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.

  • $\begingroup$ Formatting tip: open interval endpoints are () and not <>. $\endgroup$ 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

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 $(2/7,1]\cup (2,6].$

  • $\begingroup$ 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

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 $$S_1\cap S_2=\left((-\infty,-2]\cup(-\frac 12,1]\cup(2,\infty)\right)\cap\left([\frac 27,1)\cup(1,6]\right)=\color{red}{(\frac 27,1)\cup (2,6]}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.