Question 172055
Find three consecutive odd integers such that three times the second minus the third is 11 more than the first. 
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They're x, x+2 and x+4
Three times the 2nd is 3x+6
3x+6 - (x+4) = x + 11
3x #2...3rd....1st+11
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3x+6 - (x+4) = x + 11
2x + 2 = x + 11
x = 9
So they're 9, 11 and 13