Learning Objective(s)

·Use coordinate pairs to graph linear relationships.

·Graph a linear equation using x- and y- intercepts.

·Determine whether an ordered pair is a solution of an equation.

·Solve application problems involving graphs of linear equations.


Introduction

Graphing ordered pairs is only the beginning of the story. Once you know how to place points on a grid, you can use them to make sense of all kinds of mathematical relationships.


You can use a coordinate plane to plot points and to map various relationships, such as the relationship between an object's distance and the elapsed time. Many mathematical relationships are linear relationships. Let's look at what a linear relationship is.


Linear Relationships

A linear relationship is a relationship between variables such that when plotted on a coordinate plane, the points lie on a line. Let's start by looking at a series of points in Quadrant I on the coordinate plane.


Look at the five ordered pairs (and their x- and y-coordinates) below. Do you see any pattern to the location of the points? If this pattern continued, what other points could be on the line?



You probably identified that if this pattern continued the next ordered pair would be at (5, 10). This makes sense because the point (5, 10) "lines up” with the other points in the series--it is literally on the same line as the others. Applying the same logic, you may identify that the ordered pairs (6, 12) and (7, 14) would also belong if this coordinate plane were larger; they, too, will line up with the other points.


These series of points can also be represented in a table. In the table below, the x- and y-coordinates of each ordered pair on the graph is recorded.


x-coordinate

y-coordinate

0

0

1

2

2

4

3

6

4

8

5

10

6

12

7

14


Notice that each y-coordinate is twice the corresponding x-value. All of these x- and y-values follow the same pattern, and, when placed on a coordinate plane, they all line up.


Once you know the pattern that relates the x- and y-values, you can find a y-value for any x-value that lies on the line. So if the rule of this pattern is that each y-coordinate is twice the corresponding x-value, then the ordered pairs (1.5, 3), (2.5, 5), and (3.5, 7) should all appear on the line too, correct? Look to see what happens.



If you were to keep adding ordered pairs (xy) where the y-value was twice the x-value, you would end up with a graph like this.



Look at how all of the points blend together to create a line. You can think of a line, then, as a collection of an infinite number of individual points that share the same mathematical relationship. In this case, the relationship is that the y-value is twice the x-value.


There are multiple ways to represent a linear relationship--a table, a linear graph, and there is also a linear equation. A linear equation is an equation with two variables whose ordered pairs graph as a straight line.


There are several ways to create a graph from a linear equation. One way is to create a table of values for x and y, and then plot these ordered pairs on the coordinate plane. Two points are enough to determine a line. However, it's always a good idea to plot more than two points to avoid possible errors.


Then you draw a line through the points to show all of the points that are on the line. The arrows at each end of the graph indicate that the line continues endlessly in both directions. Every point on this line is a solution to the linear equation.


Example

Problem

Graph the linear equation 1.5x.

x values

1.5x

y values

0

1.5(0)

0

2

1.5(2)

3

4

1.5(4)

6

6

1.5(6)

9

Evaluate 1.5x for different values of x, and create a table of corresponding x and yvalues.

Since the coefficient of xis 1.5, it is convenient to choose multiples of 2 forx. This ensures that y is an integer, and makes the line easier to graph.

(0, 0)

(2, 3)

(4, 6)

(6, 9)

Convert the table to ordered pairs. Then plot the ordered pairs (shown below).

Draw a line through the points to indicate all of the points on the line.

Answer


Example

Problem

Graph the linear equation = 2x + 3.

x values

2x + 3

y values

0

2(0) + 3

3

1

2(1) + 3

5

2

2(2) + 3

7

3

2(3) + 3

9

Evaluate = 2+ 3 for different values of x, and create a table of corresponding x and yvalues.

(0, 3)

(1, 5)

(2, 7)

(3, 9)

Convert the table to ordered pairs.

Plot the ordered pairs (shown below).

Draw a line through the points to indicate all of the points on the line.

Answer


The linear equations graphed above were solved for y. If the equation is not stated in terms of y, it is best to first solve the equation for y. If there is no y in the equation, then solve the equation for x.


Example

Problem

Graph the linear equation y + 3x = 5.

