linear system problem solving

4.1 Solve Systems of Linear Equations with Two Variables

Learning objectives.

By the end of this section, you will be able to:

• Determine whether an ordered pair is a solution of a system of equations
• Solve a system of linear equations by graphing
• Solve a system of equations by substitution
• Solve a system of equations by elimination
• Choose the most convenient method to solve a system of linear equations

Be Prepared 4.1

Before you get started, take this readiness quiz.

For the equation y = 2 3 x − 4 , y = 2 3 x − 4 , ⓐ Is ( 6 , 0 ) ( 6 , 0 ) a solution? ⓑ Is ( −3 , −2 ) ( −3 , −2 ) a solution? If you missed this problem, review Example 3.2 .

Be Prepared 4.2

Find the slope and y -intercept of the line 3 x − y = 12 . 3 x − y = 12 . If you missed this problem, review Example 3.16 .

Be Prepared 4.3

Find the x- and y -intercepts of the line 2 x − 3 y = 12 . 2 x − 3 y = 12 . If you missed this problem, review Example 3.8 .

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations , we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations .

System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations .

In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

A linear equation in two variables, such as 2 x + y = 7 , 2 x + y = 7 , has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs ( x , y ) ( x , y ) that make both equations true. These are called the solutions of a system of equations .

Solutions of a System of Equations

The solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair ( x , y ) . ( x , y ) .

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Example 4.1

Determine whether the ordered pair is a solution to the system { x − y = −1 2 x − y = −5 . { x − y = −1 2 x − y = −5 .

ⓐ ( −2 , −1 ) ( −2 , −1 ) ⓑ ( −4 , −3 ) ( −4 , −3 )

Determine whether the ordered pair is a solution to the system { 3 x + y = 0 x + 2 y = −5 . { 3 x + y = 0 x + 2 y = −5 .

ⓐ ( 1 , −3 ) ( 1 , −3 ) ⓑ ( 0 , 0 ) ( 0 , 0 )

Determine whether the ordered pair is a solution to the system { x − 3 y = −8 − 3 x − y = 4 . { x − 3 y = −8 − 3 x − y = 4 .

ⓐ ( 2 , −2 ) ( 2 , −2 ) ⓑ ( −2 , 2 ) ( −2 , 2 )

Solve a System of Linear Equations by Graphing

In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.

Example 4.2

How to solve a system of equations by graphing.

Solve the system by graphing { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Solve the system by graphing: { x − 3 y = −3 x + y = 5 . { x − 3 y = −3 x + y = 5 .

Solve the system by graphing: { − x + y = 1 3 x + 2 y = 12 . { − x + y = 1 3 x + 2 y = 12 .

The steps to use to solve a system of linear equations by graphing are shown here.

Solve a system of linear equations by graphing.

• Step 1. Graph the first equation.
• Step 2. Graph the second equation on the same rectangular coordinate system.
• Step 3. Determine whether the lines intersect, are parallel, or are the same line.
• If the lines intersect, identify the point of intersection. This is the solution to the system.
• If the lines are parallel, the system has no solution.
• If the lines are the same, the system has an infinite number of solutions.
• Step 5. Check the solution in both equations.

In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.

Example 4.3

Solve the system by graphing: { 3 x + y = − 1 2 x + y = 0 . { 3 x + y = − 1 2 x + y = 0 .

We’ll solve both of these equations for y y so that we can easily graph them using their slopes and y -intercepts.

Solve the system by graphing: { − x + y = 1 2 x + y = 10 . { − x + y = 1 2 x + y = 10 .

Solve the system by graphing: { 2 x + y = 6 x + y = 1 . { 2 x + y = 6 x + y = 1 .

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

Example 4.4

Solve the system by graphing: { y = 1 2 x − 3 x − 2 y = 4 . { y = 1 2 x − 3 x − 2 y = 4 .

Solve the system by graphing: { y = − 1 4 x + 2 x + 4 y = − 8 . { y = − 1 4 x + 2 x + 4 y = − 8 .

Solve the system by graphing: { y = 3 x − 1 6 x − 2 y = 6 . { y = 3 x − 1 6 x − 2 y = 6 .

Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.

Example 4.5

Solve the system by graphing: { y = 2 x − 3 − 6 x + 3 y = − 9 . { y = 2 x − 3 − 6 x + 3 y = − 9 .

If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.

Solve the system by graphing: { y = − 3 x − 6 6 x + 2 y = − 12 . { y = − 3 x − 6 6 x + 2 y = − 12 .

Try It 4.10

Solve the system by graphing: { y = 1 2 x − 4 2 x − 4 y = 16 . { y = 1 2 x − 4 2 x − 4 y = 16 .

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident . Coincident lines have the same slope and same y- intercept.

Coincident Lines

Coincident lines have the same slope and same y- intercept.

The systems of equations in Example 4.2 and Example 4.3 each had two intersecting lines. Each system had one solution.

In Example 4.5 , the equations gave coincident lines, and so the system had infinitely many solutions.

The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.

A system with parallel lines, like Example 4.4 , has no solution. We call a system of equations like this inconsistent. It has no solution.

Consistent and Inconsistent Systems

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent . If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.

Let’s sum this up by looking at the graphs of the three types of systems. See below and Table 4.1 .

Example 4.6

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ { y = 3 x − 1 6 x − 2 y = 12 { y = 3 x − 1 6 x − 2 y = 12 ⓑ { 2 x + y = − 3 x − 5 y = 5 { 2 x + y = − 3 x − 5 y = 5

ⓐ We will compare the slopes and intercepts of the two lines.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Try It 4.11

ⓐ { y = −2 x − 4 4 x + 2 y = 9 { y = −2 x − 4 4 x + 2 y = 9 ⓑ { 3 x + 2 y = 2 2 x + y = 1 { 3 x + 2 y = 2 2 x + y = 1

Try It 4.12

ⓐ { y = 1 3 x − 5 x − 3 y = 6 { y = 1 3 x − 5 x − 3 y = 6 ⓑ { x + 4 y = 12 − x + y = 3 { x + 4 y = 12 − x + y = 3

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

Solve a System of Equations by Substitution

We will now solve systems of linear equations by the substitution method.

We will use the same system we used first for graphing.

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

Example 4.7

How to solve a system of equations by substitution.

Solve the system by substitution: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.13

Solve the system by substitution: { − 2 x + y = −11 x + 3 y = 9 . { − 2 x + y = −11 x + 3 y = 9 .

Try It 4.14

Solve the system by substitution: { 2 x + y = −1 4 x + 3 y = 3 . { 2 x + y = −1 4 x + 3 y = 3 .

Solve a system of equations by substitution.

• Step 1. Solve one of the equations for either variable.
• Step 2. Substitute the expression from Step 1 into the other equation.
• Step 3. Solve the resulting equation.
• Step 4. Substitute the solution in Step 3 into either of the original equations to find the other variable.
• Step 5. Write the solution as an ordered pair.
• Step 6. Check that the ordered pair is a solution to both original equations.

Be very careful with the signs in the next example.

Example 4.8

Solve the system by substitution: { 4 x + 2 y = 4 6 x − y = 8 . { 4 x + 2 y = 4 6 x − y = 8 .

We need to solve one equation for one variable. We will solve the first equation for y .

Try It 4.15

Solve the system by substitution: { x − 4 y = −4 − 3 x + 4 y = 0 . { x − 4 y = −4 − 3 x + 4 y = 0 .

Try It 4.16

Solve the system by substitution: { 4 x − y = 0 2 x − 3 y = 5 . { 4 x − y = 0 2 x − 3 y = 5 .

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d .

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

The y ’s add to zero and we have one equation with one variable.

Let’s try another one:

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2 , −2 , we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2 . −2 .

Then rewrite the system of equations.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Example 4.9

How to solve a system of equations by elimination.

Solve the system by elimination: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.17

Solve the system by elimination: { 3 x + y = 5 2 x − 3 y = 7 . { 3 x + y = 5 2 x − 3 y = 7 .

Try It 4.18

Solve the system by elimination: { 4 x + y = − 5 − 2 x − 2 y = − 2 . { 4 x + y = − 5 − 2 x − 2 y = − 2 .

The steps are listed here for easy reference.

Solve a system of equations by elimination.

• Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
• Decide which variable you will eliminate.
• Multiply one or both equations so that the coefficients of that variable are opposites.
• Step 3. Add the equations resulting from Step 2 to eliminate one variable.
• Step 4. Solve for the remaining variable.
• Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
• Step 6. Write the solution as an ordered pair.
• Step 7. Check that the ordered pair is a solution to both original equations.

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Example 4.10

Solve the system by elimination: { 4 x − 3 y = 9 7 x + 2 y = −6 . { 4 x − 3 y = 9 7 x + 2 y = −6 .

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.

Try It 4.19

Solve the system by elimination: { 3 x − 4 y = − 9 5 x + 3 y = 14 . { 3 x − 4 y = − 9 5 x + 3 y = 14 .

Try It 4.20

Solve each system by elimination: { 7 x + 8 y = 4 3 x − 5 y = 27 . { 7 x + 8 y = 4 3 x − 5 y = 27 .

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.

Example 4.11

Solve the system by elimination: { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 . { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 .

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.

Try It 4.21

Solve each system by elimination: { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 . { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 .

Try It 4.22

Solve each system by elimination: { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 . { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 .

When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.

Example 4.12

Solve the system by elimination: { 3 x + 4 y = 12 y = 3 − 3 4 x . { 3 x + 4 y = 12 y = 3 − 3 4 x .

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Try It 4.23

Solve the system by elimination: { 5 x − 3 y = 15 y = − 5 + 5 3 x . { 5 x − 3 y = 15 y = − 5 + 5 3 x .

Try It 4.24

Solve the system by elimination: { x + 2 y = 6 y = − 1 2 x + 3 . { x + 2 y = 6 y = − 1 2 x + 3 .

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Example 4.13

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

Since both equations are in standard form, using elimination will be most convenient.

Since one equation is already solved for y , using substitution will be most convenient.

Try It 4.25

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 4 x − 5 y = −32 3 x + 2 y = −1 { 4 x − 5 y = −32 3 x + 2 y = −1 ⓑ { x = 2 y − 1 3 x − 5 y = −7 { x = 2 y − 1 3 x − 5 y = −7

Try It 4.26

ⓐ { y = 2 x − 1 3 x − 4 y = − 6 { y = 2 x − 1 3 x − 4 y = − 6 ⓑ { 6 x − 2 y = 12 3 x + 7 y = −13 { 6 x − 2 y = 12 3 x + 7 y = −13

Section 4.1 Exercises

Practice makes perfect.

In the following exercises, determine if the following points are solutions to the given system of equations.

{ 2 x − 6 y = 0 3 x − 4 y = 5 { 2 x − 6 y = 0 3 x − 4 y = 5

ⓐ ( 3 , 1 ) ( 3 , 1 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

{ − 3 x + y = 8 − x + 2 y = −9 { − 3 x + y = 8 − x + 2 y = −9

ⓐ ( −5 , −7 ) ( −5 , −7 ) ⓑ ( −5 , 7 ) ( −5 , 7 )

{ x + y = 2 y = 3 4 x { x + y = 2 y = 3 4 x

ⓐ ( 8 7 , 6 7 ) ( 8 7 , 6 7 ) ⓑ ( 1 , 3 4 ) ( 1 , 3 4 )

{ 2 x + 3 y = 6 y = 2 3 x + 2 { 2 x + 3 y = 6 y = 2 3 x + 2 ⓐ ( −6 , 2 ) ( −6 , 2 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

In the following exercises, solve the following systems of equations by graphing.

