Expert: Steve Holleran - 9/15/2008

Question
I need an explanation of how to solve this question, so I know how to do others like it when they come up in my lessons for the next month or so:

Find three consecutive odd integers such that 7 times the sum of the first and the third is 8 less than 14 times the second.

I'm pretty good with normal consecutive integer problems, but the whole 'opposite' idea is really messing me up.

Answer
Hi LeeAnn,

I'm not sure what you mean by "opposite" here, but something's missing or incorrect, because if you call:

   1st integer = x-2    2nd integer = x  3rd integer = x + 2

(You could also call them x, x+2, x+4)

Then you have

 7[(x-2) + (x+2)] = 14(x) - 8

 7 (2x) = 14x - 8    so    14x = 14x - 8 and then 0 = -8 and there is no solution.

Let me know if something's not right.
Steve