y - 9
:
How old are they now?
:
We know that x and y are single digit integers
y = 3 is the only one that will satisfy
8x = 19(3) - 9
8x = 57 - 9
8x = 48
x = 6
;
Dad is 63, son is 36
:
Check in the statement:
" Nine years ago, the father was twice as old as the son."
63 - 9 = 2(36 - 9)
54 = 2(27)
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