3.9 Graph Linear Equations in Two Variables

Mike LePine; ktomanelli

3.9 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
Verify solutions to an equation in two variables
Complete a table of solutions to a linear equation
Find solutions to a linear equation in two variables
Recognize the relationship between the solutions of an equation and its graph.
Graph a linear equation by plotting points.
Graph vertical and horizontal lines.
Identify the $x$ -intercept and $y$ -intercept on a graph
Find the $x$ -intercept and $y$ -intercept from an equation of a line
Graph a line using the intercepts

Try It

Before you get started, take this readiness quiz:

1) Evaluate $x + 3$ when $x = - 1$ .
2) Evaluate $2 x - 5 y$ when $x = 3$ and $y = - 2$ .
3) Solve for $y$ : $40 - 4 y = 20$ .
4) Evaluate $3 x + 2$ when $x = - 1$ .
5) Solve $3 x + 2 y = 12$ for $y$ in general.
6) Solve: $3 \cdot 0 + 4 y = - 2$ .

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 $x y$ -plane or the ‘coordinate plane’.

The horizontal number line is called the $x$ -axis. The vertical number line is called the $y$ -axis. The $x$ -axis and the $y$ -axis together form the rectangular coordinate system. 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 3.9.1.

‘Quadrant’ has the root ‘quad,’ which means ‘four.’

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The top-right portion of the plane is labeled "I", the top-left portion of the plane is labeled "II", the bottom-left portion of the plane is labeled "III" and the bottom-right portion of the plane is labeled "IV". — Figure 3.9.1-‘Quadrant’ has the root ‘quad,’ which means ‘four.’

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.

Ordered Pair

An ordered pair, $(x, y)$ , 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.

The phrase ‘ordered pair’ means 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 $(0, 0)$ . The point $(0, 0)$ has a special name. It is called the origin.

The Origin

The point $(0, 0)$ 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 $x y$ -plane. Let’s plot the point $(1, 3)$ as an example. First, locate $1$ on the $x$ -axis and lightly sketch a vertical line through $x = 1$ . Then, locate $3$ on the $y$ -axis and sketch a horizontal line through $y = 3$ . Now, find the point where these two lines meet—that is the point with coordinates $(1, 3)$ .

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. An arrow starts at the origin and extends right to the number 2 on the x-axis. The point (1, 3) is plotted and labeled. Two dotted lines, one parallel to the x-axis, the other parallel to the y-axis, meet perpendicularly at 1, 3. The dotted line parallel to the x-axis intercepts the y-axis at 3. The dotted line parallel to the y-axis intercepts the x-axis at 1. — Figure 3.9.3

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

Example 1

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

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

Solution

The first number of the coordinate pair is the $x$ -coordinate, and the second number is the $y$ -coordinate.

a. Since $x = - 5$ , the point is to the left of the $y$ -axis. Also, since $y = 4$ , the point is above the $x$ -axis. The point $(- 5, 4)$ is in Quadrant II.
b. Since $x = - 3$ , the point is to the left of the $y$ -axis. Also, since $y = - 4$ , the point is below the $x$ -axis. The point $(- 3, - 4)$ is in Quadrant III.
c. Since $x = 2$ , the point is to the right of the $y$ -axis. Since $y = - 3$ , the point is below the $x$ -axis. The point $(2, - 3)$ is in Quadrant lV.
d. Since $x = - 2$ , the point is to the left of the $y$ -axis. Since $y = 3$ , the point is above the $x$ -axis. The point $(- 2, 3)$ is in Quadrant II.
e. Since $x = 3$ , 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 graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The points (negative 5, 4), (negative 2, 3), (negative 3, negative 4), (3, five halves), and (2, negative 3) are plotted and labeled. — Figure 3.9.4

Try It

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

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

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (negative 2, 1) is plotted and labeled "a". The point (negative 3, negative 1) is plotted and labeled "b". The point (4, negative 4) is plotted and labeled "c". The point (negative 4, negative one half) is plotted and labeled “d”. — Figure 3.9.5

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

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

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (negative 4, 1) is plotted and labeled "a". The point (negative 2, 3) is plotted and labeled "b". The point (2, negative 5) is plotted and labeled "c". The point (negative 3, 2 and one half) is plotted and labeled “d”. — Figure 3.9.6

How do the signs affect the location of the points? You may have noticed some patterns as you graphed the points in the previous example.

For the point in Figure 3.9.4 in Quadrant IV, what do you notice about the signs of the coordinates? What about the signs of the coordinates of points in the third quadrant? The second quadrant? The first quadrant?

Can you tell just by looking at the coordinates in which quadrant the point $(- 2, 5)$ is located? In which quadrant is $(2, - 5)$ located?

Quadrants

We can summarize sign patterns of the quadrants in this way.

Quadrant I	Quadrant II	Quadrant III	Quadrant IV
$(x, y)$	$(x, y)$	$(x, y)$	$(x, y)$
$(+, +)$	$(-, +)$	$(-, -)$	$(+, -)$

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. The top-right portion of the plane is labeled "I" and "ordered pair +, +", the top-left portion of the plane is labeled "II" and "ordered pair -, +", the bottom-left portion of the plane is labelled "III" "ordered pair -, -" and the bottom-right portion of the plane is labeled "IV" and "ordered pair +, -". — Figure 3.9.7

What if one coordinate is zero as shown in Figure 3.9.8? Where is the point $(0, 4)$ located? Where is the point $(- 2, 0)$ located?

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. Points (0, 4) and (negative 2, 0) are plotted and labeled. — Figure 3.9.8

The point $(0, 4)$ is on the y-axis and the point $(- 2, 0)$ 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 $(a, 0)$ .

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

Example 2

Plot each point:

a. $(0, 5)$
b. $(4, 0)$
c. $(- 3, 0)$
d. $(0, 0)$
e. $(0, - 1)$

Solution

a. Since $x = 0$ , the point whose coordinates are $(0, 5)$ is on the $y$ -axis.
b. Since $y = 0$ , the point whose coordinates are $(4, 0)$ is on the $x$ -axis.
c. Since $y = 0$ , the point whose coordinates are $(- 3, 0)$ is on the $x$ -axis.
d. Since $x = 0$ and $y = 0$ , the point whose coordinates are $(0, 0)$ is the origin.
e. Since $x = 0$ , the point whose coordinates are $(0, - 1)$ is on the $y$ -axis.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 7 to 7. The points (negative 3, 0), (0, 0), (0, negative 1), (0, 5), and (4, 0) are plotted and labeled. — Figure 3.9.9

Try It

9) Plot each point:

a. $(4, 0)$
b. $(- 2, 0)$
c. $(0, 0)$
d. $(0, 2)$
e. $(0, - 3)$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The points (4, 0), (negative 2, 0), (0, 0), (0, 2), and (0, negative 3) are plotted and labeled. — Figure 3.9.10

10) Plot each point:

a. $(- 5, 0)$
b. $(3, 0)$
c. $(0, 0)$
d. $(0, - 1)$
e. $(0, 4)$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The points (negative 5, 0), (3, 0), (0, 0), (0, negative 1), and (0, 4) are plotted and labeled. — Figure 3.9.11

In algebra, being able to identify the coordinates of a point shown on a graph is just as important as being able to plot points. To identify the $x$ -coordinate of a point on a graph, read the number on the $x$ -axis directly above or below the point. To identify the $y$ -coordinate of a point, read the number on the $y$ -axis directly to the left or right of the point. Remember, when you write the ordered pair use the correct order, $(x, y)$ .

Example 3

Name the ordered pair of each point shown in the rectangular coordinate system.

Solution

Point A is above $- 3$ on the $x$ -axis, so the $x$ -coordinate of the point is $- 3$ .

The point is to the left of $3$ on the $y$ -axis, so the $y$ -coordinate of the point is $3$ .
The coordinates of the point are $(- 3, 3)$ .

Point B is below $- 1$ on the $x$ -axis, so the $x$ -coordinate of the point is $- 1$ .

The point is to the left of $- 3$ on the $y$ -axis, so the $y$ -coordinate of the point is $- 3$ .
The coordinates of the point are $(- 1, - 3)$ .

Point C is above $2$ on the $x$ -axis, so the $x$ -coordinate of the point is $2$ .

The point is to the right of $4$ on the $y$ -axis, so the $y$ -coordinate of the point is $4$ .
The coordinates of the point are $(2, 4)$ .

Point D is below $4$ on the $x$ -axis, so the $x$ -coordinate of the point is $4$ .

The point is to the right of $- 4$ on the $y$ -axis, so the $y$ -coordinate of the point is $- 4$ .
The coordinates of the point are $(4, - 4)$ .

Point E is on the $y$ -axis at $y = - 2$ . The coordinates of point E are $(0, - 2)$ .

Point F is on the $x$ -axis at $x = 3$ . The coordinates of point F are $(3, 0)$ .

Try It

11) Name the ordered pair of each point shown in the rectangular coordinate system.

Solution

a: $(5, 1)$
b: $(- 2, 4)$
c: $(- 5, - 1)$
d: $(3, - 2)$
e: $(0, - 5)$
f: $(4, 0)$

12) Name the ordered pair of each point shown in the rectangular coordinate system.

Solution

a: $(4, 2)$
b: $(- 2, 3)$
c: $(- 4, - 4)$
d: $(3, - 5)$
e: $(- 3, 0)$
f: $(0, 2)$

Verify Solutions to an Equation in Two Variables

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. The process of solving an equation ended with a statement like $x = 4$ . (Then, you check the solution by substituting back into the equation.)

Here’s an example of an equation in one variable, and its one solution.

$\begin{array}{rcl} 3 x + 5 & = & 17 \\ 3 x & = & 12 \\ x & = & 4 \end{array}$

However, equations can have more than one variable. Equations with two variables may be of the form $A x + B y = C$ . Equations of this form are called linear equations in two variables.

Linear Equation

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

Notice the word line in linear. Here is an example of a linear equation in two variables, $x$ and $y$ .

$\begin{aligned} A x + B y & = C \\ x + 4 y & = 8 \\ A = 1, B = 4, & C = 8 \end{aligned}$

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.

