<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] If $x\ge 1$ we have $$|3x+2|\geq4|x-1|\iff 3x+2\geq 4x-4 \iff x\le 6.$$ So $[1,6]$ is solution of the first inequality. Now, if $-2/3\le x\le 1$ we have
$$|3x+2|\geq4|x-1|\iff 3x+2\geq 4-4x \iff x\ge 2/7.$$ So $[2/7,1]$ is solution of the system. Finally, if $x\le -2/3$ then 
$$|3x+2|\geq4|x-1|\iff -3x-2\geq 4-4x \iff x\ge 6,$$ which is impossible. That is, the solution set of the first inequality is $[2/7,6].$
(The idea is to solve $3x+2=0$ and $x-1=0$ and study the inequality on each region you obtain.)
Now, we will work with the second inequality. Write it as
$$\dfrac{(x-1)(x+2)}{2(x-2)(x+1/2)}\ge 0.$$ Study the sign on $(-\infty,-2),$ $(-2,-1/2),$ $(-1/2,1),$ $(1/2,1)$ and $(1,\infty).$ You should obtain that the set solution is $(-\infty,-2]\cup (-1/2,1]\cup (2,\infty).$
(The idea is to solve $x^2+x-2=0$ and $2+3x-2x^2=0$ and study the inequality on each region you obtain. Note that we can't divide by $0.$ So $x\ne-1/2$ and $x\ne 2.$)
Finally, one gets the intersection to obtain $(2/7,1]\cup (2,6].$
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