4.2 Graph Linear Equations in Two Variables

Learning Objectives

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

  • Recognize the relationship between the solutions of an equation and its graph.
  • Graph a linear equation by plotting points.
  • Graph vertical and horizontal lines.

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

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

3x+2y=6
x y \left(x,y\right)
0 3 \left(0,3\right)
2 0 \left(2,0\right)
1 \dfrac{3}{2} \left(1,\dfrac{3}{2}\right)
A graph that plots the points (0, 3), (1, three halves), and (2, 0).
Figure .1

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

Described in previous paragraph.
Figure .2

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 \left(-2,6\right) is on the line shown in (Figure 3). If you substitute x=-2 and y=6 into the equation, you find that it is a solution to the equation.

Graphs the equation 3x plus 2y equals 6. The points (negative 2, 6) and (4, 1) are plotted. The line goes through (−2, 6) but not (4, 1).
Figure .3

The figure shows a series of equations to check if the ordered pair (negative 2, 6) is a solution to the equation 3x plus 2y equals 6. The first line states “Test (negative 2, 6)”. The negative 2 is colored blue and the 6 is colored red. The second line states the two- variable equation 3x plus 2y equals 6. The third line shows the ordered pair substituted into the two- variable equation resulting in 3(negative 2) plus 2(6) equals 6 where the negative 2 is colored blue to show it is the first component in the ordered pair and the 6 is red to show it is the second component in the ordered pair. The fourth line is the simplified equation negative 6 plus 12 equals 6. The fifth line is the further simplified equation 6equals6. A check mark is written next to the last equation to indicate it is a true statement and show that (negative 2, 6) is a solution to the equation 3x plus 2y equals 6.

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

The figure shows a series of equations to check if the ordered pair (4, 1) is a solution to the equation 3x plus 2y equals 6. The first line states “What about (4, 1)?”. The 4 is colored blue and the 1 is colored red. The second line states the two- variable equation 3x plus 2y equals 6. The third line shows the ordered pair substituted into the two- variable equation resulting in 3(4) plus 2(1) equals 6 where the 4 is colored blue to show it is the first component in the ordered pair and the 1 is red to show it is the second component in the ordered pair. The fourth line is the simplified equation 12 plus 2 equals 6. A question mark is placed above the equals sign to indicate that it is not known if the equation is true or false. The fifth line is the further simplified statement 14 not equal to 6. A “not equals” sign is written between the two numbers and looks like an equals sign with a forward slash through it.

So \left(4,1\right) is not a solution to the equation 3x+2y=6. Therefore, the point \left(4,1\right) is not on the line. See (Figure 2). 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 3x+2y=6.

Graph of a linear equation

The graph of a linear equation Ax+By=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 1

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

Graphs the line 2x−3.

For each ordered pair, decide:

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

A \left(0,-3\right) B \left(3,3\right) C \left(2,-3\right) D \left(-1,-5\right)

Solution

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

  1. The figure shows a series of equations to check if the ordered pairs (0, negative 3), (3, 3), (2, negative 3), and (negative 1, negative 5) are a solutions to the equation y equals 2x negative 3. The first line states the ordered pairs with the labels A: (0, negative 3), B: (3, 3), C: (2, negative 3), and D: (negative 1, negative 5). The first components are colored blue and the second components are colored red. The second line states the two- variable equation y equals 2x minus 3. The third line shows the four ordered pairs substituted into the two- variable equation resulting in four equations. The first equation is negative 3 equals 2(0) minus 3 where the 0 is colored clue and the negative 3 on the left side of the equation is colored red. The second equation is 3 equals 2(3) minus 3 where the 3 in parentheses is colored clue and the 3 on the left side of the equation is colored red. The third equation is negative 3 equals 2(2) minus 3 where the 2 in parentheses is colored clue and the negative 3 on the left side of the equation is colored red. The fourth equation is negative 5 equals 2(negative 1) minus 3 where the negative 1 is colored clue and the negative 5 is colored red. Question marks are placed above all the equal signs to indicate that it is not known if the equations are true or false. The fourth line shows the simplified versions of the four equations. The first is negative 3 equals negative 3 with a check mark indicating (0, negative 3) is a solution. The second is 3 equals 3 with a check mark indicating (3, 3) is a solution. The third is negative 3 not equals 1 indicating (2, negative 3) is not a solution. The fourth is negative 5 equals negative 5 with a check mark indicating (negative 1, negative 5) is a solution.
  2. Plot the points A \left(0,-3\right), B \left(3,3\right), C \left(2,-3\right), and D \left(-1,-5\right).
    Graph of the equation 2x−3. The points described in the previous paragraph are plotted.

