3x + ky = 0
2x + 2y = 0

Solve 

2x + 2y = 0 for y

2x + 2y = 0  divide through by 2

  x + y = 0

      y = -x

Substitute in 

3x + ky = 0
3x + k(-x) = 0
3x - kx = 0

x(3 - k) = 0

Either x = 0 or else k = 3

If k ≠ 3 then x = 0, and y = -x = -(0) = 0, 

and (x,y) = (0,0) is the only (unique) solution.

However if k = 3, then x can be chosen as any number,
and y will equal -x, [the opposite of whatever
was chosen for x].

If k = 3, then there is not just one (unique) solution, 
since (x,y) = (a,-a) will always be a solution to the 
system as long as k = 3:

3x + 3y = 0
2x + 2y = 0

For instance, not only is (x,y)=(0,0)  a solution to that
system, but also (1,-1), (-1,1), (-2,2), (3.1416,-3.1416),
(-1000000,1000000) and infinitely many more are solutions. 

However if k is any other value besides 3, there is just one
(unique) solution, (x,y) = (0,0).  For example, these systems:

3x + 2y = 0   3x - 7y = 0    3x =  y = 0    3x + 999y = 0
2x + 2y = 0,  3x + 2y = 0,   2x + 2y = 0,   3x +   2y = 0

all have just one (unique) solution (x,y)=(0,0), since k ≠ 3


Bottom line:  
unique solution (0,0) if k ≠ 3, 
infinitely many solutions (a,-a) if k = 3, where a is any real number

Edwin