Contents

Preface
1Foundations
- Introduction
- 1.1Use the Language of Algebra
- 1.2Integers
- 1.3Fractions
- 1.4Decimals
- 1.5Properties of Real Numbers
- Chapter Review
  Key Terms
  Key Concepts
- Exercises
  Review Exercises
  Practice Test
2Solving Linear Equations
3Graphs and Functions
4Systems of Linear Equations
5Polynomials and Polynomial Functions
6Factoring
7Rational Expressions and Functions
8Roots and Radicals
9Quadratic Equations and Functions
10Exponential and Logarithmic Functions
11Conics
12Sequences, Series and Binomial Theorem
- Introduction
- 12.1Sequences
- 12.2Arithmetic Sequences
- 12.3Geometric Sequences and Series
- 12.4Binomial Theorem
- Chapter Review
  Key Terms
Answer Key
Index

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.
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.
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.

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.

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 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.

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 $x = 1 x = 1$ 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.

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

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:

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

Show solution

Solution

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 $y = \frac{5}{2}, y = \frac{5}{2},$ the point is above the x-axis. (It may be helpful to write $\frac{5}{2} \frac{5}{2}$ as a mixed number or decimal.) The point $(3, \frac{5}{2}) (3, \frac{5}{2})$ is in Quadrant I.

This figure shows points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The following points are labeled: (3, 5 divided by 2), (negative 2, 3), negative 5, 4), (negative 3, negative 4), and (2, negative 3).

Try It 3.1

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

Try It 3.2

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

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:

Quadrants

\begin{matrix} Quadrant I & Quadrant II & Quadrant III & Quadrant IV \\ (x, y) & (x, y) & (x, y) & (x, y) \\ (+, +) & (-, +) & (-, -) & (+, -) \end{matrix} \begin{matrix} Quadrant I \\ (x, y) \\ (+, +) \end{matrix}

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.

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 $y = −3 x + 5 y = −3 x + 5$ 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.

\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} \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 & 3 x + y = 5 \end{matrix}

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 $A \geq 0 A \geq 0$ 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.

This figure shows a straight line graphed on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line has arrows on both ends and goes through the points (negative 3, negative 9), (negative 2, negative 7), (negative 1, negative 5), (0, negative 3), (1, negative 1), (2, 1), (3, 3), (4, 5), (5, 7), and (6, 9). The line is labeled y plus 2 x minus 3.

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: $(−1, −5) (−1, −5)$

Show solution

Solution

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

ⓐ

Example A shows the ordered pair (0, negative 3). Under this is the equation y plus 2 x minus 3. Under this is the equation negative 3 equals 2 times 0 minus 3. The negative 3 and 0 are colored the same as the negative 3 and 0 in the ordered pair at the top. There is a question mark above the plus sign. Below this is the equation negative 3 plus negative 3. Below this is the statement (0, negative 3) is a solution. Example B shows the ordered pair (3, 3). Under this is the equation y plus 2 x minus 3. Under this is the equation 3 equals 2 times 3 minus 3. The 3 and 3 are colored the same as the 3 and 3 in the ordered pair at the top. There is a question mark above the plus sign. Below this is the equation 3 plus 3. Below this is the statement (3, 3) is a solution. Example C shows the ordered pair (2, negative 3). Under this is the equation y plus 2 x minus 3. Under this is the equation negative 3 equals 2 times 2 minus 3. The negative 3 and 2 are colored the same as the negative 3 and 2 in the ordered pair at the top. There is a question mark above the plus sign. Below this is the inequality negative 3 is not equal to 1. Below this is the statement (2, negative 3) is not a solution. Example D shows the ordered pair (negative 1, negative 5). Under this is the equation y plus 2 x minus 3. Under this is the equation negative 5 equals 2 times negative 1 minus 3. The negative 1 and negative 5 are colored the same as the negative 1 and negative 5 in the ordered pair at the top. There is a question mark above the plus sign. Below this is the equation negative 5 plus negative 5. Below this is the statement (negative 1, negative 5) is a solution.

