Find all real constants such that the system
x+3y=kx,3x+y=ky.
has a solution other than $(x,y)=(0,0).$
I'm assuming that you mean "all real constants k"
First add the two equations together.
4(x+y)=k(x+y). k = 4
Then subtract the two.
−2(x−y)=k(x−y). k = -2.
So our two solutions are k = 4, -2