$\begin{aligned} y & = - 3 x + 5 \\ Step 1: Add to both sides. & y + 3 x & = - 3 x + 5 + 3 x \\ Step 2: Simplify. & y + 3 x & = 5 \\ Step 3: Use the Commutative Property to put it in Ax+By=C form. & 3 x = y & = 5 \end{aligned}$

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.

Standard Form of Linear Equation

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

Most people prefer to have $A$ , $B$ , and $C$ be integers and $𝐴 \geq 0$ 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 $(x, y)$ . 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 $(x, y)$ is a solution of the linear equation $A x + B y = C$ , if the equation is a true statement when the $x$ and $y$ values of the ordered pair are substituted into the equation.

Example 4

Determine which ordered pairs are solutions to the equation $x + 4 y = 8$ .

a. $(0, 2)$
b. $(2, - 4)$
c. $(- 4, 3)$

Solution

Substitute the $x$ – and $y$ -values from each ordered pair into the equation and determine if the result is a true statement.

(a)	(b)	(c)
$\begin{aligned} (0, 2) \\ x = 0, y & = 2 \\ x + 4 y & = 8 \\ 0 + 4 \cdot 2 & \overset{?}{=} 8 \\ 0 + 8 & \overset{?}{=} 8 \\ 8 & = 8 ✓ \end{aligned}$	$\begin{aligned} (2, - 4) \\ x = 2, y & = - 4 \\ x + 4 y & = 8 \\ 2 + 4 \cdot - 4 & \overset{?}{=} 8 \\ 2 + (- 16) & \overset{?}{=} 8 \\ - 14 & \neq 8 \end{aligned}$	$\begin{aligned} (- 4, 3) \\ x = - 4, y & = 3 \\ x + 4 y & = 8 \\ - 4 + 4 \cdot 3 & \overset{?}{=} 8 \\ - 4 + 12 & \overset{?}{=} 8 \\ 8 & = 8 ✓ \end{aligned}$
$(0, 2)$ is a solution	$(2, - 4)$ is not a solution	$(- 4, 3)$ is a solution

Try It

13) Which of the following ordered pairs are solutions to $2 x + 3 y = 6$ ?

a. $(3, 0)$
b. $(2, 0)$
c. $(6, - 2)$

Solution

a, c

14) Which of the following ordered pairs are solutions to the equation $4 x - y = 8$ ?

a. $(0, 8)$
b. $(2, 0)$
c. $(1, - 4)$

Solution

b, c

Example 5

Which of the following ordered pairs are solutions to the equation $y = 5 x - 1$ ?

a. $(0, - 1)$
b. $(1, 4)$
c. $(- 2, - 7)$

Solution

Substitute the $x$ and $y$ values from each ordered pair into the equation and determine if it results in a true statement.

(a)	(b)	(c)
$\begin{aligned} (0, - 1) \\ x = 0, y & = - 1 \\ y & = 5 x - 1 \\ - 1 & \overset{?}{=} 5 (0) - 1 \\ - 1 & \overset{?}{=} 0 - 1 \\ - 1 & = - 1 ✓ \end{aligned}$	$\begin{aligned} (1, 4) \\ x = 1, y & = 4 \\ y & = 5 x - 1 \\ 4 & \overset{?}{=} 5 (1) - 1 \\ 4 & \overset{?}{=} 5 - 1 \\ 4 & = 4 ✓ \end{aligned}$	$\begin{aligned} (- 2, - 7) \\ x = - 2, y & = - 7 \\ y & = 5 x - 1 \\ - 7 & \overset{?}{=} 5 (- 2) - 1 \\ - 7 & \overset{?}{=} - 10 - 1 \\ 7 & \neq - 11 \end{aligned}$
$(0, - 1)$ is a solution.	$(1, 4)$ is a solution.	$(- 2, - 7)$ is not a solution.

Try It

15) Which of the following ordered pairs are solutions to the equation $y = 4 x - 3$ ?

a. $(0, 3)$
b. $(1, 1)$
c. $(- 1, - 1)$

Solution

b

16) Which of the following ordered pairs are solutions to the equation $y = - 2 x + 6$ ?

a. $(0, 6)$
b. $(1, 4)$
c. $(- 2, - 2)$

Solution

a, b

Complete a Table of Solutions to a Linear Equation in Two Variables

In the examples above, we substituted the $x$ and $y$ values of a given ordered pair to determine whether or not it was a solution to a linear equation. But how do you find the ordered pairs if they are not given? It’s easier than you might think—you can just pick a value for $x$ and then solve the equation for $y$ . Or, pick a value for $y$ and then solve for $x$ .

We’ll start by looking at the solutions to the equation $y = 5 x - 1$ that we found in Example 3.9.5. We can summarize this information in a table of solutions, as shown in the below table.

$y = 5 x - 1$
$x$	$y$	$(x, y)$
$0$	$- 1$	$(0, - 1)$
$1$	$4$	$(1, 4)$

To find a third solution, we’ll let $x = 2$ and solve for $y$ .

$\begin{aligned} y & = 5 x - 1 \\ Substitute x = 2 . & y & = 5 (2) - 1 \\ Multiply. & y & = 10 - 1 \\ Simplify. & y & = 9 \end{aligned}$

The ordered pair $(2, 9)$ is a solution to $y = 5 x - 1$ . We will add it to the below table.

$y = 5 x - 1$
$x$	$y$	$(x, y)$
$0$	$- 1$	$(0, - 1)$
$1$	$4$	$(1, 4)$
$2$	$9$	$(2, 9)$

We can find more solutions to the equation by substituting in any value of

x

or any value of

y

and solving the resulting equation to get another ordered pair that is a solution. There are infinitely many solutions to this equation.

Example 6

Complete the below table to find three solutions to the equation $y = 4 x - 2$ .

$y = 4 x - 2$
$x$	$y$	$(x, y)$
$0$
$- 1$
$2$

Solution

Substitute $x = 0$ , $x = - 1$ , and $x = 2$ into $y = 4 x - 2$

$\begin{aligned} x & = 0 & x & = - 1 & x & = 2 \\ y & = 4 x - 2 & y & = 4 x - 2 & y & = 4 x - 2 \\ y & = 4 \cdot 0 - 2 & y & = 4 (- 1) - 2 & y & = 4 \cdot 2 - 2 \\ y & = - 2 & y & = - 4 - 2 & y & = 8 - 2 \\ y & = - 2 & y & = - 6 & y & = 6 \\ (0, - 2) & (- 1, - 6) & (2, 6) \end{aligned}$

The results are summarized in the below table.

$y = 4 x - 2$
$x$	$y$	$(x, y)$
$0$	$- 2$	$(0, - 2)$
$- 1$	$- 6$	$(- 1, - 6)$
$2$	$6$	$(2, 6)$

Try It

17) Complete the table to find three solutions to this equation: $y = 3 x - 1$ .

$y = 3 x - 1$
$x$	$y$	$(x, y)$
$0$
$- 1$
$2$

Solution

$y = 3 x - 1$
$x$	$y$	$(x, y)$
$0$	$- 1$	$(0, - 1)$
$- 1$	$- 4$	$(- 1, - 4)$
$2$	$5$	$(2, 5)$

18) Complete the table to find three solutions to this equation: $y = 6 x - 1$ .

$y = 6 x - 1$
$x$	$y$	$(x, y)$
$0$
$- 1$
$2$

Solution

$y = 6 x - 1$
$x$	$y$	$(x, y)$
$0$	$- 1$	$(0, - 1)$
$- 1$	$- 7$	$(- 1, - 7)$
$2$	$11$	$(2, 11)$

Example 7

Complete the below table to find three solutions to the equation $5 x - 4 y = 20$ .

$5 x - 4 y = 20$
$x$	$y$	$(x, y)$
$0$
	$0$
	$5$

Solution

Substitute the given value into the equation $5 x - 4 y = 20$ and solve for the other variable. Then, fill in the values in the table.

$\begin{aligned} x & = 0 & y & = 0 & y & = 5 \\ 5 x - 4 y & = 20 - 2 & 5 x - 4 y & = 20 & 5 x - 4 y & = 20 \\ 5 \cdot 0 & = 20 & 5 x - 4 \cdot 0 & = 20 & 5 x - 4 \cdot 5 & = 20 \\ 0 - 4 y & = 20 & 5 x - 0 & = 20 & 5 x - 20 & = 20 \\ - 4 y & = 20 & 5 x & = 20 & 5 x & = 40 \\ y & = - 5 & x & = 4 & x & = 8 \\ (0, - 5) & (4, 0) & (8, 5) \end{aligned}$

The results are summarized in the below table.

$5 x - 4 y = 20$
$x$	$y$	$(x, y)$
$0$	$- 5$	$(0, - 5)$
$4$	$0$	$(4, 0)$
$8$	$5$	$(8, 5)$

Try It

19) Complete the table to find three solutions to this equation: $2 x - 5 y = 20$ .

$2 x - 5 y = 20$
$x$	$y$	$(x, y)$
$0$
	$0$
$- 5$

Solution

$2 x - 5 y = 20$
$x$	$y$	$(x, y)$
$0$	$- 4$	$(0, - 4)$
$10$	$0$	$(10, 0)$
$- 5$	$- 6$	$(- 5, - 6)$

20) Complete the table to find three solutions to this equation: $3 x - 4 y = 12$ .

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

Solution

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

Find Solutions to a Linear Equation in Two Variables

To find a solution to a linear equation, you really can pick any number you want to substitute into the equation for $x$ or $y$ . But since you’ll need to use that number to solve for the other variable it’s a good idea to choose a number that’s easy to work with.

When the equation is in $y$ -form, with the $y$ by itself on one side of the equation, it is usually easier to choose values of $x$ and then solve for $y$ .

Example 8

Find three solutions to the equation

y = - 3 x + 2

.

Solution

We can substitute any value we want for $x$ or any value for $y$ . Since the equation is in $y$ -form, it will be easier to substitute in values of $x$ .