ⓑ Plot the points $(0, −3), (0, −3),$ $(3, 3), (3, 3),$ $(2, −3), (2, −3),$ and $(−1, −5) . (−1, −5) .$

This figure shows the graph of the linear equation y plus 2 x minus 3 and some points graphed on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line has arrows on both ends and goes through the points (negative 1, negative 5), (0, negative 3), and (3, 3). The point (2, negative 3) is also plotted but not on the line.

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.

Try It 3.3

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 $(2, 5) (2, 5)$

Try It 3.4

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 $(3, −1) (3, −1)$ 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.

Show solution

Solution

Step 2 is to plot the points in a rectangular coordinate system. Plot: (0, 1), (1, 3), (negative 2, negative 3). The figure then shows a graph of some points plotted on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The points (0, 1), (1, 3), and (negative 2, negative 3) are plotted. Check that the points line up. If they do not, carefully check your work! Do the point line up? Yes, the points in this example line up.

Step 3 is to draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line. This line is the graph of y plus 2 x plus 1. The figure shows the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 6 to 6. The points (negative 2, negative 3), (0, 1), and (1, 3) are plotted. The straight line goes through the three points and has arrows on both ends.

Try It 3.5

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

Try It 3.6

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.

How To

Graph a linear equation by plotting points.

Step 1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
Step 2. Plot the points 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.

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

Show solution

Solution

Find three points that are solutions to the equation. Since this equation has the fraction $\frac{1}{2} \frac{1}{2}$ 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 $\frac{1}{2} \frac{1}{2}$ simplifies to a whole number

The first set of equations starts with x plus 0. Under this is the equation y plus 1 half x plus 3. Under this is the equation y plus 1 half times 0 plus 3. Below this is the equation y plus 0 plus 3. Below this is the equation y plus 3. The second set of equations starts with x plus 2. Under this is the equation y plus 1 half x plus 3. Under this is the equation y plus 1 half times 2 plus 3. Below this is the equation y plus 1 plus 3. Below this is the equation y plus 4. The third set of equations starts with x plus 4. Under this is the equation y plus 1 half x plus 3. Under this is the equation y plus 1 half times 4 plus 3. Below this is the equation y plus 2 plus 3. Below this is the equation y plus 5.

The points are shown in Table 3.1.

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

Table 3.1

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

The figure shows the graph of a straight line on the x y-coordinate plane. The x and y axes run from negative 7 to 7. The points (0, 3), (2, 4), and (4, 5) are plotted. The straight line goes through the three points and has arrows on both ends. The line is labeled y plus 1 divided by 2 times x plus 3.

Try It 3.7

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

Try It 3.8

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

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.

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

Table 3.2

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 $y = 4 . y = 4 .$ 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.

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

Table 3.3

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.

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

Show solution

Solution

ⓐ 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.

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

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 (2, 1), (2, 2), and (2, 3) are plotted. The line goes through the three points and has arrows on both ends. The line is labeled x plus 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 $−1 . −1 .$

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

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 (negative 3, negative 1), (0, negative 1), and (3, negative 1) are plotted. The line goes through the three points and has arrows on both ends. The line is labeled y plus negative 1.

Try It 3.9

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.

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 $y = 4 x y = 4 x$ 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.

Show solution

Solution

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.

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 negative 3 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, negative 3, and (1, negative 3). The fifth row has the numbers 2, negative 6, and (2, neg ative 6). The second table has 5 rows and 3 columns. The first row is a title row with the equation y plus negative 3. The second row is a header row with the headers x, y, and (x, y). The third row has the numbers 0, negative 3, and (0, negative 3). The fourth row has the numbers 1, negative 3, and (1, negative 3). The fifth row has the numbers 2, negative 3, and (2, negative 3).

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.

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.

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	$(3, 0) (3, 0)$	6	$(0, 6) (0, 6)$
Figure (b)	4	$(4, 0) (4, 0)$	$−3 −3$	$(0, −3) (0, −3)$
Figure (c)	5	$(5, 0) (5, 0)$	$−5 −5$	$(0, 5) (0, 5)$
Figure (d)	0	$(0, 0) (0, 0)$	0	$(0, 0) (0, 0)$
General Figure	a	$(a, 0) (a, 0)$	b	$(0, b) (0, b)$

Table 3.4

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.

