Systems of Linear Equations: Two Variables

Learning Objectives

Determine whether an ordered pair is a solution of a system of equations (IA 4.1.1)
Solve a system of linear equations by graphing (IA 4.1.2)

Objective: Determine whether an ordered pair is a solution of a system of equations (IA 4.1.1)

A system of linear equations is a group of two or more linear equations. For example,

\{\begin{cases} y = - 2 x + 5 \\ y = 2 x + 7 \end{cases}

is a system of linear equations

A solution to a system of linear equations is an ordered pair x,y that is a solution to every equation in the system.

Example 1

Determine whether the ordered pairs are solutions to the given system.

$\{\begin{array}{l} 2 x - 6 y = 0 \\ 3 x - y = 5 \end{array}$ at $(3, 1)$ and $(- 3, 4)$

Solution

We substitute (3, 1) into both equations:

.
$2 x - 6 y = 0$	$3 x - 4 y = 5$
$\begin{matrix} 2 (3) - 6 (1) = 0 \\ 6 - 6 = 0 \\ 0 = 0 \end{matrix}$ True	$\begin{matrix} 3 (3) - 4 (1) = 5 \\ 9 - 4 = 5 \\ 5 = 5 \end{matrix}$ True
$(3, 1)$ is a solution to $2 x - 6 y = 0$	$(3, 1)$ is a solution to $3 x - 4 y = 5$
Conclusion: since $(3, 1)$ is a solution to both equations, then it is a solution to the system $\{\begin{cases} 2 x - 6 y = 0 \\ 3 x - 4 y = 5 \end{cases}$

Next we substitute (-3, -1) into both equations:

.
$2 x - 6 y = 0$	$3 x - 4 y = 5$
$\begin{matrix} 2 (- 3) - 6 (- 1) = 0 \\ - 6 + 6 = 0 \\ 0 = 0 \end{matrix}$ True	$\begin{matrix} 3 (- 3) - 4 (- 1) = 5 \\ - 9 + 4 = 5 \\ - 5 = 5 \end{matrix}$ False
$(- 3, - 1)$ is a solution to $2 x - 6 y = 0$	$(- 3, - 1)$ is not a solution to $3 x - 4 y = 5$
Conclusion: Since $(- 3, - 1)$ is not a solution to one of the equations, then it is not a solution to the system $\{\begin{cases} 2 x - 6 y = 0 \\ 3 x - 4 y = 5 \end{cases}$

Practice Makes Perfect

Determine whether the ordered pairs are solutions to the given system.

$\{\begin{cases} 3 x + y = 0 \\ x + 2 y = - 5 \end{cases}$ at $(0, 0)$ and $(1, - 3)$

At (0, 0):

.
$3 x + y = 0$	$x + 2 y = –5$
________________________ ________________________	________________________ ________________________
________________________ ________________________	________________________ ________________________
Conclusion: ________________________

At (1, –3)

.
$3 x + y = 0$	$x + 2 y = –5$
________________________ ________________________	________________________ ________________________
________________________ ________________________	________________________ ________________________
Conclusion: ________________________

Solve a system of linear equations by graphing (IA 4.1.2)

Solve a system of linear equations by graphing

Graph the first equation.
Graph the second equation on the same rectangular coordinate system.
Determine whether the lines intersect, are parallel, or are the same line.
Identify the solution to the system.
Check the solution in both equations.

.
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 solutions.
If the lines are the same, the system has an infinite number of solutions.

Example 2

Solve the system by graphing.

$\{\begin{cases} - x + y = 1 \\ 2 x + y = 10 \end{cases}$

Solution

.
Step 1	Graph $- x + y = 1$ We can use the slope intercept – form: $y = x + 1$ Slope = 1 y-intercept: (0, 1)
Step 2	Graph $2 x + y = 10$ We can use the slope intercept – form: $y = - 2 x + 10$ Slope = –2 y-intercept: (0, 10)
Step 3	The lines intersect
Step 4	The solution is the point (3, 4)
Step 5	Let’s check the solution: $\begin{matrix} - x + y = 1 & 2 x + y = 10 \\ - 3 + 4 = 1 & 2 (3) + 4 = 10 \\ 1 = 1 & 10 = 10 \end{matrix}$ Since (3, 4) is a solution to both equations, then it is a solution to the system $- x + y = 1$ $2 x + y = 10$

Solve a system of linear equations by graphing $\{\begin{cases} y = - 2 x + 4 \\ 4 x + 2 y = 6 \end{cases}$

