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(39630)   (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(10556)  (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.
|
|
|
