TITLE: What exactly do systems of equations represent?
QUESTION [11 upvotes]: Systems of equations are taught pretty early in American curriculum. We are taught methods of substitution and methods of elimination in order to solve them. We are taught how to use matrices or graphs as alternative strategies to encode/visualize them.
There are linear systems of equations...there are non-linear systems of equations...and there are systems of equations with 1 or many variables. However, to this day, I still do not really understand what systems of equations are. I've tried to find an abstract interpretation of systems of equations but without much success. I am very interested in figuring out this "systems of equations abstraction" because there are canonical statements (e.g. the classic "You need as many equations as variables to find a solution") that I would very much like to prove.
I've tried myself to come up with some semblance of an abstraction but I have not made much progress. I will illustrate the case for one equation with one variable and two equations with two variables (which is where I run in to trouble).
One Equation - One Variable
Consider the purely arbitrary equation: $$a = b +\alpha x \ \text{ where}\  \alpha \neq 0$$
The effort to solve this equation can be rephrased as "Find me the $x$ that maps to $a$ through the function $f(x) = b + \alpha x$." In effect, therefore, this question is one requiring that we find the inverse function $f^{-1}$ that, when $a$ is given as input, $x$ will be output.

Solving for $x$ by virtue of substracting $b$ from both sides and dividing both sides by $\alpha$ effectively amounts to determining the inverse function such that: $f^{-1}(x')=\frac{x' -b}{\alpha}$...for future purposes, also note that this can be denoted as $f^{-1}\big(f(x)\big)=\frac{f(x) -b}{\alpha}$ . Plugging in $a$ for $x'$ we arrive at the solution to this equation, which is $f^{-1}(a)=\frac{a-b}{\alpha}$. So far, so good.
Two Equations - Two Variables
Consider the following two arbitrary equations of two variables:
$$a=b+\alpha x + \beta y \ \text{ where}\  \alpha, \beta \neq 0$$
$$c=d +\gamma x + \delta y \ \text{ where}\  \gamma, \delta\neq 0$$
Following the same logic of the "One Equation - One Variable" section, solving for $x$ and $y$ can be viewed as constructing the inverse functions for the two above equations, which can be recast as specific instances of:
$$g\big( (x,y) \big) = b+\alpha x + \beta y$$
$$h \big ( (x,y) \big ) = d +\gamma x + \delta y$$

As to how one solves for these inverses, I have not the faintest clue. Obviously, we could revert to the standard methods of substitution and arrive at the following bulky solutions:
$$ y = \frac{\alpha \Big ( h \big((x,y)\big) -d \Big) -\gamma g\big( (x,y) \big)+\gamma b}{\delta \alpha - \beta \gamma}$$
and
$$ x = \frac{g\big( (x,y) \big) -b - \beta y}{\alpha}$$
However, these are not inverses of the functions $g$ and $h$. In fact...I don't really even know WHAT these equations represent. If you were to plug in the value of $y$ in the final equation for $x$ (I omitted that for brevity), you can see that the $y$ equation and the $x$ equation both have $g\big ( (x,y) \big)$ and $h \big ( (x,y) \big)$ in them...so these equations are inverse-like. That is to say, determining $(x,y)$ requires information from both $g$ and $h$, which provides the first clues as to how one might prove, "You need as many equations as variables to find a solution." Linking this back to the "One Equation - One Variable" section, recall that $f^{-1}\big(f(x)\big)=x = \frac{f(x)-b}{\alpha}$ depends on only one function in order to solve.
Hopefully I did not completely botch this question and was able to sufficiently convey what I am after. Any insights would be greatly appreciated. Cheers~

REPLY [3 votes]: A system of equations can be understood as a set of equalities over a given FOL structure. It is the structure that tells you what operations are allowed, which allows you to form terms possibly with free variables, and each equality is simply a formula of the form "$t=u$" where $t,u$ are terms over the structure. You have to understand FOL in order to fully understand what equations mean.
For example, polynomial equations over a ring are simply equalities over that ring. When you solve a system of linear equations with real coefficients, what you are essentially doing is finding a tuple of reals that can be assigned as values to the free variables of those equations to make them true in the structure of reals. More generally, Gaussian elimination on a linear system of equations works over any field, not just the field of reals, because all the operations you use in Gaussian elimination are supported by any such field.
This viewpoint readily captures the most general notion of "equation system" that you are likely to encounter in practice. For example, you can have a system of differential equations as well, where the underlying structure is often a 2-sorted structure $((ℂ,F),D,...)$, where the first sort $ℂ$ is the complex numbers, and the second sort $F$ is the set of differentiable functions on some subset of $ℂ$, and $D$ is the differentiation operation on $F$, and the "$...$" are for all the usual operations that come with $ℂ$ and $F$. For example, the system $\Big\{ \frac{dx}{dt} = a·y , \frac{dy}{dt} = b·x \Big\}$ where $a,b∈ℂ$ would formally be the system $\{ \ D(f) = scale(a,g) \ , \ D(g) = scale(b,f) \ \}$ where $scale : ℂ×F→F$ is the operation representing the scalar multiplication of a complex number on a function in $F$, and $f,g$ are free variables of sort $F$.
By the way, you do not need as many equations as variables to find a solution. $\{ x+y = 0 \}$ has one equation but two variables, yet you can easily find a solution... What you probably wanted to say was "unique solution", which applies to linear systems of equations over a field and not in general. For instance, $\{ x^2+y^2 = 0 \}$ has one equation but two variables, and yet has a unique solution over the reals.
Also, as you might realize by now, the fact that solution sets for linear systems of equations over a field forms a hyperplane is quite a special one, and the general situation is much messier. Just for example, $\{ x+y = 2 , x-y = 1 \}$ has no solution over the integers.
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