直线y=kx+3与直线y=x/k-5在y=x上相交,求k值

y=x=x/k-5
(1/k-1)x=5
x=5/(1/k-1)=5k/(1-k)
y=x=5k/(1-k)
代入y=kx+3
5k/(1-k)=5k²/(1-k)+3
5k=5k²+3-3k
5k²-8k+3=0
(5k-3)(k-1)=0
k=3/5,k=1