I need help understanding the following:
Two equations are derived, each from a different set of constraints that don't enforce each other. That is, the $x, y$ relationship in equation (1) is not the same as the $x, y$ relationship in equation (2). However, the constraints do provide a complete picture of the system and $y_1(x)$ and $y_2(x)$ should represent the same quantity in the consistent picture. We must find a $y(x)$ consistent with equation (1) to solve for the set of valid $x$ solutions. (Note that functions $f_i(x)$ are consistent with both equations. Further, it may be relevant that $f_i(x)$ are non-linear functions of $x$.)
(1): $$ x = \frac{y_1(x) + f_1(x)}{1+y_1(x)f_1(x)+f_2(x)}; a \leq x \leq b $$
(2): $$ y_2(x) = \frac{x-f_1(x)+f_3(x)}{1-xf_1(x)+f_4(x)} $$
We have a vector of $x$ values on the interval $(a,b)$, so once an expression for $y(x)$ consistent with equation (1) is obtained, it can be plugged into the right-hand side of equation (1), this expression can be plotted, and the intersections of this curve with the $x$ vector indicates the solutions. To find such a $y(x)$, one rewrites equations (1) and (2) as
(1'): $$ x+xy(x)f_1(x)+xf_2(x) = y(x) + f_1(x) $$
(2'): $$ y(x)-xy(x)f_1(x)+y(x)f_4(x) = x-f_1(x)+f_3(x) $$
where $y_1(x)$ and $y_2(x)$ are both set to the same $y(x)$ (I'm assuming this is done because the picture presented by (1) and (2) is incomplete and in the consistent picture we know that both $y_1(x)$ and $y_2(x)$ are the same $y(x)$)
Then, taking the sum of the left- and right-hand parts of equations (1') and (2') we get the equation
$$ x+xy(x)f_1(x)+xf_2(x)+y(x)-xy(x)f_1(x)+y(x)f_4(x) = y(x) + f_1(x) + x-f_1(x)+f_3(x) $$
Simplifying this equation, all but one term with a factor of $y(x)$ is eliminated
$$ xf_2(x)+y(x)f_4(x) = f_3(x) $$
solving this for $y(x)$ yields
$$ y(x) = \frac{f_3(x)-xf_2(x)}{f_4(x)} $$
This expression for $y(x)$ is consistent with equation (1) and as such can be used in equation (1) to find solutions for $x$.
My question is: Why/how does adding the left- and right-hand sides of equations (1') and (2') yield an expression for $y(x)$ that allows us to correctly find solutions for $x$ using equation (1)? I know that is a rather broad question so I would be happy with any relevant terms I could Google to gain this understanding.