3.1 Graph Linear Equations in Two Variables

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.

This figure shows a square grid. A horizontal number line in the middle is labeled x. A vertical number line in the middle is labeled y. The number lines intersect at zero and together divide the square grid into 4 equally sized smaller squares. The square in the top right is labeled I. The square in the top left is labeled II. The square in the bottom left is labeled III. The square in the bottom right is labeled IV. — **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.

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

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

This figure shows a point plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The point (1, 3) is labeled. A dashed vertical line goes through the point and intersects the x-axis at xplus1. A dashed horizontal line goes through the point and intersects the y-axis at yplus3. — **Figure 3.3**

Notice that the vertical line through $x equals 1$ and the horizontal line through $y equals 3$ are not part of the graph. We just used them to help us locate the point $open parenthesis 1, 3 close parenthesis.$

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

This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The point (negative 2, 0) is labeled and lies on the x-axis. The point (0, 4) is labeled and lies on the y-axis. — **Figure 3.4**

Example 1

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

ⓐ $open parenthesis −5, 4 close parenthesis$ ⓑ $open parenthesis −3, −4 close parenthesis$ ⓒ $open parenthesis 2, −3 close parenthesis$ ⓓ $open parenthesis 0, −1 close parenthesis$ ⓔ $(3, \frac{5}{2}) .$

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 $x equals −5,$ the point is to the left of the y-axis. Also, since $y equals 4,$ the point is above the x-axis. The point $open parenthesis −5, 4 close parenthesis$ is in Quadrant II.
ⓑ Since $x equals −3,$ the point is to the left of the y-axis. Also, since $y equals −4,$ the point is below the x-axis. The point $open parenthesis −3, −4 close parenthesis$ is in Quadrant III.
ⓒ Since $x equals 2,$ the point is to the right of the y-axis. Since $y equals −3,$ the point is below the x-axis. The point $open parenthesis 2, −3 close parenthesis$ is in Quadrant IV.
ⓓ Since $x equals 0,$ the point whose coordinates are $open parenthesis 0, −1 close parenthesis$ is on the y-axis.
ⓔ Since $x equals 3,$ the point is to the right of the y-axis. Since $y = \frac{5}{2},$ the point is above the x-axis. (It may be helpful to write $\frac{5}{2}$ as a mixed number or decimal.) The point $(3, \frac{5}{2})$ is in Quadrant I.

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 $A x + B y = C .$ An equation of this form is called a linear equation in two variables.

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

This figure shows the equation A x plus B y plus C. Below this is the equation x plus 4 y plus 8. Below this are the equations A plus 1, B plus 4, C plus 8. B and 4 are the same color in all the equations. C and 8 are the same color in all the equations.

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

$\begin{matrix} y & = & −3 x + 5 \\ Add to both sides. & y + 3 x & = & −3 x + 5 + 3 x \\ Simplify. & y + 3 x & = & 5 \\ Use the Commutative Property to put it in \\ A x + B y = C form. & 3 x + y & = & 5 \end{matrix}$

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

Most people prefer to have A, B, and C be integers and $A greater than or equal to 0$ 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 $(x, y) .$ 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.

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 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.

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

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.

The figure shows two images. In the first image there are three points with a straight line going through all three. In the second image there are three points that do not all lie on a straight line.

When an equation includes a fraction as the coefficient of $x$ , 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 4

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

Find three points that are solutions to the equation. Since this equation has the fraction $\frac{1}{2}$ as a coefficient of x, we will choose values of x carefully. We will use zero as one choice and multiples of 2 for the other choices. Why are multiples of two a good choice for values of x? By choosing multiples of 2 the multiplication by $\frac{1}{2}$ simplifies to a whole number

The points are shown in Table 3.1.

Table 3.1

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

$(x, y)$

open parenthesis 0, 3 close parenthesis

open parenthesis 2, 4 close parenthesis

open parenthesis 4, 5 close parenthesis

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

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

So to make a table of values, write $−3$ 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.

Table 3.2

$x equals −3$

$(x, y)$

−3

open parenthesis −3, 1 close parenthesis

−3

open parenthesis −3, 2 close parenthesis

−3

open parenthesis −3, 3 close parenthesis

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

The figure shows the graph of a straight vertical line on the x y-coordinate plane. The x and y axes run from negative 7 to 7. The points (negative 3, 1), (negative 3, 2), and (negative 3, 3) are plotted. The line goes through the three points and has arrows on both ends. The line is labeled x plus negative 3.