{ 3 x + y = −3 2 x + 3 y = 5 { 3 x + y = −3 2 x + 3 y = 5

{ − x + y = 2 2 x + y = −4 { − x + y = 2 2 x + y = −4

{ y = x + 2 y = −2 x + 2 { y = x + 2 y = −2 x + 2

{ y = x − 2 y = −3 x + 2 { y = x − 2 y = −3 x + 2

{ y = 3 2 x + 1 y = − 1 2 x + 5 { y = 3 2 x + 1 y = − 1 2 x + 5

{ y = 2 3 x − 2 y = − 1 3 x − 5 { y = 2 3 x − 2 y = − 1 3 x − 5

{ x + y = −4 − x + 2 y = −2 { x + y = −4 − x + 2 y = −2

{ − x + 3 y = 3 x + 3 y = 3 { − x + 3 y = 3 x + 3 y = 3

{ − 2 x + 3 y = 3 x + 3 y = 12 { − 2 x + 3 y = 3 x + 3 y = 12

{ 2 x − y = 4 2 x + 3 y = 12 { 2 x − y = 4 2 x + 3 y = 12

{ x + 3 y = −6 y = − 4 3 x + 4 { x + 3 y = −6 y = − 4 3 x + 4

{ − x + 2 y = −6 y = − 1 2 x − 1 { − x + 2 y = −6 y = − 1 2 x − 1

{ − 2 x + 4 y = 4 y = 1 2 x { − 2 x + 4 y = 4 y = 1 2 x

{ 3 x + 5 y = 10 y = − 3 5 x + 1 { 3 x + 5 y = 10 y = − 3 5 x + 1

{ 4 x − 3 y = 8 8 x − 6 y = 14 { 4 x − 3 y = 8 8 x − 6 y = 14

{ x + 3 y = 4 − 2 x − 6 y = 3 { x + 3 y = 4 − 2 x − 6 y = 3

{ x = −3 y + 4 2 x + 6 y = 8 { x = −3 y + 4 2 x + 6 y = 8

{ 4 x = 3 y + 7 8 x − 6 y = 14 { 4 x = 3 y + 7 8 x − 6 y = 14

{ 2 x + y = 6 − 8 x − 4 y = −24 { 2 x + y = 6 − 8 x − 4 y = −24

{ 5 x + 2 y = 7 − 10 x − 4 y = −14 { 5 x + 2 y = 7 − 10 x − 4 y = −14

{ y = 2 3 x + 1 − 2 x + 3 y = 5 { y = 2 3 x + 1 − 2 x + 3 y = 5

{ y = 3 2 x + 1 2 x − 3 y = 7 { y = 3 2 x + 1 2 x − 3 y = 7

{ 5 x + 3 y = 4 2 x − 3 y = 5 { 5 x + 3 y = 4 2 x − 3 y = 5

{ y = − 1 2 x + 5 x + 2 y = 10 { y = − 1 2 x + 5 x + 2 y = 10

{ 5 x − 2 y = 10 y = 5 2 x − 5 { 5 x − 2 y = 10 y = 5 2 x − 5

In the following exercises, solve the systems of equations by substitution.

{ 2 x + y = −4 3 x − 2 y = −6 { 2 x + y = −4 3 x − 2 y = −6

{ 2 x + y = −2 3 x − y = 7 { 2 x + y = −2 3 x − y = 7

{ x − 2 y = −5 2 x − 3 y = −4 { x − 2 y = −5 2 x − 3 y = −4

{ x − 3 y = −9 2 x + 5 y = 4 { x − 3 y = −9 2 x + 5 y = 4

{ 5 x − 2 y = −6 y = 3 x + 3 { 5 x − 2 y = −6 y = 3 x + 3

{ − 2 x + 2 y = 6 y = −3 x + 1 { − 2 x + 2 y = 6 y = −3 x + 1

{ 2 x + 5 y = 1 y = 1 3 x − 2 { 2 x + 5 y = 1 y = 1 3 x − 2

{ 3 x + 4 y = 1 y = − 2 5 x + 2 { 3 x + 4 y = 1 y = − 2 5 x + 2

{ 2 x + y = 5 x − 2 y = −15 { 2 x + y = 5 x − 2 y = −15

{ 4 x + y = 10 x − 2 y = −20 { 4 x + y = 10 x − 2 y = −20

{ y = −2 x − 1 y = − 1 3 x + 4 { y = −2 x − 1 y = − 1 3 x + 4

{ y = x − 6 y = − 3 2 x + 4 { y = x − 6 y = − 3 2 x + 4

{ x = 2 y 4 x − 8 y = 0 { x = 2 y 4 x − 8 y = 0

{ 2 x − 16 y = 8 − x − 8 y = −4 { 2 x − 16 y = 8 − x − 8 y = −4

{ y = 7 8 x + 4 − 7 x + 8 y = 6 { y = 7 8 x + 4 − 7 x + 8 y = 6

{ y = − 2 3 x + 5 2 x + 3 y = 11 { y = − 2 3 x + 5 2 x + 3 y = 11

In the following exercises, solve the systems of equations by elimination.

{ 5 x + 2 y = 2 − 3 x − y = 0 { 5 x + 2 y = 2 − 3 x − y = 0

{ 6 x − 5 y = −1 2 x + y = 13 { 6 x − 5 y = −1 2 x + y = 13

{ 2 x − 5 y = 7 3 x − y = 17 { 2 x − 5 y = 7 3 x − y = 17

{ 5 x − 3 y = −1 2 x − y = 2 { 5 x − 3 y = −1 2 x − y = 2

{ 3 x − 5 y = −9 5 x + 2 y = 16 { 3 x − 5 y = −9 5 x + 2 y = 16

{ 4 x − 3 y = 3 2 x + 5 y = −31 { 4 x − 3 y = 3 2 x + 5 y = −31

{ 3 x + 8 y = −3 2 x + 5 y = −3 { 3 x + 8 y = −3 2 x + 5 y = −3

{ 11 x + 9 y = −5 7 x + 5 y = −1 { 11 x + 9 y = −5 7 x + 5 y = −1

{ 3 x + 8 y = 67 5 x + 3 y = 60 { 3 x + 8 y = 67 5 x + 3 y = 60

{ 2 x + 9 y = −4 3 x + 13 y = −7 { 2 x + 9 y = −4 3 x + 13 y = −7

{ 1 3 x − y = −3 x + 5 2 y = 2 { 1 3 x − y = −3 x + 5 2 y = 2

{ x + 1 2 y = 3 2 1 5 x − 1 5 y = 3 { x + 1 2 y = 3 2 1 5 x − 1 5 y = 3

{ x + 1 3 y = −1 1 3 x + 1 2 y = 1 { x + 1 3 y = −1 1 3 x + 1 2 y = 1

{ 1 3 x − y = −3 2 3 x + 5 2 y = 3 { 1 3 x − y = −3 2 3 x + 5 2 y = 3

{ 2 x + y = 3 6 x + 3 y = 9 { 2 x + y = 3 6 x + 3 y = 9

{ x − 4 y = −1 − 3 x + 12 y = 3 { x − 4 y = −1 − 3 x + 12 y = 3

{ − 3 x − y = 8 6 x + 2 y = −16 { − 3 x − y = 8 6 x + 2 y = −16

{ 4 x + 3 y = 2 20 x + 15 y = 10 { 4 x + 3 y = 2 20 x + 15 y = 10

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ { 8 x − 15 y = −32 6 x + 3 y = −5 { 8 x − 15 y = −32 6 x + 3 y = −5 ⓑ { x = 4 y − 3 4 x − 2 y = −6 { x = 4 y − 3 4 x − 2 y = −6

ⓐ { y = 7 x − 5 3 x − 2 y = 16 { y = 7 x − 5 3 x − 2 y = 16 ⓑ { 12 x − 5 y = −42 3 x + 7 y = −15 { 12 x − 5 y = −42 3 x + 7 y = −15

ⓐ { y = 4 x + 9 5 x − 2 y = −21 { y = 4 x + 9 5 x − 2 y = −21 ⓑ { 9 x − 4 y = 24 3 x + 5 y = −14 { 9 x − 4 y = 24 3 x + 5 y = −14

ⓐ { 14 x − 15 y = −30 7 x + 2 y = 10 { 14 x − 15 y = −30 7 x + 2 y = 10 ⓑ { x = 9 y − 11 2 x − 7 y = −27 { x = 9 y − 11 2 x − 7 y = −27

Writing Exercises

In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.

Solve the system of equations by substitution and explain all your steps in words: { 3 x + y = 12 x = y − 8 . { 3 x + y = 12 x = y − 8 .

Solve the system of equations by elimination and explain all your steps in words: { 5 x + 4 y = 10 2 x = 3 y + 27 . { 5 x + 4 y = 10 2 x = 3 y + 27 .

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6

ⓐ by graphing ⓑ by substitution ⓒ Which method do you prefer? Why?

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

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…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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11.1: Systems of Linear Equations - Two Variables

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Learning Objectives

• 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.

A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions.

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.

\[\begin{align*} 2x+y &= 15 \\ 3x–y &= 5 \end{align*}\]

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.

\[\begin{align*} 2(4)+(7) &=15 \text{ True} \\ 3(4)−(7) &= 5 \text{ True} \end{align*}\]

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.

TYPES OF LINEAR SYSTEMS

There are three types of systems of linear equations in two variables, and three types of solutions.

• An independent system has exactly one solution pair \((x,y)\). The point where the two lines intersect is the only solution.
• An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect.
• A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.

Figure \(\PageIndex{2}\) compares graphical representations of each type of system.

Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution

• Substitute the ordered pair into each equation in the system.
• Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.

Example \(\PageIndex{1}\): Determining Whether an Ordered Pair Is a Solution to a System of Equations

Determine whether the ordered pair \((5,1)\) is a solution to the given system of equations.

\[\begin{align*} x+3y &= 8 \\ 2x−9 &= y \end{align*}\]

Substitute the ordered pair \((5,1)\) into both equations.

\[ \begin{align*} (5)+3(1) &= 8 \\ 8 &= 8 \text{ True} \\ 2(5)−9 &= (1) \\ 1 &= 1 \text{ True} \end{align*}\]

The ordered pair \((5,1)\) satisfies both equations, so it is the solution to the system.

We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines. See Figure \(\PageIndex{3}\).

Exercise \(\PageIndex{1}\)

Determine whether the ordered pair \((8,5)\) is a solution to the following system.

\[\begin{align*} 5x−4y &= 20 \\ 2x+1 &= 3y \end{align*}\]

Not a solution.

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 \(\PageIndex{2}\): Solving a System of Equations in Two Variables by Graphing

Solve the following system of equations by graphing. Identify the type of system.

\[\begin{align*} 2x+y &= −8 \\ x−y &= −1 \end{align*}\]

Solve the first equation for \(y\).

\[\begin{align*} 2x+y &= −8 \\ y &= −2x−8 \end{align*}\]

Solve the second equation for \(y\).

\[\begin{align*} x−y &= −1 \\ y &= x+1 \end{align*}\]

Graph both equations on the same set of axes as in Figure \(\PageIndex{4}\).

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.

\[\begin{align*} 2(−3)+(−2) &= −8 \\ −8 &= −8 \text{ True} \\ (−3)−(−2) &= −1 \\ −1 &= −1 \text{ True} \end{align*}\]

The solution to the system is the ordered pair \((−3,−2)\),so the system is independent.

Exercise \(\PageIndex{2}\)

Solve the following system of equations by graphing.

\[\begin{align*} 2x−5y &= −25 \\ −4x+5y &= 35 \end{align*}\]

The solution to the system is the ordered pair \((−5,3)\).

Can graphing be used if the system is inconsistent or dependent?

Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.

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.

How to: Given a system of two equations in two variables, solve using the substitution method.

• Solve one of the two equations for one of the variables in terms of the other.
• Substitute the expression for this variable into the second equation, then solve for the remaining variable.
• Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
• Check the solution in both equations.

Example \(\PageIndex{3}\): Solving a System of Equations in Two Variables by Substitution

Solve the following system of equations by substitution.

\[\begin{align*} −x+y &= −5 \\ 2x−5y &= 1 \end{align*}\]

First, we will solve the first equation for \(y\).

\[\begin{align*} −x+y &=−5 \\ y &= x−5 \end{align*}\]

Now we can substitute the expression \(x−5\) for \(y\) in the second equation.

\[\begin{align*} 2x−5y &= 1 \\ 2x−5(x−5) &= 1 \\ 2x−5x+25 &= 1 \\ −3x &= −24 \\ x &= 8 \end{align*}\]

Now, we substitute \(x=8\) into the first equation and solve for \(y\).

\[\begin{align*} −(8)+y &= −5 \\ y &= 3 \end{align*}\]

Our solution is \((8,3)\).

Check the solution by substituting \((8,3)\) into both equations.

\[\begin{align*} −x+y &= −5 \\ −(8)+(3) &= −5 \text{ True} \\ 2x−5y &= 1 \\ 2(8)−5(3) &= 1 \text{ True} \end{align*}\]

Exercise \(\PageIndex{3}\)

\[\begin{align*} x &= y+3 \\ 4 &= 3x−2y \end{align*}\]

\((−2,−5)\)

Can the substitution method be used to solve any linear system in two variables?

Yes, but the method works best if one of the equations contains a coefficient of \(1\) or \(–1\) so that we do not have to deal with fractions.

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.

How to: Given a system of equations, solve using the addition method.

• Write both equations with x - and y -variables on the left side of the equal sign and constants on the right.
• Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
• Solve the resulting equation for the remaining variable.
• Substitute that value into one of the original equations and solve for the second variable.
• Check the solution by substituting the values into the other equation.

