unit 4 linear equations homework 3 graphing linear equations

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Chapter 3: Graphing

3.4 Graphing Linear Equations

There are two common procedures that are used to draw the line represented by a linear equation. The first one is called the slope-intercept method and involves using the slope and intercept given in the equation.

If the equation is given in the form [latex]y = mx + b[/latex], then [latex]m[/latex] gives the rise over run value and the value [latex]b[/latex] gives the point where the line crosses the [latex]y[/latex]-axis, also known as the [latex]y[/latex]-intercept.

Example 3.4.1

Given the following equations, identify the slope and the [latex]y[/latex]-intercept.

• [latex]\begin{array}{lll} y = 2x - 3\hspace{0.14in} & \text{Slope }(m)=2\hspace{0.1in}&y\text{-intercept } (b)=-3 \end{array}[/latex]
• [latex]\begin{array}{lll} y = \dfrac{1}{2}x - 1\hspace{0.08in} & \text{Slope }(m)=\dfrac{1}{2}\hspace{0.1in}&y\text{-intercept } (b)=-1 \end{array}[/latex]
• [latex]\begin{array}{lll} y = -3x + 4 & \text{Slope }(m)=-3 &y\text{-intercept } (b)=4 \end{array}[/latex]
• [latex]\begin{array}{lll} y = \dfrac{2}{3}x\hspace{0.34in} & \text{Slope }(m)=\dfrac{2}{3}\hspace{0.1in} &y\text{-intercept } (b)=0 \end{array}[/latex]

When graphing a linear equation using the slope-intercept method, start by using the value given for the [latex]y[/latex]-intercept. After this point is marked, then identify other points using the slope.

This is shown in the following example.

Example 3.4.2

Graph the equation [latex]y = 2x - 3[/latex].

First, place a dot on the [latex]y[/latex]-intercept, [latex]y = -3[/latex], which is placed on the coordinate [latex](0, -3).[/latex]

Now, place the next dot using the slope of 2.

A slope of 2 means that the line rises 2 for every 1 across.

Simply, [latex]m = 2[/latex] is the same as [latex]m = \dfrac{2}{1}[/latex], where [latex]\Delta y = 2[/latex] and [latex]\Delta x = 1[/latex].

Placing these points on the graph becomes a simple counting exercise, which is done as follows:

Once several dots have been drawn, draw a line through them, like so:

Note that dots can also be drawn in the reverse of what has been drawn here.

Slope is 2 when rise over run is [latex]\dfrac{2}{1}[/latex] or [latex]\dfrac{-2}{-1}[/latex], which would be drawn as follows:

Example 3.4.3

Graph the equation [latex]y = \dfrac{2}{3}x[/latex].

First, place a dot on the [latex]y[/latex]-intercept, [latex](0, 0)[/latex].

Now, place the dots according to the slope, [latex]\dfrac{2}{3}[/latex].

This will generate the following set of dots on the graph. All that remains is to draw a line through the dots.

The second method of drawing lines represented by linear equations and functions is to identify the two intercepts of the linear equation. Specifically, find [latex]x[/latex] when [latex]y = 0[/latex] and find [latex]y[/latex] when [latex]x = 0[/latex].

Example 3.4.4

Graph the equation [latex]2x + y = 6[/latex].

To find the first coordinate, choose [latex]x = 0[/latex].

This yields:

[latex]\begin{array}{lllll} 2(0)&+&y&=&6 \\ &&y&=&6 \end{array}[/latex]

Coordinate is [latex](0, 6)[/latex].

Now choose [latex]y = 0[/latex].

[latex]\begin{array}{llrll} 2x&+&0&=&6 \\ &&2x&=&6 \\ &&x&=&\frac{6}{2} \text{ or } 3 \end{array}[/latex]

Coordinate is [latex](3, 0)[/latex].

Draw these coordinates on the graph and draw a line through them.

Example 3.4.5

Graph the equation [latex]x + 2y = 4[/latex].

[latex]\begin{array}{llrll} (0)&+&2y&=&4 \\ &&y&=&\frac{4}{2} \text{ or } 2 \end{array}[/latex]

Coordinate is [latex](0, 2)[/latex].

[latex]\begin{array}{llrll} x&+&2(0)&=&4 \\ &&x&=&4 \end{array}[/latex]

Coordinate is [latex](4, 0)[/latex].

Example 3.4.6

Graph the equation [latex]2x + y = 0[/latex].

[latex]\begin{array}{llrll} 2(0)&+&y&=&0 \\ &&y&=&0 \end{array}[/latex]

Coordinate is [latex](0, 0)[/latex].

Since the intercept is [latex](0, 0)[/latex], finding the other intercept yields the same coordinate. In this case, choose any value of convenience.

Choose [latex]x = 2[/latex].

[latex]\begin{array}{rlrlr} 2(2)&+&y&=&0 \\ 4&+&y&=&0 \\ -4&&&&-4 \\ \hline &&y&=&-4 \end{array}[/latex]

Coordinate is [latex](2, -4)[/latex].

For questions 1 to 10, sketch each linear equation using the slope-intercept method.

