Three $\Delta$'s and a $\diamondsuit$ will balance nine $\bullet$'s.  One $\Delta$ will balance a $\diamondsuit$ and a $\bullet$.

How many $\bullet$'s will balance the two $\diamondsuit$'s in this balance?

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$
For simplicity, suppose $\Delta = a$, $\diamondsuit = b$ and $\bullet = c$.  Then,
\[3a+b=9c\]
\[a=b+c\]
and we want to know what $2b$ is in terms of $c$.  Substituting the second equation into the first, we have 
\[4b=6c\Rightarrow 2b=3c\]
Thus, we need $3$ $\bullet$'s $\rightarrow \boxed{\text{C}}$.