SOLUTION: Find three consecutive odd integers such that the product of the first and third is equal to 1 less than twice the second. It's quite a confusing problem to read, but I wasn't

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Question 1009159: Find three consecutive odd integers such that the product of the first and third is equal to 1 less than twice the second.
It's quite a confusing problem to read, but I wasn't taught this in class because I recently moved schools.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!

n is any positive integer.
2n is a positive even integer.
2n+1 is an ODD positive integer.

The description transcribed into an equation:


Solve for n;
evaluate .
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Find three consecutive odd integers such that the product of the first and third is equal to 1 less than twice the second.
It's quite a confusing problem to read, but I wasn't taught this in class because I recently moved schools.
Let the smallest be S

Then others are: S + 2, and S + 4

We then get: S(S + 4) = 2(S + 2) - 1







(S + 3)(S - 1) = 0                 

S, or smallest =         OR         

You should be able to find the other 2 

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