• [latex]y = -\dfrac{1}{4}x - 3[/latex]
• [latex]y = \dfrac{3}{2}x - 1[/latex]
• [latex]y = -\dfrac{5}{4}x - 4[/latex]
• [latex]y = -\dfrac{3}{5}x + 1[/latex]
• [latex]y = -\dfrac{4}{3}x + 2[/latex]
• [latex]y = \dfrac{5}{3}x + 4[/latex]
• [latex]y = \dfrac{3}{2}x - 5[/latex]
• [latex]y = -\dfrac{2}{3}x - 2[/latex]
• [latex]y = -\dfrac{4}{5}x - 3[/latex]
• [latex]y = \dfrac{1}{2}x[/latex]

For questions 11 to 20, sketch each linear equation using the [latex]x\text{-}[/latex] and [latex]y[/latex]-intercepts.

• [latex]x + 4y = -4[/latex]
• [latex]2x - y = 2[/latex]
• [latex]2x + y = 4[/latex]
• [latex]3x + 4y = 12[/latex]
• [latex]4x + 3y = -12[/latex]
• [latex]x + y = -5[/latex]
• [latex]3x + 2y = 6[/latex]
• [latex]x - y = -2[/latex]
• [latex]4x - y = -4[/latex]

For questions 21 to 28, sketch each linear equation using any method.

• [latex]y = -\dfrac{1}{2}x + 3[/latex]
• [latex]y = 2x - 1[/latex]
• [latex]y = -\dfrac{5}{4}x[/latex]
• [latex]y = -3x + 2[/latex]
• [latex]y = -\dfrac{3}{2}x + 1[/latex]
• [latex]y = \dfrac{1}{3}x - 3[/latex]
• [latex]y = \dfrac{3}{2}x + 2[/latex]
• [latex]y = 2x - 2[/latex]

For questions 29 to 40, reduce and sketch each linear equation using any method.

• [latex]y + 3 = -\dfrac{4}{5}x + 3[/latex]
• [latex]y - 4 = \dfrac{1}{2}x[/latex]
• [latex]x + 5y = -3 + 2y[/latex]
• [latex]3x - y = 4 + x - 2y[/latex]
• [latex]4x + 3y = 5 (x + y)[/latex]
• [latex]3x + 4y = 12 - 2y[/latex]
• [latex]2x - y = 2 - y \text{ (tricky)}[/latex]
• [latex]7x + 3y = 2(2x + 2y) + 6[/latex]
• [latex]x + y = -2x + 3[/latex]
• [latex]3x + 4y = 3y + 6[/latex]
• [latex]2(x + y) = -3(x + y) + 5[/latex]
• [latex]9x - y = 4x + 5[/latex]

Answer Key 3.4

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3.1 Graph Linear Equations in Two Variables

Learning objectives.

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

• Plot points in a rectangular coordinate system
• Graph a linear equation by plotting points
• Graph vertical and horizontal lines
• Find the x- and y-intercepts
• Graph a line using the intercepts

Be Prepared 3.1

Before you get started, take this readiness quiz.

Evaluate 5 x − 4 5 x − 4 when x = −1 . x = −1 . If you missed this problem, review Example 1.6 .

Be Prepared 3.2

Evaluate 3 x − 2 y 3 x − 2 y when x = 4 , y = −3 . x = 4 , y = −3 . If you missed this problem, review Example 1.21 .

Be Prepared 3.3

Solve for y : 8 − 3 y = 20 . 8 − 3 y = 20 . If you missed this problem, review Example 2.2 .

Plot Points on a Rectangular Coordinate System

Just like maps use a grid system to identify locations, a grid system is used in algebra to show a relationship between two variables in a rectangular coordinate system. The rectangular coordinate system is also called the xy -plane or the “coordinate plane.”

The rectangular coordinate system is formed by two intersecting number lines, one horizontal and one vertical. The horizontal number line is called the x -axis. The vertical number line is called the y -axis. These axes divide a plane into four regions, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise. See Figure 3.2 .

In the rectangular coordinate system, every point is represented by an ordered pair . The first number in the ordered pair is the x -coordinate of the point, and the second number is the y -coordinate of the point. The phrase “ordered pair” means that the order is important.

Ordered Pair

An ordered pair , ( x , y ) ( x , y ) gives the coordinates of a point in a rectangular coordinate system. The first number is the x -coordinate. The second number is the y -coordinate.

What is the ordered pair of the point where the axes cross? At that point both coordinates are zero, so its ordered pair is ( 0 , 0 ) . ( 0 , 0 ) . The point ( 0 , 0 ) ( 0 , 0 ) has a special name. It is called the origin .

The point ( 0 , 0 ) ( 0 , 0 ) is called the origin . It is the point where the x -axis and y -axis intersect.

We use the coordinates to locate a point on the xy -plane. Let’s plot the point ( 1 , 3 ) ( 1 , 3 ) as an example. First, locate 1 on the x -axis and lightly sketch a vertical line through x = 1 . x = 1 . Then, locate 3 on the y -axis and sketch a horizontal line through y = 3 . y = 3 . Now, find the point where these two lines meet—that is the point with coordinates ( 1 , 3 ) . ( 1 , 3 ) . See Figure 3.3 .

Notice that the vertical line through x = 1 x = 1 and the horizontal line through y = 3 y = 3 are not part of the graph. We just used them to help us locate the point ( 1 , 3 ) . ( 1 , 3 ) .

When one of the coordinate is zero, the point lies on one of the axes. In Figure 3.4 the point ( 0 , 4 ) ( 0 , 4 ) is on the y -axis and the point ( −2 , 0 ) ( −2 , 0 ) is on the x -axis.

Points on the Axes

Points with a y -coordinate equal to 0 are on the x -axis, and have coordinates ( a , 0 ) . ( a , 0 ) .