What if the equation has y but no x? Let’s graph the equation $y equals 4.$ 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.

Table 3.3

$y equals 4$

$(x, y)$

open parenthesis 0, 4 close parenthesis

open parenthesis 2, 4 close parenthesis

open parenthesis 4, 4 close parenthesis

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

The figure shows the graph of a straight horizontal line on the x y-coordinate plane. The x and y axes run from negative 7 to 7. The points (0, 4), (2, 4), and (4, 4) are plotted. The line goes through the three points and has arrows on both ends. The line is labeled y plus 4.

What is the difference between the equations $y equals 4 x$ and $y equals 4 ?$

The equation $y equals 4 x$ 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 $y equals 4$ 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.

This figure has two tables. The first table has 5 rows and 3 columns. The first row is a title row with the equation y plus 4 x. The second row is a header row with the headers x, y, and (x, y). The third row has the numbers 0, 0, and (0, 0). The fourth row has the numbers 1, 4, and (1, 4). The fifth row has the numbers 2, 8, and (2, 8). The second table has 5 rows and 3 columns. The first row is a title row with the equation y plus 4. The second row is a header row with the headers x, y, and (x, y). The third row has the numbers 0, 4, and (0, 4). The fourth row has the numbers 1, 4, and (1, 4). The fifth row has the numbers 2, 4, and (2, 4).

The figure shows the graphs of a straight horizontal line and a straight slanted line on the same x y-coordinate plane. The x and y axes run from negative 7 to 7. The horizontal line goes through the points (0, 4), (1, 4), and (2,4) and is labeled y plus 4. The slanted line goes through the points (0, 0), (1, 4), and (2, 8) and is labeled y plus 4 x.

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

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.

Let’s look at the graphs of the lines.

The figure shows four graphs of different equations. In example a the graph of 2 x plus y plus 6 is graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The points (0, 6) and (3, 0) are plotted and labeled. A straight line goes through both points and has arrows on both ends. In example b the graph of 3 x minus 4 y plus 12 is graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The points (0, negative 3) and (4, 0) are plotted and labeled. A straight line goes through both points and has arrows on both ends. In example c the graph of x minus y plus 5 is graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The points (0, negative 5) and (5, 0) are plotted and labeled. A straight line goes through both points and has arrows on both ends. In example d the graph of y plus negative 2 x is graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The point (0, 0) is plotted and labeled. A straight line goes through this point and the points (negative 1, 2) and (1, negative 2) and has arrows on both ends.

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.

**Table 3.4**
Figure	The line crosses the x-axis at:	Ordered pair for this point	The line crosses the y-axis at:	Ordered pair for this point
Figure (a)	3	$open parenthesis 3, 0 close parenthesis$	6	$open parenthesis 0, 6 close parenthesis$
Figure (b)	4	$open parenthesis 4, 0 close parenthesis$	$−3$	$open parenthesis 0, −3 close parenthesis$
Figure (c)	5	$open parenthesis 5, 0 close parenthesis$	$−5$	$open parenthesis 0, 5 close parenthesis$
Figure (d)	0	$open parenthesis 0, 0 close parenthesis$	0	$open parenthesis 0, 0 close parenthesis$
General Figure	a	$open parenthesis a, 0 close parenthesis$	b	$open parenthesis 0, b close parenthesis$

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 $open parenthesis a, 0 close parenthesis$ 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 $open parenthesis 0, b close parenthesis$ and is called the y-intercept of the line. The y-intercept occurs when x is zero.

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 $y equals 0$ and solve for x. To find the y-intercept, let $x equals 0$ and solve for y.

Example 8

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

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

To find the x-intercept, let $y equals 0.$

Let $y equals 0.$
Simplify.

The x-intercept is:	$open parenthesis 4, 0 close parenthesis$
To find the y-intercept, let $x equals 0.$

Let $x equals 0.$
Simplify.

The y-intercept is:	$open parenthesis 0, 8 close parenthesis$

The intercepts are the points $open parenthesis 4, 0 close parenthesis$ and $open parenthesis 0, 8 close parenthesis$ as shown in the table.

$2 x + y = 8$

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.

