If $y=a+\frac{b}{x}$, where $a$ and $b$ are constants, and if $y=1$ when $x=-1$, and $y=5$ when $x=-5$, then $a+b$ equals:
$\textbf{(A)}\ -1 \qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$
Plugging in the x and y values, we obtain the following system of equations:
\[\begin {cases} 1 = a - b \\ 5 = a - \frac{b}{5} \end {cases}\]
We can then subtract the equations to obtain the equation $4 = 0.8b$, which works out to $b = 5$.
Plugging that in to the original system of equations:
\[\begin {cases} 1 = a - 5 \\ 5 = a - \frac{5}{5} \end {cases}\]
It is easy to see that $a = 6$, and that $a+b$ is (E) 11.