SOLUTION: Find three consecutive even integers such that three times the second integer is six more than the sum of the first and third integers.

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Question 1103753: Find three consecutive even integers such that three times the second integer is six more than the sum of the first and third integers.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the consecutive even integers be:
+n+, +n+%2B+2+, and +n%2B+4+
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+3%2A%28+n+%2B+2+%29+=+n+%2B+n+%2B+4+%2B+6+
+3n+%2B+6+=+2n+%2B+10+
+n+=+4+
and
+n+%2B+2+=+6+
+n+%2B+4+=+8+
----------------------------------
The even integers are 4, 6, and 8
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check:
+3%2A%28+n+%2B+2+%29+=+2n+%2B+10+
+3%2A6+=+8+%2B+10+
+18+=+18+
OK