Let’s pick $x = 0$ , $x = 1$ , and $x = - 1$ .

\begin{aligned} x & = 0 \\ y & = - 3 x + 2 \end{aligned}

\begin{aligned} x & = 1 \\ y & = - 3 x + 2 \end{aligned}

\begin{aligned} x & = - 1 \\ y & = - 3 x + 2 \end{aligned}

Step 1: Substitute the value into the equation.

y = - 3 \times 0 + 2

y = - 3 \times 1 + 2

y = - 3 \times (- 1) + 2

Step 2: Simplify.

y = 0 + 2

y = - 3 + 2

y = 3 + 2

Step 3: Simplify.

y = 2

y = - 1

y = 5

Step 4: Write the ordered pair.

(0, 2)

(1, - 1)

(- 1, 5)

Step 5: Check.

\begin{array}{rcl} y & = & - 3 x + 2 \\ 2 & \overset{?}{=} & - 3 (0) + 2 \\ 2 & = & 2 ✓ \end{array}

\begin{array}{rcl} y & = & - 3 x + 2 \\ - 1 & \overset{?}{=} & - 3 (1) + 2 \\ - 1 & = & - 1 ✓ \end{array}

\begin{array}{rcl} y & = & - 3 x + 2 \\ 5 & \overset{?}{=} & - 3 (- 1) + 2 \\ 5 & = & 5 ✓ \end{array}

So, $(0, 2)$ , $(1, - 1)$ and $(- 1, 5)$ are all solutions to $y = - 3 x + 2$ . We show them in the below table.

$y = - 3 x + 2$
$x$	$y$	$(x, y)$
$0$	$2$	$(0, 2)$
$1$	$- 1$	$(1, - 1)$
$- 1$	$5$	$(- 1, 5)$

Try It

21) Find three solutions to this equation: $y = - 2 x + 3$ .

Solution

Answers will vary.

22) Find three solutions to this equation: $y = - 4 x + 1$ .

Solution

Answers will vary.

We have seen how using zero as one value of $x$ makes finding the value of $y$ easy. When an equation is in standard form, with both the $x$ and $y$ on the same side of the equation, it is usually easier to first find one solution when $x = 0$ find a second solution when $y = 0$ , and then find a third solution.

Example 9

Find three solutions to the equation $3 x + 2 y = 6$ .

Solution

We can substitute any value we want for $x$ or any value for $y$ . Since the equation is in standard form, let’s pick first $x = 0$ , then $y = 0$ , and then find a third point.

x = 0

y = 0

x = 1

Step 1: Substitute the value into the equation.

3 x + 2 y = 6

3 x + 2 y = 6

3 x + 2 y = 6

Step 2. Simplify.

3 (0) + 2 y = 6

3 x + 2 (0) = 6

3 (1) + 2 y = 6

Step 3: Solve.

\begin{array}{rcl} 0 + 2 y & = & 6 \\ 2 y & = & 6 \\ y & = & 3 \end{array}

\begin{array}{rcl} 3 x + 0 & = & 6 \\ 3 x & = & 6 \\ x & = & 2 \end{array}

\begin{array}{r} 3 + 2 y = 6 \\ 2 y = 3 \\ y = \frac{3}{2} \end{array}

Step 4: Write the ordered pair.

(0, 3)

(2, 0)

(1, \frac{3}{2})

Step 5: Check.

\begin{array}{rcl} 3 x + 2 y & = & 6 \\ 3 (0) + 2 (3) & \overset{?}{=} & 6 \\ 0 + 6 & = & 6 \\ 6 & = & 6 ✓ \end{array}

\begin{array}{rcl} 3 x + 2 y & = & 6 \\ 3 (2) + 2 (0) & \overset{?}{=} & 6 \\ 6 + 0 & = & 6 \\ 6 & = & 6 ✓ \end{array}

\begin{array}{rcl} 3 x + 2 y & = & 6 \\ 3 (1) + 2 (\frac{3}{2}) & \overset{?}{=} & 6 \\ 3 + 3 & \overset{?}{=} & 6 \\ 6 & = & 6 ✓ \end{array}

So $(0, 3)$ , $(2, 0)$ , and $(1, \frac{3}{2})$ are all solutions to the equation $3 x + 2 y = 6$ . We can list these three solutions in the below table.

$3 x + 2 y = 6$
$x$	$y$	$(x, y)$
$0$	$3$	$(0, 3)$
$2$	$0$	$(2, 0)$
$1$	$\frac{3}{2}$	$(1, \frac{3}{2})$

Try It

23) Find three solutions to the equation $2 x + 3 y = 6$ .

Solution

Answers will vary.

24) Find three solutions to the equation $4 x + 2 y = 8$ .

Solution

Answers will vary.

Recognize the Relationship Between the Solutions of an Equation and its Graph

In the previous section, we found several solutions to the equation $3 x + 2 y = 6$ . They are listed in the table below. So, the ordered pairs $(0, 3)$ , $(2, 0)$ , and $(1, \frac{3}{2})$ are some solutions to the equation $3 x + 2 y = 6$ . We can plot these solutions in the rectangular coordinate system as shown in the below table.

$3 x + 2 y = 6$
$x$	$y$	$(x, y)$
$0$	$3$	$(0, 3)$
$2$	$0$	$(2, 0)$
$1$	$\frac{3}{2}$	$(1, \frac{3}{2})$

The figure shows four points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the four points at (0, 3), (1, three halves), (2, 0), and (4, negative 3). The four points appear to line up along a straight line. — Figure 3.9.15

Notice how the points line up perfectly? We connect the points with a line to get the graph of the equation $3 x + 2 y = 6$ . See Figure 3.9.16. Notice the arrows on the ends of each side of the line. These arrows indicate the line continues.

The figure shows a straight line drawn through four points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the four points at (0, 3), (1, three halves), (2, 0), and (4, negative 3). A straight line with a negative slope goes through all four points. The line has arrows on both ends pointing to the edge of the figure. The line is labeled with the equation 3x plus 2y equals 6. — Figure 3.9.16

Every point on the line is a solution of the equation. Also, every solution of this equation is a point on this line. Points not on the line are not solutions.

Notice that the point whose coordinates are $(- 2, 6)$ is on the line shown in Figure 3.9.17. If you substitute $x = - 2$ and $y = 6$ into the equation, you find that it is a solution to the equation.

The figure shows a straight line and two points and on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the two points and are labeled by the coordinates “(negative 2, 6)” and “(4, 1)”. The straight line goes through the point (negative 2, 6) but does not go through the point (4, 1). — Figure 3.9.17

$\begin{aligned} Test (- 2, & 6) \\ 3 x + 2 y & = 6 \\ 3 (- 2) + 2 (6) & = 6 \\ - 6 + 12 & = 6 \\ 6 & = 6 ✓ \end{aligned}$

So the point $(- 2, 6)$ is a solution to the equation $3 x + 2 y = 6$ . (The phrase “the point whose coordinates are $(- 2, 6)$ ” is often shortened to “the point $(- 2, 6)$ .”)

$\begin{aligned} What about (4, & 1) \\ 3 x + 2 y & = 6 \\ 3 (4) + 2 (1) & = 6 \\ 12 + 2 & \overset{?}{=} 6 \\ 14 & \neq 6 ✓ \end{aligned}$

So $(4, 1)$ is not a solution to the equation $3 x + 2 y = 6$ . Therefore, the point $(4, 1)$ is not on the line. See Figure 3.9.17. This is an example of the saying, “A picture is worth a thousand words.” The line shows you all the solutions to the equation. Every point on the line is a solution of the equation. And, every solution of this equation is on this line. This line is called the graph of the equation $3 x + 2 y = 6$ .

Graph of a Linear Equation

The graph of a linear equation $A x + B y = C$ is a line.

Every point on the line is a solution of the equation.
Every solution of this equation is a point on this line.

Example 10

The graph of $y = 2 x - 3$ is shown.

The figure shows a straight line on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line has a positive slope and goes through the y-axis at the (0, negative 3). The line is labeled with the equation y equals 2x negative 3. — Figure 3.9.18

For each ordered pair, decide:

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

a. $(0, - 3)$
b. $(3, 3)$
c. $(2, - 3)$
d. $(- 1, - 5)$

Solution

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

a.

A:	B:	C:	D:
$\begin{aligned} (0, - 3) \\ y & = 2 x - 3 \\ - 3 & \overset{?}{=} 2 (0) - 3 \\ - 3 & = - 3 ✓ \end{aligned}$	$\begin{aligned} (3, 3) \\ y & = 2 x - 3 \\ 3 & \overset{?}{=} 2 (3) - 3 \\ 3 & = 3 ✓ \end{aligned}$	$\begin{aligned} (2, - 3) \\ y & = 2 x - 3 \\ - 3 & \overset{?}{=} 2 (2) - 3 \\ - 3 & \neq 1 \end{aligned}$	$\begin{aligned} (- 1, - 5) \\ y & = 2 x - 3 \\ - 5 & \overset{?}{=} 2 (- 1) - 3 \\ - 5 & = - 5 ✓ \end{aligned}$
$(0, - 3)$ is a solution.	$(3, 3)$ is a solution	$(2, - 3)$ is not a solution	$(- 1, - 5)$ is a solution

b. Plot the points A $(0, 3)$ , B $(3, 3)$ , C $(2, - 3)$ , and D $(- 1, - 5)$ .

The points $(0, 3)$ , $(3, 3)$ , and $(- 1, - 5)$ are on the line $y = 2 x - 3$ , and the point $(2, - 3)$ is not on the line.

The points that are solutions to $y = 2 x - 3$ are on the line, but the point that is not a solution is not on the line.

Try It

25) Use the graph of $y = 3 x - 1$ to decide whether each ordered pair is:

a solution to the equation.
on the line.

a. $(0, - 1)$
b. $(2, 5)$

Solution

a. yes, yes
b. yes, yes

26) Use graph of $y = 3 x - 1$ to decide whether each ordered pair is:

a solution to the equation
on the line

a. $(3, - 1)$
b. $(- 1, - 4)$

Solution

a. no, no
b. yes, yes

Graph a Linear Equation by Plotting Points

Several methods can be used to graph a linear equation. The method we used to graph $3 x + 2 y = 6$ is called plotting points, or the Point–point-plotting method.

Example 11

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

Solution

Step 1: Find three points whose coordinates are solutions to the equation.

You can choose any values for $x$ or $y$ .

In this case, since $y$ is isolated on the left side of the equation, it is easier to choose values for $x$ .

