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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.

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  • $\begingroup$ Formatting tip: open interval endpoints are () and not <>. $\endgroup$ – Parcly Taxel Oct 23 '16 at 10:44
  • $\begingroup$ Just one more question, how would you solve this by not studying the regions, in other words, how would you find the intersection of $((-\infty, -2]\cup[1,\infty]$) and ($(-\infty, -1/2)\cup(2,\infty)$)? $\endgroup$ – lmc Oct 23 '16 at 13:16
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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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  • $\begingroup$ Thanks, I appreciate your help! $\endgroup$ – lmc Oct 23 '16 at 13:08
  • $\begingroup$ @mfl: I think you have to edit your answer, dear friend, discarding the point $x=1$. Or am I wrong?. Regards. $\endgroup$ – Piquito Oct 23 '16 at 13:22
  • $\begingroup$ @Piquito Why do you want to discard $x=1?$ It is $|3\cdot 1+2|=5\ge 0=4 |1-1|$ and $\dfrac{1^2+1-2}{2+3\cdot 1 -2 \cdot 1^2}=\dfrac{0}{3}=0\le 0.$ $\endgroup$ – mfl Oct 23 '16 at 16:18
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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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