11.3: Graphing Linear Equations (2024)

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Learning Objectives

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

Recognize the relation between the solutions of an equation and its graph
Graph a linear equation by plotting points
Graph vertical and horizontal lines

Be Prepared 11.4

Before you get started, take this readiness quiz.

Evaluate: $3 x + 2$ when $x = −1 .$
If you missed this problem, review Example 3.56.

Be Prepared 11.5

Solve the formula: $5 x + 2 y = 20$ for $y .$
If you missed this problem, review Example 9.62.

Be Prepared 11.6

Simplify: $\frac{3}{8} (−24) .$
If you missed this problem, review Example 4.28.

Recognize the Relation Between the Solutions of an Equation and its Graph

In Use the Rectangular Coordinate System, we found a few solutions to the equation $3 x + 2 y = 6$ . They are listed in the table below. So, the ordered pairs $(0, 3)$ , $(2, 0)$ , $(1, \frac{3}{2})$ , $(4, - 3)$ , are some solutions to the equation $3 x + 2 y = 6$ . We can plot these solutions in the rectangular coordinate system as shown on the graph at right.

Notice how the points line up perfectly? We connect the points with a straight line to get the graph of the equation $3 x + 2 y = 6$ . Notice the arrows on the ends of each side of the line. These arrows indicate the line continues.

Every point on the line is a solution of the equation. Also, every solution of this equation is a point on this line. Points not on the line are not solutions!

Notice that the point whose coordinates are $(- 2, 6) Figure 11.8. If you substitute x = - 2 and y = 6 into the equation, you find that it is a solution to the equation.$

So $(4, 1)$ is not a solution to the equation $3 x + 2 y = 6$ . Therefore the point $(4, 1)$ is not on the line.

This is an example of the saying,” A picture is worth a thousand words.” The line shows you all the solutions to the equation. Every point on the line is a solution of the equation. And, every solution of this equation is on this line. This line is called the graph of the equation $3 x + 2 y = 6$ .

Graph of a Linear Equation

The graph of a linear equation $A x + B y = C$ is a straight line.

Every point on the line is a solution of the equation.
Every solution of this equation is a point on this line.

Example 11.15

The graph of $y = 2 x - 3$ is shown below.

For each ordered pair decide

ⓐ Is the ordered pair a solution to the equation?
ⓑ Is the point on the line?

(a) $(0, 3)$
(b) $(3, 3)$
(c) $(2, - 3)$
(d) $(- 1, - 5)$

Answer

Substitute the $x$ - and $y$ -values into the equation to check if the ordered pair is a solution to the equation.

ⓐ

ⓑ Plot the points A: $(0, - 3)$ B: $(3, 3)$ C: $(2, - 3)$ and D: $(- 1, - 5)$ .
The points $(0, - 3)$ , $(3, 3)$ , and $(- 1, - 5)$ are on the line $y = 2 x - 3$ , and the point $(2, - 3)$ is not on the line.

The points which are solutions to $y = 2 x - 3$ are on the line, but the point which is not a solution is not on the line.

Try It 11.29

The graph of $y = 3 x - 1$ is shown.

For each ordered pair, decide

ⓐ is the ordered pair a solution to the equation?
ⓑ is the point on the line?

$(0, - 1)$
$(2, 2)$
$(3, - 1)$
$(- 1, - 4)$

Graph a Linear Equation by Plotting Points

There are several methods that can be used to graph a linear equation. The method we used at the start of this section to graph is called plotting points, or the Point-Plotting Method.

Let’s graph the equation $y = 2 x + 1$ by plotting points.

We start by finding three points that are solutions to the equation. We can choose any value for $x$ or $y,$ and then solve for the other variable.

Since $y$ is isolated on the left side of the equation, it is easier to choose values for $x .$ We will use $0, 1,$ and $-2$ for $x$ for this example. We substitute each value of $x$ into the equation and solve for $y .$

We can organize the solutions in a table. See Table 11.2.

$y = 2 x + 1$
$x$	$y$	$(x, y)$
$0$	$1$	$(0, 1)$
$1$	$3$	$(1, 3)$
$−2$	$−3$	$(−2, −3)$

Table 11.2

Now we plot the points on a rectangular coordinate system. Check that the points line up. If they did not line up, it would mean we made a mistake and should double-check all our work. See Figure 11.9.

Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line. The line is the graph of $y = 2 x + 1 .$

How To

Graph a linear equation by plotting points.

Step 1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
Step 2. Plot the points on a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
Step 3. Draw the line through the points. Extend the line to fill the grid and put arrows on both ends of the line.

It is true that it only takes two points to determine a line, but it is a good habit to use three points. If you plot only two points and one of them is incorrect, you can still draw a line but it will not represent the solutions to the equation. It will be the wrong line. If you use three points, and one is incorrect, the points will not line up. This tells you something is wrong and you need to check your work. See Figure 11.11.

