SOLUTION: The product of two positive consecutive odd integers is 1 less than 5 times their sum. Find the integers

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Question 845227: The product of two positive consecutive odd integers is 1 less than 5 times their sum.
Find the integers
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
n-1= the smallest of the two consecutive odd integers.
n%2B1= the next odd integer (the other one of the two consecutive odd integers).
%28n-1%29%2B%28n%2B1%29=2n= the sum of the two consecutive odd integers.
%28n-1%29%2A%28n%2B1%29=n%5E2-1= the product of the two consecutive odd integers.
The equation is
n%5E2-1=5%2A%282n%29-1
Solving:
n%5E2-1=5%2A%282n%29-1
n%5E2-1=10n-1
n%5E2=10n
Since n%3C%3E0, I can divide both sides by n and get
n%5E2%2Fn=10n%2Fn
n=10
So n-1=10-1 --> n-1=highlight%289%29 , and
n%2B1=10%2B1 --> n%2B1=highlight%2811%29
Verification:
The numbers 9 and 11 are odd.
Their sum is 9%2B11=20 .
Five times their sum is 5%2A20=100 .
One less than 5 times their sum is 100-1=99 .
Their product is 9%2A11=99 and that is exactly "1 less than 5 times their sum."

NOTES:
I could have called the two consecutive odd integers x and x%2B2 ,
but I thought that would make the calculations more complicated.
Someone could think to call the two odd integers 2k%2B1 and 2k%2B3 ,
or 2k-1 and 2k%2B1 ,
but that was going to make it more complicated.
If there are consecutive odd integers, or consecutive even integers,
just call them m and m%2B2 ,
or m-1 and m%2B1 .
Those expressions do not say that the numbers are even or odd,
but they say that they are the same kind (both odd or both even),
and when you find the solution you can verify if it is the right kind.