SOLUTION: find 3 consecutive odd integers such that are twice the product of the first two is 7 more than the product of the last two.

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Question 886344: find 3 consecutive odd integers such that are twice the product of the first two is 7 more than the product of the last two.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39625)   (Show Source): You can put this solution on YOUR website!
Put that into symbols.

n is any whole number.
The numbers needed are 2n+1, 2n+3, 2n+5.

The problem you are describing:


Solve for n, and compute or evaluate each of the numbers.
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

find 3 consecutive odd integers such that are twice the product of the first two is 7 more than the product of the last two.


Let first integer be F

Then 2nd is: F + 2, and 3rd is: F + 4

Therefore, 2(F)(F + 2) = (F + 2)(F + 4) + 7





Solve for F, the 1st integer, and then determine the other 2 integers. Note that 2 values will ensue for

the 1st integer, F, thus 2 values for the 2nd and 3rd integers. 

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