Question 453181: Five years from now, a father's age will be three times his son's age, and five years ago, he was seven times as old as his son was. What are their present ages?
Found 2 solutions by stanbon, pedjajov:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Five years from now, a father's age will be three times his son's age, and five years ago, he was seven times as old as his son was. What are their present ages?
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Equation(s):
(f+5) = 3(s+5)
(f-7) = 7(s-7)
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Rearrange:
f - 3s = 10
f - 7s = -42
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Subtract and solve for "s":
4s = 52
s = 13 (son's age now)
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Solve for "f":
f -3s = 10
f - 39 = 10
f = 49 (father's age now)
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Cheers,
Stan H.
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Answer by pedjajov(51) (Show Source):
You can put this solution on YOUR website! f - father's age
s - son's age
:
In five years they will be f+5 and s+5
Five years ago they were f-5 and s-5
:
Five years from now it will be
(f+5)=3(s+5)
:
Five years ago it was
(f-5)=7(s-5)
:
These two form system of two equations with two variables, f and s:
(f+5)=3(s+5)
(f-5)=7(s-5)
:
f+5=3s+15
f-5=7s-35, solve for f -> f=7s-30
:
Substitute f in first equation:
7s-30+5=3s+15
7s-25=3s+15, subtract 3s and add 25 to both sides
4s=40
s=10, this is current age for the son
:
f=7s-30
f=7*10-30
f=40, this is current age of the father
:
check
in 5 years, f=40+5=45, s=10+5=15 -> f=3s, ok
5 years ago,f=40-5=35, s=10-5=5 -> f=7s, ok
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