SOLUTION: Find three consecutive even integers such that 5 times the sum of the first and the third is 16 greater than 9 times the second

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Question 966628: Find three consecutive even integers such that 5 times the sum of the first and the third is 16 greater than 9 times the second
Answer by ramkikk66(644) About Me  (Show Source):
You can put this solution on YOUR website!
Find three consecutive even integers such that 5 times the sum of the first and the third is 16 greater than 9 times the second 
Let the 3 integers be (n-2), n and (n+2)

Sum of 1st and 3rd = n+-+2+%2B+n+%2B+2+=+2%2An

9 times second = 9*n

So the equation is 5+%2A+2%2An+=+9%2An+%2B+16

or 

10%2An+=+9%2An+%2B+16 i.e. n+=+16

Check: If n = 16, the 3 numbers are 14, 16 and 18.
5 times sum of 14 and 18 = 32*5 = 160
9 times 16 = 144
160 is 16 more than 144. Correct!

Hope it's clear!