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11.3: Graphing Linear Equations

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• Page ID 115029

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

Be Prepared 11.5

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

Be Prepared 11.6

Simplify: 3 8 ( −24 ) . 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 3 x + 2 y = 6 . They are listed in the table below. So, the ordered pairs ( 0 , 3 ) ( 0 , 3 ) , ( 2 , 0 ) ( 2 , 0 ) , ( 1 , 3 2 ) ( 1 , 3 2 ) , ( 4 , − 3 ) ( 4 , − 3 ) , are some solutions to the equation 3 x + 2 y = 6 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 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 x = − 2 and y = 6 y = 6 into the equation, you find that it is a solution to the equation.

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

Graph of a Linear Equation

The graph of a linear equation A x + B y = C 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 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 ) ( 0 , 3 )
• (b) ( 3 , 3 ) ( 3 , 3 )
• (c) ( 2 , − 3 ) ( 2 , − 3 )
• (d) ( − 1 , − 5 ) ( − 1 , − 5 )

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

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

The points which are solutions to y = 2 x − 3 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 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 ) ( 0 , − 1 )
• ( 2 , 2 ) ( 2 , 2 )
• ( 3 , − 1 ) ( 3 , − 1 )
• ( − 1 , − 4 ) ( − 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 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 x or y , y , and then solve for the other variable.

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

We can organize the solutions in a table. See 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 . y = 2 x + 1 .

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 . y = −3 x .

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

List the points in a table.

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 . y = −4 x .

Try It 11.31

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

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

Example 11.17

Graph the equation y = 1 2 x + 3 . y = 1 2 x + 3 .

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

The points are shown in the table.

Try It 11.32

Graph the equation: y = 1 3 x − 1 . y = 1 3 x − 1 .

Try It 11.33

Graph the equation: y = 1 4 x + 2 . y = 1 4 x + 2 .

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

Example 11.18

Graph the equation x + y = 5 . x + y = 5 .

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

We list the points in a table.

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

Try It 11.34

Graph the equation: x + y = −2 . x + y = −2 .

Try It 11.35

Graph the equation: x − y = 6 . 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 . 2 x + y = 3 . If y y is 0 , 0 , what is the value of x ? x ?

The solution is the point ( 3 2 , 0 ) . ( 3 2 , 0 ) . This point has a fraction for the x x -coordinate. While we could graph this point, it is hard to be precise graphing fractions. Remember in the example y = 1 2 x + 3 , y = 1 2 x + 3 , we carefully chose values for x x so as not to graph fractions at all. If we solve the equation 2 x + y = 3 2 x + y = 3 for y , y , it will be easier to find three solutions to the equation.

2 x + y = 3 2 x + y = 3

y = −2 x + 3 y = −2 x + 3

Now we can choose values for x x that will give coordinates that are integers. The solutions for x = 0 , x = 1 , x = 0 , x = 1 , and x = −1 x = −1 are shown.

Example 11.19

Graph the equation 3 x + y = −1 . 3 x + y = −1 .

Find three points that are solutions to the equation.

First, solve the equation for y . y .

3 x + y = −1 y = −3 x − 1 3 x + y = −1 y = −3 x − 1

We’ll let x x be 0 , 1 , 0 , 1 , and −1 −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.

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 - x - and y y -axes are the same, the graphs match.

Try It 11.36

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

Try It 11.37

Graph each equation: 4 x + y = −3 . 4 x + y = −3 .

Graph Vertical and Horizontal Lines

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

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

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

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 . x = a .

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

Example 11.20

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

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

Plot the points and connect them as shown.

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

Try It 11.38

Graph the equation: x = 5 . x = 5 .

Try It 11.39

Graph the equation: x = −2 . x = −2 .

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

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

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

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

Horizontal Line

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

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

Example 11.21

Graph the equation y = −1 . y = −1 .

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

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

Try It 11.40

Graph the equation: y = −4 . y = −4 .

Try It 11.41

Graph the equation: y = 3 . y = 3 .

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

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

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

The graph shows both equations.

