How can I solve the linear congruence $7x+3 \equiv 4x+1 \pmod{10}$?
I got:
$7x+3 \equiv 4x+1 \pmod {10} :\iff 10\mid (3x+2) \implies \exists k\in \mathbb{Z} : 10k-3x=2$
I want to apply Bézout's identity to find $x$, but therefore I have to get rid of the $2$ and instead have a $1$. Can I just divide by $2$?
All the equivalences are mod $10$:
$$3x+2 \equiv 0 \Leftrightarrow 3x \equiv 8 \Leftrightarrow x \equiv 8(3)^{-1} \equiv 56 \equiv 6.$$
$3$ is invertible because it is coprime to $10$ and it has inverse $7$. So any integer $x$ that is equal to $6$ mod $10$ satisfies what you want.