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Systems of Linear Equations: Two Variables
Solve systems of equations by graphing.
Systems of Linear Equations: Two Variables
- Determine whether an ordered pair is a solution of a system of equations (IA 4.1.1)
- Solve a system of linear equations by graphing (IA 4.1.2)
A system of linear equations is a group of two or more linear equations. For example,
\[\left\{\begin{array}{l}y=-2x+5 \\ y=2x+7\end{array}\right\]is a system of linear equations
A solution to a system of linear equations is an ordered pair x,y that is a solution to every equation in the system.
Example
Determine whether the ordered pairs are solutions to the given system.
Try it.
\(\left\{\begin{array}{l}2x-6y=0 \\ 3x-y=5\end{array}\right\) at \((3,1)\) and \((-3,4)\)
Solution
We substitute (3, 1) into both equations:
| \(2x-6y=0\) | \(3x-4y=5\) |
| \(\begin{array}{l}2(3)-6(1)=0 \\ 6-6=0 \\ 0=0\end{array}\) True | \(\begin{array}{l}3(3)-4(1)=5 \\ 9-4=5 \\ 5=5\end{array}\) True |
| \((3,1)\) is a solution to \(2x-6y=0\) | \((3,1)\) is a solution to \(3x-4y=5\) |
| Conclusion: since \((3,1)\) is a solution to both equations, then it is a solution to the system \(\left\{\begin{array}{l}2x-6y=0 \\ 3x-4y=5\end{array}\right\) |
Next we substitute (-3, -1) into both equations:
| \(2x-6y=0\) | \(3x-4y=5\) |
| \(\begin{array}{l}2(-3)-6(-1)=0 \\ -6+6=0 \\ 0=0\end{array}\) True | \(\begin{array}{l}3(-3)-4(-1)=5 \\ -9+4=5 \\ -5=5\end{array}\) False |
| \((-3,-1)\) is a solution to \(2x-6y=0\) | \((-3,-1)\) is not a solution to \(3x-4y=5\) |
| Conclusion: Since \((-3,-1)\) is not a solution to one of the equations, then it is not a solution to the system \(\left\{\begin{array}{l}2x-6y=0 \\ 3x-4y=5\end{array}\right\) |
Determine whether the ordered pairs are solutions to the given system.
\(\left\{\begin{array}{l}3x+y=0 \\ x+2y=-5\end{array}\right\) at \((0,0)\) and \((1,-3)\)
Try it.
At (0, 0):
| \(3x+y=0\) | \(x+2y=-5\) |
| ________________________ ________________________ | ________________________ ________________________ |
| ________________________ ________________________ | ________________________ ________________________ |
| Conclusion: ________________________ |
Try it.
At (1, –3)
| \(3x+y=0\) | \(x+2y=-5\) |
| ________________________ ________________________ | ________________________ ________________________ |
| ________________________ ________________________ | ________________________ ________________________ |
| Conclusion: ________________________ |
Condensed — the full section is in OpenStax College Algebra 2e.
Introduction to Systems of Equations
In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.
In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.
\[\begin{array}{l}2x+y=\ 15 \\ 3x-y=\ 5\end{array}\]The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.
\[\begin{array}{l}2(4)+(7)=15\ \text{True} \\ 3(4)-(7)=5\ \text{True}\end{array}\]In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.
Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different y-intercepts. There are no points common to both lines; hence, there is no solution to the system.
Condensed — the full section is in OpenStax College Algebra 2e.
Solving Systems of Equations by Graphing
There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.
Example
Try it.
Solve the following system of equations by graphing. Identify the type of system.
\[\begin{array}{l}2x+y=-8 \\ x-y=-1\end{array}\]Solution
Solve the first equation for \(y.\)
\[\begin{array}{l}2x+y=-8 \\ y=-2x-8\end{array}\]Solve the second equation for \(y.\)
\[\begin{array}{l}x-y=-1 \\ y=x+1\end{array}\]Graph both equations on the same set of axes as in .
The lines appear to intersect at the point \((-3,-2).\) We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.
\[\begin{array}{ll}2(-3)+(-2)=-8 & \\ -8=-8 & \text{True} \\ (-3)-(-2)=-1 & \\ -1=-1 & \text{True}\end{array}\]The solution to the system is the ordered pair \((-3,-2),\) so the system is independent.
Solving Systems of Equations by Substitution
Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.
Example
Try it.
Solve the following system of equations by substitution.
\[\begin{array}{l}-x+y=-5 \\ 2x-5y=1\end{array}\]Solution
First, we will solve the first equation for \(y.\)
\[\begin{array}{l}-x+y=-5 \\ y=x-5\end{array}\]Now we can substitute the expression \(x-5\) for \(y\) in the second equation.
\[\begin{array}{l}2x-5y=1 \\ 2x-5(x-5)=1 \\ 2x-5x+25=1 \\ -3x=-24 \\ x=8\end{array}\]Now, we substitute \(x=8\) into the first equation and solve for \(y.\)
\[\begin{array}{l}-(8)+y=-5 \\ y=3\end{array}\]Our solution is \((8,3).\)
Check the solution by substituting \((8,3)\) into both equations.
\[\begin{array}{llll}-x+y=-5 & & & \\ -(8)+(3)=-5 & & & \text{True} \\ 2x-5y=1 & & & \\ 2(8)-5(3)=1 & & & \text{True}\end{array}\]Solving Systems of Equations in Two Variables by the Addition Method
A third method of solving systems of linear equations is the addition method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.