Points with an x -coordinate equal to 0 are on the y -axis, and have coordinates ( 0 , b ) . ( 0 , b ) .

Example 3.1

Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located:

ⓐ ( −5 , 4 ) ( −5 , 4 ) ⓑ ( −3 , −4 ) ( −3 , −4 ) ⓒ ( 2 , −3 ) ( 2 , −3 ) ⓓ ( 0 , −1 ) ( 0 , −1 ) ⓔ ( 3 , 5 2 ) . ( 3 , 5 2 ) .

The first number of the coordinate pair is the x -coordinate, and the second number is the y -coordinate. To plot each point, sketch a vertical line through the x -coordinate and a horizontal line through the y -coordinate. Their intersection is the point. ⓐ Since x = −5 , x = −5 , the point is to the left of the y -axis. Also, since y = 4 , y = 4 , the point is above the x -axis. The point ( −5 , 4 ) ( −5 , 4 ) is in Quadrant II. ⓑ Since x = −3 , x = −3 , the point is to the left of the y -axis. Also, since y = −4 , y = −4 , the point is below the x -axis. The point ( −3 , −4 ) ( −3 , −4 ) is in Quadrant III. ⓒ Since x = 2 , x = 2 , the point is to the right of the y -axis. Since y = −3 , y = −3 , the point is below the x -axis. The point ( 2 , −3 ) ( 2 , −3 ) is in Quadrant IV. ⓓ Since x = 0 , x = 0 , the point whose coordinates are ( 0 , −1 ) ( 0 , −1 ) is on the y -axis. ⓔ Since x = 3 , x = 3 , the point is to the right of the y -axis. Since y = 5 2 , y = 5 2 , the point is above the x -axis. (It may be helpful to write 5 2 5 2 as a mixed number or decimal.) The point ( 3 , 5 2 ) ( 3 , 5 2 ) is in Quadrant I.

Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located: ⓐ ( −2 , 1 ) ( −2 , 1 ) ⓑ ( −3 , −1 ) ( −3 , −1 ) ⓒ ( 4 , −4 ) ( 4 , −4 ) ⓓ ( −4 , 4 ) ( −4 , 4 ) ⓔ ( −4 , 3 2 ) ( −4 , 3 2 )

Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located: ⓐ ( −4 , 1 ) ( −4 , 1 ) ⓑ ( −2 , 3 ) ( −2 , 3 ) ⓒ ( 2 , −5 ) ( 2 , −5 ) ⓓ ( −2 , 5 ) ( −2 , 5 ) ⓔ ( −3 , 5 2 ) ( −3 , 5 2 )

The signs of the x -coordinate and y -coordinate affect the location of the points. You may have noticed some patterns as you graphed the points in the previous example. We can summarize sign patterns of the quadrants in this way:

Up to now, all the equations you have solved were equations with just one variable. In almost every case, when you solved the equation you got exactly one solution. But equations can have more than one variable. Equations with two variables may be of the form A x + B y = C . A x + B y = C . An equation of this form is called a linear equation in two variables.

Linear Equation

An equation of the form A x + B y = C , A x + B y = C , where A and B are not both zero, is called a linear equation in two variables.

Here is an example of a linear equation in two variables, x and y .

The equation y = −3 x + 5 y = −3 x + 5 is also a linear equation. But it does not appear to be in the form A x + B y = C . A x + B y = C . We can use the Addition Property of Equality and rewrite it in A x + B y = C A x + B y = C form.

By rewriting y = −3 x + 5 y = −3 x + 5 as 3 x + y = 5 , 3 x + y = 5 , we can easily see that it is a linear equation in two variables because it is of the form A x + B y = C . A x + B y = C . When an equation is in the form A x + B y = C , A x + B y = C , we say it is in standard form of a linear equation .

Standard Form of Linear Equation

A linear equation is in standard form when it is written A x + B y = C . A x + B y = C .

Most people prefer to have A , B , and C be integers and A ≥ 0 A ≥ 0 when writing a linear equation in standard form, although it is not strictly necessary.

Linear equations have infinitely many solutions. For every number that is substituted for x there is a corresponding y value. This pair of values is a solution to the linear equation and is represented by the ordered pair ( x , y ) . ( x , y ) . When we substitute these values of x and y into the equation, the result is a true statement, because the value on the left side is equal to the value on the right side.

Solution of a Linear Equation in Two Variables

An ordered pair ( x , y ) ( x , y ) is a solution of the linear equation A x + B y = C , A x + B y = C , if the equation is a true statement when the x - and y -values of the ordered pair are substituted into the equation.

Linear equations have infinitely many solutions. We can plot these solutions in the rectangular coordinate system. The points will line up perfectly in a straight line. We connect the points with a straight line to get the graph of the equation. We put arrows on the ends of each side of the line to indicate that the line continues in both directions.

A graph is a visual representation of all the solutions of the equation. It is an example of the saying, “A picture is worth a thousand words.” The line shows you all the solutions to that 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. Points not on the line are not solutions!

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 3.2

The graph of y = 2 x − 3 y = 2 x − 3 is shown.

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

ⓑ Plot the points ( 0 , −3 ) , ( 0 , −3 ) , ( 3 , 3 ) , ( 3 , 3 ) , ( 2 , −3 ) , ( 2 , −3 ) , and ( −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 that are solutions to y = 2 x − 3 y = 2 x − 3 are on the line, but the point that is not a solution is not on the line.

