How come these two inequalities have different solution sets? The following inequality has a solution set of $\{x \in \mathbb{R} | \frac{3}{2} < x < 5, x<\frac{-1}{4}\}$:
$$\frac{2x + 1}{2x - 3}>\frac{x + 1}{x - 5}$$
However this inequality has a different solution set of $\{x \in \mathbb{R} | x > 5, \frac{-1}{4} < x < \frac{3}{2}\}$:
$$\frac{2x + 1}{2x - 3} - \frac{x + 1}{x - 5}>0$$
How come this happens? I am trying to find the values of x that make the first inequality true, and need to do it algebraically so I tried to make it easier by putting everything to one side but I see that there are different solutions now.
 A: The solution is $\{x\in\mathbb{R}\mid x<-\frac{1}{4}\text{ OR }\frac{3}{2} < x < 5\}$. I.e., your first solution is the correct one. Subtracting one term from both sides doesn't change the inequality, so you must have assumed that the sign change works differently. 
(I worked it below)

For $x>5$ or $x < 3/2$, both inequalities reduce to $x<-1/4$. That's because if either of these are true, then $x-5$ and $2x-3$ are both positive or both negative, and you can reduce the inequality algebraically without flipping the inequality.
However, if you have an $x$ between these, then $x-5$ is negative and $2x-3$ is positive, so reducing means that you have to flip the inequality to get $x>-1/4$.
So if $x>5$, you can't have $x<-1/4$, so that's out. If $x<3/2$, then the inequality is only true if $x<-1/4$, which already implies $x<3/2$, so we can just say $x<-1/4$.
$3/2 < x < 5$ implies that $x>-1/4$, so we can just say $3/2 < x < 5$.
A: Let $f(x)=\frac{2x+1}{2x-3}-\frac{x+1}{x-5}=-\frac{8x+2}{(2x-3)(x-5)}$. This is $0$ at $x=-1/4$.
The only places where $f(x)$ could conceivably change sign are at a root of $f(x)=0$, or at a place where $f(x)$ is undefined, in this case at $x=3/2$ and $x=5$.
Now for each of the intervals $(-\infty,-1/4)$, $(-1/4,3/2)$, $(3/2,5)$, and $(5,\infty)$ we can use a convenient  test point to determine the sign of $f(x)$ in that interval. This bypasses the risk of an incorrect handling of the inequalities when clearing denominators.   
