4.1 Solve Systems of Equations by Graphing

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

In Solving Linear Equations and Applications of Linear Functions we learned how to solve linear equations with one variable. Remember that the solution of an equation is a value of the variable that makes a true statement when substituted into the equation.

Now we will work with systems of linear equations, two or more linear equations grouped together.

We will focus our work here on systems of two linear equations in two unknowns. Later, you may solve larger systems of equations.

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, like [latex]2x+y=7[/latex], 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 [latex](x,y)[/latex] that make both equations true. These are called the solutions to a system of equations.

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.

Let’s consider the system below:

Is the ordered pair [latex]\left(2,-1\right)[/latex] a solution?

We substitute [latex]{\color{red}{x}}{\color{red}{\;}}{\color{red}{=}}{\color{red}{\;}}{\color{red}{2}}[/latex] into both equations.

[latex]\begin{eqnarray*}3{\color{red}{x}}\;-\;{\color{blue}{y}}\;&=&\;7\\3({\color{red}{2}})\;-\;({\color{blue}{-}}{\color{blue}{1}})\;&\overset?=&\;7\\7\;&=&\;7\;\checkmark\\\\{\color{red}{x}}\;-\;2{\color{blue}{y}}\;&=&\;4\\{\color{red}{2}}\;-\;2({\color{blue}{-}}{\color{blue}{1}})\;&\overset?=&\;4\\4\;&=&\;4\;\checkmark\end{eqnarray*}[/latex]

The ordered pair [latex](2, −1)[/latex] made both equations true. Therefore [latex](2, −1)[/latex] is a solution to this system.

Let’s try another ordered pair. Is the ordered pair [latex](3, 2)[/latex] a solution?

We substitute [latex]{\color{red}{x}}{\color{red}{\;}}{\color{red}{=}}{\color{red}{\;}}{\color{red}{3}}[/latex] into both equations.

The ordered pair [latex](3, 2)[/latex] made one equation true, but it made the other equation false. Since it is not a solution to both equations, it is not a solution to this system.

Solve a System of Linear Equations by Graphing

In this chapter 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 in Figure 4.1.1:

For the first example of solving a system of linear equations in this section and in the next two sections, we will solve the same system of two linear equations. But we’ll use a different method in each section. After seeing the third method, you’ll decide which method was the most convenient way to solve this system.

Example 4.1.2

Solve the system by graphing: [latex]\{\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}[/latex]

Solution

Step 1: Graph the first equation.

To graph the first line, write the equation in slope-intercept form.

[latex]\begin{align*}{\color{red}{2x+y}}&{\color{red}{\;=7}}\\ y&=-2x+7\\ m&=-2\\ b&=7 \end{align*}[/latex]

Step 2: Graph the second equation on the same rectangular coordinate system.

To graph the second line, use intercepts.

[latex]x-2y=6[/latex]

[latex](0,-3)\;\;(6,0)[/latex]

Step 3: Determine whether the lines intersect, are parallel, or are the same line.

Look at the graph of the lines.

The lines intersect.

Step 4: Identify the solution to the system.

If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. 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.

Since the lines intersect, find the point of intersection.

Check the point in both equations.

The lines intersect at [latex](4, -1)[/latex]

[latex]\begin{align*} 2x+y&=7\\ 2(4)+(-1)&\overset?=7\\ 8-1&\overset?=7\\ 7&=7\checkmark\\[3ex] x-2y&=6\\ 4-2(-1)&\overset?=6\\ 6&=6\checkmark \end{align*}[/latex]

The solution is [latex](4,-1)[/latex].

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

Example 4.1.3

Solve the system by graphing: [latex]\{\begin{array}{c}y=2x+1\hfill \\ y=4x-1\hfill \end{array}[/latex]

Solution

Both of the equations in this system are in slope-intercept form, so we will use their slopes and [latex]y[/latex]-intercepts to graph them. [latex]\{\begin{array}{c}y=2x+1\hfill \\ y=4x-1\hfill \end{array}[/latex]

Step 1: Find the slope and [latex]y[/latex]-intercept of the first equation.

[latex]\begin{align*}y &= 2x + 1\\m &=2\\ &= 1\end{align*}[/latex]

Step 2: Find the slope and [latex]y[/latex]-intercept of the second equation.

[latex]\begin{align*}y &= 4x - 1\\m &= 4\\ &= -1\end{align*}[/latex]

Step 3: Graph the two lines.

Step 4: Determine the point of intersection.

The lines intersect at [latex](1, 3)[/latex].

Step 5: Check the solution in both equations.

