Find all values of k, such that the system x + 3y = kx, 3x + y = ky has a solution other than (x,y) = (0,0).

Guest Feb 16, 2020

#1**+1 **

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.

Guest Feb 16, 2020

#2**+1 **

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

Guest Feb 16, 2020