<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] $$\left(\frac{x-3}{x+1}\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 \Rightarrow \\ \left(\left|\frac{x-3}{x+1}\right|\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 $$
$$\Delta=\left(-7\right)^2-4 \cdot 10=49-40=9$$
$$\left|\frac{x-3}{x+1}\right|_{1,2}=\frac{-\left(-7\right) \pm \sqrt{\Delta}}{2}=\frac{7 \pm \sqrt{9}}{2}=\frac{7 \pm 3}{2}$$
$\left(\left|\frac{x-3}{x+1}\right|\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 \Rightarrow \left|\frac{x-3}{x+1}\right| \in \left(\frac{7-3}{2},\frac{7+3}{2}\right)=\left(2,5\right)$
$$\left|\frac{x-3}{x+1}\right|\gt 2 \Rightarrow \frac{x-3}{x+1}\lt -2 \text{ OR } \frac{x-3}{x+1}\gt 2 $$
and
$$\left|\frac{x-3}{x+1}\right|\lt 5 \Rightarrow -5\lt \frac{x-3}{x+1}\lt 5 $$
Solve at each case for $x$.
Can you continue?
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