Misael Matthews

2022-07-01

Solve the system:
$|3x+2|\ge 4|x-1|$
$\frac{{x}^{2}+x-2}{2+3x-2{x}^{2}}\le 0$

Elianna Douglas

Expert

If $x\ge 1$ we have
$|3x+2|\ge 4|x-1|\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3x+2\ge 4x-4\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x\le 6.$
So $\left[1,6\right]$ is solution of the first inequality. Now, if $-2/3\le x\le 1$ we have
$|3x+2|\ge 4|x-1|\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3x+2\ge 4-4x\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x\ge 2/7.$
So $\left[2/7,1\right]$ is solution of the system. Finally, if $x\le -2/3$ then
$|3x+2|\ge 4|x-1|\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}-3x-2\ge 4-4x\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x\ge 6,$
which is impossible. That is, the solution set of the first inequality is $\left[2/7,6\right].$ (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
$\frac{\left(x-1\right)\left(x+2\right)}{2\left(x-2\right)\left(x+1/2\right)}\ge 0.$
Study the sign on $\left(-\mathrm{\infty },-2\right),$, $\left(-2,-1/2\right),$, $\left(-1/2,1\right),$, $\left(1/2,1\right)$ and $\left(1,\mathrm{\infty }\right).$ You should obtain that the set solution is $\left(-\mathrm{\infty },-2\right]\cup \left(-1/2,1\right]\cup \left(2,\mathrm{\infty }\right).$ (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 $\left(2/7,1\right]\cup \left(2,6\right].$

April Bush

Expert