Example
Try it.
Solve the given system of equations by addition.
\[\begin{array}{l}x+2y=-1 \\ -x+y=3\end{array}\]Solution
Both equations are already set equal to a constant. Notice that the coefficient of \(x\) in the second equation, –1, is the opposite of the coefficient of \(x\) in the first equation, 1. We can add the two equations to eliminate \(x\) without needing to multiply by a constant.
\[\frac{\begin{array}{l} \\ x+2y=-1 \\ -x+y=3\end{array}}{\ \ \ 3y=2}\]Now that we have eliminated \(x,\) we can solve the resulting equation for \(y.\)
\[\begin{array}{l}3y=2 \\ y=\frac{2}{3}\end{array}\]Then, we substitute this value for \(y\) into one of the original equations and solve for \(x.\)
\[\begin{array}{l}-x+y=3 \\ -x+\frac{2}{3}=3 \\ -x=3-\frac{2}{3} \\ -x=\frac{7}{3} \\ x=-\frac{7}{3}\end{array}\]The solution to this system is \((-\frac{7}{3},\frac{2}{3}).\)
Check the solution in the first equation.
\[\begin{array}{llll}x+2y=-1 & & & \\ (-\frac{7}{3})+2(\frac{2}{3})= & & & \\ -\frac{7}{3}+\frac{4}{3}= & & & \\ -\frac{3}{3}= & & & \\ -1=-1 & & & \text{True}\end{array}\]Condensed — the full section is in OpenStax College Algebra 2e.
Identifying Inconsistent Systems of Equations Containing Two Variables
Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different \(y\) -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as \(12=0.\)
Example
Try it.
Solve the following system of equations.
\[\begin{array}{l}x=9-2y \\ x+2y=13\end{array}\]Solution
We can approach this problem in two ways. Because one equation is already solved for \(x,\) the most obvious step is to use substitution.
\[\begin{array}{l}\ x+2y=13 \\ (9-2y)+2y=13 \\ 9+0y=13 \\ 9=13\end{array}\]Clearly, this statement is a contradiction because \(9\ne 13.\) Therefore, the system has no solution.
The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.
\[\begin{array}{l}\ x=9-2y \\ 2y=-x+9 \\ y=-\frac{1}{2}x+\frac{9}{2}\end{array}\]We then convert the second equation expressed to slope-intercept form.
\[\begin{array}{l}x+2y=13 \\ 2y=-x+13 \\ y=-\frac{1}{2}x+\frac{13}{2}\end{array}\]Comparing the equations, we see that they have the same slope but different y-intercepts. Therefore, the lines are parallel and do not intersect.
\[\begin{array}{l}\begin{array}{l} \\ y=-\frac{1}{2}x+\frac{9}{2}\end{array} \\ y=-\frac{1}{2}x+\frac{13}{2}\end{array}\]Expressing the Solution of a System of Dependent Equations Containing Two Variables
Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as \(0=0.\)
Example
Try it.
Find a solution to the system of equations using the addition method.
\[\begin{array}{l}\ x+3y=2 \\ 3x+9y=6\end{array}\]Solution
With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating \(x.\) If we multiply both sides of the first equation by \(-3,\) then we will be able to eliminate the \(x\) -variable.
\[\begin{array}{l}\ x+3y=2 \\ (-3)(x+3y)=(-3)(2) \\ -3x-9y=-6\end{array}\]Now add the equations.
\[\begin{array}{lll}\underset{______________}{\begin{array}{ll}\ -3x-9y & =-6 \\ +\ 3x+9y & =6\end{array}} \\ \begin{array}{ll}\ 0 & =0\end{array}\end{array}\]We can see that there will be an infinite number of solutions that satisfy both equations.
Using Systems of Equations to Investigate Profits
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation \(R=xp,\) where \(x=\) quantity and \(p=\) price. The revenue function is shown in orange in .
The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in . The \(x\) -axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars.
The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.
The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as \(P(x)=R(x)-C(x).\) Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.
Example
Try it.
Given the cost function \(C(x)=0.85x+35,000\) and the revenue function \(R(x)=1.55x,\) find the break-even point and the profit function.
Solution
Write the system of equations using \(y\) to replace function notation.
\[\begin{array}{l}\begin{array}{l} \\ y=0.85x+35,000\end{array} \\ y=1.55x\end{array}\]Substitute the expression \(0.85x+35,000\) from the first equation into the second equation and solve for \(x.\)
\[\begin{array}{l}0.85x+35,000=1.55x \\ 35,000=0.7x \\ 50,000=x\end{array}\]Then, we substitute \(x=50,000\) into either the cost function or the revenue function.
\[1.55(50,000)=77,500\]The break-even point is \((50,000,77,500).\)
The profit function is found using the formula \(P(x)=R(x)-C(x).\)
\[\begin{array}{l}P(x)=1.55x-(0.85x+35,000) \\ \ =0.7x-35,000\end{array}\]The profit function is \(P(x)=0.7x-35,000.\)
Condensed — the full section is in OpenStax College Algebra 2e.
Key Concepts
- A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.
- The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See .
- Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
- One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See .
- Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See .
- A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See .
- It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See , , and .
- Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See .
- The solution to a system of dependent equations will always be true because both equations describe the same line. See .
- Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See and .
Systems of Linear Equations: Two Variables
- Determine whether an ordered pair is a solution of a system of equations (IA 4.1.1)
- Solve a system of linear equations by graphing (IA 4.1.2)
A system of linear equations is a group of two or more linear equations. For example,
\[\left\{\begin{array}{l}y=-2x+5 \\ y=2x+7\end{array}\right\]is a system of linear equations
A solution to a system of linear equations is an ordered pair x,y that is a solution to every equation in the system.
Example
Determine whether the ordered pairs are solutions to the given system.
Try it.
\(\left\{\begin{array}{l}2x-6y=0 \\ 3x-y=5\end{array}\right\) at \((3,1)\) and \((-3,4)\)
Solution
We substitute (3, 1) into both equations:
| \(2x-6y=0\) | \(3x-4y=5\) |
| \(\begin{array}{l}2(3)-6(1)=0 \\ 6-6=0 \\ 0=0\end{array}\) True | \(\begin{array}{l}3(3)-4(1)=5 \\ 9-4=5 \\ 5=5\end{array}\) True |
| \((3,1)\) is a solution to \(2x-6y=0\) | \((3,1)\) is a solution to \(3x-4y=5\) |
| Conclusion: since \((3,1)\) is a solution to both equations, then it is a solution to the system \(\left\{\begin{array}{l}2x-6y=0 \\ 3x-4y=5\end{array}\right\) |
Next we substitute (-3, -1) into both equations:
| \(2x-6y=0\) | \(3x-4y=5\) |
| \(\begin{array}{l}2(-3)-6(-1)=0 \\ -6+6=0 \\ 0=0\end{array}\) True | \(\begin{array}{l}3(-3)-4(-1)=5 \\ -9+4=5 \\ -5=5\end{array}\) False |
| \((-3,-1)\) is a solution to \(2x-6y=0\) | \((-3,-1)\) is not a solution to \(3x-4y=5\) |
| Conclusion: Since \((-3,-1)\) is not a solution to one of the equations, then it is not a solution to the system \(\left\{\begin{array}{l}2x-6y=0 \\ 3x-4y=5\end{array}\right\) |
Determine whether the ordered pairs are solutions to the given system.
\(\left\{\begin{array}{l}3x+y=0 \\ x+2y=-5\end{array}\right\) at \((0,0)\) and \((1,-3)\)
Try it.
At (0, 0):
| \(3x+y=0\) | \(x+2y=-5\) |
| ________________________ ________________________ | ________________________ ________________________ |
| ________________________ ________________________ | ________________________ ________________________ |
| Conclusion: ________________________ |
Try it.
At (1, –3)
| \(3x+y=0\) | \(x+2y=-5\) |
| ________________________ ________________________ | ________________________ ________________________ |
| ________________________ ________________________ | ________________________ ________________________ |
| Conclusion: ________________________ |
Condensed — the full section is in OpenStax Precalculus 2e.
Introduction to Systems of Equations
In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.
In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.
\[\begin{array}{l}2x+y=\ 15 \\ 3x-y=\ 5\end{array}\]The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.
\[\begin{array}{l}2(4)+(7)=15\ \text{True} \\ 3(4)-(7)=5\ \text{True}\end{array}\]In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.
Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different y-intercepts. There are no points common to both lines; hence, there is no solution to the system.
Condensed — the full section is in OpenStax Precalculus 2e.
Solving Systems of Equations by Graphing
There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.
Example
Try it.
Solve the following system of equations by graphing. Identify the type of system.
\[\begin{array}{l}2x+y=-8 \\ x-y=-1\end{array}\]Solution
Solve the first equation for \(y.\)
\[\begin{array}{l}2x+y=-8 \\ y=-2x-8\end{array}\]Solve the second equation for \(y.\)
\[\begin{array}{l}x-y=-1 \\ y=x+1\end{array}\]Graph both equations on the same set of axes as in .
The lines appear to intersect at the point \((-3,-2).\) We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.
\[\begin{array}{ll}2(-3)+(-2)=-8 & \\ -8=-8 & \text{True} \\ (-3)-(-2)=-1 & \\ -1=-1 & \text{True}\end{array}\]The solution to the system is the ordered pair \((-3,-2),\) so the system is independent.
Solving Systems of Equations by Substitution
Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.
Example
Try it.
Solve the following system of equations by substitution.
\[\begin{array}{l}-x+y=-5 \\ 2x-5y=1\end{array}\]Solution
First, we will solve the first equation for \(y.\)
\[\begin{array}{l}-x+y=-5 \\ y=x-5\end{array}\]Now we can substitute the expression \(x-5\) for \(y\) in the second equation.
\[\begin{array}{l}2x-5y=1 \\ 2x-5(x-5)=1 \\ 2x-5x+25=1 \\ -3x=-24 \\ x=8\end{array}\]Now, we substitute \(x=8\) into the first equation and solve for \(y.\)
\[\begin{array}{l}-(8)+y=-5 \\ y=3\end{array}\]Our solution is \((8,3).\)
Check the solution by substituting \((8,3)\) into both equations.
\[\begin{array}{llll}-x+y=-5 & & & \\ -(8)+(3)=-5 & & & \text{True} \\ 2x-5y=1 & & & \\ 2(8)-5(3)=1 & & & \text{True}\end{array}\]Solving Systems of Equations in Two Variables by the Addition Method
A third method of solving systems of linear equations is the addition method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.
