How to solve $\frac{|x+2|}{x-1}>\frac{x+1}{2x+1}$?

Question

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?

Answer 1

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 $x>1$

Answer 2

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.

Answer 3

We need to consider 2 cases

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

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

2. 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.