SOLUTION: The product of 2 consecutive odd integers is 7 more than their sum. What are the integers? I got: n(n+2) = 7 + n(n+2) nē+2n = 7 + nē+2n nē-nē+2n

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: The product of 2 consecutive odd integers is 7 more than their sum. What are the integers? I got: n(n+2) = 7 + n(n+2) nē+2n = 7 + nē+2n nē-nē+2n      Log On


   

Question 616250: The product of 2 consecutive odd integers is 7 more than their sum. What are the integers?
I got:
n(n+2) = 7 + n(n+2)
nē+2n = 7 + nē+2n
nē-nē+2n-2n-7 = 0
BUT...this is not correct...please help and thank you!
Answer by dragonwalker(73) About Me  (Show Source):
You can put this solution on YOUR website!
You were almost there, but on the right side you accidentally retyped the formula for their product rather than their sum i.e. instead of 7+n(n+2) you should have typed 7+n+(n+2) or just 7+n+n+2
So:
n(n+2)=n+n+2+7
n%5E2%2B2n=2n%2B9
Solve for n and put all numbers containing n to the left (remember to change the sign if moved):
n%5E2%2B2n-2n=9
n%5E2=9
n=3
so if n=3 this is the first integer and the second is 5.