SOLUTION: show an algebraic solution for 3 consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third.

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Question 1051982: show an algebraic solution for 3 consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third.
Found 3 solutions by advanced_Learner, MathTherapy, ikleyn:
Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
x-2,x and x+2
%28x-2%29%2B2x=%28x%2B2%29%2B12%29
%283x-2%29=%28x%2B14%29%29
%282x-2%29=%2814%29%29
%282x%29=%2814%2B2%29%29
%28x%29=%2816%2F2%29%29
%28x%29=%288%29%29
other numbers are %286%29 and %2810%29%29
the numbers are 6,8 and 10.
found the arithmetic mistake and corrected accordingly.
check
6+2(8)=10+12
checks

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
show an algebraic solution for 3 consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third. 
Integers: highlight_green%28matrix%281%2C6%2C+6%2C+%22%2C%22%2C+8%2C+%22%2C%22%2C+and%2C+10%29%29

By the way, why would anyone name the integers: x - 2, x, and x + 2. The question DOESN'T ask for the MIDDLE integer, so why do this?

It makes absolutely no sense to me. Just name them: x, x + 2, and x + 4. 

Why do people who try to help always seem to further confuse the people who ask for help? It's beyond me!

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
x-2,x and x+2
%28x-2%29%2B2x=%28x%2B2%29%2B12%29
%283x-2%29=%28x%2B14%29%29
%282x-2%29=%2814%29%29
%282x%29=%2814%2B2%29%29
%28x%29=%2816%2F2%29%29
%28x%29=%288%29%29
other numbers are %286%29 and %2810%29%29
the numbers are 6,8,10
check
6+2*8 = 6+16 = 22 = 2*10 + 2.
checks

The solution by the tutor "advanced_Learner" was wrong.