#### Answer

$k$ can take any value exept $\pm1$ ($k\in R/\{-1,1\}$)

#### Work Step by Step

If we want the solution to be unique these two equations have to be independent of each other. Since their right sides are equal, in order to be independent their left sides must not look identical (they must not have the same coefficients multiplying $x$ and $y$).
The only case where that would happen is if we choose $k=\pm1$. Then we would have infinitely many solutions since the equations would be identical if we choose $+$ or the second equation would be the first one multiplied by $-1$ if we choose $-$. So $k\neq \pm1$ if we want the unique solution.