$\begin{aligned} y & = 2 x + 1 \\ x & = 0 \\ y & = 2 x + 1 \\ y & = 2 \cdot 0 + 1 \\ y & = 0 + 1 \\ y & = 1 \\ x & = 1 \\ y & = 2 x + 1 \\ y & = 2 \cdot 1 + 1 \\ y & = 2 + 1 \\ y & = 3 \\ x & = - 2 \\ y & = 2 x + 1 \\ y & = 2 \cdot - 2 + 1 \\ y & = - 4 + 1 \\ y & = - 3 \end{aligned}$

Organize the solutions in a table.

Put the three solutions in a table.

y=2x+1
x	y	(x,y)
0	1	(0,1)
1	3	(1,3)
-2	-3	(-2.-3)

Step 2: Plot the points in a rectangular coordinate system.

Plot: (0,1), (1, 3), (-2,-3).

A graph shows the three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points at (0, 1), (1, 3), and (negative 2, negative 3). — Figure 3.9.21

Check that the points line up. If they do not, carefully check your work!

Do the points line up? Yes, the points line up.

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.

This line is the graph of $y = 2 x + 1$ .

A graph shows a straight line drawn through three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points at (0, 1), (1, 3), and (negative 2, negative 3). A straight line goes through all three points. The line has arrows on both ends pointing to the edge of the figure. The line is labeled with the equation y equals 2x plus 1. — Figure 3.9.22

Try It

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

Solution

28) Graph the equation by plotting points: $y = - 2 x + 4$ .

Solution

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

How to

Graph a linear equation by plotting points.

Find three points whose coordinates are solutions to the equation. Organize them in a table.
Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
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 part (a) and part (b) in Figure 3.9.25.

Figure a shows three points with a straight line going through them. Figure b shows three points that do not lie on the same line. — Figure 3.9.25

Let’s do another example. This time, we’ll show the last two steps all on one grid.

Example 12

Graph the equation $y = - 3 x$ .

Solution

Step 1: Find three points that are solutions to the equation.

Here, again, it’s easier to choose values for $x$ . Do you see why?

$\begin{aligned} x & = 0 & x & = 1 & x & = - 2 \\ y & = - 3 x & y & = - 3 & y & = - 3 \\ y & = - 3 \cdot 0 & y & = - 3 \cdot 1 & y & = - 3 (- 2) \\ y & = 0 & y & = - 3 & y & = 6 \end{aligned}$

Step 2: List the points in the following table.

$y = - 3 x$
$x$	$y$	$(x, y)$
$0$	$0$	$(0, 0)$
$1$	$- 3$	$(1, - 3)$
$- 2$	$6$	$(- 2, 6)$

Step 3: Plot the points, check that they line up, and draw the line.

The figure shows a straight line drawn through three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points which are labeled by their ordered pairs (negative 2, 6), (0, 0), and (1, negative 3). A straight line goes through all three points. The line has arrows on both ends pointing to the outside of the figure. The line is labeled with the equation y equals negative 3x. — Figure 3.9.26

Try It

29) Graph the equation by plotting points: $y = - 4 x$ .

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The straight line goes through the points (negative 2, 8), (0, 0), and (2, negative 8). The line has arrows on both ends pointing to the outside of the figure. — Figure 3.9.27

30) Graph the equation by plotting points: $y = x$ .

Solution

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

Example 13

Graph the equation

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

Solution

Step 1: 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 2 a good choice for values of $x$ ?

$\begin{aligned} x & = 0 & x & = 2 & x & = 4 \\ y & = \frac{1}{2} x + 3 & y & = \frac{1}{2} x + 3 & y & = \frac{1}{2} x + 3 \\ y & = \frac{1}{2} (0) + 3 & y & = \frac{1}{2} (2) + 3 & y & = \frac{1}{2} (4) + 3 \\ y & = 0 + 3 & y & = 1 + 3 & y & = 2 + 3 \\ y & = 3 & y & = 4 & y & = 5 \end{aligned}$

The points are shown in the below table.

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

Step 2: Plot the points, check that they line up, and draw the line.

Try It

31) Graph the equation

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

Solution

32) Graph the equation

y = \frac{1}{4} x + 2

.

Solution

So far, all the equations we graphed had $y$ given in terms of $x$ . Now we’ll graph an equation with $x$ and $y$ on the same side. Let’s see what happens in the equation $2 x + y = 3$ . If $y = 0$ what is the value of $x$ ?

$\begin{aligned} y & = 0 \\ 2 x + y & = 3 \\ 2 x + 0 & = 3 \\ 2 x & = 3 \\ x & = \frac{3}{2} \\ (\frac{3}{2}, 0) \end{aligned}$

This point has a fraction for the $x$ – coordinate and, while we could graph this point, it is hard to be precise graphing fractions. Remember in the example $y = \frac{1}{2} x + 3$ , we carefully chose values for $x$ so as not to graph fractions at all. If we solve the equation $2 x + y = 3$ for $y$ , it will be easier to find three solutions to the equation.

\begin{aligned} 2 x + y & = 3 \\ y & = - 2 x + 3 \end{aligned}

The solutions for $x = 0$ , $x = 1$ , and $x = - 1$ are shown in the below table. The graph is shown in the Figure 3.9.32.

$2 x + y = 3$
$x$	$y$	$(x, y)$
$0$	$3$	$(0, 3)$
$1$	$1$	$(1, 1)$
$- 1$	$5$	$(- 1, 5)$

Can you locate the point $(\frac{3}{2}, 0)$ , which we found by letting $y = 0$ , on the line?

Example 14

Graph the equation $3 x + y = - 1$ .

Solution

Step 1: Find three points that are solutions to the equation.

$3 x + y = - 1$

Step 2: First, solve the equation for $y$ .

$y = - 3 x - 1$

Step 3: We’ll let $x$ be $0$ , $1$ , and $- 1$ to find 3 points.

The ordered pairs are shown in the below table. Plot the points, check that they line up, and draw the line. See Figure 3.9.33.

$3 x + y = - 1$
$x$	$y$	$(x, y)$
$0$	$- 1$	$(0, - 1)$
$1$	$- 4$	$(1, - 4)$
$- 1$	$2$	$(- 1, 2)$

Try It

33) Graph the equation

2 x + y = 2

.

Solution

34) Graph the equation $4 x + y = - 3$ .

Solution

If you can choose any three points to graph a line, how will you know if your graph matches the one shown in the answers in the book? If the points where the graphs cross the $x$ – and $y$ -axis are the same, the graphs match!

The equation in Example 3.9.14, was written in standard form, with both $x$ and $y$ on the same side. We solved that equation for $y$ in just one step. But for other equations in standard form, it is not that easy to solve for $y$ , so we will leave them in standard form. We can still find a first point to plot by letting $x = 0$ and solving for $y$ . We can plot a second point by letting $y = 0$ and then solving for $x$ . Then we will plot a third point by using some other value for $x$ or $y$ .

Example 15

Graph the equation $2 x - 3 y = 6$

Solution

Step 1: Find three points that are solutions to the equation.

$2 x - 3 y = 6$

Step 2: First, let $x = 0$ .

$2 (0) - 3 y = 6$

Step 3: Solve for $y$ .

$\begin{aligned} - 3 y & = 6 \\ y & = - 2 \end{aligned}$

Step 4: Now let $y = 0$ .

$2 x - 3 (0) = 6$

Step 5: Solve for $x$ .

$\begin{aligned} 2 x & = 6 \\ x & = 3 \end{aligned}$

Step 6: We need a third point. Remember, we can choose any value for $x$ or $y$ . We’ll let $x = 6$ .

$2 (6) - 3 y = 6$

Step 7: Solve for $y$ .

$\begin{aligned} 12 - 3 y & = 6 \\ - 3 y & = - 6 \\ y & = 2 \end{aligned}$

We list the ordered pairs in the below table. Plot the points, check that they line up, and draw the line. See Figure 3.9.36

$2 x - 3 y = 6$
$x$	$y$	$(x, y)$
$0$	$- 2$	$(0, - 2)$
$3$	$0$	$(3, 0)$
$6$	$2$	$(6, 2)$

Try It

35) Graph the equation $4 x + 2 y = 8$ .

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The straight line goes through the points (negative 1, 6), (0, 4), (1, 2), (2, 0), (3, negative 2), and (4, negative 4). The line has arrows on both ends pointing to the outside of the figure. — Figure 3.9.37

36) Graph the equation $2 x - 4 y = 8$ .

Solution

Graph Vertical and Horizontal Lines

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

Let’s consider the equation $x = - 3$ . 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 $y$ is, 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 the table below.

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

Plot the points from the above table and connect them with a straight line. Notice in Figure 3.9.39 that we have graphed a vertical line.

The figure shows a vertical straight line drawn through three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points which are labeled by their ordered pairs (negative 3, 1), (negative 3, 2), and (negative 3, 3). A vertical straight line goes through all three points. The line has arrows on both ends pointing to the outside of the figure. The line is labeled with the equation x equals negative 3. — Figure 3.9.39

Vertical Line

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

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

Example 16

Graph the equation $x = 2$ .

Solution

The equation has only one variable, $x$ , and $x$ is always equal to $2$ . We create the table below 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 $2$ . See Figure 3.9.40.

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

The figure shows a straight vertical line drawn through three points on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. Dots mark off the three points which are labeled by their ordered pairs (2, 1), (2, 2), and (2, 3). A vertical straight line goes through all three points. The line has arrows on both ends pointing to the outside of the figure. The line is labeled with the equation x equals 2. — Figure 3.9.40

Try It

37) Graph the equation $x = 5$ .

Solution

The figure shows a straight vertical line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The straight line goes through the points (5, 1), (5, 2), (5, 3), and all other points with first coordinate 5. The line has arrows on both ends pointing to the outside of the figure. — Figure 3.9.41

38) Graph the equation $x = - 2$ .

Solution

What if the equation has $y$ but no $x$ ? Let’s graph the equation $y = 4$ . 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 the below table and then choose any values for $x$ . We’ll use $0$ , $2$ , and $4$ for the $x$ -coordinates.

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

The graph is a horizontal line passing through the $y$ -axis at $4$ . See Figure 3.9.43.