Use graph of y = 3 x − 1 . y = 3 x − 1 . For each ordered pair, decide:

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

A ( 0 , −1 ) ( 0 , −1 ) B ( 2 , 5 ) ( 2 , 5 )

A ( 3 , −1 ) ( 3 , −1 ) B ( −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 first method we will use is called plotting points, or the Point-Plotting Method. We find three points whose coordinates are solutions to the equation and then plot them in a rectangular coordinate system. By connecting these points in a line, we have the graph of the linear equation.

Example 3.3

How to graph a linear equation by plotting points.

Graph the equation y = 2 x + 1 y = 2 x + 1 by plotting points.

Graph the equation by plotting points: y = 2 x − 3 . y = 2 x − 3 .

Graph the equation by plotting points: y = −2 x + 4 . y = −2 x + 4 .

The steps to take when graphing a linear equation by plotting points are summarized here.

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 in 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 three 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 only plot 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. Look at the difference between these illustrations.

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

Example 3.4

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 , we will choose values of x carefully. We will use zero as one choice and multiples of 2 for the other choices. Why are multiples of two a good choice for values of x ? By choosing multiples of 2 the multiplication by 1 2 1 2 simplifies to a whole number

The points are shown in Table 3.1 .

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

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

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

Graph Vertical and Horizontal Lines

Some linear equations have only one variable. They may have just x and no y , or just y without an x . This changes how we make a table of values to get the points to plot.

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

So to make a table of values, write −3 −3 in for all the x -values. Then choose any values for y . Since x does not depend on y , you can choose any numbers you like. But to fit the points on our coordinate graph, we’ll use 1, 2, and 3 for the y -coordinates. See Table 3.2 .

Plot the points from the table and connect them with a straight line. Notice that we have graphed a vertical line .

What if the equation has y but no x ? Let’s graph the equation y = 4 . y = 4 . This time the y- value is a constant, so in this equation, y does not depend on x . Fill in 4 for all the y ’s in Table 3.3 and then choose any values for x . We’ll use 0, 2, and 4 for the x -coordinates.

In this figure, we have graphed a horizontal line passing through the y -axis at 4.

Vertical and Horizontal Lines

A vertical line is the graph of an equation of the form x = a . x = a .

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

A horizontal line is the graph of an equation of the form y = b . y = b .

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

Example 3.5

Graph: ⓐ x = 2 x = 2 ⓑ y = −1 . y = −1 .

ⓐ The equation has only one variable, x , and x is always equal to 2. We create a table where x is always 2 and then put in any values for y . The graph is a vertical line passing through the x -axis at 2.

ⓑ Similarly, the equation y = −1 y = −1 has only one variable, y . The value of y is constant. All the ordered pairs in the next table have the same y -coordinate. The graph is a horizontal line passing through the y -axis at −1 . −1 .

Graph the equations: ⓐ x = 5 x = 5 ⓑ y = −4 . y = −4 .

Try It 3.10

Graph the equations: ⓐ x = −2 x = −2 ⓑ y = 3 . y = 3 .

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 and y . The value of y depends on the value of x , so the y -coordinate changes according to the value of x . The equation y = 4 y = 4 has only one variable. The value of y is constant, it does not depend on the value of x , so the y -coordinate is always 4.

Notice, in the graph, the equation y = 4 x y = 4 x gives a slanted line, while y = 4 y = 4 gives a horizontal line.

Example 3.6

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

We notice that the first equation has the variable x , while the second does not. We make a table of points for each equation and then graph the lines. The two graphs are shown.

Try It 3.11

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

Try It 3.12

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

Find x - and y -intercepts

Every linear equation can be represented by a unique line that shows all the solutions of the equation. We have seen that when graphing a line by plotting points, you can use any three solutions to graph. This means that two people graphing the line might use different sets of three points.

At first glance, their two lines might not appear to be the same, since they would have different points labeled. But if all the work was done correctly, the lines should be exactly the same. One way to recognize that they are indeed the same line is to look at where the line crosses the x -axis and the y -axis. These points are called the intercepts of a line .

Intercepts of a Line

The points where a line crosses the x -axis and the y -axis are called the intercepts of the line .

Let’s look at the graphs of the lines.

First, notice where each of these lines crosses the x -axis. See Table 3.4 .

Now, let’s look at the points where these lines cross the y -axis.

Do you see a pattern?

For each line, the y -coordinate of the point where the line crosses the x -axis is zero. The point where the line crosses the x -axis has the form ( a , 0 ) ( a , 0 ) and is called the x-intercept of the line. The x -intercept occurs when y is zero.

In each line, the x - coordinate of the point where the line crosses the y -axis is zero. The point where the line crosses the y -axis has the form ( 0 , b ) ( 0 , b ) and is called the y-intercept of the line. The y -intercept occurs when x is zero.

x -intercept and y -intercept of a Line

The x -intercept is the point ( a , 0 ) ( a , 0 ) where the line crosses the x -axis.

The y -intercept is the point ( 0 , b ) ( 0 , b ) where the line crosses the y -axis.

Example 3.7

Find the x - and y -intercepts on each graph shown.

