<s>[INST] Solving $\frac{x-1}{x+3}>\frac{x}{x-2}$ I'm having little trouble solving$$\frac{x-1}{x+3}>\frac{x}{x-2}$$
What steps should I take?
Need this to write the topological spaces of the set defined by this inequation.
 [/INST] $$\frac{x-1}{x+3}>\frac{x}{x-2}$$
$$\frac{x-1}{x+3}-\frac{x}{x-2}>0$$
$$
\frac{(x-1)(x-2)}{(x+3)(x-2)}-\frac{x(x+3)}{(x-2)(x+3)}>0
$$
$$
\frac{(x-1)(x-2)-x(x+3)}{(x-2)(x+3)}>0
$$
$$
\frac{(x^2 -3x +2)-(x^2+3x)}{(x-2)(x+3)}>0
$$
$$
\frac{-6x +2}{(x-2)(x+3)}>0
$$
$$
\frac{x -(1/3)}{(x-2)(x+3)}<0
$$
This fraction undergoes sign changes at $-3$, $1/3$, and $2$.  So ascertain in each of the four intervals whether it is positive or negative on that interval.
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