SOLUTION: Help please Find three consecutive odd positive integers such that 2 times the sum of all three is 7 more than the product of the first and second integers. The smallest odd in

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Question 1040942: Help please
Find three consecutive odd positive integers such that 2 times the sum of all three is 7 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!
2n+1, 2n+3, 2n+5 are these integers; n is any positive integer.

2%28%282n%2B1%29%2B%282n%2B3%29%2B%282n%2B5%29%29=7%2B%282n%2B1%29%282n%2B3%29
Simplify and solve for n;
compute or evaluate the three integers.

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

Help please
Find three consecutive odd positive integers such that 2 times the sum of all three is 7 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 ____
Let the smallest integer be S

Then other two are: S + 2, and S + 4

We then get: 2(S + S + 2 + S + 4) = S(S + 2) + 7

2%283S+%2B+6%29+=+S%5E2+%2B+2S+%2B+7

Solve this equation for S: the smallest integer. Add 2 to S and 4 to S to get the other 2 integers. You should get 2 values for S, the smallest integer