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

Example 11.22

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

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

Try It 11.42

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

Try It 11.43

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

ACCESS ADDITIONAL ONLINE RESOURCES

• Use a Table of Values
• Graph a Linear Equation Involving Fractions
• Graph Horizontal and Vertical Lines

Section 11.2 Exercises

Practice makes perfect.

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

y = x + 2 y = x + 2

• ( 0 , 2 ) ( 0 , 2 )
• ( 1 , 2 ) ( 1 , 2 )
• ( − 1 , 1 ) ( − 1 , 1 )
• ( − 3 , 1 ) ( − 3 , 1 )

y = x − 4 y = x − 4

• ( 0 , − 4 ) ( 0 , − 4 )
• ( 1 , − 5 ) ( 1 , − 5 )

y = 1 2 x − 3 y = 1 2 x − 3

• ( 0 , − 3 ) ( 0 , − 3 )
• ( 2 , − 2 ) ( 2 , − 2 )
• ( − 2 , − 4 ) ( − 2 , − 4 )
• ( 4 , 1 ) ( 4 , 1 )

y = 1 3 x + 2 y = 1 3 x + 2

• ( 3 , 3 ) ( 3 , 3 )
• ( − 3 , 2 ) ( − 3 , 2 )
• ( − 6 , 0 ) ( − 6 , 0 )

In the following exercises, graph by plotting points.

y = 3 x − 1 y = 3 x − 1

y = 2 x + 3 y = 2 x + 3

y = −2 x + 2 y = −2 x + 2

y = −3 x + 1 y = −3 x + 1

y = x − 3 y = x − 3

y = − x − 3 y = − x − 3

y = − x − 2 y = − x − 2

y = 2 x y = 2 x

y = 3 x y = 3 x

y = −4 x y = −4 x

y = −2 x y = −2 x

y = 1 2 x + 2 y = 1 2 x + 2

y = 1 3 x − 1 y = 1 3 x − 1

y = 4 3 x − 5 y = 4 3 x − 5

y = 3 2 x − 3 y = 3 2 x − 3

y = − 2 5 x + 1 y = − 2 5 x + 1

y = − 4 5 x − 1 y = − 4 5 x − 1

y = − 3 2 x + 2 y = − 3 2 x + 2

y = − 5 3 x + 4 y = − 5 3 x + 4

x + y = 6 x + y = 6

x + y = 4 x + y = 4

x + y = −3 x + y = −3

x + y = −2 x + y = −2

x − y = 2 x − y = 2

x − y = 1 x − y = 1

x − y = −1 x − y = −1

x − y = −3 x − y = −3

− x + y = 4 − x + y = 4

− x + y = 3 − x + y = 3

− x − y = 5 − x − y = 5

− x − y = 1 − x − y = 1

3 x + y = 7 3 x + y = 7

5 x + y = 6 5 x + y = 6

2 x + y = −3 2 x + y = −3

4 x + y = −5 4 x + y = −5

2 x + 3 y = 12 2 x + 3 y = 12

3 x − 4 y = 12 3 x − 4 y = 12

1 3 x + y = 2 1 3 x + y = 2

1 2 x + y = 3 1 2 x + y = 3

Graph Vertical and Horizontal lines

In the following exercises, graph the vertical and horizontal lines.

x = 4 x = 4

x = 3 x = 3

x = −2 x = −2

x = −5 x = −5

y = 3 y = 3

y = 1 y = 1

y = −5 y = −5

y = −2 y = −2

x = 7 3 x = 7 3

x = 5 4 x = 5 4

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

y = − 1 2 x y = − 1 2 x and y = − 1 2 y = − 1 2

y = − 1 3 x y = − 1 3 x and y = − 1 3 y = − 1 3

y = 2 x y = 2 x and y = 2 y = 2

y = 5 x y = 5 x and y = 5 y = 5

Mixed Practice

In the following exercises, graph each equation.

