How to solve $\frac{|x+2|}{x-1}>\frac{x+1}{2x+1}$? I am working on the following problem:
$$\frac{|x+2|}{x-1}>\frac{x+1}{2x+1}$$
Here's what I have done so far:
$$|x+2|>\frac{x+1}{2x+1}\times(x-1)$$
$$-\left(\frac{(x+1)(x-1)}{(2x+1)}\right)<x+2<\frac{(x+1)(x-1)}{(2x+1)}$$
This is where I stopped. I am not entirely sure how to go about solving this type of inequality. Am I using the correct approach?
 A: Hint: You must do case work:
Case 1: $$x>1$$ then we get $$2x+1>0$$ and $|x+2|=x+2$ and we have
$$(x+2)(2x+1)>x^2-1$$ so $2x^2+5x+2>x^2-1$.
Can you proceed?
For your Control:
The result is given by
$\frac{-5}{2}+\frac{1}{2}\sqrt{3}<x<-\frac{1}{2}$ or $x>1$
A: Your last step is incorrect as we know nothing about the sign of $x+1$, and also if $|a|>b$, then $a<-b, a>b$, not $-b<a<b$.
What you need to do is consider each case between the points $x=-1, -\frac12,1$ as they determine whether the inequality should be flipped over, and whether you could drop the absolute value.
A: We need to consider 2 cases

  
*
  
*for $x+2\ge 0 \implies x\ge -2$ we need to solve
  

$$\frac{x+2}{x-1}>\frac{x+1}{2x+1}$$


  
*for $x+2< 0 \implies x< -2$ we need to solve
  

$$\frac{-x-2}{x-1}>\frac{x+1}{2x+1}$$
then the final solution is given by the union of the solution obtained for each case.
For case 1 we can proceed as follow
$$\frac{x+2}{x-1}>\frac{x+1}{2x+1}\iff \frac{x+2}{x-1}-\frac{x+1}{2x+1}>0\iff \frac{(x+2)(2x+1)-(x+1)(x-1)}{(x-1)(2x+1)}>0$$
$$\iff \frac{x^2+5x+3}{(x-1)(2x+1)}>0\iff \frac{\left(x-\frac{5+\sqrt 13}{2}\right)\left(x-\frac{5-\sqrt 13}{2}\right)}{(x-1)(2x+1)}>0$$
then we ca easily find the solutions under the condition $x\ge -2$.
In a similar way we can study case 2.
