SOLUTION: The sum of three consecutive odd integers is .11 more than twice the second one. Find all three consecutive integers. Show your work.

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Question 1047511: The sum of three consecutive odd integers is .11 more than twice the second one. Find all three consecutive integers. Show your work.
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
The sum of three consecutive odd integers is .11 more than twice the second one. Find all three consecutive integers. Show your work.
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1st:: 2x-1
2nd:: 2x+1
3rd:: 2x+3
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Sum = 6x+3
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Equation:
6x+3 = 2(2x+1) + 11
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6x+3 = 4x +13
2x = 10
x = 5
------
1st: 2x-1 = 9
2nd: 11
3rd: 13
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Cheers,
Stan H.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

The sum of three consecutive odd integers is .11 more than twice the second one. Find all three consecutive integers. Show your work. 
Let first integer be F

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

We then get: F + F + 2 + F + 4 = 2(F + 2) + 11

3F + 6 = 2F + 4 + 11

3F - 2F = 15 - 6

F, or 1st integer = 

2nd: 9 + 2

3rd: 9 + 4

It's very EASY.....far from being complex! 

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