ⓐ The graph crosses the x -axis at the point ( 4 , 0 ) . ( 4 , 0 ) . The x- intercept is ( 4 , 0 ) . ( 4 , 0 ) . The graph crosses the y -axis at the point ( 0 , 2 ) . ( 0 , 2 ) . The y -intercept is ( 0 , 2 ) . ( 0 , 2 ) . ⓑ The graph crosses the x -axis at the point ( 2 , 0 ) . ( 2 , 0 ) . The x -intercept is ( 2 , 0 ) . ( 2 , 0 ) . The graph crosses the y -axis at the point ( 0 , −6 ) . ( 0 , −6 ) . The y -intercept is ( 0 , −6 ) . ( 0 , −6 ) . ⓒ The graph crosses the x -axis at the point ( −5 , 0 ) . ( −5 , 0 ) . The x -intercept is ( −5 , 0 ) . ( −5 , 0 ) . The graph crosses the y -axis at the point ( 0 , −5 ) . ( 0 , −5 ) . The y -intercept is ( 0 , −5 ) . ( 0 , −5 ) .

Try It 3.13

Find the x - and y -intercepts on the graph.

Try It 3.14

Recognizing that the x -intercept occurs when y is zero and that the y -intercept occurs when x is zero, gives us a method to find the intercepts of a line from its equation. To find the x -intercept, let y = 0 y = 0 and solve for x . To find the y -intercept, let x = 0 x = 0 and solve for y .

Find the x - and y -intercepts from the Equation of a Line

Use the equation of the line. To find:

• the x -intercept of the line, let y = 0 y = 0 and solve for x .
• the y -intercept of the line, let x = 0 x = 0 and solve for y .

Example 3.8

Find the intercepts of 2 x + y = 8 . 2 x + y = 8 .

We will let y = 0 y = 0 to find the x -intercept, and let x = 0 x = 0 to find the y -intercept. We will fill in a table, which reminds us of what we need to find.

The intercepts are the points ( 4 , 0 ) ( 4 , 0 ) and ( 0 , 8 ) ( 0 , 8 ) as shown in the table.

Try It 3.15

Find the intercepts: 3 x + y = 12 . 3 x + y = 12 .

Try It 3.16

Find the intercepts: x + 4 y = 8 . x + 4 y = 8 .

Graph a Line Using the Intercepts

To graph a linear equation by plotting points, you need to find three points whose coordinates are solutions to the equation. You can use the x- and y- intercepts as two of your three points. Find the intercepts, and then find a third point to ensure accuracy. Make sure the points line up—then draw the line. This method is often the quickest way to graph a line.

Example 3.9

How to graph a line using the intercepts.

Graph – x + 2 y = 6 – x + 2 y = 6 using the intercepts.

Try It 3.17

Graph using the intercepts: x – 2 y = 4 . x – 2 y = 4 .

Try It 3.18

Graph using the intercepts: – x + 3 y = 6 . – x + 3 y = 6 .

The steps to graph a linear equation using the intercepts are summarized here.

Graph a linear equation using the intercepts.

• Let y = 0 y = 0 and solve for x .
• Let x = 0 x = 0 and solve for y .
• Step 2. Find a third solution to the equation.
• Step 3. Plot the three points and check that they line up.
• Step 4. Draw the line.

Example 3.10

Graph 4 x − 3 y = 12 4 x − 3 y = 12 using the intercepts.

Find the intercepts and a third point.

We list the points in the table and show the graph.

Try It 3.19

Graph using the intercepts: 5 x − 2 y = 10 . 5 x − 2 y = 10 .

Try It 3.20

Graph using the intercepts: 3 x − 4 y = 12 . 3 x − 4 y = 12 .

When the line passes through the origin, the x -intercept and the y -intercept are the same point.

Example 3.11

Graph y = 5 x y = 5 x using the intercepts.

This line has only one intercept. It is the point ( 0 , 0 ) . ( 0 , 0 ) . To ensure accuracy, we need to plot three points. Since the x - and y -intercepts are the same point, we need two more points to graph the line.

The resulting three points are summarized in the table.

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

Try It 3.21

Graph using the intercepts: y = 4 x . y = 4 x .

Try It 3.22

Graph the intercepts: y = − x . y = − x .

Section 3.1 Exercises

Practice makes perfect.

Plot Points in a Rectangular Coordinate System

In the following exercises, plot each point in a rectangular coordinate system and identify the quadrant in which the point is located.

ⓐ ( −4 , 2 ) ( −4 , 2 ) ⓑ ( −1 , −2 ) ( −1 , −2 ) ⓒ ( 3 , −5 ) ( 3 , −5 ) ⓓ ( −3 , 0 ) ( −3 , 0 ) ⓔ ( 5 3 , 2 ) ( 5 3 , 2 )

ⓐ ( −2 , −3 ) ( −2 , −3 ) ⓑ ( 3 , −3 ) ( 3 , −3 ) ⓒ ( −4 , 1 ) ( −4 , 1 ) ⓓ ( 4 , −1 ) ( 4 , −1 ) ⓔ ( 3 2 , 1 ) ( 3 2 , 1 )

ⓐ ( 3 , −1 ) ( 3 , −1 ) ⓑ ( −3 , 1 ) ( −3 , 1 ) ⓒ ( −2 , 0 ) ( −2 , 0 ) ⓓ ( −4 , −3 ) ( −4 , −3 ) ⓔ ( 1 , 14 5 ) ( 1 , 14 5 )

ⓐ ( −1 , 1 ) ( −1 , 1 ) ⓑ ( −2 , −1 ) ( −2 , −1 ) ⓒ ( 2 , 0 ) ( 2 , 0 ) ⓓ ( 1 , −4 ) ( 1 , −4 ) ⓔ ( 3 , 7 2 ) ( 3 , 7 2 )

