Answer:
k = 3 or k = -4
(Anyone can correct me if I'm wrong)
Step-by-step explanation:
[tex]2x=ky+2\to\ eq1\\(k+1)x=6y-3\to\ eq2\\$Change to this form: $y=mx+c\\ky=2x-2\\y=\frac{2x-2}{k}\\$gradient of eq1: $\frac{2}{k}\\6y=(k+1)x+3\\y=\frac{(k+1)x+3}{6}\\y=\frac{k+1}{6} x+\frac{3}{6}\\y= \frac{k+1}{6} x+\frac{1}{2}\\$gradient of eq1: $\frac{k+1}{6}\\$Since gradient, m, is the same for both lines,$\\\frac{2}{k}=\frac{k+1}{6}\\12=k^{2}+k\\k^{2}+k-12=0\\(k-3)(k+4)=0\\k=3 $ or $ k=-4[/tex]
