Misael Matthews

Answered

2022-07-01

Solve the system:

$|3x+2|\ge 4|x-1|$

$\frac{{x}^{2}+x-2}{2+3x-2{x}^{2}}\le 0$

$|3x+2|\ge 4|x-1|$

$\frac{{x}^{2}+x-2}{2+3x-2{x}^{2}}\le 0$

Answer & Explanation

Elianna Douglas

Expert

2022-07-02Added 23 answers

If $x\ge 1$ we have

$|3x+2|\ge 4|x-1|\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x+2\ge 4x-4\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x\le 6.$

So $[1,6]$ is solution of the first inequality. Now, if $-2/3\le x\le 1$ we have

$|3x+2|\ge 4|x-1|\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x+2\ge 4-4x\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x\ge 2/7.$

So $[2/7,1]$ is solution of the system. Finally, if $x\le -2/3$ then

$|3x+2|\ge 4|x-1|\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}-3x-2\ge 4-4x\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x\ge 6,$

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

$\frac{(x-1)(x+2)}{2(x-2)(x+1/2)}}\ge 0.$

Study the sign on $(-\mathrm{\infty},-2),$, $(-2,-1/2),$, $(-1/2,1),$, $(1/2,1)$ and $(1,\mathrm{\infty}).$ You should obtain that the set solution is $(-\mathrm{\infty},-2]\cup (-1/2,1]\cup (2,\mathrm{\infty}).$ (The idea is to solve ${x}^{2}+x-2=0$ and $2+3x-2{x}^{2}=0$ 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 $(2/7,1]\cup (2,6].$

$|3x+2|\ge 4|x-1|\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x+2\ge 4x-4\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x\le 6.$

So $[1,6]$ is solution of the first inequality. Now, if $-2/3\le x\le 1$ we have

$|3x+2|\ge 4|x-1|\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x+2\ge 4-4x\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x\ge 2/7.$

So $[2/7,1]$ is solution of the system. Finally, if $x\le -2/3$ then

$|3x+2|\ge 4|x-1|\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}-3x-2\ge 4-4x\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x\ge 6,$

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

$\frac{(x-1)(x+2)}{2(x-2)(x+1/2)}}\ge 0.$

Study the sign on $(-\mathrm{\infty},-2),$, $(-2,-1/2),$, $(-1/2,1),$, $(1/2,1)$ and $(1,\mathrm{\infty}).$ You should obtain that the set solution is $(-\mathrm{\infty},-2]\cup (-1/2,1]\cup (2,\mathrm{\infty}).$ (The idea is to solve ${x}^{2}+x-2=0$ and $2+3x-2{x}^{2}=0$ 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 $(2/7,1]\cup (2,6].$

April Bush

Expert

2022-07-03Added 6 answers

Thanks, I appreciate your help!

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