x is divided by 3 and 4 implies its also divided by 6 I want to show that if $[3$ divides $x$ and $4$ divides $x]$ then $[6$ divides $x]$.

I guess my starting point is something like
$x=3n$
$x=4m$

But how do i show that there is also a $c$ with
$x=6c$?

Is this the correct approach?
You can do it like this, which is generalisable:

$x=3n$ implies $4x=12n$

$x=4m$ implies $3x=12m$

Subtract to obtain $x=12(n-m)$



If $p$ and $q$ are both factors of $x$ we can write:

$x=pm=qn$

Now suppose that $p$ and $q$ are coprime. There is a theorem that we can find integers $a, b$ such that $ap+bq=1$.

We multiply this by $x$ to obtain $apx+bqx=x$ and we substitute carefully for $x$ on the left-hand side so that $$apqn+bpqm=pq(an+bm)=x$$And we see that $x$ is divisible by $pq$. You don't need to be able to prove the theorem to use it in a particular case - it tells you that if you look carefully to will find the $a$ and $b$ you need.
Since $3$ and $4$ divide $x$, $x$ must therefore equal $4 * 3 * n$ = $2*2*3*n$ for some $n$ (since 3 and 4 are relatively prime). Taking a new integer $m = 2n$, it is easy to see that $x = 2*3*m = 6*m$ for the integer $m$. Therefore, 6 divides $x$.