Distributive Law

Date: 8/2/96 at 17:50:40
From: Anonymous
Subject: Distribution of Subtraction

I am an English teacher trying to teach Algebra to high school 
students and I need a lot of help.  

Regarding the following equation:

[(2y)(squared) + 8] - (y (squared) - y) = 3y sq

The 1st step of the solution in my text is:
 (4y sq + 8) - y sq + y = 3y sq

My question: Why does "-(y sq - y)" become "-y sq + y?"  Why does the 
subtraction symbol change to addition?

Thank you.

Date: 8/5/96 at 0:6:44
From: Doctor Mike
Subject: Re: Distribution of Subtraction

Hello,
   
You are NOT the first person in history to get hung up on this point.
This is a form of (or really a consequence of) the Distributive Law.
  
The most common version of the Distributive Law is: 

               A*(B + C) = A*B + A*C 

It says that to multiply a sum of 2 numbers by A, you multiply each
one of them individually by A, and then add those 2 results together.
  
Another common version of the Distributive Law is: 

               A*(B - C) = A*B - A*C

This looks different from the first one above, but it really isn't a
new and different property of numbers.  Here's why it follows : 

Since "B - C" (B minus C) is the same as B + -C (B plus the negative
of C), it is true that "A*(B - C)" is the same as "A*(B + -C)".  
So using the first version of the Distributive Law from above, 
   
               A*(B - C) = A*(B + -C) 
                         = A*B + A*(-C)
                         = A*B - A*C 
 
Now, finally, what you have been waiting for.  An expression such as
-(B - C) is the same as (-1)*(B - C), so 

               -(B - C) = (-1)*(B - C)
                        = (-1)*B - (-1)*C
                        = -B -(-C)
                        = -B + C
  
By the way, the reason for that last step was that the "negative of 
the negative of C" is just C.  See?  
   
In your particular example "-(y sq - y)", B is y squared and C is y. 
  
I hope this gets you and your class started in the right direction.
  
-Doctor Mike,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/