y = 4 x y = 4 x

y = − 1 2 x + 3 y = − 1 2 x + 3

y = 1 4 x − 2 y = 1 4 x − 2

y = − x y = − x

y = x y = x

x − y = 3 x − y = 3

x + y = − 5 x + y = − 5

4 x + y = 2 4 x + y = 2

2 x + y = 6 2 x + y = 6

y = −1 y = −1

y = 5 y = 5

2 x + 6 y = 12 2 x + 6 y = 12

5 x + 2 y = 10 5 x + 2 y = 10

x = −4 x = −4

Everyday Math

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

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

Writing Exercises

Explain how you would choose three x -values x -values to make a table to graph the line y = 1 5 x − 2 . y = 1 5 x − 2 .

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

ⓐ 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?

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• Problem Solving with Linear Graphs

Key Questions

I assume that the equation you would be creating is in slope-intercept form , or #y= mx + b# .

If so, there is a simple answer. You can tell that you need to create a linear equation by the information the problem gives you. The problem should list the Y- intercept, a starting amount of something and a slope , or a rate of change.

For example, here is a problem:

Maddie and Cindy are starting their very own babysitting business. They charge parents $5 dollars right when they come in and $2 for every hour they need to babysit a child. How can they calculate how much they will charge for an evening of babysitting?

Your formula would be: #y= 2x + 5#

Your #x# value would be hour(s) and your #y# value would be total cost. After you write your equation you can simply solve it.

Problem Solving with Linear Graphs

January 25, 2021.

Students are challenged to solve a range of problems involving straight line graphs.

This lesson is aimed at students aiming for grades 4 to 7 on the Foundation or Higher GCSE course.

There are five problems that link to area, midpoints, gradients and solving equations.

Linear Graphs  |  Foundation GCSE Scheme of Work | Higher GCSE Scheme of Work

Mr Mathematics Blog

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Planes of Symmetry in 3D Shapes for Key Stage 3/GCSE students.

Use isometric paper for hands-on learning and enhanced understanding.

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Master GCSE Math: Get key SOH-CAH-TOA tips, solve triangles accurately, and tackle area tasks. Ideal for students targeting grades 4-5.

Regions in the Complex Plane

Explore Regions in the Complex Plane with A-Level Further Maths: inequalities, Argand diagrams, and geometric interpretations.

[FREE] Fun Math Games & Activities Packs

Always on the lookout for fun math games and activities in the classroom? Try our ready-to-go printable packs for students to complete independently or with a partner!

In order to access this I need to be confident with:

Ratio and Proportions

Proportions

Coordinate Graph

Linear graph

Here you will learn about plotting linear graphs, interpreting linear graphs, and determining whether they represent proportional relationships.

Students first learn about first quadrant linear graphs in 6th grade with their work in ratio and proportional relationships. They expand this knowledge in pre-algebra and algebra.

What is a linear graph?

A linear graph is a straight line graph that shows a relationship between the x -coordinate and the y -coordinate. Some linear graphs show a proportional relationship between the x -coordinate and the y -coordinate.

For example,

Take a look at this linear graph. Notice how it passes through the origin, (0, 0).

Now, let’s write the ratio of each of the points in the form \, \cfrac{y}{x} .

\begin{aligned} & (1,1) \rightarrow \cfrac{1}{1} \\\\ & (2,2) \rightarrow \cfrac{2}{2}=\cfrac{2 \, \div \, 2}{2 \, \div \, 2}=\cfrac{1}{1} \\\\ & (3,3) \rightarrow \cfrac{3}{3}=\cfrac{3 \, \div \, 3}{3 \, \div \, 3}=\cfrac{1}{1} \end{aligned}

(0,0) \rightarrow \cfrac{0}{0} \, (the origin is not a good point to use because division by 0 does not exist)

You can conclude that this linear graph represents a proportional relationship because:

The line passes through the origin.

• There is a common ratio of each of the points, which is \, \cfrac{1}{1} .

Notice that the movement from one point to the next is the same as the common ratio, up one unit and right one unit.

Let’s look at another example of a linear relationship that is proportional. The table represents three points on a coordinate plane.

