SOLUTION: One positive integer is 3 less than twice another. The sum of their squares is 117.
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Question 1052208: One positive integer is 3 less than twice another. The sum of their squares is 117.
Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
larger integer = x
smaller integer = y
x = 2y - 3
x^2 + y^2 = 117
Use substitution.
(2y - 3)^2 + y^2 = 117
4y^2 - 12y + 9 + y^2 = 117
5y^2 - 12y - 108 = 0
(y - 6)(5y + 18) = 0
y is an integer, so y = 6.
The integers are x = 9 and y = 6.
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