Solve the system: | 3 x + 2 | ≥ 4 | x − 1...

If $x\ge 1$ we have
$|3x+2|\ge 4|x-1|⟺3x+2\ge 4x-4⟺x\le 6.$
So $[1,6]$ is solution of the first inequality. Now, if $-2/3\le x\le 1$ we have
$|3x+2|\ge 4|x-1|⟺3x+2\ge 4-4x⟺x\ge 2/7.$
So $[2/7,1]$ is solution of the system. Finally, if $x\le -2/3$ then
$|3x+2|\ge 4|x-1|⟺-3x-2\ge 4-4x⟺x\ge 6,$
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
${\displaystyle \frac{(x-1)(x+2)}{2(x-2)(x+1/2)}}\ge 0.$
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 $2+3x-2x^2=0$ 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 $(2/7,1]\cup (2,6].$

Thanks, I appreciate your help!