SOLUTION: I do not know how to solve this. Find three consecutive even intergers such that four times the middle interger is equal to two less than the sum of the first and third intgers.

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Question 135065: I do not know how to solve this.
Find three consecutive even intergers such that four times the middle interger is equal to two less than the sum of the first and third intgers.
Equation:?
Set up
1st no. = n
2nd no. = n+2
3rd no n+4
I have no clue how to solve this.

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
You have more than a clue. Don't fret, You are part way there already.
Your solution uses the number n. But there is nothing that ensures that 'n' is even. So let's use a different set of numbers.
Let's assume n is even, if so, then n/2 is still an integer.
Our three numbers will be n/2, (n+4)/2 and (n+8)/2
Add in one more given " four times the middle integer is equal to two less than the sum of the first and third integers."
so 4%2A%28%28n%2B4%29%2F2%29+=+%28%28n%2F2+%2B+%28n%2B8%29%2F2%29%29+-+2+%29
Left left side is four times the middle number. The right side is the (sum of the first and third) minus 2
Now solve
4%2A%28%28n%2B4%29%2F2%29+=+%28%28n%2F2+%2B+%28n%2B8%29%2F2%29%29+-+2+%29
4%2A%28n%2B4%29+=+%28%28n+%2B+%28n%2B8%29%29+-+4+%29
4n%2B16%29+=+%282n+%2B+8+-+4+%29
4n%2B16%29+=+2n+%2B+4+
2n+=+-12+
n = -6
Check : is 4(-4) = (-6 + -2) - 2?
Nope, so I think there is not a solution to this one