Example 11.16

Graph the equation $y = −3 x .$

Answer

Find three points that are solutions to the equation. It’s easier to choose values for $x,$ and solve for $y .$ Do you see why?

List the points in a table.

$y = −3 x$
$x$	$y$	$(x, y)$
$0$	$0$	$(0, 0)$
$1$	$3$	$(1, −3)$
$−2$	$6$	$(−2, 6)$

Plot the points, check that they line up, and draw the line as shown.

Try It 11.30

Graph the equation by plotting points: $y = −4 x .$

Try It 11.31

Graph the equation by plotting points: $y = x .$

When an equation includes a fraction as the coefficient of $x,$ we can substitute any numbers for $x .$ But the math is easier if we make ‘good’ choices for the values of $x .$ This way we will avoid fraction answers, which are hard to graph precisely.

Example 11.17

Graph the equation $y = \frac{1}{2} x + 3 .$

Answer

Find three points that are solutions to the equation. Since this equation has the fraction $\frac{1}{2}$ as a coefficient of $x,$ we will choose values of $x$ carefully. We will use zero as one choice and multiples of $2$ for the other choices.

The points are shown in the table.

$y = \frac{1}{2} x + 3$
$x$	$y$	$(x, y)$
$0$	$3$	$(0, 3)$
$2$	$4$	$(2, 4)$
$4$	$5$	$(4, 5)$

Plot the points, check that they line up, and draw the line as shown.

Try It 11.32

Graph the equation: $y = \frac{1}{3} x - 1 .$

Try It 11.33

Graph the equation: $y = \frac{1}{4} x + 2 .$

So far, all the equations we graphed had $y$ given in terms of $x .$ Now we’ll graph an equation with $x$ and $y$ on the same side.

Example 11.18

Graph the equation $x + y = 5 .$

Answer

Find three points that are solutions to the equation. Remember, you can start with any value of $x$ or $y .$

We list the points in a table.

$x + y = 5$
$x$	$y$	$(x, y)$
$0$	$5$	$(0, 5)$
$1$	$4$	$(1, 4)$
$4$	$1$	$(4, 1)$

Then plot the points, check that they line up, and draw the line.

Try It 11.34

Graph the equation: $x + y = −2 .$

Try It 11.35

Graph the equation: $x - y = 6 .$

In the previous example, the three points we found were easy to graph. But this is not always the case. Let’s see what happens in the equation $2 x + y = 3 .$ If $y$ is $0,$ what is the value of $x ?$

$y = −2 x + 3$
$0$	$3$	$(0, 3)$
$1$	$1$	$(1, 1)$
$−1$	$5$	$(- 1, 5)$

Example 11.19

Graph the equation $3 x + y = −1 .$

Answer

Find three points that are solutions to the equation.

First, solve the equation for $y .$

$\begin{matrix} 3 x + y & = & −1 \\ y & = & −3 x - 1 \end{matrix}$

We’ll let $x$ be $0, 1,$ and $−1$ to find three points. The ordered pairs are shown in the table. Plot the points, check that they line up, and draw the line.

$y = −3 x - 1$
$x$	$y$	$(x, y)$
$0$	$−1$	$(0, −1)$
$1$	$−4$	$(1, −4)$
$−1$	$2$	$(−1, 2)$

If you can choose any three points to graph a line, how will you know if your graph matches the one shown in the answers in the book? If the points where the graphs cross the $x -$ and $y$ -axes are the same, the graphs match.

Try It 11.36

Graph each equation: $2 x + y = 2 .$

Try It 11.37

Graph each equation: $4 x + y = −3 .$

Graph Vertical and Horizontal Lines

Can we graph an equation with only one variable? Just $x$ and no $y,$ or just $y$ without an $x ?$ How will we make a table of values to get the points to plot?

Let’s consider the equation $x = −3 .$ The equation says that $x$ is always equal to $−3,$ so its value does not depend on $y .$ No matter what $y$ is, the value of $x$ is always $−3 .$

To make a table of solutions, we write $−3$ for all the $x$ values. Then choose any values for $y .$ Since $x$ does not depend on $y,$ you can chose any numbers you like. But to fit the size of our coordinate graph, we’ll use $1, 2,$ and $3$ for the $y$ -coordinates as shown in the table.

$x = −3$
$x$	$y$	$(x, y)$
$−3$	$1$	$(−3, 1)$
$−3$	$2$	$(−3, 2)$
$−3$	$3$	$(−3, 3)$

Then plot the points and connect them with a straight line. Notice in Figure 11.12 that the graph is a vertical line.

Vertical Line

A vertical line is the graph of an equation that can be written in the form $x = a .$

The line passes through the $x$ -axis at $(a, 0)$ .