Horizontal Line

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

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

Example 17

Graph the equation $y = - 1$

Solution

The equation $y = - 1$ has only one variable, $y$ . The value of $y$ is constant. All the ordered pairs in the below table have the same $y$ -coordinate. The graph is a horizontal line passing through the $y$ -axis at $- 1$ , as shown in Figure 3.9.44.

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

Try It

39) Graph the equation $y = - 4$ .

Solution

The figure shows a straight horizontal line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. The straight line goes through the points (negative 4, negative 4), (0, negative 4), (4, negative 4), and all other points with second coordinate negative 4. The line has arrows on both ends pointing to the outside of the figure. — Figure 3.9.45

40) Graph the equation $y = 3$ .

Solution

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

The equation $y = 4 x$ has both $x$ and $y$ . The value of $y$ depends on the value of $x$ . The $y$ -coordinate changes according to the value of $x$ . The equation $y = 4$ has only one variable. The value of $y$ is constant. The $y$ -coordinate is always $4$ . It does not depend on the value of $x$ . See the table below.

$y = 4 x$			$y = 4$
$x$	$y$	$(x, y)$	$x$	$y$	$(x, y)$
$0$	$0$	$(0, 0)$	$0$	$4$	$(0, 4)$
$1$	$4$	$(1, 4)$	$1$	$4$	$(1, 4)$
$2$	$8$	$(2, 8)$	$2$	$4$	$(2, 4)$

The figure shows a two straight lines drawn on the same x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. One line is a straight horizontal line labeled with the equation y equals 4. The other line is a slanted line labeled with the equation y equals 4x. — Figure 3.9.47

Notice, in Figure 3.9.47, the equation $y = 4 x$ gives a slanted line, while $y = 4$ gives a horizontal line.

Example 18

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

Solution

Notice that the first equation has the variable $x$ , while the second does not. See the below table. The two graphs are shown in Figure 3.9.48.

$y = - 3 x$			$y = - 3$
$x$	$y$	$(x, y)$	$x$	$y$	$(x, y)$
$0$	$0$	$(0, 0)$	$0$	$- 3$	$(0, - 3)$
$1$	$- 3$	$(1, - 3)$	$1$	$- 3$	$(1, - 3)$
$2$	$- 6$	$(2, - 6)$	$2$	$- 3$	$(2, - 3)$

Try It

41) Graph

y = - 4 x

and

y = - 4

in the same rectangular coordinate system.

Solution

The figure shows a two straight lines drawn on the same x y-coordinate plane. The x-axis of the plane runs from negative 12 to 12. The y-axis of the plane runs from negative 12 to 12. One line is a straight horizontal line going through the points (negative 4, negative 4), (0, negative 4), (4, negative 4), and all other points with second coordinate negative 4. The other line is a slanted line going through the points (negative 2, 8), (negative 1, 4), (0, 0), (1, negative 4), and (2, negative 8). — Figure 3.9.49

42) Graph $y = 3$ and $y = 3 x$ in the same rectangular coordinate system.

Solution

Identify the x-Intercept and y-Intercept from an Equation and its Graph

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 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 the line.

Intercepts of a Line

The points where a line crosses the

x

-axis and the

y

-axis are called the intercepts of a line.

Let’s look at the graphs of the lines in Figure 3.9.51.

Examples of graphs crossing the

x

-negative axis.

Four figures, each showing a different straight line on the x y- coordinate plane. The x- axis of the planes runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Figure a shows a straight line crossing the x- axis at the point (3, 0) and crossing the y- axis at the point (0, 6). The graph is labeled with the equation 2x plus y equals 6. Figure b shows a straight line crossing the x- axis at the point (4, 0) and crossing the y- axis at the point (0, negative 3). The graph is labeled with the equation 3x minus 4y equals 12. Figure c shows a straight line crossing the x- axis at the point (5, 0) and crossing the y- axis at the point (0, negative 5). The graph is labeled with the equation x minus y equals 5. Figure d shows a straight line crossing the x- axis and y- axis at the point (0, 0). The graph is labeled with the equation y equals negative 2x. — Figure 3.9.51

First, notice where each of these lines crosses the $x$ negative axis. See the table below.

Figure	The line crosses the x-axis at:	Ordered pair of this point
Figure (a)	$3$	$(3, 0)$
Figure (b)	$4$	$(4, 0)$
Figure (c)	$5$	$(5, 0)$
Figure (d)	$0$	$(0, 0)$

Do you see a pattern?

For each row, 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 $(a, 0)$ and is called the $x$ -intercept of a line. The $x$ -intercept occurs when $y$ is zero.

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

Figure	The line crosses the y-axis at:	Ordered pair for this point
Figure (a)	$6$	$(0, 6)$
Figure (b)	$- 3$	$(0, - 3)$
Figure (c)	$- 5$	$(0, 5)$
Figure (d)	$0$	$(0, 0)$

What is the pattern here?

In each row, 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 $(0, b)$ and is called the $y$ –intercept of the line. The $y$ -intercept occurs when $x$ is zero.

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

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

x

y

The

x

-intercept occurs when y is zero.

a

0

The

y

-intercept occurs when x is zero

0

b

Example 19

Find the $x$ -intercept and $y$ -intercept on each graph.

Three figures, each showing a different straight line on the x y- coordinate plane. The x- axis of the planes runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Figure a shows a straight line going through the points (negative 6, 5), (negative 4, 4), (negative 2, 3), (0, 2), (2, 1), (4, 0), and (6, negative 1). Figure b shows a straight line going through the points (0, negative 6), (1, negative 3), (2, 0), (3, 3), and (4, 6). Figure c shows a straight line going through the points (negative 6, 1), (negative 5, 0), (negative 4, negative 1), (negative 3, negative 2), (negative 2, negative 3), (negative 1, negative 4), (0, negative 5), and (1, negative 6). — Figure 3.9.52

Solution

a. The graph crosses the $x$ -axis at the point $(4, 0)$ . The $x$ -intercept is $(4, 0)$ .
The graph crosses the $y$ -axis at the point $(0, 2)$ . The $y$ -intercept is $(0, 2)$ .

b. The graph crosses the $x$ -axis at the point $(2, 0)$ . The $x$ -intercept is $(2, 0)$ .
The graph crosses the $y$ -axis at the point $(0, - 6)$ . The $y$ -intercept is $(0, - 6)$ .

c. The graph crosses the $x$ -axis at the point $(- 5, 0)$ . The $x$ -intercept is $(- 5, 0)$ .
The graph crosses the $y$ -axis at the point $(0, - 5)$ . The $y$ -intercept is $(0, - 5)$ .

Try It

43) Find the $x$ -intercept and $y$ -intercept on the graph.

A figure showing a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 10 to 10. The y- axis of the planes runs from negative 10 to 10. The straight line goes through the points (negative 8, negative 10), (negative 6, negative 8), (negative 4, negative 6), (negative 2, negative 4), (0, negative 2), (2, 0), (4, 2), (6, 4), (8, 6), and (10, 8). — Figure 3.9.53

Solution

$x$ -intercept: $(2, 0)$ ; $y$ -intercept: $(0, - 2)$

44) Find the $x$ -intercept and $y$ -intercept on the graph.

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 10 to 10. The y- axis of the planes runs from negative 10 to 10. The straight line goes through the points (negative 9, 8), (negative 6, 6), (negative 3, 4), (0, 2), (3, 0), (6, negative 2), and (9, negative 4). — Figure 3.9.54

Solution

$x$ -intercept: $(3, 0)$ , $y$ -intercept: $(0, 2)$

Intercepts from an Equation of a Line

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

Find the $x$ -intercept and $y$ -intercept from the Equation of a Line

Use the equation of the line. To find:

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

Example 20

Find the intercepts of $2 x + y = 6$ .

Solution

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

$2 x + y = 6$

$x$

$y$

0

x

-intercept

0

y

-intercept

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

Step 1: Let $y = 0$ .

$\begin{aligned} 2 x + y & = 6 \\ 2 x + 0 & = 6 \end{aligned}$

Step 2: Simplify.

$\begin{array}{rcl} 2 x & = & 6 \\ x & = & 3 \end{array}$

Step 3: The $x$ -intercept is.

$(3, 0)$

Step 4: To find the $y$ -intercept, let $x = 0$ .

Step 5: Let $x = 0$ .

$\begin{array}{rcl} 2 x + y & = & 6 \\ 2 \times 0 + y & = & 6 \end{array}$

Step 6: Simplify.

$\begin{array}{rcl} 0 + y & = & 6 \\ y & = & 6 \end{array}$

Step 7: The $y$ -intercept is.

$(0, 6)$

The intercepts are the points $(3, 0)$ and $(0, 6)$ as shown in the below table.

$2 x + y = 6$

$x$

$y$

3

0

0

6

Try It

45) Find the intercepts of $3 x + y = 12$ .

Solution

$x$ -intercept: $(4, 0)$ , $y$ -intercept: $(0, 12)$

46) Find the intercepts of $x + 4 y = 8$ .

Solution

$x$ -intercept: $(8, 0)$ , $y$ -intercept: $(0, 2)$

Example 21

Find the intercepts of $4 x - 3 y = 12$ .

Solution

Step 1: To find the $x$ -intercept, let $y = 0$ .

$\begin{aligned} 4 x - 3 y & = 12 \\ 4 x - 3 \times (0) & = 12 \end{aligned}$

Step 3: Simplify.

$\begin{array}{rcl} 4 x - 0 & = & 12 \\ 4 x & = & 12 \\ x & = & 3 \end{array}$

Step 4: The $x$ -intercept is:

$(3, 0)$

Step 5: To find the $y$ -intercept, let $x = 0$ .

Step 6: Let $x = 0$ .

$\begin{aligned} 4 x - 3 y & = 12 \\ 4 \times 0 - 3 y & = 12 \end{aligned}$

Step 7: Simplify.

$\begin{array}{rcl} 0 - 3 y & = & 12 \\ - 3 y & = & 12 \\ y & = & - 4 \end{array}$

Step 8: The $y$ -intercept is:

$(0, - 4)$

The intercepts are the points $(3, 0)$ and $(0, - 4)$ as shown in the following table.