In the following exercises, for each ordered pair, decide

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

y = x + 2 ; y = x + 2 ; A: ( 0 , 2 ) ; ( 0 , 2 ) ; B: ( 1 , 2 ) ; ( 1 , 2 ) ; C: ( −1 , 1 ) ; ( −1 , 1 ) ; D: ( −3 , −1 ) . ( −3 , −1 ) .

y = x − 4 ; y = x − 4 ; A: ( 0 , −4 ) ; ( 0 , −4 ) ; B: ( 3 , −1 ) ; ( 3 , −1 ) ; C: ( 2 , 2 ) ; ( 2 , 2 ) ; D: ( 1 , −5 ) . ( 1 , −5 ) .

y = 1 2 x − 3 ; y = 1 2 x − 3 ; A: ( 0 , −3 ) ; ( 0 , −3 ) ; B: ( 2 , −2 ) ; ( 2 , −2 ) ; C: ( −2 , −4 ) ; ( −2 , −4 ) ; D: ( 4 , 1 ) ( 4 , 1 )

y = 1 3 x + 2 ; y = 1 3 x + 2 ; A: ( 0 , 2 ) ; ( 0 , 2 ) ; B: ( 3 , 3 ) ; ( 3 , 3 ) ; C: ( −3 , 2 ) ; ( −3 , 2 ) ; D: ( −6 , 0 ) . ( −6 , 0 ) .

In the following exercises, graph by plotting points.

y = x + 2 y = x + 2

y = x − 3 y = x − 3

y = 3 x − 1 y = 3 x − 1

y = −2 x + 2 y = −2 x + 2

y = − x − 3 y = − x − 3

y = − x − 2 y = − x − 2

y = 2 x y = 2 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

Graph Vertical and Horizontal lines

In the following exercises, graph each equation.

ⓐ x = 4 x = 4 ⓑ y = 3 y = 3

ⓐ x = 3 x = 3 ⓑ y = 1 y = 1

ⓐ x = −2 x = −2 ⓑ y = −5 y = −5

ⓐ x = −5 x = −5 ⓑ y = −2 y = −2

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

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

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

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

Find x- and y- Intercepts

In the following exercises, find the x - and y -intercepts on each graph.

In the following exercises, find the intercepts for each equation.

x − y = 5 x − y = 5

x − y = −4 x − y = −4

3 x + y = 6 3 x + y = 6

x − 2 y = 8 x − 2 y = 8

4 x − y = 8 4 x − y = 8

5 x − y = 5 5 x − y = 5

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

3 x − 2 y = 12 3 x − 2 y = 12

In the following exercises, graph using the intercepts.

− x + 4 y = 8 − x + 4 y = 8

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

x + y = −3 x + y = −3

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

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

3 x − y = −6 3 x − y = −6

2 x − y = −8 2 x − y = −8

2 x + 4 y = 12 2 x + 4 y = 12

3 x − 2 y = 6 3 x − 2 y = 6

2 x − 5 y = −20 2 x − 5 y = −20

3 x − 4 y = −12 3 x − 4 y = −12

y = 5 x y = 5 x

y = x y = x

y = − x y = − x

Mixed Practice

y = 3 2 x y = 3 2 x

y = − 2 3 x y = − 2 3 x

y = − 1 2 x + 3 y = − 1 2 x + 3

y = 1 4 x − 2 y = 1 4 x − 2

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

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

y = −1 y = −1

x = 3 x = 3

Writing Exercises

Explain how you would choose three 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?

Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation 4 x + y = −4 ? 4 x + y = −4 ? Why?

Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation y = 2 3 x − 2 ? y = 2 3 x − 2 ? Why?

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

ⓑ If most of your checks were:

Confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

With some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

No, I don’t get it. This is a warning sign and you must address it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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This Linear Equations Unit Bundle contains guided notes, homework assignments, three quizzes, study guide and a unit test that cover the following topics:

• Slope from a Graph

• Slope from Ordered Pairs (The Slope Formula)

• Linear Equations: Slope Intercept Form vs. Standard Form

• Graphing by Slope Intercept Form

• Writing Linear Equations Given a Graph

• Graphing by Intercepts

• Vertical vs. Horizontal Lines

• Writing Linear Equations given Point and Slope

• Writing Linear Equations given Two Points

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• Parallel vs. Perpendicular Lines

• Scatter Plots & Line of Best Fit

• Linear Regression

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(3) Google Slides Version of the PDF: The second page of the Video links document contains a link to a Google Slides version of the PDF. Each page is set to the background in Google Slides. There are no text boxes;  this is the PDF in Google Slides.  I am unable to do text boxes at this time but hope this saves you a step if you wish to use it in Slides instead! 

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Unit 6 – Exponents and Exponential Functions

Unit 7 – Polynomials & Factoring

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Unit 12 – Statistics

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4.2.2.1: Using Linear Equations

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Before we start practicing calculating all of the variables in a regression line equation, let's work a little with just the equation on it's own.

Regression Line Equations

As we just learned, linear regression for two variables is based on a linear equation:

\[\widehat{\mathrm{Y}}=\mathrm{a}+(\mathrm{b}*{X}) \nonumber \]

where \(a\) and \(b\) are constant numbers. What this means is that for every sample, the intercept (a) and the slope (b) will be the same for every score. The X score will change, and that affects Y (or predicted Y, or \(\widehat{\mathrm{Y}}\)). Some consider the predictor variable (X) as an IV and the outcome variable (Y) as the DV, but be careful that you aren't confusing prediction with causation!