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

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

TRY IT 1.1

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

  • a solution to the equation.
  • on the line.

a) \left(0,-1\right) b) \left(2,5\right)

Graph of the equation y = 3x−1.

Show answer

a) yes, yes b) yes, yes

TRY IT 1.2

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

  • a solution to the equation
  • on the line

a) \left(3,-1\right) b) \left(-1,-4\right)

Graph of the equation y = 3x−1.

Show answer

a) no, no b) yes, yes

Graph a Linear Equation by Plotting Points

There are several methods that can be used to graph a linear equation. The method we used to graph 3x+2y=6 is called plotting points, or the Point–Plotting Method.

EXAMPLE 2

How To Graph an Equation By Plotting Points

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

Solution

The figure shows the three step procedure for graphing a line from the equation using the example equation y equals 2x minus 1. The first step is to “Find three points whose coordinates are solutions to the equation. Organize the solutions in a table”. The remark is made that “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”. The work for the first step of the example is shown through a series of equations aligned vertically. From the top down, the equations are y equals 2x plus 1, x equals 0 (where the 0 is blue), y equals 2x plus 1, y equals 2(0) plus 1 (where the 0 is blue), y equals 0 plus 1, y equals 1, x equals 1 (where the 1 is blue), y equals 2x plus 1, y equals 2(1) plus 1 (where the 1 is blue), y equals 2 plus 1, y equals 3, x equals negative 2 (where the negative 2 is blue), y equals 2x plus 1, y equals 2(negative 2) plus 1 (where the negative 2 is blue), y equals negative 4 plus 1, y equals negative 3. The work is then organized in a table. The table has 5 rows and 3 columns. The first row is a title row with the equation y equals 2x plus 1. The second row is a header row and it labels each column. The first column header is “x”, the second is “y” and the third is “(x, y)”. Under the first column are the numbers 0, 1, and negative 2. Under the second column are the numbers 1, 3, and negative 3. Under the third column are the ordered pairs (0, 1), (1, 3), and (negative 2, negative 3).The second step is to “Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work!” For the example the points are (0, 1), (1, 3), and (negative 2, negative 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). The question “Do the points line up?” is stated and followed with the answer “Yes, the points line up.”The third step of the procedure is “Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.” 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. The statement “This line is the graph of y equals 2x plus 1” is included next to the graph.

TRY IT 2.1

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

Show answer
Graph of the equation y = 2x−3.

TRY IT 2.2

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

Show answer
Graph of the equation y = −2+4.

HOW TO: Graph a linear equation by plotting points.

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

  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.

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 4).

Figure a shows three points with a straight line through them. Figure b shows three points that do not lie on the same line.
Figure .4

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

EXAMPLE 3

Graph the equation y=-3x.

Solution

Find three points that are solutions to the equation. Here, again, it’s easier to choose values for x. Do you see why?

The figure shows three sets of equations used to determine ordered pairs from the equation y equals negative 3x. The first set has the equations: x equals 0 (where the 0 is blue), y equals negative 3x, y equals negative 3(0) (where the 0 is blue), y equals 0. The second set has the equations: x equals 1 (where the 1 is blue), y equals negative 3x, y equals negative 3(1) (where the 1 is blue), y equals negative 3. The third set has the equations: x equals negative 2 (where the negative 2 is blue), y equals negative 3x, y equals negative 3(negative 2) (where the negative 2 is blue), y equals 6.