Example \(\PageIndex{4}\): Solving a System by the Addition Method

Solve the given system of equations by addition.

\[\begin{align*} x+2y &= −1 \\ −x+y &=3 \end{align*}\]

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.

\[\begin{align*} x+2y &= -1 \\ \underline{-x+y}& = \underline{3} \\ 3y&= 2 \\ \end{align*}\]

Now that we have eliminated \(x\), we can solve the resulting equation for \(y\).

\[\begin{align*} 3y &= 2 \\ y &=\dfrac{2}{3} \end{align*}\]

Then, we substitute this value for \(y\) into one of the original equations and solve for \(x\).

\[\begin{align*} −x+y &= 3 \\ −x+\dfrac{2}{3} &= 3 \\ −x &= 3−\dfrac{2}{3} \\ −x &= \dfrac{7}{3} \\ x &= −\dfrac{7}{3} \end{align*}\]

The solution to this system is \(\left(−\dfrac{7}{3},\dfrac{2}{3}\right)\).

Check the solution in the first equation.

\[\begin{align*} x+2y &= −1 \\ \left(−\dfrac{7}{3}\right)+2\left(\dfrac{2}{3}\right) &= \\ −\dfrac{7}{3}+\dfrac{4}{3} &= −\dfrac{3}{3} \\ −1 &= −1 \;\;\;\;\;\;\;\; \text{True} \end{align*}\]

We gain an important perspective on systems of equations by looking at the graphical representation. See Figure \(\PageIndex{6}\) to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.

Example \(\PageIndex{5}\): Using the Addition Method When Multiplication of One Equation Is Required

Solve the given system of equations by the addition method.

\[\begin{align*} 3x+5y &= −11 \\ x−2y &= 11 \end{align*}\]

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.

\[\begin{align*} x−2y &= 11 \\ −3(x−2y) &=−3(11) \;\;\;\;\;\;\;\; \text{Multiply both sides by }−3. \\ −3x+6y &= −33 \;\;\;\;\;\;\;\;\; \text{Use the distributive property.} \end{align*}\]

Now, let’s add them.

\[\begin{align*} 3x+5y &= -11 \\ \underline{-3x+6y }& = \underline{-33} \\ 11y&= -44 \\ y&= -4 \end{align*}\]

For the last step, we substitute \(y=−4\) into one of the original equations and solve for \(x\).

\[\begin{align*} 3x+5y &= −11 \\ 3x+5(−4) &= −11 \\ 3x−20 &= −11 \\ 3x &= 9 \\ x &= 3 \end{align*}\]

Our solution is the ordered pair \((3,−4)\). See Figure \(\PageIndex{7}\). Check the solution in the original second equation.

\[\begin{align*} x−2y &= 11 \\ (3)−2(−4) &= 3+8 \\ &= 11 \;\;\;\;\;\;\;\;\;\; \text{True} \end{align*}\]

Exercise \(\PageIndex{4}\)

Solve the system of equations by addition.

\[\begin{align*} 2x−7y &= 2 \\ 3x+y &= −20 \end{align*}\]

\((−6,−2)\)

Example \(\PageIndex{6}\): Using the Addition Method When Multiplication of Both Equations Is Required

Solve the given system of equations in two variables by addition.

\[\begin{align*} 2x+3y &= −16 \\ 5x−10y &= 30 \end{align*}\]

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 \(2\).

\[\begin{align*} −5(2x+3y) &= −5(−16) \\ −10x−15y &= 80 \\ 2(5x−10y) &= 2(30) \\ 10x−20y &= 60 \end{align*}\]

Then, we add the two equations together.

\[\begin{align*} -10x-15y &= 80 \\ \underline{10x-20y}& = \underline{60} \\ -35y&= 140 \\ y&= -4 \end{align*}\]

Substitute \(y=−4\) into the original first equation.

\[ \begin{align*} 2x+3(−4) &=−16 \\ 2x−12 &= −16 \\ 2x &= −4 \\ x &=−2 \end{align*}\]

The solution is \((−2,−4)\). Check it in the other equation.

\[\begin{align*} 5x−10y &= 30 \\ 5(−2)−10(−4) &= 30 \\ −10+40 &= 30 \\30 &=30 \end{align*}\]

See Figure \(\PageIndex{8}\).

Example \(\PageIndex{7}\): Using the Addition Method in Systems of Equations Containing Fractions

\[ \begin{align*} \dfrac{x}{3}+\dfrac{y}{6} &= 3 \\ \dfrac{x}{2}−\dfrac{y}{4} &= 1 \end{align*}\]

First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.

\[\begin{align*} 6\left(\dfrac{x}{3}+\dfrac{y}{6}\right) &= 6(3) \\ 2x+y &= 18 \\ 4\left(\dfrac{x}{2}−\dfrac{y}{4}\right) &= 4(1) \\ 2x−y &= 4 \end{align*}\]

Now multiply the second equation by \(−1\) so that we can eliminate the x -variable.

\[\begin{align*} −1(2x−y) &= −1(4) \\ −2x+y &= −4 \end{align*}\]

Add the two equations to eliminate the \(x\)-variable and solve the resulting equation.

\[\begin{align*} 2x+y &= 18 \\ −2x+y &= −4 \\ 2y &= 14 \\ y &=7 \end{align*}\]

Substitute \(y=7\) into the first equation.

\[\begin{align*} 2x+(7) &= 18 \\ 2x &= 11 \\ x &= \dfrac{11}{2} \\ &= 7.5 \end{align*}\]

The solution is \(\left(\dfrac{11}{2},7\right)\). Check it in the other equation.

\[\begin{align*} \dfrac{x}{2}−\dfrac{y}{4} &= 1 \\ \dfrac{\dfrac{11}{2}}{2}−\dfrac{7}{4} &=1 \\ \dfrac{11}{4}−\dfrac{7}{4} &=1 \\ \dfrac{4}{4} &=1 \end{align*}\]

Exercise \(\PageIndex{5}\)

\[\begin{align*} 2x+3y &= 8 \\ 3x+5y &= 10 \end{align*}\]

\((10,−4)\)

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 \(\PageIndex{8}\): Solving an Inconsistent System of Equations

Solve the following system of equations.

\[\begin{align*} x &= 9−2y \\ x+2y &= 13 \end{align*}\]

We can approach this problem in two ways. Because one equation is already solved for \(x\), the most obvious step is to use substitution.

\[\begin{align*} x+2y &= 13 \\ (9−2y)+2y &= 13 \\ 9+0y &= 13 \\ 9 &= 13 \end{align*}\]

Clearly, this statement is a contradiction because \(9≠13\). 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.

\[\begin{align*} x &= 9−2y \\ 2y &= −x+9 \\ y &= −\dfrac{1}{2}x+\dfrac{9}{2} \end{align*}\]

We then convert the second equation expressed to slope-intercept form.

\[\begin{align*} x+2y &= 13 \\ 2y &= −x+13 \\ y &= −\dfrac{1}{2}x+\dfrac{13}{2} \end{align*}\]

Comparing the equations, we see that they have the same slope but different \(y\)-intercepts. Therefore, the lines are parallel and do not intersect.

\[\begin{align*} y &= −\dfrac{1}{2}x+\dfrac{9}{2} \\ y &= −\dfrac{1}{2}x+\dfrac{13}{2} \end{align*}\]

Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown in Figure \(\PageIndex{9}\).

Exercise \(\PageIndex{6}\)

Solve the following system of equations in two variables.

\[\begin{align*} 2y−2x &= 2 \\ 2y−2x &= 6 \end{align*}\]

No solution. It is an inconsistent system.

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 \(\PageIndex{9}\): Finding a Solution to a Dependent System of Linear Equations

Find a solution to the system of equations using the addition method .

\[\begin{align*} x+3y &= 2 \\ 3x+9y &= 6 \end{align*}\]

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.

\[\begin{align*} x+3y &= 2 \\ (−3)(x+3y) &= (−3)(2) \\ −3x−9y &= −6 \end{align*}\]

Now add the equations.

\[\begin{align*} -3x-9y &= -6 \\ \underline{+\space 3x+9y}& = \underline{6} \\ 0&= 0 \\ \end{align*}\]

We can see that there will be an infinite number of solutions that satisfy both equations.

If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form.

\[ \begin{align*} x+3y &= 2 \\ 3y &= −x+2 \\ y &= −\dfrac{1}{3}x+\dfrac{2}{3} \\ 3x+9y &= 6 \\ 9y &=−3x+6 \\ y &= −\dfrac{3}{9}x+\dfrac{6}{9} \\ y &= −\dfrac{1}{3}x+\dfrac{2}{3} \end{align*}\]

See Figure \(\PageIndex{10}\). Notice the results are the same. The general solution to the system is \(\left(x, −\dfrac{1}{3}x+\dfrac{2}{3}\right)\).

Exercise \(\PageIndex{7}\)

\[\begin{align*} y−2x &= 5 \\ −3y+6x &= −15 \end{align*}\]

The system is dependent so there are infinite solutions of the form \((x,2x+5)\).

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 Figure \(\PageIndex{11}\).

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 Figure \(\PageIndex{11}\). 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 \(\PageIndex{10}\): Finding the Break-Even Point and the Profit Function Using Substitution

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.

Write the system of equations using \(y\) to replace function notation.

\[\begin{align*} y &= 0.85x+35,000 \\ y &= 1.55x \end{align*}\]

Substitute the expression \(0.85x+35,000\) from the first equation into the second equation and solve for \(x\).

\[\begin{align*} 0.85x+35,000 &= 1.55x \\ 35,000 &= 0.7x \\ 50,000 &= x \end{align*}\]

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 \(P(x)=R(x)−C(x)\).

\[\begin{align*} P(x) &= 1.55x−(0.85x+35,000) \\ &=0.7x−35,000 \end{align*}\]

The profit function is \(P(x)=0.7x−35,000\).

The cost to produce \(50,000\) units is \($77,500\), and the revenue from the sales of \(50,000\) units is also \($77,500\). To make a profit, the business must produce and sell more than \(50,000\) units. See Figure \(\PageIndex{12}\).

We see from the graph in Figure \(\PageIndex{13}\) that the profit function has a negative value until \(x=50,000\), when the graph crosses the \(x\)-axis. Then, the graph emerges into positive \(y\)-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is \(0\). The area to the left of the break-even point represents operating at a loss.

Example \(\PageIndex{11}\): Writing and Solving a System of Equations in Two Variables

The cost of a ticket to the circus is \($25.00\) for children and \($50.00\) for adults. On a certain day, attendance at the circus is \(2,000\) and the total gate revenue is \($70,000\). How many children and how many adults bought tickets?

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 \($25.00\) by the number of children, \(25c\). The revenue from all adults can be found by multiplying \($50.00\) by the number of adults, \(50a\). The total revenue is \($70,000\). 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.

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 \(a\).

\[\begin{align*} c+a &= 2,000 \\ a &= 2,000−c \end{align*}\]

Substitute the expression \(2,000−c\) in the second equation fora a and solve for \(c\).

\[\begin{align*} 25c+50(2,000−c) &= 70,000 \\ 25c+100,000−50c &= 70,000 \\ −25c &= −30,000 \\ c &= 1,200 \end{align*}\]

Substitute \(c=1,200\) into the first equation to solve for \(a\).

\[\begin{align*} 1,200+a &= 2,000 \\ a &= 800 \end{align*}\]

We find that \(1,200\) children and \(800\) adults bought tickets to the circus that day.

Exercise \(\PageIndex{8}\)

Meal tickets at the circus cost \($4.00\) for children and \($12.00\) for adults. If \(1,650\) meal tickets were bought for a total of, \($14,200\), how many children and how many adults bought meal tickets?

\(700\) children, \(950\) adults

Access these online resources for additional instruction and practice with systems of linear equations.

• Solving Systems of Equations Using Substitution
• Solving Systems of Equations Using Elimination
• Applications of Systems of Equations

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What are systems of equations?

A system of equations is a set of one or more equations involving a number of variables..

The solutions to systems of equations are the variable mappings such that all component equations are satisfied—in other words, the locations at which all of these equations intersect. To solve a system is to find all such common solutions or points of intersection.

Systems of linear equations are a common and applicable subset of systems of equations. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. The system is said to be inconsistent otherwise, having no solutions. Systems of linear equations involving more than two variables work similarly, having either one solution, no solutions or infinite solutions (the latter in the case that all component equations are equivalent).

More general systems involving nonlinear functions are possible as well. These possess more complicated solution sets involving one, zero, infinite or any number of solutions, but work similarly to linear systems in that their solutions are the points satisfying all equations involved. Going further, more general systems of constraints are possible, such as ones that involve inequalities or have requirements that certain variables be integers.

Solving systems of equations is a very general and important idea, and one that is fundamental in many areas of mathematics, engineering and science.

Module 5: Systems of Linear Equations

5.2 – applications of systems of linear equations, learning objectives.

• Specify what the variables in a cost/ revenue system of linear equations represent
• Determine and apply an appropriate method for solving the system

(5.2.2) – Solve value problems with a system of linear equations

(5.2.3) – solve mixture problems with a system of linear equations, (5.2.4) – solve uniform motion problems with a system of linear equations.

A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible?

