\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Precalculus - Systems of Equations

Algebra

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Trigonometry

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Applications

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Calculus 1 Practice

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Types of Systems of Equations

Linear with the same number of equations as unknowns - covered on this page

Linear with fewer equations than unknowns - covered on the dependent system page

Non-Linear - covered on the non-linear system page

Systems of equations are made up of a set of two or more equations that contain multiple variables. Linear systems have all linear terms, i.e. all the variables have a power of one and none of the variables are multiplied together. Nonlinear systems have at least one variable with a power other than one, like a square root or a squared term, or have variables multiplied together.

Linear Systems of Equations - 3 Cases

For example, here is a linear system of two equations and two unknowns, x and y.
\( \begin{array}{rcrcl} 2x & + & y & = & 3 \\ x & - & 3y & = & 0 \end{array} \)
Notice that we have two variables and two equations. In this case, we have three possible types of solutions.
1. We have one, unique solution. If we think of these equations as lines in the plane, there is one point where they intersect.
2. We have no solution. In this case, the lines would be parallel and never intersect.
3. We have an infinite number of solutions, i.e. the lines completely overlap and the two equations represent the same line. When we have an infinite number of solutions or we have fewer equations than variables, then we have what is called a dependent system of equations. We can get some equations that 'solve' the system, i.e. we can get a set of equations that represent the infinite set of solutions. We cover that idea on a separate page. On this page, we will stick with linear systems where we have the same number of equations as variables.

Before we go on, let's watch a quick video clip about some terminology related to these three cases of linear systems.

Thinkwell - Three Cases for Linear Systems [1min-10secs]

video by Thinkwell

Graphically, the three possible (2-dimensional) cases are shown here. The idea is to find where the lines intersect.

1. One, Unique Solution

2. Infinite Solutions

3. No Solution

Case 1: One, Unique Solution
In this case, the two lines intersect at exactly one point. This is the easiest, most common and the best case possible. The other two cases are usually considered special situations.

Case 2: Infinite Solutions
In this case, the two lines are exactly the same. In the plot above, it looks like we have only one line. However, when we have two equations and we plot them, the lines are the same and so in graphs it looks like there is only one line. Since the lines are the same, the lines intersect at every point and, therefore, we say there are an infinite number of solutions.

Case 3: No Solution
When the lines are parallel, they will never intersect. So we say there is no solution since there is no point where they intersect.

Note: Some books and instructors switch the two cases above and call case 2 no solution and case 3 infinite solutions. The order does not matter, so it is best to follow your instructor and book.

Here is another good video showing all three cases, side-by-side including graphs.

Khan Academy - Solving systems of equations (2x2), the special solution cases [9min-11secs]

video by Khan Academy

Overview of Techniques

Solving systems of equations can be intuitively thought of as finding where the equations intersect. This is a great way to get a feel for what is happening. There are 4 main ways to solve systems of equations.
1. graphing
2. substitution
3. elimination
4. using matrices: Gaussian Elimination/Row Reduction, inverse matrices and Cramer's Rule.

This introduction video goes into a little more detail on these 4 techniques. The presenter actually crosses out the use of matrices option but on this site we cover that technique on other pages.

Brightstorm - Introduction to Systems of Equations [3min-14secs]

video by Brightstorm

Important Last Step: Checking Your Answers

Before we discuss specific techniques, let's talk about how to make sure you have the correct answer. I know what you are thinking, you are busy and it takes time to check your answer. Besides, you were careful when you worked the problem. That's true. However, think about it this way. If you made a mistake, you will not get full credit for your work and some of your time was wasted. And you have a way to check your answer (not always possible in higher math) and if you don't check it, you will regret it later. If only I had checked my answer! I might have gotten an A! It takes just a few seconds and if you do find an error, you may learn how to not repeat it and improve your grade on your next exam. So it is well worth the time to check.

To check your answer, just plug the point into all of the original equations. Using the original equations is extremely important, since there could be a mistake at any point. And make sure and check all of the equations, since your answer may work in one but not all of them.

Okay, let's start with solving these problems by discussing 3 basic techniques, graphing, substitution and elimination.