The table has 3 rows and 2 columns. The first row is a header row with the headers x and y. The second row contains a and 0. The third row contains 0 and b.

Example 3.7

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

The figure has three graphs. Figure a 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 8, 6), (negative 4, 4), (0, 2), (4, 0), (8, negative 2). Figure b 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, negative 6), (2, 0), and (4, 6). Figure c 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 5, 0), (negative 3, negative 3), (0, negative 5), (1, negative 6), and (2, negative 7).

Show solution

Solution

ⓐ 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.

This figure a shows a straight line graphed on the x y-coordinate plane. The x and y axes run from negative 10 to 10. The line goes through the points (negative 6, negative 8), (negative 4, negative 6), (negative 2, negative 4), (0, negative 2), (2, 0), (4, 2), (6, 4), (8, 6).

Try It 3.14

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

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 $2 x + y = 8 . 2 x + y = 8 .$

Show solution

Solution

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 figure has a table with 4 rows and 2 columns. The first row is a title row with the equation 2 x plus y plus 8. The second row is a header row with the headers x and y. The third row is labeled x-intercept and has the first column blank and a 0 in the second column. The fourth row is labeled y-intercept and has a 0 in the first column and the second column blank.

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

Let $y = 0 . y = 0 .$
Simplify.

The x-intercept is:	$(4, 0)$
To find the y-intercept, let $x = 0 . x = 0 .$

Let $x = 0 . x = 0 .$
Simplify.

The y-intercept is:	$(0, 8)$

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

$2 x + y = 8 2 x + y = 8$
x	y
4	0
0	8

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.

Show solution

Solution

Step 1 is to find the x and y-intercepts of the line. To find the x-intercept let y plus 0 and solve for x. The equation negative x plus 2 y plus 6 becomes negative x plus 2 times 0 plus 6. This simplifies to negative x plus 6. This is equivalent to x plus negative 6. The x-intercept is (negative 6, 0). To find the y-intercept let x plus 0 and solve for y. The equation negative x plus 2 y plus 6 becomes negative 0 plus 2 y plus 6. This simplifies to negative 2 y plus 6. This is equivalent to y plus 3. The y-intercept is (0, 3).

Step 3 is to plot the three points. The figure shows a table with 4 rows and 3 columns. The first row is a header row with the headers x, y, and (x, y). The second row contains negative 6, 0, and (negative 6, 0). The third row contains 0, 3, and (0, 3). The fourth row contains 2, 4, and (2, 4). The figure also has a graph of the three points on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The three points (negative 6, 0), (0, 3), and (2, 4) are plotted and labeled.

Step 4 is to draw the line. The figure shows a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 6 to 6. The straight line goes through the points (negative 6, 0), (0, 3), and (2, 4).

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.

How To

Graph a linear equation using the intercepts.

Step 1.
Find the x- and y-intercepts of the line.
- 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.

Show solution

Solution

Find the intercepts and a third point.

To find the x-intercept let y plus 0 and solve for x. The equation 4 x minus 3 y plus 12 becomes 4 x minus 3 times 0 plus 12. This simplifies to negative 4 x plus 12. This is equivalent to x plus 3. To find the y-intercept let x plus 0 and solve for y. The equation 4 x minus 3 y plus 12 becomes 4 times 0 minus 3 y plus 12. This simplifies to negative 3 y plus 12. This is equivalent to y plus negative 4. To find the third point let y plus 4 and solve for x. The equation 4 x minus 3 y plus 12 becomes 4 x minus 3 times 4 plus 12. This simplifies to negative 4 x plus 24. This is equivalent to x plus 6.

