SOLUTION: 2. Twice the sum of two consecutive odd integers is three less than the product of two less than the smaller odd integer and two more than the larger odd integer

Question 973166: 2. Twice the sum of two consecutive odd integers is three less than the product of two less than the smaller odd integer and two more than the larger odd integer
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Solving the equation you need is easier than transcribing the description into that equation.

The two integers with n being any integer:
2n+1, 2n+3.

Find n, and use it to evaluate the two consecutive odd integers.
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

2. Twice the sum of two consecutive odd integers is three less than the product of two less than the smaller odd integer and two more than the larger odd integer

Let the smaller integer be S
Then larger is: S + 2
Therefore, we get: 2(S + S + 2) = (S � 2) * (S + 2 + 2) � 3
2(2S + 2) = (S � 2)(S + 4) � 3

(S - 5)(S + 3) = 0
S, or smaller integer = 5 OR S = - 3
If the smaller integer is 5, then integers are:
However, if the smaller integer is - 3, then integers are:
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