Solve by Graphing

If you have only two equations and two unknowns, graphing can sometimes be used to solve the system of equations. We say 'sometimes' since this technique is not very exact and works only if you can graph very, very accurately and read the graph accurately. But for starters it is good to learn to do this to get a feel for what you are actually doing with the equations.
Here is a good video to get you started. He explains how to solve by graphing while doing an example. He also shows how to check your answer, an important step when working these kinds of problems.

Khan Academy - Graphing Systems of Equations [6min-34secs]

video by Khan Academy

Okay, let's work some practice problems before we go on.

Practice - Solve By Graphing

Solve these linear systems by graphing.

\(5x-y=6\)
\(2x+y=8\)

Problem Statement

\(5x-y=6\)
\(2x+y=8\)

Solution

1705 video

video by PatrickJMT

close solution

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\(y=-2x+2\)
\(y=-6x-2\)

Problem Statement

\(y=-2x+2\)
\(y=-6x-2\)

Solution

1706 video

video by MIP4U

close solution

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\(y=-x+3\)
\(2x-2y=10\)

Problem Statement

\(y=-x+3\)
\(2x-2y=10\)

Solution

1708 video

video by MIP4U

close solution

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\(2x+y=4\)
\(y=-2x-5\)

Problem Statement

\(2x+y=4\)
\(y=-2x-5\)

Solution

1709 video

video by MIP4U

close solution

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\( y=x+3 \)
\( y=-x+3 \)

Problem Statement

\( y=x+3 \)
\( y=-x+3 \)

Solution

2128 video

video by Khan Academy

close solution

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\(y=3x-6\)
\(y=-x+6\)

Problem Statement

\(y=3x-6\)
\(y=-x+6\)

Solution

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Solve by Substitution

A second technique to solve systems of equations is substitution. Just like graphing, this technique is limited to two equations with two unknowns since higher order systems become quite complicated very quickly. Also this technique will often produce fractions very quickly, making the algebra more difficult. However, this is a handy technique to learn since you will use it later on with more effect.
Here is a good video showing this technique using an example. He doesn't check his answer in this video, which is not something you want to skip. Also, notice that his answer contains fractions, so graphing would not be the way to solve this problem.

MathOnPoint - Solving Linear Systems Substitution Method [7min-58secs]

video by MathOnPoint

Let's work some practice problems.

Practice - Solve Using Substitution

Solve these linear systems using substitution.

\(2x+4y=4\)
\(y=x-2\)

Problem Statement

\(2x+4y=4\)
\(y=x-2\)

Solution

1710 video

video by PatrickJMT

close solution

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\(x+3y=6\)
\(2x+6y=-12\)

Problem Statement

\(x+3y=6\)
\(2x+6y=-12\)

Solution

1711 video

video by PatrickJMT

close solution

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\(2x-3y=6\)
\(4x-6y=12\)

Problem Statement

\(2x-3y=6\)
\(4x-6y=12\)

Solution

1712 video

video by PatrickJMT

close solution

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\(x+2y=9\)
\(3x+5y=20\)

Problem Statement

\(x+2y=9\)
\(3x+5y=20\)

Solution

1713 video

video by Khan Academy

close solution

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\(y=3x\)
\(2x+y=-10\)

Problem Statement

\(y=3x\)
\(2x+y=-10\)

Solution

1715 video

video by MIP4U

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\(x-2y=1\)
\(3x-6y=-18\)

Problem Statement

\(x-2y=1\)
\(3x-6y=-18\)

Solution

1716 video

video by MIP4U

close solution

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\(x-2y=7\)
\(2y-3x=-1\)

Problem Statement

\(x-2y=7\)
\(2y-3x=-1\)

Solution

1717 video

video by MIP4U

close solution

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\(2x-y=-5\)
\(4x-2y=-10\)

Problem Statement

\(2x-y=-5\)
\(4x-2y=-10\)

Solution

1718 video

video by MIP4U

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\(x+3y=12; 2x+y=6\)

Problem Statement

\(x+3y=12; 2x+y=6\)

Solution

2129 video

video by MathOnPoint

close solution

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Here is a little more difficult problem that consists of 3 equations and 3 unknowns. Try substitution here as well.

