<s>[INST] Solve the system of inequalities Solve the system: 
$$|3x+2|\geq4|x-1|$$
$$\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$$ 
So for $|3x+2|\geq4|x-1|$ I got the solution $x=6$ and $x=2/7$ and Wolframalpha agrees with me. I'm having troubles writing the final solution for $\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$.
$1)$ $x^{2}+x-2\leq0$ and $-2x^{2}+3x+2>0$
$x^{2}+x-2\leq0$ $\Rightarrow x\in [-2,1]$ 
$-2x^{2}+3x+2>0$ $\Rightarrow x\in (-1/2,2)$ 
Now I need the intersection of those two sets, which is $(-1/2,1]$. From the first inequality I got $x=6$ and $x=2/7$ so the final solution is $x=2/7$. 
$2)$ $x^{2}+x-2\geq 0$ and $-2x^{2}+3x+2<0$ 
$x^{2}+x-2\geq 0$ $\Rightarrow x\in(-\infty, -2]\cup[1,\infty]$
$-2x^{2}+3x+2<0$ $\Rightarrow x\in(-\infty, -1/2)\cup(2,\infty)$ 
Now I'm having troubles finding the intersection of those sets.
 [/INST] Consider $$f(x)=\frac{x^{2}+x-2}{2+3x-2x^{2}}=\frac{(x-1)(x+2)}{-(2x+1)(x-2)}$$ It follows $$f(x)=\frac{-(x-1)(x+2)}{(2x+1)(x-2)}\le0\iff\frac{(x-1)(x+2)}{(2x+1)(x-2)}\ge0$$
One has at once the solution set for $f(x)\le 0$ is $$S_1=(-\infty,-2]\cup(-\frac 12,1]\cup(2,\infty)$$
Now consider $$g(x)=\left|\frac{3x+2}{x-1}\right|=\left|\frac{3+\frac 2x}{1-\frac 1x}\right|$$ it follows $$\begin{cases}\lim_{x\to\pm\infty}=3\\\lim_{x\to-1}g(x)=\infty\\g\text { decreasing on } (-\infty,-\frac 23)\cup(-1,\infty)\\g\text { increasing on } (-\frac 23,1)\end{cases}$$ Furthermore $$\left|\frac{3x+2}{x-1}\right|=4\iff\begin{cases}3x+2=4(x-1)\iff x=6\\3x+2=-4(x-1)\iff x=\frac 27\end{cases}$$ Hence the solution set for $g(x)\ge 4$ is $$S_2=[\frac 27,1)\cup(1,6]$$
Thus our solution is $$S_1\cap S_2=\left((-\infty,-2]\cup(-\frac 12,1]\cup(2,\infty)\right)\cap\left([\frac 27,1)\cup(1,6]\right)=\color{red}{(\frac 27,1)\cup (2,6]}$$
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