SOLUTION: Find three consecutive odd integers such that twice the first is 7 more than the third

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Question 712167: Find three consecutive odd integers such that twice the first is 7 more than the third
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Start this way:

The numbers are: 2n+1, 2n+3, 2n+5

2(2n+1)=7+2n+5
First solve for n. Then compute each of the three numbers.