+566 Find all real constants such that the system
\(x+3y=kx,\)
\(3x+y=ky\)
has a solution other than (x, y) = (0, 0)
+36916 From first equation x = -3y/(1-k) sub this in to the second equation
3 (-3y)/(1-k) + y(1-k) = 0
-9y + y (1-k)^2 = 0
y ( k^2-2k-8) = 0 already stated y cannot be 0
(k+2)(k-4) = 0 k = -2, 4 (if I did the math correctly !) 
