Right, but that's what you've got to prove in order to show that you've got a vector space. You're going to have to look at the specifics of your problem. What does $\displaystyle r\mathbf{x}$ look like (referencing my previous post)?
y is the same kind of number (a real number) as k, and it's the same kind of number as x. So why not let y=kx? Then it does, indeed, look like a vector in your set.