Plot the points and write the ratio of each coordinate in the form of \, \cfrac{y}{x} .

\begin{aligned} & (1,3) \rightarrow \cfrac{3}{1}=3 \\\\ & (2,6) \rightarrow \cfrac{6}{2}=\cfrac{6 \, \div \, 2}{2 \, \div \, 2}=\cfrac{3}{1}=3 \\\\ & (3,9) \rightarrow \cfrac{9}{3}=\cfrac{9 \, \div \, 3}{3 \, \div \, 3}=\cfrac{3}{1}=3 \end{aligned}

The ratios are equal which means they are proportional where the common ratio (unit rate) is \, \cfrac{3}{1} \, .

Also, notice the line passes through the origin, (0, 0).

• There is a common ratio of each of the points, which is \, \cfrac{3}{1} .

Like with the previous example, notice that the movement from one point to the next is the same as the common ratio, up three units and right one unit.

In this next example, does the linear graph represent a proportional relationship?

Notice that the line does not pass through the origin. Write the points as a ratio in the form \, \cfrac{y}{x} .

(0,1) \rightarrow \cfrac{1}{0} \, ( 0 in the denominator does not exist, so not a good point to use)

\begin{aligned} & (2,3) \rightarrow \cfrac{3}{2} \\\\ & (4,5) \rightarrow \cfrac{5}{4} \\\\ & (6,7) \rightarrow \cfrac{7}{6} \end{aligned}

\cfrac{3}{2} \, ≠ \, \cfrac{3}{2} \, ≠ \, \cfrac{5}{4} \, ≠ \, \cfrac{7}{6}

You can conclude that this linear graph does not represent a proportional relationship because:

• The line does not pass through the origin.
• The ratio of the points is not equal.

Equations that represent linear graphs

Linear graphs can be represented by equations. You can check to see if an equation represents a linear relationship by making a table of values.

For example, let’s see if the equation y=2x represents a linear graph.

Since x is the independent variable, you can select any number to substitute for x. Substitute the value for x in the equation and solve for the y value.

Plot the points from the table on a linear graph.

The points form a line, so y=2x is a linear relationship.

The ratio of the points in the form of \, \cfrac{y}{x} \, is \, \cfrac{2}{1} \, .

\begin{aligned} & \cfrac{4}{2}=\cfrac{4 \, \div \, 2}{2 \, \div \, 2}=\cfrac{2}{1} \\\\ & \cfrac{2}{1} \\\\ & \cfrac{-2}{-1}=\cfrac{-2 \, \div \, -1}{-1 \, \div \, -1}=\cfrac{2}{1} \\\\ & \cfrac{-4 \, \div \, -2}{-2 \, \div \, -2}=\cfrac{2}{1} \\\\ & \cfrac{4}{2}=\cfrac{2}{1}=\cfrac{-2}{-1}=\cfrac{-4}{-2} \end{aligned}

You can conclude that y=2x is also a proportional relationship because:

• There is a common ratio (unit rate) of the points which is \, \cfrac{2}{1} \, .

Notice that the common ratio (unit rate) is \, \cfrac{2}{1} \, and the coefficient of x in the equation y=2x is also 2 or \, \cfrac{2}{1} \, .

In equations that represent linear graphs that are proportional, the common ratio will be equal to the coefficient of the x term in the equation.

k=\cfrac{y}{x} \, and y=k x

k represents the common ratio. Also notice the movement from one point to the next on the graph of y=2x is up 2 units and to the right 1 unit.

Let’s look at the equation y=-x+1. Does it represent a linear relationship?

Like in the last example, let’s make a table of values.

Since x is the independent variable, you can select any number to substitute for x.

Substitute the value for x in the equation and solve for the y value.

Plot the points on a coordinate graph.

The points form a line so y=-x +1 is a linear relationship. However, the line does not pass through the origin.

