How does this technique for solving simultaneous congruences work? 
Find $x\in \Bbb Z$ with

*

*$x\equiv 3 \mod 7$

*$x\equiv 9 \mod 11$

*$x\equiv 1 \mod 5$

So here's what I do: I first find $r_1\in \Bbb Z$ with $r_1\equiv 1 \mod 7$ and $r_1\equiv 0 \mod(11\cdot 5)$, e.g. $r_1=330$.
Then I find $r_2\in \Bbb Z$ with $r_2\equiv 1\mod 11$ and $r_2\equiv 0 \mod(7\cdot 5)$, e.g. $r_2=175$.
Then I find $r_3\in \Bbb Z$ with $r_3\equiv 1\mod 5$ and $r_3\equiv 0 \mod (7\cdot 11)$, e.g. $r_3=154$.
Now if we set $x=3\cdot r_1+9\cdot r_2+1\cdot r_3$ this does the job, right?
It kind of makes sense that this would work but can someone give me a better reason than 'kind of makes sense' ?
 A: Sure: What's the remainder of $x$ on division by $7$? Well, it's the sum of the remainders of the three terms, mod 7. The remainder of the first, mod 7, is $3 \cdot 1$. The remainder of the second is $0$; so is the remainder of the third. So the total remainder, mod 7, is just 3. The same argument applies to the other two remainders. 
There's one subtle point:
Why, knowing that $r_2$ is $0$ mod $7 \cdot 5$, do I know that it's zero $\bmod 7$? Straight from the definitions: Because if $7\cdot 5$ divides evenly into $r_2$ (say $r_2 = (7 \cdot 5) k$), then $7$ also does, since $r_2 = 7 \cdot (5 \cdot k)$.  (Thanks to @Bill Dubuque for straightening me out on this!)
A: Hint $\ $ Those elements span the set of all values $\,(a_1,a_2,a_3)\, \pmod{7,11,5},\,$ i.e.
$$\begin{align} r_1 \equiv (1,0,0)\pmod{7,11,5}\\
r_2 \equiv (0,1,0)\pmod{7,11,5}\\
r_3 \equiv (0,0,1)\pmod{7,11,5}
\end{align}$$
$$\begin{align}\Rightarrow\quad\ &\,\ \ a_1 r_1 + a_2 r_2 + a_3r_3\\ 
\equiv&\,\ (a_1,0,0)+(0,a_2,0)+(0,0,a_3)\\ \equiv&\,\ (a_1,a_2,a_3)\end{align}$$
This will become clearer when you learn the ring-theoretic view of CRT, which here says that $\,\Bbb Z/(7\cdot 11\cdot 5)\,\cong\,  \Bbb Z/7 \times \Bbb Z/11\times \Bbb Z/5,\,$  where $\,\Bbb Z/n = $ integers mod $\,n.\,$  Informally this means these triples $\,(a_1,a_2,a_3)\,$ with the natural induced componentwise addition and multiplication have the same arithmetical (ring-theoretic) structure as the integers mod $\,385= 7\cdot 11\cdot 5.$
For example $\,r_1 + r_2 + r_3 = (1,1,1)\,$ which is the identity element in the product ring, which maps to  $\,330-175-154 = 1\,$ in the integers mod $\,385$ (I corrected the signs on your $\,r_2,r_3)$
This means that we can solve arithmetical problems in the integers mod $385$ by mapping them into a triple of corresponding problems in the smaller rings of integers mod $\,7,11,5,\,$ and then, by CRT, lift that triple of solutions to the corresponding solution in the integers mod $385.\,$ For example, we can solve polynomial equations in this manner. For example see this answer which computes square roots in that manner.