Example
Try it.
Solve the given system of equations by addition.
\[\begin{array}{l}x+2y=-1 \\ -x+y=3\end{array}\]Solution
Both equations are already set equal to a constant. Notice that the coefficient of \(x\) in the second equation, –1, is the opposite of the coefficient of \(x\) in the first equation, 1. We can add the two equations to eliminate \(x\) without needing to multiply by a constant.
\[\frac{\begin{array}{l} \\ x+2y=-1 \\ -x+y=3\end{array}}{\ \ \ 3y=2}\]Now that we have eliminated \(x,\) we can solve the resulting equation for \(y.\)
\[\begin{array}{l}3y=2 \\ y=\frac{2}{3}\end{array}\]Then, we substitute this value for \(y\) into one of the original equations and solve for \(x.\)
\[\begin{array}{l}-x+y=3 \\ -x+\frac{2}{3}=3 \\ -x=3-\frac{2}{3} \\ -x=\frac{7}{3} \\ x=-\frac{7}{3}\end{array}\]The solution to this system is \((-\frac{7}{3},\frac{2}{3}).\)
Check the solution in the first equation.
\[\begin{array}{llll}x+2y=-1 & & & \\ (-\frac{7}{3})+2(\frac{2}{3})= & & & \\ -\frac{7}{3}+\frac{4}{3}= & & & \\ -\frac{3}{3}= & & & \\ -1=-1 & & & \text{True}\end{array}\]Condensed — the full section is in OpenStax Precalculus 2e.
Identifying Inconsistent Systems of Equations Containing Two Variables
Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different \(y\) -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as \(12=0.\)
Example
Try it.
Solve the following system of equations.
\[\begin{array}{l}x=9-2y \\ x+2y=13\end{array}\]Solution
We can approach this problem in two ways. Because one equation is already solved for \(x,\) the most obvious step is to use substitution.
\[\begin{array}{l}\ x+2y=13 \\ (9-2y)+2y=13 \\ 9+0y=13 \\ 9=13\end{array}\]Clearly, this statement is a contradiction because \(9\ne 13.\) Therefore, the system has no solution.
The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.
\[\begin{array}{l}\ x=9-2y \\ 2y=-x+9 \\ y=-\frac{1}{2}x+\frac{9}{2}\end{array}\]We then convert the second equation expressed to slope-intercept form.
\[\begin{array}{l}x+2y=13 \\ 2y=-x+13 \\ y=-\frac{1}{2}x+\frac{13}{2}\end{array}\]Comparing the equations, we see that they have the same slope but different y-intercepts. Therefore, the lines are parallel and do not intersect.
\[\begin{array}{l}\begin{array}{l} \\ y=-\frac{1}{2}x+\frac{9}{2}\end{array} \\ y=-\frac{1}{2}x+\frac{13}{2}\end{array}\]Expressing the Solution of a System of Dependent Equations Containing Two Variables
Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as \(0=0.\)
Example
Try it.
Find a solution to the system of equations using the addition method.
\[\begin{array}{l}\ x+3y=2 \\ 3x+9y=6\end{array}\]Solution
With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating \(x.\) If we multiply both sides of the first equation by \(-3,\) then we will be able to eliminate the \(x\) -variable.
\[\begin{array}{l}\ x+3y=2 \\ (-3)(x+3y)=(-3)(2) \\ -3x-9y=-6\end{array}\]Now add the equations.
\[\begin{array}{lll}\underset{______________}{\begin{array}{ll}\ -3x-9y & =-6 \\ +\ 3x+9y & =6\end{array}} \\ \begin{array}{ll}\ 0 & =0\end{array}\end{array}\]We can see that there will be an infinite number of solutions that satisfy both equations.
Using Systems of Equations to Investigate Profits
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation \(R=xp,\) where \(x=\) quantity and \(p=\) price. The revenue function is shown in orange in .
The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in . The \(x\) -axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars.
The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.
The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as \(P(x)=R(x)-C(x).\) Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.
Example
Try it.
Given the cost function \(C(x)=0.85x+35,000\) and the revenue function \(R(x)=1.55x,\) find the break-even point and the profit function.
Solution
Write the system of equations using \(y\) to replace function notation.
\[\begin{array}{l}\begin{array}{l} \\ y=0.85x+35,000\end{array} \\ y=1.55x\end{array}\]Substitute the expression \(0.85x+35,000\) from the first equation into the second equation and solve for \(x.\)
\[\begin{array}{l}0.85x+35,000=1.55x \\ 35,000=0.7x \\ 50,000=x\end{array}\]Then, we substitute \(x=50,000\) into either the cost function or the revenue function.
\[1.55(50,000)=77,500\]The break-even point is \((50,000,77,500).\)
The profit function is found using the formula \(P(x)=R(x)-C(x).\)
\[\begin{array}{l}P(x)=1.55x-(0.85x+35,000) \\ \ =0.7x-35,000\end{array}\]The profit function is \(P(x)=0.7x-35,000.\)
Condensed — the full section is in OpenStax Precalculus 2e.
