Systems of Absolute Inequalities Answer Format

Question

I was doing some online questions and ran into this problem: (the following is a system)

$|3-2x|\ge1$

$x^2/(x+2)\le0$

The computer wouldn't accept the answer $x<-2$ nor would it accept the answer $x<-2, x \neq 2$.

Please advise on this with the answer to this.

Answer 1

$x=0$ is also a possible solution, the solution set is therefore $x<-2 \text{ or } x=0$.

Here's the proof:

We first focus on the second inequality: $\frac{x^2}{x+2}\leq 0$ implies that either the nominator or the denominator are negative (not both). Or that the nominator is equal to 0. The denominator is never negative. Hence we are left with $x=0$ (nominator is zero). Or $x<-2$ (denominator negative). We can plug in all these into the first inequality and see that they all hold.