This is a general question about the case when the characteristic equations are such that are not (easily) solvable. The following is just an example; in particular, the right-hand side need not be zero.
Suppose we have an equation like $yu_x(x,y)+xu_y=0$ subject to some appropriate condition (e.g. $u(0,x)=h(x)$ for some function $h$).
As this is a quasilinear first-order PDE, we can try and solve it using the method of characteristics. So, we set $x=x(s,t)$, $y=y(s,t)$, and $z(s,t)=u(x(s,t),y(s,t))$. We have to solve $$\frac{dx}{dt}=y,\quad \frac{dy}{dt}=x,\quad \frac{dz}{dt}=0$$ subject to some conditions. The third equation gives $z(s,t)=A(s)$ for some function $A(s)$ to be found using the conditions. However, the first two equations lead nowhere because to solve, say, the first one, you have to have an expression for $y$ in terms of $s$ and $t$. But you don't have this because in order to solve the second equation you have to have an expression for $x$ in terms of $s$ and $t$.
(We could perhaps write $y=\int x\,dt + C(s)$ for some $C$ and then substitute in $$\frac{dx}{dt}=\int x\,dt+C(s)$$ but then how do solve for $x$ from this?)
So my question is: how does the method of characteristics apply in this situation? Is it perhaps that we can solve these equations but I just don't see why?