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Question 149859: Three years ago, a man was eight times as old as his son. In two years, he will be four times as old as the son. Find their present ages.
Thank you.: Three years ago, a man was eight times as old as his son. In two years, he will be four times as old as the son. Find their present ages.
Thank you.
Answer by mangopeeler07(428) About Me  (Show Source):
You can put this solution on YOUR website!
man=m
son=s

m-3=8(s-3)
m+2=4(s+2)

Solve for m in the first equation

m=8(s-3)+3
Distribute
m=8s-24+3
m=8s-21

Plug that into the second equation
8s-21+2=4(s+2)

Distribute the 4
8s-21+2=4s+8

Combine like terms
8s-19=4s+8

Add 19 to both sides
8s=4s+27

Subtract 4s from both sides
4s=27

Divide by 4
s=27/4 0r 6 and 3/4

m-3=8(s-3)
m-3=8(27/4-3)
m-3=54-24
m-3=30
m=33

son=6 years and 9 months
man=33 years