Why is this the general solution to this system of linear partial differential equations? For a system of PDE's given by
$$
\begin{cases}
\frac{\partial u}{\partial x} + 6 \frac{\partial u}{\partial y} = 0 \\
\frac{\partial^2 u}{\partial x^2} - 6 \frac{\partial^2 u}{\partial y^2} = 0
\end{cases}
$$
I have been told that the general solution is given by
$$
u = f(6x-y) \hspace{8mm} \text{and} \hspace{8mm} u' = g(6x+y)
$$
Can anyone please tell me why this is?
(Note: This is a more concise wording - and a more specific example - of the question asked here, which recieved no answers)
 A: Claim:

For a system of PDEs of the form
  $$
\begin{cases}
a \frac{\partial u}{\partial x} + b \frac{\partial u}{\partial y} = 0 \\
c \frac{\partial^2 u}{\partial x^2} + d \frac{\partial^2 u}{\partial y^2} = 0
\end{cases}
$$
  the general solution is given by
  $$
u^{(1)} = f(bx-ay) \hspace{8mm} \text{and} \hspace{8mm} u^{(2)} = g(cx-dy)
$$

Proof: Start by considering the first PDE in this new system. This has characteristic equation 
$$
\frac{dy}{dx} = \frac{b}{a}
$$
which has solution
$$
y = \frac{bx}{a} + c \hspace{6mm} bx - ay = d \hspace{14mm} \left( c \in \mathbb{R}, \hspace{3mm} d = -ac \right)
$$
So, we can solve this PDE by using the substitution
$$
\epsilon = x \hspace{10mm} \eta = bx - ay \hspace{10mm} \omega = u
$$
which, by the chain rule gives
$$
\frac{\partial \omega}{\partial x} = \frac{\partial \omega}{\partial \epsilon}\frac{\partial \epsilon}{\partial x} + \frac{\partial \omega}{\partial \eta}\frac{\partial \eta}{\partial x} = \frac{\partial \omega}{\partial \epsilon} + b \frac{\partial \omega}{\partial \eta} \\
\frac{\partial \omega}{\partial y} = \frac{\partial \omega}{\partial \epsilon}\frac{\partial \epsilon}{\partial y} + \frac{\partial \omega}{\partial \eta}\frac{\partial \eta}{\partial y} = -a \frac{\partial \omega}{\partial \eta}
$$
Substituting the above into the origional PDE and simplifying, we get
$$
\frac{\partial \omega}{\partial \eta} = 0
$$
Taking the partial anti-derivative of the above, we get
$$
\omega = f(\eta)
$$
where $f(\eta)$ is some arbitrary function. Finally, substitution back in our original variables, we get
$$
u = u(x,y) = f(bx - ay)
$$
The second result follows in the same way.
