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
