The lines $x=\frac{1}{4}y+a$ and $y=\frac{1}{4}x+b$ intersect at the point $(1,2)$. What is $a+b$?
$\textbf{(A) } 0\qquad \textbf{(B) } \frac{3}{4}\qquad \textbf{(C) } 1\qquad \textbf{(D) } 2\qquad \textbf{(E) } \frac{9}{4}$
Since $(1,2)$ is a solution to both equations, plugging in $x=1$ and $y=2$ will give the values of $a$ and $b$. 
$1 = \frac{1}{4} \cdot 2 + a$
$a = \frac{1}{2}$
$2 = \frac{1}{4} \cdot 1 + b$
$b = \frac{7}{4}$
So $a+b = \frac{1}{2} + \frac{7}{4} = \boxed{\textbf{(E) }\frac{9}{4}}$.