Question 529700: Two numbers, which differ by 3, have a product of 88. Find them.
Answer by unlockmath(1688)   (Show Source):
You can put this solution on YOUR website! Hello,
Let's set up one number as n and the other as n+3, now we can write an equation as:
n(n+3)=88
Expand this:
n^2+3n=88
Subtract 88:
n^2+3n-88=0
Factor:
(n-8)(n+11)=0
Solve for n:
n = 8
n= -11
From this we see that 8 times 11
or -11 times -8 works.
Make sense?
RJ
www.math-unlock.com
|
|
|
