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Question 553478: find the three consecutive integers such that the sum of the first and twice the second is one hundred ten minus three times the third?
Answer by TutorDelphia(193)   (Show Source):
You can put this solution on YOUR website! First off, consecutive integers can be written as n, n+1 and n+2 (try putting a number in for n if it doesn't make sense yet).
The sum of the first and twice the second would be n+2(n+1)
so our equation becomes
n+2(n+1)=110-3(n+2)
n+2n+2=110-3n-6
3n+2=104-3n
6n+2=104
6n=102
n=17
So our three terms are 17,18 and 19
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