4.2 Graph linear equations in two variables (Page 2/7)

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How to graph an equation by plotting points

Graph the equation $y = 2 x + 1$ 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.

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

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

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 goes through the points (negative 1, 6), (0, 4), (1, 2), (2, 0), (3, negative 2), (4, negative 4), and (5, negative 6). There are arrows at the ends of the line pointing to the outside of the figure.

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

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 [link] .

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.

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

Graph the equation $y = −3 x$ .

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 [link] .

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

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

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

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.

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

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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.

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

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 [link] .

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

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

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

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

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 12, negative 1), (negative 8, 0), (negative 4, 1), (0, 2), (4, 3), (8, 4), and (12, 5). The line has arrows on both ends pointing to the outside of the figure.

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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 $x$ ?

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 $y = \frac{1}{2} x + 3$ , 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 $x = 0$ , $x = 1$ , and $x = −1$ are shown in the [link] . The graph is shown in [link] .

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

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.

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

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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