SOLUTION: Jim is a year older than Colleen and the difference of the squares of their ages is 97. How old is each?

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Question 1134351: Jim is a year older than Colleen and the difference of the squares of their ages is 97. How old is each?
Found 3 solutions by Boreal, MathLover1, greenestamps:
Answer by Boreal(15235) About Me  (Show Source):
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Colleen is x
Jim is x+1
(x+1)^2-x^2=97
x^2+2x+1-x^2=97
2x+1=97
2x=96
x=48 Colleen's age, and square is 2304
x+1=49 Jim's age, and square is 2401
difference is 97

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

let Jim's age be J and Colleens's age C
if Jim is a year older than Colleen, we have J=C%2B1
and the difference of the squares of their ages is 97, we have
J%5E2-C%5E2=97...solve for C
C%5E2=J%5E2-97...........substitute J=C%2B1
C%5E2=%28C%2B1%29%5E2-97.......solve for C
C%5E2=C%5E2%2B2C%2B1-97
C%5E2=C%5E2%2B2C-96
C%5E2-C%5E2-2C%2B96=0
-2C%2B96=0
2C=96...simplify
C=48->Colleens's age
J=48%2B1->J=49->Jim's age


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The difference between the squares of consecutive integers is equal to the sum of the two integers:

5^2-4^2 = 25-16 = 9 = 5+4
13^2-12^2 = 169-144 = 25 = 13+12
(n+1)^2-n^2 = (n^2+2n+1)-n^2 = 2n+1 = (n+1)+n

Since you know their ages are consecutive integers, the difference of the squares of their ages is the sum of their two ages. Since the sum of their ages is 97 and their ages are consecutive integers, their ages are 49 and 48.