Find all real constants k such that the system
3x+y=ky has a solution other than (x,y)=(0,0)
oops this wasn't supposed to be off topic
If you take x = y = 1, then you get 4 = k, so k = 4 is the only solution.
Hmm! What about k = -2?
Rearrange to get
3y = (k-1)x and
(k-1)y = 3x
Divide one by the other, the x and y cancel, and you will be left with a quadratic in k.