$4 x - 3 y = 12$

$x$

$y$

3

0

0

- 4

Try It

47) Find the intercepts of $3 x - 4 y = 12$ .

Solution

$x$ -intercept: $(4, 0)$ , $y$ -intercept: $(0, - 3)$

48) Find the intercepts of $2 x - 4 y = 8$ .

Solution

$x$ -intercept: $(4, 0)$ , $y$ -intercept: $(0, - 2)$

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-intercept and $y$ -intercept 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.

How to

Graph a linear equation using the intercepts.

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

Find the $x$ -intercept and $y$ -intercept of the line.
- Let $y = 0$ and solve for $x$ .
- Let $x = 0$ and solve for $y$ .
Find a third solution to the equation.
Plot the three points and check that they line up.
Draw the line.

Example 22

Graph $- x + 2 y = 6$ using the intercepts.

Solution

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

Let $y = 0$ and solve for $x$ .

Let $x = 0$ and solve for $y$ .

Find the $x$ -intercept.

$\begin{aligned} Let y & = 0 \\ - x + 2 y & = 6 \\ - x + 2 (0) & = 6 \\ - x & = 6 \\ x & = 6 \\ The x-intercept is (-6,0). \end{aligned}$

Find the $y$ -intercept.

$\begin{aligned} Let x & = 0 \\ - x + 2 y & = 6 \\ - 0 + 2 y & = 6 \\ 2 y & = 6 \\ y & = 3 \\ The y-intercept is (0,3). \end{aligned}$

Step 2: Find another solution to the equation.

We’ll use $x = 2$ .

$\begin{aligned} Let x & = 2 \\ - x + 2 y & = 6 \\ - 2 + 2 y & = 6 \\ 2 y & = 8 \\ y & = 4 \\ A third point is (2,4). \end{aligned}$

Step 3: Plot the three points.

Check that the points line up.

$x$	$y$	$(x, y)$
-6	0	(-6,0)
0	3	(0,3)
2	4	(2,4)

The graph has three points on the x- y coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Three points are marked at (negative 6, 0), (0, 3), and (2, 4). — Figure 3.9.55

Step 4: Draw the line.

a graph of a straight line going through three points on the x y- coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Three points are marked at (negative 6, 0), (0, 3), and (2, 4). The straight line is drawn through the points (negative 6, 0), (negative 4, 1), (negative 2, 2), (0, 3), (2, 4), (4, 5), and (6, 6). — Figure 3.9.56

See the graph.

Try It

49) Graph $x - 2 y = 4$ using the intercepts.

Solution

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 12 to 12. The y- axis of the planes runs from negative 12 to 12. The straight line goes through the points (negative 10, negative 7), (negative 8, negative 6), (negative 6, negative 5), (negative 4, negative 4), (negative 2, negative 3), (0, negative 2), (2, negative 1), (4, 0), (6, 1), (8, 2), and (10, 3). — Figure 3.9.57

50) Graph $- x + 3 y = 6$ using the intercepts.

Solution

Example 23

Graph $4 x - 3 y = 12$ using the intercepts.

Solution

$\begin{aligned} x-intercept, let & y = 0 & y-intercept, let x & = 0 & third point, let y & = 4 \\ 4 x - 3 y & = 12 & 4 x - 3 y & = 12 & 4 x - 3 y & = 12 \\ 4 x - 3 (0) & = 12 & 4 (0) - 3 y & = 12 & 4 x - 3 (4) & = 12 \\ 4 x & = 12 & - 3 y & = 12 & 4 x - 12 & = 12 \\ x & = 3 & y & = - 4 & 4 x & = 24 \\ x & = 6 \end{aligned}$

Find the intercepts and a third point. We list the points in the table below and show the graph below.

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

The figure shows the graph of a straight line going through three points on the x y- coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. Three points are marked at (0, negative 4), (3, 0), and (6, 4). The straight line is drawn through the points (0, negative 4), (3, 0), and (6, 4). — Figure 3.9.59

Try It

51) Graph $5 x - 2 y = 10$ using the intercepts.

Solution

The figure shows the graph of a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. The straight line goes through the points (0, negative 5), (2, 0), and (4, 5). — Figure 3.9.60

52) Graph $3 x - 4 y = 12$ using the intercepts.

Solution

Example 24

Graph $y = 5 x$ using the intercepts.

Solution

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

$\begin{aligned} x-intercept & y-intercept \\ Let y & = 0 & Let x & = 0 \\ y & = 5 x & y & = 5 x \\ 0 & = 5 x & y & = 5 \cdot 0 \\ 0 & = x & y & = 0 \\ (0, & 0) & (0, 0) \end{aligned}$

This line has only one intercept. It is the point $(0, 0)$ .

Step 2: To ensure accuracy we need to plot three points. Since the $x$ -intercept and $y$ -intercept are the same point, we need two more points to graph the line.

$\begin{aligned} Let x & = 1 & Let x & = - 1 \\ y & = 5 x & y & = 5 x \\ y & = 5 \cdot (1) & y & = 5 (- 1) \\ y & = 5 & y & = - 5 \end{aligned}$

See table below.

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

Step 3: Plot the three points, check that they line up, and draw the line.

The figure shows the graph of a straight line going through three points on the x y- coordinate plane. The x- axis of the plane runs from negative 10 to 10. The y- axis of the planes runs from negative 10 to 10. Three points are marked and labeled with their coordinates at (negative 1, negative 5), (0, 0), and (1, 5). The straight line is drawn through the points (negative 1, negative 5), (0, 0), and (1, 5). — Figure 3.9.362

Try It

53) Graph $y = 4 x$ using the intercepts.

Solution

54) Graph $y = - x$ the intercepts.

Solution

Key Concepts

Sign Patterns of the Quadrants

Quadrant I	Quadrant II	Quadrant III	Quadrant IV
$(x, y)$	$(x, y)$	$(x, y)$	$(x, y)$
$(+, +)$	$(-, +)$	$(-, -)$	$(+, -)$

Points on the Axes
- On the $x$ -axis, $y = 0$ . Points with a $y$ -coordinate equal to $0$ are on the $x$ -axis, and have coordinates $(a, 0)$ .
- On the $y$ -axis, $x = 0$ . Points with an $x$ -coordinate equal to $0$ are on the $y$ -axis, and have coordinates $(0, b)$ .
Solution of a Linear Equation
- An ordered pair $(x, y)$ is a solution of the linear equation $A x + B y = C$ , if the equation is a true statement when the $x$ and $y$ values of the ordered pair are substituted into the equation.
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.
Find the x-intercept and y-Intercept from the Equation of a Line
- Use the equation of the line to find the $x$ -intercept of the line, let $y = 0$ and solve for $x$ .
- Use the equation of the line to find the $y$ -intercept of the line, let $x = 0$ and solve for $y$ .

Graph a Linear Equation using the Intercepts
1. Find the $x$ -intercept and $y$ -intercept of the line.
  Let $y = 0$ and solve for $x$ .
  Let $x = 0$ and solve for $y$ .
2. Find a third solution to the equation.
3. Plot the three points and then check that they line up.
4. Draw the line.
Strategy for Choosing the Most Convenient Method to Graph a Line:
- Consider the form of the equation.
- If it only has one variable, it is a vertical or horizontal line.
  $x = a$ is a vertical line passing through the $x$ -axis at $a$ .
  $y = b$ is a horizontal line passing through the $y$ -axis at $b$ .
- If $y$ is isolated on one side of the equation, graph by plotting points.
- Choose any three values for $x$ and then solve for the corresponding $y$ -values.
- If the equation is of the form $A x + B y = C$ , find the intercepts. Find the $x$ -intercept and $y$ -intercept, then a third point.

Glossary

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

ordered pair: An ordered pair $(x, y)$ gives the coordinates of a point in a rectangular coordinate system.

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

quadrant: The $x$ -axis and the $y$ -axis divide a plane into four regions, called quadrants.

rectangular coordinate system: A grid system is used in algebra to show a relationship between two variables; also called the $x y$ -plane or the ‘coordinate plane’.

x-coordinate: The first number in an ordered pair $(x, y)$ .

y-coordinate: The second number in an ordered pair $(x, y)$ .

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.

horizontal line: A horizontal line is the graph of an equation of the form $y = b$ . The line passes through the $y$ -axis at $(0, b)$ .

vertical line: A vertical line is the graph of an equation of the form $x = a$ . The line passes through the $x$ -axis at $(a, 0)$ .

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

x-intercept: The point $(a, 0)$ where the line crosses the $x$ -axis; the $x$ –intercept occurs when $y$ is zero.

y-intercept: The point $(0, b)$ where the line crosses the $y$ -axis; the $y$ -intercept occurs when $x$ is zero.

Exercises: Plot Points in a Rectangular Coordinate System

Instructions: For questions 1-8, plot each point in a rectangular coordinate system and identify the quadrant in which the point is located.

1.

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

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (negative 4, 2) is plotted and labeled "a". The point (negative 1, negative 2) is plotted and labeled "b". The point (3, negative 5) is plotted and labeled "c". The point (negative 3, 5) is plotted and labeled “d”. The point (5 thirds, 2) is plotted and labeled “e”. — Figure P3.9.1

2.

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

3.

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

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (3, negative 1) is plotted and labeled "a". The point (negative 3, 1) is plotted and labeled "b". The point (negative 2, 2) is plotted and labeled "c". The point (negative 4, negative 3) is plotted and labeled “d”. The point (1, 14 fifths) is plotted and labeled “e”. — Figure P3.9.2

4.

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

5.

a. $(- 2, 0)$
b. $(- 3, 0)$
c. $(0, 0)$
d. $(0, 4)$
e. $(0, 2)$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (negative 2, 0) is plotted and labeled "a". The point (negative 3, 0) is plotted and labeled "b". The point (0, 0) is plotted and labeled "c". The point (0, 4) is plotted and labeled “d”. The point (0, 3) is plotted and labeled “e”. — Figure 3P.9.3

6.

a. $(0, 1)$
b. $(0, - 4)$
c. $(- 1, 0)$
d. $(0, 0)$
e. $(5, 0)$

7.

a. $(0, 0)$
b. $(0, - 3)$
c. $(- 4, 0)$
d. $(1, 0)$
e. $(0, - 2)$

Solution

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (0, 0) is plotted and labeled "a". The point (0, negative 3) is plotted and labeled "b". The point (negative 4, 0) is plotted and labeled "c". The point (1, 0) is plotted and labeled “d”. The point (0, negative 2) is plotted and labeled “e”. — Figure 3P.9.4

8.

a. $(- 3, 0)$
b. $(0, 5)$
c. $(0, - 2)$
d. $(2, 0)$
e. $(0, 0)$

Exercises: Name Ordered Pairs in a Rectangular Coordinate System

Instructions: For questions 9-12, name the ordered pair of each point shown in the rectangular coordinate system.