.
Step 1	Graph $y = - 2 x + 4$ Slope = ________ y-intercept = ________
Step 2	Graph $4 x + 2 y = 6$ Slope = ________ y-intercept = ________
Step 3	Do the lines intersect? ________
Step 4	Read from the graph the point of intersection. ________
Step 5	Check the solution in both equations. ________

Practice Makes Perfect

Determine whether the ordered pair is a solution to the given system $\{\begin{cases} x - 3 y = - 8 \\ - 3 x - y = 4 \end{cases}$

Solve the following system by graphing. $\{\begin{cases} y = - \frac{1}{4} x + 2 \\ x + 4 y = 8 \end{cases}$
An empty Cartesian coordinate plane with a grid. The x-axis is labeled from -5 to 5, and the y-axis is labeled from -2 to 6.

A skateboarder catches air in a concrete skatepark with the ocean and a sunset horizon in the background. Other people are seen relaxing on the beach and in the skatepark. — Figure 1 (credit: Thomas Sørenes)

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{matrix} 2 x + y = 15 \\ 3 x - y = 5 \end{matrix}

The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.

\begin{array}{l} 2 (4) + (7) = 15 True \\ 3 (4) - (7) = 5 True \end{array}

In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.

Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different y-intercepts. There are no points common to both lines; hence, there is no solution to the system.

Example 3

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{array}{l} x + 3 y = 8 \\ 2 x - 9 = y \end{array}

Solution

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

\begin{array}{l} (5) + 3 (1) = 8 \\ 8 = 8 & True \\ 2 (5) - 9 = (1) \\ 1=1 & True \end{array}

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

Analysis

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 3.

A graph displays two intersecting lines representing the equations x + 3y = 8 (red) and 2x - 9 = y (blue), with their intersection point clearly marked at (5, 1). — Figure 3

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 4

Solving a System of Equations in Two Variables by Graphing

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

\begin{matrix} 2 x + y = −8 \\ x - y = −1 \end{matrix}

Solution

Solve the first equation for $y .$

\begin{matrix} 2 x + y = −8 \\ y = −2 x −8 \end{matrix}

Solve the second equation for $y .$

\begin{matrix} x - y = −1 \\ y = x + 1 \end{matrix}

Graph both equations on the same set of axes as in Figure 4.

This image displays a Cartesian coordinate system with grid lines, an x-axis, and a y-axis. Two linear equations are plotted. The first line is red and labeled "y = x + 1". The second line is blue and labeled "y = -2x - 8". The two lines intersect at a single point, which is marked with a black dot and explicitly labeled with the coordinates "(-3, -2)". — Figure 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{array}{l} 2 (−3) + (−2) = −8 \\ −8 = −8 & True \\ (−3) - (−2) = −1 \\ −1 = −1 & True \end{array}

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

Solving Systems of Equations by Substitution

Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.

Example 5

Solving a System of Equations in Two Variables by Substitution

Solve the following system of equations by substitution.

\begin{array}{l} - x + y = −5 \\ 2 x - 5 y = 1 \end{array}

Solution

First, we will solve the first equation for $y .$

\begin{array}{l} - x + y = −5 \\ y = x −5 \end{array}

Now we can substitute the expression $x −5$ for $y$ in the second equation.

\begin{array}{l} 2 x - 5 y = 1 \\ 2 x - 5 (x - 5) = 1 \\ 2 x - 5 x + 25 = 1 \\ - 3 x = −24 \\ x = 8 \end{array}

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

\begin{array}{l} - (8) + y = −5 \\ y = 3 \end{array}

Our solution is $(8, 3) .$

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

\begin{array}{l} - x + y = - 5 \\ - (8) + (3) = - 5 & True \\ 2 x - 5 y = 1 \\ 2 (8) - 5 (3) = 1 & True \end{array}

Solving Systems of Equations in Two Variables by the Addition Method

A third method of solving systems of linear equations is the addition method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.

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 6

Solving a System by the Addition Method

Solve the given system of equations by addition.

\begin{array}{l} x + 2 y = −1 \\ - x + y = 3 \end{array}

Solution

Both equations are already set equal to a constant. Notice that the coefficient of $x$ in the second equation, –1, is the opposite of the coefficient of $x$ in the first equation, 1. We can add the two equations to eliminate $x$ without needing to multiply by a constant.

