<s>[INST] Why does the CRT formula yield a solution of a congruence system? I understand there is a method for solving simultaneous modular equations. For example;
$$x = 2 \mod{3}$$
$$x = 3 \mod{5}$$
$$x = 2 \mod{7}$$
We find numbers equal to the product of every given modulo except one of them - giving $5 \cdot 7$, $3 \cdot 7$ and $3 \cdot 5$. We then find the multiplicative inverses of these numbers with modulo equal to the number missing from the product. The numbers found are then 2, 1 and 1 in this case. The value of x is then given by:
$$x = 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1 = 233 = 23 \mod{3\cdot5\cdot7}$$
But I do not understand how this method correctly gives the value of $x$. I understand that the Chinese remainder theorem proves that there is a unique value of $0\le x \lt 3\cdot5\cdot7 \mod{3\cdot5\cdot7}$ but can someone please explain why this method finds this value of x?
 [/INST] Taking Bill Dubuque's graphic answer and graphically expanding on it:
$x = 2 \cdot\overbrace{ (5 \cdot 7) \cdot 2}^{\equiv 1 \pmod 3\\ \equiv 0 \pmod 5\\ \equiv 0 \pmod 7} + 3 \cdot \overbrace{(3 \cdot 7) \cdot 1}^{\equiv 0 \pmod 3\\ \equiv 1 \pmod 5\\ \equiv 0 \pmod 7} + 2 \cdot \overbrace{(3 \cdot 5) \cdot 1}^{\equiv 0 \pmod 3\\ \equiv 0 \pmod 5\\ \equiv 1\pmod 7}\equiv\, \overbrace{2,\,3,\,2\pmod{3,5,7}}^{\equiv 2 + 0 +0\pmod 3\\ \equiv0+3+0 \pmod 5\\ \equiv 0+0+2\pmod 7}$
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Think about what you, yourself just stated.  
If take  this sum $x = 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1$ and $\mod 3$ it, then $(5\cdot 7)$ and $2$ are inverses so $2\cdot[(5\cdot 7)\cdot 2]\pmod 3\equiv 2\cdot 1\pmod 3 \equiv 2 \pmod 3$.  ANd the other terms are multiples of $3$ so they are $\equiv 0 \pmod 3$.  So  $x\equiv 2 \pmod 3$.  
If you take that term $x = 2 \cdot (5 \cdot 7) \cdot 2 + 3 \cdot (3 \cdot 7) \cdot 1 + 2 \cdot (3 \cdot 5) \cdot 1$ and $\mod 5$ it, then $3\cdot 7$ and $1$ are inverses so $3\cdot[(3\cdot 7) \cdot 1] \equiv 3 \cdot 1 \equiv 3 \pmod 5$.  ANd the other terms are multiples of $5$.  So the sum $x \equiv 3 \pmod 5$.
And so on.
....
If you want to solve
$x \equiv a \pmod m$
$x \equiv b \pmod n$
$x \equiv c \pmod v$ then
And assuming you were able find $(nv)^{-1}\mod m; (mv)^{-1}\mod n; $and $(nm)^{-1}\mod v$
Then Let $K = a(nv)^{-1}(nv) + b (mv)^{-1}(mv) + c(nm)^{-1}(nm)$
Note:  $K \pmod m \equiv$
$a(nv)^{-1}(nv) + b (mv)^{-1}(mv) + c(nm)^{-1}(nm)\pmod m\equiv$
$a*1 + [b(mv)^{-1}v]m + [c(nm)^{-1}n]m \pmod m\equiv$
$a*1 + 0 + 0 \equiv a\pmod m$.
And likewise:
$K \pmod n \equiv$
$a(nv)^{-1}(nv) + b (mv)^{-1}(mv) + c(nm)^{-1}(nm)\pmod n\equiv$
$[a(nv)^{-1}v]n + b*1 + [c(nm)^{-1}m]n \pmod n\equiv$
$0 + b*1 + 0 \equiv b\pmod n$.
And
$a(nv)^{-1}(nv) + b (mv)^{-1}(mv) + c(nm)^{-1}(nm)\pmod v\equiv$
$[a(nv)^-1n]v + [b(mv)^{-1}m]v + c*1 \pmod v\equiv$
$0 + 0 + c \equiv c\pmod m$.
So $K$ is A solution.
If $m,n,v$ are pairwise relative prime then $K$ is a unique solution upto $\mod nmv$.
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