Simplifying algebraic fractions.

Question

I cannot simplify the following expression. Please need help.

$\large \frac{3}{x+1}-\frac{1}{x+3}+\frac{3}{1-x}-\frac{1}{3-x}$

Thanks in advance. Regards !

Answer 1

Just a little re-arrangement is needed to make things easier. $$\frac{3}{x+1} + \frac{3}{1-x} + \frac{1}{x-3} - \frac{1}{x+3}$$ $$3(\frac{1-x+1+x}{1-x^2}) + \frac{x+3-x+3}{x^2-9}$$ $$6(\frac{1}{1-x^2}+\frac{1}{x^2-9})$$ I guess you can take it from here.

Answer 2

$$\frac 3{x+1}-\frac 1{x+3}+\frac 1{x-3}-\frac 3{x-1}$$

$$=\frac {2x+8}{x^2+4x+3}-\frac {2x-8}{x^2-4x+3}$$

$$=\frac {(8-(-8))\cdot 3}{(x^2+3)^2-16x^2}=\frac {48}{x^4-10x^2+9}$$

There's some steps glossed over.