<s>[INST] Solve $\dfrac{x}{x-2}>2$ by first rewriting it in the form $\dfrac{P(x)}{Q(x)}>0$ Edit: So then is this the correct final solution? $x<4,(\infty,4), x\ne2$
I am asked to do this:
Solve $\dfrac{x}{x-2}>2$ by first rewriting it in the form $\dfrac{P(x)}{Q(x)}>0$
$$\dfrac{x}{x-2}>2$$
$$\dfrac{x}{x-2}-\dfrac{2(x-2)}{1(x-2)}>0$$
$$\dfrac{x-2x+4}{x-2}>0$$
$$\dfrac{-1(-x+4)}{x-2}>0$$
$$\dfrac{x-4}{x-2}<0(x-2)$$
$$x-4<0$$
$$x<4$$
 [/INST] Your method not is correct because: if
$$\frac{4-x}{x-2}>0\equiv-1\cdot\frac{(4-x)}{x-2}<(-1)\cdot0\equiv\frac{x-4}{x-2}<0$$
then $(x-4)>0$ and $(x-2)<0$, or $(x-4)<0$ and $(x-2)>0$
solutions $2<x<4$
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