Question

Just need help here pls

Answer 1

Answer:

$k=6, k=-1$

Step-by-step explanation:

                         $\dfrac{k+3}{3}-\dfrac{2}{k-5}=1$

$\implies \dfrac{(k+3)(k-5)}{3(k-5)}-\dfrac{3 \cdot 2}{3(k-5)}=1$

$\implies \dfrac{k^2-2k-15-6}{3(k-5)}=1$

$\implies k^2-2k-21=3k-15$

$\implies k^2-5k-6=0$

$\implies (k-6)(k+1)=0$

$\implies k=6, k=-1$