Solving system of congurences with the Chinese Remainder Theorem 
Solve the system of congruences
  \begin{cases} x \equiv 1\ (\textrm{mod}\ 3) \\ x \equiv 4\ (\textrm{mod}\ 5) \\ x \equiv 6\ (\textrm{mod}\ 7)\end{cases}

I'm trying to learn about the Chinese Remainder Theorem and tried some problems as this.
I started with $x \equiv 6\ (\textrm{mod}\ 7)$ implying that $x=7k+6$ for some $k$. Then substituting this for $x \equiv 4\ (\textrm{mod}\ 5)$ I would get $7k+6 \equiv 4\ (\textrm{mod}\ 5)$. However here I got stuck, the proposed solution stated that I would have to solve
$$7k+6 \equiv 4\ (\textrm{mod}\ 5)$$
for $k$ and that it would result in $k\equiv 4\ (\textrm{mod}\ 5).$ I don't see how this would be  possible. Solving $7k+6 \equiv 4\ (\textrm{mod}\ 5)$ for $k$ would result in $k\equiv \frac{-2}{7}\ (\textrm{mod}\ 5)$?
 A: I like to use Bezout coefficients and isomorphisms as in the Chinese remainder theorem.
$-3\cdot3+2\cdot5=1$.  Thus for the first two we get $x\cong -9\cdot4+10\cdot1\cong{-26}\cong4\pmod{15}$.
Then $1\cdot15-2\cdot7=1$.
So $x\cong15\cdot6-14\cdot4\cong34\pmod{105}$.
A: Yes it results , and $\ k\equiv \frac{-2}{7}\ \equiv \frac{-2}{7-5}\ =-1 \equiv 4\ (\textrm{mod}\ 5)$ 
So, $$x \equiv 34\ (\textrm{mod}\ 35)$$
Also, $$x \equiv 1\ \equiv 34\ (\textrm{mod}\ 3)$$
Hence, $$x \equiv 34\ (\textrm{mod}\ 105)$$
A: If you're not a fan of substituting in modular arithmetic, there is an explicit way of solving these kinds of problems, which goes like this: given the system
$$\begin{cases} 
x \equiv a_1\ (\textrm{mod}\ m_1) \\ 
\quad \vdots \\ 
x \equiv a_r\ (\textrm{mod}\ m_r)
\end{cases}$$
Define the full modulus $M=\prod^{r}_{i=1} m_i$ and the reduced modulus $M_i=M/m_i$, then the solution is
$$x=\sum^r_{i=1}a_iM_iN_i\qquad(\!\!\!\!\!\mod\!\!M)$$
where $N_iM_i=1\;(\!\!\!\mod m_i)$ $-$ or, in plain English, the $N_i$ are the inverses to the reduced moduli $M_i$ in modulo $m_i$, which you can find either by trial-and-error or by using the Euclidean algorithm.
This shifts the weight from solving modular equations to calculating a few products, using the Euclidean algorithm $r$ times, and doing some addition at the end.
