SOLUTION: 3 consecutive odd integers such that 3 times the sum of all three is 26 less than the product of the first and second integer

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Question 885476: 3 consecutive odd integers such that 3 times the sum of all three is 26 less than the product of the first and second integer
Answer by lwsshak3(11628) About Me  (Show Source):
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3 consecutive odd integers such that 3 times the sum of all three is 26 less than the product of the first and second integer
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let x=1st consecutive odd integer
let x+2=2nd consecutive odd integer
let x+4=3rd consecutive odd integer
..
3(x+x+2+x+4)=x(x+2)-26
3(3x+6)=x^2+2x-26
9x+18=x^2+2x-26
x^2-7x-44=0
(x-11)(x+4)=0
x=11
x+2=13
x=4=15
1st consecutive odd integer=11
2nd consecutive odd integer=13
3rd consecutive odd integer=15