+0  
 
0
44
2
avatar

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

 Feb 16, 2020
 #1
avatar
+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.

 Feb 16, 2020
 #2
avatar
+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

 Feb 16, 2020

51 Online Users

avatar
avatar