(credit: Thomas Sørenes)

(5.2.1) – Solve cost and revenue problems

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 [latex]R=xp[/latex], where [latex]x=[/latex] quantity and [latex]p=[/latex] price. The revenue function is shown in orange in the graph below.

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 graph below. The [latex]x[/latex] -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 [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.

A business wants to manufacture bike frames. Before they start production, they need to make sure they can make a profit with the materials and labor force they have. Their accountant has given them a cost equation of [latex]y=0.85x+35,000[/latex] and a revenue equation of [latex]y=1.55x[/latex]:

• Interpret x and y for the cost equation
• Interpret x and y for the revenue equation

Cost: [latex]y=0.85x+35,000[/latex]

Revenue:[latex]y=1.55x[/latex]

The cost equation represents money leaving the company, namely how much it costs to produce a given number of bike frames. If we use the skateboard example as a model, x would represent the number of frames produced (instead of skateboards) and y would represent the amount of money it would cost to produce them (the same as the skateboard problem).

The revenue equation represents money coming into the company, so in this context x still represents the number of bike frames manufactured, and y now represents the amount of money made from selling them.  Let’s organize this information in a table:

Example: Finding the Break-Even Point and the Profit Function Using Substitution

Given the cost function [latex]C\left(x\right)=0.85x+35,000[/latex] and the revenue function [latex]R\left(x\right)=1.55x[/latex], find the break-even point and the profit function.

Write the system of equations using [latex]y[/latex] to replace function notation.

[latex]\begin{array}{l}\begin{array}{l}\\ y=0.85x+35,000\end{array}\hfill \\ y=1.55x\hfill \end{array}[/latex]

Substitute the expression [latex]0.85x+35,000[/latex] from the first equation into the second equation and solve for [latex]x[/latex].

[latex]\begin{array}{c}0.85x+35,000=1.55x\\ 35,000=0.7x\\ 50,000=x\end{array}[/latex]

Then, we substitute [latex]x=50,000[/latex] into either the cost function or the revenue function. [latex]1.55\left(50,000\right)=77,500[/latex]

The break-even point is [latex]\left(50,000,77,500\right)[/latex].

The profit function is found using the formula [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex].

[latex]\begin{array}{l}P\left(x\right)=1.55x-\left(0.85x+35,000\right)\hfill \\ \text{ }=0.7x - 35,000\hfill \end{array}[/latex]

The profit function is [latex]P\left(x\right)=0.7x - 35,000[/latex].

Analysis of the Solution

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.

We see from the graph below that the profit function has a negative value until [latex]x=50,000[/latex], when the graph crosses the x -axis. Then, the graph emerges into positive y -values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.

It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.

How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

1) Identify unknown quantities in a problem represent them with variables.

2) Write a system of equations which models the problem’s conditions.

3) Solve the system.

4) Check proposed solution.

Now let’s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.

Example: Writing and Solving a System of Equations in Two Variables

The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?

Let c = the number of children and a = the number of adults in attendance.

The total number of people is [latex]2,000[/latex]. We can use this to write an equation for the number of people at the circus that day.

[latex]c+a=2,000[/latex]

The revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.

[latex]25c+50a=70,000[/latex]

We now have a system of linear equations in two variables.

[latex]\begin{array}{c}c+a=2,000\\ 25c+50a=70,000\end{array}[/latex]

In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[/latex] or [latex]a[/latex]. We will solve for [latex]a[/latex].

[latex]\begin{array}{c}c+a=2,000\\ a=2,000-c\end{array}[/latex]

Substitute the expression [latex]2,000-c[/latex] in the second equation for [latex]a[/latex] and solve for [latex]c[/latex].

[latex]\begin{array}{l} 25c+50\left(2,000-c\right)=70,000\hfill \\ 25c+100,000 - 50c=70,000\hfill \\ \text{ }-25c=-30,000\hfill \\ \text{ }c=1,200\hfill \end{array}[/latex]

Substitute [latex]c=1,200[/latex] into the first equation to solve for [latex]a[/latex].

[latex]\begin{array}{l}1,200+a=2,000\hfill \\ \text{ }\text{}a=800\hfill \end{array}[/latex]

We find that [latex]1,200[/latex] children and [latex]800[/latex] adults bought tickets to the circus that day.

In this video example we show how to set up a system of linear equations that represents the total cost for admission to a museum.

Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets?

700 children, 950 adults

Sometimes, a system can inform a decision.  In our next example, we help answer the question, “Which truck rental company will give the best value?”

Example: Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile. [1] When will Keep on Trucking, Inc. be the better choice for Jamal?

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

A linear function is of the form [latex]f\left(x\right)=mx+b[/latex]. Using the rates of change and initial charges, we can write the equations

[latex]\begin{array}{l}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{array}[/latex]

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\left(d\right)<M\left(d\right)[/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\left(d\right)[/latex] function is smaller.

These graphs are sketched above, with K ( d ) in blue.

To find the intersection, we set the equations equal and solve:

[latex]\begin{array}{l}K\left(d\right)=M\left(d\right)\hfill \\ 0.59d+20=0.63d+16\hfill \\ 4=0.04d\hfill \\ 100=d\hfill \\ d=100\hfill \end{array}[/latex]

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that [latex]K\left(d\right)[/latex] is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d>100[/latex].

One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution.  A solution is a mixture of two or more different substances like water and salt or vinegar and oil.  Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand.  There are many other disciplines that use solutions as well.

The concentration or strength of a liquid solution is often described  as a percentage.  This number comes from the ratio of how much mass is in a specific volume of liquid.  For example if you have 50 grams of salt in a 100mL of water you have a 50% salt solution based on the following ratio:

[latex]\frac{50\text{ grams }}{100\text{ mL }}=0.50\frac{\text{ grams }}{\text{ mL }}=50\text{ % }[/latex]

Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths.  In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.

We will use the following table to help us solve mixture problems:

To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is 120mL of a 9% solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.

First, find the total mass of solids for each solution by multiplying the volume by the concentration.

Next we add the new volumes and new masses.

Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution.  In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.

A chemist has 70 mL of a 50% methane solution. How much of an 80% solution must she add so the final solution is 60% methane?

Let’s use the problem solving process outlined in Module 1 to help us work through a solution to the problem.

Read and Understand:  We are looking for a new amount – in this case a volume –  based on the words “how much”.  We know two starting  concentrations and the final concentration, as well as one volume.

Define and Translate:  Solution 1 is the 70 mL of 50% methane and solution 2 is the unknown amount with 80% methane.  We can call our unknown amount x.

Write and Solve:   Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.

Multiply amount by concentration to get total, be sure to distribute on the last row: [latex]\left(70 + x\right)0.6[/latex]Add the entries in the amount column to get final amount. The concentration for this amount is 0.6 because we want the final solution to be 60% methane.

Add the total mass for solution 1 and solution 2 to get the total mass for the 60% solution. This is our equation for finding the unknown volume.

[latex]35+0.8x=42+0.6x[/latex]

[latex]\begin{array}{c}35+0.8x=42+0.6x\\\underline{-0.6x}\,\,\,\,\,\,\,\underline{-0.6x}\\35+0.2x=42\\\end{array}[/latex]

Subtract 35 from both sides

[latex]\begin{array}{c}35+0.2x=42\\\underline{-35}\,\,\,\,\,\,\,\underline{-35}\\0.2x=7\end{array}[/latex]

Divide both sides by 0.2

[latex]\begin{array}{c}0.2x=7\\\frac{0.2x}{0.2}=\frac{7}{0.2}\end{array}[/latex] [latex]x=35[/latex]

35mL must be added to the original 70 mL to gain a solution with a concentration of 60%

The above problem illustrates how we can use the mixture table to define an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.

A farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons of 20% butterfat?

Read and Understand:  We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is 42 gallons. There are two unknowns in this problem.

Define and Translate:  We will call the unknown volume of the  24% solution x, and the unknown volume of the 18% solution y.

Write and Solve:  Fill in the table with the information we know.

Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from

When you sum the amount column you get one equation: [latex]x+ y = 42[/latex] When you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[/latex]

Use elimination to find a value for [latex]x[/latex], and [latex]y[/latex].

Multiply the first equation by [latex]-0.18[/latex]

[latex]\begin{array}{cc}-0.18(x+y) &= (42)(-0.18) \\ -0.18x-0.18y &= -7.56 \end{array}[/latex]

Now our system of equations looks like this:

[latex]\begin{array}{cc} -0.18x-0.18y &= -7.56\\0.24x + 0.18y &= 8.4 \end{array}[/latex]

Adding the two equations together to eliminate the y terms gives this equation:

[latex]0.06x = 8.4[/latex]

Divide by 0.06 on each side:

[latex]x = 14[/latex]

Now substitute the value for x into one of the equations in order to solve for y.

[latex]\begin{array}{cc} (14) + y &= 42\\ y &= 28 \end{array}[/latex]

This can be interpreted as 14 gallons of 24% butterfat milk added to 28 gallons of 18% butterfat milk will give 42 gallons of 20% butterfat milk.

In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water [latex]b[/latex] and the speed of the river current [latex]c[/latex].

The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is [latex]b+c[/latex].

Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is [latex]b-c[/latex].

We’ll put some numbers to this situation in the next example.

Translate to a system of equations and then solve.

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Read the problem: This is a uniform motion problem and a picture will help us visualize the situation.

Identify  what we are looking for: We are looking for the speed of the ship in still water and the speed of the current.

Name  what we are looking for:

Let [latex]s=[/latex] the rate of the ship in still water.

Let [latex]c=[/latex] the rate of the current.

A chart will help us organize the information. The ship goes downstream and then upstream. Going downstream, the current helps the ship and so the ship’s actual rate is [latex]s+c[/latex]. Going upstream, the current slows the ship and so the actual rate is [latex]s-c[/latex]. Downstream it takes 4 hours. Upstream it takes 5 hours. Each way the distance is 60 miles.

Translate  into a system of equations. Since rate times time is distance, we can write the system of equations.

 [latex]\begin{array}{c}4(s+c)=60 \\ 5(s-c) = 60\end{array}[/latex]

Solve  the system of equations. Distribute to put both equations in standard form, then solve by elimination.

 [latex]\begin{array}{c}4s+4c=60 \\ 5s-5c = 60\end{array}[/latex]

Multiply the top equation by 5 and the bottom equation by 4. Add the equations, then solve for [latex]s[/latex].

[latex]\begin{array}{cc}20s+20c &= 300 \\ 20s-20c &= 240 \\ \hline \\ 40s &= 540 \\ s &= 13.5 \end{array}[/latex]

Substitute [latex]s=13.5[/latex] into one of the original equations.

[latex]\begin{array}{cc}4(s+c) &= 60 \\ 4(13.5+c) &= 60 \\ 54 + 4c &= 60 \\ 4c &= 6 \\ c &= 1.5 \end{array}[/latex]

Check  the answer in the problem. The downstream rate would be:

[latex]13.5+1.5 = 15[/latex] mph

In 4 hours the ship would travel:

[latex]15 \cdot 4 = 60[/latex] miles.

The upstream rate would be

[latex]13.5 - 1.5 = 12[/latex] mph.

In 5 hours the ship would travel

[latex]12\cdot 5[/latex] miles.

Answer  the question. The rate of the ship is 13.5 mph and the rate of the current is 1.5 mph.

In the next video, we present another example of a uniform motion problem which can be solved with a system of linear equations.

• Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ ↵
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System of Linear Equations (Simultaneous Equations)

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• Hemang Agarwal
• Andrew Ellinor
• Thaddeus Abiy
• Yuxuan Seah
• Mahindra Jain
• Aditya Virani
• A Former Brilliant Member
• Abhishu Brahmecha
• Lii Gang Hah
• Chung Kevin

A system of linear equations is a collection of linear equations which involve the same set of variables. As an example,

\[ \begin{align} x+2y & =2 \\ -x+y & =1 \end{align} \]

is a system of equations that has two variables \(x\) and \(y.\) The solution to a linear system is an assignment of numbers to the variables that satisfy every equation in the system. For a given system, we could have one solution, no solutions or infinitely many solutions.

Substitution Method

Elimination method, system of linear equations - more variables, elementary row operation or gaussian elimination method, system of linear equations word problems - basic, system of linear equations word problems - intermediate, system of linear equations - problem solving.

The system of linear equations are shown in the figure bellow:\[\]

• Inconsistent: If a system of linear equations has no solution, then it is called inconsistent.
• Consistent: If a system of linear equations has at least one solution, then it is called consistent.
• Homogeneous system of equations: If the constant term of a system of linear equations is zero, i.e. the value after the "=" sign is zero, then it is called the homogeneous system of equations.
• Trivial and non-trivial solutions: Every homogeneous system of equations has a common solution and that is zero because they have a common solution for all variables, which is 0. This solution is called the trivial solution. If there are other solutions, then they are called non-trivial solutions.