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

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

Key Concepts

Points on the Axes
- Points with a y-coordinate equal to 0 are on the x-axis, and have coordinates $open parenthesis a, 0 close parenthesis.$
- Points with an x-coordinate equal to $0$ are on the y-axis, and have coordinates $open parenthesis 0, b close parenthesis.$
Quadrant
$\begin{matrix} Quadrant I & Quadrant II & Quadrant III & Quadrant IV \\ (x, y) & (x, y) & (x, y) & (x, y) \\ (+, +) & (-, +) & (-, -) & (+, -) \end{matrix}$
Graph of a Linear Equation: The graph of a linear equation $A x + B y = C$ is a straight line.
Every point on the line is a solution of the equation.
Every solution of this equation is a point on this line.
How to graph a linear equation by plotting points.
1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
2. Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
3. Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.
x-intercept and y-intercept of a Line
- The x-intercept is the point $open parenthesis a, 0 close parenthesis$ where the line crosses the x-axis.
- The y-intercept is the point $open parenthesis 0, b close parenthesis$ where the line crosses the y-axis.
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 $y equals 0$ and solve for x.
  the y-intercept of the line, let $x equals 0$ and solve for y.
How to graph a linear equation using the intercepts.
1. Find the x- and y-intercepts of the line.
  Let $y equals 0$ and solve for x.
  Let $x equals 0$ and solve for y.
2. Find a third solution to the equation.
3. Plot the three points and check that they line up.
4. Draw the line

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.

ⓐ $open parenthesis −4, 2 close parenthesis$ ⓑ $open parenthesis −1, −2 close parenthesis$ ⓒ $open parenthesis 3, −5 close parenthesis$ ⓓ $open parenthesis −3, 0 close parenthesis$ ⓔ $(\frac{5}{3}, 2)$

This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The point labeled a is 4 units to the left of the origin and 2 units above the origin and is located in quadrant II. The point labeled b is 1 unit to the left of the origin and 2 units below the origin and is located in quadrant III. The point labeled c is 3 units to the right of the origin and 5 units below the origin and is located in quadrant IV. The point labeled d is 3 units to the left of the origin and 5 units above the origin and is located in quadrant II. The point labeled e is 1 and a half units to the right of the origin and 2 units above the origin and is located in quadrant I.

ⓐ $open parenthesis −2, −3 close parenthesis$ ⓑ $open parenthesis 3, −3 close parenthesis$ ⓒ $open parenthesis −4, 1 close parenthesis$ ⓓ $open parenthesis 4, −1 close parenthesis$ ⓔ $(\frac{3}{2}, 1)$

ⓐ $open parenthesis 3, −1 close parenthesis$ ⓑ $open parenthesis −3, 1 close parenthesis$ ⓒ $open parenthesis −2, 0 close parenthesis$ ⓓ $open parenthesis −4, −3 close parenthesis$ ⓔ $(1, \frac{14}{5})$

This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The point labeled a is 3 units to the right of the origin and 1 unit below the origin and is located in quadrant IV. The point labeled b is 3 units to the left of the origin and 1 unit above the origin and is located in quadrant II. The point labeled c is 2 units to the left of the origin and 2 units above the origin and is located in quadrant II. The point labeled d is 4 units to the left of the origin and 3 units below the origin and is located in quadrant III. The point labeled e is 1 unit to the right of the origin and 3 and 4 fifths units above the origin and is located in quadrant I.

ⓐ $open parenthesis −1, 1 close parenthesis$ ⓑ $open parenthesis −2, −1 close parenthesis$ ⓒ $open parenthesis 2, 0 close parenthesis$ ⓓ $open parenthesis 1, −4 close parenthesis$ ⓔ $(3, \frac{7}{2})$

In the following exercises, for each ordered pair, decide

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

$y = x + 2;$
A: $open parenthesis 0, 2 close parenthesis;$ B: $open parenthesis 1, 2 close parenthesis;$ C: $open parenthesis −1, 1 close parenthesis;$ D: $open parenthesis −3, −1 close parenthesis.$

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 3, negative 1), (negative 2, 0), (negative 1, 1), (0, 2), (1, 3), (2, 4), and (3, 5).

