Types of Systems of Equations 

Basics of linear systems with the same number of equations as unknowns  covered on the main linear systems page 
Linear with fewer independent equations than unknowns  covered on this page 
NonLinear  covered on the nonlinear system page 
Consistent System With Infinite Solutions 

When we have a (consistent) linear system with infinitely many solutions or we have a linear system with fewer equations than unknowns, we have what is known as a dependent system. These two cases are essential the same, since in a consistent system we have the same number of equations as unknowns but one of the equations does not give us any more information than one of the other equations. So we can eliminate the duplicate equation and then solve the system as if we had fewer equations than unknowns.
The key to this technique is to know how to describe the answer since we have infinitely many solutions. Many instructors just tell you to stop your work once you realize this but knowing how to describe the answer is a valuable technique. So we show you how to do this here.
Let's start with the case where you are solving a consistent linear system and you realize that you have infinitely many solutions. There are a couple of ways to know this based on which technique you are using. They are equivalent but they look a bit different.
1. If you are using substitution or basic elimination, you will add two equations together and get an equation with no variables which is true all the time, like \(3=3\) or \(0=0\).
2. If you are using gaussian elimination (also called row reduction), you will get a row with all zeros.
In both of these cases, you have eliminated the redundent information and you now have fewer equations than unknowns. Take a few seconds to write out your remaining equations and you will see that the next section can now be applied to your system.
Okay, time for some practice problems.
Problem Statement 

\(3x2y+4z=1 \text{ and } x+y2z=3 \text{ and } 2x3y+6z=8\)
Solution 

video by mattemath 

close solution

Problem Statement 

\(x+2y7z=4 \text{ and } 2x+3y+z=5 \text{ and } 3x+7y36z=25\)
Solution 

video by mattemath 

close solution

Solve \( y = 2x+1; \) \( 4x+2y=2 \)
Fewer Equations Than Unknowns  NonSquare Systems 

At this point you have fewer equations than unknowns. That means there are infinitely many solutions to your system.
The first thing we do is to count the number of variables in the system and subtract the number of independent equations. This will give us the number of variables that we cannot account for. To learn how to handle this, let's watch this video. This guy explains the technique very well while he works a couple of examples.
video by Thinkwell 

So the idea is determine the number of variables that we cannot account for, usually just one, and solve the other variables in terms of the unknowns. It may seem strange to replace one variable with another one but this is the standard, so we will go with it.
Okay, time for some more practice problems.
Problem Statement 

\(xy+z=6 \text{ and } 2x+2y6z=4\)
Solution 

video by Thinkwell 

close solution

Problem Statement 

\(4x2y+6z=5 \text{ and } 2xy+3z=2\)
Solution 

video by Thinkwell 

close solution

Problem Statement 

\(xy+4z=3 \text{ and } 4xz=0\)
Solution 

video by mattemath 

close solution

Problem Statement 

\(12x+5y+z=0 \text{ and } 23x+2yz=0\)
Final Answer 

Problem Statement 

\(12x+5y+z=0 \text{ and } 23x+2yz=0\)
Solution 

video by mattemath 

Final Answer 

\((a/13, 5a/13,a)\) 
close solution

Problem Statement 

\(9x+5y+9z=3 \text{ and } 45x+10y+27z=6\)
Solution 

video by PatrickJMT 

close solution

Problem Statement 

\(10x6y+2z=0 \text{ and } 19x5yz=0\)
Final Answer 

Problem Statement 

\(10x6y+2z=0 \text{ and } 19x5yz=0\)
Solution 

video by mattemath 

Final Answer 

\((a/4,3a/4,a)\) 
close solution

Solve \( x + 2y  3z  4w = 10; \) \( x + 3y  3z  4w = 15; \) \( 2x + 2y  6z  8w = 10 \)
Problem Statement 

Solve \( x + 2y  3z  4w = 10; \) \( x + 3y  3z  4w = 15; \) \( 2x + 2y  6z  8w = 10 \)
Solution 

close solution

Solve \( xyz = 1; \) \( x+2y3z = 4 ;\) \( 3x2y7z = 0 \)
Problem Statement 

Solve \( xyz = 1; \) \( x+2y3z = 4 ;\) \( 3x2y7z = 0 \)
Solution 

close solution

Solve \( 2x3y+5z=4; \) \( x+y+2z=4; \) \( 3x2y+7z=8 \)
Problem Statement 

Solve \( 2x3y+5z=4; \) \( x+y+2z=4; \) \( 3x2y+7z=8 \)
Solution 

close solution

Problem Statement 

\(x+3y+2z=4 \text{ and } 2x+7yz=5\)
Final Answer 

Problem Statement 

\(x+3y+2z=4 \text{ and } 2x+7yz=5\)
Solution 

video by Jon Anderson  Learning Algebra 

Final Answer 

\((17z+13, 5z3,z)\) 
close solution

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