11.2 Graphing Linear Equations

Learning objectives

Learning objectives

• Recognize the relation between the solutions of an equation and its graph

• Graph a linear equation by plotting points

• Graph vertical and horizontal lines

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

In Use the Rectangular Coordinate System, we found a few solutions to the equation $3x+2y=6$. They are listed in the table below. So, the ordered pairs $(0,3)$, $(2,0)$, $(1,\frac{3}{2})$, $(4,-3)$, are some solutions to the equation$3x+2y=6$. We can plot these solutions in the rectangular coordinate system as shown on the graph at right.

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

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 11.8. If you substitute $x=-2$ and $y=6$ into the equation, you find that it is a solution to the equation.

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

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 $3x+2y=6$.

Example 11.15

Solution

1. Frame the two tests. An ordered pair is a solution exactly when substituting its coordinates into $y=2x-3$ gives the stated $y$-value, and such a point must lie on the graphed line.
2. Test point A. For $(0,-3)$, $2(0)-3=-3$, so the pair is a solution and point A lies on the line.
3. Test point B. For $(3,3)$, $2(3)-3=6-3=3$, so the pair is a solution and point B lies on the line.
4. Test point C. For $(2,-3)$, $2(2)-3=4-3=1$, which is not $-3$, so the pair is not a solution and point C is not on the line.
5. Test point D. For $(-1,-5)$, $2(-1)-3=-2-3=-5$, so the pair is a solution and point D lies on the line.
6. Compare the results. The three successful substitutions match the three points shown on the line, while the failed substitution identifies the point off the line.

Takeaway: Substitution answers both questions at once: $(0,-3)$, $(3,3)$ and $(-1,-5)$ satisfy $y=2x-3$ and lie on its line, while $(2,-3)$ does neither.

Graph a Linear Equation by Plotting Points

There are several methods that can be used to graph a linear equation. The method we used at the start of this section to graph is called plotting points, or the Point-Plotting Method.

Let’s graph the equation $y=2x+1$ by plotting points.

We start by finding three points that are solutions to the equation. We can choose any value for $x$ or $y,$ and then solve for the other variable.

Since $y$ is isolated on the left side of the equation, it is easier to choose values for $x.$ We will use $0,1,$ and $-2$ for $x$ for this example. We substitute each value of $x$ into the equation and solve for $y.$

We can organize the solutions in a table. See Table 11.2.

Table

11.2

Now we plot the points on a rectangular coordinate system. Check that the points line up. If they did not line up, it would mean we made a mistake and should double-check all our work. See Figure 11.9.

Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line. The line is the graph of $y=2x+1.$

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 plot only 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. See Figure 11.11.

Example 11.16

Solution

1. Choose convenient $x$-values. Use $x=0$, $x=1$, and $x=-2$ so the products in $y=-3x$ are easy to calculate without fractions.
2. Evaluate the first pair. Substituting $x=0$ gives $y=-3(0)=0$, so one solution is $(0,0)$.
3. Evaluate the other pairs. Substituting $x=1$ gives $y=-3$, while substituting $x=-2$ gives $y=-3(-2)=6$, producing $(1,-3)$ and $(-2,6)$.
4. Plot and connect the points. Plot $(0,0)$, $(1,-3)$, and $(-2,6)$; because they form a straight line, draw the line through all three points.

Takeaway: Choosing $x=0,1,-2$ in $y=-3x$ gives $(0,0)$, $(1,-3)$ and $(-2,6)$; three points that line up determine the graph.

When an equation includes a fraction as the coefficient of $x,$ we can 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 11.17

Solution

1. Choose multiples of the denominator. Use $x=0$, $x=2$, and $x=4$ so multiplying by $\frac{1}{2}$ produces whole-number $y$-values.
2. Evaluate the first pair. Substituting $x=0$ gives $y=\frac{1}{2}(0)+3=3$, so one point is $(0,3)$.
3. Evaluate the other pairs. Substituting $x=2$ gives $y=4$, and substituting $x=4$ gives $y=5$, producing $(2,4)$ and $(4,5)$.
4. Plot and connect the points. Plot $(0,3)$, $(2,4)$, and $(4,5)$; they form a straight line, so draw the graph through all three points.

Takeaway: When the coefficient of $x$ is a fraction such as $\frac{1}{2}$, choose multiples of the denominator: $x=0,2,4$ give $(0,3)$, $(2,4)$ and $(4,5)$ with no fractions.

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.

Example 11.18

Solution

1. Choose convenient $x$-values. Use $x=0$, $x=1$, and $x=4$, which make the corresponding values of $y$ easy to find from $y=5-x$.
2. Evaluate the first pair. Substituting $x=0$ gives $y=5-0=5$, so one solution is $(0,5)$.
3. Evaluate the other pairs. Substituting $x=1$ gives $y=4$, and substituting $x=4$ gives $y=1$, producing $(1,4)$ and $(4,1)$.
4. Plot and connect the points. Plot $(0,5)$, $(1,4)$, and $(4,1)$; since they lie on a straight line, draw the line through all three points.

Takeaway: Any convenient $x$ works in $x+y=5$, so $x=0,1,4$ give $(0,5)$, $(1,4)$ and $(4,1)$, and the three points fix the line.

In the previous example, the three points we found were easy to graph. But this is not always the case. Let’s see what happens in the equation $2x+y=3.$ If $y$ is $0,$ what is the value of $x?$

The solution is the point $(\frac{3}{2},0).$ This point has a fraction for the $x$-coordinate. 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 $2x+y=3$ for $y,$ it will be easier to find three solutions to the equation.

Now we can choose values for $x$ that will give coordinates that are integers. The solutions for $x=0,x=1,$ and $x=-1$ are shown.

