SOLUTION: find three consecutive integers such that 5 times the sum of the first and third is 14 greater than 8 times the second.

Algebra ->  Probability-and-statistics -> SOLUTION: find three consecutive integers such that 5 times the sum of the first and third is 14 greater than 8 times the second.      Log On


   

Question 122332: find three consecutive integers such that 5 times the sum of the first and third is 14 greater than 8 times the second.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!

Let the first of three consecutive integers be n.

Then the second is n%2B1 and the third is n%2B2.


The sum of the first and third is n%2B%28n%2B2%29 or 2n%2B2.
5 times that is 5%282n%2B2%29=10n%2B10

This quantity is (meaning equals) 14 greater than 8 times the second.
8 times the second is 8%28n%2B1%29=8n%2B8, and 14 more than that is
8n%2B8%2B14=8n%2B22, so we can write:


10n%2B10=8n%2B22


Add -8n to both sides
10n-8n%2B10=8n-8n%2B22
2n%2B10=22


Add -10 to both sides
2n%2B10-10=22-10
2n=12


Multiply by 1%2F2
%281%2F2%292n=%281%2F2%2912
n=6


Therefore the first number is 6, the second is 7 and the third is 8.



Check the answer


6 plus 8 = 14.   5 times 14 = 70
8 times 7 = 56.  56 plus 14 = 70

Answer checks