We list the points in the table below.

y=-3x
x y \left(x,y\right)
0 0 \left(0,0\right)
1 -3 \left(1,-3\right)
-2 6 \left(-2,6\right)

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

Graph of the equation y = −3x. The points listed in the previous table are plotted.

 

TRY IT 3.1

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

Show answer
A graph of the equation y = −4x.

EXAMPLE 3.2

Graph the equation by plotting points: y=x.

Show answer
A graph of the equation y = x.

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 4

Graph the equation y=\dfrac{1}{2}x+3.

Solution

Find three points that are solutions to the equation. Since this equation has the fraction \dfrac{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?

The figure shows three sets of equations used to determine ordered pairs from the equation y equals (one half)x plus 3. The first set has the equations: x equals 0 (where the 0 is blue), y equals (one half)x plus 3, y equals (one half)(0) plus 3 (where the 0 is blue), y equals 0 plus 3, y equals 3. The second set has the equations: x equals 2 (where the 2 is blue), y equals (one half)x plus 3, y equals (one half)(2) plus 3 (where the 2 is blue), y equals 1 plus 3, y equals 4. The third set has the equations: x equals 4 (where the 4 is blue), y equals (one half)x plus 3, y equals (one half)(4) plus 3 (where the 4 is blue), y equals 2 plus 3, y equals 5.

The points are shown in the table below.

y=\dfrac{1}{2}x+3
x y \left(x,y\right)
0 3 \left(0,3\right)
2 4 \left(2,4\right)
4 5 \left(4,5\right)

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

The points listed in the previous table are plotted. The equation y = 1 half x + 3 is graphed.

TRY IT 4. 1

Graph the equation y=\dfrac{1}{3}x-1.

Show answer
A graph of the equation y = 1 third x−1.

TRY IT 4.2

Graph the equation y=\dfrac{1}{4}x+2.

Show answer
A graph of the equation y = 1 fourth + 2.

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 2x+y=3. If y=0 what is the value of x?

The figure shows a set of equations used to determine an ordered pair from the equation 2x plus y equals 3. The first equation is y equals 0 (where the 0 is red). The second equation is the two- variable equation 2x plus y equals 3. The third equation is the onenegative variable equation 2x plus 0 equals 3 (where the 0 is red). The fourth equation is 2x equals 3. The fifth equation is x equals three halves. The last line is the ordered pair (three halves, 0).

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=\dfrac{1}{2}x+3, we carefully chose values for x so as not to graph fractions at all. If we solve the equation 2x+y=3 for y, it will be easier to find three solutions to the equation.

\begin{array}{ccc}\hfill 2x+y& =\hfill & 3\hfill \\ \hfill y& =\hfill & -2x+3\hfill \end{array}

The solutions for x=0, x=1, and x=-1 are shown in the table below. The graph is shown in (Figure 5).

2x+y=3
x y \left(x,y\right)
0 3 \left(0,3\right)
1 1 \left(1,1\right)
-1 5 \left(-1,5\right)

 

The points listed in the previous table are plotted. The equation 2x + y = 3 is graphed.
Figure .5

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

EXAMPLE 5

Graph the equation 3x+y=-1.

Solution
Find three points that are solutions to the equation. 3x+y\phantom{\rule{0.5em}{0ex}}=\phantom{\rule{0.5em}{0ex}}-1
First, solve the equation for y. y\phantom{\rule{0.5em}{0ex}}=\phantom{\rule{0.5em}{0ex}}-3x-1

We’ll let x be 0, 1, and -1 to find 3 points. The ordered pairs are shown in the table below. Plot the points, check that they line up, and draw the line. See (Figure 6).

3x+y=-1
x y \left(x,y\right)
0 -1 \left(0,-1\right)
1 -4 \left(1,-4\right)
-1 2 \left(-1,2\right)
The points listed in the previous table are plotted. The equation 3x+y = −1 is graphed.
Figure .6

EXAMPLE 5.1

Graph the equation 2x+y=2.

Show answer
Graph of the equation 2 x + y = 2.

TRY IT 5.2

Graph the equation 4x+y=-3.

Show answer
Graph of the equation 4 x + y = −3.

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 5) 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 6

Graph the equation 2x-3y=6.