Example 11.19

Solution

1. Rewrite the equation for $y$. Subtracting $3x$ from $3x+y=-1$ gives $y=-3x-1$, which makes the output easier to calculate for chosen $x$-values.
2. Evaluate the first pair. Substituting $x=0$ gives $y=-1$, so one point is $(0,-1)$.
3. Evaluate the other pairs. Substituting $x=1$ gives $y=-4$, and substituting $x=-1$ gives $y=2$, producing $(1,-4)$ and $(-1,2)$.
4. Plot the three points. Mark $(0,-1)$, $(1,-4)$, and $(-1,2)$ on the coordinate plane.
5. Check and draw. The points form a straight line, and its intercepts agree with the displayed graph, so draw the line through all three points.

Takeaway: Solving $3x+y=-1$ for $y$ first turns the work into three substitutions: $x=0,1,-1$ give $(0,-1)$, $(1,-4)$ and $(-1,2)$.

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 $x=-3.$ 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 $-3.$

To make a table of solutions, we write $-3$ 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 size of our coordinate graph, we’ll use $1,2,$ and $3$ for the $y$-coordinates as shown in the table.

Then plot the points and connect them with a straight line. Notice in Figure 11.12 that the graph is a vertical line.

Example 11.20

Solution

1. Recognize the constant $x$-coordinate. In $x=2$, every solution has $x=2$, while the value of $y$ may be any number.
2. Build a solution table. Choose $y=1$, $y=2$, and $y=3$, giving the ordered pairs $(2,1)$, $(2,2)$, and $(2,3)$.
3. Plot the points. Mark the three points directly above $x=2$ on the coordinate plane.
4. Identify the line. The points have equal $x$-coordinates and lie in a vertical column, so the graph is a vertical line through $(2,0)$.

Takeaway: An equation in $x$ alone, such as $x=2$, allows every $y$ but only one $x$, so its graph is the vertical line through $(2,0)$.

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 $x.$

To make a table of solutions, write $4$ for all the $y$ values and then choose any values for $x.$

We’ll use $0,2,$ and $4$ for the $x$-values.

Plot the points and connect them, as shown in Figure 11.13. This graph is a horizontal line passing through the $y\mathrm{-axis}$ at $4.$

Example 11.21

Solution

1. Recognize the constant $y$-coordinate. In $y=-1$, every solution has $y=-1$, while the value of $x$ may be any number.
2. Build a solution table. Choose $x=-3$, $x=0$, and $x=3$, giving the ordered pairs $(-3,-1)$, $(0,-1)$, and $(3,-1)$.
3. Plot the points. Mark the three points on the horizontal line $1$ unit below the $x$-axis.
4. Identify the line. All three points have the same $y$-coordinate, so the graph is a horizontal line crossing the $y$-axis at $(0,-1)$.

Takeaway: An equation in $y$ alone, such as $y=-1$, forces every solution to share the same $y$-value, so its graph is the horizontal line through $(0,-1)$.

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 $y\mathrm{-coordinate}$ changes according to the value of $x.$

The equation $y=4$ has only one variable. The value of $y$ is constant. The $y\mathrm{-coordinate}$ is always $4.$ It does not depend on the value of $x.$

The graph shows both equations.

Notice that the equation $y=4x$ gives a slanted line whereas $y=4$ gives a horizontal line.

Example 11.22

Solution

1. Choose points for $y=-3x$. Setting $x=-1,0,1$ gives $y=3,0,-3$, so the line passes through $(-1,3)$, $(0,0)$, and $(1,-3)$.
2. Choose points for $y=-3$. The $y$-coordinate remains $-3$ for every $x$, so points include $(-1,-3)$, $(0,-3)$, and $(1,-3)$; the $y$-intercept is $(0,-3)$ and there is no $x$-intercept.
3. Compare the lines. The first line rises over run by $\frac{-3}{1}=-3$ from $(0,0)$ to $(1,-3)$, while the second has rise $0$ over run $1$, so it is horizontal at $y=-3$.
4. Draw both graphs. Plot all six points and draw a descending line through $(-1,3)$, $(0,0)$, and $(1,-3)$, plus a horizontal line through the three points with $y=-3$.

Takeaway: Choose method from the equation's form, and remember that a constant $y$-value produces a horizontal line with no $x$-intercept.

Section 11.2 Exercises

Section 11.2 Exercises

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

In each of the following exercises, an equation and its graph is shown. For each ordered pair, decide

1. ⓐ is the ordered pair a solution to the equation?

2. ⓑ is the point on the line?

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

Graph Vertical and Horizontal lines

In the following exercises, graph the vertical and horizontal lines.

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

Mixed Practice

In the following exercises, graph each equation.

Everyday Math

Writing Exercises

Self Check

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

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

1. $x-y=3$

1. $y=-x$

1. $4x+y=2$

1. $y=-1$

1. $x=3$

1. The graph of $y=3x-1$ is shown.

For each ordered pair, decide

1. ⓐ is the ordered pair a solution to the equation?

2. ⓑ is the point on the line?

1. $(0,-1)$

2. $(2,2)$

3. $(3,-1)$

4. $(-1,-4)$

1. $y=x+2$

1. $(0,2)$

2. $(1,2)$

3. $(-1,1)$

4. $(-3,1)$

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

1. $(0,-3)$

2. $(2,-2)$

3. $(-2,-4)$

4. $(4,1)$

1. $2x+6y=12$

1. Motor home cost The Robinsons rented a motor home for one week to go on vacation. It cost them $\text{\textdollar 594}$ plus $\text{\textdollar 0.32}$ 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 $400,800,\,\mathrm{and}\,1,200$ miles, and then graph the line.

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