Solve the inequality $\frac{3-x}{x^2-2x-3}\le\frac{3-x}{x^2+2x-3}$ Solve the inequality $$\dfrac{3-x}{x^2-2x-3}\le\dfrac{3-x}{x^2+2x-3}$$
We have $D: \begin{cases}x^2-2x-3\ne0\Rightarrow x\ne -1;3 \\ x^2+2x-3\ne0\Rightarrow x\ne-3;1\end{cases}$
Is the given equality equivalent (in $D$) to $$x^2-2x-3\ge x^2+2x-3\\\iff x\le0$$ So the solutions are $x\le0\cap D$? It seems like the inequalities I wrote aren't equivalent, because I don't get the answer.

hint
For $ x\notin\{-1,1,-3,3\}$, the inequation is equivalent to
$$(3-x)\Bigl(\frac{1}{(x-1)(x+3)}-\frac{1}{(x+1)(x-3)}\Bigr)\ge 0$$
$$\iff x(x-1)(x+1)(x+3)\ge 0$$
You will find that the solution is
$$S=(-\infty,-3)\cup(-1,0]\cup(1,3)\cup(3,+\infty)$$
