<s>[INST] inequality method of solution Im looking for an efficent method of solving the following inequality: $$\left(\frac{x-3}{x+1}\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 <0$$
I've tried first determining when the absolute value will be positive or negative etc, and than giving it the signing in accordance to range it is in, bur it turned out to be quite complex and apperently also wrong. Are there any other ways?
 [/INST] When $t \equiv \frac{x-3}{x+1} > 0$, the solution to $ t^2 - 7t + 10 <0$ is $ 2<t<5$. (Factor the quadratic.)
When $t < 0$, the solution to $ t^2 + 7t + 10 <0$ is $ -2<t<-5$. 
On the $t < 0$ branch, we need to solve $-2 < \frac{x-3}{x+1} < 5$. We break this up into two possibilities, $x < -1$ and $x > -1$, because when we multiply through by $x+1$ in the  $x < -1$case we have to flip the sense of the the inequality.
When  $x < -1$ we then get on one side $-2(x+1) < x-3$, which gives $x > +\frac{1}{3}$ and this does not work.  But for  $x > -1$ we get 
$$\begin{array}{c}
-5(x+1) < x-3 < 2(x-1) \\
6x > -2 \rightarrow x > -\frac{1}{3} \\
3x < +1 \rightarrow x < +\frac{1}{3}
\end{array}
$$
which has solution 
$$
-\frac{1}{3} < x < \frac{1}{3}
$$
On the $t > 0$ branch, we need to solve $2 < \frac{x-3}{x+1} < 5$. We again break this up into two possibilities, $x < -1$ and $x > -1$.
When  $x < -1$ we then get on one side $2(x+1) > x-3$, which gives $x > -2$, and $5(x+1) < x-3$, which gives $x > -5$.  For  $x > -1$ we get $2(x+1) < x-3$, which holds if $x < -5$, which contradicts $x > -1$ so that case does not give any solutions.
Thus the answer combines the two solution regions:
$$
\left( -5 < x < 2 \right) \bigcup \left( -\frac{1}{3} < x < \frac{1}{3} \right)
$$
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