SOLUTION: three consecutive odd integers such that seven times the sum of the first two integers is three more than nine times the third integer.

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Question 224169: three consecutive odd integers such that seven times the sum of the first two integers is three more than nine times the third integer.
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
Let x, x+2 & x+4 be the 3 odd integers.
7(x+x+2)=9(x+4)+3
7(2x+2)=9x+36+3
14x+14=9x+39
14x-9x=39-14
5x=25
x=25/5
x=5 the smallest integer.
5+2=7 the middle integer.
5+4=9 the largest integer.
Proof:
7(5+7)=9*9+3
7*12=81+3
84=84