Algebra with sets 2

If $A\cap C = \{1,5\}$ then $5\in C$ , so $a-b=5$ . Therefore $\sum\limits_{k=1}^i (a_k-b_k) = 5i$ , where $i$ is the number of solutions.

And if $A\setminus B = \{1,5\}$ then $4\in B$ , so $a^2+b=4$ . Substituting the first equation into the second give $a^2+a-9=0$ . Since $1+36>0$ , there are two solutions and we arrive at a final answer of $5\times 2 = \boxed{10}$ .