SOLUTION: One interger is 5 less than another. The sum of their squares is 97. Find the intergers.

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Question 801327: One interger is 5 less than another. The sum of their squares is 97. Find the intergers.
Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
x = y - 5
x^2 + y^2 = 97

Substitute x = y - 5 into the second equation.

(y - 5)^2 + y^2 = 97
y^2 - 10y + 25 + y^2 = 97
2y^2 - 10y - 72 = 0
y^2 - 5y - 36 = 0
(y - 9)(y + 4) = 0
y = 9, y = -4

The integers are (x, y) = (4, 9) and (-9, -4)