Find all real constants k such that the system

x+3y=kx

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.

Exactly right!