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Chapter 16 The Method of Elimination

Section 16.1 Make it Go Away

We have seen that we can solve systems of linear equations using substitution, but that one of the challenges of this method was the computational problem of manipulating fractions. The method of elimination is another approach to solving systems of equations that mostly sidesteps the issue (at least at this level), which often makes it a more efficient approach.
The core concept of this method is a slightly different use of axioms of equality that were introduced in Section 1.1 mixed in with a substitution. We can add the same value to both sides of an equation without losing the equality. The only twist is that we use two different representations in the addition step. Here is how it looks:
\begin{equation*} \text{If $a = b$ and $c = d$, then $a + c = b + d$.} \end{equation*}
The challenge is to find useful equations so that adding the equations together gives a helpful result. Here are two examples to compare:
First, notice that the addition in columns is simply an organizational tool. By putting like terms together, it simplifies the process of combining them and avoids errors. Second, notice that the final equation on the right only has one variable even though the two initial equations had two variables. The goal of the method of elimination is to eliminate one of the variables. This allows you to solve for the remaining variable.

Activity 16.1. Solving a System by Elimination (Part 1).

Here is the worked solution to the system of equations on the right (see above):
Try it!
Solve the system of equations below using the method of elimination.
\begin{equation*} \left\{ \begin{array}{rcrcr} 3x \amp + \amp y \amp = \amp 3 \\ -3x \amp + \amp 3y \amp = \amp -7 \\ \end{array} \right. \end{equation*}
Solution.

Activity 16.2. Solving a System by Elimination (Part 2).

Not all systems of equations are in a form that immediately lends itself to the method of elimination. Let’s take another look at the example on the left from earlier:
\begin{equation*} \left\{ \begin{array}{rcrcr} 2x \amp + \amp 3y \amp = \amp 3 \\ 3x \amp + \amp 2y \amp = \amp 5 \\ \end{array} \right. \end{equation*}
If we add these equations together directly, nothing will cancel out. But we can multiply either equation by any non-zero number to change its form. And if this is done correctly, it will lead to cancellation.
From here, we can plug in \(y = -\frac{1}{5}\) into either of the original equations, but that will lead to fraction manipulations. If we wanted to avoid those, we can just use the same idea to eliminate the terms instead.
This leads to the solution \((x,y) = ( \frac{9}{5}, -\frac{1}{5} )\text{.}\)
Try it!
Solve the system of equations below using the method of elimination.
\begin{equation*} \left\{ \begin{array}{rcrcr} 2x \amp + \amp 3y \amp = \amp -7 \\ -3x \amp + \amp y \amp = \amp 3 \\ \end{array} \right. \end{equation*}
Solution.
There’s no real "trick" to figuring out the numbers that you need to get the terms to cancel, but there is a general heuristic. Once you’ve identified the variable you want to eliminate, find the least common multiple of the coefficients, and multiply the equations by the appropriate value to make match, and then make sure that the signs are opposite each other. While some instructors suggest that you can "subtract" the equations, this ends up causing a lot of computational errors, and so it’s usually better to use addition.

Section 16.2 Worksheets

PDF Version of these Worksheets
 1 
external/worksheets/16-Worksheets.pdf

Worksheet Worksheet 1

1.
Solve the system of equations using the method of elimination.
\begin{equation*} \left\{ \begin{array} {rcrcr} x \amp - \amp y \amp = \amp 5 \\ x \amp + \amp y \amp = \amp 1 \end{array} \right. \end{equation*}
2.
Solve the system of equations using the method of elimination.
\begin{equation*} \left\{ \begin{array} {rcrcr} 2x \amp - \amp 2y \amp = \amp 8 \\ -2x \amp + \amp 3y \amp = \amp -6 \end{array} \right. \end{equation*}

