Question 507801
Find three consecutive odd integers such that three times the middle integer is one more than the sum of the first and third. 
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All you have to realize is that "odd integers" are a distance of 2 apart.
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Let x = first of three odd integers
then
x+2 = middle odd integer
x+4 = third odd integer
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then from: "three times the middle integer is one more than the sum of the first and third" we get
3(x+2) = (x + x+4) + 1
3x+6 = (2x+4) + 1
3x+6 = 2x+5
x+6 = 5
x = -1 (first odd integer)
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middle odd integer:
x+2 = -1+2 = 1
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third odd integer:
x+4 = -1+4 = 3
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Answer: -1, 1, 3