1. ## Complex fraction simplification

Good afternoon forum I am hav[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-3.jpg[/IMG]ing problems with the below problem

[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-4.jpg[/IMG]

2. Hello, AlgebraicallyChallenged!

Simplify: . $\frac{\dfrac{x+2}{x^2-1} + \dfrac{1}{x+1}} {\dfrac{x}{2x^2-x-1} + \dfrac{1}{x-1}}$
Factor all denominators: . $\frac{\dfrac{x+2}{(x-1)(x+1)} + \dfrac{1}{x+1}} {\dfrac{x}{(x-1)(2x+1)} + \dfrac{1}{x-1}}$

The LCD is $(x-1)(x+1)(2x+1)$

Multiply top and bottom by the LCD:

. . $\frac{(x-1)(x+1)(2x+1)\left[\dfrac{x+2}{(x-1)(x+1)} + \dfrac{1}{x+1}\right]} {(x-1)(x+1)(2x+1)\left[\dfrac{x}{(x-1)(2x+1)} + \dfrac{1}{x-1}\right]}$ . $= \;\frac{(2x+1)(x+2) + (x-1)(2x+1)}{x(x+1) + (x+1)(2x+1)}$

. . $= \;\frac{2x^2 + 5x + 2 + 2x^2 - x - 1}{x^2+x + 2x^2+3x+1} \;=\;\frac{4x^2 + 4x + 1}{3x^2+4x+1} \;=\;\boxed{\frac{(2x+1)^2}{(x+1)(3x+1)}}$ . . . doesn't reduce

3. Originally Posted by AlgebraicallyChallenged
Good afternoon forum I am hav[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-3.jpg[/IMG]ing problems with the below problem

[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-4.jpg[/IMG]
$\frac{ \frac{x+2}{x^2+1}+\frac{1}{x+1} }{\frac{x}{2x^2-x-1}+\frac{1}{x-1}}$

$= \frac{ \frac{(x+2)(x+1)}{(x+1)(x^2+1)}+\frac{(x^2+1}{(x^2 +1)(x+1)} }{\frac{x(x-1)}{(2x^2-x-1)(x-1)}+\frac{2x^2-x-1}{(2x^2-x-1)(x-1)}}$

$= \frac{\frac{(x+2)(x+1)+x^2+1}{(x^2+1)(x+1)}}{\frac {x(x-1) + 2x^2-x-1}{(2x^2-x-1)(x-1)}}$

$= \frac{(x+2)(x+1)+x^2+1}{(x^2+1)(x+1)} \times \frac{(2x^2-x-1)(x-1)}{x(x-1) + 2x^2-x-1}$

How's that?

[Beaten to it]