Answer:
[tex]k=6, k=-1[/tex]
Step-by-step explanation:
[tex]\dfrac{k+3}{3}-\dfrac{2}{k-5}=1[/tex]
[tex]\implies \dfrac{(k+3)(k-5)}{3(k-5)}-\dfrac{3 \cdot 2}{3(k-5)}=1[/tex]
[tex]\implies \dfrac{k^2-2k-15-6}{3(k-5)}=1[/tex]
[tex]\implies k^2-2k-21=3k-15[/tex]
[tex]\implies k^2-5k-6=0[/tex]
[tex]\implies (k-6)(k+1)=0[/tex]
[tex]\implies k=6, k=-1[/tex]
