SOLUTION: find 3 consecutive odd integers such that the sum of the largest and twice the smallest is 12 more than the middle number

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Question 1094709: find 3 consecutive odd integers such that the sum of the largest and twice the smallest is 12 more than the middle number
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
x = smallest odd integer of the three 
x+2 = middle odd integer of the three
x+4 = largest odd integer of the three
the sum of the largest and twice the smallest is 12 more than the middle number


 (x+4)  + 2*(x) = (x+2) + 12
  x+4   + 2x    =  x+2  + 12
        3x + 4  =  x + 14
            2x  =  10
             x  =  5

 x = smallest odd integer of the three = 5 
x+2 = middle odd integer of the three = 5+2 = 7
x+4 = largest odd integer of the three = 5+4 = 9

We check the numbers in the words:
3 consecutive odd integers
5, 7, and 9 are all odd and consecutive. That checks.
such that the sum of the largest and twice the smallest
The largest is 9.  Twice the smallest is two time 5, or 10. 
Their sum, when we add those, 9+10 is 19
is 12 more than the middle number.
Indeed! The middle number is 7, and 19 is 12 more than 7.
So we know we are right.

Edwin