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?
 A: No, something is wrong with your result. Note that the left-hand side is negative in $(-12,3)$ whereas the right-hand side is always positive.
Hint. The inequality is equivalent to
$$\frac{15}{x-3}=\frac{x+12}{x-3} -1\gt |x-1|.$$
Now  if $x\leq 3$ the inequality does not hold (why?).
It remains to consider the case when $x>3$ which implies that $|x-1|=x-1$. So just solve
$$15>(x-3)(x-1).$$
Can you take it from here?
A: We can rearrange this equation to give us 
\begin{align}\frac{x+12}{x-3}&>|x-1|+1\\
\frac{x+12}{x-3}-1&>|x-1|\\
\frac{x+12}{x-3}-\frac{x-3}{x-3}&>|x-1|\\
\frac{x+12-(x-3)}{x-3}&>|x-1|\\
\frac{15}{x-3}&>|x-1|\end{align}
Clearly $x\neq 3$
We now need to consider three intervals: $(-\infty,1]$, $[1,3)$, $(3,\infty)$
When $x\in(-\infty,1]$ then we have $x-3<0$, and $x-1<0$ so:  \begin{align}\frac{15}{x-3}&>-(x-1)\\
15&<(x-3)(1-x)\\
15&<-x^2+4x-3\\
0&<-x^2+4x-18\end{align}
This has no solutions, so we now know that $x>1$, although we can't say anything more specific yet.
Now when $x\in[1,3)$ then we have $x-3<0$ and $x-1>0$ so:
\begin{align}\frac{15}{x-3}&>x-1\\
15&<(x-1)(x-3)\\
15&<x^2-4x+3\\
0&<x^2-4x-12\\
0&<(x+2)(x-6)\end{align}
In the interval $[1,3)$, $(x+2)(x-6)$ is always negative, and thus the inequality is never satisfied. Now we can say that we know $x>3$, although we still cannot say anything more specific.
We turn to the final interval, when $x\in(3,\infty)$ then $x-3>0$ and $x-1>0$ so:
\begin{align}\frac{15}{x-3}&>x-1\\
15&>(x-1)(x-3)\\
15&>x^2-4x+3\\
0&>x^2-4x-12\\
0&>(x+2)(x-6)\end{align}
Remembering we solved this with the assumption that $x>3$, then the solutions to this are $3<x<6$ (without the assumption, the solutions are $-2<x<6$).
Now we need to combine the answers from each of the three intervals. The first two did not have any solutions, so the solution to the initial question is simply $$3<x<6$$
We can plot each side of the inequality and see that this is the case too.
A: 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
