SOLUTION: How do I find three consecutive odd integers such that three times the second minus the third is 11 more than the first?

Question 564240: How do I find three consecutive odd integers such that three times the second minus the third is 11 more than the first?

Found 2 solutions by nerdybill, stanbon:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
How do I find three consecutive odd integers such that three times the second minus the third is 11 more than the first?
Let x = first (smallest) of three consecutive odd integers
then
x+2 = second
x+4 = third
.
3(x+2)-(x+4) = x+11
(3x+6)-(x+4) = x+11
3x+6-x-4 = x+11
2x+6-4 = x+11
2x+2 = x+11
x+2 = 11
x = 9 (first)
.
second:
x+2 = 9+2 = 11
.
third:
x+4 = 9+4 = 13
.
answer: 9,11,13
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How do I find three consecutive odd integers such that three times the second minus the third is 11 more than the first?
-------------------------------------------------
1st: 2x-1
2nd: 2x+1
3rd: 2x+3
-----
Equation:
2(2x+1)-(2x+3) = 2x-1+11
------------------------------
4x+2-2x-3 = 2x+10
2x - 1 = 2x + 10
-1 = 10
=================
The conditions are contradictory.
No solution.
=====================
Cheers,
Stan H.
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