ⓐ A: yes, B: no, C: yes, D: yes ⓑ A: yes, B: no, C: yes, D: yes

$y = x - 4;$
A: $open parenthesis 0, −4 close parenthesis;$ B: $open parenthesis 3, −1 close parenthesis;$ C: $open parenthesis 2, 2 close parenthesis;$ D: $open parenthesis 1, −5 close parenthesis.$

$y = \frac{1}{2} x - 3;$
A: $open parenthesis 0, −3 close parenthesis;$ B: $open parenthesis 2, −2 close parenthesis;$ C: $open parenthesis −2, −4 close parenthesis;$ D: $open parenthesis 4, 1 close parenthesis$

ⓐ A: yes, B: yes, C: yes, D: no ⓑ A: yes, B: yes, C: yes, D: no

$y = \frac{1}{3} x + 2;$
A: $open parenthesis 0, 2 close parenthesis;$ B: $open parenthesis 3, 3 close parenthesis;$ C: $open parenthesis −3, 2 close parenthesis;$ D: $open parenthesis −6, 0 close parenthesis.$

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

$y = x + 2$

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 3, negative 1), (negative 2, 0), (negative 1, 1), (0, 2), (1, 3), (2, 4), and (3, 5).

$y = x - 3$

$y = 3 x - 1$

$y = −2 x + 2$

$y = − x - 3$

$y = − x - 2$

$y equals 2 x$

$y equals −2 x$

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

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

$y = \frac{4}{3} x - 5$

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

$y = - \frac{2}{5} x + 1$

$y = - \frac{4}{5} x - 1$

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

$y = - \frac{5}{3} x + 4$

Graph Vertical and Horizontal lines

In the following exercises, graph each equation.

ⓐ $x equals 4$ ⓑ $y equals 3$

ⓐ

This figure shows a vertical straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (4, negative 1), (4, 0), and (4, 1).

ⓑ

This figure shows a horizontal straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 12 to 12. The line goes through the points (negative 1, 3), (0, 3), and (1, 3).

ⓐ $x equals 3$ ⓑ $y equals 1$

ⓐ $x equals −2$ ⓑ $y equals −5$

ⓐ

ⓑ

ⓐ $x equals −5$ ⓑ $y equals −2$

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

$y equals 2 x$ and $y equals 2$

$y equals 5 x$ and $y equals 5$

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

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

Find x- and y-Intercepts

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

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 6, 9), (negative 3, 6), (0, 3), (3, 0), and (6, negative 3).

$open parenthesis 3, 0 close parenthesis, open parenthesis 0, 3 close parenthesis$

$open parenthesis 5, 0 close parenthesis, open parenthesis 0, −5 close parenthesis$

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

$x - y = 5$

$open parenthesis 5, 0 close parenthesis, open parenthesis 0, −5 close parenthesis$

$x - y = −4$

$3 x + y = 6$

$open parenthesis 2, 0 close parenthesis, open parenthesis 0, 6 close parenthesis$

$x - 2 y = 8$

$4 x - y = 8$

$open parenthesis 2, 0 close parenthesis, open parenthesis 0, −8 close parenthesis$

$5 x - y = 5$

$2 x + 5 y = 10$

$open parenthesis 5, 0 close parenthesis, open parenthesis 0, 2 close parenthesis$

$3 x - 2 y = 12$

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

$- x + 4 y = 8$

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (negative 8, 0), (0, 2), (4, 3), and (8, 4).

$x + 2 y = 4$

$x + y = −3$

$x - y = −4$

$4 x + y = 4$

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 12 to 12. The line goes through the points (0, 4), (1, 0), and (2, negative 4).

$3 x + y = 3$

$3 x - y = −6$

$2 x - y = −8$

$2 x + 4 y = 12$

The figure shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (0, 3), (2, 2), and (6, 0).

$3 x - 2 y = 6$

$2 x - 5 y = −20$

$3 x - 4 y = −12$

$y equals −2 x$

$y equals 5 x$

$y = x$

$y equals − x$

Mixed Practice

In the following exercises, graph each equation.

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

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

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

$y = \frac{1}{4} x - 2$

$4 x + y = 2$

$5 x + 2 y = 10$

$y equals −1$

The figure shows a horizontal straight line graphed on the x y-coordinate plane. The x and y axes run from negative 8 to 8. The line goes through the points (negative 2, negative 1), (0, negative 1), and (1, negative 1).

$x equals 3$

Writing Exercises

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

Answers will vary.

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 $4 x + y = −4 ?$ Why?

Answers will vary.

Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation $y = \frac{2}{3} x - 2 ?$ Why?

Self Check

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

This table has 6 rows and 4 columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “Confidently”, the third is “With some help”, and the fourth is “No, I don’t get it”. Under the first column are the phrases “plot points on a rectangular coordinate system”, “graph a linear equation by plotting points”, “graph vertical and horizontal lines”, “find x and y intercepts”, and “graph a line using intercepts”. The other columns are left blank so that the learner may indicate their mastery level for each topic.

ⓑ 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.