Substitution with functional equations 
I've found this nice introduction worksheet that I started to work through with the goal to get a better understanding of functions and finding them in equations. I've gotten so far but in this one solution there is a few steps that is REALLY unclear. I'd appreciate if someone can check the edited screenshot and give me some pointers. Thank you!
 A: We begin with the expression 
$$f(x)+f\left(\frac{x-1}{x}\right)=1+x \tag 1$$
The first substitution is to let $t=\frac{x-1}{x}\implies x=\frac{1}{1-t}$ from which $(1)$ becomes
$$f\left(\frac{1}{1-t}\right)+f\left(t\right)=1+\frac{1}{1-t} \tag2$$
Then, the development merely switches back from the dummy variable $t$ to $x$ to arrive at Equation $(1.1)$ from the posted question, namely 
$$f\left(\frac{1}{1-x}\right)+f\left(x\right)=1+\frac{1}{1-x} \tag{1.1}$$
Next comes a second substitution in $(1)$.  Here, let $t=\frac{1}{1-x}\implies x=\frac{t-1}{t}$ so that $\frac{x-1}{x}=\frac{1}{-t}\frac{t}{t-1}=\frac{1}{1-t}$. Therefore, $(1)$ becomes
$$f\left(\frac{t-1}{t}\right)+f\left(\frac{1}{1-t}\right)=1+\frac{t-1}{t} \tag 3$$
Changing dummy variables in $(3)$ from $t$ to $x$ yields Equation $(1.2)$ from the posted question
$$f\left(\frac{t-1}{t}\right)+f\left(\frac{1}{1-t}\right)=1+\frac{t-1}{t} \tag{1.2}$$
If we enforce the second substitution $t=\frac{1}{1-x}\implies x=\frac{t-1}{t}$ into $(1.1)$, and switch dummy variables back, from $t$ to $x$, we obtain Equation $(1.3)$ 
$$f\left(x\right)+f\left(\frac{x-1}{x}\right)=1+x \tag{1.3}$$
The remaining part of the development uses $(1.1)-(1.3)$ to eliminate the terms with arguments $\frac{1}{1-x}$ and $\frac{x-1}{x}$ to result in an expression for $f(x)$ alone.  In the posted question, this is done in two steps.  A more efficient way forward is to note simply that if we add the both sides of  $(1.1)$ and $(1.3)$, subtract from the result $(1.2)$, and finally divide by a factor of $2$, the resulting left-hand side if $f(x)$.