Solution
Find three points that are solutions to the equation. \begin{array}{ccc}\hfill 2x-3y& =\hfill & 6\hfill \end{array}
First, let x=0. \begin{array}{ccc}\hfill 2\left(0\right)-3y& =\hfill & 6\hfill \end{array}
Solve for y. \begin{array}{ccc}\hfill -3y& =\hfill & 6\hfill \\ \hfill y& =\hfill & -2\hfill \end{array}
Now let y=0. \begin{array}{ccc}\hfill 2x-3\left(0\right)& =\hfill & 6\hfill \end{array}
Solve for x. \begin{array}{ccc}\hfill 2x& =\hfill & 6\hfill \\ \hfill x& =\hfill & 3\hfill \end{array}
We need a third point. Remember, we can choose any value for x or y. We’ll let x=6. \begin{array}{ccc}\hfill 2\left(6\right)-3y& =\hfill & 6\hfill \end{array}
Solve for y. \begin{array}{ccc}\hfill 12-3y& =\hfill & 6\hfill \\ \hfill -3y& =\hfill & -6\hfill \\ \hfill y& =\hfill & 2\hfill \end{array}

We list the ordered pairs in the table below. Plot the points, check that they line up, and draw the line. See (Figure 7).

2x-3y=6
x y \left(x,y\right)
0 -2 \left(0,-2\right)
3 0 \left(3,0\right)
6 2 \left(6,2\right)
The points listed in previous table are plotted. The equation 2x − 3y = 6 is plotted.
Figure .7

TRY IT 6.1

Graph the equation 4x+2y=8.

Show answer
Graph of the equation 4x + 2y = 8.

TRY IT 6.2

Graph the equation 2x-4y=8.

Show answer
Graph of the equation 2x − 3y = 8.

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 \left(x,y\right)
-3 1 \left(-3,1\right)
-3 2 \left(-3,2\right)
-3 3 \left(-3,3\right)

Plot the points from the table and connect them with a straight line. Notice in (Figure 8) that we have graphed a vertical line.

The points listed in the previous table are plotted. The equation x = −3 is graphed. The resulting line is vertical.
Figure .8

Vertical line

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

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

EXAMPLE 7

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 9).

x=2
x y \left(x,y\right)
2 1 \left(2,1\right)
2 2 \left(2,2\right)
2 3 \left(2,3\right)
The points listed in the previous table are plotted. The equation x = 2 is graphed. The resulting line is vertical.
Figure .9

TRY IT 7.1

Graph the equation x=5.

Show answer
Graph of the equation x = 5. The resulting line is vertical.

TRY IT 7.2

Graph the equation x=-2.

Show answer
Graph of the equation x = −2. The resulting line is vertical.

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 table below and then choose any values for x. We’ll use 0, 2, and 4 for the x-coordinates.

y=4
x y \left(x,y\right)
0 4 \left(0,4\right)
2 4 \left(2,4\right)
4 4 \left(4,4\right)

The graph is a horizontal line passing through the y-axis at 4. See (Figure 10).

The points listed in the previous table are plotted. The equation y = 4 is graphed. The resulting line is horizontal.
Figure .10

Horizontal line

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

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

EXAMPLE 8

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 table below have the same y-coordinate. The graph is a horizontal line passing through the y-axis at -1, as shown in (Figure 11).

y=-1
x y \left(x,y\right)
0 -1 \left(0,-1\right)
3 -1 \left(3,-1\right)
-3 -1 \left(-3,-1\right)
The points listed in the previous table are plotted. The equation y = −1 is graphed. The resulting line is horizontal.
Figure .11

TRY IT 8.1

Graph the equation y=-4.

Show answer
Graph of the equation y = −4. The resulting line is horizontal.

TRY IT 8.2

Graph the equation y=3.

Show answer
Graph of the equation y = 3. The resulting line is horizontal.

