Question: I'm studying Boolean algebra: I know this is a straightforward answer.
I'm a bit confused about if $B'C + BC'$ would be opposites in this case and would follow the rule that $A + A' = 1$ Or how can that simplify?

Thanks so much!

Answer: 
There is no way to reduce $B'C + BC'$. However, you can see that $B'C + BC'$ is *not* equal to $1$. Try drawing a truth table for all values of $B$ and $C$.

> 
> $$\begin{array}{|c|c|c|}\hline B & C & B'C & BC' & B'C + BC' \\\hline 0 & 0 & 0 & 0 & 0 \\\hline 0 & 1 & 1 & 0 & 1\\\hline 1 & 0 & 0 & 1 & 1 \\\hline 1 & 1 & 0 & 0 & 0\\\hline\end{array}$$
> Does this remind you of a specific gate?
> 
> 
>
If both $B$ and $C$ are true (or false) then the equation is false.

Indeed recall DeMorgan's law 
\begin{eqnarray\*}
(B^{'} C)^{'} =B + C' \neq BC'.
\end{eqnarray\*}