matxxpshsn matxxpshsn
  • 13-03-2022
  • Mathematics
contestada

Just need help here pls

Just need help here pls class=

Respuesta :

semsee45
semsee45 semsee45
  • 13-03-2022

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]

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