SOLUTION: Find three consecutive odd integers such that four times the sum of the last two is two greater than ten times the first.

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Question 281171: Find three consecutive odd integers such that four times the sum of the last two is two greater than ten times the first.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The consecutive odd integers are:
n
n%2B2
n+%2B+4
given:
4%2A%28n%2B2+%2B+n+%2B+4%29+=+10n+%2B+2
4%2A%282n+%2B+6%29+=+10n+%2B+2
8n+%2B+24+=+10n+%2B+2
2n+=+22
n+=+11
n+%2B+2+=+13
n+%2B+4+=+15
The numbers are 11,13, and 15
check:
4%2A%28n%2B2+%2B+n+%2B+4%29+=+10n+%2B+2
4%2A%2811%2B2+%2B+11+%2B+4%29+=+10%2A11+%2B+2
4%2A28+=+110+%2B+2
112+=+112
OK