\frac{\begin{array}{l} x + 2 y = - 1 \\ - x + y = 3 \end{array}}{3 y = 2}

Now that we have eliminated $x,$ we can solve the resulting equation for $y .$

\begin{array}{l} 3 y = 2 \\ y = \frac{2}{3} \end{array}

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

\begin{array}{l} - x + y = 3 \\ - x + \frac{2}{3} = 3 \\ - x = 3 - \frac{2}{3} \\ - x = \frac{7}{3} \\ x = - \frac{7}{3} \end{array}

The solution to this system is $(- \frac{7}{3}, \frac{2}{3}) .$

Check the solution in the first equation.

\begin{array}{l} x + 2 y = −1 \\ (- \frac{7}{3}) + 2 (\frac{2}{3}) = \\ - \frac{7}{3} + \frac{4}{3} = \\ - \frac{3}{3} = \\ −1 = −1 & True \end{array}

Analysis

We gain an important perspective on systems of equations by looking at the graphical representation. See Figure 5 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.

This graph displays a system of two linear equations plotted on a coordinate plane. The blue line represents the equation x + 2y = -1, while the red line represents the equation -x + y = 3. The point where these two lines intersect, indicated by a black dot, is the solution to the system of equations and is labeled as (-7/3, 2/3). The x-axis ranges from -6 to 6, and the y-axis ranges from -5 to 5, with a grid background to aid in reading the coordinates. — Figure 5

Example 7

Using the Addition Method When Multiplication of One Equation Is Required

Solve the given system of equations by the addition method.

\begin{array}{l} 3 x + 5 y = −11 \\ x - 2 y = 11 \end{array}

Solution

Adding these equations as presented will not eliminate a variable. However, we see that the first equation has $3 x$ in it and the second equation has $x .$ So if we multiply the second equation by $−3,$ the x-terms will add to zero.

\begin{array}{l} x −2 y = 11 \\ −3 (x −2 y) = −3 (11) & Multiply both sides by −3. \\ −3 x + 6 y = −33 & Use the distributive property . \end{array}

Now, let’s add them.

\begin{array}{l} \underset{_______________}{\begin{array}{l} \begin{array}{l} 3 x + 5 y = −11 \\ −3 x + 6 y = −33 \end{array} \end{array}} \\ 11 y = −44 \\ y = −4 \end{array}

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

\begin{matrix} 3 x + 5 y = - 11 \\ 3 x + 5 (- 4) = - 11 \\ 3 x - 20 = - 11 \\ 3 x = 9 \\ x = 3 \end{matrix}

Our solution is the ordered pair $(3, −4) .$ See Figure 6. Check the solution in the original second equation.

\begin{array}{l} x - 2 y = 11 \\ (3) - 2 (- 4) = 3 + 8 \\ 11 = 11 & True \end{array}

A graph shows two linear equations, 3x + 5y = -11 (red) and x - 2y = 11 (blue), intersecting at the point (3, -4) on a Cartesian coordinate system. — Figure 6

Example 8

Using the Addition Method When Multiplication of Both Equations Is Required

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

\begin{matrix} 2 x + 3 y = −16 \\ 5 x −10 y = 30 \end{matrix}

Solution

One equation has $2 x$ and the other has $5 x .$ The least common multiple is $10 x$ so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate $x$ by multiplying the first equation by $−5$ and the second equation by $2.$

\begin{array}{l} - 5 (2 x + 3 y) = - 5 (−16) \\ - 10 x - 15 y = 80 \\ 2 (5 x - 10 y) = 2 (30) \\ 10 x - 20 y = 60 \end{array}

Then, we add the two equations together.

\begin{array}{l} \underset{________________}{\begin{array}{l} \begin{array}{l} −10 x −15 y = 80 \\ 10 x −20 y = 60 \end{array} \end{array}} \\ −35 y = 140 \\ y = −4 \end{array}

Substitute $y = −4$ into the original first equation.

\begin{matrix} 2 x + 3 (−4) = −16 \\ 2 x - 12 = −16 \\ 2 x = −4 \\ x = −2 \end{matrix}

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

\begin{array}{r} 5 x −10 y = 30 \\ 5 (−2) −10 (−4) = 30 \\ −10 + 40 = 30 \\ 30 = 30 \end{array}

See Figure 7.

