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Five times the second of three consecutive even integers is six more than twice the sum of the first and third integer.

Find the middle even integer.

I assume:

\(\text{Let the first even integer $=2(n-1)$ } \\ \text{Let the middle even integer $=2n$ } \\ \text{Let the third even integer $=2(n+1)$ }\)

\(\begin{array}{|rcll|} \hline 5(2n) &=& 6+2\Big( 2(n-1)+2(n+1) \Big) \\ 10n &=& 6+2( 2n-2+2n+2 ) \\ 10n &=& 6+2*4n \\ 10n &=& 6+8n \\ 10n-8n &=& 6 \\ \mathbf{2n} &=& \mathbf{6} \\ \hline \end{array}\)

The middle even integer is 6