SOLUTION: This is an extra credit problem I'm having trouble with. A boy on being asked the age of himself and his sister replied: three years ago I was seven times as old as my sister; tw

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Question 69978: This is an extra credit problem I'm having trouble with.
A boy on being asked the age of himself and his sister replied: three years ago I was seven times as old as my sister; two years ago I was four times as old; last year I was three times as old; and this year I am two and a half times as old. What are their ages now?
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!

Let x=his sister's age now
x-3=sister's age 3 years ago
x-2=sister's age 2 years ago
x-1=sister's age 1 year ago
7(x-3)=boy's age 3 years ago
4(x-2)=boy's age 2 years ago
3(x-1)=boy's age 1 year ago
(2.5)x=boy's age now
Now we know that if we add 1 year to the boy's age three years ago, then we'll have his age two years ago and so forth. There's several equations that will yield the correct answer. Lets do the 3 year one. You can do the others for drill:
7(x-3)+1=4(x-2) get rid of parens
7x-21+1=4x-8 or
7x-20=4x-8 add 20 to and subtract 4x from both sides
7x-4x=20-8
3x=12
x=4 years old ----------------------sister's age now
2.5x=2.5*4=10 years old---------------- boy's age now
CK
Boy's age 1 year ago: 3(x-1)=3*4-3=9; 10-1=9
Boy's age 2 years ago: 4(x-2)=4*4-8=8; 10-2=8
Boy's age 3 years ago: 7(x-3)=7*4-21=7; 10-3=7

Hope this helps----ptaylor