Systems of Linear Equations: Two Variables

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.

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.

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 $\left(x,y\right).$ 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 2 compares graphical representations of each type of system.

Figure 2

Example 1

Determining Whether an Ordered Pair Is a Solution to a System of Equations

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

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

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Solution

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

$$\begin{array}{ll}(5)+3(1)=8\hfill & \hfill \\ 8=8\hfill & \text{True}\hfill \\ 2(5)-9=(1)\hfill & \hfill \\ \text{1=1}\hfill & \text{True}\hfill \end{array}$$

The ordered pair $\left(5,1\right)$ 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.

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 2

Solving a System of Equations in Two Variables by Graphing

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

$$\begin{array}{c}2x+y=\mathrm{-8}\\ x-y=\mathrm{-1}\end{array}$$

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Solution

Solve the first equation for $y.$

$$\begin{array}{c}2x+y=\mathrm{-8}\\ y=\mathrm{-2}x\mathrm{-8}\end{array}$$

Solve the second equation for $y.$

$$\begin{array}{c}x-y=\mathrm{-1}\\ y=x+1\end{array}$$

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

Figure 4

The lines appear to intersect at the point $\left(\mathrm{-3,}\mathrm{-2}\right).$ We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.

$$\begin{array}{ll}2(\mathrm{-3})+(\mathrm{-2})=\mathrm{-8}\hfill & \hfill \\ \mathrm{-8}=\mathrm{-8}\hfill & \text{True}\hfill \\ (\mathrm{-3})-(\mathrm{-2})=\mathrm{-1}\hfill & \hfill \\ \mathrm{-1}=\mathrm{-1}\hfill & \text{True}\hfill \end{array}$$

The solution to the system is the ordered pair $\left(\mathrm{-3,}\mathrm{-2}\right),$ 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.

Solving Systems of Equations in Two Variables by the Addition Method

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

Example 4

Solving a System by the Addition Method

Solve the given system of equations by addition.

$$\begin{array}{l}x+2y=\mathrm{-1}\hfill \\ -x+y=3\hfill \end{array}$$

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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.

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

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

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

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

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

Check the solution in the first equation.

$$\begin{array}{llll}x+2y=\mathrm{-1}\hfill & \hfill & \hfill & \hfill \\ \left(-\frac{7}{3}\right)+2\left(\frac{2}{3}\right)=\hfill & \hfill & \hfill & \hfill \\ -\frac{7}{3}+\frac{4}{3}=\hfill & \hfill & \hfill & \hfill \\ -\frac{3}{3}=\hfill & \hfill & \hfill & \hfill \\ \mathrm{-1}=\mathrm{-1}\hfill & \hfill & \hfill & \text{True}\hfill \end{array}$$

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.

Figure 5

Example 5

Using the Addition Method When Multiplication of One Equation Is Required

Solve the given system of equations by the addition method.

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

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Solution

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 $\mathrm{-3},$ the x-terms will add to zero.

$$\begin{array}{llll}x\mathrm{-2}y=11\hfill & \hfill & \hfill & \hfill \\ \mathrm{-3}(x\mathrm{-2}y)=\mathrm{-3}(11)\hfill & \hfill & \hfill & \text{Multiply both sides by }\mathrm{-3.}\hfill \\ \mathrm{-3}x+6y=\mathrm{-33}\hfill & \hfill & \hfill & \text{Use the distributive property}.\hfill \end{array}$$

Now, let’s add them.

$$\begin{array}{l}\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}{\begin{array}{l}\hfill \\ \begin{array}{l}3x+5y=\mathrm{-11}\hfill \\ \mathrm{-3}x+6y=\mathrm{-33}\hfill \end{array}\hfill \end{array}}\hfill \\ 11y=\mathrm{-44}\hfill \\ y=\mathrm{-4}\hfill \end{array}$$

For the last step, we substitute $y=\mathrm{-4}$ into one of the original equations and solve for $x.$

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

Our solution is the ordered pair $\left(3,\mathrm{-4}\right).$ See Figure 6. Check the solution in the original second equation.

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

Figure 6

Example 6

Using the Addition Method When Multiplication of Both Equations Is Required

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

$$\begin{array}{c}2x+3y=\mathrm{-16}\\ 5x\mathrm{-10}y=30\end{array}$$

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Solution

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 $\mathrm{-5}$ and the second equation by $2.$

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

Then, we add the two equations together.

$$\begin{array}{l}\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}{\begin{array}{l}\hfill \\ \begin{array}{l}\mathrm{-10}x\mathrm{-15}y=80\hfill \\ 10x\mathrm{-20}y=60\hfill \end{array}\hfill \end{array}}\hfill \\ \mathrm{-35}y=140\hfill \\ y=\mathrm{-4}\hfill \end{array}$$

Substitute $y=\mathrm{-4}$ into the original first equation.

$$\begin{array}{c}2x+3(\mathrm{-4})=\mathrm{-16}\\ 2x-12=\mathrm{-16}\\ 2x=\mathrm{-4}\\ x=\mathrm{-2}\end{array}$$

The solution is $\left(\mathrm{-2},\mathrm{-4}\right).$ Check it in the other equation.

$$\begin{array}{r}\hfill 5x\mathrm{-10}y=30\\ \hfill 5(\mathrm{-2})\mathrm{-10}(\mathrm{-4})=30\\ \hfill \mathrm{-10}+40=30\\ \hfill 30=30\end{array}$$

See Figure 7.

Figure 7

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 8

Solving an Inconsistent System of Equations

Solve the following system of equations.

$$\begin{array}{l}x=9\mathrm{-2}y\hfill \\ x+2y=13\hfill \end{array}$$

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Solution

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

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

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

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

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

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

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

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

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

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.

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 9

Finding a Solution to a Dependent System of Linear Equations

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

$$\begin{array}{c}x+3y=2\\ 3x+9y=6\end{array}$$

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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 $\mathrm{-3},$ then we will be able to eliminate the $x$ -variable.

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

Now add the equations.

$$\begin{array}{l}\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_}{\begin{array}{ll}-3x-9y\hfill & =\mathrm{-6}\hfill \\ +3x+9y\hfill & =6\hfill \end{array}}\hfill \\ \begin{array}{ll}0\hfill & =0\hfill \end{array}\hfill \end{array}$$

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.

$$\begin{array}{l}x+3y=2\hfill \\ 3y=-x+2\hfill \\ y=-\frac{1}{3}x+\frac{2}{3}\hfill \\ 3x+9y=6\hfill \\ 9y=\mathrm{-3}x+6\hfill \\ y=-\frac{3}{9}x+\frac{6}{9}\hfill \\ y=-\frac{1}{3}x+\frac{2}{3}\hfill \end{array}$$

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

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 $R=xp,$ 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.

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 10

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

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

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Solution

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

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

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

$$\begin{array}{c}0.85x+\mathrm{35,000}=1.55x\\ \mathrm{35,000}=0.7x\\ \mathrm{50,000}=x\end{array}$$

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

$$1.55\left(\mathrm{50,000}\right)=\mathrm{77,500}$$

The break-even point is $\left(\mathrm{50,000},\mathrm{77,500}\right).$

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

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

The profit function is $P(x)=0.7x\mathrm{-35,000.}$

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.

Figure 11

We see from the graph in Figure 12 that the profit function has a negative value until $x=\mathrm{50,000},$ 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.

Figure 12