Prove why the 'elimination method' of solving simultaneous equations works. [This has been deleted due to me stressing out over something obvious that I didn't see; apologies if I seemed aggressive, I was merely stressed. Thanks.
 A: The rationale of the elimination method is rather obvious: as you can't solve an equation in two unknowns directly, you transform the system in a way to get an equation in a single unknown, which is easily solved.
You'll soon learn that this process generalizes to a system of $n$ equations in $n$ unknowns, where you transform a square system in a triangular one. The approach is very smart and very efficient.
The method works because you are allowed to replace an equation by a linear combination of itself and other equations (as you combine equal expressions, the equality $LHS=RHS$ isn't lost).

Any system of equations (linear or not) can be written as equalities to zero of some functions of the unknowns.
$$f(x,y)=ax+by-n=0,\\g(x,y)=cx+dy-m=0.$$
You are "allowed" to transform the system into another, provided you don't introduce new solutions nor discard valid ones.
For instance, omitting the arguments $(x,y)$, the single equation $$f^2+g^2=0$$ has the same solution set, $f=0,g=0$.
In particular, you can form linear combinations, such as
$$\alpha f+\beta g=0,\\\gamma f+\delta g=0,$$ provided that the corresponding system isn't indeterminate, i.e. if $\alpha\delta-\beta\gamma\ne0$.
A special case is that used in elimination, where you add an equation to another and keep the other unchanged,
$$f=0,\\\gamma f+g=0.$$
It obviously fulfills the criterion.
This new system is equivalent to
$$ax+by=n,\\(\gamma a +c)x+(\gamma b+d)y=\gamma n+m,$$
and you are free to choose $\gamma$ so that the term in $y$ vanishes.
A: I've been searching around the internet looking for an explanation for my Algebra students as to why one can solve a system by elimination. It's not immediately clear as to why you can just "combine" equations, resulting in a new equation (which if graphed, creates a new line, but happens to go through the same solution point!)
My students are comfortable with the idea of substitution (sure, if something is equivalent, then I can replace it somewhere else in the equation with what it's equivalent to!). The intuition is lost when we get to solving by elimination. Well, finally I've come across an accessible answer for these 7th and 8th graders! These are my notes:
Question: Why does the elimination method work?
My two equations are:


*

*$x-2y=1$

*$3x+2y=11$
We know that we can add something to both sides of an equation and maintain equality. For example I can take
$3x+2y=11$
and add 1 to both sides to get
$3x+2y+1=12$
And from the first equation, we know that $x-2y=1$, which we can substitute back in.
$3x+2y+1=12$ becomes $3x+2y+x-2y=12$
Combine like-terms to get $4x=12$. Solve for $x$, then solve for $y$. 
Then I guess I would show my students how doing it in columns saves a lot of time, and also makes it more obvious of multiplying the equations by some constant to ensure they will eventually eliminate a variable.
A: Lets have:
$$5x + 2y = 3$$
$$7x -3y = 1$$
You can sum something at both sides of an equation (See this link for discussion: Why we can add an element on both sides like this?)
Now we add something to the first equation:
$$5x + 2y + (7x -3y) = 3 + (7x -3y)$$
But this something equals to 1, then:
$$5x + 2y + (7x -3y) = 3 + (1)$$
And rearranging:
$$5x + 7x + 2y -3y = 3 + (1)$$
Finally:
$$12x - y = 4$$
This new equation is called a linear combination of both original equations. Since we added something at both sides in the first equation of the system, the new equation is like the first one but transformed and has the same solutions, the same way that:
$$x + y = 3$$
and
$$ 2x + 2y = 6$$
(In this last case we multiplied a something by another something)
By the way, I think this is not so obvious. I started to think about it studying the first chaper of Hoffman-Kunze book. 
A: We have:
$$x=\dfrac{n-by}{a}$$
$$x=\dfrac{m-dy}{c}$$
so we eliminate $x$ by writing:
$$\dfrac{n-by}{a}=\dfrac{m-dy}{c}$$
$$cn-bcy=am-ady$$
$$y(ad-bc)=am-cn$$
$$y=\dfrac{am-cn}{ad-bc}$$
A: I think maybe something's not being said here that's really important. (This is really a long comment, and only kind of an answer.)
You say things like "the rule seems arbitrary," and "has been thrown in for no reason." This is a rather strange way of looking at mathematics, as if for any given situation put in front of you, there are some rules for how to solve it and you just follow them. If this was the case, computers would just do all math and there'd be no mathematicians.
But this isn't what mathematics is like at all. Instead, we have some starting points we call axioms. These are arbitrary, but in some sense they are designed to conform to our expectations. Commutativity, associativity, distributivity, all of these are basic building blocks which we intuitively understand as true.
From here, we go about proving theorems, and solving problems. By this I mean, using the basic axioms and definitions to prove properties of certain things, like equations or geometric objects or algebraic systems. This method produces something new which we didn't know before, and in the case of your particular question, is how the solution is worked out. In algebra of real numbers, we have a rule for cancelling and a rule for adding/subtracting/multiplying/dividing equally to both sides of an equation. These are the axioms for algebra with real numbers. It turns out that by just using these basic properties, we can determine what the solution to the problem is. In short, it isn't some rule which is just made up arbitrarily, it is the result of thinking through the difficulties and using the properties to reach a solution.
To get an intuitive picture of what all this means, think of it like this. You have two equations in two variables, each has infinitely many possible solutions since they are like lines, but you want to find only the solutions that satisfy both. That is to say, you want to restrict your search to only look for solutions of one guy that also satisfy the other. To do this, you must somehow insert as a restriction the requirement that the other equation hold. That is to say, you must combine the two relationships between $y$ and $x$ so that you can find a solution to both. 
So what you do, is solve one equation for one of the variables. This is just rewriting the same relationship in different terms. But once one of the equations is solved in terms of one variable, substituting that equation into the first is the restriction we need. We are saying "this equation satisfies the relationship it expresses itself, but it also has this other relation we have just inserted into it." Now you can do algebra in one variable to solve, and then use this information to find the other variable.
So really we are not applying any arbitrary rules, we are thinking through the problem and applying the basic concepts we know to reach a solution.
