SOLUTION: Find three consecutive odd positive integers such that 5 times the sum of all three is 60 more than the product of the first and second integers. The smallest odd integer is ___

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Question 1062176: Find three consecutive odd positive integers such that 5 times the sum of all three is 60 more than the product of the first and second integers.
The smallest odd integer is ___.
The next consecutive odd integer is ____.
The third consecutive odd integer is ____.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
n is a positive integer.

The numbers: system%282n%2B1%2C2n%2B3%2C2n%2B5%29

Equation for the description: 5%28%282n%2B1%29%2B%282n%2B3%29%2B%282n%2B5%29%29=60%2B%282n%2B1%29%282n%2B3%29

Simplify and solve for n;
use n to evaluate the three consecutive odd positive integers.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Find three consecutive odd positive integers such that 5 times the sum of all three is 60 more than the product of the first and second integers.
The smallest odd integer is ___.
The next consecutive odd integer is ____.
The third consecutive odd integer is ____. 
With the smallest integer being S, the following equation is formed:

5(S + S + 2 + S + 4) = S(S + 2) + 60

5%283S+%2B+6%29+=+S%5E2+%2B+2S+%2B+60

Continue to solve for S, the smallest integer. Add 2 to S to get the middle integer, and then 4 to S to get the largest integer.

You will get 2 possible answers. Based on the given info. determine which is correct.

It's that simple.....nothing COMPLEX!!