We also just learned that the graph of a linear equation of the form \(\widehat{\mathrm{Y}}=\mathrm{a}+(\mathrm{b}*{X}) \nonumber \) is a straight line.

Exercise \(\PageIndex{1}\)

Is the following an example of a linear equation? Why or why not?

No, the graph is not a straight line; therefore, it is not a linear equation.

The minimum criterion for using a linear regression formula is that there be a linear relationship between the predictor and the criterion (outcome) variables.

Exercise \(\PageIndex{2}\)

What statistic shows us whether two variables are linearly related?

Pearson's r (correlation).

If two variables aren’t linearly related, then you can’t use linear regression to predict one from the other! The stronger the linear relationship (larger the Pearson’s correlation), the more accurate will be the predictions based on linear regression.

Slope and Y-Intercept of a Linear Equation

As we learned previously, \(b =\) slope and \(a = y\)-intercept. From algebra recall that the slope is a number that describes the steepness of a line, and the \(y\)-intercept is the \(y\) coordinate of the point \((0, a)\) where the line crosses the \(y\)-axis. Figure \(\PageIndex{2}\) shows t​hree possible graphs of the regression equation (\(y = a + b\text{x}\)). Panel (a) shows what the regression line looks like if the slope is positive (\(b > 0\)), the line slopes upward to the right. Panel (b) shows what the regression line looks like if there's no slope (\(b = 0\)); the line is horizontal. Finally, Panel (c) shows what the regression line looks like if the slope is negative (\(b < 0\)), the line slopes downward to the right.

I get it, everything has been pretty theoretical so far. So let's get practical. Let's try constructing the regression line equation even when you don't have the scores for either of the variables. First, we'll start by identifying the variables in the examples.

Example \(\PageIndex{1}\)

Svetlana tutors to make extra money for college. For each tutoring session, she charges a one-time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total amount of money Svetlana earns for each session she tutors is \(y = 25 + 15\text{x}\).

What are the predictor and criterion (outcome) variables? What is the \(y\)-intercept and what is the slope? Answer using complete sentences.

The predictor variable, \(x\), is the number of hours Svetlana tutors each session. The criterion (outcome) variable, \(y\), is the amount, in dollars, Svetlana earns for each session.

The \(y\)-intercept is the constant, the one time fee of $25 (\(a = 25\)). The slope is 15 (\(b = 15\)) because Svetlana earns $15 for each hour she tutors.

Although it doesn't make sense in these examples, the y-intercept (a) is determined when \(x = 0\). I guess with Svetlana, you could say that she gets $25 for any sessions that you miss or don't cancel ahead of time. But geometrically and mathematically, the y-intercept is based on when the predictor variable (x) has a value of zero.

Exercise \(\PageIndex{3}\)

Jamal repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Jamal earns per visit is \(y = 25 + 20\text{x}\).

The predictor variable, \(x\), is the number of hours Jamal works each visit. he criterion (outcome) variable, \(y\), is the amount, in dollars, Jamal earns for each visit.

The y -intercept is 25 (\(a = 25\)). At the start of a visit, Jamal charges a one-time fee of $25 (this is when \(x = 0\)). The slope is 20 (\(b = 20\)). For each visit, Jamal earns $20 for each hour he works.

Now, we can start constructing the regression line equations.

Example \(\PageIndex{2}\)

Alejandra's Word Processing Service (AWPS) does word processing. The rate for services is $32 per hour plus a $31.50 one-time charge. The total cost to a customer depends on the number of hours it takes to complete the job.

Find the equation that expresses the total cost in terms of the number of hours required to complete the job. For this example,

• \(x =\) the number of hours it takes to get the job done.
• \(y =\) the total cost to the customer.

The $31.50 is a fixed cost. This is the number that you add after calculating the rest, so it must be the intercept (a).

If it takes \(x\) hours to complete the job, then \((32)(x)\) is the cost of the word processing only.

Thus, the total cost is: \(y = 31.50 + 32\text{x}\)

Let's try another example of constructing the regression line equation.

Exercise \(\PageIndex{4}\)

Elektra's Extreme Sports hires hang-gliding instructors and pays them a fee of $50 per class as well as $20 per student in the class. The total cost Elektra pays depends on the number of students in a class. Find the equation that expresses the total cost in terms of the number of students in a class.

For this example,

• \(x =\) number of students in class
• \(y =\) the total cost

The constant is $50 per class, so that must be the intercept (a).

So $20 per student is the slope (b).

The resulting regression equation is: \(y = 50 + 20\text{x}\)

You can also use the regression equation to graph the line if you input scores from your X variable and your Y variable into the equation. Let's see what that might look like in Figure \(\PageIndex{3}\) for the equation: \(y = -1 + 2\text{x}\)

In the example in Figure \(\PageIndex{3}\), the intercept (a) is replaced by -1 and the slope (b) is replaced by 2 to get the regression equation (\(y = -1 + 2\text{x}\)). Right now, you are being provided these constants. Soon, you'll be calculating them yourself!

