6.3 Graphing a Line from Its Equation

Graphing a Line from Its Equation

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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 Table 6.1. So, the ordered pairs $open parenthesis 0, 3 close parenthesis$ , $open parenthesis 2, 0 close parenthesis$ , 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 Figure 6.4.

Table 6.1

$3 x + 2 y = 6$

$x$

$y$

$(x, y)$

open parenthesis 0, 3 close parenthesis

open parenthesis 2, 0 close parenthesis

\frac{3}{2}

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

Three points lie on a coordinate grid with labeled ticks from -6 to 6 on both axes: (0, 3), (1, 3/2), and (2, 0). They line up; no fourth point is drawn. — **Figure 6.4**

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 6.5. Notice the arrows on the ends of each side of the line. These arrows indicate the line continues.

The line 3x + 2y = 6 passes through four marked points: (0, 3), (1, 3/2), (2, 0), and (4, -3). Labeled ticks run from -6 to 6 on both axes. Arrows extend the line in both directions. — **Figure 6.5**

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.

The straight line y = 2x + 1 in dark blue, with a single red point sitting on it at (1, 3). The point is held to the line, so wherever along it the point is moved its coordinates still satisfy the equation — the line is the picture of every solution, not only the three the book tabulates. Adjustable parameter: x-value of the point (h) = 1. Viewing window: x from -9.5 to 9.5, y from -5.87 to 5.88. — The book graphs y = 2x + 1 by plotting the points (0, 1), (1, 3), and (2, 5) from its table and drawing the line through them. The marked point rides the line: at x = 1 it sits at (1, 3), and 3 = 2·1 + 1 checks. Drag the x-value anywhere — at x = −2.5 the point is (−2.5, −4) and −4 = 2·(−2.5) + 1 still checks — because the line is the picture of all the equation's solutions, not just the three plotted ones.

Notice that the point whose coordinates are $open parenthesis −2, 6 close parenthesis$ is on the line shown in Figure 6.6. If you substitute $x equals −2$ and $y equals 6$ into the equation, you find that it is a solution to the equation.

The line 3x + 2y = 6 and two marked points are shown. (-2, 6) lies on the line; (4, 1) does not. Labeled ticks run from -6 to 6 on both axes. — **Figure 6.6**

Checking if ordered pair negative 2, 6 solves 3x + 2y = 6. Substituting gives 3 times negative 2 plus 2 times 6 equals 6. Simplifying yields negative 6 plus 12 equals 6, then 6 equals 6. A checkmark indicates the solution is correct.

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

Checking if (4, 1) solves 3x + 2y = 6. Substituting gives 3(4) + 2(1) = 6, which simplifies to 12 + 2 = 6. This further simplifies to 14 does not equal 6, so the ordered pair is not a solution.

So $open parenthesis 4, 1 close parenthesis$ is not a solution to the equation $3 x + 2 y = 6$ . Therefore, the point $open parenthesis 4, 1 close parenthesis$ is not on the line. See Figure 6.5. 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 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 $3 x + 2 y = 6$ is called plotting points, or the Point–Plotting Method.

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

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 6.7.

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 6.7**

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

Example 3

Graph the equation $y equals −3 x$ .

Show solution

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

We list the points in Table 6.2.

Table 6.2

$y equals −3 x$

$x$

$y$

$(x, y)$

open parenthesis 0, 0 close parenthesis

−3

open parenthesis 1, −3 close parenthesis

−2

open parenthesis −2, 6 close parenthesis

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

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

Show solution

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 $x$ ?

The points are shown in Table 6.3.

Table 6.3

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

$x$

$y$

$(x, y)$

open parenthesis 0, 3 close parenthesis

open parenthesis 2, 4 close parenthesis

open parenthesis 4, 5 close parenthesis

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

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 $y equals 0$ 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 one-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 $y = \frac{1}{2} x + 3$ , 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.

$\begin{matrix} 2 x + y & = & 3 \\ y & = & −2 x + 3 \end{matrix}$

The solutions for $x equals 0$ , $x equals 1$ , and $x equals −1$ are shown in the Table 6.4. The graph is shown in Figure 6.8.

Table 6.4

$2 x + y = 3$

$x$

$y$

$(x, y)$

open parenthesis 0, 3 close parenthesis

open parenthesis 1, 1 close parenthesis

−1

open parenthesis −1, 5 close parenthesis

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 1, 5), (0, 3), and (1, 1). 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 2x plus y equals 3. — **Figure 6.8**

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

Example 5

Graph the equation $3 x + y = −1$ .

Show solution

**Table 6.5**
Find three points that are solutions to the equation.	$3 x + y = −1$
First, solve the equation for $y$ .	$y = −3 x - 1$

We’ll let $x$ be 0, 1, and $−1$ to find 3 points. The ordered pairs are shown in Table 6.6. Plot the points, check that they line up, and draw the line. See Figure 6.9.

Table 6.6

$3 x + y = −1$

$x$

$y$

$(x, y)$

−1

open parenthesis 0, −1 close parenthesis

−4

open parenthesis 1, −4 close parenthesis

−1

open parenthesis −1, 2 close parenthesis

A straight line passes through the points negative 1 comma 2, 0 comma negative 1, and 1 comma negative 4 on a coordinate plane. The line's equation is three x plus y equals negative one, indicating it continues infinitely in both directions. — **Figure 6.9**

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? Compare two distinct points on each line. If both lines contain the same two distinct points, the graphs match. Matching x- and y-intercepts determine the same line only when the intercepts are distinct; lines through the origin can have different slopes.