Worksheet Worksheet 2

1.
Solve the system of equations using the method of elimination.
\begin{equation*} \left\{ \begin{array} {rcrcr} 3x \amp + \amp 2y \amp = \amp -5 \\ 2x \amp + \amp y \amp = \amp 2 \end{array} \right. \end{equation*}
2.
Solve the system of equations using the method of elimination.
\begin{equation*} \left\{ \begin{array} {rcrcr} 4x \amp + \amp 5y \amp = \amp 0 \\ 2x \amp - \amp 3y \amp = \amp 3 \end{array} \right. \end{equation*}

Worksheet Worksheet 3

1.
Solve the system of equations using the method of elimination.
\begin{equation*} \left\{ \begin{array} {rcrcr} -x \amp + \amp 2y \amp = \amp 6 \\ 3x \amp - \amp 4y \amp = \amp 1 \end{array} \right. \end{equation*}
2.
Solve the system of equations using the method of elimination.
\begin{equation*} \left\{ \begin{array} {rcrcr} -3x \amp + \amp 5y \amp = \amp 3 \\ -2x \amp + \amp 3y \amp = \amp -2 \end{array} \right. \end{equation*}

Worksheet Worksheet 4

1.
Attempt to solve the system of equations using the method of elimination. Explain what happened and the conclusion that you can draw from it.
\begin{equation*} \left\{ \begin{array} {rcrcr} 2x \amp + \amp 4y \amp = \amp 8 \\ -x \amp - \amp 2y \amp = \amp -4 \end{array} \right. \end{equation*}
2.
Attempt to solve the system of equations using the method of elimination. Explain what happened and the conclusion that you can draw from it.
\begin{equation*} \left\{ \begin{array} {rcrcr} x \amp - \amp 3y \amp = \amp 2 \\ -3x \amp + \amp 9y \amp = \amp 5 \end{array} \right. \end{equation*}

Worksheet Worksheet 5

1.
Solve the system of equations using the method of elimination.
\begin{equation*} \left\{ \begin{array} {rcrcr} ax \amp + \amp by \amp = \amp c \\ dx \amp + \amp ey \amp = \amp f \end{array} \right. \end{equation*}
2.
Use the formula you derived from the previous problem to solve the system of equations. Do you find it easier to use the formula or to use the method of elimination?
\begin{equation*} \left\{ \begin{array} {rcrcr} -2x \amp + \amp 3y \amp = \amp 5 \\ 3x \amp - \amp 5y \amp = \amp -6 \end{array} \right. \end{equation*}

Section 16.3 Deliberate Practice: Solving by Elimination

Algebra is a skill, which means it requires practice to become proficient. But it will take more than rote repetition to get there. Deliberate practice is the thoughtful repetition of a task. For each of these sections, you will be given a list of specific skills or ideas to focus on as you practice thinking through the problems.
Focus on these skills:
  • Write the original system of equations.
  • Indicate the manipulations of the equations in the process of eliminating the variables.
  • Present your work legibly.

Worksheet Worksheet

Instructions: Solve the system of equations by elimination. If the system has either zero or infinitely many solutions, explain how you can determine this information from your calculations.
1.
\(\left\{ \begin{array}{rcrcr} x \amp + \amp 2y \amp = \amp 7 \\ -2x \amp + \amp y \amp = \amp 1 \end{array} \right.\)
2.
\(\left\{ \begin{array}{rcrcr} x \amp - \amp 2y \amp = \amp 4 \\ -3x \amp + \amp 6y \amp = \amp -12 \end{array} \right.\)
3.
\(\left\{ \begin{array}{rcrcr} 3x \amp - \amp 2y \amp = \amp -8 \\ -x \amp + \amp 3y \amp = \amp 5 \end{array} \right.\)
4.
\(\left\{ \begin{array}{rcrcr} -2x \amp + \amp 3y \amp = \amp -8 \\ -x \amp + \amp 2y \amp = \amp -5 \end{array} \right.\)
5.
\(\left\{ \begin{array}{rcrcr} 2x \amp - \amp 3y \amp = \amp 1 \\ 4x \amp - \amp 6y \amp = \amp 4 \end{array} \right.\)
6.
\(\left\{ \begin{array}{rcrcr} 3x \amp + \amp 4y \amp = \amp 5 \\ 2x \amp + \amp 2y \amp = \amp 3 \end{array} \right.\)
7.
\(\left\{ \begin{array}{rcrcr} 2x \amp - \amp y \amp = \amp -4 \\ x \amp - \amp 2y \amp = \amp 5 \end{array} \right.\)
8.
\(\left\{ \begin{array}{rcrcr} -3x \amp - \amp 5y \amp = \amp -3 \\ x \amp + \amp y \amp = \amp 6 \end{array} \right.\)
9.
\(\left\{ \begin{array}{rcrcr} 3x \amp + \amp 3y \amp = \amp 4 \\ -3x \amp - \amp 2y \amp = \amp -1 \end{array} \right.\)
10.
\(\left\{ \begin{array}{rcrcr} 4x \amp + \amp 2y \amp = \amp -1 \\ 3x \amp + \amp 2y \amp = \amp 2 \end{array} \right.\)

