SOLUTION: find three consecutive odd integers such that 3 times the sum of all three is 28 more than the product of the first and second integers.

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Question 1029126: find three consecutive odd integers such that 3 times the sum of all three is 28 more than the product of the first and second integers.
Found 2 solutions by Boreal, MathLover1:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
They are x, x+2 and x+4.
add them to get 3x+6
3(3x+6) -28 = x (x+2)=x^2+2x
9x+18-28=9x-10=x^2+2x
x^2-7x+10=0
(x-5)(x-2)=0
x=5 and 2, but only 5 is odd.
The numbers are 5,7,9
They add to 21
3*21-28=35
That is the product of the first two integers.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Let x,%28x%2B2%29,%28x%2B4%29 represent the three consecutive odd integers.
if 3 times the sum of all three is 28 more than the product of the first and second integers, we have:
3%28x%2B%28x%2B2%29%2B%28x%2B4%29%29=x%28x%2B2%29%2B28.....solve for x
3%28x%2Bx%2B2%2Bx%2B4%29=x%5E2%2B2x%2B28
3%283x%2B6%29=x%5E2%2B2x%2B28
9x%2B18=x%5E2%2B2x%2B28
x%5E2%2B2x-9x-18%2B28=0
x%5E2-7x%2B10=0
x%5E2-5x-2x%2B10=0
%28x%5E2-5x%29-%282x-10%29=0
x%28x-5%29-2%28x-5%29=0
%28x-5%29+%28x-2%29+=+0

solutions:
%28x-5%29+=+0->x=5
or
%28x-2%29+=+0->x=2->even
so, your numbers are:5,7,9