How do you solve $\frac{|x^2+5x+6|}{|x|-3} = 1$ How do you solve 
$ \frac{|x^2+5x+6|}{|x|-3} = 1 $   ?
I have tried rearranging, polynomial division, multiplying both sides by a a fraction to simplify to no avail for the last hour.
 A: $$ \frac{|x^2+5x+6|}{|x|-3} = 1 \implies |x^2+5x+6|=|x|-3$$
Solving Algebraically:
Case$1$: When, $x <-3$,
$$ x^2+5x+6=-x-3 \implies x^2+6x+9=0 \implies x=-3 \text{ [No solution]. }$$
Case$2$: When, $-3 < x \leq -2$,
$$ -x^2-5x-6=-x-3 \implies x^2+4x+3=0 \implies (x+3)(x+1)=0 \implies x=-3,-1 \text{ [No solution from this case]. }$$
Case$3$: When, $-2 < x \leq 0$,
$$ x^2+5x+6=-x-3 \implies x^2+6x+9=0 \implies x=-3 \text{ [No solution from this case]. }$$
Case$4$: When, $x > 0$ but $x \neq 3$,
$$ x^2+5x+6=x+3 \implies x^2+4x+3=0 \implies \implies (x+3)(x+1)=0 \implies x=-3,-1 \text{ [No solution from this case]. }$$
So, the answer is no solution.
Solving Geometrically:
As mentioned by  Laars Helenius.
You can plot both $|x^2+5x+6|$ and $(|x|-3)$ on the same plot and check for the intersecting point, and that will be your answer. But you have to check whether those solution points in your domain. In this case, you will get one point $-3$, but your function domain is $\mathbb{R}\setminus \{-3,3\}$. So, no solution.

A: Since you want $\frac{|x^2+5x+6|}{|x|-3}=1$, that implies $|x^2+5x+6|=|x|-3$ as long as $|x|-3\ne 0$. Graph both equations simultaneously, and you will see they intersect at exactly $1$ spot: $(-3,0)$. Unfortunately we must reject the solution since it makes $|x|-3=0$. Thus there is no solution.
By graphing them, you will also be able to see how to chop up the domain in order to evaluate all the possibilities alluded to in the hints and other solutions provided.
A: Hint: You can rewrite the numerator as $|(x+2)(x+3)|$.  From there you can figure out the cases for which the expression inside the absolute value is negative and for which it is positive (and rewrite the numerator appropriately, i.e. as either $x^2 +5x+6$ or $-(x^2 +5x+6)$), which will also help you decide whether to rewrite $|x|$ as either $x$ or $-x$.