A graph displays two intersecting lines: a red line representing 2x + 3y = -16 and a blue line representing 5x - 10y = 30. Their intersection point is labeled as (-2, -4). — Figure 7

Example 9

Using the Addition Method in Systems of Equations Containing Fractions

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

\begin{array}{l} \frac{x}{3} + \frac{y}{6} = 3 \\ \frac{x}{2} - \frac{y}{4} = ​ 1 \end{array}

Solution

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

\begin{array}{l} 6 (\frac{x}{3} + \frac{y}{6}) = 6 (3) \\ 2 x + y = 18 \\ 4 (\frac{x}{2} - \frac{y}{4}) = 4 (1) \\ 2 x - y = 4 \end{array}

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

\begin{array}{l} −1 (2 x - y) = −1 (4) \\ −2 x + y = −4 \end{array}

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

\begin{array}{l} 2 x + y = 18 \\ \underset{_____________}{−2 x + y = −4} \\ 2 y = 14 \\ y = 7 \end{array}

Substitute $y = 7$ into the first equation.

\begin{array}{l} 2 x + (7) = 18 \\ 2 x = 11 \\ x = \frac{11}{2} \\ = 5.5 \end{array}

The solution is $(\frac{11}{2}, 7) .$ Check it in the other equation.

\begin{matrix} \frac{x}{2} - \frac{y}{4} = 1 \\ \frac{\frac{11}{2}}{2} - \frac{7}{4} = 1 \\ \frac{11}{4} - \frac{7}{4} = 1 \\ \frac{4}{4} = 1 \end{matrix}

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 10

Solving an Inconsistent System of Equations

Solve the following system of equations.

\begin{array}{l} x = 9 −2 y \\ x + 2 y = 13 \end{array}

Solution

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

\begin{array}{l} x + 2 y = 13 \\ (9 - 2 y) + 2 y = 13 \\ 9 + 0 y = 13 \\ 9 = 13 \end{array}

Clearly, this statement is a contradiction because $9 \neq 13.$ Therefore, the system has no solution.

The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.

\begin{array}{l} x = 9 −2 y \\ 2 y = - x + 9 \\ y = - \frac{1}{2} x + \frac{9}{2} \end{array}

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

\begin{array}{l} x + 2 y = 13 \\ 2 y = - x + 13 \\ y = - \frac{1}{2} x + \frac{13}{2} \end{array}

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

\begin{array}{l} \begin{array}{l} y = - \frac{1}{2} x + \frac{9}{2} \end{array} \\ y = - \frac{1}{2} x + \frac{13}{2} \end{array}

Analysis

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 8.

A Cartesian coordinate system shows two distinct parallel lines. The red line, labeled with the equation y = -1/2x + 9/2, has a negative slope and a y-intercept of 4.5. The blue line, labeled with the equation y = -1/2x + 13/2, also has a negative slope of -1/2 and a y-intercept of 6.5. Both lines extend across the grid, with the x-axis ranging from -12 to 12 and the y-axis ranging from -12 to 12, marked with increments of 2. — Figure 8

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 11

Finding a Solution to a Dependent System of Linear Equations

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

\begin{matrix} x + 3 y = 2 \\ 3 x + 9 y = 6 \end{matrix}

Solution

With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating $x .$ If we multiply both sides of the first equation by $−3,$ then we will be able to eliminate the $x$ -variable.

\begin{array}{l} x + 3 y = 2 \\ (−3) (x + 3 y) = (−3) (2) \\ −3 x - 9 y = - 6 \end{array}

Now add the equations.

\begin{array}{l} \underset{______________}{\begin{array}{l} - 3 x - 9 y & = −6 \\ + 3 x + 9 y & = 6 \end{array}} \\ \begin{array}{l} 0 & = 0 \end{array} \end{array}

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

Analysis

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{array}{l} x + 3 y = 2 \\ 3 y = - x + 2 \\ y = - \frac{1}{3} x + \frac{2}{3} \\ 3 x + 9 y = 6 \\ 9 y = −3 x + 6 \\ y = - \frac{3}{9} x + \frac{6}{9} \\ y = - \frac{1}{3} x + \frac{2}{3} \end{array}

See Figure 9. Notice the results are the same. The general solution to the system is $(x, - \frac{1}{3} x + \frac{2}{3}) .$

A coordinate plane displays the graphs of two linear equations. The first equation, x + 3y = 2, is represented by a red line. The second equation, 3x + 9y = 6, is represented by a blue line. Both lines are coincident, meaning they overlap perfectly, indicating that the two equations are equivalent and have infinitely many solutions. The x-axis is labeled from -5 to 5, and the y-axis is labeled from -5 to 5. The lines pass through points such as (2, 0) and (-4, 2). — Figure 9

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 $R = x p,$ where $x =$ quantity and $p =$ price. The revenue function is shown in orange in Figure 10.

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 10. The $x$ -axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars.

A break-even graph plots Money (in hundreds of dollars) vs. Quantity (in hundreds of units), showing Cost, Revenue, and the break-even point at (7, 33), with profit/loss regions. — Figure 10

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 12

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

Given the cost function $C (x) = 0.85 x + 35,000$ and the revenue function $R (x) = 1.55 x,$ find the break-even point and the profit function.

