SOLUTION: Find 3 consecutive even integers so that the first plus twice the second is twice the third

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Question 9473: Find 3 consecutive even integers so that the first plus twice the second is twice the third
Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
If they are 3 consecutive even integers, they'll be n, n + 2, and n + 4. When they told you that they have to be even, this means that your answer at the end should be the even numbers. You don't have to do anything fancy with your choice of variable.

"The first plus twice the second": n + 2(n+2)

"Is twice the third": = 2(n+4)

+n+%2B+2%28n%2B2%29+=+2%28n%2B4%29+ <---- start here

+n+%2B+2n+%2B+4+=+2n+%2B+8+ <----- distribute

+3n+%2B+4+=+2n+%2B+8+ <---- combine like terms

+n+%2B+4+=+8+ <----- subtract 2n from both sides

+n+=+4+ <----- subtract 4 from both sides. If n was the first number and it was 4, then n + 2 would be 6, and then n + 4 would be 8. So those three numbers turned out to be 4, 6, and 8.

Tip: When they tell you some thing like "three consecutive even (or odd) integers", ALWAYS start with n, n + 2, and n + 4. If they told you that n is odd, so are n + 2 and n + 4. Don't let the +2 and the +4 throw you off thinking that the quantity is even. It could either be even or odd!