Writing the ratio of each of the points in the form of \, \cfrac{y}{x} \, does not give you a common ratio.

\begin{aligned} & (3,-2) \rightarrow \cfrac{-2}{3} \\\\ & (2,-1) \rightarrow \cfrac{-1}{2} \\\\ & (1,0) \rightarrow \cfrac{0}{1} \end{aligned}

(0,1) \rightarrow \cfrac{1}{0} \, ( 0 in the denominator does not exist so not a good point to use)

(-2,3) \rightarrow \cfrac{3}{-2}

\cfrac{3}{-2} \, ≠ \, \cfrac{2}{-1} \, ≠ \, \cfrac{1}{0} \, ≠ \, \cfrac{0}{1} \, ≠ \, \cfrac{-1}{2}

You can conclude that y=-x+1 is not a proportional relationship because:

• There is no common ratio (unit rate) of the points.

What are linear graphs?

Common Core State Standards

How does this relate to 6th grade math, 7th grade math, and 8th grade math?

• Grade 6 – Expressions and Equations (6.EE.C.9) Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation.
• Grade 7 – Ratios and Proportional Relationships (7.RP.A.2.b) Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships.
• Grade 7 – Ratios and Proportional Relationships (7.RP.A.2.d) Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) , where r is the unit rate.
• Grade 8 – Expressions and Equations (8.EE.B.5) Graph proportional relationships, interpreting the unit rate as the slope of the graph.

How to work with a linear graph

In order to determine if a linear graph represents a proportional relationship:

Plot the points on the graph, connect them, and make sure it forms a straight line.

Look at the graph to see if it passes through the origin. If it does, go to step 3.

If it goes through the origin, find the common ratio (unit rate) in the form of \, \cfrac{\textbf{y}}{\textbf{x}} \, .

Explain what the common ratio (unit rate) means on the graph.

In order to determine if an equation represents a linear proportional relationship:

• Make a table of values to represent the equation and select a minimum of three values for \textbf{x} .

Substitute values for \textbf{x} in the equation to find the \textbf{y} value.

Plot the points, connect them, and make sure it forms a straight line.

• Determine if the graph represents a linear relationship. If it is, go to step 5.
• Determine if the linear graph passes through the origin and find the common ratio in the form of \, \cfrac{\textbf{y}}{\textbf{x}} \, .

[FREE] Types of Graphs Check for Understanding Quiz (Grade 5 to 8)

Use this quiz to check your grade 5 to 8 students’ understanding of type of graphs. 15+ questions with answers covering a range of 5th to 8th grade type of graphs topics to identify areas of strength and support!

Linear graph examples

Example 1: determine if a table of values represents a linear, proportional relationship.

Determine if the points represented in the table are a linear, proportional relationship.

2 Look at the graph to see if it passes through the origin. If it does, go to step 3.

3 If it goes through the origin, find the common ratio (unit rate) in the form of \, \cfrac{\textbf{y}}{\textbf{x}} \, .

The ratio of the points.

The common ratio is \, \cfrac{3}{1} \, .

4 Explain what the common ratio (unit rate) means on the graph.

The common ratio (unit rate) is \, \cfrac{2}{1} \, which means the movement from one point to the next is 2 units up and 1 unit to the right.

Example 2: determine if a table represents a linear, proportional relationship

Do the points in the table represent a proportional relationship? If so, what is the common ratio (unit rate)?

The line passes through the origin, so it represents a proportional relationship.

The common ratio is \, \cfrac{-5}{1} \,.

The common ratio (unit rate) is \, \cfrac{-5}{1} \, which means the movement from one point to the next is 5 units down and 1 unit to the right.

Example 3: determine if a linear graph is a proportional relationship from a table

Do the points in the table represent a proportional relationship?

The line does not go through the origin so it does not represent a proportional relationship.

Example 4: real-world scenario linear, proportional relationship

The linear graph shows a relationship between the pounds of potting soil bought and the cost of the potting soil.

Does this relationship show a proportional relationship and what is the cost of 6 \, lbs of potting soil?

Points already plotted and connected to form a straight line.

The line goes through the origin because it contains the point (0, 0). So, it does represent a proportional relationship.

