SOLUTION: Find 3 consecutive odd integers such that 3 times the 2nd minus the 3rd is 31 more than the first.

Question 210394: Find 3 consecutive odd integers such that 3 times the 2nd minus the 3rd is 31 more than the first.
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find 3 consecutive odd integers such that 3 times the 2nd minus the 3rd is 31 more than the first.
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1st: 2x-1
2nd: 2x+1
3rd: 2x+3
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Equation:
3*(2x+1)-(2x+3) = (2x-1)+31
------------------------------
6x-3 - 2x-3 = 2x+30
---
4x-6 = 2x+30
2x = 36
-----------------
1st: 2x-1 = 35
2nd: 2x+1 = 37
3rd: 2x+3 = 39
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Cheers,
Stan H.
Answer by MathTherapy(10555) (Show Source): You can put this solution on YOUR website!
Find 3 consecutive odd integers such that 3 times the 2nd minus the 3rd is 31 more than the first.

Let the 1st integer be F, then the 2nd is F + 2, and the 3rd is F + 4

Since 3 times the 2nd minus the 3rd is 31 more than the first, we'll have:

3(F + 2) - (F + 4) = F + 31

3F + 6 - F - 4 = F + 31

3F - F - F = 31 - 6 + 4

F = 29

Therefore, the 3 consecutive odd integers are:

---------------
Check
---------------
3(31) - 33 = 29 + 31

93 - 33 = 60

60 = 60 (TRUE)

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