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When we teach linear equations to our ninth graders in Algebra I (or whatever your school system calls the course formerly known as "Algebra I"), we always have to prepare our students to get swamped with boring formulas and tons of memorization that they will find useless and that we will have to re-teach in the course formerly known as Algebra II.
What we want to introduce is a new way of approaching this generally boring topic and bring out a method that your students will appreciate a great deal more than what you're currently doing. So now you're asking, "what is so great about this method as opposed to the method I'm already teaching?"
How does the phrase "No Fractions," or maybe "minimal fractions" sound to you?
Think about what I just said: Minimal Fractions!!! That should have really excited you.
I hope you got excited when you thought about how many fractions you have to deal with in this concept. Slopes with fractions (which still may be unavoidable); y-intercepts with fractions (very avoidable); and solutions with fractions (also very avoidable). Still think that's not a big deal? Just watch this presentation...
Let's look at this table of values:
If we assume that these are points that lie on a line, we can find the equation of this line without having to use the vaunted point-slope formula or any other kind of silly formula.
We need to assume that when we deal with these linear equations, they will come in the form By = Ax + C. (yes, this looks an awful like standard form for linear equations...we will get to that). Forget about the A, the B, and the C for just a second. Let's look at this as
Next we need to find out what goes in the box above. To find this number is very simple. Look at your 'y' column in the table above. Do you see what the difference is between each y term? Hopefully, you notice that this difference is a 2. The 2 is what goes in the box. Ok, you don't believe me. You try checking and seeing if this is correct. How do you check? Take a value out of the x column, multiply it by two, and see if the product you get is what is in the y column. You should have noticed, regardless of which pair you used, that any of those pairs fits the equation y = 2 x.
I know that seemed awfully easy, and there have to be some of those that don't work (what ever "doesn't work" means). Let's now take a look at another set of values:
The greatest temptation now is to take the x values, try to multiply them by some constant (or some fraction for that matter), and come up with what is in the corresponding y entry. Let's try our method once again and see what happens.
Let's keep in mind that we want an equation that will come up in the form y = x. The way we found this number before was to find out what the difference was between each y value. Ok, let's try that again: The difference from 1 to 4 is 3, and the difference between 4 and 7 is 3, and so forth. So the number that goes in the box must be a 3, right? Right! But, you will notice that when you go back and test this by plugging the values from the table into the equation y = 3x, they won't quite work.
There is a reason why we are having a problem. The reason is that we lost track from the beginning what form we wanted the equations to come in. We want them to come in the form By = Ax + C. So what we really have done is just find out what goes in A's place. We didn't need anything to go in the C place in the first example; so we assumed that C was equal to 0. Now things change a little bit. All we know is that we have y = 3x + . How do we now find out what goes in the box? We plugged numbers in before to test whether we had the correct value in front of the x; now we need to find out what goes in the box (C). Take one of our sets of solutions from the table above and plug in what we have. Try the first line, where x = 1 and y = 1. We should have the following:
This simplifies into 1 = 3 + . Hopefully, we can all see that the number in the box is a -2. The linear equation that we have derived from the table above is y = 3x - 2 (note: if the value of what is in the box is negative, the plus that was assumed to be before the C term will turn to a minus).
If you clicked on the worksheet 1, hopefully you noticed a few things. Most importantly, you probably noticed that the values in the y column don't necessarily go from lower number to higher numbers. Let's look at an example where this is the case and examine what that does to our equation:
What is the difference between each of the y-terms? The "common difference" (a term that we will use more as this essay goes along) is -3. If we look at the difference from 5 to 2, we need to think "what is 2 - 5?" Then think, "what is the difference between -1 - 2? Of course, it's -3 again. Given that these points are the solutions of a linear equation, we should see the common difference remaining the same from value to value in the y column. Knowing this much, we have enough information to find the equation of the line with the values above as solutions.
We remember that the form we are looking for is y = -3x + . What goes in the box?
