Question 803383
 2x-5=ky and (k+1)x=6y-3 have same slopes


 2x-5=ky
y= (2/k)x-5/k



(k+1)x=6y-3

(k+1)x+3 =6y

(k+1)/6 x +1/2 = y


the slopes of both lines are equal

2/k  =(k+1)/6

12=k(k+1)

12=k^2+k

k^2+k-12=0
k^2+4k-3k-12=
k(k+4)-3(k+4)=0
(k+4)(k-3)=0

k=4 OR 3