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**HallsofIvy** The way I would have done it is this: $\displaystyle \frac{1}{x-3}> 4 $ is the same as $\displaystyle 0< 4- \frac{1}{x- 3}= \frac{4x- 13}{x- 3}$. Now, that can change from "> 0" to "< 0" only where the numerator, 4x- 13, is equal to 0 (so the fraction is 0) or where x- 3= 0 (where the fraction is "discontinuous"). The sign is constant on the intervals $\displaystyle (-\infty, 3)$, $\displaystyle (3, 13/4)$, and $\displaystyle (13/4, \infty)$. If x= 0, which is less than 3, $\displaystyle \frac{1}{0-3}= -\frac{1}{3}< 4$. If x= 4, which is larger than 13/4, $\displaystyle \frac{1}{4-3}= 1< 4$. If x= 3.1, which is between 3 and $\displaystyle \frac{13}{4}= 3\frac{1}{4}$, $\displaystyle \frac{1}{3.1- 3}= \frac{1}{0.1}= 10> 4$.

Therefore, the inequality is true on the interval $\displaystyle \left(3, \frac{13}{4}\right)$.