Solution

Write the system of equations using $y$ to replace function notation.

\begin{array}{l} \begin{array}{l} y = 0.85 x + 35,000 \end{array} \\ y = 1.55 x \end{array}

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

\begin{matrix} 0.85 x + 35,000 = 1.55 x \\ 35,000 = 0.7 x \\ 50,000 = x \end{matrix}

Then, we substitute $x = 50,000$ into either the cost function or the revenue function.

1.55 (50,000) = 77,500

The break-even point is $(50,000, 77,500) .$

The profit function is found using the formula $P (x) = R (x) - C (x) .$

\begin{array}{l} P (x) = 1.55 x - (0.85 x + 35, 000) \\ = 0.7 x - 35, 000 \end{array}

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

Analysis

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 11.

A line graph plots 'Dollars' on the y-axis against 'Quantity' on the x-axis. The blue line represents the Revenue function, R(x) = 1.55x, starting from the origin. The red line represents the Cost function, C(x) = 0.85x + 35,000, starting from a y-intercept of 35,000. The two lines intersect at a point labeled 'Break-even point' with coordinates (50,000, 77,500). The shaded area where the revenue line is above the cost line, to the right of the break-even point, is labeled 'Profit'. The x-axis ranges from 0 to 100,000, and the y-axis ranges from 0 to 100,000. — Figure 11

We see from the graph in Figure 12 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.

A line graph showing profit P(x) = 0.7x - 35,000. The x-axis is quantity, y-axis is dollars profit. The break-even point is at (50,000, 0), where profit is zero. — Figure 12

Example 13

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?

Solution

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

25 c + 50 a = 70,000

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

\begin{matrix} c + a = 2,000 \\ 25 c + 50 a = 70,000 \end{matrix}

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{matrix} c + a = 2,000 \\ a = 2,000 - c \end{matrix}

Substitute the expression $2,000 - c$ in the second equation for $a$ and solve for $c .$

\begin{array}{l} 25 c + 50 (2,000 - c) = 70,000 \\ 25 c + 100,000 - 50 c = 70,000 \\ - 25 c = −30,000 \\ c = 1,200 \end{array}

Substitute $c = 1,200$ into the first equation to solve for $a .$

\begin{array}{l} 1,200 + a = 2,000 \\ a = 800 \end{array}

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

Key Concepts

A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.
The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See Example 3.
Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See Example 4.
Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See Example 5.
A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See Example 6.
It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See Example 7, Example 8, and Example 9.
Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See Example 10.
The solution to a system of dependent equations will always be true because both equations describe the same line. See Example 11.
Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See Example 12 and Example 13.

Section Exercises

Verbal

Exercise 1

Can a system of linear equations have exactly two solutions? Explain why or why not.

Solution

No, you can either have zero, one, or infinitely many. Examine graphs.

Exercise 2

If you are performing a break-even analysis for a business and their cost and revenue equations are dependent, explain what this means for the company’s profit margins.

Exercise 3

If you are solving a break-even analysis and get a negative break-even point, explain what this signifies for the company?

Solution

This means there is no realistic break-even point. By the time the company produces one unit they are already making profit.

Exercise 4

If you are solving a break-even analysis and there is no break-even point, explain what this means for the company. How should they ensure there is a break-even point?

Exercise 5

Given a system of equations, explain at least two different methods of solving that system.

Solution

You can solve by substitution (isolating $x$ or $y$ ), graphically, or by addition.

Algebraic

For the following exercises, determine whether the given ordered pair is a solution to the system of equations.

Exercise 6

$\begin{matrix} 5 x - y = 4 \\ x + 6 y = 2 \end{matrix}$ and $(4, 0)$

Exercise 7

$\begin{array}{l} −3 x - 5 y = 13 \\ - x + 4 y = 10 \end{array}$ and $(−6, 1)$

Solution

Yes

Exercise 8

$\begin{matrix} 3 x + 7 y = 1 \\ 2 x + 4 y = 0 \end{matrix}$ and $(2, 3)$

Exercise 9

$\begin{array}{l} −2 x + 5 y = 7 \\ 2 x + 9 y = 7 \end{array}$ and $(−1, 1)$

Solution

Yes

Exercise 10

$\begin{matrix} x + 8 y = 43 \\ 3 x −2 y = −1 \end{matrix}$ and $(3, 5)$

For the following exercises, solve each system by substitution.

