<s>[INST] Solve $\frac{x+12}{x-3} \gt |x-1| + 1$ Solve $\frac{x+12}{x-3} \gt |x-1| + 1$
I finally got $\frac{(x-6)(x+2)}{x-3} \lt 0$
That means the solutions are $3\lt x \le 6$ and $x\lt -2$ Am I right?
 [/INST] at first we note that $$x\ne 3$$ and then we do case work:
$$x\geq 1$$ then we have $$\frac{x+12}{x-3}>x$$ then we have to solve
$$\frac{-x^2+4x+12}{x-3}>0$$
solving this we obatin $$3<x<6$$
in the second case we assume
2) $$x<1$$ and this is equivalent to
$$\frac{x+12}{x-3}>2-x$$
can you finish?
the last inequation is equivalent to
$$\frac{x^2-4x+18}{x-3}>0$$
this gives a contradiction
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