Example 11.20

Graph the equation $x = 2 .$ What type of line does it form?

Answer

The equation has only variable, $x,$ and $x$ is always equal to $2 .$ We make a table where $x$ is always $2$ and we put in any values for $y .$

$x = 2$
$x$	$y$	$(x, y)$
$2$	$1$	$(2, 1)$
$2$	$2$	$(2, 2)$
$2$	$3$	$(2, 3)$

Plot the points and connect them as shown.

The graph is a vertical line passing through the $x$ -axis at $2 .$

Try It 11.38

Graph the equation: $x = 5 .$

Try It 11.39

Graph the equation: $x = −2 .$

What if the equation has $y$ but no $x$ ? Let’s graph the equation $y = 4 .$ This time the $y$ -value is a constant, so in this equation $y$ does not depend on $x .$

To make a table of solutions, write $4$ for all the $y$ values and then choose any values for $x .$

We’ll use $0, 2,$ and $4$ for the $x$ -values.

$y = 4$
$x$	$y$	$(x, y)$
$0$	$4$	$(0, 4)$
$2$	$4$	$(2, 4)$
$4$	$4$	$(4, 4)$

Plot the points and connect them, as shown in Figure 11.13. This graph is a horizontal line passing through the $y -axis$ at $4 .$

Horizontal Line

A horizontal line is the graph of an equation that can be written in the form $y = b .$

The line passes through the $y -axis$ at $(0, b) .$

Example 11.21

Graph the equation $y = −1 .$

Answer

The equation $y = −1$ has only variable, $y .$ The value of $y$ is constant. All the ordered pairs in the table have the same $y$ -coordinate, $−1$ . We choose $0, 3,$ and $−3$ as values for $x .$

$y = −1$
$x$	$y$	$(x, y)$
$−3$	$−1$	$(−3, −1)$
$0$	$−1$	$(0, −1)$
$3$	$−1$	$(3, −1)$

The graph is a horizontal line passing through the $y$ -axis at $–1$ as shown.

Try It 11.40

Graph the equation: $y = −4 .$

Try It 11.41

Graph the equation: $y = 3 .$

The equations for vertical and horizontal lines look very similar to equations like $y = 4 x .$ What is the difference between the equations $y = 4 x$ and $y = 4 ?$

The equation $y = 4 x$ has both $x$ and $y .$ The value of $y$ depends on the value of $x .$ The $y -coordinate$ changes according to the value of $x .$

The equation $y = 4$ has only one variable. The value of $y$ is constant. The $y -coordinate$ is always $4 .$ It does not depend on the value of $x .$

The graph shows both equations.

Notice that the equation $y = 4 x$ gives a slanted line whereas $y = 4$ gives a horizontal line.

Example 11.22

Graph $y = −3 x$ and $y = −3$ in the same rectangular coordinate system.

Answer

Find three solutions for each equation. Notice that the first equation has the variable $x,$ while the second does not. Solutions for both equations are listed.

The graph shows both equations.

Try It 11.42

Graph the equations in the same rectangular coordinate system: $y = −4 x$ and $y = −4 .$

Try It 11.43

Graph the equations in the same rectangular coordinate system: $y = 3$ and $y = 3 x .$

Media

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Section 11.2 Exercises

Practice Makes Perfect

Recognize the Relation Between the Solutions of an Equation and its Graph

In each of the following exercises, an equation and its graph is shown. For each ordered pair, decide

ⓐ is the ordered pair a solution to the equation?
ⓑ is the point on the line?

39.

$y = x + 2$

$(0, 2)$
$(1, 2)$
$(- 1, 1)$
$(- 3, 1)$

40.

$y = x - 4$

$(0, - 4)$
$(3, - 1)$
$(2, 2)$
$(1, - 5)$

41.

$y = \frac{1}{2} x - 3$

$(0, - 3)$
$(2, - 2)$
$(- 2, - 4)$
$(4, 1)$

42.

$y = \frac{1}{3} x + 2$

$(0, 2)$
$(3, 3)$
$(- 3, 2)$
$(- 6, 0)$

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

43.

$y = 3 x - 1$

44.

$y = 2 x + 3$

45.

$y = −2 x + 2$

46.

$y = −3 x + 1$

47.

$y = x + 2$

Everyday Math

113.

Motor home cost The Robinsons rented a motor home for one week to go on vacation. It cost them $$594$ plus $$0.32$ per mile to rent the motor home, so the linear equation $y = 594 + 0.32 x$ gives the cost, $y,$ for driving $x$ miles. Calculate the rental cost for driving $400, 800, and 1,200$ miles, and then graph the line.

114.

