solving $\frac{x^2-7|x|+10}{x^2-5x+6}<0$ 
Solve $$\frac{x^2-7|x|+10}{x^2-5x+6}<0$$

I started off by making a system of equations where 
$$\begin{cases} \dfrac{x^2-7x+10}{x^2-5x+6}<0 &\text{if } x\geq0\\
\dfrac{x^2+7x+10}{x^2-5x+6}<0 &\text{if } x<0\end{cases}$$
By solving the first one I arrived to the conclusion that $x\in(2,3)\cup(3,5)$
By solving the second one I concluded that $x\in(-5,-2)$
By intersecting both sets $x\in \emptyset$
But by putting it in WolframAlpha it said that $x\in (-5,2) \cup (3,5) \cup(2,3)$
So it's obvious that they calculated the union of both sets why is that?
 A: You have two inequality since $x$ could be positive ($|x|=x$) or $x$ could be negative ($|x|=-x$). As you wrote with your system. Both case are disjoint, so the answer is the union of both answers, not the intersection.
A: The solution set is the set of all solutions. 
Every $x \in (-5,-2)$ is a solution. Also, every $x \in (2,3) \cup (3,5)$ is a solution. And there are no other solutions.
Therefore, $x$ is a solution if and only if $x \in (-5,-2)$ OR $x \in (2,3) \cup (3,5)$, which (by definition of the union) is equivalent to $x \in  (-5,-2) \underbrace{\cup}_{\text{*not* $\cap$}} ((2,3) \cup (3,5))$.
If that OR were instead an AND then the intersection operator $\cap$ would be appropriate. But in this situation it's not appropriate.
A: Another way, you want to find when
$$\frac{(|x|-2)(|x|-5)}{(x-2)(x-3)}< 0$$
Sign changes can occur only when crossing the points $|x|=2, x=3, |x|=5$, i.e. when $x\in \{-5, -2, 2, 3, 5\}$.  As $x\to \pm \infty$ the value is positive, it is enough to evaluate at any point in each of the open intervals bounded by these values of $x$, and you get $x \in (-5, -2) \cup (2, 3) \cup (3, 5)$ for the inequality to hold.
P.S. Drawing a number line with these points marked and putting +/- signs in the line segments in between helps visualise why union and not intersection is relevant here.
