SOLUTION: This is a problem from my math text.  
Find all sets of three consecutive multiples of 3 whose sum is between -134 and -109. 
This is how I set it up:
-134 < 3n + 3(n+1) + 3(n
Algebra.Com
Question 250147:  This is a problem from my math text.  
Find all sets of three consecutive multiples of 3 whose sum is between -134 and -109. 
This is how I set it up:
-134 < 3n + 3(n+1) + 3(n+2) < -109
But it can't be right because the solution is very messy. Can you set this up for me?
Thanks! 
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
 I am not sure I know how to do this mathematically but your three consecutive multiples of three have to be n, n+3 and n+6
So that we have -134>3n+9<-109
33+36+39=108 too low
36+39+42=117 lowest triplet
39+42+45=126 highest triplet
42+45+48=135 too high
 
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