Question 213442: I havea problem and couldn't find a similar problem:
2 years ago a father was 4 times as old as his son. 3 years from now the father will be only 3 times as old as the son. how old is each now?
can you find a problem at least similar to this one?
thank you,
Destiny
Found 2 solutions by stanbon, rapaljer:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 2 years ago a father was 4 times as old as his son. 3 years from now the father will be only 3 times as old as the son. how old is each now?
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Let father's age NOW be "f":
Let son's age NOW be "s"
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Equation:
(f-2)=4(s-2)
Equation:
(f+3)= 3(s+3)
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Simplify both equations:
f = 4s -8+2
f = 4s-6
and
f = 3s+9-3
f = 3s+6
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Equate the two expressions for "f":
4s-6 = 3s+6
s = 12 yrs old (son's age NOW)
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Substitute into f = 3s+6 to solve for "f"
f = 3*12+6 = 42 yrs old (father's age NOW)
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Cheers,
Stan H.
Answer by rapaljer(4671)  (Show Source):
You can put this solution on YOUR website! Let x = age of son now
y= age of father now
x-2= age of son 2 years ago
y-2 = age of father 2 years ago
x+3 = age of father 3 years from now
y+3= age of father 3 years from now
y-2=4(x-2)
y+3=3(x+3)
y-2=4x-8, so y = 4x-6
y+3=3x+9, so y=3x+6
Now, since
y=4x-6 and y=3x+6,
4x-6=3x+6
x=12 Age of son now
y=3x+6
y=3(12)+6
y=36+6=42 Age of father now
Check:
2 years ago, son was 10 and father was 40--(four times as old!)
3 years from now, the son will be 15 and father will be 45--(three times as old!!)
I would love to recommend my website to you on this, but I just don't have anything like this on it!!!
R^2
Dr. Robert J. Rapalje, Retired
Seminole Community College
Altamonte Springs Campus
Florida
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