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**Prove It** First thing you should note is that $\displaystyle \displaystyle x \neq 3$ because that will give a $\displaystyle \displaystyle 0$ denominator.

$\displaystyle \displaystyle \frac{x}{3-x} < 2$

$\displaystyle \displaystyle \frac{x - 3 + 3}{3-x} < 2$

$\displaystyle \displaystyle \frac{3}{3-x} - \frac{3-x}{3-x} < 2$

$\displaystyle \displaystyle \frac{3}{3-x} - 1 < 2$

$\displaystyle \displaystyle \frac{3}{3-x} < 3$

$\displaystyle \displaystyle \frac{1}{3-x} < 1$.

In order for this inequality to hold, we require that the denominator be larger than the numerator, or for the denominator to be negative.

Case 1: $\displaystyle \displaystyle 3 - x > 1$

$\displaystyle \displaystyle 3 > x + 1$

$\displaystyle \displaystyle 2 > x$.

Case 2: $\displaystyle \displaystyle 3 - x < 0$

$\displaystyle \displaystyle 3 < x$.

So the solution is $\displaystyle \displaystyle x \in (-\infty, 2) \cup (3, \infty)$.