Key Concepts
- A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.
- The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See .
- Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
- One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See .
- Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See .
- A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See .
- It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See , , and .
- Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See .
- The solution to a system of dependent equations will always be true because both equations describe the same line. See .
- Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See and .
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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\(\left\{\begin{array}{l}2x-6y=0 \\ 3x-y=5\end{array}\right\) at \((3,1)\) and \((-3,4)\)
答えを明らかにしろ
We substitute (3, 1) into both equations:
\(2x-6y=0\) \(3x-4y=5\) \(\begin{array}{l}2(3)-6(1)=0 \\ 6-6=0 \\ 0=0\end{array}\) True \(\begin{array}{l}3(3)-4(1)=5 \\ 9-4=5 \\ 5=5\end{array}\) True \((3,1)\) is a solution to \(2x-6y=0\) \((3,1)\) is a solution to \(3x-4y=5\) Conclusion: since \((3,1)\) is a solution to both equations, then it is a solution to the system \(\left\{\begin{array}{l}2x-6y=0 \\ 3x-4y=5\end{array}\right\) Next we substitute (-3, -1) into both equations:
\(2x-6y=0\) \(3x-4y=5\) \(\begin{array}{l}2(-3)-6(-1)=0 \\ -6+6=0 \\ 0=0\end{array}\) True \(\begin{array}{l}3(-3)-4(-1)=5 \\ -9+4=5 \\ -5=5\end{array}\) False \((-3,-1)\) is a solution to \(2x-6y=0\) \((-3,-1)\) is not a solution to \(3x-4y=5\) Conclusion: Since \((-3,-1)\) is not a solution to one of the equations, then it is not a solution to the system \(\left\{\begin{array}{l}2x-6y=0 \\ 3x-4y=5\end{array}\right\) -
At (0, 0):
\(3x+y=0\) \(x+2y=-5\) ________________________
________________________________________________
________________________________________________
________________________________________________
________________________Conclusion: ________________________ -
At (1, –3)
\(3x+y=0\) \(x+2y=-5\) ________________________
________________________________________________
________________________________________________
________________________________________________
________________________Conclusion: ________________________ -
\(\left\{\begin{array}{l}-x+y=1 \\ 2x+y=10\end{array}\right\)
答えを明らかにしろ
Step 1 Graph \(-x+y=1\)
We can use the slope intercept – form: \(y=x+1\)
Slope = 1
y-intercept: (0, 1)
Step 2 Graph \(2x+y=10\)
We can use the slope intercept – form: \(y=-2x+10\)
Slope = –2
y-intercept: (0, 10)
Step 3 The lines intersect Step 4 The solution is the point (3, 4) Step 5 Let’s check the solution:
\(\begin{array}{ll}-x+y=1 & 2x+y=10 \\ -3+4=1 & 2(3)+4=10 \\ 1=1 & 10=10\end{array}\)
Since (3, 4) is a solution to both equations, then it is a solution to the system
\(-x+y=1\)
\(2x+y=10\) -
Determine whether the ordered pair is a solution to the given system \(\left\{\begin{array}{l}x-3y=-8 \\ -3x-y=4\end{array}\right\)
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Solve the following system by graphing. \(\left\{\begin{array}{l}y=-\frac{1}{4}x+2 \\ x+4y=8\end{array}\right\)
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Determine whether the ordered pair \((5,1)\) is a solution to the given system of equations.
\[\begin{array}{l}\ x+3y=8 \\ 2x-9=y\end{array}\]答えを明らかにしろ
Substitute the ordered pair \((5,1)\) into both equations.
\[\begin{array}{ll}(5)+3(1)=8 & \\ 8=8 & \text{True} \\ 2(5)-9=(1) & \\ \text{1=1} & \text{True}\end{array}\]The ordered pair \((5,1)\) satisfies both equations, so it is the solution to the system.
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Determine whether the ordered pair \((8,5)\) is a solution to the following system.
\[\begin{array}{l}5x-4y=20 \\ 2x+1=3y\end{array}\]答えを明らかにしろ
Not a solution.
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Solve the following system of equations by graphing. Identify the type of system.
\[\begin{array}{l}2x+y=-8 \\ x-y=-1\end{array}\]答えを明らかにしろ
Solve the first equation for \(y.\)
\[\begin{array}{l}2x+y=-8 \\ y=-2x-8\end{array}\]Solve the second equation for \(y.\)
\[\begin{array}{l}x-y=-1 \\ y=x+1\end{array}\]Graph both equations on the same set of axes as in .
The lines appear to intersect at the point \((-3,-2).\) We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.
\[\begin{array}{ll}2(-3)+(-2)=-8 & \\ -8=-8 & \text{True} \\ (-3)-(-2)=-1 & \\ -1=-1 & \text{True}\end{array}\]The solution to the system is the ordered pair \((-3,-2),\) so the system is independent.
