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bulgar [2K]
3 years ago
7

Just need help here pls

Mathematics
1 answer:
julia-pushkina [17]3 years ago
7 0

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

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