Exercise 11

$\begin{array}{l} x + 3 y = 5 \\ 2 x + 3 y = 4 \end{array}$

Solution

$(−1, 2)$

Exercise 12

$\begin{array}{l} 3 x −2 y = 18 \\ 5 x + 10 y = −10 \end{array}$

Exercise 13

$\begin{array}{l} 4 x + 2 y = −10 \\ 3 x + 9 y = 0 \end{array}$

Solution

$(−3, 1)$

Exercise 14

$\begin{array}{l} 2 x + 4 y = −3.8 \\ 9 x −5 y = 1.3 \end{array}$

Exercise 15

$\begin{array}{l} - 2 x + 3 y = 1.2 \\ - 3 x - 6 y = 1.8 \end{array}$

Solution

$(- \frac{3}{5}, 0)$

Exercise 16

$\begin{array}{l} x −0.2 y = 1 \\ −10 x + 2 y = 5 \end{array}$

Exercise 17

$\begin{array}{l} 3 x + 5 y = 9 \\ 30 x + 50 y = −90 \end{array}$

Solution

No solutions exist.

Exercise 18

$\begin{array}{l} −3 x + y = 2 \\ 12 x −4 y = −8 \end{array}$

Exercise 19

$\begin{array}{l} \frac{1}{2} x + \frac{1}{3} y = 16 \\ \frac{1}{6} x + \frac{1}{4} y = 9 \end{array}$

Solution

$(\frac{72}{5}, \frac{132}{5})$

Exercise 20

$\begin{array}{l} - \frac{1}{4} x + \frac{3}{2} y = 11 \\ - \frac{1}{8} x + \frac{1}{3} y = 3 \end{array}$

For the following exercises, solve each system by addition.

Exercise 21

$\begin{array}{l} −2 x + 5 y = −42 \\ 7 x + 2 y = 30 \end{array}$

Solution

$(6, −6)$

Exercise 22

$\begin{array}{l} 6 x −5 y = −34 \\ 2 x + 6 y = 4 \end{array}$

Exercise 23

$\begin{array}{l} 5 x - y = −2.6 \\ −4 x −6 y = 1.4 \end{array}$

Solution

$(- \frac{1}{2}, \frac{1}{10})$

Exercise 24

$\begin{array}{l} 7 x −2 y = 3 \\ 4 x + 5 y = 3.25 \end{array}$

Exercise 25

$\begin{array}{l} −x + 2 y = −1 \\ 5 x −10 y = 6 \end{array}$

Solution

No solutions exist.

Exercise 26

$\begin{array}{l} 7 x + 6 y = 2 \\ −28 x −24 y = −8 \end{array}$

Exercise 27

$\begin{array}{l} \frac{5}{6} x + \frac{1}{4} y = 0 \\ \frac{1}{8} x - \frac{1}{2} y = - \frac{43}{120} \end{array}$

Solution

$(- \frac{1}{5}, \frac{2}{3})$

Exercise 28

$\begin{array}{l} \frac{1}{3} x + \frac{1}{9} y = \frac{2}{9} \\ - \frac{1}{2} x + \frac{4}{5} y = - \frac{1}{3} \end{array}$

Exercise 29

$\begin{array}{l} \begin{array}{l} −0.2 x + 0.4 y = 0.6 \\ x −2 y = −3 \end{array} \end{array}$

Solution

$(x, \frac{x + 3}{2})$

Exercise 30

$\begin{array}{l} \begin{array}{l} −0.1 x + 0.2 y = 0.6 \end{array} \\ 5 x −10 y = 1 \end{array}$

For the following exercises, solve each system by any method.

Exercise 31

$\begin{array}{l} 5 x + 9 y = 16 \\ x + 2 y = 4 \end{array}$

Solution

$(−4, 4)$

Exercise 32

$\begin{array}{l} 6 x −8 y = −0.6 \\ 3 x + 2 y = 0.9 \end{array}$

Exercise 33

$\begin{array}{l} 5 x −2 y = 2.25 \\ 7 x −4 y = 3 \end{array}$

Solution

$(\frac{1}{2}, \frac{1}{8})$

Exercise 34

$\begin{array}{l} x - \frac{5}{12} y = - \frac{55}{12} \\ −6 x + \frac{5}{2} y = \frac{55}{2} \end{array}$

Exercise 35

$\begin{array}{l} 7 x −4 y = \frac{7}{6} \\ 2 x + 4 y = \frac{1}{3} \end{array}$

Solution

$(\frac{1}{6}, 0)$

Exercise 36

$\begin{array}{l} 3 x + 6 y = 11 \\ 2 x + 4 y = 9 \end{array}$

Exercise 37

$\begin{array}{l} \frac{7}{3} x - \frac{1}{6} y = 2 \\ - \frac{21}{6} x + \frac{3}{12} y = −3 \end{array}$