[latex]\begin{array}{cccc}\begin{array}{ccc}\hfill y& =\hfill & 2x+1\hfill \\ \hfill 3& \stackrel{?}{=}\hfill & 2\cdot 1+1\hfill \\ \hfill 3& =\hfill & 3✓\hfill \end{array}& & & \begin{array}{ccc}\hfill y& =\hfill & 4x-1\hfill \\ \hfill 3& \stackrel{?}{=}\hfill & 4\cdot 1-1\hfill \\ \hfill 3& =\hfill & 3✓\hfill \end{array}\end{array}[/latex]

Step 6: Write solution as point [latex](x,y)[/latex].

The solution is [latex](1, 3)[/latex].

Both equations in Example 4.1.3 were given in slope–intercept form. This made it easy for us to quickly graph the lines. In the next example, we’ll first re-write the equations into slope–intercept form.

Example 4.1.4

Solve the system by graphing: [latex]\{\begin{array}{c}3x+y=-1\hfill \\ 2x+y=0\hfill \end{array}[/latex]

Solution

We’ll solve both of these equations for [latex]y[/latex] so that we can easily graph them using their slopes and [latex]y[/latex]-intercepts. [latex]\{\begin{array}{c}3x+y=-1\hfill \\ 2x+y=0\hfill \end{array}[/latex]

Step 1: Solve the first equation for [latex]y[/latex].

Step 2: Find the slope and [latex]y[/latex]-intercept.

[latex]\begin{array}{c}\begin{array}{ccc}\hfill 3x+y& =\hfill & -1\hfill \\ \hfill y& =\hfill & -3x-1\hfill \\ \\ \hfill m& =\hfill & -3\hfill \\ \hfill b& =\hfill & -1\hfill \\ \\ \\ \hfill 2x+y& =\hfill & 0\hfill \\ \hfill y& =\hfill & -2x\hfill \\ \\ \hfill m& =\hfill & -2\hfill \\ \hfill b& =\hfill & 0\hfill \\ \hfill \end{array}\end{array}[/latex]

Step 3: Solve the second equation for [latex]y[/latex].

Step 4: Find the slope and [latex]y[/latex]-intercept.

Step 5: Graph the lines.

Step 6: Determine the point of intersection.

The lines intersect at [latex](−1, 2)[/latex].

Step 7: Check the solution in both equations.

[latex]\begin{array}{cccc}\begin{array}{ccc}\hfill 3x+y& =\hfill & -1\hfill \\ \hfill 3\left(-1\right)+2& \stackrel{?}{=}\hfill & -1\hfill \\ \hfill -1& =\hfill & -1✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 2x+y& =\hfill & 0\hfill \\ \hfill 2\left(-1\right)+2& \stackrel{?}{=}\hfill & 0\hfill \\ \hfill 0& =\hfill & 0✓\hfill \end{array}\end{array}[/latex]

The solution is [latex](−1, 2)[/latex].

Usually when equations are given in standard form, the most convenient way to graph them is by using the intercepts. We’ll do this in Example 4.1.5.

Example 4.1.5

Solve the system by graphing: [latex]\{\begin{array}{c}x+y=2\hfill \\ x-y=4\hfill \end{array}[/latex]

Solution

We will find the [latex]x[/latex]– and [latex]y[/latex]-intercepts of both equations and use them to graph the lines.

 [latex]x + y = 2[/latex]

Step 1:To find the intercepts, let [latex]x = 0[/latex] and solve for [latex]y[/latex], then let [latex]y = 0[/latex] and solve for [latex]x[/latex].

[latex]\begin{array}{cccc}\begin{array}{ccc}\hfill x+y& =\hfill & 2\hfill \\ 0+y& =\hfill & 2\hfill \\ y& =\hfill & 2\hfill \end{array}& & & \begin{array}{ccc}\hfill x+y& =\hfill & 2\hfill \\ x+0& =\hfill & 2\hfill \\ x& =\hfill & 2\hfill \end{array}\end{array}[/latex]

[latex]x[/latex]	[latex]y[/latex]
[latex]0[/latex]	[latex]2[/latex]
[latex]2[/latex]	[latex]0[/latex]

Step 2: To find the intercepts, let [latex]x = 0[/latex] then let [latex]y = 0[/latex].

[latex]\begin{array}{cccc}\begin{array}{ccc}\hfill x-y& =\hfill & 4\hfill \\ \hfill 0-y& =\hfill & 4\hfill \\ \hfill -y& =\hfill & 4\hfill \\ \hfill y& =\hfill & -4\hfill \end{array}& & & \begin{array}{ccc}\hfill x-y& =\hfill & 4\hfill \\ \hfill x-0& =\hfill & 4\hfill \\ \hfill x& =\hfill & 4\hfill \\ \\ \\ \end{array}\end{array}[/latex]

[latex]x[/latex]	[latex]y[/latex]
[latex]0[/latex]	[latex]-4[/latex]
[latex]4[/latex]	[latex]0[/latex]

Step 3: Graph the line.

