SOLUTION: what are three consecutive odd positive integers such that 5 times the sum of all three is 72 more than the product of the first and second integers?

SOLUTION: what are three consecutive odd positive integers such that 5 times the sum of all three is 72 more than the product of the first and second integers?

Algebra.Com

Question 967903: what are three consecutive odd positive integers such that 5 times the sum of all three is 72 more than the product of the first and second integers?

Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
consecutive odd integers: (x-2), x, (x+2)

5((x-2) + x + (x+2)) = 72 + (x-2)(x)
5(3x) = 72 + x^2 - 2x
0 = x^2 - 17x + 72
0 = (x - 8)(x - 9)
x = 8 or 9

The integers are odd, so x = 9 and the integers are 7, 9, and 11.
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