Inequality with absolute value and variable on both sides I'm not able to understand how to get through this
$$\dfrac{|x+6|}{x+1} \leq x-2$$
$x$ can't, of course, be equal to $-1$.
I'm at a point where I'm not sure how to deal with the absolute value after reaching
$$|x+6| \leq (x-2)(x+1)$$
Thanks!
 A: Lets look at this case by case.
Case 1: $x\leq-6$
Then you had $x+1\leq-5$, and so when multiplying across, the inequality sign changes. Also, $|x+6|=6-x$, so: $$6-x\geq(x-2)(x+1)=x^2-x-2\implies x^2\leq8$$ This is not possible since $x\leq -6$.
Case 2: $-6\leq x<-1$
Then you still had $x+1<0$, but now $|x+6|=x+6$, so
$$x+6\geq x^2-x-2\implies x^2-2x-8\leq0\implies-2\leq x\leq 4$$
From this range, only $-2\leq x<-1$ works.
Case 3: $x>-1$
Then $$x+6\leq x^2-x-2\implies x^2-2x-8\geq0\implies x\leq-2\text{ or }x\geq4$$
From this range, only $x\geq4$ works.

Final solution: $x\in[-2,-1)$ or $x\in[4,\infty)$.
A: $$\dfrac{|x+6|}{x+1} \leq x-2$$
Let's start with $\dfrac{|x+6|}{x+1} = x-2$.
\begin{array}{rclcrcl}
     |x+6|  &= &(x+1)(x-2)  &| \\
   x^2-x-2  &= &|x+6|       &|\\
   x^2-x-2  &= &x+6         &\text{or}  & x^2-x-2 &= &-x-6 \\
   x^2-2x-8 &= &0           &\text{or}  & x^2+4 &= &0 \\
          x &\in &\{-2,4\}  &\text{or}  &     x &\in &\emptyset \\
\end{array}
So we have three critical points: zeros at $x=-2$ and $x=4$ and a pole at $x=-1$. The critical points divide the real number line into four regions:
$(-\infty, -2), (-2,-1), (-1, 4)$, and $(4, \infty).$ Each region is either all solutions or all no solutions. We test each region.
\begin{array}{c}
   \text{region} & \text{sample point} & \text{Is solution?} \\
\hline
   (-\infty, -2) & -10  & \text{no} \\
   (-2,-1)       & -1.5 & \text{yes} \\
   (-1, 4)       &    0 & \text{no}\\
   (4, \infty)   &   12 & \text{yes} \\
\hline
\end{array}
So the solution set is $[-2,-1) \cup [4, \infty)$.
