This is a form of a transport equation, standard first semester PDE problem. Probably in a chapter about characteristics?
First, the "divinely inspired" solution; consider a solution of the form $u(t,x) = f(x-t)$. Then, $u_t=-f'(x-t)$ and $u_x=f'(x-t)$, implying $u_t=-u_x$. With that said, your initial conditions yield $f(x-0)=f(x)=0$. In other words, your solution is identically zero (at least on the domain of interest).
Why is this? Well, you write a linear PDE with 0 assigned to the side conditions. This sort of means that your solution must be 0 -- its a homogeneous equation, so 0 is definitely a solution (existence), and 0 also happens to satisfy your side conditions. Since we have uniqueness (think back to ODEs, similar sort of idea), 0 is the solution.
Second, more interestingly, how did I know so "guess" a form of $u(t,x)=f(x-t)$? Recall from... high school physics (?) that sometimes problems have a "natural frame of reference". (Back in AP Physics, that was often rotating your axes 45 degrees). In first order PDEs, the same idea (but different in the details, of course) is often at play -- our transport equation has a "natural" frame of reference -- that is, moving along the curve $z=x-t$. You'll learn in your class how to identify such natural frames of reference; it really is a powerful technique.