SOLUTION: The ages of a man and his son are 40 and 12 respectively.Find: > how many years ago was the man's age eight times the age of his son?

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Question 147217: The ages of a man and his son are 40 and 12 respectively.Find:
> how many years ago was the man's age eight times the age of his son?
> how many years hence will the age of man be thrice the age of his son?
Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website!
(1) Let x=number of years ago that the man's age was eight times his son's age
Our equation to solve is:
40-x=8(12-x) get rid of parens
40-x=96-8x subtract 40 from and add 8x to both sides
40-40-x+8x=96-40-8x+8x collect like terms
7x=56 divide each side by 7
x=8
CK
40-8=8(12-8)
32=8*4
32=32
(2) Let y=the number of years hence that the man will be three times the age of his son
Our equation to solve:
40+y=3(12+y) get rid of parens
40+y=36+3y subtract 36 and also y from each side
40-36+y-y=36-36+3y-y collect like terms
4=2y divide each side by 2
y=2
CK
40+2=3(12+2)
42=3*14
42=42

Hope this helps---ptaylor

SOLUTION: The ages of a man and his son are 40 and 12 respectively.Find: > how many years ago was the man's age eight times the age of his son? > how many years hence will the age of ma