We will explore how to solve a system of linear equations using the substitution and elimination methods.

In this method, we find a relation that isolates one of the variables by changing the subject ; substitute the relation into the other equation(s) to reduce the number of variables by 1; repeat until we are left with a single variable, and solve for it; substitute the solved value back into the relations; state the complete solution.

Let's follow the above steps to solve the following system of equations:

\[\begin{align} x+y &= 5 \\ x-y &= 3. \end{align}\]

Step 1: Isolate variable \(x\) using the second equation: \( x = 3 + y \).

Step 2. Substitute the relation into the other equation: \( (3 + y ) + y = 5 \Rightarrow 3 + 2y = 5 \).

Step 3. Repeat, and then solve. Since we now have only 1 variable, solve for it: \( 3 + 2y = 5 \Rightarrow y = \frac{5-3}{2} = 1 \).

Step 4. Substitute back into the relations: \( x = 3 + y = 3 + 1 = 4 \).

Step 5. State the complete solution: \( (x,y) = (4,1) \).

Solve the system of equations \[ \begin{align} 2x + y &= 10\\ 2y &= 8 . \end{align}\] In this example, because we are already given \( 2y = 8 \) in the second equation, we know that \( y = 4 \). Substituting this into the first equation, we get \( 2x + 4 = 10 \), and thus \( x = 3 \). \(_\square\) Note: It matters which equation we use and which variable we isolate out. In this example, even if we obtained \( x = \frac{ 10 - y } { 2 } \) from the first equation, we find that we are unable to substitute it into the second equation as there is no \(x\) term.

Solve the system of equations \[ \begin{align} 2x + y &= 4\\ 3x + 2y &= 7. \end{align}\] Solving the first equation for \(y\) gives \(y=4-2x\). Substituting this into the second equation gives \[\begin{align} 3x + 2(-2x+4) &= 7\\ 3x -4x + 8 &= 7\\ -x & = -1\\ x & = 1. \end{align}\] Then substituting this value into the first equation, we have \[y=-2x + 4= -2(1) + 4 = -2 + 4 = 2.\] Therefore, the solution to the system of equations is \(x=1, y=2\). \(_\square\)

Using the process of substitution​ may not be the quickest nor the easiest approach for a given system of linear equations. However, we are always guaranteed to find the solution, if we work through the entire process. The word "system" indicates that the equations are to be considered collectively, rather than individually. Thus the solution must not lose validity for any of the equations. Select your options so that your calculations are simple and use any method that suits you.

The elimination method multiplies the given \(n\) equations with suitable constants so that when the modified equations are added, one of the variables is eliminated. Once this is done, the system will have effectively been reduced by one variable and one equation. This process is repeated until one variable and one equation remain (namely, the value of the variable). From there, the obtained value is substituted into the equation with 2 variables, allowing a solution to be found for the second variable. The process is repeated until the values of all \(n\) variables are found.

Method Find two equations that have the same variable. Multiply each equation by a number such that their coefficients are equal. Subtract the two equations. Repeat until we are left with a single variable, and solve for it. Substitute the solved value back into the original equations to solve for the remaining variables.

\[\begin{align} 3x - 4y &= 0 \\ 9x - 8 y &= 12. \end{align}\]

Step 1: Multiply each equation by a number such that the coefficients of a variable are the same. Let's say that we want to eliminate the variable \(x\). Multiply the first equation by 3 and the second equation by 1, and we obtain \[ 9 x - 12 y = 0, 9 x - 8 y = 12. \]

Step 2: Subtract the two equations: \[\begin{align} 9x - 12y &= 0 \\ - (9x - 8 y &= 12) \\ \hline -4 y & = - 12. \end{align} \]

Step 3: Repeat and solve. We are already down to one variable. Solving it gives us \( y = \frac{ -12}{-4} = 3 \).

Step 4: Substitute the solved value back into the relations. Substituting into the first equation, we get \( 3x - 4 \times 3 = 0 \Rightarrow x = \frac{12}{3} = 4 \).

When we have more variables to work with, we just have to remember to stick to a particular method, and keep on reducing the number of equations or variables.

We will solve the following system of equations using both approaches:

\[ \begin{align} x + 3y - z &= 6\\ 2x - y + 2z &= 1\\ 3x + 2y - z &= 2. \end{align}\]

Substitution method We begin with the first of the above three equations: Step 1: The first equation gives \(x+3y-z=6 \Rightarrow x=6-3y+z\). Step 2: Substituting for \(x\) in the second equation, we obtain \[ \begin{align} 2(6-3y+z) - y + 2z &= 1\\ 12 - 6y + 2z - y + 2z &= 1\\ -7y + 4z &= -11. \qquad (1) \end{align}\] Substituting for \(x\) in the third equation, we obtain \[ \begin{align} 3(6-3y+z) + 2y - z & = 2\\ 18 - 9y + 3z + 2y - z &= 2\\ 2z - 7y &= -16. \qquad (2) \end{align}\] Step 3. We need to repeat until we have only one equation. Now, we begin with \((1)\) of the two equations \((1)\) and \((2)\) above: Step 1: The equation \((1)\) gives us \( y = \frac{ 4z+11}{7} \). Step 2: Substituting for \(y\) in \((2)\) gives \[ \begin{align} 2z - 7 \frac{ 4z+11}{7}&= -16\\ 2z - ( 4z + 11 ) & = - 16 \\ -2z & = -5. \end{align}\] Step 3: We are now down to one equation. Solving it gives us \( z = \frac{-5}{-2} = \frac{5}{2} \). Step 4: Substitute \( z = \frac{5}{2} \) into \( y = \frac{ 4z + 11} { 7} \) to obtain \[ y = \frac{ 4z+ 11}{ 7} = \frac{ 4 \times \frac{5}{2} + 11 } { 7} = \frac{ 21}{7} = 3. \] Now, substitute \( z = \frac{5}{2}\) and \(y = 3 \) into \( x = 6 - 3y + z \) to obtain \[ x = 6 - 3 \times 3 + \frac{5}{2} = - \frac{1}{2}. \] Step 5: Hence, the solution is \[ (x,y,z) = \left( - \frac{1}{2}, 3, \frac{5}{2} \right). \ _\square \]

Elimination method We are given the same system of linear equations as above: \[ \begin{align} x + 3y - z &= 6\\ 2x - y + 2z &= 1\\ 3x + 2y - z &= 2. \end{align}\] Step 1: Let's eliminate \(x\) from the equations. Step 2. Twice of the first equation minus the second is \[ \begin{align} 2(x + 3y - z &= 6)\\ -(2x - y + 2z &= 1)\\ \hline 7y - 4z & = 11. \qquad (4) \\\end{align} \] Thrice of the first equation minus the third is \[ \begin{align} 3(x + 3y - z &= 6)\\ -(3x + 2y - z &= 2)\\ \hline 7y - 2z & = 16. \qquad (5) \\ \end{align} \] We need to repeat till we just have one variable. Step 1. Let's eliminate \( y \) from the equations. Step 2. The fourth equation minus the fifth is \[ \begin{align} 7y - 4z & = 11 \\ - ( 7y - 2z & = 16 ) \\ \hline -2z & = - 5. \qquad (6) \\ \end{align} \] Step 3. We are now down to one variable. Solving it gives us \( z = \frac{-5}{-2} = \frac{5}{2} \). Step 4. Substitute \( z = \frac{5}{2} \) into the fourth equation \( 7y - 4z = 11 \) to obtain \[ y = \frac{ 4z+ 11}{ 7} = \frac{ 4 \times \frac{5}{2} + 11 } { 7} = \frac{ 21}{7} = 3. \] Now, substitute \( z = \frac{5}{2} , y = 3 \) into the first equation \( x = 6 - 3y + z \) to obtain \[ x = 6 - 3 \times 3 + \frac{5}{2} = - \frac{1}{2}. \] Step 5. Hence, the solution is \[ (x,y,z) = \left( - \frac{1}{2}, 3, \frac{5}{2} \right).\ _\square \]

Solve the following system of equations: \[\] \[ \begin{align} x + 2y - 3z &= -3\\ 2x - 5y + 4z &= 13\\ 5x + 4y - z&= 5. \\ \end{align} \] Let us start with the last equation. Solving for \(z\), we obtain \[z = 5x + 4y - 5.\] Substituting this into the second equation gives \[ \begin{align} 2x - 5y + 4(5x + 4y - 5 ) &= 13 \\ 22x + 11y - 33 &= 0 \\ 2x + y - 3 &= 0. \end{align} \] Substituting this into the first equation gives \[ \begin{align} x + 2y - 3(5x + 4y - 5) &= -3 \\ x + 2y - 15x -12y + 15 + 3 &= 0 \\ -14x - 10y + 18 &= 0\\ -7x - 5y + 9 &= 0. \end{align} \] Thus, we now have reduced our system into a pair of equations with two variables: \[ \begin{align} 2x + y - 3 &= 0\\ -7x - 5y + 9 &= 0. \end{align} \] Solving for \(y\) in the first equation, we get \[ y = 3 - 2x. \] Substituting this into the second equation gives \[ \begin{align} -7x - 5(3-2x)+9&=0 \\ -7x-15 + 10x + 9 &= 0\\ 3x - 6 &= 0 \\ x &= 2. \end{align} \] Hence, \[y = 3 - 2\cdot2 = -1, z = 5\cdot2 + 4\cdot(-1) -5 = 1. \] Thus, the values of \(x, y,\) and \(z\) which satisfy the given system of equations are \((2,-1,1). \ _\square \)

Main Article: Solving Linear Systems Using Matrices

Elementary row operation or Gaussian elimination is a popular method for solving system of linear equations. By this method, everyone can solve system of linear equations only by matrix row operations that you already know. The terms we need to know for this section are explained below.

Augmented Matrix: An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation and each column represents all the coefficients for a single variable. Suppose the system of linear equation is \[\begin{align} x_{1} + 2x_{2} + 5x_{3}&=2\\ 3x_{1} + x_{2} + 4x_{3}&=1\\ 2x_{1} - 7x_{2} + x_{3}&=5. \end{align}\] Then the augmented matrix of this system of linear equations is \(\begin{pmatrix} 1 & 2 & 5 &2 \\ 3 & 1 & 4 & 1\\ 2 & -7 & 1 & 5 \end{pmatrix}.\)

What are row echelon form, reduced row echelon form, and leading 1?

The matrix is in reduced row echelon form \(\begin{pmatrix} 1 & 0 & 0 &1 \\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 2 \end{pmatrix}.\) To be of this form, a matrix must have the following properties: If a row does not consist entirely of zeros, then the first nonzero number in the row is a 1. We call this a leading 1. If there are any rows that consist entirely of zeros, then they are grouped together at the bottom of the matrix. In any two successive rows that do not consist entirely of zeros, the leading 1 in the lower row occurs farther to the right than the leading 1 in the higher row. Each column that contains a leading 1 has zeros everywhere else in that column. A matrix that has the first three properties is said to be in row echelon form. If that matrix has the \(4^\text{th}\) property, then it's called reduced row echelon form.

Changing a matrix into row echelon form by elementary row operation:

A matrix can be changed to its reduced row echelon form, or row reduced to its reduced row echelon form using the elementary row operations, which are as follows: Interchange one row of the matrix with another of the matrix. Multiply one row of the matrix by a nonzero scalar constant. Replace the one row with the one row plus a constant times another row of the matrix. Suppose that a system of linear equation is \[\begin{align*} x_{1} + 2x_{2} + 2x_{3}&=4\\ x_{1} + 3x_{2} + 3x_{3}&=5\\ 2x_{1} + 6x_{2} + 5x_{3}&=6, \end{align*}\] and its augmented matrix is \(\begin{pmatrix} 1 & 2 & 2 &4 \\ 1 & 3 & 3 & 5\\ 2 & 6 & 5 & 6 \end{pmatrix}.\) Now, we'll apply elementary row operation on this matrix and convert it to reduced row echelon form: \[\begin{align} &\Rightarrow\begin{pmatrix} 1 & 2 & 2 &4 \\ 0 & 1 & 1 & 1\\ 2 & 6 & 5 & 6 \end{pmatrix}-1R_{1}+R_{2}\\ \\ &\Rightarrow \begin{pmatrix} 1 & 2 & 2 &4 \\ 0 & 1 & 1 & 1\\ 0 & 2 & 1 & -2 \end{pmatrix}-2R_{1}+R_{3}\\ \\ &\Rightarrow\begin{pmatrix} 1 & 2 & 2 &4 \\ 0 & 1 & 1 & 1\\ 0 & 0 & -1 & -4 \end{pmatrix}-2R_{2}+R_{3}\\ \\ &\Rightarrow\begin{pmatrix} 1 & 2 & 2 &4 \\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 4 \end{pmatrix}-1R_{3}. \end{align}\]

Finally the matrix is in the row echelon form: \(\begin{pmatrix} 1 & 2 & 2 &4 \\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 4 \end{pmatrix}. \)

So, at last we get the reduced form of the system of linear equation, which is \[\begin{align*} x_{1} +2x_{2}+2x_{3} &= 4\\ x_{2}+x_{3}&=1 \\ x_{3} &= 4. \end{align*}\] By solving this, we get \[\begin{align*} x_{1}&= 2\\ x_{2} &=-3 \\ x_{3} &= 4. \end{align*}\]

Md Nur Uddin is a student of Mathematics Discipline Of Khulna University. Now, you have to find his student ID by solving the following system of linear equations:

\[\begin{align*} -3x+2y-6z &=6 \\ 5x+7y-5z &=6 \\ x+4y-2z &=8. \end{align*}\]

If the solution satisfies \(x+y=z,\) what is the value of \(x+y,\) or equivalently \(z?\)

What weight will the fourth scale display?

