SOLUTION: Find three consecutive odd intergers such that three times the second minus the third is 11 more than the first

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Question 51917: Find three consecutive odd intergers such that three times the second minus the third is 11 more than the first
Answer by checkley71(8403) About Me  (Show Source):
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x+(x+2)+(x+4) are the three consecutive intergers then
3[(x+2)-(x+4)]=x+11 or 3(x+2-x-4)=x+11 or-6=x+11 or x=-17
-17,-15,-13 Proof 3[-15-(-13)]=-17+11 or 3[-15+13]=-6 or 3*-2=- or -6=-6