Needing assistance concerning mult. Rational expressions

[Ravingsofthesane]

First off , thank you to everyone who has given me assistance in the past . I appreciate the help


My problem is concerning the multiplcation of rational expressions, I know this is most likely a simple solution but I am a bit confused on how to address this.

3 - R R - 6
----- x -----
R - 3 6 - R

I know that solution is recipocials of one another

(3-R)(R-6)
-----------
(R-3)(6-R)

now I understand that to fix this I need to multiply by a -1 to get the correct numbers so I can simplify the order, however where and how is this applied?

The final anwser is 1 , but I am just not seeing where to apply the -1 to make this work.

as always any help is appreciated.

 

[pka]

First note that −(a−b)= −a+b=b−a.
Therefore, a−b and b−a are the negatives of each other.
This means that [a−b]/[b−a] = −1.

 

[soroban]

Hello, Ravings!

Simplify: \(\displaystyle \L \frac{3\,-\,R}{R\,-\,3}\:\times\:\frac{R\,-\,6}{6\,-\,R}\)


To make the factor "match", factor out a -1:

. . . \(\displaystyle \L \frac{-1(R\,-\,3)}{R\,-\,3}\:\times\:\frac{R\,-\,6}{-(R\,-\,6)}\) . . . then cancel.

 

[Ravingsofthesane]

Hello, Ravings!

Simplify: \(\displaystyle \L \frac{3\,-\,R}{R\,-\,3}\:\times\:\frac{R\,-\,6}{6\,-\,R}\)


To make the factor "match", factor out a -1:

. . . \(\displaystyle \L \frac{-1(R\,-\,3)}{R\,-\,3}\:\times\:\frac{R\,-\,6}{-(R\,-\,6)}\) . . . then cancel.

Oh ok .. So you factor out the -1 so the final anwser is actually -1 over -1 or 1


thanks .. that is what i needed to see