Mathematics for Teaching Solving systems of equations by elimination – why it works

## Solving systems of equations by elimination – why it works

Mathematical knowledge is only powerful to the extent to which it is understood conceptually, not just procedurally. For example, students are taught the three ways of solving a system of linear equation: by graphing, by substitution and by elimination. Of these three methods, graphing is the one that would easily make sense to many students. Substitution, which involves expressing the equations in terms of one of the variables and then equating them is based on the principle of transitive property: if a = c and b = c then a = b. But, what about the elimination method, what is the idea behind it? Why does it work?

While the elimination method seems to be the most efficient of the three methods especially for linear equations of the form ax + by = c, the principle behind it is not easily accessible to most students.

Example: Solve the system (1) 3x + y = 12 , (2) x – 2y = -2.

To solve the system by the method of elimination by eliminating y we multiply equation (1) by 2. This gives the equation (3), 6x + 2y = 24. Thus we have the resulting system,

6x + 2y = 24
x – 2y = -2.

The procedure for elimination tells us that we should add the two equations. This gives us a fourth equation (4), 7x = 22. We can then solve for x and then for y. But we have actually introduced 2 more equations, (3) and (4) in this process. Why is it ok to ‘mix’ these equations with the original equations in the system?

Equation (3) is easy to explain. Just graph 3x + y = 12 and 6x + 2y = 24. The graph of these two equations coincide which means they are equal. But what about equation (4), why is it correct to add to any of the equations? The figure below shows that equation (4) will intersect(1) and (2) at the same point.

Is this always the case? Think of any two linear equations A and B and then graph them. Take the sum or difference of A and B and graph the resulting equation C. What do you notice? This is the principle behind the procedure for the elimination method. But before students can do this investigation, they need to have some fluency on creating equation passing through a given point. The following problem can thus be given before introducing them to elimination method.

Is there a systematic way of generating other equations passing through (3,1)? This will lead to the discovery that when two linear equations A and B intersect at (p,q), A+B will also pass through (p,q). With little help, students can even discover the elimination method for solving systems of linear equations themselves from this. This problem is again another example of a task that can be used for teaching mathematics through problem solving . The task also links algebra and geometry. Click this link for a proposed introductory activity for teaching systems of equation by elimination method.

## 5 thoughts on “Solving systems of equations by elimination – why it works”

1. Alfonso says:

How does this apply to the nonlinear case? Thanks.

2. thekla manning says:

Excellent post! A great approach to teaching solving by elimination – too often it is just magic…works, but WHY?

3. certificate of deposit interest rates says:

solid post , really good view on the subject and very well written, this certainly has put a spin on my day, many thanks from the USA and observe up the good work