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.

-

$$\frac{x-1}{x+3}>\frac{x}{x-2}$$ $$\frac{x-1}{x+3}-\frac{x}{x-2}>0$$ $$\frac{(x-1)(x-2)}{(x+3)(x-2)}-\frac{x(x+3)}{(x-2)(x+3)}>0$$ $$\frac{(x-1)(x-2)-x(x+3)}{(x-2)(x+3)}>0$$ $$\frac{(x^2 -3x +2)-(x^2+3x)}{(x-2)(x+3)}>0$$ $$\frac{-6x +2}{(x-2)(x+3)}>0$$ $$\frac{x -(1/3)}{(x-2)(x+3)}<0$$ 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.
You can multiply by $(x+3)(x-2)$ to clear the fractions. You need to split into cases, as when $-3 \lt x \lt 2$ you need to reverse the sign of the inequality as you have multiplied by a negative number. The cubic terms will cancel, leaving you with quadratics.