Answer
The denominator in the fraction limits part of the solution set. However, since we have a greater than sign, we want only those solutions that make the two functions true.
Work Step by Step
The second function needs values of $x$ greater than 3 to be true. (Otherwise, the function is zero or negative). Likewise, the fraction needs $x > 3$ for the denominator to be positive. When $x>3$, the numerator will be positive, and the expression will be true.
