Question 190743
    find three consecutive odd integers so that the sum of three times the sum of the first and the third is 28 less than two times the second
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Let x = first of three consecutive odd integer
then
x+2 = second odd integer
x+4 = third odd integer
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From: "the sum of three times the sum of the first and the third is 28 less than two times the second" we get:
3(x + x+4) = 2(x+2) - 28
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Solving for 'x':
3(x + x+4) = 2(x+2) - 28
3(2x+4) = (2x+4) - 28
6x+12 = 2x-24
4x+12 = -24
4x = -36
x = -9 (first odd integer)
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second:
x+2 = -9+2 = -7
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third:
x+4 = -9+4 = -5
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The three odd integers:
-9, -7, and -5