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Solve the following system of equations by graphing.
\[\begin{array}{l}2x-5y=-25 \\ -4x+5y=35\end{array}\]答えを明らかにしろ
The solution to the system is the ordered pair \((-5,3).\)
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Solve the following system of equations by substitution.
\[\begin{array}{l}-x+y=-5 \\ 2x-5y=1\end{array}\]答えを明らかにしろ
First, we will solve the first equation for \(y.\)
\[\begin{array}{l}-x+y=-5 \\ y=x-5\end{array}\]Now we can substitute the expression \(x-5\) for \(y\) in the second equation.
\[\begin{array}{l}2x-5y=1 \\ 2x-5(x-5)=1 \\ 2x-5x+25=1 \\ -3x=-24 \\ x=8\end{array}\]Now, we substitute \(x=8\) into the first equation and solve for \(y.\)
\[\begin{array}{l}-(8)+y=-5 \\ y=3\end{array}\]Our solution is \((8,3).\)
Check the solution by substituting \((8,3)\) into both equations.
\[\begin{array}{llll}-x+y=-5 & & & \\ -(8)+(3)=-5 & & & \text{True} \\ 2x-5y=1 & & & \\ 2(8)-5(3)=1 & & & \text{True}\end{array}\] -
Solve the following system of equations by substitution.
\[\begin{array}{l}x=y+3 \\ 4=3x-2y\end{array}\]答えを明らかにしろ
\((-2,-5)\)
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Solve the given system of equations by addition.
\[\begin{array}{l}x+2y=-1 \\ -x+y=3\end{array}\]答えを明らかにしろ
Both equations are already set equal to a constant. Notice that the coefficient of \(x\) in the second equation, –1, is the opposite of the coefficient of \(x\) in the first equation, 1. We can add the two equations to eliminate \(x\) without needing to multiply by a constant.
\[\frac{\begin{array}{l} \\ x+2y=-1 \\ -x+y=3\end{array}}{\ \ \ 3y=2}\]Now that we have eliminated \(x,\) we can solve the resulting equation for \(y.\)
\[\begin{array}{l}3y=2 \\ y=\frac{2}{3}\end{array}\]Then, we substitute this value for \(y\) into one of the original equations and solve for \(x.\)
\[\begin{array}{l}-x+y=3 \\ -x+\frac{2}{3}=3 \\ -x=3-\frac{2}{3} \\ -x=\frac{7}{3} \\ x=-\frac{7}{3}\end{array}\]The solution to this system is \((-\frac{7}{3},\frac{2}{3}).\)
Check the solution in the first equation.
\[\begin{array}{llll}x+2y=-1 & & & \\ (-\frac{7}{3})+2(\frac{2}{3})= & & & \\ -\frac{7}{3}+\frac{4}{3}= & & & \\ -\frac{3}{3}= & & & \\ -1=-1 & & & \text{True}\end{array}\] -
Solve the given system of equations by the addition method.
\[\begin{array}{l}3x+5y=-11 \\ \\ x-2y=11\end{array}\]答えを明らかにしろ
Adding these equations as presented will not eliminate a variable. However, we see that the first equation has \(3x\) in it and the second equation has \(x.\) So if we multiply the second equation by \(-3,\) the x-terms will add to zero.
\[\begin{array}{llll}x-2y=11 & & & \\ -3(x-2y)=-3(11) & & & \text{Multiply both sides by }-3. \\ -3x+6y=-33 & & & \text{Use the distributive property}.\end{array}\]Now, let’s add them.
\[\begin{array}{l}\underset{_______________}{\begin{array}{l} \\ \begin{array}{l}\ 3x+5y=-11 \\ -3x+6y=-33\end{array}\end{array}} \\ \ 11y=-44 \\ \ y=-4\end{array}\]For the last step, we substitute \(y=-4\) into one of the original equations and solve for \(x.\)
\[\begin{array}{l}\ 3x+5y=-11 \\ 3x+5(-4)=-11 \\ 3x-20=-11 \\ 3x=9 \\ x=3\end{array}\]Our solution is the ordered pair \((3,-4).\) See . Check the solution in the original second equation.
\[\begin{array}{llll}\ x-2y=11 & & & \\ (3)-2(-4)=3+8 & & & \\ 11=11 & & & \text{True}\end{array}\] -
Solve the system of equations by addition.
\[\begin{array}{l}2x-7y=2 \\ 3x+y=-20\end{array}\]答えを明らかにしろ
\((-6,-2)\)
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Solve the given system of equations in two variables by addition.
\[\begin{array}{l}2x+3y=-16 \\ 5x-10y=30\end{array}\]答えを明らかにしろ
One equation has \(2x\) and the other has \(5x.\) The least common multiple is \(10x\) so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate \(x\) by multiplying the first equation by \(-5\) and the second equation by \(2.\)
\[\begin{array}{l}-5(2x+3y)=-5(-16) \\ \ -10x-15y=80 \\ \ 2(5x-10y)=2(30) \\ \ 10x-20y=60\end{array}\]Then, we add the two equations together.
\[\begin{array}{l}\underset{________________}{\begin{array}{l} \\ \begin{array}{l}-10x-15y=80\ \\ 10x-20y=60\end{array}\end{array}} \\ -35y=140 \\ y=-4\end{array}\]Substitute \(y=-4\) into the original first equation.
\[\begin{array}{l}2x+3(-4)=-16 \\ 2x-12=-16 \\ 2x=-4 \\ x=-2\end{array}\]The solution is \((-2,-4).\) Check it in the other equation.
\[\begin{array}{l}\ 5x-10y=30 \\ 5(-2)-10(-4)=30 \\ \ -10+40=30 \\ \ 30=30\end{array}\]See .
