SOLUTION: What is the answer to The sum of three consecutive even numbers is the same as twice the difference of the smallest number and 14

The tutor above interpreted the problem wrong.  He interpreted it as if it had read:

But it's this:

Instead of doing yours for you, I'll do this one instead which is
done exactly the same:

x = the smallest
x+2 = next to smallest
x+4 = next to largest
x+6 = largest

 
So we add them all up by putting in plusses 

x + (x+2) + (x+4) + (x+6)

 
That means to put an equal sign after it.
So we have:

x + (x+2) + (x+4) + (x+6) =

 
So we write 3∙ after the equal sign, and we have:

x + (x+2) + (x+4) + (x+6) = 3∙

 
That means to subtract them in that order.  Since the next to the smallest is
(x+2) is mentioned first and 20 second, we subtract them in that order and put
[(x+2)-20] after the 3∙

So the equation is

x + (x+2) + (x+4) + (x+6) = 3∙[(x+2)-20]

x + x + 2 + x + 4 + x + 6 = 3∙[x + 2 - 20]

4x + 12 = 3∙[x - 18]

4x + 12 = 3x - 54

      x = -66

  x = the smallest = -66
x+2 = next to smallest = -66+2 = -64
x+4 = next to largest = -66+4 = -62
x+6 = largest = -66+6 = -60

Now do yours, which has only 3 consecutive numbers.

[Your answers will be negative too]

Edwin

The answer the other person dkppathak(420) gave you makes ABSOLUTELY no SENSE!!

Let the first integer be F
Then other 2 are: F + 2, and F + 4
As these numbers are INTEGERS, the CORRECT equation should NOT be: F + F + 2 + F + 4 = 2(14 - F)

Instead, it should be: F + F + 2 + F + 4 = 2(F - 14)
3F + 6 = 2F - 28
3F - 2F = - 28 - 6
First integer, or 

Now, it shouldn't be too difficult to find the other 2 integers!