+27 The sum of two consecutive even integers is at most seven more than half the sum of the next two consecutive even integers.
Yeah. I have no clue either.
+118677 The sum of two consecutive even integers is at most seven more than half the sum of the next two consecutive even integers.
\((x-2)+x \le \frac{1}{2}(x+2+x+4)+7\\ 2x-2 \le \frac{1}{2}(2x+6)+7\\ 2x-2 \le (x+3)+7\\ 2x-2 \le x+10\\ x \le 12\\\)
So the first 2 numbers have to be less than or equal to 10 and 12
+118677 ok you should have said so int the first place then maybe I would already have tried to explain better.
The sum of two consecutive even integers is at most seven more than half the sum of the next two consecutive even integers.
Let the 4 consecutative EVEN numbers be \(x-2,\;\;x, \;\;x+2,\; \;and\;\; x+4 \)
\(x\) is the second smallest. (I didn't have to make it the second smallest, I could have made it any of them)
So the sum of the 2 smallest numbers are
\((x-2) \quad + \quad x\quad\\= 2x-2\)
Half the sum of the 2 bigger ones is
\(\frac{1}{2}(x+2\;\;+\;\; x+4)\\ =\frac{1}{2}(2x+6)\\ =\frac{1}{2}*2(x+3)\\ =\frac{1}{\not{2}}*\not{2}(x+3)\\ =x+3\)
Now the first answer has to be at most 7 more than the second answer
So
The first answer has to be less than or equal to the second answer +7 more
so
\(2x-2\le x+3+7\\ 2x-2\le x+10\\ x-2\le 10\\ x\le 12\\ \)
So 12 is the giggest possibility for the second number
so the two original even numbers must be less than or equal to 10 and 12.
Maybe that will make more sense 
