Stuck : Using inverses to solve linear congruences? 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.
 A: $x\equiv-8\pmod7$ comes from Step 2.
A: $-6$ and $1$ are in the same equivalence-class in $\mathbb{Z}_7$. In modulo arithmetic in the ring $\mathbb{Z}/n\mathbb{Z}$ (Z modulo nZ)  you can add the modulus to any number and the numbers are still congruent to each other since they have the same residue when divided by $n$ because we've only added multiples of $n$ to the number. Thus,
$$-6 \equiv -6 + 7 \equiv 1 \mod 7$$ Also then,
$$-8 \equiv -8 + 7 \equiv -1 + 7 \equiv 6 \mod 7$$
I find this to be a very weird way to solve the eqation
$$3x \equiv 4 \mod 7$$
We just have to find the inverse of $3$ in $\mathbb{Z}_7$, and since $7$ is prime, we know that $3^{-1} \mod 7$ exists. We would then use the extended eucledian algorithm (EEA) to compute that inverse.
\begin{aligned}
\text{I} \quad 7 &\quad 0 && \\
\text{II} \quad 3 &\quad 1  &&\\
\text{I-2II} \quad1 &\quad 0-2\cdot 1 = -2 \equiv 5 \mod 7 
\end{aligned}
This we have 
$$3\cdot5 \equiv 1 \mod 7$$, thus $$3^{-1} \equiv 5 \mod 7$$. Multiplying both sides with the multiplicative inverse gives
$$3x \equiv 4 \mod 7 \quad |\cdot 3^{-1}$$
$$x \equiv 4\cdot 5\mod 7 $$
$$x \equiv 20 \mod 7$$
Now just reduce 20 to the smallest residue modulo 7 (subtract 7 two times) and we get 
$$x \equiv 6 \mod 7$$
A: $$\begin{eqnarray} \overbrace{(-2)\,3}^{\large{\equiv -6\equiv\color{#c00}{\bf 1}}}x &\equiv& \overbrace{(-2)\,4}^{\large\equiv -8\equiv\color{#0a0}{\bf  -1}}\pmod 7\\
\Rightarrow\quad\ \ \color{#c00}{\bf 1}\cdot x&\equiv& \color{#0a0}{\bf -1}\equiv 6\ \,  \pmod 7   
\end{eqnarray}$$
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.
A: In step 2, you have 
$$-2 \cdot 3x \equiv -2 \cdot 4 \pmod{7}$$
Since $-2 \cdot 3 = -6$ and $-2 \cdot 4 = -8$, we have 
$$-6x \equiv -8 \pmod{7}$$
Since $-6 \equiv 1 \pmod{7}$ and $-8 \equiv 6 \pmod{7}$, we may substitute $1$ for $-6$ and $6$ for $-8$ in the equivalence $-6x \equiv -8 \pmod{7}$ to obtain
$$x \equiv 6 \pmod{7}$$
A: From equation 1,
$$\begin{align*}
-6&\equiv 1&\pmod7\\
-6x&\equiv x&\pmod 7
\end{align*}$$
From step 2,
$$\begin{align*}
-2\cdot3x&\equiv -2\cdot4&\pmod7\\
-6x&\equiv -8&\pmod 7
\end{align*}$$
Combining with equation 2: $-8\equiv6\pmod7$, all of the below four are equivalent:
$$x\equiv-6x\equiv-8\equiv 6\pmod 7$$
