Understanding the last step in solving a linear congruence?

Question

So I want to solve the linear congruence $$17x \equiv 3\mod 29$$

The inverse of $17 (mod 29)$ is 12, from here on, I have no clue how to solve for $x$.

I get the results

$$12*17 \equiv 1\mod29$$ $$12*17x \equiv 36\mod29$$

But I dont understand how to solve of x, could someone please explain?

Answer 1

$12\times 17x \equiv 36\mod29$

$1\times x \equiv 36\mod29$

$x \equiv 36 \equiv7\mod29$

So, the inverse is used to get rid of the number, in front of $x$. Then, the new number ($36$) at the right-hand side is calculated, modulo $29$

Answer 2

We use here one of the arithmetic rules of "$\equiv$", namely: $$ a\equiv b\,(\mathrm{mod}\ m) \Rightarrow ac\equiv bc\,(\mathrm{mod}\ m)\qquad(\ast) $$ for all $c$.

Therefore: if $17x\equiv 3\,(\mathrm{mod}\ 29)$, then by $(\ast)$ $$ 12\cdot 17x\equiv 12\cdot3\,(\mathrm{mod}\ 29). $$ But $17\cdot 12\equiv 1\,(\mathrm{mod}\ 29)$, so by again by $(\ast)$ $17\cdot 12\cdot x\equiv 1\cdot x\,(\mathrm{mod}\ 29)$. Thus $$ 1\cdot x\equiv 12\cdot3\,(\mathrm{mod}\ 29), $$ that is $x\equiv 36\,(\mathrm{mod}\ 29)$. Since $36\equiv 7\,(\mathrm{mod}\ 29)$, it follows that your solution is $x\equiv 7\,(\mathrm{mod}\ 29)$.