SOLUTION: find three consecutive odd positive integers such that 3 times the sum of all 3 is 26 less than product of the first and second integers

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Question 1061592: find three consecutive odd positive integers such that 3 times the sum of all 3 is 26 less than product of the first and second integers
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
find three consecutive odd positive integers such that 3 times the sum of all 3 is 26 less than product of the first and second integers
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1st:: 2x-3
2nd:: 2x-1
3rd:: 2x+1
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Equations:
3(6x-3) = (2x-3)(2x-1)-26
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18x - 9 = 4x^2-8x+3-26
4x^2 - 26x -14 = 0
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x = 7 or x = -2
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To get positive solutions let x = 7.
Then, 1st = 2x-3 = 11
2nd:: 13
3rd:: 15
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Cheers,
Stan H.
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Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

find three consecutive odd positive integers such that 3 times the sum of all 3 is 26 less than product of the first and second integers
Let smallest be S

Then others are: S + 2, and S + 4

We then get: 3(S + S + 2 + S + 4) = S(S + 2) - 26









(S - 11)(S + 4) = 0

S, or        OR         S = - 4 (ignore) 

       

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