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.

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)   (Show Source): You can put this solution on YOUR website!
x-2,x and x+2
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=
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other numbers are and
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)   (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:
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(52792)   (Show Source): You can put this solution on YOUR website!
.
x-2,x and x+2
=
=
=
=
=
=
other numbers are and
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.

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