\cfrac{3}{2} \, is the common ratio (unit rate).

The common ratio (unit rate) means that for every 3 \, lbs of potting soil, the cost is \$2.

So, if you buy 6 pounds of potting soil, the cost will be \$4.

Example 5: determine if the equation represents a proportional relationship.

Graph the equation y=\cfrac{1}{2} \, x by making a table of values.

Then determine if it is a proportional relationship.

Make a table of values to represent the equation and select a minimum of \bf{3} values for \textbf{x} .

Determine if the linear graph is proportional. If it is, state the common ratio and how it relates to the equation.

The linear graph goes through the origin, so it represents a proportional relationship.

The common ratio is \, \cfrac{1}{2} \, because that is the coefficient of x in the equation, and it is the ratio of the points.

Teaching tips for linear graphs

• Have students make comparisons on their own between the points and the visual graph so that they can formulate understanding for themselves as to what a proportional linear graph looks like versus a non-proportional linear graph.
• Infuse activities from the free graphing platform, Desmos, for students to explore linear graphs.
• Have students create their own proportional linear graphs and provide an explanation.
• Resources such as Khan Academy are useful for reinforcement of concepts but do not replace the authentic investigation that takes place in the classroom.

Easy mistakes to make

• Confusing the \textbf{x} coordinate and the \textbf{y} coordinate. For example, plotting the point (1, 2) as (2, 1).
• Not knowing where the origin of the coordinate graph is located.
• Reading a table of values incorrectly. For example, mixing up the x and y coordinates.
• When making a table of values, plugging in values for \textbf{y} and solving for \textbf{x} .

Related types of graphs lessons

• Types of graphs
• x and y axis
• Interpreting graphs
• Coordinate plane
• Plot points on a graph
• Independent and dependent variables
• Direct variation

Practice linear graph questions

1. Which linear graph represents a proportional relationship?

A linear graph is considered to be proportional if it goes through the origin, which is the point (0, 0). The graph goes through the origin so it is a proportional relationship.

2. The linear graph is proportional. What is the common ratio of the points?

It’s a proportional relationship because the line goes through the origin. Write the ratio of each point in the form of \, \cfrac{y}{x} \, .

The common ratio is \cfrac{1}{4}. This means the movement from one point to the next is 1 unit up and 4 units right.

3. The linear graph shows a proportional relationship between how many pencils purchased and the cost of the pencils. What is the cost of 12 pencils?

The graph is a proportional relationship. So the ratio of y to x for each of the points is equal.

Use the point (12, 9) which represents 12 pencils for \$9. So, the answer is \$9.

4. Which graph shows a proportional relationship that passes through the point (3, 1)?

The linear graph has to pass through the origin (0, 0) in order for it to have a proportional relationship.

There is only one linear graph that passes through (0, 0) and contains the point (3, 1).

5. Which table of points shows a proportional relationship where the common ratio is \, \cfrac{1}{2} \, ?

Write the ratio of each of the points in the form of \, \cfrac{y}{x} \, .

\cfrac{3}{6}=\cfrac{4}{8}=\cfrac{5}{10} \, , where the common ratio is \, \cfrac{1}{2} \, .

6. Which table shows a proportional relationship where the common ratio is 3?

\cfrac{6}{2}=\cfrac{9}{3}=\cfrac{12}{4} \, , where the common ratio is \, \cfrac{3}{1}=3 .

7. Which linear equation represents a proportional relationship with a unit rate of \, \cfrac{1}{2} \, ?

Make a table of values selecting three values for x.

Graph the points to form a linear graph.

The linear graph goes through the origin proving that it is a proportional relationship.

Linear graph FAQS

No, only linear graphs that pass through the origin are proportional.

Yes, as you move through pre-algebra and algebra you will learn that the unit rate is the slope of the line.

The x -intercept is the point where the line crosses the x -axis. The y -intercept is the point where the line crosses the y -axis.

In pre-algebra and algebra, you will learn standard ways to write linear equations, which are called slope-intercept form of a line, point-slope form of a line, and standard form of a line.

