SOLUTION: 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

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
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
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
.
Let x = first of three consecutive odd integer
then
x+2 = second odd integer
x+4 = third odd integer
.
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
.
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)
.
second:
x+2 = -9+2 = -7
.
third:
x+4 = -9+4 = -5
.
The three odd integers:
-9, -7, and -5

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