[tex](k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12[/tex]
So in order to get all the fractions to have a common denominator, we need to multiply [tex]\dfrac k{k-2}[/tex] by [tex]\dfrac{k-6}{k-6}[/tex], and [tex]\dfrac1{k-6}[/tex] by [tex]\dfrac{k-2}{k-2}[/tex]:
[tex]\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}[/tex]
[tex]\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}[/tex]
Now,
[tex]\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}[/tex]
As long as [tex]k\neq2[/tex] and [tex]k\neq6[/tex] (which we can't have because otherwise [tex]k^2-8k+12=0[/tex]), we can cancel [tex]k^2-8k+12[/tex] in the denominators on both sides:
[tex]4=(k^2-6k)+(k-2)[/tex]
[tex]4=k^2-5k-2[/tex]
[tex]0=k^2-5k-6[/tex]
We can factorize the right side:
[tex]0=(k-6)(k+1)[/tex]
which tells us that [tex]k=6[/tex] and [tex]k=-1[/tex] are solutions.
