Question 807080: Find three consecutive odd positive integers such that 2 times the sum of all three is 7 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 positive integers: (x-2), x, (x+2)
2((x-2) + x + (x+2)) = 7 + (x-2)(x)
6x = 7 + x^2 - 2x
0 = x^2 - 8x + 7
0 = (x - 1)(x - 7)
The numbers are all positive, so x = 7 and the numbers are 7, 9, and 11.
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