SOLUTION: Eight times the first of three consecutive odd integers is ten more than twice the second.

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Question 92593: Eight times the first of three consecutive odd integers is ten more than twice the second.
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Eight times the first of three consecutive odd integers is ten more than twice the second.
ANY INTEGER IS N
EVEN NUMBER IS 2N
ODD INTEGER IS 2N+1
NEXT ODD INTEGER IS 2N+1+2=2N+3
THE NEXT ODD INTEGER IS 2N+3+2=2N+5
8 TIMES THE I = 8(2N+1)=16N+8
TWICE THE SECOND = 2(2N+3)=4N+6
TEN MORE =4N+6+10=4N+16.............HENCE
16N+8=4N+16
16N-4N=16-8
12N=8
N=8/12=......THERE IS NO SOLUTION.......PLEASE CHECK THE DATA IN THE PROBLEM
FOR ANY MISTAKES IN TYPING.