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.

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.
