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 
NonLinear  covered on the nonlinear 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.
video by Thinkwell 

Graphically, the three possible (2dimensional) 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, sidebyside including graphs.
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.
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.
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.
\(5xy=6\)
\(2x+y=8\)
Problem Statement 

\(5xy=6\)
\(2x+y=8\)
Solution 

video by PatrickJMT 

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\(y=2x+2\)
\(y=6x2\)
Problem Statement 

\(y=2x+2\)
\(y=6x2\)
Solution 

video by MIP4U 

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\(y=x+3\)
\(2x2y=10\)
Problem Statement 

\(y=x+3\)
\(2x2y=10\)
Solution 

video by MIP4U 

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\(2x+y=4\)
\(y=2x5\)
Problem Statement 

\(2x+y=4\)
\(y=2x5\)
Solution 

video by MIP4U 

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\( y=x+3 \)
\( y=x+3 \)
Problem Statement 

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

video by Khan Academy 

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\(y=3x6\)
\(y=x+6\)
Problem Statement 

\(y=3x6\)
\(y=x+6\)
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.
video by MathOnPoint 

Let's work some practice problems.
Practice  Solve Using Substitution
Solve these linear systems using substitution.
\(2x+4y=4\)
\(y=x2\)
Problem Statement 

\(2x+4y=4\)
\(y=x2\)
Solution 

video by PatrickJMT 

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\(x+3y=6\)
\(2x+6y=12\)
Problem Statement 

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

video by PatrickJMT 

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\(2x3y=6\)
\(4x6y=12\)
Problem Statement 

\(2x3y=6\)
\(4x6y=12\)
Solution 

video by PatrickJMT 

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

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

video by Khan Academy 

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\(y=3x\)
\(2x+y=10\)
Problem Statement 

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

video by MIP4U 

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\(x2y=1\)
\(3x6y=18\)
Problem Statement 

\(x2y=1\)
\(3x6y=18\)
Solution 

video by MIP4U 

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\(x2y=7\)
\(2y3x=1\)
Problem Statement 

\(x2y=7\)
\(2y3x=1\)
Solution 

video by MIP4U 

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\(2xy=5\)
\(4x2y=10\)
Problem Statement 

\(2xy=5\)
\(4x2y=10\)
Solution 

video by MIP4U 

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

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

video by MathOnPoint 

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Here is a little more difficult problem that consists of 3 equations and 3 unknowns. Try substitution here as well.
\( 7x4y2z=7; 8x2y8z=1; x+2y6z=6 \)
Problem Statement 

\( 7x4y2z=7; 8x2y8z=1; x+2y6z=6 \)
Solution 

video by PatrickJMT 

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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.
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 

video by PatrickJMT 

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\(x3y=6\)
\(4x3y=10\)
Problem Statement 

\(x3y=6\)
\(4x3y=10\)
Solution 

video by PatrickJMT 

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\(4x2y=16\)
\(5x+2y=11\)
Problem Statement 

\(4x2y=16\)
\(5x+2y=11\)
Solution 

video by MIP4U 

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\(4x+3y=8\)
\(x3y=7\)
Problem Statement 

\(4x+3y=8\)
\(x3y=7\)
Solution 

video by MIP4U 

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

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

video by MIP4U 

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\(2x3y=1\)
\(4x+6y=5\)
Problem Statement 

\(2x3y=1\)
\(4x+6y=5\)
Solution 

video by MIP4U 

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\(5x+2y=4\)
\(5x+2y=2\)
Problem Statement 

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

video by MIP4U 

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

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

video by MIP4U 

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\(2x=6y+8\)
\(3x5y=2\)
Problem Statement 

\(2x=6y+8\)
\(3x5y=2\)
Final Answer 

Problem Statement 

\(2x=6y+8\)
\(3x5y=2\)
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.
\(4x4y+8z=20; 8x+4y4z=4; 12x8y12z=40\)
Problem Statement 

\(4x4y+8z=20; 8x+4y4z=4; 12x8y12z=40\)
Solution 

video by PatrickJMT 

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\(2xy+z=3; 5x+2y3z=1; 2x+yz=2\)
Problem Statement 

\(2xy+z=3; 5x+2y3z=1; 2x+yz=2\)
Solution 

video by PatrickJMT 

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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
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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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