-
Solve the given system of equations in two variables by addition.
\[\begin{array}{l}\frac{x}{3}+\frac{y}{6}=3 \\ \frac{x}{2}-\frac{y}{4}=\text{}\ 1\end{array}\]答えを明らかにしろ
First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.
\[\begin{array}{l}6(\frac{x}{3}+\frac{y}{6})=6(3) \\ \ 2x+y=18 \\ 4(\frac{x}{2}-\frac{y}{4})=4(1) \\ \ 2x-y=4\end{array}\]Now multiply the second equation by \(-1\) so that we can eliminate the x-variable.
\[\begin{array}{l}-1(2x-y)=-1(4) \\ \ -2x+y=-4\end{array}\]Add the two equations to eliminate the x-variable and solve the resulting equation.
\[\begin{array}{l}2x+y=18 \\ \underset{_____________}{-2x+y=-4} \\ 2y=14 \\ y=7\end{array}\]Substitute \(y=7\) into the first equation.
\[\begin{array}{l}2x+(7)=18 \\ \ 2x=11 \\ \ x=\frac{11}{2} \\ \ =5.5\end{array}\]The solution is \((\frac{11}{2},7).\) Check it in the other equation.
\[\begin{array}{l}\ \frac{x}{2}-\frac{y}{4}=1 \\ \frac{\frac{11}{2}}{2}-\frac{7}{4}=1 \\ \frac{11}{4}-\frac{7}{4}=1 \\ \frac{4}{4}=1\end{array}\] -
Solve the system of equations by addition.
\[\begin{array}{lll}2x+3y & = & 8 \\ 3x+5y & = & 10\end{array}\]答えを明らかにしろ
\((10,-4)\)
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Solve the following system of equations.
\[\begin{array}{l}x=9-2y \\ x+2y=13\end{array}\]答えを明らかにしろ
We can approach this problem in two ways. Because one equation is already solved for \(x,\) the most obvious step is to use substitution.
\[\begin{array}{l}\ x+2y=13 \\ (9-2y)+2y=13 \\ 9+0y=13 \\ 9=13\end{array}\]Clearly, this statement is a contradiction because \(9\ne 13.\) Therefore, the system has no solution.
The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.
\[\begin{array}{l}\ x=9-2y \\ 2y=-x+9 \\ y=-\frac{1}{2}x+\frac{9}{2}\end{array}\]We then convert the second equation expressed to slope-intercept form.
\[\begin{array}{l}x+2y=13 \\ 2y=-x+13 \\ y=-\frac{1}{2}x+\frac{13}{2}\end{array}\]Comparing the equations, we see that they have the same slope but different y-intercepts. Therefore, the lines are parallel and do not intersect.
\[\begin{array}{l}\begin{array}{l} \\ y=-\frac{1}{2}x+\frac{9}{2}\end{array} \\ y=-\frac{1}{2}x+\frac{13}{2}\end{array}\] -
Solve the following system of equations in two variables.
\[\begin{array}{l}2y-2x=2 \\ 2y-2x=6\end{array}\]答えを明らかにしろ
No solution. It is an inconsistent system.
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Find a solution to the system of equations using the addition method.
\[\begin{array}{l}\ x+3y=2 \\ 3x+9y=6\end{array}\]答えを明らかにしろ
With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating \(x.\) If we multiply both sides of the first equation by \(-3,\) then we will be able to eliminate the \(x\) -variable.
\[\begin{array}{l}\ x+3y=2 \\ (-3)(x+3y)=(-3)(2) \\ -3x-9y=-6\end{array}\]Now add the equations.
\[\begin{array}{lll}\underset{______________}{\begin{array}{ll}\ -3x-9y & =-6 \\ +\ 3x+9y & =6\end{array}} \\ \begin{array}{ll}\ 0 & =0\end{array}\end{array}\]We can see that there will be an infinite number of solutions that satisfy both equations.
-
Solve the following system of equations in two variables.
\[\begin{array}{l}\begin{array}{l} \\ \ \ y-2x=5\end{array} \\ -3y+6x=-15\end{array}\]答えを明らかにしろ
The system is dependent so there are infinite solutions of the form \((x,2x+5).\)
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Given the cost function \(C(x)=0.85x+35,000\) and the revenue function \(R(x)=1.55x,\) find the break-even point and the profit function.
答えを明らかにしろ
Write the system of equations using \(y\) to replace function notation.
\[\begin{array}{l}\begin{array}{l} \\ y=0.85x+35,000\end{array} \\ y=1.55x\end{array}\]Substitute the expression \(0.85x+35,000\) from the first equation into the second equation and solve for \(x.\)
\[\begin{array}{l}0.85x+35,000=1.55x \\ 35,000=0.7x \\ 50,000=x\end{array}\]Then, we substitute \(x=50,000\) into either the cost function or the revenue function.
\[1.55(50,000)=77,500\]The break-even point is \((50,000,77,500).\)
The profit function is found using the formula \(P(x)=R(x)-C(x).\)
\[\begin{array}{l}P(x)=1.55x-(0.85x+35,000) \\ \ =0.7x-35,000\end{array}\]The profit function is \(P(x)=0.7x-35,000.\)
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The cost of a ticket to the circus is \(\text{\$}25.00\) for children and \(\text{\$}50.00\) for adults. On a certain day, attendance at the circus is \(2,000\) and the total gate revenue is \(\text{\$}70,000.\) How many children and how many adults bought tickets?
