Simplifying complex fractions and finding restrictions

[grfulford]

What is the simplified version of (6/x-3)-(8/x+4) and what are the restrictions
What is the simplified version of (-49/x^2-7x)+(x/x-7) and what are the restrictions
What is the simplified version of (1/5x)+(x/5x^2-x) and what are the restrictions
What is the simplified version of ((3/x-4)/6+(6/x-4)) and what are the restrictions

 

[pka]

Are you going to follow posting guidelines and post your efforts and/or your doubts?

 

[topsquark]

What is the simplified version of (6/x-3)-(8/x+4) and what are the restrictions

I really hope you mean what is 6/(x - 3) - 8/(x + 4). What you wrote was [math]\left ( \dfrac{6}{x} - 3 \right ) - \left ( \dfrac{8}{x} + 4 \right )[/math].

For this kind of problem you want to find a common denominator and add or subtract as usual. But your assignment seems to be dealing with complex fractions of the form
[math]\dfrac{ 3 + \dfrac{1}{x - 2} }{ \dfrac{x + 2}{x - 6} }[/math]
What you need to do here is clear the denominators. I'll do a simpler example, the process is the same in all cases.
[math]\dfrac{1 + \dfrac{x - 2}{x + 3} }{ x + 1}[/math]
We need to get rid of the fraction in the numerator. So multiply both the top and bottom of the fraction by x + 3:
[math]\dfrac{1 + \dfrac{x - 2}{x + 3} }{ x + 1} = \dfrac{1 + \dfrac{x - 2}{x + 3} }{ x + 1} \cdot \dfrac{x + 3}{x + 3}[/math]
[math]= \dfrac{ \left ( 1 + \dfrac{x - 2}{x + 3} \right ) (x + 3)}{ (x + 1) \cdot (x + 3)}[/math]
[math]= \dfrac{(x + 3) + (x - 2)}{(x+1)(x + 3)}[/math]and simplify as usual.

See what you can do with this. If you are still having problems post your work and we'll help you out.

-Dan

 

[Steven G]

...and also find the restrictions.