The gradient of the line is the movement from one point to the next on a linear graph. It is also known as the slope.

Yes, all straight line graphs are linear graphs.

Yes, non-linear graphs form curves such as u-shape figures called parabolas and curves formed by polynomials.

There are several methods to use when graphing linear equations. One way is to make a table of values, but there are other ways you will learn that are quicker, such as graphing linear equations using slope-intercept.

A linear function is a linear equation that is represented as a straight line on a graph.

Yes, in algebra, you will learn how to graph linear inequalities.

The next lessons are

• Number patterns
• Constant of proportionality

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Linear Functions Problems with Solutions

Linear functions are highly used throughout mathematics and are therefore important to understand. A set of problems involving linear functions , along with detailed solutions, are presented. The problems are designed with emphasis on the meaning of the slope and the y intercept.

Problem 1: f is a linear function. Values of x and f(x) are given in the table below; complete the table.

Problem 2: A family of linear functions is given by

Solution to Problem 2: a)

Problem 3: A high school had 1200 students enrolled in 2003 and 1500 students in 2006. If the student population P ; grows as a linear function of time t, where t is the number of years after 2003. a) How many students will be enrolled in the school in 2010? b) Find a linear function that relates the student population to the time t. Solution to Problem 3: a) The given information may be written as ordered pairs (t , P). The year 2003 correspond to t = 0 and the year 2006 corresponds to t = 3, hence the 2 ordered pairs (0, 1200) and (3, 1500) Since the population grows linearly with the time t, we use the two ordered pairs to find the slope m of the graph of P as follows m = (1500 - 1200) / (6 - 3) = 100 students / year The slope m = 100 means that the students population grows by 100 students every year. From 2003 to 2010 there are 7 years and the students population in 2010 will be P(2010) = P(2003) + 7 * 100 = 1200 + 700 = 1900 students. b) We know the slope and two points, we may use the point slope form to find an equation for the population P as a function of t as follows P - P1 = m (t - t1) P - 1200 = 100 (t - 0) P = 100 t + 1200

Problem 4: The graph shown below is that of the linear function that relates the value V (in $) of a car to its age t, where t is the number of years after 2000.

Problem 5: The cost of producing x tools by a company is given by

Problem 6: A 500-liter tank full of oil is being drained at the constant rate of 20 liters par minute. a) Write a linear function V for the number of liters in the tank after t minutes (assuming that the drainage started at t = 0). b) Find the V and the t intercepts and interpret them. e) How many liters are in the tank after 11 minutes and 45 seconds? Solution to Problem 6: After each minute the amount of oil in the tank deceases by 20 liters. After t minutes, the amount of oil in the tank decreases by 20*t liters. Hence if at the start there 500 liters, after t minute the amount V of oil left in the tank is given by V = 500 - 20 t b) To find the V intercept, set t = 0 in the equation V = 500 - 20 t. V = 500 liters : it is the amount of oil at the start of the drainage. To find the t intercept, set V = 0 in the equation V = 500 - 20 t and solve for t. 0 = 500 - 20 t t = 500 / 20 = 25 minutes : it is the total time it takes to drain the 500 liters of oil. c) Convert 11 minutes 45 seconds in decimal form. t = 11 minutes 45 seconds = 11.75 minutes Calculate V at t = 11.75 minutes. V(11.75) = 500 - 20*11.75 = 265 liters are in the tank after 11 minutes 45 seconds of drainage.

Problem 7: A 50-meter by 70-meter rectangular garden is surrounded by a walkway of constant width x meters.

