SOLUTION: five times the smallest of three consectutive integers is 17 less than twice the sum of the integers

Question 1015672: five times the smallest of three consectutive integers is 17 less than twice the sum of the integers
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
You did not ask a question. I assume you want to know the integers.
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S=smallest integer; M=middle integer=S+1; L=largest integer=S+2
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5S=2(S+M+L)-17
5S=2(S+S+1+S+2)-17
5S=2(3S+3)-17
5S=6S+6-17
5S=6S-11
11=S
ANSWER 1: The smallest integer is 11.
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M=S+1=11+1=12
ANSWER 2: The middle integer is 12.
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L=S+2=11+2=13
ANSWER 3: The largest integer is 13.
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CHECK:
5S=2(S+M+L)-17
5(11)=2(11+12+13)-17
55=2(36)-17
55=72-17
55=55
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