Does a function $f$ exist such that $(y+z)/q =x$, $(z+q)/y = f(x)$ where $f$ is independent of $y,z,q$?

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I came across this problem when deriving a relationship for electromagnetic waves moving through 3 dielectric mediums. Is there a function that can be applied to x such that

$${y+z\over q} = x \quad\text{and}\quad {z+q\over y} = f(x)$$

Where $f$ is independent of $y, z$ and $q$ (this is not necessary but would be nice).

$y, z$ and $x$(edit q not x) are vector fields that vary in both time and space (electric fields to be exact). However, I believe they can be treated as constants in this case as it is a limit case.

I am in undergraduate engineering and don't really know where to start looking for an answer to this. I would love if you can send me in the right direction, even if you don't know of an answer.

^^^^ This may be the wrong question...

$${y+z\over q} = x \quad\text{and}\quad {z+q\over y} = ?$$

is there a way to re-arange to get this, what is "?"

thanks.

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If all the variables are scalars, $f$ can be made "independent" of exactly one of $y$, $z$, and $q$.

Solve the first equation for the one you want to be independent and plug it into left hand side of the second equations:

Independent of $y = qx - z$: $f(x; q, z) = (z + q)/(qx - z)$.

Independent of $z = qx - y$: $f(x; q, y) = (qx - y + q) / y = q(1+x)/y - 1$.

Independent of $q = (y+z)/x$: $f(x; y, z) = (z + (y+z)/x)/y = (y + z + xz)/xy$.

Note that these answers are not truly independent, it's just different forms of writing the same function on a constrained surface where the first equation holds. In particular, none of them agree if you're not on that surface, and have questionable validity outside that surface.