After taking the first set of points (1, 5) and plugging that into for x and y in the equation above, we see that we have 5 = -3 (1) + . This simplifies to 5 = -3 + . The number that goes in the box has to be an 8 to make the statement true. So our linear equation is:
After these examples, hopefully, worksheet 1 is a bit easier for you.
Let's now take a look at a new table of values that is slightly different than any that we have seen so far.
The difference between this example and the examples above is that the x column is not the counting numbers from 1 to 5 anymore. Now our students are getting worried because they've dealt with the change in the y column, but when you switch both columns up, a bit of panic arises. But, let's keep a few things in mind. We know how to handle the changes in the y column; so let's work with it alone. No matter what, we are going to have y = x + . Let's first concern ourselves with what goes in the box before the x.
Remembering what we did before, we see that the common difference in the y column is 2; so the number in front of the x is a 2. To update what we have:
Before we deal with the second box, let's now look at the common difference in the X column. As you can see, the common difference for the x's is 2. This 2 is what goes in front of the y! Now I know at this point you're getting angry with me, but remember what we said from the beginning: the linear equations that we are dealing with are in the form By = Ax + C. Recently, we have dealt with the missing A and C terms. The common difference in the x column is what goes in for B. So the equation, based on what we have found so far is:
In the same manner we found the missing c term before, fill in the x and y terms of our equation with a solution from our table above. I will choose the point (2, 1). So after plugging in, we have 2 (1) = 2 (2) + . This simplifies to 2 = 4 + ; so it is quite clear that the number that goes in the box is -2. Our final linear equation is 2y = 2x - 2.
In case you missed what we just discussed, let's take a look at another example. Take note in the changes in the x's and y's. These are the keys to finding the coefficients of the x and y in our linear equation.
To build from what we know so far, we are hoping to find a linear equation in the form By = Ax + C. To find B, find the common difference in x column. After subtracting 3 - (-4), we see that the common difference on the left-hand side (the x column) is 7. In the y column, we find the common difference to be -2 - 3 = -5. So now, we can find the first part of our equation. We know now that equation looks like 7y = -5x + . Let's go and find what goes in the box (the c term). After filling in the x and y in our equation with one of the solutions above, in the table, we see that the = 1. The linear equation with these points, in the table, as solutions is:
If you are a teacher, I'm sure you have already noticed the idea I'm about to give you, but the overzealous student may not have seen it yet. In the tables above, you have noticed that the common difference from one x to the next x is the same each time, as is the common difference for the y's. But the question I want to pose now, especially for any students who are reading, is how many solutions do you need to be able to come up with the linear equations you have seen so far? Remember the old geometry adage that you need two points to make a line? Well, there is the answer to my question. In the tables that we have seen as examples (above and on the worksheets), there have been five points on each of those lines. We really don't need but two solutions. So teachers, perhaps you may want to make up some worksheets of your own that only include two points on the line, like this:
Even from this mini-table, you should be able to easily find the common difference in the x column as well as the common difference in the y column. After finding the difference in the x's to be -4 and the difference in the y's to be -2, the equation that we will have looks like -4y = -2x + . Plug in one of your solutions above, and you find that the equation is -4y = -2x - 10.
Ok, so now we have beaten the heck out of working with the solutions of linear equations in tabular form. Now, perhaps it's a good idea to work in the opposite direction. Let's start with a linear equation, and work to fill in a table.
For example, can we find a series points on the line 2y = 4x + 8. What we typically like to do when we teach this is let the students struggle with this problem before we start making it any easier for them. Eventually, the students figure out that they need to let x = 0, thereby making y = 4. So the first point that they discover is on this line is (0, 4). Quite possibly, the students can use this same idea to figure out that another point on this line is (-2, 0) by letting y = 0.
But we can make this process a lot easier than setting one variable equal to zero, then repeating the process for the other variable. What I recommend your doing (as a teacher) is to start putting these linear equations into standard form. Standard form just means putting both variables on the left of the equal sign. So you equation should be in the form of Ax + By = C. The rules we usually see for standard form are that the A, B, and C terms cannot be fractions or decimals, and depending on how strictly you enforce this one, A should always be non-negative.