9.

Solution

A: $(- 4, 1)$
B: $(- 3, - 4)$
C: $(1, - 3)$
D: $(4, 3)$

10.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 10 to 10. The point (negative 4, 2) is plotted and labeled “A”. The point (3, 5) is plotted and labeled “B”. The point (negative 3, negative 2) is plotted and labeled “C”. The point (5, negative 1) is plotted and labeled “D”. — Figure 3P.9.6

11.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (0, negative 2) is plotted and labeled “A”. The point (negative 2, 0) is plotted and labeled “B”. The point (0, 5) is plotted and labeled “C”. The point (5, 0) is plotted and labeled “D”. — Figure 3P.9.7

Solution

A: $(0, - 2)$
B: $(- 2, 0)$
C: $(0, 5)$
D: $(5, 0)$

12.

The graph shows the x y-coordinate plane. The x- and y-axes each run from negative 6 to 6. The point (0, negative 1) is plotted and labeled “A”. The point (negative 1, 0) is plotted and labeled “B”. The point (4, 0) is plotted and labeled “C”. The point (0, 4) is plotted and labeled “D”. — Figure 3P.9.8

Exercises: Verify Solutions to an Equation in Two Variables

Instructions: For questions 13-20, which ordered pairs are solutions to the given equations?

13. $2 x + y = 6$

a. $(1, 4)$
b. $(3, 0)$
c. $(2, 3)$

Solution

a, b

14. $x + 3 y = 9$

a. $(0, 3)$
b. $(6, 1)$
c. $(- 3, - 3)$

15. $4 x - 2 y = 8$

a. $(3, 2)$
b. $(1, 4)$
c. $(0, - 4)$

Solution

a, c

16. $3 x - 2 y = 12$

a. $(4, 0)$
b. $(2, - 3)$
c. $(1, 6)$

17. $y = 4 x + 3$

a. $(4, 3)$
b. $(- 1, - 1)$
c. $(\frac{1}{2}, 5)$

Solution

b, c

18. $y = 2 x - 5$

a. $(0, - 5)$
b. $(2, 1)$
c. $(\frac{1}{2}, - 4)$

19. $y = \frac{1}{2} x - 1$

a. $(2, 0)$
b. $(- 6, - 4)$
c. $(- 4, - 1)$

Solution

a, b

20. $y = \frac{1}{3} x + 1$

a. $(- 3, 0)$
b. $(9, 4)$
c. $(- 6, - 1)$

Exercises: Complete a Table of Solutions to a Linear Equation

Instructions: For questions 21-32, complete the table to find solutions to each linear equation.

21. $y = 2 x - 4$

$x$	$y$	$(x, y)$
$0$
$2$
$- 1$

Solution

$x$	$y$	$(x, y)$
$0$	$- 4$	$(0, - 4)$
$2$	$0$	$(2, 0)$
$- 1$	$- 6$	$(- 1, - 6)$

22. $y = 3 x - 1$

$x$	$y$	$(x, y)$
$0$
$2$
$- 1$

23. $y = - x + 5$

$x$	$y$	$(x, y)$
$0$
$3$
$- 2$

Solution

$x$	$y$	$(x, y)$
$0$	$5$	$(0, 5)$
$3$	$2$	$(3, 2)$
$- 2$	$7$	$(- 2, 7)$

24. $y = - x + 2$

$x$	$y$	$(x, y)$
$0$
$3$
$- 2$

25. $y = \frac{1}{3} x + 1$

$x$	$y$	$(x, y)$
$0$
$3$
$6$

Solution

$x$	$y$	$(x, y)$
$0$	$1$	$(0, 1)$
$3$	$2$	$(3, 2)$
$6$	$3$	$(6, 3)$

26. $y = \frac{1}{2} x + 4$

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

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

$x$	$y$	$(x, y)$
$0$
$2$
$- 2$

Solution

$x$	$y$	$(x, y)$
$0$	$- 2$	$(0, - 2)$
$2$	$- 5$	$(2, - 5)$
$- 2$	$1$	$(- 2, 1)$

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

$x$	$y$	$(x, y)$
$0$
$3$
$- 3$

29. $x + 3 y = 6$

$x$	$y$	$(x, y)$
$0$
$3$
	$0$

Solution

$x$	$y$	$(x, y)$
$0$	$2$	$(0, 2)$
$3$	$1$	$(3, 1)$
$6$	$0$	$(6, 0)$

30. $x + 2 y = 8$

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

31. $2 x - 5 y = 10$

$x$	$y$	$(x, y)$
$0$
$10$
	$0$

Solution

$x$	$y$	$(x, y)$
$0$	$- 2$	$(0, - 2)$
$10$	$2$	$(10, 2)$
$5$	$0$	$(5, 0)$

32. $3 x - 4 y = 12$

$x$	$y$	$(x, y)$
$0$
$8$
	$0$

Exercises: Find Solutions to a Linear Equation

Instructions: For questions 33-48, find three solutions to each linear equation.

33. $y = 5 x - 8$

Solution

Answers will vary.

34. $y = 3 x - 9$

35. $y = - 4 x + 5$

Solution

Answers will vary.

36. $y = - 2 x + 7$

37. $x + y = 8$

Solution

Answers will vary.

38. $x + y = 6$

39. $x + y = - 2$

Solution

Answers will vary.

40. $x + y = - 1$

41. $3 x + y = 5$

Solution

Answers will vary.

42. $2 x + y = 3$

43. $4 x - y = 8$

Solution

Answers will vary.

44. $5 x - y = 10$

45. $2 x + 4 y = 8$

Solution

Answers will vary.

46. $3 x + 2 y = 6$

47. $5 x - 2 y = 10$

Solution

Answers will vary.

48. $4 x - 3 y = 12$

Exercises: Recognize the Relationship Between the Solutions of an Equation and its Graph

Instructions: For questions 49-52, for each ordered pair, decide:

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

49. $y = x + 2$

a. $(0, 2)$
b. $(1, 2)$
c. $(- 1, 1)$
d. $(- 3, - 1)$

Solution

a. yes; yes
b. no; no
c. yes; yes
d. yes; yes

50. $y = x - 4$

a. $(0, - 4)$
b. $(3, - 1)$
c. $(2, 2)$
d. $(1, - 5)$

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

a. $(0, - 3)$
b. $(2, - 2)$
c. $(- 2, - 4)$
d. $(4, 1)$

Solution

a. yes; yes
b. yes; yes
c. yes; yes
d. no; no

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

a. $(0, 2)$
b. $(3, 3)$
c. $(- 3, 2)$
d. $(- 6, 0)$

Exercises: Graph a Linear Equation by Plotting Points

Instructions: For questions 53-96, graph by plotting points.

53. $y = 3 x - 1$

Solution

54. $y = 2 x + 3$

55. $y = - \frac{9}{4} x + 3$

Solution

56. $y = - 3 x + 1$

57. $y = x + 2$

Solution

58. $y = x - 3$

59. $y = - x - 3$

Solution

60. $y = - x - 2$

61. $y = 2 x$

Solution

62. $y = 3 x$

63. $y = - 4 x$

Solution

64. $y = - 2 x$

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

Solution

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

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

Solution

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

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

Solution

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

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

Solution

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

73. $x + y = 6$

Solution

74. $x + y = 4$

75. $x + y = - 3$

Solution

76. $x + y = - 2$

77. $x - y = 2$

Solution

78. $x - y = 1$

79. $x - y = - 1$

Solution

80. $x - y = - 3$

81. $3 x + y = 7$

Solution

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to -7. The equation 3 x plus y equals 7 is graphed. — Figure 3P.9.27

82. $5 x + y = 6$

83. $2 x + y = - 3$

Solution

84. $4 x + y = - 5$

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

Solution

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

87. $\frac{2}{5} x - y = 4$

Solution

88. $\frac{3}{4} x - y = 6$

89. $2 x + 3 y = 12$

Solution

90. $4 x + 2 y = 12$

91. $3 x - 4 y = 12$

Solution

92. $2 x - 5 y = 10$

93. $x - 6 y = 3$

Solution

94. $x - 4 y = 2$

95. $5 x + 2 y = 4$

Solution

96. $3 x + 5 y = 5$

Exercises: Graph Vertical and Horizontal Lines

Instructions: For questions 97-108, graph each equation.

97. $x = 4$

Solution

98. $x = 3$

99. $x = - 2$

Solution

100. $x = - 5$

101. $y = 3$

Solution

102. $y = 1$

103. $y = - 5$

Solution

104. $y = - 2$

105. $x = \frac{7}{3}$

Solution

106. $x = \frac{5}{4}$

107. $y = - \frac{15}{4}$

Solution

The figure shows a straight horizontal line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The horizontal line goes through the points (0, negative 15/4), (1, negative 15/4), (2, negative 15/4) and all points with second coordinate negative 15/4. — Figure 3P.9.40

108. $y = - \frac{5}{3}$

Exercises: Graph a Pair of Equations in the Same Rectangular Coordinate System

Instructions: For questions 109-112, graph each pair of equations in the same rectangular coordinate system.

109. $y = 2 x$ and $y = 2$

Solution

110. $y = 5 x$ and $y = 5$

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

Solution

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

Exercises: Mixed Practice

Instructions: For questions 113-128, graph each equation.

