TITLE: Stuck : Using inverses to solve linear congruences?
QUESTION [2 upvotes]: Question : What are the solutions of the linear congruence 3x ≡ 4 (mod 7)?
Step 1 - We know that −2 is an inverse of 3 modulo 7. 
Step 2 - Multiplying both sides of the congruence by −2 shows that
−2·3x ≡−2·4(mod7).
Step 3 - Because 
−6 ≡ 1 (mod 7) - Equation 1
and −8 ≡ 6 (mod 7) - Equation 2
it follows that if x is a solution, 
then x ≡ −8 ≡ 6 (mod 7).

In Step 3, 
I am unable to understand how Equation 1 and Equation 2 lead to the statement 
x ≡ −8 ≡ 6 (mod 7).
Here are the conclusions I was able to derive from these facts,
-6 mod 7 = 1 mod 7
-8 mod 7 = 6 mod 7
(-1) is the inverse of 6 modulo 7
It'd be great if you can help me figure out what other conclusion I'm missing.

REPLY [1 votes]: $$\begin{eqnarray} \overbrace{(-2)\,3}^{\large{\equiv -6\equiv\color{#c00}{\bf 1}}}x &\equiv& \overbrace{(-2)\,4}^{\large\equiv -8\equiv\color{#0a0}{\bf  -1}}\pmod 7\\
\Rightarrow\quad\ \ \color{#c00}{\bf 1}\cdot x&\equiv& \color{#0a0}{\bf -1}\equiv 6\ \,  \pmod 7   
\end{eqnarray}$$
Above we applied the following fundamental
Congruence Product Rule $\rm\quad\ \,  A\!\equiv a\ \Rightarrow\ Ab\equiv ab\   \pmod{\!n}\ \  \ $ 
when we made the inference  $\ (-2)3 \equiv 1\,\Rightarrow\, (-2)3x \equiv 1\cdot x$
and also we used that congruence is an equivalence relation.
{"set_name": "stack_exchange", "score": 2, "question_id": 1776602}