The most basic type of association is a linear association. This type of relationship can be defined algebraically by the equations used, numerically with actual or predicted data values, or graphically from a plotted. Algebraically, a linear equation typically takes the form \(y = mx + b\), where \(m\) and \(b\) are constants, \(x\) is the independent variable, \(y\) is the dependent variable. In a statistical context, a linear equation is written in the form \(y = a + bx\), where \(a\) and \(b\) are the constants. This form is used to help readers distinguish the statistical context from the algebraic context. In the equation \(y = a + b\text{x}\), the constant b that multiplies the \(x\) variable (\(b\) is called a coefficient) is called the slope. The constant a is called the \(y\)-intercept.

The slope of a line is a value that describes the rate of change between the two quantitative variables. The slope tells us how the criterion variable (\(y\)) changes for every one unit increase in the predictor (\(x\)) variable, on average. The \(y\)-intercept is used to describe the criterion variable when the predictor variable equals zero.

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3.1: Linear Equations

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Linear regression for two variables is based on a linear equation with one independent variable. The equation has the form:

\[y = a + b\text{x}\nonumber \]

where \(a\) and \(b\) are constant numbers. The variable \(x\) is the independent variable, and \(y\) is the dependent variable. Typically, you choose a value to substitute for the independent variable and then solve for the dependent variable.

Example \(\PageIndex{1}\)

The following examples are linear equations.

\[y = 3 + 2\text{x}\nonumber \]

\[y = -0.01 + 1.2\text{x}\nonumber \]

Exercise \(\PageIndex{1}\)

Is the following an example of a linear equation?

\[y = -0.125 - 3.5\text{x}\nonumber \]

The graph of a linear equation of the form \(y = a + b\text{x}\) is a straight line . Any line that is not vertical can be described by this equation.

Example \(\PageIndex{2}\)

Graph the equation \(y = -1 + 2\text{x}\).

Exercise \(\PageIndex{2}\)

Is the following an example of a linear equation? Why or why not?

No, the graph is not a straight line; therefore, it is not a linear equation.

Example \(\PageIndex{3}\)

Aaron's Word Processing Service (AWPS) does word processing. The rate for services is $32 per hour plus a $31.50 one-time charge. The total cost to a customer depends on the number of hours it takes to complete the job.

Find the equation that expresses the total cost in terms of the number of hours required to complete the job.

Let \(x =\) the number of hours it takes to get the job done.

Let \(y =\) the total cost to the customer.

The $31.50 is a fixed cost. If it takes \(x\) hours to complete the job, then \((32)(x)\) is the cost of the word processing only. The total cost is: \(y = 31.50 + 32\text{x}\)

Exercise \(\PageIndex{3}\)

Emma’s Extreme Sports hires hang-gliding instructors and pays them a fee of $50 per class as well as $20 per student in the class. The total cost Emma pays depends on the number of students in a class. Find the equation that expresses the total cost in terms of the number of students in a class.

\(y = 50 + 20\text{x}\)

Slope and Y-Intercept of a Linear Equation

For the linear equation \(y = a + b\text{x}\), \(b =\) slope and \(a = y\)-intercept. From algebra recall that the slope is a number that describes the steepness of a line, and the \(y\)-intercept is the \(y\) coordinate of the point \((0, a)\) where the line crosses the \(y\)-axis.

Example \(\PageIndex{4}\)

Svetlana tutors to make extra money for college. For each tutoring session, she charges a one-time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total amount of money Svetlana earns for each session she tutors is \(y = 25 + 15\text{x}\).

What are the independent and dependent variables? What is the \(y\)-intercept and what is the slope? Interpret them using complete sentences.

The independent variable (\(x\)) is the number of hours Svetlana tutors each session. The dependent variable (\(y\)) is the amount, in dollars, Svetlana earns for each session.

The \(y\)-intercept is 25 (\(a = 25\)). At the start of the tutoring session, Svetlana charges a one-time fee of $25 (this is when \(x = 0\)). The slope is 15 (\(b = 15\)). For each session, Svetlana earns $15 for each hour she tutors.

Exercise \(\PageIndex{4}\)

Ethan repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Ethan earns per visit is \(y = 25 + 20\text{x}\).

The independent variable (\(x\)) is the number of hours Ethan works each visit. The dependent variable (\(y\)) is the amount, in dollars, Ethan earns for each visit.

The y -intercept is 25 (\(a = 25\)). At the start of a visit, Ethan charges a one-time fee of $25 (this is when \(x = 0\)). The slope is 20 (\(b = 20\)). For each visit, Ethan earns $20 for each hour he works.

The most basic type of association is a linear association. This type of relationship can be defined algebraically by the equations used, numerically with actual or predicted data values, or graphically from a plotted curve. (Lines are classified as straight curves.) Algebraically, a linear equation typically takes the form \(y = mx + b\), where \(m\) and \(b\) are constants, \(x\) is the independent variable, \(y\) is the dependent variable. In a statistical context, a linear equation is written in the form \(y = a + bx\), where \(a\) and \(b\) are the constants. This form is used to help readers distinguish the statistical context from the algebraic context. In the equation \(y = a + b\text{x}\), the constant b that multiplies the \(x\) variable (\(b\) is called a coefficient) is called the slope . The constant a is called the \(y\)-intercept.

The slope of a line is a value that describes the rate of change between the independent and dependent variables. The slope tells us how the dependent variable (\(y\)) changes for every one unit increase in the independent (\(x\)) variable, on average. The \(y\) -intercept is used to describe the dependent variable when the independent variable equals zero.

Formula Review

\(y = a + b\text{x}\) where a is the \(y\)-intercept and \(b\) is the slope. The variable \(x\) is the independent variable and \(y\) is the dependent variable.