SOLUTION: Three consecutive intergers are such that the first plus one-half the second plus seven less than twice the third is 2101. What are the integers? Help!!!

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Question 31102: Three consecutive intergers are such that the first plus one-half the second plus seven less than twice the third is 2101. What are the integers? Help!!!
Found 2 solutions by venugopalramana, ikdeep:
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Three consecutive intergers are such that the first plus one-half the second plus seven less than twice the third is 2101. What are the integers? Help!!!
LET THE 3 CONSEUTIVE INTEGERS BE X,X+1,X+2
FIRST NUMBER =X................................1
HALF OF SECOND NUMBER =(X+1)/2...............2
TWICE THE THIRD =2(X+2)=2X+4
SEVEN LESS THAN THIS =2X+4-7=2X-3...........3
SUM OF THESE 3 IS
X+(X+1)/2 +2X-3=2101.....MULTIPLY BY 2 THROUGH OUT
2X+X+1+4X-6=4202
7X=4202-1+6=4207
X=4207/7=601
SO THE INTEGERS ARE X=601,X+1=602,X+2=603.

Answer by ikdeep(226) About Me  (Show Source):
You can put this solution on YOUR website!
let first consecutive integer = x
let second consecutive integer = (x+1)
let third consecutive integer = (x+2)
According to the question we get the equation....
(x) + (x+1)/2 + 2(x+2) - 7 = 2101...

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