<s>[INST] 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.
 [/INST] $$ \frac{|x^2+5x+6|}{|x|-3} = 1 \implies |x^2+5x+6|=|x|-3$$
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.

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