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

The equation y=4x 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=4x y=4
x y \left(x,y\right) x y \left(x,y\right)
0 0 \left(0,0\right) 0 4 \left(0,4\right)
1 4 \left(1,4\right) 1 4 \left(1,4\right)
2 8 \left(2,8\right) 2 4 \left(2,4\right)
The equations y = 4 and y = 4x are graphed and labelled.
Figure .12

Notice, in (Figure 12), the equation y=4x gives a slanted line, while y=4 gives a horizontal line.

EXAMPLE 9

Graph y=-3x 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 table below. The two graphs are shown in (Figure 13).

y=-3x y=-3
x y \left(x,y\right) x y \left(x,y\right)
0 0 \left(0,0\right) 0 -3 \left(0,-3\right)
1 -3 \left(1,-3\right) 1 -3 \left(1,-3\right)
2 -6 \left(2,-6\right) 2 -3 \left(2,-3\right)
The equations y = −3 and y = −3x are graphed and labelled. The equation y = −3x is a slanted line while y = −3 is horizontal.
Figure .13

TRY IT 9.1

Graph y=-4x and y=-4 in the same rectangular coordinate system.

Show answer
The equations y = −4 and y = −4x are graphed and labelled. The equation y = −4x is a slanted line while y = −4 is horizontal.

TRY IT 9.2

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

Show answer
The equations y = 3 and y = 3x are graphed and labelled. The equation y = 3x is a slanted line while y = 3 is horizontal.

Key Concepts

  • 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.

Glossary

graph of a linear equation
The graph of a linear equation Ax+By=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 \left(0,b\right).
vertical line
A vertical line is the graph of an equation of the form x=a. The line passes through the x-axis at \left(a,0\right).

Practice Makes Perfect

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

In the following exercises, for each ordered pair, decide:

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

1. y=x+2

a) \left(0,2\right)
b) \left(1,2\right)
c) \left(-1,1\right)
d) \left(-3,-1\right)

Graph of the equation y = x + 2.

2. y=x-4

a) \left(0,-4\right)
b) \left(3,-1\right)
c) \left(2,2\right)
d) \left(1,-5\right)

Graph of the equation y = x − 4.

3. y=\dfrac{1}{2}x-3

a) \left(0,-3\right)
b) \left(2,-2\right)
c) \left(-2,-4\right)
d) \left(4,1\right)

Graph of the equation y = 1 half x − 3.

4. y=\dfrac{1}{3}x+2

a) \left(0,2\right)
b) \left(3,3\right)
c) \left(-3,2\right)
d) \left(-6,0\right)

Graph of the equation y = 1 third x + 2.

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

5. y=3x-1 6. y=2x+3
7. y=-3x+3 8. y=-3x+1
9. y=x+2 10. y=x-3
11. y=-x-3 12. y=-x-2
13. y=2x 14. y=3x
15. y=3x 16. y=-2x
17. y=\dfrac{1}{2}x+2 18. y=\dfrac{1}{3}x-1
19. y=\dfrac{4}{3}x-5 20. y=\dfrac{3}{2}x-3
21. y=-\dfrac{2}{5}x+1 22. y=-\dfrac{4}{5}x-1
23. y=-\dfrac{3}{2}x+2 24. y=-\dfrac{5}{3}x+4
25. x+y=6 26. x+y=4
27. x+y=-3 28. x+y=-3
29. x-y=2 30. x-y=1
31. x-y=-1 32. x-y=-3
33. 3x+y=7 34. 5x+y=6
35. 2x+y=-3 36. 4x+y=-5
37. \dfrac{1}{3}x+y=2 38. \dfrac{1}{2}x+y=3
39. \dfrac{2}{5}x+y=-4 40. \dfrac{3}{4}x-y=6
41. 2x+3y=12 42. 4x+2y=12
43. 3x-4y=12 44. 2x-5y=10
45. x-6y=3 46. x-4y=2
47. 3x+y=2 48. 3x+5y=5

Graph Vertical and Horizontal Lines

In the following exercises, graph each equation.

49. x=4 50. x=3
51. x=-2 52. x=-5
53. y=3 54. y=1
55. y=-5 56. y=-2
57. x=\dfrac{7}{3} 58. x=\dfrac{5}{4}
59. y=-\dfrac{15}{4} 60. y=-\dfrac{5}{3}

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

61. y=2x and y=2 62. y=5x and y=5
63. y=-\dfrac{1}{2}x and y=-\dfrac{1}{2} 64. y=-\dfrac{1}{3}x and y=-\dfrac{1}{3}

Mixed Practice

In the following exercises, graph each equation.