113. $y = 4 x$

Solution

114. $y = 2 x$

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

Solution

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

117. $y = - x$

Solution

118. $y = x$

119. $x - y = 3$

Solution

120. $x + y = - 5$

121. $4 x + y = 2$

Solution

122. $2 x + y = 6$

123. $y = - 1$

Solution

124. $y = 5$

125. $2 x + 6 y = 12$

Solution

126. $5 x + 2 y = 10$

127. $x = 3$

Solution

The figure shows a straight vertical line drawn on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 7 to 7. The vertical line goes through the points (3, 0), (3, 1), (3, 2) and all points with first coordinate 3. — Figure 3P.9.50

128. $x = - 4$

Exercises: Identify the x and y-Intercepts on a Graph

Instructions: For questions 129-140, find the $x$ and $y$ -intercepts on each graph.

129.

Solution

$(3, 0), (0, 3)$

130.

131.

Solution

$(5, 0), (0, - 5)$

132.

133.

Solution

$(- 2, 0), (0, - 2)$

134.

135.

Solution

$(- 1, 0), (0, 1)$

136.

137.

Solution

$(6, 0), (0, 3)$

138.

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from negative 7 to 7. The y- axis of the planes runs from negative 7 to 7. The straight line goes through the points (negative 6, 5), (negative 4, 4), (negative 2, 3), (0, 2), (2, 1), (4, 0), and (6, negative 1). — Figure 3P.9.60

139.

Solution

$(0, 0)$

140.

Exercises: Find the x and y-Intercepts from an Equation of a Line

Instructions: For questions 141-168, find the intercepts for each equation.

141. $x + y = 4$

Solution

$(4, 0), (0, 4)$

142. $x + y = 3$

143. $x + y = - 2$

Solution

$(- 2, 0), (0, - 2)$

144. $x + y = - 5$

145. $x - y = 5$

Solution

$(5, 0), (0, - 5)$

146. $x - y = 1$

147. $x - y = - 3$

Solution

$(- 3, 0), (0, 3)$

148. $x - y = - 4$

149. $x + 2 y = 8$

Solution

$(8, 0), (0, 4)$

150. $x + 2 y = 10$

151. $3 x + y = 6$

Solution

$(2, 0), (0, 6)$

152. $3 x + y = 9$

153. $x - 3 y = 12$

Solution

$(12, 0), (0, - 4)$

154. $x - 2 y = 8$

155. $4 x - y = 8$

Solution

$(2, 0), (0, - 8)$

156. $5 x - y = 5$

157. $2 x + 5 y = 10$

Solution

$(5, 0), (0, 2)$

158. $2 x + 3 y = 6$

159. $3 x - 2 y = 12$

Solution

$(4, 0), (0, - 6)$

160. $3 x - 5 y = 30$

161. $y = \frac{1}{3} x + 1$

Solution

$(- 3, 0), (0, 1)$

162. $y = \frac{1}{4} x - 1$

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

Solution

$(- 10, 0), (0, 2)$

164. $y = \frac{1}{3} x + 4$

165. $y = 3 x$

Solution

$(0, 0)$

166. $y = - 2 x$

167. $y = - 4 x$

Solution

$(0, 0)$

168. $y = 5 x$

Exercises: Graph a Line Using the Intercepts

Instructions: For questions 169-194, graph using the intercepts.

169. $- x + 5 y = 10$

Solution

170. $- x + 4 y = 8$

171. $x + 2 y = 4$

Solution

172. $x + 2 y = 6$

173. $x + y = 2$

Solution

174. $x + y = 5$

175. $x + y = - 3$

Solution

176. $x + y = - 1$

177. $x - y = 1$

Solution

178. $x - y = 2$

179. $x - y = - 4$

Solution

180. $x - y = - 3$

181. $4 x + y = 4$

Solution

182. $3 x + y = 3$

183. $2 x + 4 y = 12$

Solution

184. $3 x + 2 y = 12$

185. $3 x - 2 y = 6$

Solution

186. $5 x - 2 y = 10$

187. $2 x - 5 y = - 20$

Solution

188. $3 x - 4 y = - 12$

189. $3 x - y = - 6$

Solution

190. $2 x - y = - 8$

191. $y = - 2 x$

Solution

192. $y = - 4 x$

193. $y = x$

Solution

194. $y = 3 x$

Exercises: Everyday Math

Instructions: For questions 195-200, answer the given everyday math word problems.

195. Weight of a baby. Mackenzie recorded her baby’s weight every two months. The baby’s age, in months, and weight, in pounds, are listed in the table below, and shown as an ordered pair in the third column.

a. Plot the points on a coordinate plane.

Coordinate plane with x-axis ranging from zero to twelve and y-axis ranging from zero to 25. — Figure 3P.9.76

b. Why is only Quadrant I needed?

Age $x$	Weight $y$	$(x, y)$
$0$	$7$	$(0, 7)$
$2$	$11$	$(2, 11)$
$4$	$15$	$(4, 15)$
$6$	$16$	$(6, 16)$
$8$	$19$	$(8, 19)$
$10$	$20$	$(10, 20)$
$12$	$21$	$(12, 21)$

Solution

a.

The graph shows the x y-coordinate plane. The x- and y-axes each run from 0 to 25. The points (0, 7), (2, 11), (4, 15), (6, 16), (8, 19), (10, 20) and (12, 21) are plotted and labeled. — Figure 3P.9.77

b. Age and weight are only positive.

196. Weight of a child. Latresha recorded her son’s height and weight every year. His height, in inches, and weight, in pounds, are listed in the table below, and shown as an ordered pair in the third column.

a. Plot the points on a coordinate plane.

A coordinate plane is displayed with an x-axis ranging from zero to fifty and a y-axis ranging from zero to 50. — Figure 3P.9.78

b. Why is only Quadrant I needed?

Height $x$	Weight $y$	$(x, y)$
$28$	$22$	$(28, 22)$
$31$	$27$	$(31, 27)$
$33$	$33$	$(33, 33)$
$37$	$35$	$(37, 35)$
$40$	$41$	$(40, 41)$
$42$	$45$	$(42, 45)$

197. Motor home cost. The Robinsons rented a motor home for one week to go on vacation. It cost them $$ 594$ plus $$ 0.32$ per mile to rent the motor home, so the linear equation $y = 594 + 0.32 x$ gives the cost, $y$ , for driving $x$ miles. Calculate the rental cost for driving $400$ , $800$ , and $1200$ miles, and then graph the line.

Solution

$$ 722$ , $$ 850$ , $$ 978$

The figure shows a straight line drawn on the x y-coordinate plane. The x-axis of the plane runs from 0 to 1200 in increments of 100. The y-axis of the plane runs from 0 to 1000 in increments of 100. The straight line starts at the point (0, 594) and goes through the points (400, 722), (800, 850), and (1200, 978). The right end of the line has an arrow pointing up and to the right. — Figure 3P.9.79

198. Weekly earnings. At the art gallery where he works, Salvador gets paid $$ 200$ per week plus $15 %$ of the sales he makes, so the equation $y = 200 + 0.15 x$ gives the amount, $y$ , he earns for selling $x$ dollars of artwork. Calculate the amount Salvador earns for selling $$ 900$ , $$ 1600$ , and $$ 2000$ , and then graph the line.

199. Road trip. Damien is driving from Chicago to Denver, a distance of 1000 miles. The $x$ -axis on the graph below shows the time in hours since Damien left Chicago. The $y$ -axis represents the distance he has left to drive.

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from 0 to 16. The y- axis of the planes runs from 0 to 1200 in increments of 200. The straight line goes through the points (0, 1000), (3, 800), (6, 600), (9, 400), (12, 200), and (15, 0). The points (0, 1000) and (15, 0) are marked and labeled with their coordinates. — Figure 3P.9.80

a. Find the $x$ and $y$ -intercepts.
b. Explain what the $x$ and $y$ -intercepts mean for Damien.

Solution

a. $(0, 1000), (15, 0)$
b. At $(0, 1000)$ , he has been gone $0$ hours and has $1000$ miles left. At $(15, 0)$ , he has been gone $15$ hours and has $0$ miles left to go.

200. Road trip. Ozzie filled up the gas tank of his truck and headed out on a road trip. The $x$ -axis on the graph below shows the number of miles Ozzie drove since filling up. The $y$ -axis represents the number of gallons of gas in the truck’s gas tank.

The figure shows a straight line on the x y- coordinate plane. The x- axis of the plane runs from 0 to 350 in increments of 50. The y- axis of the planes runs from 0 to 18 in increments of 2. The straight line goes through the points (0, 16), (150, 8), and (300, 0). The points (0, 16) and (300, 0) are marked and labeled with their coordinates — Figure 3P.9.81

a. Find the $x$ and $y$ -intercepts.
b. Explain what the $x$ and $y$ -intercepts mean for Ozzie.

Exercises: Writing Exercises

Instructions: For questions 201-210, answer the given writing exercises.

201. Explain in words how you plot the point $(4, - 2)$ in a rectangular coordinate system.

Solution

Answers will vary.

202. How do you determine if an ordered pair is a solution to a given equation?

203. Is the point $(- 3, 0)$ on the $x$ -axis or $y$ -axis? How do you know?

Solution

Answers will vary.

204. Is the point $(0, 8)$ on the $x$ -axis or $y$ -axis? How do you know?

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

Solution

Answers will vary.

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

207. How do you find the $x$ -intercept of the graph of $3 x - 2 y = 6$ ?

Solution

Answers will vary.

208. 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?

209. 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?

Solution

Answers will vary.

210. Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation $y = 6$ ? Why?

License

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Plot Points on a Rectangular Coordinate System

Ordered Pair

The Origin

Quadrants

Points on the Axes

Verify Solutions to an Equation in Two Variables

Linear Equation

Standard Form of Linear Equation

Solution of a Linear Equation in Two Variables

Complete a Table of Solutions to a Linear Equation in Two Variables

Find Solutions to a Linear Equation in Two Variables

Recognize the Relationship Between the Solutions of an Equation and its Graph

Graph of a Linear Equation

Graph a Linear Equation by Plotting Points

Graph a linear equation by plotting points.

Graph Vertical and Horizontal Lines

Vertical Line

Horizontal Line

Identify the x-Intercept and y-Intercept from an Equation and its Graph

Intercepts of a Line

Intercepts from an Equation of a Line

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

Graph a Line Using the Intercepts

Graph a linear equation using the intercepts.

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Find the $x$ -intercept and $y$ -intercept from the Equation of a Line