Let's assume $x \in \mathbb{R}$ and $t \ge 0$. I am asked to find d'Alembert's formula to the wave equation $$\begin{cases}u_{tt}-u_{xx}=0 & \text{in } \mathbb{R} \times (0,\infty) \\ u = g, u_t = h &\text{on } \mathbb{R} \times \{t=0\} \end{cases}$$ by changing the variables like this: $$\xi=x+t \text{ and }\eta = x-t.$$
At some point in my work (see below), I arrive at $$u(\xi,\eta)=F(\xi)+G(\eta).$$ When we back-substitute the variables, we have $$u(x,t)=F(x+t)+G(x-t).$$
This may be a really simple question to ask, but why is it not $u(\color{#00FF00}{x+t,x-t})=F(x+t)+G(x-t)$?
My work so far:
Solving $u_{xy}=0$ gives the general solution $u(x,y)=F(x)+G(y)$, where $F,G$ are arbitrary functions.
Now system solving $\xi=x+t$ and $\eta=x-t$ for $x,t$ gives $x=\frac{\xi+\eta}2$ and $t=\frac{\xi-\eta}2$. So I set $$v(\xi,\eta):=u\left(\frac{\xi+\eta}2,\frac{\xi-\eta}2 \right)=u(x,t)$$
Differentiating twice, I get $$v_{tt}=v_{\xi \xi} - 2v_{\xi \eta}+v_{\eta \eta} \quad \text{and} \quad v_{xx}=v_{\xi \xi} + 2v_{\xi \eta}+v_{\eta \eta}.$$
Hence, $v_{tt}-v_{xx}=0$ gives $v_{\xi \eta}=0$.
Finally, we have $v(\xi,\eta)=F(\xi)+G(\eta)$. Hence, $u(x,t)=F(x+t)+G(x-t)$.
Define a function $v$ by $v(x+t,x-t)=u(x,t).$ Defining $\xi=x+t,$ $\eta=x-t,$ we can write $v(\xi,\eta)=u((\xi+\eta)/2,(\xi-\eta)/2).$ I will take the notation $u_x$ to mean the derivative of $u$ with respect to its first argument and $u_t$ to mean the derivative of $u$ with respect to its second argument.
So $$ u_t(x,t)=\frac{\partial}{\partial t}u(x,t)=\frac{\partial}{\partial t}v(\xi,\eta)=\frac{\partial}{\partial\xi}v(\xi,\eta)\frac{\partial\xi}{\partial t}+\frac{\partial}{\partial\eta}v(\xi,\eta)\frac{\partial\eta}{\partial t}=v_\xi(\xi,\eta)-v_\eta(\xi,\eta). $$ I will not use the notations $u_\xi,$ $u_\eta,$ $v_t,$ $v_x$ since I find them somewhat ambiguous. The notation $v_\xi$ means the derivative of $v$ with respect to its first argument; similarly, $v_\eta$ means the derivative of $v$ with respect to its second argument. Continuing in the same way, we compute $$ u_x(x,t)=v_\xi(\xi,\eta)+v_\eta(\xi,\eta), $$ and then $$ \begin{aligned} u_{tt}(x,t)&=v_{\xi\xi}(\xi,\eta)-2v_{\xi\eta}(\xi,\eta)+v_{\eta\eta}(\xi,\eta),\\ u_{xx}(x,t)&=v_{\xi\xi}(\xi,\eta)+2v_{\xi\eta}(\xi,\eta)+v_{\eta\eta}(\xi,\eta). \end{aligned} $$ From $u_{tt}(x,t)-u_{xx}(x,t)=0$ we conclude that $-4v_{\xi\eta}(\xi,\eta)=0.$ This implies we may write $v(\xi,\eta)=F(\xi)+G(\eta).$ Therefore $$ u(x,t)=v(x+t,x-t)=F(x+t)+G(x-t). $$