2 variables, 1 equation?

$kx-12=3k \\ kx=3k+12 \\ x=\dfrac{3k+12}{k} \\ x=\dfrac{12}{k}+3$

For $x$ to be an integer , $k \ | \ 12$ .

Such positive $k$ possible are: $1,2,3,4,6,12$ which are $\boxed{6}$ in number.

$kx-12=3k$ $k(x-3)=12$

Since, $x$ needs to be an integer and so does $k$ , we can say that $k$ is a factor of 12. Thus, $k=1,2,3,4,6,12$ . Hence, required solution is $\boxed{6}$ .

Rearranging the Equation, We get:-

$kx-3k=12$

$\rightarrow k(x-3)=12$

$k= \dfrac {12}{(x-3)}$

For k to have an integer value, x-3 must be a factor of 12.

Now, $12= {2}^{2} \times 3$

Therefore, 12 has $(2+1)(1+1)= 6$ factors.

Hence, For 6 values of k, x has an integral value.

Answer:- 6

Cheers!