Find all values of k, such that the system x + 3y = kx, 3x + y = ky has a solution other than (x,y) = (0,0).
For k = 4, all values of x and y such that x = y work. For all other values of k, the only solution is x = 0 and y = 0, so the unique value of k is 4.
x + 3y = kx, 3x + y = ky
Solve for x:
3x + y = ky
3x = ky - y
x = ky/3 - y/3
Subsitute into other equation:
x + 3y = kx
ky/3 - y/3 + 3y = kky/3 - ky/3
y(k/3 - 1/3 + 3) = y (kk/3 - k/3)
k/3 + 3 - 1/3 = kk/3 - k/3
k + 9 - 1 = kk - k
k + 8 = kk - k
8 = kk - 2k
k^2 - 2k - 8 = 0
Use Quadratic Formula.
x = 4, -2