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 $x equals 0$ and solving for $y$ . We can plot a second point by letting $y equals 0$ 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 $2 x - 3 y = 6$ .

Show solution

**Table 6.7**
Find three points that are solutions to the equation.	$\begin{matrix} 2 x - 3 y & = & 6 \end{matrix}$
First, let $x equals 0$ .	$\begin{matrix} 2 (0) - 3 y & = & 6 \end{matrix}$
Solve for $y$ .	$\begin{matrix} −3 y & = & 6 \\ y & = & −2 \end{matrix}$
Now let $y equals 0$ .	$\begin{matrix} 2 x - 3 (0) & = & 6 \end{matrix}$
Solve for $x$ .	$\begin{matrix} 2 x & = & 6 \\ x & = & 3 \end{matrix}$
We need a third point. Remember, we can choose any value for $x$ or $y$ . We'll let $x equals 6$ .	$\begin{matrix} 2 (6) - 3 y & = & 6 \end{matrix}$
Solve for $y$ .	$\begin{matrix} 12 - 3 y & = & 6 \\ −3 y & = & −6 \\ y & = & 2 \end{matrix}$

We list the ordered pairs in Table 6.8. Plot the points, check that they line up, and draw the line. See Figure 6.10.

Table 6.8

$2 x - 3 y = 6$

$x$

$y$

$(x, y)$

−2

open parenthesis 0, −2 close parenthesis

open parenthesis 3, 0 close parenthesis

open parenthesis 6, 2 close parenthesis

Practice Makes Perfect

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

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

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

$y = x + 2$

ⓐ $open parenthesis 0, 2 close parenthesis$
ⓑ $open parenthesis 1, 2 close parenthesis$
ⓒ $open parenthesis −1, 1 close parenthesis$
ⓓ $open parenthesis −3, −1 close parenthesis$

The figure shows a straight line drawn on the x y-coordinate plane. Labeled ticks run from -6 to 6 on both axes. The straight line goes through the points (negative 6, negative 4), (negative 5, negative 3), (negative 4, negative 2), (negative 3, negative 1), (negative 2, 0), (negative 1, 1), (0, 2), (1, 3), (2, 4), (3, 5), (4, 6), and (5, 7).

Show solution

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

$y = x - 4$

ⓐ $open parenthesis 0, −4 close parenthesis$
ⓑ $open parenthesis 3, −1 close parenthesis$
ⓒ $open parenthesis 2, 2 close parenthesis$
ⓓ $open parenthesis 1, −5 close parenthesis$

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

ⓐ $open parenthesis 0, −3 close parenthesis$
ⓑ $open parenthesis 2, −2 close parenthesis$
ⓒ $open parenthesis −2, −4 close parenthesis$
ⓓ $open parenthesis 4, 1 close parenthesis$

Show solution

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

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

ⓐ $open parenthesis 0, 2 close parenthesis$
ⓑ $open parenthesis 3, 3 close parenthesis$
ⓒ $open parenthesis −3, 2 close parenthesis$
ⓓ $open parenthesis −6, 0 close parenthesis$

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

$y = 3 x - 1$

Show 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 3, negative 10), (negative 2, negative 7), (negative 1, negative 4), (0, negative 1), (1, 2), (2, 5), and (3, 8).

$y = 2 x + 3$

$y = −2 x + 2$

Show solution

$y = −3 x + 1$

$y = x + 2$

Show solution

$y = x - 3$

$y = − x - 3$

Show solution

$y = − x - 2$

$y equals 2 x$

Show solution

$y equals 3 x$

$y equals −4 x$

Show solution

$y equals −2 x$

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

Show solution

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

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

Show solution

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

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

Show solution

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

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

Show solution

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

$x + y = 6$

Show solution

$x + y = 4$

$x + y = −3$

Show solution

$x + y = −2$

$x - y = 2$

Show solution

$x - y = 1$

$x - y = −1$

Show solution

$x - y = −3$

$3 x + y = 7$

Show solution

$5 x + y = 6$

$2 x + y = −3$

Show solution

The figure shows a straight line drawn on the x y-coordinate plane. Labeled ticks run from -8 to 8 on both axes. The straight line goes through the points (negative 5, 7), (negative 4, 5), (negative 3, 3), (negative 2, 1), (negative 1, negative 1), (0, negative 3), (1, negative 5), and (2, negative 7).

$4 x + y = −5$

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

Show solution

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

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

Show solution

The line (2/5)x - y = 4 rises from left to right, passing through (-5, -6), (0, -4), and (5, -2). Labeled ticks run from -8 to 8 on both axes.

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

$2 x + 3 y = 12$

Show solution

$4 x + 2 y = 12$

$3 x - 4 y = 12$

Show solution

$2 x - 5 y = 10$

$x - 6 y = 3$

Show solution

$x - 4 y = 2$

$5 x + 2 y = 4$

Show solution

$3 x + 5 y = 5$

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