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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.

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$$\frac{x-1}{x+3}>\frac{x}{x-2}$$ $$\frac{x-1}{x+3}-\frac{x}{x-2}>0$$ $$ \frac{(x-1)(x-2)}{(x+3)(x-2)}-\frac{x(x+3)}{(x-2)(x+3)}>0 $$ $$ \frac{(x-1)(x-2)-x(x+3)}{(x-2)(x+3)}>0 $$ $$ \frac{(x^2 -3x +2)-(x^2+3x)}{(x-2)(x+3)}>0 $$ $$ \frac{-6x +2}{(x-2)(x+3)}>0 $$ $$ \frac{x -(1/3)}{(x-2)(x+3)}<0 $$ 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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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.

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