SOLUTION: Find three consecutive odd integers such that three times the second minus the third is 11 more than the first. This is what I have, 2x+1, 2x+3, 2x+5, but from there I need help

Question 171069: Find three consecutive odd integers such that three times the second minus the third is 11 more than the first.
This is what I have, 2x+1, 2x+3, 2x+5, but from there I need help please.
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Well, that's a good start!
Three times the second can be expressed as: 3(2x+3)
Eleven more than the first can be expressed as: (2x+1)+11
So, putting it all together, you get:
3(2x+3)-(2x+5) = (2x+1)+11 Simplify this to solve for x.
6x+9-2x-5 = 2x+12 Combine like-terms.
4x+4 = 2x+12 Subtract 2x from both sides.
2x+4 = 12 Subtract 4 from both sides.
2x = 8 Divide both sides by 2.
x = 4, so...
2x+1 = 8+1 = 9
2x+3 = 8+3 = 11
2x+5 = 8+5 = 13
Check:
3(11)-13 = 9+11
33-13 = 20
20 = 20 OK!
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