The system has infinitely many solutions or has no solutions when:
[tex] \dfrac{9}{k} = \dfrac{k}{1} =(\neq) \dfrac{9}{-3} , \\ k^2=9, \\ k=\pm 3[/tex].
1. k=3, then [tex]9x+3y=9 \\ 3x+y=-3[/tex] has no solutions, because
[tex]\dfrac{9}{3} = \dfrac{3}{1}\neq \dfrac{9}{-3} [/tex].
2. k=-3, then [tex]9x-3y=9 \\ -3x+y=-3[/tex] has infinetely many solutions, because
[tex]\dfrac{9}{-3} = \dfrac{-3}{1}= \dfrac{9}{-3} [/tex].
When [tex]k\neq \pm3[/tex], the system has unique solution.