65. y=4x 66. y=2x
67. y=-\dfrac{1}{2}x+3 68. y=\dfrac{1}{4}x-2
69. y=-x 70. y=x
71. x-y=3 72. x+y=-5
73. 4x+y=2 74. 2x+y=6
75. y=-1 76. y=5
77. 2x+6y=12 78. 5x+2y=10
79. x=3 80. x=-4

Everyday Math

81. Motor home cost. The Stonechilds 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+0.32x gives the cost, y, for driving x miles. Calculate the rental cost for driving 400, 800, and 1200 miles, and then graph the line. 82. Weekly earnings. At the art gallery where he works, Archisma gets paid $200 per week plus 15% of the sales he makes, so the equation y=200+0.15x gives the amount, y, he earns for selling x dollars of artwork. Calculate the amount Archisma earns for selling $900, $1600, and $2000, and then graph the line.

Writing Exercises

83. Explain how you would choose three x– values to make a table to graph the line y=\dfrac{1}{5}x-2. 84. What is the difference between the equations of a vertical and a horizontal line?

Answers

1. a) yes; no b) no; no c) yes; yes d) yes; yes 3. a) yes; yes b) yes; yes c) yes; yes d) no; no
5.

Graph of the equation y = 3x − 1.

 

7.

Graph of the equation y = −3x + 3.

9.

Graph of the equation y = x + 2.

11.

Graph of the equation y = −x − 3.

 

13.

Graph of the equation y = 2x.

15. Click on the graph to see a larger version

Graph of line y=3x

17.

Graph of the equation y = 1 half x + 2.

19. Click on the graph to see a larger version

Graph of the equation y = 4 thirds x − 5.

21.

Graph of the equation y = − 2 fifths x + 1.

23.

Graph of the equation y = − 3 halves x + 2.

25.

Graph of the equation x + y = 6.

27.

Graph of the equation x + y = −3.

29.

Graph of the equation x − y = 2.

31.

Graph of the equation x − y = −1.

33. Click on the graph to see a larger version

Graph of the equation 3x + y = 7.

35.

Graph of the equation 2x + y = −3.

 

37.

Graph of the equation 1 third x + y = 2.

39. Click on the graph to see a larger version

Graph of the equation 2 fifths x + y = -4

41.

Graph of the equation 2x + 3y = 12.

43.

Graph of the equation 3x − 4y = 12.

45.

Graph of the equation x − 6y = 3.

47.

Graph of the equation 3x + y = 2.

49.

Graph of the equation x = 4. The resulting line is vertical.

51.

Graph of the equation x = −2. The resulting line is vertical.

53.

Graph of the line y = 3. The resulting line is horizontal.

55.

Graph of the line y = −5. The resulting line is horizontal.

57.

Graph of the equation x = 7 thirds. The resulting line is vertical.

59.

Graph of the equation y = − 15 fourths. The resulting line is horizontal.

61.

The equations y= 2x and y = 2 are graphed. The equation y = 2x is a slanted line while y = 2 is horizontal.

63.

The equations y = − 1 half x and y = − 1 half are graphed. The equation y = − 1 half x is a slanted line while y = − 1 half is horizontal.

65.

Graph of the equation y = 4x.

67.

Graph of the equation y = − 1 half x + 3.

69.

Graph of the equation y = − x.

71.

graph of the equation x − y = 3.

73.

Graph of the equation 4x + y = 2.

75.

Graph of the equation y = −1.

77.

Graph of the equation 2x + 6y = 12.

79.

Graph of the equation x = 3.

81. $722, $850, $978
Graph of the equation y = 594 + 0.32x.
83. Answers will vary.

Attributions

This chapter has been adapted from “Graph Linear Equations in Two Variables” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.

License

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Intermediate Algebra II Copyright © 2021 by Pooja Gupta is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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