Problem 8: A driver starts a journey with 25 gallons in the tank of his car. The car burns 5 gallons for every 100 miles. Assuming that the amount of gasoline in the tank decreases linearly, a) write a linear function that relates the number of gallons G left in the tank after a journey of x miles. b) What is the value and meaning of the slope of the graph of G? c) What is the value and meaning of the x intercept? Solution to Problem 8: a) If 5 gallons are burnt for 100 miles then (5 / 100) gallons are burnt for 1 mile. Hence for x miles, x * (5 / 100) gallons are burnt. G is then equal to the initial amount of gasoline decreased by the amount gasoline burnt by the car. Hence G = 25 - (5 / 100) x b) The slope of G is equal to 5 / 1000 and it represent the amount of gasoline burnt for a distance of 1 mile. c) To find the x intercept, we set G = 0 and solve for x. 25 - (5 / 100) x = 0 x = 500 miles : it is the distance x for which all 25 gallons of gasoline will be burnt.

Problem 9: A rectangular wire frame has one of its dimensions moving at the rate of 0.5 cm / second. Its width is constant and equal to 4 cm. If at t = 0 the length of the rectangle is 10 cm,

Linear Equations

A linear equation is an equation for a straight line

These are all linear equations:

Let us look more closely at one example:

Example: y = 2x + 1 is a linear equation:

The graph of y = 2x+1 is a straight line

• When x increases, y increases twice as fast , so we need 2x
• When x is 0, y is already 1. So +1 is also needed
• And so: y = 2x + 1

Here are some example values:

Check for yourself that those points are part of the line above!

Different Forms

There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:

• Exponents (like the 2 in x 2 )
• Square roots , cube roots , etc

Examples: These are NOT linear equations:

Slope-intercept form.

The most common form is the slope-intercept equation of a straight line :

Example: y = 2x + 1

• Slope: m = 2
• Intercept: b = 1

Point-Slope Form

Another common one is the Point-Slope Form of the equation of a straight line:

Example: y − 3 = (¼)(x − 2)

It is in the form y − y 1 = m(x − x 1 ) where:

General Form

And there is also the General Form of the equation of a straight line:

Example: 3x + 2y − 4 = 0

It is in the form Ax + By + C = 0 where:

There are other, less common forms as well.

As a Function

Sometimes a linear equation is written as a function , with f(x) instead of y :

And functions are not always written using f(x):

The Identity Function

There is a special linear function called the "Identity Function":

And here is its graph:

It is called "Identity" because what comes out is identical to what goes in:

Constant Functions

Another special type of linear function is the Constant Function ... it is a horizontal line:

No matter what value of "x", f(x) is always equal to some constant value.

Using Linear Equations

You may like to read some of the things you can do with lines:

• Finding the Midpoint of a Line Segment
• Finding Parallel and Perpendicular Lines
• Finding the Equation of a Line from 2 Points

Drawing Linear Graphs Practice Questions

Click here for questions, click here for answers.

xy table, straight line

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• \frac{3}{4}x+\frac{5}{6}=5x-\frac{125}{3}
• \sqrt{2}x-\sqrt{3}=\sqrt{5}
• 7y+5-3y+1=2y+2
• \frac{x}{3}+\frac{x}{2}=10
• What is a linear equation?
• A linear equation represents a straight line on a coordinate plane. It can be written in the form: y = mx + b where m is the slope of the line and b is the y-intercept.
• How do you find the linear equation?
• To find the linear equation you need to know the slope and the y-intercept of the line. To find the slope use the formula m = (y2 - y1) / (x2 - x1) where (x1, y1) and (x2, y2) are two points on the line. The y-intercept is the point at which x=0.
• What are the 4 methods of solving linear equations?
• There are four common methods to solve a system of linear equations: Graphing, Substitution, Elimination and Matrix.
• How do you identify a linear equation?
• Here are a few ways to identify a linear equation: Look at the degree of the equation, a linear equation is a first-degree equation. Check if the equation has two variables. Graph the equation.
• What is the most basic linear equation?
• The most basic linear equation is a first-degree equation with one variable, usually written in the form of y = mx + b, where m is the slope of the line and b is the y-intercept.

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Course: Algebra 1   >   Unit 4

• Slope, x-intercept, y-intercept meaning in context
• Slope and intercept meaning in context
• Relating linear contexts to graph features
• Using slope and intercepts in context
• Slope and intercept meaning from a table
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