Section 16.4 Closing Ideas

This section created a new toolbox of ideas to solve the types of equations that you already knew how to solve. In that sense, this section is extraneous. Why learn a new way to do something you already know how to do? But in another sense, this is a critical section because it teaches you that sometimes the way you’ve done things in the past isn’t necessarily the way you want to do things in the future.
The method of substitution works for a wide range of problems, and it follows a very straightforward pattern: Solve for one of the variables and plug it into the other equation. The method of elimination requires some intuition and intellectual flexibility. You have to actively assess the situation and determine what choice of multiplication values will cause the elimination to happen. But if you choose wisely, the calculations are much simpler. In fact, with a bit of practice you can solve some of these problems mentally (as long as the numbers aren’t too big). You would find that to be extremely difficult if you were thinking about using the method of substitution.
So consider this section to be one about simply expanding your toolbox of techniques. Instead of being confined to just one way of looking at these problems, you now have two. You have the option of looking at the equations to decide whether substitution or elimination makes more sense, and you have the freedom to pick the method that suits you the best. As you get further in your mathematical studies, you will find more and more situations where there are multiple approaches to solve a problem, and that you’ll be making more active decisions about the methods and techniques that you apply.

Section 16.5 Going Deeper: Three Dimensional Systems

In the last couple sections, we’ve looked at systems of two linear equations with two variables. We saw that there was a geometric aspect to understanding these systems by looking at the graphs of the lines, and we also saw that there was an algebraic aspect by looking at the methods of solving these systems. We are going to take a look at what happens when we increase the complexity by looking at systems of three equations and three variables. Here is an example:
\begin{equation*} \left\{ \begin{array}{rcrcrcr} 2x \amp - \amp y \amp - \amp z \amp = \amp 1 \\ x \amp + \amp 2y \amp + \amp z \amp = \amp 4 \\ x \amp + \amp y \amp - \amp z \amp = \amp 4 \end{array} \right. \end{equation*}
We will first wrap our minds around the geometric understanding of these equations. It turns out that these formulas form planes and that planes are the natural shape to generalize the "linear-ness" of lines in two variables. When we think about three simultaneous systems of equations in three variables, we’re looking for a point that lies on all three planes. Since the number of dimensions has increased, it makes sense that the number of possible arrangements would also increase. However, we still have the three same basic classifications that we had before. There will either be zero solutions, one solution, or infinitely many solutions.
In each of the systems above, there are no solutions. This may seem unusual at first, since there are clearly some points of intersection. However, we are working with simultaneous systems of equations, which means that we’re looking for points that lie on all three planes, not just pairs of planes. Notice that we cannot use the simple heuristic of just thinking about whether the planes are parallel, because the diagram on the right shows that it’s possible to have no solutions even though none of the planes are parallel to each other.
A key observation about these diagrams is that when there are intersections between planes, it forms a line. This is always true of the intersection of two planes, which is an idea we’ll come back to when we look at the algebra.
This is an example of a system with one solution. Although everything is drawn to be at right angles with each other, there is a lot of flexibility in terms of the relative positions of the planes. You should be able to visualize each pair of planes forming a line, and that each of those lines cross each other in exactly one place.
Lastly, we have these configurations to get infinitely many points of intersection. The diagram on the left has all three planes overlapping each other. The middle diagram has two planes overlapping each other with the third plane cutting across them. The last diagram has all three planes intersecting along the same line. Even though these all have infinitely many solutions, there is a distinction between the first one and the last two. Intuitively, what we’re seeing are different numbers of dimensions in the overlap. The intersection of the first diagram is a two-dimensional plane, whereas the intersection of the last two are a one-dimensional line.
When it comes to solving these systems algebraically, it’s just the methodical application of either substitution or elimination to reduce the number of variables. Conceptually, we’re going to find the intersection of planes to get lines, then find the intersection of lines to get points. When working through the process, it’s much more about keeping the work organized so that you don’t get lost.