A bottle that fully contains honey weighs 1500 grams. The bottle with half of the honey weighs 900 grams. What is the weight of the empty bottle in grams?

Today, in a 10-member committee, an old member was replaced by a young member. As such, the average age is the same today as it was 4 years ago.

What is the (positive) difference in ages between the new member and the replaced old member?

The brothers Luiz and Lucio bought a land surrounded by a wall of \(340 \text{ meters} \). They built an inner wall to divide the land into two parts. Now, the part of Luiz is surrounded by a wall of \(260 \text{ meters} ,\) and the part of Lucio by a wall of \(240 \text{ meters} \). What is the length of the inner wall?

Today, there are 4 times as many days to the end of my exams, as there are to the start of my exams. Tomorrow, there are 5 times as many days to the end of my exams, as there are to the start of my exams.

How many days do my exams last for?

There was a thief who went to a wine shop to steal wine. He went in at 10 pm, stole 15 L of wine and added 15 L of water to top up the barrel. He came back at 1 am, stole 15 L of the mixture from the same barrel and added 15 L of water to top it up. He came back at 4 am, stole 15 L of the mixture from the same barrel and added 15 L of water to top it up. In the morning, the ratio of wine to water in the barrel was 343:169. Find the initial amount of wine in the barrel.

The solutions of the system of equations \[3x−y=a, \qquad x+y=5\] are the same as those of the system of equations \[2x+y=24, \qquad x−3y=b.\] What is the value of \(a+b\)? Since the two systems of equations have the same solutions, we first use substitution to find the solutions to the simultaneous equations \[x+y=5, \qquad 2x+y=24.\] Solving the first equation for \(y\) gives \(y=5-x\) and plugging into the second equation gives \(2x + (5-x) = 24\), or \(x = 19\). Plugging this into the first equation gives \(19+y=5\), or \(y=-14\). Thus, \(x=19\) and \( y=−14. \) Substituting these values into \(3x−y=a\) gives \[a=3x−y=3×19−(−14)=71.\] Similarly, substituting into \(x−3y=b\) gives \[b=x−3y=19−3×(−14)=61.\] Therefore, \[a+b=71+61=132. \ _\square\]

If the numbers alongside each column and row are the total of the values of the symbols within that column and row, then which number should replace the question mark?

Given that \(x=y=0\) is not the only solution to the following system of linear equations, determine all the possible values of \(k:\) \[\begin{align} x+2y &= kx \\ 2x+y &= ky . \end{align}\]

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Math Made Easy

Systems of Linear Equations and Word Problems

Note that we saw how to solve linear inequalities here in the Coordinate System and Graphing Lines section . Note also that we solve Algebra Word Problems without Systems here , and we solve systems using matrices in the Matrices and Solving Systems with Matrices  section here.

Introduction to Systems

“Systems of equations” just means that we are dealing with more than one equation and variable. So far, we’ve basically just played around with the equation for a line, which is $ y=mx+b$. Let’s say we have the following situation:

Now, you can always do “guess and check” to see what would work, but you might as well use algebra! It’s much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra way will let you solve any algebra problem – even the really complicated ones.

The first trick in problems like this is to figure out what we want to know. This will help us decide what variables (unknowns) to use. What we want to know is how many pairs of jeans we want to buy (let’s say “$ j$”) and how many dresses we want to buy (let’s say “$ d$”). Always write down what your variables will be:

Let $ j=$ the number of jeans you will buy Let $ d=$ the number of dresses you’ll buy

Like we did before, let’s translate word-for-word from math to English:

Now we have the 2 equations as shown below. Notice that the $ j$ variable is just like the $ x$ variable and the $ d$ variable is just like the $ y$. It’s easier to put in $ j$   and $ d$ so we can remember what they stand for when we get the answers.

This is what we call a system, since we have to solve for more than one variable – we have to solve for 2 here. The cool thing is to solve for 2   variables , you typically need 2   equations , to solve for 3   variables , you need 3   equations , and so on. That’s easy to remember, right?

We need to get an answer that works in both equations ; this is what we’re doing when we’re solving; this is called solving simultaneous systems , or solving system simultaneously . There are several ways to solve systems; we’ll talk about graphing first.

Solving Systems by Graphing

Remember that when you graph a line, you see all the different coordinates (or $ x/y$ combinations) that make the equation work. In systems, you have to make both equations work, so the intersection of the two lines shows the point that fits both equations (assuming the lines do in fact intersect; we’ll talk about that later).  The points of intersections satisfy both equations simultaneously. 

Put these equations into the $ y=mx+b$ ($ d=mj+b$) format, by solving for the $ d$ (which is like the $ y$):

$ \displaystyle j+d=6;\text{ }\,\text{ }\text{solve for }d:\text{ }d=-j+6\text{ }$

$ \displaystyle 25j+50d=200;\text{ }\,\,\text{solve for }d:\text{ }d=\frac{{200-25j}}{{50}}=-\frac{1}{2}j+4$

Now graph both lines:

Note that with non-linear equations, there will most likely be more than one intersection; an example of how to get more than one solution via the Graphing Calculator can be found in the  Exponents and Radicals in Algebra section. Also, t here are some examples of systems of inequality  here in the Coordinate System and Graphing Lines section .

Solving Systems with Substitution

Substitution is the favorite way to solve for many students! It involves exactly what it says: substituting one variable in another equation so that you only have one variable in that equation. Here is the same problem:

You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50 .  You really, really want to take home 6 items of clothing because you “need” that many new things. How many pairs of jeans and how many dresses you can buy so you use the whole $200 (tax not included)?

Below are our two equations, and let’s solve for “$ d$” in terms of “$ j$” in the first equation. Then, let’s substitute what we got for “$ d$” into the next equation. Even though it doesn’t matter which equation you start with, remember to always pick the “easiest” equation first (one that we can easily solve for a variable) to get a variable by itself.

We could buy 4 pairs of jeans and 2 dresses . Note that we could have also solved for “$ j$” first; it really doesn’t matter. You’ll want to pick the variable that’s most easily solved for. Let’s try another substitution problem that’s a little bit different:

Solving Systems with Linear Combination or Elimination

Probably the most useful way to solve systems is using linear combination, or linear elimination. The reason it’s most useful is that usually in real life we don’t have one variable in terms of another (in other words, a “$ y=$” situation).

The main purpose of the linear combination method is to add or subtract the equations so that one variable is eliminated. We can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below. We are using the Additive Property of Equality , Subtraction Property of Equality , Multiplicative Property of Equality , and/or Division Property of Equality that we saw here in the Types of Numbers and Algebraic Properties section :

If we have a set of 2 equations with 2 unknowns, for example, we can manipulate them by adding, multiplying or subtracting (we usually prefer adding) so that we get one equation with one variable. Let’s use our previous problem:

We could buy 4 pairs of jeans and 2 dresses .

Here’s another example:

Types of equations

In the example above, we found one unique solution to the set of equations. Sometimes, however, for a set of equations, there are no solutions (when lines are parallel) or an infinite number of solutions or infinitely many solutions (when the two lines are actually the same line, and one is just a “multiple” of the other).

When there is at least one solution , the equations are consistent equations , since they have a solution. When there is only one solution, the system is called independent , since they cross at only one point. When equations have infinite solutions, they are the same equation, are consistent , and are called dependent or coincident (think of one just sitting on top of the other).

When equations have no solutions , they are called inconsistent equations , since we can never get a solution . 

Here are graphs of inconsistent and dependent equations that were created on a graphing calculator:

Systems with Three Equations

Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns to solve for.

Let’s say at the same store, they also had pairs of shoes for $20 and we managed to get $60 more to spend! Now we have a new problem. To spend the even $260 , how many pairs of jeans, dresses, and pairs of shoes should we get if want, for example, exactly 10 total items (Remember that jeans cost $25 each and dresses cost $50 each).

Let’s let $ j=$ the number of pair of jeans, $ d=$ the number of dresses, and $ s=$ the number of pairs of shoes we should buy. So far, we’ll have the following equations:

$ \displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+\,20s=260\end{array}$

We’ll need another equation, since for three variables, we need three equations (otherwise, we theoretically might have infinite ways to solve the problem). In this type of problem, you would also need something like this: We want twice as many pairs of jeans as pairs of shoes . Now, since we have the same number of equations as variables , we can potentially get one solution for the system of equations. Here are the three equations:

We’ll learn later how to put these in our calculator to easily solve using matrices (see the  Matrices and Solving Systems with Matrices section). For now, we can use two equations at a time to eliminate a variable (using substitution and/or elimination), and keep doing this until we’ve solved for all variables. These can get really difficult to solve, but remember that in “real life”, there are computers to do all this work!

Remember again, that if we ever get to a point where we end up with something like this, it means there are an infinite number of solutions : $ 4=4$  (variables are gone and a number equals another number and they are the same). And if we up with something like this, it means there are no solutions : $ 5=2$ (variables are gone and two numbers are left and they don’t equal each other).

Let’s solve our system:      $ \displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+20s=260\\j=2s\end{array}$ :

We could buy 6 pairs of jeans, 1 dress, and 3 pairs of shoes .

Here’s one more example of a three-variable system of equations, where we’ll only use linear elimination:

$ \displaystyle \begin{align}5x-6y-\,7z\,&=\,7\\6x-4y+10z&=\,-34\\2x+4y-\,3z\,&=\,29\end{align}$

I know – this is really difficult stuff! But if you do it step-by-step and keep using the equations you need with the right variables, you can do it. Think of it like a puzzle – you may not know exactly where you’re going, but do what you can in baby steps, and you’ll get there (sort of like life sometimes, right?!). And we’ll learn much easier ways to do these types of problems.

Algebra Word Problems with Systems

Let’s do more word problems; you’ll notice that many of these are the same type that we did earlier in the Algebra Word Problems  section , but now we can use more than one variable. This will actually make the problems easier! Again, when doing these word problems:

• If you’re wondering what the variables (or unknowns) should be when working on a word problem, look at what the problem is asking. These are usually (but not always) what your variables are!
• If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!

Here are some problems:

Investment Word Problem

We also could have set up this problem with a table:

Mixture Word Problems

Here’s a mixture word problem . With mixture problems, remember if the problem calls for a pure solution or concentrate , use 100% (if the percentage is that solution) or 0% (if the percentage is another solution).

Let’s do the math (use substitution )!

$ \displaystyle \begin{array}{c}x\,\,+\,\,y=10\\.01x+.035y=10(.02)\end{array}$          $ \displaystyle \begin{array}{c}\,y=10-x\\.01x+.035(10-x)=.2\\.01x\,+\,.35\,\,-\,.035x=.2\\\,-.025x=-.15;\,\,\,\,\,x=6\\\,y=10-6=4\end{array}$

We would need 6 liters of the 1% milk, and 4 liters of the 3.5% milk.

Here’s another mixture problem:

$ \displaystyle \begin{array}{c}x+y=50\\8x+4y=50\left( {6.4} \right)\end{array}$                   $ \displaystyle \begin{array}{c}y=50-x\\8x+4\left( {50-x} \right)=320\\8x+200-4x=320\\4x=120\,;\,\,\,\,x=30\\y=50-30=20\\8x+4y=50(6.4)\end{array}$

We would need 30 pounds of the $8 coffee bean, and 20 pounds of the $4 coffee bean. See how similar this problem is to the one where we use percentages?

Distance Word Problem:

Here’s a distance word problem using systems ; distance problems have to do with an object’s speed, time, and distance. Note that, as well as the distance word problem here in the Algebra Word Problems section , there’s an example of a Parametric Distance Problem here in the Parametric Equations section .

Which Plumber Problem

Many word problems you’ll have to solve have to do with an initial charge or setup charge, and a charge or rate per time period. In these cases, the initial charge will be the $ \boldsymbol {y}$ -intercept , and the  rate will be the slope . Here is an example:

Geometry Word Problem:

Many times, we’ll have a geometry problem as an algebra word problem; these might involve perimeter, area, or sometimes angle measurements (so don’t forget these things!). Let’s do one involving angle measurements.

