Quotients of polynomial rings by way of evaluation $R[x]/(x-a,f(x),g(x)\ldots) \cong R/(f(a),g(a),\ldots)$ 
Exercise III.4.11 (Aluffi): Let $R$ be a commutative ring, $a \in R$ and $f_1(x),...,f_r(x) \in R[x]$.
  
  
*
  
*Prove the equality of ideals
  \begin{equation}
(f_1(x),...,f_r(x),x-a) = (f_1(a),...,f_r(a),x-a).
\end{equation}
  
*Prove the useful substitution trick
  \begin{equation}
\frac{R[x]}{(f_1(x),...,f_r(x),x-a)} \cong \frac{R}{(f_1(a),...,f_r(a))}.
\end{equation}

So here is my current logic, I'm having trouble formulating my ideas/want to make sure I'm going down the correct path.
For the first equality it is clear that 
\begin{equation}
(f_1(x),...,f_r(x), x-a) = (f_1(x)) + \dots (f_r(x)) + (x-a)
\end{equation}
and 
\begin{equation}
(f_1(a),...,f_r(a),x-a) = (f_1(a)) + \dots (f_r(a)) + (x-a).
\end{equation}
Thus, my "modding out" by the ideal $(x-a)$ it would follow that equality holds correct? As all that is left are the polynomials generated by $f_1(x),...,f_r(x)$ where $f(a) \neq 0$. Is this logic correct?
Second, assuming the first equality holds. Recall that 
\begin{equation}
\frac{R[x]}{(x-a)} \cong R.
\end{equation} 
Thus I feel like the substitution trick should follow quickly after consideration of this fact as well as the first equality. However, I'm a little lost with regards to formulating a concrete proof. 
Any hints/help will be greatly appreciated!
 A: Here are some hints:
For the first equality, you're on the right track! Recall the correspondence theorem for Rings (page 142 of Aluffi). In particular, for ideals $I$ and $J$ containing $\mathfrak{a}$, we have $I=J$ if and only if $I / \mathfrak{a} = J / \mathfrak{a}$ in $R/\mathfrak{a}$.
Now if we consider $I = \langle f_i(x) \rangle$, $J = \langle f_i(a) \rangle$, and $\mathfrak{a} = \langle x-a \rangle$, we will have
$$
\langle f_i(x), x-a \rangle = \langle f_i(a), x-a \rangle 
\iff 
I + \mathfrak{a} = J + \mathfrak{a}
\iff
I/\mathfrak{a} = J/\mathfrak{a}
$$
Of course, as you have noted, it is clear that $I/\mathfrak{a} = J/\mathfrak{a}$ since $x=a$ in the quotient ring, so $f_i(x) = f_i(a)$.
For the second equality, you are also on the right track. Notice:
$$
R[x]/(I + \mathfrak{a})
\cong 
(R[x]/\mathfrak{a})/(I/\mathfrak{a})
\cong
(R[x]/\mathfrak{a})/(J/\mathfrak{a})
\cong
R/(J/\mathfrak{a})
$$
Of course, it is easily checked that $J/\mathfrak{a} = \langle f_i(a) \rangle$.
Feel free to comment if there's anything you want me to clarify!

I hope this helps ^_^