Subsection 16.5.1 Solving Three Dimensional Systems by Substitution

\begin{equation*} \left\{ \begin{array}{rcrcrcr} 2x \amp - \amp y \amp - \amp z \amp = \amp 1 \\ x \amp + \amp 2y \amp + \amp z \amp = \amp 4 \\ x \amp + \amp y \amp - \amp z \amp = \amp 4 \end{array} \right. \end{equation*}
We will solve the first equation for \(z\text{:}\)
\begin{equation*} \begin{aligned} 2x - y - z \amp = 1 \\ z \amp = 2x - y - 1 \end{aligned} \end{equation*}
Then we will substitute this into the other equations and simplify:
\begin{equation*} \begin{aligned} x + 2y + z \amp = 4 \\ x + 2y + (2x - y - 1) \amp = 4 \\ 3x + y = \amp 5 \end{aligned} \qquad\qquad \begin{aligned} x + y - z \amp = 4 \\ x + y - (2x - y - 1) \amp = 4 \\ -x + 2y \amp = 3 \end{aligned} \end{equation*}
We can now solve for \(y\) in the first of the new equations:
\begin{equation*} \begin{aligned} 3x + y \amp = 5 \\ y \amp = -3x + 5 \end{aligned} \end{equation*}
And plug that into the second of the new equations and solve for \(x\text{:}\)
\begin{equation*} \begin{aligned} -x + 2y \amp = 3 \\ -x + 2(-3x + 5) \amp = 3 \\ -7x + 10 \amp = 3 \\ x \amp = 1 \end{aligned} \end{equation*}
We can now work backwards to get the remaining variables.
\begin{equation*} \begin{aligned} y \amp = -3x + 5 \\ \amp = -3(1) + 5 \\ \amp = 2 \end{aligned} \qquad\qquad \begin{aligned} z \amp = 2x - y - 1 \\ \amp = 2(1) - (2) - 1 \\ \amp = -1 \end{aligned} \end{equation*}
The solution of the system of equations is \((x,y,z) = (1,2,-1)\text{.}\)

Subsection 16.5.2 Solving Three Dimensional Systems by Elimination

\begin{equation*} \left\{ \begin{array}{rcrcrcr} 2x \amp - \amp y \amp - \amp z \amp = \amp 1 \\ x \amp + \amp 2y \amp + \amp z \amp = \amp 4 \\ x \amp + \amp y \amp - \amp z \amp = \amp 4 \end{array} \right. \end{equation*}
We will first eliminate \(z\) twice by first combining the first two equations then the last two equations:
We will now combine these equations to eliminate \(x\text{:}\)
Then combine them again to eliminate \(y\text{:}\)
While we could be stubborn and go through this process again to solve for \(z\) using elimination, it makes much more sense to simply substitute the known values into one of the original equations to solve for \(z\text{.}\) We will use the first equation.
\begin{equation*} \begin{aligned} 2x - y - z \amp = 1 \\ 2(1) - (2) - z \amp = 1 \\ -z \amp = 1 \\ z \amp = -1 \end{aligned} \end{equation*}
The solution of the system of equations is \((x,y,z) = (1,2,-1)\text{.}\)