SOLUTION: Find three consecutive odd integers such that the sum of the first and twice the second is equal to 15 more than twice the third.

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Question 339697: Find three consecutive odd integers such that the sum of the first and twice the second is equal to 15 more than twice the third.
Found 2 solutions by josmiceli, mananth:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the smallest integer n
The 3 integers are n, n%2B2, and n+%2B+4
given:
n+%2B+2%2A%28n+%2B+2%29+=+2%2A%28n+%2B+4%29+%2B+15
n+%2B+2n+%2B+4+=+2n+%2B+8+%2B+15
3n+%2B+4+=+2n+%2B+23
n+=+23+-+4
n+=+19
n+%2B+2+=+21
n+%2B+4+=+23
The consecutive odd integers are 19, 21, 23
check:
n+%2B+2%2A%28n+%2B+2%29+=+2%2A%28n+%2B+4%29+%2B+15
19+%2B+2%2A21+=+2%2A23+%2B+15
61+=+61
OK

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
x, x+2 , x+4 be the odd integers
x+2(x+1) = 2(x+4)+15
x+2x+2 = 2x+8 +15
3x+2 = 2x+23
3x-2x=23-2
x=21
..
21,23,25