You can begin your switching equations to standard form at any point of the lesson, but it may be easier to let the students get comfortable with creating the equations from the tables in Ay = Bx + C form, then switch their results to standard form. You may find that teaching in the order that this unit is laid out will allow your students to master one idea at a time without having to learn too many concepts simultaneously, thus avoiding a lot of confusion.
Let's practice switching a couple of these equations around to standard form.
Example: What is the equation -3y = 2x + 6 in standard form?
Let's look at something a bit more dramatic.
Example: What is (-3/5)y = (-1/3)x - 1 in standard form?
Maybe, at this point it would be best if we have a worksheet to practice our ability to change from By = Ax + C form to standard form, and vice versa. Click here for Worksheet 3.
Now that we have a couple of equations in standard form, let's work on getting some points that are on that line.
How many points are there on a line? "Infinite," your students answer. Correct. How many points are there that we can find? Hopefully you get the same answer you got for the last question.
At this point, we're in a position where we can find some points that lie on the lines that were our examples a moment ago.
Let's first look at the first equation we put in standard form: 2x + 3y = -6
If you recall, we can find a pair of points on this line quite easily by, literally, covering up one of the terms with variables. So cover up the 2x. What does that leave us? 3y = -6. After solving for y, we see that y = -2. But when does y = -2? Answer that question by thinking about what you did to find that y = -2. Remember, you "covered up" 2x. What does it mean to cover up a variable? It means that you are setting it equal to zero. So one of your points on this line is definitely (0, -2).
To find a second point (vital to graphing a line), we can repeat the process above, except this time, we want to cover up the y term, thereby making y = 0. Then we have 2x = -6 remaining, and it is easy to see that when y = 0, then x = -3. So the other point that we need to find to sketch the graph of a line is (-3, 0). When we have these two points, we go to the coordinate plane and plot the points. This is what our graph looks like:
There really wasn't a lot of magic involved in graphing that equation because we were able to just cover up both variables and find our two points. As Bert Green at Parkview High School calls it, these types of equations are called "easy's," from their ease in graphing.
Next, let's look at an equation that Bert would call a "middle easy" -- not too hard, but not quite as easy as the first one. The equation to look at now is 5x - 9y = -15 (the second equation changed into standard form).
In this example, it is easy to find one point that is on this line. Simply covering the -9y leaves us a mini-equation of 5x = -15; so it is easy to see that when y = 0, then x = -3. The problem here is that when we try to cover up the 5x term, we end up with a mini-equation of -9y = -15 -- and we are about to break the fundamental rule that we stated above: NO FRACTIONS. Yes, if we were in an enormous hurry and got completely stuck with the methods that we will try, we could succumb to the evil fractions and say that when x = 0, then y = 15/9 or 5/3. I will leave this to the reader to graph these to points; have fun graphing y = 5/3!
So what do we do? We're stuck, right? No, not at all.
Remember what we were doing at the beginning of this lesson? We were looking at points on a line in tabular form. What did we do then? We found that we could find the linear equation that generated these points.
So why can't we use the linear equation to generate some points? Well, in this equation, we've already gotten one point taken care of. Let's look at what our table looks like so far:
We now need something to fill in the blanks. Well, do you remember how we found out what went in front of the variables when we were forming the equation itself? The equation we have now looks a little different than the one we have now. The equation that we are currently working on is 5x - 9y = -15 and is in standard form. Maybe for the comfort of your students, it may be best to switch this equation back to By = Ax + C. After doing that, we find that the equation is -9y = -5x - 15. The number on the left of the x (the coefficient) is -5. This -5 is what the common difference on the y column was. The number on the left of the y is -9: the common difference of the x column. From the two common differences we can fill in the rest of the table with the following:
The reason I bolded the first two points on the table was to emphasize the fact that only two points are required to graph our line.
