Question 949818: A father is three times as old as his son was n years ago. At that same time (n years ago), the father was twice as old as his son will be two years from now. Find the present age of each is they sum to 55.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! f = father's present age
s = son's present age
:
Write an equation for each statement
:
A father is three times as old as his son was n years ago.
f = 3(s - n)
f = 3s - 3n
f + 3n = 3s
At that same time (n years ago), the father was twice as old as his son will be two years from now.
f - n = 2(s + 2)
f - n = 2s + 4
multiply this by 3, add to the 1st equation
3f - 3n = 6s + 12
f + 3n = 3s
----------------adding in eliminates n
4f = 9s + 12
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Find the present age of each is they sum to 55.
f + s = 55
f = (55-s)
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Replace f with (55-s) in the 4f = 9s + 12 equation
4(55-s) = 9s + 12
220 - 4s = 9s + 12
220 -12 = 9s + 4s
208 = 13s
s = 208/13
s = 16 is the son's age
then
55 - 16 = 39 is father's age
:
;
Check this out, find n
f = 3(s - n)
39 = 3(16-n)
39 = 48 - 3n
3n = 48 - 39
n = 9/3
n = 3
"At that same time (n years ago), the father was twice as old as his son will be two years from now."
39 - 3 = 2(18)
36 = 36 confirms our solutions
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