Here is my first logical steps to solve 10x ≡ 6mod7
1. 20mod7 ≡ 6mod7, which evaluate to 6
2. 10 $\times$ 2 = 20
3. Therefore, multiply both sides of congruence by 2.
After these steps I have: 20x ≡ 12mod7
My questions is: how could this evaluate to x=2. What are the next steps from here. Any help is appreciated.
So we have $20x \equiv 12 \ (mod\ 7)$. We can reduce this to $$6x \equiv 5 \ (mod\ 7)\ (*)$$ Note that $gcd(6, 7)=1$ and $1|5$, so a solution does exist. Here's one way you can solve this:
Mod the left-hand side by $7$: $6x \equiv 5 \ (mod\ 7) \Rightarrow -x \equiv 5\ (mod\ 7)$.
Multiply by $-1$: $$(-1)(-x) \equiv (-1)5\ (mod\ 7) \Rightarrow x \equiv -5 \equiv 2\ (mod\ 7). $$
Thus, $x \equiv 2\ (mod\ 7)$.
Check: $10x \equiv 6\ (mod\ 7) \Rightarrow 10(2) =20 \equiv 6\ (mod\ 7)$. It works!