Day 2:

Graph a linear equation using a table of values

Objectives:

1) Learn the difference between a solution and a graph of an equation.

2) Learn to verify solutions to equations

3) Be able to graph a linear equation using a table of values

1) The first thing for the students to understand is that a solution to an equation is not the graph of the equation. A solution is an ordered pair that makes the equation true. The graph of the equation is the set of all points (ordered pairs) that are solutions to the equation. So a particular solution is just one ordered pair where as the graph is all the ordered pairs that makes the equation true.

2) When students are asked to verify that a solution to an equation is correct, they are generally given an ordered pair to begin with. They are then supposed to substitute the x and y values found in the ordered pair into the equation to see if the values make the equation true. For example:

Does the points (5,3) and (0,4) make the equation 2x + y = 4 a true statement? To answer this, each ordered pair needs to be substituted into the equation, let's do (5,3) first.

2x + y = 4

2(5) + 3 = 4

10 + 3 = 4

13 = 4

This is obviously not a true statement so the ordered pair (5,3) is not a solution to this equation. Now, let's try (0,4).

2x + y = 4

2(0) + 4 = 4

0 + 4 = 4

4 = 4

This is obviously a true statement therefore we can say that the ordered pair (0,4) is a solution to the given equation.

For practice: Given the ordered pairs (5,5), (2,2), (1,2), (3,7), tell which ordered pairs are solutions for which equations.

i) 3x + 2y = 23

ii) 2x + 5y = 12

iii) x - y = 0

iv) 4x - 2y = 4

3) In the above, the ordered pairs were given and the students were to just verify which points were solutions to the equation. When given an equation and asked to graph it, one method is to use a table of values. With this, you must find the needed ordered pairs. When given an equation like 2x + 4y = 12 and asked to graph this, the first thing to do, if using a table of values is to pick a few values for x and place them in a table similar to the following:

2x + 4y = 12

 x y 2 1 0 -1 -2

Notice the numbers I chose. I believe that is always interesting to see what happens when x is equal to zero so I always choose this number. I also think that it is a good idea to see what goes on on either side of zero. I chose to select 5 values for x. Some feel that 3 is enough and still other teachers want to see 7. I like the middle road of those two paths so I choose 5. Now how do I find the y-values. Well, simply substitute the x values, one at a time, into the equation and find the y value. For example when x = 2, the y value can be found in the following way:

i) 2x + 4y = 12

ii) 2(2) + 4y = 12

iii) 4 + 4y = 12

iv) 4y = 8

v) y = 2

In step ii, the value of x, 2, was substituted into the equation. To get to step iv, 4 was subtracted from both sides of the equation. And lastly, to get to step v, both sides of the equation was divided by 4. Now, I do the same similar steps to obtain the remaining y values and then take this information and put it into my table from above.

2x + 4y = 12

 x y 2 2 1 2.5 0 3 -1 3.5 -2 4

Now I have 5 ordered pairs that I know are solutions to this equation and I can plot them on the coordinate plane. After I do this, I can connect the points and extend beyond them to make the graph of the equation.

This is a great opportunity for the students to use their graphing calculators or to get them into the computer lab and use some graphing software. There are a number of computer applications that will allow the students to try this knowledge out. A teacher could use some graphing software, graph a line, and then ask the students to verify if certain ordered pairs were solutions. The possibilities are almost endless here.

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