Solution

$(x, 2 (7 x −6))$

Exercise 38

$\begin{array}{l} \frac{1}{2} x + \frac{1}{3} y = \frac{1}{3} \\ \frac{3}{2} x + \frac{1}{4} y = - \frac{1}{8} \end{array}$

Exercise 39

$\begin{array}{l} 2.2 x + 1.3 y = −0.1 \\ 4.2 x + 4.2 y = 2.1 \end{array}$

Solution

$(- \frac{5}{6}, \frac{4}{3})$

Exercise 40

$\begin{array}{l} 0.1 x + 0.2 y = 2 \\ 0.35 x −0.3 y = 0 \end{array}$

Graphical

For the following exercises, graph the system of equations and state whether the system is consistent, inconsistent, or dependent and whether the system has one solution, no solution, or infinite solutions.

Exercise 41

$\begin{array}{l} 3 x - y = 0.6 \\ x −2 y = 1.3 \end{array}$

Solution

Consistent with one solution

Exercise 42

$\begin{array}{l} - x + 2 y = 4 \\ 2 x −4 y = 1 \end{array}$

Exercise 43

$\begin{array}{l} x + 2 y = 7 \\ 2 x + 6 y = 12 \end{array}$

Solution

Consistent with one solution

Exercise 44

$\begin{array}{l} 3 x −5 y = 7 \\ x −2 y = 3 \end{array}$

Exercise 45

$\begin{array}{l} 3 x −2 y = 5 \\ −9 x + 6 y = −15 \end{array}$

Solution

Dependent with infinitely many solutions

Technology

For the following exercises, use the intersect function on a graphing device to solve each system. Round all answers to the nearest hundredth.

Exercise 46

$\begin{array}{l} 0.1 x + 0.2 y = 0.3 \\ −0.3 x + 0.5 y = 1 \end{array}$

Exercise 47

$\begin{array}{l} −0.01 x + 0.12 y = 0.62 \\ 0.15 x + 0.20 y = 0.52 \end{array}$

Solution

$(−3.08, 4.91)$

Exercise 48

$\begin{array}{l} 0.5 x + 0.3 y = 4 \\ 0.25 x −0.9 y = 0.46 \end{array}$

Exercise 49

$\begin{array}{l} 0.15 x + 0.27 y = 0.39 \\ −0.34 x + 0.56 y = 1.8 \end{array}$

Solution

$(−1.52, 2.29)$

Exercise 50

$\begin{array}{l} −0.71 x + 0.92 y = 0.13 \\ 0.83 x + 0.05 y = 2.1 \end{array}$

Extensions

For the following exercises, solve each system in terms of $A, B, C, D, E,$ and $F$ where $A - F$ are nonzero numbers. Note that $A \neq B$ and $A E \neq B D .$

Exercise 51

$\begin{array}{l} x + y = A \\ x - y = B \end{array}$

Solution

$(\frac{A + B}{2}, \frac{A - B}{2})$

Exercise 52

$\begin{array}{l} x + A y = 1 \\ x + B y = 1 \end{array}$

Exercise 53

$\begin{array}{l} A x + y = 0 \\ B x + y = 1 \end{array}$

Solution

$(\frac{−1}{A - B}, \frac{A}{A - B})$

Exercise 54

$\begin{array}{l} A x + B y = C \\ x + y = 1 \end{array}$

Exercise 55

$\begin{array}{l} A x + B y = C \\ D x + E y = F \end{array}$

Solution

$(\frac{C E - B F}{B D - A E}, \frac{A F - C D}{B D - A E})$

Real-World Applications

For the following exercises, solve for the desired quantity.

Exercise 56

A stuffed animal business has a total cost of production $C = 12 x + 30$ and a revenue function $R = 20 x .$ Find the break-even point.

Exercise 57

An Ethiopian restaurant has a cost of production $C (x) = 11 x + 120$ and a revenue function $R (x) = 5 x .$ When does the company start to turn a profit?

Solution

They never turn a profit.

Exercise 58

A cell phone factory has a cost of production $C (x) = 150 x + 10, 000$ and a revenue function $R (x) = 200 x .$ What is the break-even point?

Exercise 59

A musician charges $C (x) = 64 x + 20,000$ where $x$ is the total number of attendees at the concert. The venue charges $80 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point?

Solution

$(1, 250, 100, 000)$

Exercise 60

A guitar factory has a cost of production $C (x) = 75 x + 50,000.$ If the company needs to break even after 150 units sold, at what price should they sell each guitar? Round up to the nearest dollar, and write the revenue function.

