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# Consecutive even integers.

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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.

Oct 17, 2019

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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

Oct 17, 2019
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Thank You!

veraguitars1234  Oct 17, 2019
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Do you understand?

Melody  Oct 17, 2019
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nope.

veraguitars1234  Oct 17, 2019
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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

Melody  Oct 17, 2019