Why does elimination work?

Teachers typically emphasize two techniques (though there are certainly others) for solving systems of equations: substitution and elimination. The other day I asked my class of prospective middle and secondary teachers to explain why elimination works. I asked this because as I was sitting at my kitchen table that morning, My students were not able to answer my question. I realized I too had never thought about why this commonly taught procedure works. Solving systems of linear equations is a pretty important topic in secondary mathematics, so it is definitely a question worth exploring.

Consider the system of equations below.

2x + 3y = 14

-2x − 5y = -2

As long as the lines are not parallel or the same line, they will intersect at just one point (x, y). Solving this system means that you are searching for that particular coordinate point that satisfies both equations. This is the goal of solving a system of linear equations, to determine whether or not there are points that satisfy both equations.

Elimination is one method to do this. With this method, you would add the two equations together in order to eliminate one of the variables so that you are left with a single equation with a single variable. So in the example above, you would get the following:

(2x + 3y) + (-2x − 5y) = 14 + -2

-2y = 12

y = -6

You can then substitute in this y value into either of the original equations to determine the related value of x, like so.

2x + 3y = 14

2x + 3(-6) = 14

2x + -18 = 14

2x = 32

x = 16

Thus, the solution to the system is the point (16, -6).

Ordinarily, you cannot add a term from one equation to another because they are not indeed like terms. That is to say, an x in one equation is not necessarily the same x from another equation. For example, consider x + 3 = 5 and x − 2 = 7. The value of x in the first equation is 2 and the value of x in the second equation is 9. Therefore, you can’t simply add the equations together to get x + 3 + x − 2 = 5 + 7 because the x‘s are not like terms.

Technically, you have just added 7 to both sides of the x + 3 = 5 equation, however, then you would also have added x + x but, as stated above, the x‘s don’t represent the same unknown value. If you ignore this fact, that would mean, that you would have 2x + 1 = 12, but that doesn’t make sense because you just added an x with a value of 2 (from the first equation) to an x with a value of 9 (from the second equation). If you add those x’s, you should have gotten 11 (2 + 9), however, the 2x means it should be double the value of x, so if you take it as the x from the first equation, this would mean 2x should be 4, but if you use the value of x from the second equation, this would mean 2x would be 18. You cannot have 2x simultaneously equal 18 and 4, and this is indeed the wrong answer based on the actual values of each of the x‘s. It would be okay to add these together if you, for example, keep the x from the first equation as x and then rename the x from the second equation to something else like y.  This would result in the following: x + 3 + y − 2 = 5 + 7.

Back to elimination, we do indeed keep the x‘s the same and add them as if they were like terms because they are indeed like terms. This is because we are looking for the solution for which those x‘s are equal. If this were not the case, we could not add the x‘s together. This is an important point that should be emphasized with students.

So in summary, elimination works because we are looking for the values of x and y that are the same in both equations. When we add the two equations together, we are really adding the same value to both sides of one of the equations, which is allowed. If we were to annotate the steps in the process above it would look like this:

To solve this system of linear equations

2x + 3y = 14

-2x − 5y = -2

Add -2 (in the form of -2x − 5y on the left) to both sides, which is okay because we are looking for the point where x and y are the same for both equations.

(2x + 3y) + (-2x − 5y) = 14 + -2

Simplify the new equation and solve for y.

-2y = 12

y = -6

Substitute the value of y into either of the original equations to determine the corresponding x-value.

2x + 3y = 14

2x + 3(-6) = 14

2x + -18 = 14

2x = 32

x = 16

Thus, the solution to the system is the point (16, -6).

Teachers should encourage their students to discuss these steps and the rationale as they solve to make sure they understand why this works. I hope this was helpful!