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- Homework Statement
- Solve : ##\mathbf{\dfrac{x}{x+2}\le \dfrac{1}{|x|}}##

- Relevant Equations
- (1) By definition, ##|x| = x## if ##x\ge 0## and ##|x| = -x## if ##x \le 0##.

(2) ##\frac{1}{x}## is not defined for ##x = 0##.

(3) If ##|x| > a\Rightarrow x>a\; \text{or}\; x<-a##

**I copy and paste the problem from the text to the right.
Problem Statement : **

**Attempt (mine) :**Given the inequality ##\dfrac{x}{x+2}\le \dfrac{1}{|x|}##. We see immediately that ##x\ne 0, -2##. At the same time, since ##|x|\ge 0\Rightarrow \frac{x}{x+2}\ge 0##.

Now if ##\frac{x}{x+2}\le \frac{1}{|x|}\Rightarrow \frac{1}{|x|}\ge \frac{x}{x+2}##, then we have either ##\frac{1}{x}\ge \frac{x}{x+2}## or ##\frac{1}{x}\le -\frac{x}{x+2}##.

**1.**The first inequality above reduces to ##\frac{1}{x}-\frac{x}{x+2}\ge 0\Rightarrow \frac{x+2-x^2}{x(x+2)}\ge 0\Rightarrow \frac{x^2-x-2}{x(x+2)}\le 0\Rightarrow \frac{(x-2)(x+1)}{x(x+2)}\le 0## which lead to the answers : ##\boxed{-2<x<-1\; \text{OR}\; 0<x<2}## (note the equality condition is forbidden due to considerations right at the beginning). This can be found by the number line

I attach the pictorial pneumonic to the right for those who are used to it.

**2.**The second inquality above reduces to ##\frac{1}{x}+\frac{x}{x+2}\le 0\Rightarrow \frac{x^2+x+2}{x(x+2}\le 0##. Now we can show using the method of "squaring a quadratic equation" that the numerator ##x^2+x+2>0 \;\;\forall x##. Thus focus falls on the numerator, which yields ##\boxed{-2<x<0}##.

Putting both cases

**1**and

**2**together and caring for overlaps, my solution for the proble reads : ##\boxed{-2<x<-1\;\;\;\text{OR}\;\;\; 0<x<2}## .

**Solution (Text) :**I could neither follow, nor agree, with the solution in the text. I copy and paste it below.

I have

__what I believe to be the solutions. So ##x=0?!##. Clearly that is not correct. However, the text is right when I take samples from its solutions that are less than zero, for instance ##x = -1, -0.5## etc., solutions which I did not get. Surely my method is mistaken.__

**underlined****A hint or suggestion would be welcome.**