\( 7x-4y-2z=7; 8x-2y-8z=1; x+2y-6z=-6 \)

Problem Statement

\( 7x-4y-2z=7; 8x-2y-8z=1; x+2y-6z=-6 \)

Solution

1714 video

video by PatrickJMT

close solution

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Solve by Elimination

This is the best technique to solve systems of equations since it works all the time, you can control the numbers to avoid fractions until near the end of the solution and it will prepare you for the fourth technique using matrices. So it is important to learn this technique well.
Here is a good video explaining this technique while working an example, AND he explains why this technique works.

Khan Academy - Solving Systems of Equations by Elimination [12min-43secs]

video by Khan Academy

Practice - Solve Using Elimination

Solve these linear systems using elimination.

\(2x+3y=4\)
\(-2x+7y=16\)

Problem Statement

\(2x+3y=4\)
\(-2x+7y=16\)

Solution

1719 video

video by PatrickJMT

close solution

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\(x-3y=6\)
\(4x-3y=10\)

Problem Statement

\(x-3y=6\)
\(4x-3y=10\)

Solution

1720 video

video by PatrickJMT

close solution

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\(4x-2y=16\)
\(5x+2y=11\)

Problem Statement

\(4x-2y=16\)
\(5x+2y=11\)

Solution

1721 video

video by MIP4U

close solution

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\(4x+3y=8\)
\(x-3y=7\)

Problem Statement

\(4x+3y=8\)
\(x-3y=7\)

Solution

1724 video

video by MIP4U

close solution

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\(3x+5y=4\)
\(-2x+3y=10\)

Problem Statement

\(3x+5y=4\)
\(-2x+3y=10\)

Solution

1725 video

video by MIP4U

close solution

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\(2x-3y=-1\)
\(-4x+6y=5\)

Problem Statement

\(2x-3y=-1\)
\(-4x+6y=5\)

Solution

1726 video

video by MIP4U

close solution

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\(5x+2y=4\)
\(5x+2y=-2\)

Problem Statement

\(5x+2y=4\)
\(5x+2y=-2\)

Solution

1727 video

video by MIP4U

close solution

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\(3x+y=-10; 7x+5y=-18\)

Problem Statement

\(3x+y=-10; 7x+5y=-18\)

Solution

1728 video

video by MIP4U

close solution

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\(2x=-6y+8\)
\(3x-5y=2\)

Problem Statement

\(2x=-6y+8\)
\(3x-5y=2\)

Final Answer

\((13/7,5/7)\)

Problem Statement

\(2x=-6y+8\)
\(3x-5y=2\)

Solution

The instructor in this video runs out of time and does not finish the problem. He gets \(y=5/7\) and then stops. Using his work, here is how to get the final answer.
Substituting \(y=5/7\) into the first original equation, we have
\(\begin{array}{rcl} 2x & = & -6y+8 \\ 2x & = & -6(5/7)+8 \\ 2x & = & -30/7+56/7 \\ 2x & = & 26/7 \\ x & = & 13/7 \end{array} \)

Final Answer

\((13/7,5/7)\)

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Practice - 3x3 Elimination

Here are a few 3x3 systems but elimination will work with these systems as well.

\(4x-4y+8z=20; 8x+4y-4z=4; 12x-8y-12z=-40\)

Problem Statement

\(4x-4y+8z=20; 8x+4y-4z=4; 12x-8y-12z=-40\)

Solution

1722 video

video by PatrickJMT

close solution

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\(2x-y+z=3; 5x+2y-3z=1; 2x+y-z=2\)

Problem Statement

\(2x-y+z=3; 5x+2y-3z=1; 2x+y-z=2\)

Solution

1723 video

video by PatrickJMT

close solution

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Next

If you haven't already, work some practice problems from the sections above. After that, your next step is to learn how to use matrices to solve systems of equations. Gaussian Elimination is the best place to start. You will learn some general information about matrices, what they are and how to work with them. Then we show you how to solve systems of equations using basic matrix operations. Much of what you learned on this page will help you, so make sure you are comfortable with these techniques, especially elimination, before moving on.

Really UNDERSTAND Precalculus

Topics You Need To Understand For This Page

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Topics Listed Alphabetically

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Differential Equations

Precalculus

Engineering

Circuits

Semiconductors

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