GED Mathematical Reasoning: Graphing A Line

Recall that a line is a straight path made up of points. When we draw a line, we use arrows on each end to denote that a line extends in both directions forever. Furthermore, a line is a visual representation of a linear equation which is an equation where the highest power of the variable or variables is 1.

The relationship between an equation and its line is that the x- and y-coordinate of each point on the line satisfies the equation. Meaning, the x- and y-coordinate of each point making up the line, when substituted into the equation, will make the equation TRUE.

And this provides the foundation for graphing a line using a table of values.

Geometrically speaking, at least two points are needed to form a line. So to graph a linear equation on the coordinate plane, we will find at least two ordered pairs that satisfy the equation and use those points to sketch the line. Often, a third ordered pair is used as a “check” because if all three points lie in a straight line, then we can be confident the graph is correct.

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Graphing a line

  • Although it is not necessary to graph a equation, I encourage you to solve the equation for y before going any further. It will make completing the next step much easier.
  • The next step of the process is to choose three values for  and substitute each into the equation separately to solve for the corresponding y-value.
  • Our third and final step is to plot these three points, verify that they form a straight line, and then sketch the line.

 

Example 1

Graph the line using a table of values: y = 2x + 1

 

The equation you see in example one is a linear equation. Notice that the highest power of “x” and “y” is one.

This equation is solved for y, meaning that it has “y” on the left side of the equal sign and everything else on the right side. While it is not necessary to do so, I encourage you to put an equation in this form for graphing; at least for now – while you’re just beginning to master the graphing process. It will make completing Step 2 much easier.

The graph of this linear equation consists of all the ordered pair points that make this equation true and there are an infinite number of them. We only really need two points to sketch the line and we’ll use a third point as a “check.”

Since just about every line spans all x- and y-values, you may choose whichever x-values you’d like. It is common, though, to choose values close to zero. I’ll go with the x-values:  zero, 1 and 2. And a table is a great way to organize our work.

First, we’ll let “x” equal zero. To find the corresponding y-value, we just need to simplify the right side of the equation.  y=2(0) +1 = 1.  So the corresponding y-value is 1 and the ordered pair is:  (0, 1)

Next, we’ll let “x” equal one.  y=2(1) +1 = 3.  So the corresponding y-value is 3 and the ordered pair is:  (1, 3)

Finally, we’ll let “x” equal two.  y=2(2) +1 = 5.  So the corresponding y-value is 5 and the ordered pair is:  (2, 5)

x Substitute x and solve for y y Ordered Pair (Point)
0 y = 2(0) + 1 = 0 + 1 = 1 1 (0, 1)
1 y = 2(1) + 1 = 2 + 1 = 3 3 (1, 3)
2 y = 2(2) + 1 = 4 + 1 = 5 5 (2, 5)

We now have three ordered pairs that satisfy the equation and represent three points on the line.

Our third and final step is to plot these three points, verify that they form a straight line, and then sketch the line.

We’ll plot the point (0, 1) by starting at the origin, moving NO units left or right and moving ONE unit UP.

We’ll plot the second point (1, 3) by starting at the origin, moving ONE unit to the RIGHT and then THREE units UP.

We’ll plot the third point (2, 5) by starting at the origin, moving TWO units to the RIGHT and then FIVE units UP.

Since these points lie in a straight line, we can finish the process by sketching the line. Be sure to include arrows at each end to denote that the line goes on forever in each direction.

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And remember, this line is a visual representation of all the ordered pair solutions to the equation: y = 2x + 1. Meaning, we can pick a point anywhere on the line, substitute the x- and y-values into the equation, and a true statement should result.

 

Example 2

Graph the line using a table of values: 3y - x = 6

 

We can verify that the equation shown in example 2 is a linear equation, since the highest power of “x” and “y” is one.

Step 1: Solve for “y”

3y - x = 6

3y - x + x = 6 + x

3y = 6 + x

3y \cdot \frac{1}{3} = (6 + x) \cdot \frac{1}{3}

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

So the resulting equation solved for “y” is:  y = \frac{1}{3}x + 2

Our next step is to choose three values for  and substitute each into the equation separately to solve for the corresponding y-value.

Although we can choose any x-values, as previously discussed, this is a great example to show that sometimes we must select x-values wisely. In this case when we substitute a value for “x,” we will be multiplying it by one-third which is the same as dividing by three. Which means it will be best to choose x-values that are divisible by three so that the corresponding y-values are not fractions or decimals. After all, as you might imagine, plotting a point containing a fraction or decimal coordinate would not be very exact. With all that being said, let’s use x-values:  -3, 0 and 3

First, we’ll let “x” equal -3.  y = \frac{1}{3}(-3) + 2 = 1  and the ordered pair is:  (-3, 1)

Next, we’ll let “x” equal 0.  y = \frac{1}{3}(0) + 2 = 2  and the ordered pair is:  (0, 2)

Finally, we’ll let “x” equal 3.  y = \frac{1}{3}(3) + 2 = 3  and the ordered pair is:  (3, 3)

x Substitute x and solve for y y Ordered Pair (Point)
-3 y = \frac{1}{3}(-3) + 2 = -1 + 2 = 1 1 (-3, 1)
0 y = \frac{1}{3}(0) + 2 = 0 + 2 = 2 2 (0, 2)
3 y = \frac{1}{3}(3) + 2 = 1 + 2 = 3 3 (3, 3)

Our final step is to plot these three points, verify that they form a straight line, and then sketch the line.

We’ll plot the point (-3, 1) by starting at the origin, moving THREE units to the LEFT and moving ONE unit UP.

We’ll plot the second point (0, 2) by starting at the origin, moving NO units left or right and then moving TWO units UP

We’ll plot the third point (3, 3) by starting at the origin, moving THREE units to the RIGHT and then THREE units UP.

Since these points lie in a straight line, we’ll finish the process by sketching the line, remembering to include arrows at each end to denote that the line goes on forever in each direction.

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