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

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Let the 3 integers be (n-2), n and (n+2)

Sum of 1st and 3rd = {{{n - 2 + n + 2 = 2*n}}}

9 times second = 9*n

So the equation is {{{5 * 2*n = 9*n + 16}}}

or 

{{{10*n = 9*n + 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!

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