Solving simultaneous PDEs Given the equations (1):$$\frac{\partial u}{\partial t}+g\frac{\partial \eta}{\partial x}=0$$ 
and (2):$$\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}=0$$
can we combine the two together 
to form $$\frac{\partial^{2}\eta}{\partial t^{2}}-gH\frac{\partial^{2}\eta}{\partial x^{2}}=
0$$ by substituting $\frac{\partial}{\partial t}.(2)$ into (1). 
I should note this was given as an example but the working was not 
shown, I have tried a few manipulations but cannot get the result, 
am I missing something?
 A: Differentiating the first equation with respect to $x$:$$\frac{\partial u}{\partial t} +g\frac{\partial \eta}{\partial x}=0\implies\frac{\partial^2 u}{\partial t\partial x}+g\frac{\partial^2 \eta}{\partial x^2}=0$$
and the second with respect to $t$:
$$\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}=0\implies \frac{\partial^2\eta}{\partial t^2}+H\frac{\partial^2 u}{\partial x \partial t}=0$$
Eliminating the mixed derivative $\frac{\partial^2 u}{\partial x \partial t}$ between these two gives :
$$\frac{\partial^2\eta}{\partial t^2}+H\left(-g\frac{\partial^2 \eta}{\partial x^2}\right)=0$$ i.e.
$$\frac{\partial^{2}\eta}{\partial t^{2}}-gH\frac{\partial^{2}\eta}{\partial x^{2}}=
0$$
A: $$\frac{\partial u}{\partial t}+g\frac{\partial \eta}{\partial x}=0$$
$$\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}=0$$
This implies that
$$ \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial t}+g\frac{\partial \eta}{\partial x}\right)$$
$$= \frac{\partial^2 u}{\partial t\partial x}+g\frac{\partial^2 \eta}{\partial x^2}=0$$
And
$$ \frac{\partial}{\partial t} \left(\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}\right)$$
$$=\frac{\partial^2\eta}{\partial t^2}+H\frac{\partial^2 u}{\partial t\partial x}=0$$
Using the fact that
$$\frac{\partial^2 u}{\partial t\partial x}=-g\frac{\partial^2 \eta}{\partial x^2}$$
We have
$$\frac{\partial^2\eta}{\partial t^2}+H\frac{\partial^2 u}{\partial t\partial x}$$
$$= \frac{\partial^2\eta}{\partial t^2}+H\left(-g\frac{\partial^2 \eta}{\partial x^2}\right) $$
$$=\frac{\partial^{2}\eta}{\partial t^{2}}-gH\frac{\partial^{2}\eta}{\partial x^{2}}=
0$$
