# Check answers of rational expressions

• Sep 4th 2010, 10:20 AM
dudeman893
Check answers of rational expressions
Hi,

I am in an Algebra II class. I already took it years ago, but I am taking it again as a refresher. I have done two problems that I uploaded for this community to check to see if I have the right answer. I wrote out the whole equation. This will probably be very easy for the members here.

1.) This is a subtraction and simplifying of a rational expression.
http://i53.tinypic.com/24nh2zn.jpg

2.) This is a find the domain problem...
http://i51.tinypic.com/5p0eog.jpg
• Sep 4th 2010, 10:37 AM
yeKciM
Quote:

Originally Posted by dudeman893
Hi,

I am in an Algebra II class. I already took it years ago, but I am taking it again as a refresher. I have done two problems that I uploaded for this community to check to see if I have the right answer. I wrote out the whole equation. This will probably be very easy for the members here.

1.) This is a subtraction and simplifying of a rational expression.
http://i53.tinypic.com/24nh2zn.jpg

2.) This is a find the domain problem...
http://i51.tinypic.com/5p0eog.jpg

for the first one :D

$\displaystyle \frac {8xy}{x^2-y^2} - \frac {x-y}{x+y}$

$\displaystyle \frac {8xy}{(x-y)(x+y)} - \frac {x-y}{x+y}$

$\displaystyle \frac {8xy - (x-y)(x-y)}{(x-y)(x+y)}$

........... and so on :D

for the secon one

$\displaystyle \frac {6}{x(x-7)}$

$x(x-7) =0 \Rightarrow x_1=0 ; x_2 = 7$

so it's $x \in (-\infty , 0) \cup (0 , 7 ) \cup ( 7, +\infty)$

that means that function is not defined just for x=0 , and x=7 :D