For the quotient to be $≤0$, one of the values $x-3$ and $x+1$ has to be $≥0$, the other has to be $≤0$, and $x+1$ cannot be zero. Consider the case when $x-3\geq 0$ and $x+1$ is negative.
$$x-3\geq 0$$ $$x+1<0$$
This is false for any $x$.
$$x\in \emptyset$$
Consider the case when $x-3\leq 0$ and $x+1$ is positive.
$$x-3\leq 0$$ $$x+1>0$$
The solution satisfying both inequalities is $x\in \left(-1,3\right]$.
$$x\in (-1,3]$$
The final solution is the union of the obtained solutions.