Solve. y + 3x = 5 for y

y + 3x - 3x = 5 - 3x

                 y = 5 - 3x

values

5 - 3x

y values

0

5 - 3(0)

5

1

5 - 3(1)

2

2

5 - 3(2)

1

3

5 - 3(3)

4

Evaluate y = 5 - 3x for different values of x, and create a table of corresponding x and yvalues.

(0, 5)

(1, 2)

(2, 1)

(3, 4)

Plot the ordered pairs (shown below).

Draw a line through the points to indicate all of the points on the line.

Answer


The linear equations x = 2 and y = 3 only have one variable in each of them. However, because these are linear equations, then they will graph on a coordinate plane just as the linear equations above do. Just think of the equation = 2 as = 0y + 2 and think of y = 3 as y = 0x - 3.


Example

Problem

Graph y = 3.

x values

0- 3

y values

0

0(0) - 3

3

1

0(1) - 3

3

2

0(2) - 3

3

3

0(3) - 3

3

Write y = 3 as

y = 0x - 3, and evaluate ywhen x has several values. Or just realize thaty = 3 means every yvalue will be 3, no matter what x is.

(0, 3)

(1, 3)

(2, 3)

(3, 3)

Plot the ordered pairs (shown below).

Draw a line through the points to indicate all of the points on the line.

Answer


Notice that y = 3 graphs as a horizontal line.


Which table of values could be generated by the equation 2y - 5x = 10?

A)

x

y

5

0

7.5

1

10

2

B)

x

y

1

5

2

6

3

7

C)

x

y

1

15

2

20

3

25

D)

x

y

0

5

1

7.5

2

10



x- and y- Intercepts

The intercepts of a line are the points where the line intercepts, or crosses, the horizontal and vertical axes. To help you remember what "intercept” means, think about the word "intersect”. The two words sound alike and in this case mean the same thing.


The straight line on the graph below intercepts the two coordinate axes. The point where the line crosses the x-axis is called the x-intercept. The y-intercept is the point where the line crosses the y-axis.



The x-intercept above is the point (2, 0). The y-intercept above is the point (0, 2).

Notice that the y-intercept always occurs where x = 0, and the x-intercept always occurs where y = 0.

To find the x- and y-intercepts of a linear equation, you can substitute 0 for y and for x respectively.

For example, the linear equation 3y + 2x = 6 has an x intercept when y = 0, so 3(0) + 2x = 6.

   2x = 6

     x = 3

The x-intercept is (3, 0).


Likewise the y-intercept occurs when x = 0.

            3y + 2(0) = 6

                   3y     = 6

                         y = 2

The y-intercept is (0, 2).


What is the y-intercept of a line with the equation y = 5x - 4?

A) 

B) (4, 0)

C) (0, 4)

D) (5, 4)


Using Intercepts to Graph Linear Equations

You can use intercepts to graph linear equations. Once you have found the two intercepts, draw a line through them.


Let's do it with the equation 3y + 2x = 6. You figured out that the intercepts of the line this equation represents are (0, 2) and (3, 0). That's all you need to know.



Example

Problem

Graph 5y + 3x = 30.

5y + 3x = 30

When an equation is in Ax +By = C form, you can easily find the x- and y-intercepts and then graph.

5y + 3x = 30

5y + 3(0) = 30

5y + 0 = 30

5y = 30

y = 6

y-intercept: (0, 6)

To find the y-intercept, set

= 0 and solve for y.

5y + 3x = 30

5(0) + 3x = 30

0 + 3x = 30

3x = 30

x = 10

x-intercept: (10, 0)

To find the x-intercept, set

= 0 and solve for x.

Answer


Ordered Pairs as Solutions

So far, you have considered the following ideas about lines: a line is a visual representation of a linear equation, and the line itself is made up of an infinite number of points (or ordered pairs). The picture below shows the line of the linear equation y = 2x - 5 with some of the specific points on the line.



Every point on the line is a solution to the equation y = 2x - 5. You can try any of the points that are labeled like the ordered pair, (1, 3).

            y = 2x - 5

3 = 2(1) - 5

3 = 2 - 5

3 = 3                       This is true.


You can also try ANY of the other points on the line. Every point on the line is a solution to the equation y = 2x - 5. All this means is that determining whether an ordered pair is a solution of an equation is pretty straightforward. If the ordered pair is on the line created by the linear equation, then it is a solution to the equation. But if the ordered pair is not on the line--no matter how close it may look--then it is not a solution to the equation.


