Inequalities giving incorrect solution

Question

Question:

Find the solution set for:$$\frac{|x|-1}{|x|-2} \geq 0$$ $x\not=\pm2$

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My attempt:

Let $|x| = y$, then inequality becomes $(y-1)(y-2)>=0$

Implies that:

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#1. $y-1\geq0$ and $y-2\geq0$

Then, $y\geq1$ and $y\geq2$, which means $y\geq2$.

Substituting, $|x|=y$, it becomes: $x\geq2$ and $x\leq-2 \Rightarrow$ $2\leq x \leq -2$

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#2. $y-1<0$ and $y-2<0$

Then, $y<1$ and $y<2$, which means $y<2$

Substituting, $|x|=y$, it becomes: $x<2$ and $x>-2$. $\Rightarrow -2 < x < 2$

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As it can be seen, #1 is definitely false, so answer is:

$$x \in (-2,2)$$

But the answer given in my textbook is $x \in (-\infty, -2)\cup(-1,1)\cup(2,\infty)$

I need help regarding where I did go wrong.

Answer 1

You mix different notions of "and". In case 1, $x$ is a solution if $|x|> 2$ (becasue $|x|=2$ is forbidden anyway), that is, $x$ is a solution if $x>2$, and $x$ is also a solution if $x<-2$. That is different from saying that $x$ is a solution in the rare occasions where both $x>2$ and $x<-2$. So you should find "$x>2$ or $x<-2$" instead of the contradictory $2<x<-2$.

In case 2 note that a number $y$ that is both $<1$ and $<2$ is best described as a number $<1$. (Yes, "if $y<1$ and $y<2$, then $y<2$", but we should look for "if and only if" to precisely determine all solutions).

Answer 2

Consider the two cases :

$1)$ $x \geq 0 \implies |x|=x:$ $$\frac{x-1}{x-2} \geq 0$$ Solution set would be, $x\geq1$ and $x>2$ or $x\leq1$ and $x<2$. $$x \in (2,\infty) \cup (-\infty,1]$$

$2)$ $x\leq0 \implies |x|=-x:$ $$\frac{x+1}{x+2} \geq 0$$ Solution set would be, $x\geq-1$ and $x>-2$ or $x\leq-1$ and $x<-2$. $$x \in [-1,\infty) \cup (-\infty,-2)$$

So taking the common solution from the two cases we get: $$x \in (-\infty,-2) \cup [-1,1] \cup (2,\infty)$$