See – these are getting easier! Here’s one that’s a little tricky though:

Work Problem : 

Let’s do a “ work problem ” that is typically seen when studying Rational Equations (fraction with variables in them) and can be found here in the Rational Functions, E quations and Inequalities  section .

Note that there’s also a simpler version of this problem here in the Direct, Inverse, Joint and Combined Variation section .

Three Variable Word Problem:

Let’s do one more with three equations and three unknowns:

The “Candy” Problem

Sometimes we get lucky and can solve a system of equations where we have more unknowns (variables) then equations. (Actually, I think it’s not so much luck, but having good problem writers!) Here’s one like that:

There are more Systems Word Problems in the  Matrices and Solving Systems with Matrices section , Linear Programming section , and Right Triangle Trigonometry section .

Understand these problems, and practice, practice, practice!

For Practice : Use the Mathway  widget below to try a  Systems of Equations  problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve by Substitution or Solve by Addition/Elimination  to see the answer .

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps , or Click Here , you can register at Mathway for a free trial , and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Algebraic Functions, including Domain and Range   – you’re ready! 

Problem Set

Solving systems of linear equations.

1.  Solve the following system of equations by elimination.

Answer: x = .5; y = 1.67

Rewrite in order to align the x and y terms

Add the second equation to the first equation and solve for x.

Substitute the value obtained for x into either of the original equations.

            or

2.  Solve the following system of equations by elimination.

Answer: x = -2; y = 5

Multiply the first equation by 2.

Subtract the first equation from the second equation and solve for y.

Substitute the value obtained for y into either of the original equations.

3.  Solve the following system of equations by elimination.

Answer: x 1 = 2; x 2 = 1; x 3 = -2

Part A First eliminate x 3 .

Step 1. Add the 1st original equation and the 3rd original equation.

                        +

                        =

Step 2. Multiply the 2nd original equation by 2, multiply the 3rd original equation by 5 and add the 2nd original equation to the 3rd original equation.

                       +

                       =

Part B. From Part A there are two new equations.

                              1st new equation

                          2nd new equation

Step 1. Multiply the 1st new equation by 7, subtract the second new equation from the first new equation, and solve for x 2 .

                     _

Part C. Solve for x 1 by substituting the value received for x 2 in either of the new equations in Part B.

                                1st new equation

           or

                               2nd new equation

Part D. Solve for x 3 by substituting the values obtained for x 1 and x 2 in any of the original equations.

4.  A manufacturer produces two items which are sold at prices of p 1 dollars and p 2 dollars each.  The supply and demand functions in units for the items are as follows:

            P(s 1 ) = the supply function for the first item.

            P(d 1 ) = the demand function for the first item.

            P(s 2 ) = the supply function for the second item.

            P(d 2 ) = the demand function for the second item.

How should prices be set for each item to equate supply and demand?  What are the equilibrium quantities for each item?

Answer:   The price of the first item should be set at $1.80 and the price of the second item should be set at $1.50.  The equilibrium quantity for the first item is 82 units and the equilibrium quantity for the second item is 181 units.

Step 1. Set the supply function for item 1 equal to the demand function for item 1 and collect terms.

           P(s 1 )  = P(d 1 )

Step 2. Set the supply function for item 2 equal to the demand function for item 2 and collect terms.

            P(s 2 )  = P(d 2 )

Part B. From Part A the following system of equations has been obtained.

Solve this system for the prices by elimination.

                             1st equation from Part A

Step 1. Multiply the 1st equation by 7 ,add the 1st and second equations, and solve for p 1 .

Step 2 . Substitute the value obtained for p 1 in either equation from Part A and solve for p 2 .

                            1st equation from Part A

Part C. Determine equilibrium for each item by substituting the prices obtained in Part B in either the demand function or the supply function.

• Linear systems – word problems

When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. Once you do that, these linear systems are solvable just like other linear systems . The same rules apply. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you accustomed to finding elements of linear systems inside of word problems. Example 1 : A farmhouse shelters 16 animals. Some of them are chickens and the others are cows. Altogether these animals have 60 legs. How many chickens and how many cows are in the farmhouse?

First, to make the calculations clearer, we will choose symbols to represent the number of cows and the number of chickens. Let us say that the chickens will be represented with x and the cows with y. Now, this task gave us enough information to make two equations. The first one is that the sum of the number of chickens (x) and the number of cows (y) is 16, since there are only 16 animals in the farmhouse. That equation should look like this: x + y = 16

The second piece of information we have is that the total number of legs in the farmhouse is 60. Since we know that cows have four legs each and chickens have two legs each, we have enough information to make another equation. This one will look like this: 2*x + 4*y = 60

Now we have a system of linear equations with two equations and two variables. The only thing left to do now is to solve the system. We will solve it here for you, but if you need to remind yourself how to do that step by step, read the article called Systems of linear equations . x = 16 – y 32 – 2y + 4y = 60 2y = 28 y = 14 x = 2

We can now see that there are two chickens and 14 cows in the farmhouse. The next example will be a bit harder,

Example 2: Rodney’s Kitchen Supplies makes and sells spoons and forks. It costs the store $2 to buy the supplies needed to make a fork, and $1 for the supplies needed to make a spoon. The store sells the forks for $4 and the spoons for 5$. Last month Rodney’s Kitchen Supplies spent $39 on supplies and sold the all of the forks and spoons that were made last month using those supplies for $93. How many forks and spoons did they make?

As we did in the first example, we will first designate symbols to available variables. So, the number of forks made will be represented with x and the number of spoons with y. Again, we have enough information to make two equations. The total cost of making a particular number of forks (x), which cost $2 to make each, and a particular number of spoons (y), which cost $1 to make each, is $39. So that will be our first equation and it will look like this: 2*x + y = 39 The other piece of information tells us that if we sell that number of forks (x) for $4 each and that number of spoons (y) for $5 each, we will make $93. And that will be our second equation: 4*x + 5*y = 93 This was the hard part. Now all we have to do is to solve this linear system to find how many spoons and how many forks did we make last month. y = 39 – 2x 4x + 5*(39 – 2x) = 93 4x + 195 – 10x = 93 -6x = 93 – 195 -6x = -102 |: (-6) x = 17 y = 5

We can see that last month the store made and sold 17 forks and five spoons.

Although they can seem complicated, mastery and understanding of linear systems and associated word problems will come with a bit of practice. With experience you will be able to recognize their elements and solve even complicated systems with ease. Feel free to use the math worksheets below to practice solving this type of linear systems.

Linear systems – word problems exams for teachers

Linear systems – word problems worksheets for studets.

Basic Mathematical Operations

• Addition and subtraction
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• Greatest common factor
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< 14.4 Solutions to Systems of Linear Equations | Contents | 14.6 Matrix Inversion >

Solve Systems of Linear Equations in Python ¶

Though we discussed various methods to solve the systems of linear equations, it is actually very easy to do it in Python. In this section, we will use Python to solve the systems of equations. The easiest way to get a solution is via the solve function in Numpy.

TRY IT! Use numpy.linalg.solve to solve the following equations.

We can see we get the same results as that in the previous section when we calculated by hand. Under the hood, the solver is actually doing a LU decomposition to get the results. You can check the help of the function, it needs the input matrix to be square and of full-rank, i.e., all rows (or, equivalently, columns) must be linearly independent.

TRY IT! Try to solve the above equations using the matrix inversion approach.

We can also get the \(L\) and \(U\) matrices used in the LU decomposition using the scipy package.

TRY IT! Get the \(L\) and \(U\) for the above matrix A.

We can see the \(L\) and \(U\) we get are different from the ones we got in the last section by hand. You will also see there is a permutation matrix \(P\) that returned by the lu function. This permutation matrix record how do we change the order of the equations for easier calculation purposes (for example, if first element in first row is zero, it can not be the pivot equation, since you can not turn the first elements in other rows to zero. Therefore, we need to switch the order of the equations to get a new pivot equation). If you multiply \(P\) with \(A\) , you will see that this permutation matrix reverse the order of the equations for this case.

TRY IT! Multiply \(P\) and \(A\) and see what’s the effect of the permutation matrix on \(A\) .

How To Perform Linear Algebra In Julia

Linear algebra is foundational in data science and many other computational fields. In this article, we'll guide developers through essential linear algebra techniques using Julia—a dynamic programming language renowned for its efficiency in numerical computing.

Linear algebra forms the backbone of numerous computational tasks. With Julia's efficient syntax and computational capabilities, tackling these tasks becomes significantly more approachable. Let's explore how you can seamlessly integrate these mathematical operations into your Julia projects.

Setting Up Julia For Linear Algebra

Basic vector operations, matrix operations and manipulations, decompositions and factorizations, solving linear systems, eigenvalues and eigenvectors, implementing specialized linear algebra techniques, performance tips for large scale computations, frequently asked questions, installing julia, adding required packages, basic linear algebra operations, matrix multiplication, determinant and inverse, eigenvalues and eigenvectors, linear equations.

To dive into linear algebra with Julia, you first need to ensure you have Julia installed on your machine. Visit the official Julia website to download the latest version suitable for your operating system.

Julia has an extensive package ecosystem. For linear algebra, some essential packages are LinearAlgebra (which comes with the standard library) and others like Arpack .

With the required setup done, let's explore basic linear algebra operations in Julia.

Matrix Creation

In Julia, creating matrices is straightforward. Use the following syntax:

Multiply matrices using the * operator. Ensure the matrices are conformable for multiplication.

Compute the determinant and inverse of a matrix using det() and inv() respectively.

To determine the eigenvalues and eigenvectors of a matrix, utilize the eigen() function.

To solve a system of linear equations, use the backslash operator \ .

This method leverages LU decomposition for solving. Ensure the coefficient matrix is non-singular.

Creating Vectors

Vector addition and subtraction, scalar multiplication, dot product, cross product, vector magnitude, unit vector.

In Julia, vectors are a subtype of Array . They can be created using simple square brackets.

Vector addition and subtraction are elementary. Ensure vectors are of the same size for these operations.

Multiply a vector by a scalar simply using the * operator.

Computing the dot product (or scalar product) between two vectors is a common operation in linear algebra.

The cross product produces a vector orthogonal to the two input vectors. It's defined only for 3D vectors.

The magnitude (or length) of a vector can be computed using the norm() function.

A unit vector has a magnitude of one and points in the same direction as the original vector.

When you normalize a vector like v1 , you get its unit vector, which retains the original's direction but has a length of one.

Matrix Addition and Subtraction

Matrix-scalar multiplication, matrix transposition, accessing elements and slicing, matrix determinant, matrix rank, matrix inversion.

In Julia, matrices can be created using square brackets. Elements are organized row by row.

Add or subtract matrices of the same dimension using + and - operators, respectively.

Multiply every element of a matrix by a scalar with ease.

Matrix multiplication in Julia is done using the * operator. Ensure matrices are conformable.

Switching rows and columns of a matrix is termed as transposition .

Retrieve specific elements or slices of a matrix using indexing.

Calculating the determinant is pivotal for various mathematical operations.

The rank of a matrix provides insight into its linear independence.

An inverse exists for square matrices that are non-singular.

When a matrix is multiplied by its inverse, the result is the identity matrix.

LU Decomposition

Qr decomposition, singular value decomposition (svd), cholesky decomposition, eigendecomposition, schur decomposition, hessenberg form.

LU decomposition separates a matrix into a product of a lower triangular and an upper triangular matrix.

This factorization decomposes a matrix into an orthogonal matrix (Q) and an upper triangular matrix (R).

SVD decomposes a matrix into three other matrices: an orthogonal matrix (U), a diagonal matrix (Σ), and the transpose of an orthogonal matrix (Vᵀ).

For hermitian, positive-definite matrices, Cholesky decomposition factorizes the matrix into a lower triangular matrix and its conjugate transpose.

This decomposition revolves around finding the eigenvalues and eigenvectors of a matrix.

This factorization breaks down a matrix into a unitary matrix and an upper triangular matrix.

Converting a matrix into an upper Hessenberg form is an essential step for some algorithms.

The matrix H is almost triangular, and Q is orthogonal. This form is a waypoint to other decompositions, especially in eigenvalue algorithms.

Direct Methods

Gaussian elimination, lu factorization, iterative methods, jacobi method, conjugate gradient method, condition number, checking solution validity.

Direct methods aim for an exact solution, typically in a finite number of steps.

A popular method for solving linear systems is Gaussian elimination . This method transforms the system to an equivalent upper triangular form.

LU factorization aids in solving linear systems by decomposing the matrix, which is then used for solution derivation.

Iterative methods produce approximations that converge to the exact solution.

A simple iterative method, the Jacobi method , updates the solution based on matrix rows.

The Conjugate Gradient method is suitable for symmetric positive definite matrices.

