Prove why the 'elimination method' of solving simultaneous equations works.
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.
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.