SOLUTION: Find three consecutive odd integers such that three times the first minus the second is one more than the third

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Question 1003638: Find three consecutive odd integers such that three times the first minus the second is one more than the third
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Before finding them, 2n+1, 2n+3, 2n+5.

The description means 3%282n%2B1%29-%282n%2B3%29=1%2B%282n%2B5%29.

Solve for n and evaluate the three consecutive odd integers.

6n%2B3-2n-3=1%2B2n%2B5

6n-2n=2n%2B5%2B1

4n-2n=6

2n=6

highlight%28n=3%29

The integers are 7, 9, 11.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Find three consecutive odd integers such that three times the first minus the second is one more than the third
Let the 1st integer be F

Then 2nd and 3rd are: F + 2, and F + 4, respectively

We then get: 3F - (F + 2) = F + 4 + 1

Solve for F, the 1st integer (Make sure that this is AN ODD INTEGER)

Add 2 to the 1st integer to get the 2nd

Add 4 to the 1st to get the 3rd