A man walked a certain distance at a constant rate. If he had gone $\textstyle\frac{1}{2}$ mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone $\textstyle\frac{1}{2}$ mile per hour slower, he would have been $2\textstyle\frac{1}{2}$ hours longer on the road. The distance in miles he walked was
$\textbf{(A) }13\textstyle\frac{1}{2}\qquad \textbf{(B) }15\qquad \textbf{(C) }17\frac{1}{2}\qquad \textbf{(D) }20\qquad  \textbf{(E) }25$
We can make three equations out of the information, and since the distances are the same, we can equate these equations.
\[\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{5}{2})(x-\frac{1}{2})\]
where $x$ is the man's rate and $t$ is the time it takes him.  
Looking at the first two parts of the equations, 
\[\frac{4t}{5}(x+\frac{1}{2})=xt\]
we note that we can solve for $x$. Solving for $x$, we get $x=2.$
Now we look at the last two parts of the equation:
\[xt=(t+\frac{5}{2})(x-\frac{1}{2})\]
we note that we can solve for $t$ and we get $t=\frac{15}{2}.$  We want the find the distance, which is $xt= \boxed{15}.$
-edited for readability