The condition number of a matrix gauges how a system's output changes due to small variations in input.

After finding a solution, it's crucial to verify its correctness.

A small residual indicates the found solution is close to being correct.

Eigenvalues and eigenvectors provide profound insight into the properties and behavior of linear transformations. These concepts play a crucial role in various fields like quantum mechanics, control theory, and data science.

Computing Eigenvalues and Eigenvectors

Properties of eigenvalues, power iteration method, inverse iteration method.

In Julia, the eigen function computes both eigenvalues and eigenvectors of a matrix.

The sum of eigenvalues equals the trace of a matrix, and the product of eigenvalues equals its determinant .

A useful technique for finding the largest eigenvalue in magnitude is the power iteration method .

To find the smallest eigenvalue in magnitude, the inverse iteration method is handy.

Invoke the method using inverse_iteration(A) . This function seeks the smallest eigenvalue and its corresponding eigenvector.

Sparse Matrix Operations

Toeplitz matrix, kronecker product, qr factorization with column pivoting, householder reflections.

Handling large matrices can be computationally demanding. Sparse matrices , where most entries are zero, offer memory and computational advantages.

Toeplitz matrices have constant diagonals. Exploiting this structure can speed up operations.

The Kronecker product is a special matrix multiplication that can be efficiently computed.

For symmetric positive definite matrices, the Cholesky decomposition provides a way to simplify matrix operations.

QR factorization decomposes a matrix into orthogonal and upper triangular matrices. Adding column pivoting improves numerical stability.

Householder reflections are transformation techniques that zero out matrix elements below a diagonal.

The result H is a Householder matrix. It's particularly useful in QR factorization and other decomposition methods.

Use Native Functions

Avoid global variables, pre-allocate memory, use in-place operations, parallelize computations, profile your code, optimize loops.

Julia's native functions are highly optimized. Always prefer them over custom implementations. For instance, use sum() instead of writing a loop to sum elements.

Global variables can slow down the computation. Always prefer local variables and pass them as function arguments.

For large-scale computations, allocate memory beforehand. This avoids frequent memory reallocations which are costly.

In-place operations modify existing variables instead of creating new ones. This reduces memory overhead and boosts performance.

Leverage Julia's parallel computing capabilities for heavy computations. This distributes the workload across multiple CPU cores.

The @profile macro helps identify performance bottlenecks. After profiling, use the Profile module to analyze results.

Unroll loops and avoid using dynamic arrays inside loops. This reduces overheads and speeds up execution.

Maintain a tight loop for the best performance.

What is Julia?

Julia is a high-performance programming language for technical computing, with syntax that is familiar to users of other technical computing environments.

How does Julia compare to Python or MATLAB for linear algebra tasks?

Julia boasts speed comparable to languages like C or Fortran, making it highly efficient for linear algebra operations. Its syntax is also concise, making it easier to write and understand complex mathematical operations compared to some other languages.

Is there a package in Julia for linear algebra?

Yes, Julia has a built-in package called LinearAlgebra that provides essential functionalities for linear algebra operations.

How do I install the LinearAlgebra package in Julia?

The LinearAlgebra module is part of Julia's standard library. You can use it by simply adding using LinearAlgebra at the beginning of your code.

Let’s test your knowledge!

Which Julia function is commonly used for matrix multiplication?

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Introduction to computers and engineering problem solving, course description.

This course presents the fundamentals of object-oriented software design and development, computational methods and sensing for engineering, and scientific and managerial applications. It cover topics, including design of classes, inheritance, graphical user interfaces, numerical methods, streams, threads, sensors, and …

This course presents the fundamentals of object-oriented software design and development, computational methods and sensing for engineering, and scientific and managerial applications. It cover topics, including design of classes, inheritance, graphical user interfaces, numerical methods, streams, threads, sensors, and data structures. Students use Java ® programming language to complete weekly software assignments.

How is 1.00 different from other intro programming courses offered at MIT?

1.00 is a first course in programming. It assumes no prior experience, and it focuses on the use of computation to solve problems in engineering, science and management. The audience for 1.00 is non-computer science majors. 1.00 does not focus on writing compilers or parsers or computing tools where the computer is the system; it focuses on engineering problems where the computer is part of the system, or is used to model a physical or logical system.

1.00 teaches the Java programming language, and it focuses on the design and development of object-oriented software for technical problems. 1.00 is taught in an active learning style. Lecture segments alternating with laboratory exercises are used in every class to allow students to put concepts into practice immediately; this teaching style generates questions and feedback, and allows the teaching staff and students to interact when concepts are first introduced to ensure that core ideas are understood. Like many MIT classes, 1.00 has weekly assignments, which are programs based on actual engineering, science or management applications. The weekly assignments build on the class material from the previous week, and require students to put the concepts taught in the small in-class labs into a larger program that uses multiple elements of Java together.

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Commentary | David Trone: 90 days in solitary won’t solve…

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Commentary | David Trone: 90 days in solitary won’t solve America’s criminal justice problem | GUEST COMMENTARY

“In solitary, they are trying to break you without realizing that most people in [prison] are there because someone else already tried to break them.” That’s how Paul Torres, a returning citizen who spent 90 days in solitary at the Metropolitan Detention Center in Los Angeles described his experience in solitary confinement to me last August.

Paul was sent into solitary for not having his shirt on in his cell. It was a rule that wasn’t clear or fully enforced, and for that, he paid the price. Stuck in a tiny cell — about six paces across — with a concrete bed, window facing a blank wall, plastic mattress, one blanket, a towel, a small travel toothbrush, and no clock in sight, he said he felt stripped of his humanity. 

As one of the cruelest forms of punishment, solitary confinement causes lasting mental and physical harm, yet the United States continues to use it as standard practice in its prisons. Last year, a report released by nonprofit watchdog Solitary Watch and the Unlock the Box Campaign to end solitary confinement indicated that 122,840 people were placed in solitary, also known as restrictive housing, in our prisons and jails. That’s about 6% of the total U.S. prison and jail population whose mental health and well-being were jeopardized. 

To make matters even worse, recent reports show that while Black individuals only make up 38% of the Bureau of Prisons’ (BOP) population, they account for a disproportionate percentage — around 59%  — of those placed in solitary, underscoring the systemic racism plaguing our criminal justice system. 

As someone with over 25 years of experience working with the American Civil Liberties Union and the co-founder and co-chair of the Second Chance Task Force , I’m working in Congress to reform our criminal justice system. Bringing the inhumane conditions of solitary to light is a core component of that reform, which is why I’m proud to partner with Paul to drive real change. 

For Paul, this is personal. When talking about his time in solitary, he says that it changed him. ” The isolation stripped me of my ability to interact appropriately with other people. When I was put back into the regular population, I felt threatened, overstimulated, and overwhelmed,” he said. “I hear keys, and I feel anxiety.”

I’ve always believed that our criminal justice system should be focused on rehabilitation, but what Paul and so many others like him have experienced is the polar opposite of rehabilitation: it’s torture. 

Those placed in solitary are 15%  more likely to be convicted of a future crime and often exhibit more violent behavior after release. One study even showed that people who had spent any time in solitary confinement were 24%  more likely to die in the first year after their release from prison compared to those who had been incarcerated but not placed in solitary confinement. 

We’ve seen study after study confirm that our prison system is long overdue for serious reform and come to the same conclusion: solitary confinement must be restricted. In fact, since 2014, two reports by the nonpartisan Government Accountability Office provided the BOP with 87 recommendations for how to reform solitary confinement, yet they’ve failed to implement 54 of them still to this day. That’s why I’m joining Paul in calling on the BOP to implement all 87 recommendations to ensure humane treatment in America’s prisons because we can and must do better.

After release, Paul became an Anti-Recidivism Coalition member and was given a real second chance. He says, “I was treated like the person I could be instead of like an animal. I was given resources to help me reenter. I felt valued and invested in. That feeling was everything. It changed the course of my life.” We need to start treating folks in prison that same way, and limiting the use of solitary confinement is an integral part of getting there. I hope you’ll join us this Second Chance Month as we work to reform solitary confinement and build a criminal justice system that is actually “just.” 

U.S. Rep. David Trone (trone.house.gov/contact/) is the co-founder and co-chair of the Second Chance Task Force  and the  Bipartisan Mental Health and Substance Use Disorder Task Force.

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Computer Science > Computer Vision and Pattern Recognition

Title: training-free linear image inverses via flows.

Abstract: Solving inverse problems without any training involves using a pretrained generative model and making appropriate modifications to the generation process to avoid finetuning of the generative model. While recent methods have explored the use of diffusion models, they still require the manual tuning of many hyperparameters for different inverse problems. In this work, we propose a training-free method for solving linear inverse problems by using pretrained flow models, leveraging the simplicity and efficiency of Flow Matching models, using theoretically-justified weighting schemes, and thereby significantly reducing the amount of manual tuning. In particular, we draw inspiration from two main sources: adopting prior gradient correction methods to the flow regime, and a solver scheme based on conditional Optimal Transport paths. As pretrained diffusion models are widely accessible, we also show how to practically adapt diffusion models for our method. Empirically, our approach requires no problem-specific tuning across an extensive suite of noisy linear inverse problems on high-dimensional datasets, ImageNet-64/128 and AFHQ-256, and we observe that our flow-based method for solving inverse problems improves upon closely-related diffusion-based methods in most settings.

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Solving women’s health issues through engineering focus of course

Women’s health through the lifespan has been getting a new focus in recent years from the local to the federal level, with President Joe Biden recently launching initiatives to boost federally funded research in this long-overlooked area. That focus is also active in the McKelvey School of Engineering at Washington University in St. Louis, where a new Department of Biomedical Engineering elective course is filled with students interested in how they can use engineering to solve problems in women’s health. 

The “Engineering for Women’s Health” course (BME 4780/5780), available to undergraduate and graduate students, is taught by Michelle Oyen, an associate professor of biomedical engineering and director of the Center for Women’s Health Engineering. Initially, Oyen offered 16 seats in the course, but demand quickly led her to increase the number of students to 48. 

The course includes lectures on women’s reproductive anatomy and physiology, guest speakers from FemTech (technology focused on women’s health) companies and startups, research scientists from the School of Medicine, case studies and panel discussions. 

“When we pull in the engineers from these startup companies, students can talk to the people who work in real jobs in the field,” Oyen said. “The FemTech sector has a lot of startups in women’s health, which is a trillion-dollar market. There are a lot of opportunities for young engineers to work in this space.” 

One class meeting recently featured Christine O’Brien, an assistant professor of biomedical engineering and co-founder and chief scientific officer of Armor Medical Inc., and Kelsey Mayo, co-founder and CEO of Armor Medical. O’Brien and Mayo presented students with their investor pitch and details about how the company got started. They also shared information about their device, a wrist-worn early monitoring system for obstetric hemorrhage, severe blood loss that occurs in 5% of all births and is 90% preventable if detected and treated early. 

Mayo, who survived hemorrhage, told students the company is her “passion project” and encouraged them to find their own. 

“Your day job should be where your passion meets the world’s needs,” she said. “That’s a cool thing for biomedical engineers to consider.” 

Oyen said focusing on the FemTech sector is important for students’ education and gives them an idea of what they can do after they graduate. 

“Part of what has happened over the past decade is shining a light on the fact that there are all these issues that were once only whispered about by women and not brought out into the open,” Oyen said. “People have now realized this is really important because there are unmet needs in health care, which has huge implications. The two most common surgeries for women are a C-section and a hysterectomy. Half of women have a hysterectomy by age 65, which is huge.” 

In the course, students are completing group projects to address needs in women’s health, including menopause, osteoporosis, hip fractures and lack of muscle mass. 

Annika Avula, a dual-degree student earning a bachelor’s and a master’s in biomedical engineering, said she wanted to take this course after taking O’Brien’s course “Quantitative Physiology II” (BME 301B) last fall.

“It really opened so many people’s eyes to things that can be done for women’s health,” Avula said. “This course has introduced me to a new way of thinking. I’m more interested in problems and more called to ask questions.” 

Ella Hanson, a senior earning a bachelor’s and a master’s in biomedical engineering, said she took the course because of her interest in women’s health problems. 

“Dr. Oyen and Dr. O’Brien are addressing the huge gap in BME in the women’s health field,” Hanson said. “We are learning that there are more opportunities out there in engineering beyond orthopedics and other biomedical problems.”

Avula’s group project is looking at the effects of menopause, and Hanson’s group project is looking at endometriosis, a chronic disease that affects 11% of women of reproductive age in the U.S. and for which there is no cure. 

Oyen said that while it took her some time to get comfortable using words about women’s health in front of audiences, her students are open to talking about women’s health issues and using the vocabulary, which she finds encouraging for future generations. 

“I realized that I’m not going to be the one to change the world, but I’m going to teach the ones who will,” Oyen said. 

Originally published on the McKelvey School of Engineering website

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