Identifying Solutions

To find out whether an ordered pair is a solution of a linear equation, you can do the following:

oGraph the linear equation, and graph the ordered pair. If the ordered pair appears to be on the graph of a line, then it is a possible solution of the linear equation. If the ordered pair does not lie on the graph of a line, then it is not a solution.

oSubstitute the (xy) values into the equation. If the equation yields a true statement, then the ordered pair is a solution of the linear equation. If the ordered pair does not yield a true statement then it is not a solution.


Example

Problem

Determine whether (2, 4) is a solution to the equation 4y+ 5x = 3.

4y + 5x = 3

4(4) + 5(2) = 3

For this problem, you will use the substitution method. Substitute =

2 and = 4 into the equation.

16 + (10) = 3

6 = 3

Evaluate.

The statement is not true, so (2, 4) is not a solution to the equation 4+ 5= 3.

Answer

(−2, 4) is not a solution to the equation 4+ 5= 3.


Application Problems

Linear equations can be used to model a number of real-life problems, like how much money you make over time, or the distance that a bicyclist will travel given a steady rate of pedaling. Graphing these relationships on a coordinate plane can often help you think about (and find solutions to) the problem.


Consider this problem.


Eilene drives 20 miles from her house to the train station, and then boards a non-stop train to New York. The train travels at 55 miles per hour for the entire journey. After 2 hours on the train, how far is she from her house? After about how many hours on the train will Eilene be 300 miles from her house?


Let x = the time (in hours) that Eilene traveled on the train.

You know that the rate is 55 mph on the train.

So, the distance on the train is d = rt, or 55x

She already traveled 20 miles, so her total distance is 55x + 20.

Let y be the total distance, so y = 55x + 20.

By substituting in some values for x, you can find out the corresponding values for y.


x, Time

(hours)

y, Distance from House (miles)

0

20

1

75

2

130

3

185

4

240


Once you have calculated a few ordered pairs, you can use a graph to model the situation. (Notice that this graph does not go through the origin--when Eilene boards the train, she is already 20 miles away from her home since she lives 20 miles from the train station!)

Also, keep the graph in Quadrant I, as you are constrained to positive distance and positive time.



The first part of the question can be solved by looking at the table of values or at the graph. When x = 2, y = 130; this means that Eilene will be 130 miles away from home after 2 hours on the train.


Now think about the second part of the question: after how many hours on the train will Eilene be 300 miles from her house? Look to see what the x-coordinate is when y = 300. It is a little more than 5, so she will be 300 miles away after about 5 hours (and a few minutes!) on the train. Problem solved.


Here's another problem.

Morgan is buying a laptop for $1,080 to use for school. Morgan is going to use the computer store's finance plan to make this purchase--she'll pay $45 per month for 24 months. She wants to know how much she will still owe after each month of the plan.


Morgan can keep track of her debt by making a graph. The x-axis will be the number of months and the y-axis will represent the amount of money she still owes on the computer.


Morgan knows two points in her pay-off schedule: the day she buys the computer she'll be at 0 months and $1,080 owed, and the day she pays it off completely, she'll be at 24 months and $0 owed. With these two points, she can draw a line, running from the y-intercept at (0, 1080) to the x-intercept at (24, 0).



Morgan can now use this graph to figure out how much money she still owes after any number of months. For example, after 6 months, it looks like Morgan owes $800. (And if she calculated it exactly, she would find that $810 remained on her balance.)


Summary

When graphed on a coordinate plane, a linear relationship will be a line. Examples of linear relationships are linear equations such as y = x+ 3, 2x - 5y = 8, and x = 4. In order to graph the equation, you can find sets of ordered pairs to plot by substituting numbers for one variable and finding the other. Usually it will be easiest to find order pairs if you solve the equation for y first, or if there is no y in the equation then solve the equation for x. You can also graph the equation by using the x- and y-intercepts to find two points to plot. In either case, you draw a line to indicate that all of the points on the line are ordered pairs that satisfy the linear equation. While two points can determine a line, it is always a good idea to check at least one other point.


Permissions

This reading is taken from the Developmental Math Open Program created by The NROC Project. It is available under a Creative Commons license. 

Modifié le: jeudi 25 août 2016, 09:26