For the following exercises, use a system of linear equations with two variables and two equations to solve.

Exercise 61

Find two numbers whose sum is 28 and difference is 13.

Solution

The numbers are 7.5 and 20.5.

Exercise 62

A number is 9 more than another number. Twice the sum of the two numbers is 10. Find the two numbers.

Exercise 63

The startup cost for a restaurant is $120,000, and each meal costs $10 for the restaurant to make. If each meal is then sold for $15, after how many meals does the restaurant break even?

Solution

24,000

Exercise 64

A moving company charges a flat rate of $150, and an additional $5 for each box. If a taxi service would charge $20 for each box, how many boxes would you need for it to be cheaper to use the moving company, and what would be the total cost?

Exercise 65

A total of 1,595 first- and second-year college students gathered at a pep rally. The number of first-years exceeded the number of second-years by 15. How many students from each year group were in attendance?

Solution

790 second-year students, 805 first-year students

Exercise 66

276 students enrolled in an introductory chemistry class. By the end of the semester, 5 times the number of students passed as failed. Find the number of students who passed, and the number of students who failed.

Exercise 67

There were 130 faculty at a conference. If there were 18 more women than men attending, how many of each gender attended the conference?

Solution

56 men, 74 women

Exercise 68

A jeep and a pickup truck enter a highway running east-west at the same exit heading in opposite directions. The jeep entered the highway 30 minutes before the pickup did, and traveled 7 mph slower than the pickup. After 2 hours from the time the pickup entered the highway, the cars were 306.5 miles apart. Find the speed of each car, assuming they were driven on cruise control and retained the same speed.

Exercise 69

If a scientist mixed 10% saline solution with 60% saline solution to get 25 gallons of 40% saline solution, how many gallons of 10% and 60% solutions were mixed?

Solution

10 gallons of 10% solution, 15 gallons of 60% solution

Exercise 70

An investor earned triple the profits of what they earned last year. If they made $500,000.48 total for both years, how much did the investor earn in profits each year?

Exercise 71

An investor invested 1.1 million dollars into two land investments. On the first investment, Swan Peak, her return was a 110% increase on the money she invested. On the second investment, Riverside Community, she earned 50% over what she invested. If she earned $1 million in profits, how much did she invest in each of the land deals?

Solution

Swan Peak: $750,000, Riverside: $350,000

Exercise 72

If an investor invests a total of $25,000 into two bonds, one that pays 3% simple interest, and the other that pays $2 \frac{7}{8} %$ interest, and the investor earns $737.50 annual interest, how much was invested in each account?

Exercise 73

If an investor invests $23,000 into two bonds, one that pays 4% in simple interest, and the other paying 2% simple interest, and the investor earns $710.00 annual interest, how much was invested in each account?

Solution

$12,500 in the first account, $10,500 in the second account.

Exercise 74

Blu-rays cost $5.96 more than regular DVDs at All Bets Are Off Electronics. How much would 6 Blu-rays and 2 DVDs cost if 5 Blu-rays and 2 DVDs cost $127.73?

Exercise 75

A store clerk sold 60 pairs of sneakers. The high-tops sold for $98.99 and the low-tops sold for $129.99. If the receipts for the two types of sales totaled $6,404.40, how many of each type of sneaker were sold?

Solution

High-tops: 45, Low-tops: 15

Exercise 76

A concert manager counted 350 ticket receipts the day after a concert. The price for a student ticket was $12.50, and the price for an adult ticket was $16.00. The register confirms that $5,075 was taken in. How many student tickets and adult tickets were sold?

Exercise 77

Admission into an amusement park for 4 children and 2 adults is $116.90. For 6 children and 3 adults, the admission is $175.35. Assuming a different price for children and adults, what is the price of the child’s ticket and the price of the adult ticket?

Solution

Infinitely many solutions. We need more information.

addition method

an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable

break-even point

the point at which a cost function intersects a revenue function; where profit is zero

consistent system

a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system

cost function

the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable costs

dependent system

a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system

inconsistent system

a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common

independent system

a system of linear equations with exactly one solution pair

(x, y)

profit function

the profit function is written as

P (x) = R (x) - C (x),

revenue minus cost

revenue function

the function that is used to calculate revenue, simply written as

R = x p,

where

x =

quantity and

p =

price

substitution method

an algebraic technique used to solve systems of linear equations in which one of the two equations is solved for one variable and then substituted into the second equation to solve for the second variable

system of linear equations

a set of two or more equations in two or more variables that must be considered simultaneously.