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

$4 x - 3 y = 12 4 x - 3 y = 12$
x	y	$(x, y) (x, y)$
3	0	$(3, 0) (3, 0)$
0	$−4 −4$	$(0, −4) (0, −4)$
6	4	$(6, 4) (6, 4)$

The figure shows a graph of the equation 4 x minus 3 y plus 12 on the x y-coordinate plane. The x and y-axes run from negative 7 to 7. The straight line goes through the points (0, negative 4), (3, 0), and (6, 4).

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.

Show solution

Solution

To find the x-intercept let y plus 0 and solve for x. The equation y plus 5 x becomes 0 plus 5 x. This simplifies to 0 plus x. The x-intercept is (0, 0). To find the y-intercept let x plus 0 and solve for y. The equation y plus 5 x becomes y plus 5 times 0. This simplifies to y plus 0. The y-intercept is also (0, 0).

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.

To find a second point let x plus 1 and solve for y. The equation y plus 5 x becomes y plus 5 times 1. This simplifies to y plus 5. To find a third point let x plus negative 1 and solve for y. The equation y plus 5 x becomes y plus 5 times negative 1. This simplifies to y plus negative 5

The resulting three points are summarized in the table.

$y = 5 x y = 5 x$
x	y	$(x, y) (x, y)$
0	0	$(0, 0) (0, 0)$
1	5	$(1, 5) (1, 5)$
$−1 −1$	$−5 −5$	$(−1, −5) (−1, −5)$

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

The figure shows a graph of the equation y plus 5 x on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The straight line goes through the points (negative 1, negative 5), (0, 0), and (1, 5).

Try It 3.21

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

Try It 3.22

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

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.

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

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

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

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

In the following exercises, for each ordered pair, decide

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

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

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).

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

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

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

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

$y = x + 2 y = x + 2$

10.

$y = x - 3 y = x - 3$

11.

$y = 3 x - 1 y = 3 x - 1$

12.

$y = −2 x + 2 y = −2 x + 2$

13.

$y = − x - 3 y = − x - 3$

14.

$y = − x - 2 y = − x - 2$

15.

$y = 2 x y = 2 x$

16.

$y = −2 x y = −2 x$

17.

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

18.

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

19.

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

20.

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

21.

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

22.

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

23.

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

24.

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

Graph Vertical and Horizontal lines

In the following exercises, graph each equation.

25.

ⓐ $x = 4 x = 4$ ⓑ $y = 3 y = 3$

26.

ⓐ $x = 3 x = 3$ ⓑ $y = 1 y = 1$

27.

ⓐ $x = −2 x = −2$ ⓑ $y = −5 y = −5$

28.

ⓐ $x = −5 x = −5$ ⓑ $y = −2 y = −2$

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

29.

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

30.

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

31.

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

32.

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

Find x- and y-Intercepts

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

33.

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).

34.

35.

36.

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

37.

$x - y = 5 x - y = 5$

38.

$x - y = −4 x - y = −4$

39.

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

40.

$x - 2 y = 8 x - 2 y = 8$

41.

$4 x - y = 8 4 x - y = 8$

42.

$5 x - y = 5 5 x - y = 5$

43.

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

44.

$3 x - 2 y = 12 3 x - 2 y = 12$

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

45.

$- x + 4 y = 8 - x + 4 y = 8$

46.

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

47.

$x + y = −3 x + y = −3$

48.

$x - y = −4 x - y = −4$

49.

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

50.

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

51.

$3 x - y = −6 3 x - y = −6$

52.

$2 x - y = −8 2 x - y = −8$

53.

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

54.

$3 x - 2 y = 6 3 x - 2 y = 6$

55.

$2 x - 5 y = −20 2 x - 5 y = −20$

56.

$3 x - 4 y = −12 3 x - 4 y = −12$

57.

$y = −2 x y = −2 x$

58.

$y = 5 x y = 5 x$

59.

$y = x y = x$

60.

$y = − x y = − x$

Mixed Practice

In the following exercises, graph each equation.

61.

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

62.

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

63.

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

64.

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

65.

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

66.

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

67.

$y = −1 y = −1$

68.

$x = 3 x = 3$

Writing Exercises

69.

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

70.

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

71.

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?

72.

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