Weekly earning At the art gallery where he works, Salvador gets paid $$200$ per week plus $15%$ of the sales he makes, so the equation $y = 200 + 0.15 x$ gives the amount $y$ he earns for selling $x$ dollars of artwork. Calculate the amount Salvador earns for selling $$900, $1,600, and $2,000,$ and then graph the line.

Writing Exercises

115.

Explain how you would choose three $x -values$ to make a table to graph the line $y = \frac{1}{5} x - 2 .$

116.

What is the difference between the equations of a vertical and a horizontal line?

Self Check

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

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

FAQs

How do you find the answer to a linear equation? ›

The steps for solving linear equations are:

Simplify both sides of the equation and combine all same-side like terms.
Combine opposite-side like terms to obtain the variable term on one side of the equal sign and the constant term on the other.
Divide or multiply as needed to isolate the variable.
Check the answer.

Oct 6, 2021

Tell Me More ›

Why is graphing linear equations so hard? ›

You cannot really picture a linear algebra equation on a graph on your head, and cannot see it going through fluctuations to yield different results. Since there is hardly any clear visualization, the concepts are difficult to grasp and practice.

Discover More ›

Can you always check your answers when solving a linear equation? ›

Anytime you are solving linear equations, you can always check your answer by substituting it back into the equation. If you get a true statement, then the answer is correct. This isn't 100% necessary for every problem, but it is a good habit so we will do it for our equations.

How many answers can a linear equation have? ›

A linear equation has exactly one of three possible amounts of solutions: 1, 0, or infinite. Think of this as two lines x↦0 and x↦ax+b, then the question is whether or not the lines intersect at exactly one point.

Find Out More ›

How to solve linear equations by graphing? ›

To 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. If the lines intersect, identify the point of intersection.

Apr 22, 2020

Learn More Now ›

What is the hardest math course? ›

1. Real Analysis: This is a rigorous course that focuses on the foundations of real numbers, limits, continuity, differentiation, and integration. It's known for its theoretical, proof-based approach and can be a paradigm shift for students used to computation-heavy math courses.

Find Out More ›

Is linear algebra or Calc harder? ›

Calculus is the hardest mathematics subject and only a small percentage of students reach Calculus in high school or anywhere else. Linear algebra is a part of abstract algebra in vector space. However, it is more concrete with matrices, hence less abstract and easier to understand.

Show Me More ›

Is linear algebra 1 hard? ›

Linear Algebra can seem tough at first because it involves abstract ideas like vectors and matrices. However, it gets easier with the right approach. Start with the basics and practice regularly. Use online resources, join study groups, and try applying what you learn to real-life problems.

Read On ›

Do linear equations have 2 answers? ›

Most linear systems you will encounter will have exactly one solution. However, it is possible that there are no solutions, or infinitely many. (It is not possible that there are exactly two solutions.)

Tell Me More ›

How to verify the answer? ›

Verify (a Solution) Verifying a solution ensures the solution satisfies any equation or inequality by using substitution. Verify whether or not x = 3 is a solution to the conditional equation 2x - 3 = 6 - x. Substitute x = 3 into 2x - 3 = 6 - x to see if a true or false statement results.

Tell Me More ›

What are the rules for solving linear equations? ›

The following steps provide a good method to use when solving linear equations.

Simplify each side of the equation by removing parentheses and combining like terms.
Use addition or subtraction to isolate the variable term on one side of the equation.
Use multiplication or division to solve for the variable.

Learn More ›

How do you determine the number of solutions to a linear equation? ›

If solving an equation yields a statement that is true for a single value for the variable, like x = 3, then the equation has one solution. If solving an equation yields a statement that is always true, like 3 = 3, then the equation has infinitely many solutions.

What does "no solutions" look like? ›

What does no solution look like on a graph? A graph with no solution will have functions that do not all intersect at any point. If the system consists of two functions, then there will be no points of intersection. A system of two linear equations has no solution if the lines are parallel.

Discover More Details ›

How to find the solution to a linear system? ›

A system of linear equations consists of the equations of two lines. The solution to a system of linear equations is the point which lies on both lines. In other words, the solution is the point where the two lines intersect.

Get More Info ›

How do you solve a linear function equation? ›

Solving Linear Functions. A linear function is a function with the form f(x) = ax' + b. It looks like a regular linear equation, but instead of using y, the linear function notation is f(x). To solve a linear function, you would be given the value of f(x) and be asked to find x.

What is the formula for finding the linear equation? ›

The standard form of linear equations in two variables is expressed as, Ax + By = C; where A, B and C are any real numbers, and x and y are the variables.

Explore More ›

How to find the solution of an equation? ›

Bring the variable terms to one side of the equation and the constant terms to the other side using the addition and subtraction properties of equality. Make the coefficient of the variable as 1, using the multiplication or division properties of equality. isolate the variable and get the solution.

View Details ›