Modular equations: where did I make a mistake?

Question

I want to solve the simultaneous congruences $$\begin{cases} 2x \equiv 4 \mod 8 \\ x \equiv 2 \mod 6 \end{cases} $$

My solution: $$2x \equiv 4 \mod 8 \iff x = 4l + 2 $$ $$x \equiv 2 \mod 6 \iff 4l + 2 \equiv 2 \mod 6 \iff 4l = 6k \iff x = 6k + 2 $$ And the correct answer is: $12n + 2$. Where did I made a mistake?

Answer 1

$4l = 6k \iff l = 3n, k = 2n$ so $4l = 6k \iff x = 12n + 2$

Answer 2

$$2x\equiv8\pmod8\iff x\equiv2\pmod4\iff x-2\equiv0$$

and $$x\equiv2\pmod6\iff x-2\equiv0$$

So, lcm$(4,6)\mid(x-2)$

Answer 3

You already know

• $x-2 = 4l$
• $x-2 = 6k$

Therefore, $x-2$ is a multiple of both $4$ and $6$. Then, it must be a multiple of $12$ and we get $x-2 = 12n$.

$$x=12n+2$$