Let $t$ be TNYWR.

If $n=\frac{\left(4 t^{2}-10 t-2\right)-3\left(t^{2}-t+3\right)+\left(t^{2}+5 t-1\right)}{(t+7)+(t-13)}$, what is the value of $n ?$
Simplifying,

$$
\begin{aligned}
n & =\frac{\left(4 t^{2}-10 t-2\right)-3\left(t^{2}-t+3\right)+\left(t^{2}+5 t-1\right)}{(t+7)+(t-13)} \\
& =\frac{4 t^{2}-10 t-2-3 t^{2}+3 t-9+t^{2}+5 t-1}{2 t-6} \\
& =\frac{2 t^{2}-2 t-12}{2 t-6} \\
& =\frac{t^{2}-t-6}{t-3} \\
& =\frac{(t-3)(t+2)}{t-3} \\
& =t+2
\end{aligned}
$$

assuming that $t \neq 3$.

Since the answer to (a) is 10 , then $t=10$, and so $n=t+2=12$.