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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.

Malcolm Regan
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  • I'm assuming the $f_i$ functions are given to you, right? If that is so, then why do you say you need to do all of that to "find $y(x)$"? Equation $(2)$ already gives you $y(x)$ explicitly. Or am I misunderstanding things? – Paulo Mourão Jun 29 '19 at 20:10
  • Yes, $f_i$ are just functions with known values at each value of $x$. However, one can't plug the expression for $y(x)$ given by equation (2) into equation (1) because equations (1) and (2) are derived from different sets of constraints, ie the $y(x)$ of equation (1) is not compatible with the $y(x)$ of equation (2). For some reason, adding the left hand and right hand sides of equations (1') and (2') yields a $y(x)$ that is compatible with equation (1). I do not understand why. – Malcolm Regan Jun 29 '19 at 20:21
  • Also, i should add that the $y(x)$ obtained from adding the left hand and right hand sides of equations (1') and (2') is very different from the $y(x)$ of equation (2). – Malcolm Regan Jun 29 '19 at 20:29
  • I don't understand what you mean by not being compatible. Equation $(2)$ gives you an expression for $y(x)$. Obviously, you shouldn't expect all of those pairs of points $(x,y(x))$ to be solutions of $(1)$. But by plugging in $y(x)$ in $(1)$, you get an equation for $x$ that gives you such solutions, I don't see what should can be wrong with that – Paulo Mourão Jun 29 '19 at 20:30
  • I don't know the formal math terms for the properties of the constraints that make eq (1) and (2) describe different x, y relationships (part of my asking this question is hoping that someone will recognize these properties and tell me what they are). I assure you that this is whats happening, though. The man whose calculations I'm following doesn't make mistakes. Hah. I suspect it has to do with something slightly more advanced than simple algebra alone but I just don't know what I don't know at this point. I do appreciate your taking the time to respond. – Malcolm Regan Jun 29 '19 at 20:35
  • Yes, they probably yield different relationships between $x$ and $y$. If they didn't, your two equations would be redundant, and you would only need one. But still, if I understood correctly, your goal is to find values of $x$ that satisfy both equations (i.e. they represent different relations between $x$ and $y$ and you need them both to be satisfied). If that is the case, then just plugging in $(2)$ in $(1)$ and solving gives you the right solution(s). The whole process of summing both sides and then plugging in doesn't change the system and you can show that it gives you the same answers. – Paulo Mourão Jun 29 '19 at 20:41
  • Maybe summing both sides and then plugging in would make the final equation for $x$ (the one you get after plugging in) easier to solve, and that could be an argument for doing this. – Paulo Mourão Jun 29 '19 at 20:43
  • Let us continue this discussion in chat. – Paulo Mourão Jun 29 '19 at 20:44
  • edited question to clarify that the $y(x)$'s of equations (1) and (2) are not the same – Malcolm Regan Jun 30 '19 at 16:34

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