Step 4: Determine the point of intersection.

The lines intersect at [latex](3, −1)[/latex].

Step 5: Check the solution in both equations.

[latex]\begin{array}{cccccccc}\hfill x+y& =\hfill & 2\hfill & & & \hfill x-y& =\hfill & 4\hfill \\ 3+\left(-1\right)\hfill & \stackrel{?}{=}\hfill & 2\hfill & & & \hfill 3-\left(-1\right)& \stackrel{?}{=}\hfill & 4\hfill \\ \hfill 2& =\hfill & 2✓\hfill & & & \hfill 4& =\hfill & 4✓\hfill \end{array}[/latex]

The solution is [latex](3, −1)[/latex].

There will be times when we will want to know how many solutions there will be to a system of linear equations, but we might not actually have to find the solution. It will be helpful to determine this without graphing.

We have seen that two lines in the same plane must either intersect or are parallel. The systems of equations in Example 4.1.2 through Example 4.1.6 all had two intersecting lines. Each system had one solution.

A system with parallel lines, like Example 4.1.7, has no solution. What happened in Example 4.1.8? The equations have coincident lines, and so the system had infinitely many solutions.

We’ll organize these results in Figure 4.2.10 below:

Parallel lines have the same slope but different [latex]y[/latex]-intercepts. So, if we write both equations in a system of linear equations in slope–intercept form, we can see how many solutions there will be without graphing! Look at the system we solved in Example 4.1.7.

The two lines have the same slope but different [latex]y[/latex]-intercepts. They are parallel lines.

Figure 4.1.11 shows how to determine the number of solutions of a linear system by looking at the slopes and intercepts.

Let’s take one more look at our equations in Example 4.1.7 that gave us parallel lines.

When both lines were in slope-intercept form we had:

Do you recognize that it is impossible to have a single ordered pair [latex]\left(x,y\right)[/latex] that is a solution to both of those equations?

We call a system of equations like this an inconsistent system. It has no solution.

A system of equations that has at least one solution is called a consistent system.

We also categorize the equations in a system of equations by calling the equations independent or dependent. If two equations are independent equations, 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 Figure 4.1.12 and Figure 4.2.13.

We will use the same problem solving strategy we used in Math Models to set up and solve applications of systems of linear equations. We’ll modify the strategy slightly here to make it appropriate for systems of equations.

Step 5 is where we will use the method introduced in this section. We will graph the equations and find the solution.

Example 4.1.12

Sondra is making [latex]10[/latex] quarts of punch from fruit juice and club soda. The number of quarts of fruit juice is [latex]4[/latex] times the number of quarts of club soda. How many quarts of fruit juice and how many quarts of club soda does Sondra need?

Solution

Step 1: Read the problem.

Step 2: Identify what we are looking for.

We are looking for the number of quarts of fruit juice and the number of quarts of club soda that Sondra will need.

Step 3: Name what we are looking for. Choose variables to represent those quantities.

Let [latex]f=[/latex] number of quarts of fruit juice.
Let [latex]c=[/latex] number of quarts of club soda

Step 4: Translate into a system of equations.

[latex]\text{The}\;\underbrace{\text{number of quarts of fruit juice}}_{f}\;\underbrace{\text{and}}_{+}\;\text{the}\;\underbrace{\text{number of quarts of club soda}}_{c}\;\underbrace{\text{is}\;10}_{=10}[/latex]

[latex]\text{The}\;\underbrace{\text{number of quarts of fruit juice}}_{f}\;\underbrace{\text{is}}_{=}\;\underbrace{\text{four times the number of quarts of club soda}}_{4c}[/latex]

We now have the system. [latex]\{\begin{array}{c}f+c=10\hfill \\ f=4c\hfill \end{array}[/latex]

Step 5: Solve the system of equations using good algebra techniques.

The point of intersection [latex](2, 8)[/latex] is the solution. This means Sondra needs [latex]2[/latex] quarts of club soda and [latex]8[/latex] quarts of fruit juice.

Step 6: Check the answer in the problem and make sure it makes sense.

Does this make sense in the problem?

Yes, the number of quarts of fruit juice, [latex]8[/latex] is [latex]4[/latex] times the number of quarts of club soda, [latex]2[/latex].

Yes, [latex]10[/latex] quarts of punch is [latex]8[/latex] quarts of fruit juice plus [latex]2[/latex] quarts of club soda.

Step 7: Answer the question with a complete sentence.

Sondra needs [latex]8[/latex] quarts of fruit juice and [latex]2[/latex] quarts of soda.