答えを明らかにしろ
Let c = the number of children and a = the number of adults in attendance.
The total number of people is \(2,000.\) We can use this to write an equation for the number of people at the circus that day.
\[c+a=2,000\]The revenue from all children can be found by multiplying \(\text{\$}25.00\) by the number of children, \(25c.\) The revenue from all adults can be found by multiplying \(\text{\$}50.00\) by the number of adults, \(50a.\) The total revenue is \(\text{\$}70,000.\) We can use this to write an equation for the revenue.
\[25c+50a=70,000\]We now have a system of linear equations in two variables.
\[\begin{array}{l}\ c+a=2,000 \\ 25c+50a=70,000\end{array}\]In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either \(c\) or \(a.\) We will solve for \(a.\)
\[\begin{array}{l}c+a=2,000 \\ a=2,000-c\end{array}\]Substitute the expression \(2,000-c\) in the second equation for \(a\) and solve for \(c.\)
\[\begin{array}{l}25c+50(2,000-c)=70,000 \\ 25c+100,000-50c=70,000 \\ -25c=-30,000 \\ c=1,200\end{array}\]Substitute \(c=1,200\) into the first equation to solve for \(a.\)
\[\begin{array}{l}1,200+a=2,000 \\ a=800\end{array}\]We find that \(1,200\) children and \(800\) adults bought tickets to the circus that day.
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Meal tickets at the circus cost \(\text{\$}4.00\) for children and \(\text{\$}12.00\) for adults. If \(1,650\) meal tickets were bought for a total of \(\text{\$}14,200,\) how many children and how many adults bought meal tickets?
答えを明らかにしろ
700 children, 950 adults
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Can a system of linear equations have exactly two solutions? Explain why or why not.
答えを明らかにしろ
No, you can either have zero, one, or infinitely many. Examine graphs.
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If you are performing a break-even analysis for a business and their cost and revenue equations are dependent, explain what this means for the company’s profit margins.
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If you are solving a break-even analysis and get a negative break-even point, explain what this signifies for the company?
答えを明らかにしろ
This means there is no realistic break-even point. By the time the company produces one unit they are already making profit.
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If you are solving a break-even analysis and there is no break-even point, explain what this means for the company. How should they ensure there is a break-even point?
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Given a system of equations, explain at least two different methods of solving that system.
答えを明らかにしろ
You can solve by substitution (isolating \(x\) or \(y\) ), graphically, or by addition.
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\(\begin{array}{l}5x-y=4\ \\ x+6y=2\end{array}\) and \((4,0)\)
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\(\begin{array}{l}-3x-5y=13 \\ -x+4y=10\end{array}\) and \((-6,1)\)
答えを明らかにしろ
Yes
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\(\begin{array}{l}3x+7y=1\ \\ 2x+4y=0\end{array}\) and \((2,3)\)
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\(\begin{array}{l}-2x+5y=7 \\ 2x+9y=7\end{array}\) and \((-1,1)\)
答えを明らかにしろ
Yes
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\(\begin{array}{l}x+8y=43\ \\ 3x-2y=-1\end{array}\) and \((3,5)\)
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\(\begin{array}{l}\ x+3y=5 \\ 2x+3y=4\end{array}\)
答えを明らかにしろ
\((-1,2)\)
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\(\begin{array}{l}\ 3x-2y=18 \\ 5x+10y=-10\end{array}\)
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\(\begin{array}{l}4x+2y=-10 \\ 3x+9y=0\end{array}\)
答えを明らかにしろ
\((-3,1)\)
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\(\begin{array}{l}2x+4y=-3.8 \\ 9x-5y=1.3\end{array}\)
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\(\begin{array}{l}-2x+3y=1.2 \\ -3x-6y=1.8\end{array}\)
答えを明らかにしろ
\((-\frac{3}{5},0)\)
Symbols used here
Inequalities that allow equality; < and > exclude it.
The two sides are different.
A quantity with magnitude and direction; a column of numbers.
A rectangular array of numbers; a linear map.
Scaling factor of area/volume under A; zero means singular.
The matrix that undoes A; A with rows and columns swapped.
The factor by which an eigenvector is stretched: Av = λv.
Σ u_i v_i; the length of v, √(v·v).
How to: Systems of Linear Equations: Two Variables
- Solve systems of equations by graphing.
- Solve systems of equations by substitution.
- Solve systems of equations by addition.
- Identify inconsistent systems of equations containing two variables.
- Express the solution of a system of dependent equations containing two variables.
- Determine whether an ordered pair is a solution of a system of equations (IA 4.1.1)
- Solve a system of linear equations by graphing (IA 4.1.2)
- Graph the first equation.
Questions people ask
What does a determinant mean geometrically?
It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.
What is an eigenvector?
A direction the matrix does not turn — it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.
Why is matrix multiplication not commutative?
Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.
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Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
ここに Linear Algebra
DeterminantsMatrix inverseEigenvalues and eigenvectorsMatrix multiplicationRow reductionVector spaces, span and linear independenceOrthogonality, projections and least squaresDiagonalisation and matrix powersLinear transformations and change of basis