Solving $3x\equiv 4\pmod 7$ I'm trying to learn about linear congruences of the form ax = b(mod m). In my book, it's written that if $\gcd(a, m) = 1$ then there must exist an integer $a'$ which is an inverse of $a \pmod{m}$. I'm trying to solve this example:

$$3x \equiv 4 \pmod 7$$

First I noticed $\gcd(3, 7) = 1$.
Therefore, there must exist an integer which is the multiplicative inverse of $3 \pmod 7$.
According to Bezout's Theorem, if $\gcd(a, m) = 1$ then there are integers $s$ and $t$ such that $sa+tm=1$
where $s$ is the multiplicative inverse of $a\pmod{m}$.
Using that theorem:

$\begin{align}7 = 3\cdot2 +1\\7 - 3\cdot2 = 1 \\-2\cdot3 + 7 = 1\end{align}$

$s=-2$ in the above equation so $-2$ is the inverse of $3 \pmod{7}$.
The book says that the next step to solve $3x \equiv 4 \pmod{7}$ is to multiply $-2$ on both sides.
By doing that I get:

$\begin{align}-2\cdot3x \equiv -2\cdot4 \pmod 7\\-6x\equiv -8 \pmod 7\end{align}$

What should I do after that?
I am working on this problem for hours.
Thanks :)

You have arrived at
$$-6x=-8\pmod{7}.$$
Now:
$$-6x=-8\pmod{7} \underbrace{\iff}_{\mathrm{add}\: 7x=0\pmod{7}} 7x-6x=-8\pmod{7}\\ \iff x=-8\pmod{7}\underbrace{=}_{\mathrm{add}\: 14=0\pmod{7}}(2\cdot 7-8)\pmod{7}=6\pmod{7}.$$