I know that it appears cumbersome to graph the point (-12, -5), but we do have a way of fixing that problem. Hopefully you noticed when we kept working our way downward while filling in the table, we have the ability to go upward as well if that will help our graphing endeavors.
So let's now shift the values in the table downward and look above the point (-3, 0). If the common difference on the y column was -5 going downward, we will need to add 5 to go upward. And if the common difference on the x column was -9 going downward, then we will add 9 to go back upward. So our new table looks like the following:
As you may have noticed, it is much easier to graph the first two points here, than to have to graph (-12, -5). The graph of this equation is to follow:
Now it is time to step up with the big boys. We have dealt with "easy's" and " middle easy's" but now, it's time we try some "hard's."
We already know the criteria for being an 'easy' (being able to set either variable equal to zero and still getting a point on the line that is not a fraction) and a 'middle easy' (being able to set one of the variables equal to zero and getting a point on the line that is not a fraction).
A "hard" is where we can't put our fingers on either variable (setting one variable equal to zero). We see this type of a problem with the following example:
It would appear that this example will not have any solutions that aren't fractions, but don't give up just yet. In order to find some non-fractional points on the line, we need to do something a bit unconventional.
Let's take our A and B terms (in this example, A = 3 and B = 4) and create a new table that is comprised of those two numbers and some of their multiples. Our new table is shown below:
As odd as this may read, all we need to do is find a number in the A column (under the 3) and a number under the B column (under the 4) to be able to add or subtract to get the number on the right of the equal sign (in this example, 10). We can take one number from one column (either positive or negative) and the other number from the other column (also positive or negative), add/subtract the two numbers, and get the number to the right of the equal sign. The table below shows the pair we chose:
As you see, in red, we choose the pair 6 and 4 to add to get 10. The significance is that if we have an equation like our example, 3x + 4y = 10. If we associate the 6 with the 3x term and the 4 with the 4y term, we need to think about what we need to multiply the 3 by to get 6 and what we need to multiply 4 by to get 4. More simply put, what does x have to be to get 3x = 6? What does y need to be to get 4y = 4?
As you can see x has to equal 2 and y has to equal 1; so our point will be (2, 1). But take note, we need to go back and check to make sure that when we plug (2, 1) into our original equation to make sure that 3(2) + 4(1) = 10. In this case, (2, 1) worked great. Sometimes we may have to change one or both signs of our solution to make the statement true.
Let's not lose sight of what we are trying to do. We found one point that we can use to graph the original equation. Knowing one point and our equation, finding the second point is the same as all of the other examples. The equation can be changed to 4y = -3x + 10; the first entry on our table is the point (2, 1) use the 4 as the common difference of the x column and the -3 as the common difference of the y column. A portion of the table is shown below (keep in mind we only need one other point to graph the line):
It may be safe to assume that we need another example of a 'hard.' Let's look at the equation
You can see by trying to cover up terms that this isn't an "easy" or a "middle easy." Now that we've determined this is a "hard," let's make a table that includes 4 and 5 and their multiples:
The number to the right of the equal sign is -14; so adding two numbers from the table (from separate columns, of course) will not bring us to -14. If we take 24 from the left column and 10 from the right column, we can subtract these 10 - 24 to get our -14. So we have now found a pair that will give us a non-fractional point on the line.
So our equation was 4x - 5y = -14; so we now think about what x has to equal to give us 4x = 24, and what y has to equal to give us -5y = 10. We see that x = 6 and y = -2 are our solutions, respectively.
We now know that (+/- 6, +/- 2) will be the point on the line we are looking for. Let's plug this point into our equation and determine the signs.
The point (6, 2) does not fit because the left-hand side of the equation does not equal the right-hand side. We can probably see that the x will have to be equal to -6 to make the left-hand side negative; so we will try (-6, 2). This gives us:
This solution does not work either because the left-hand side is equal to -34; so now we will change the 2 to -2 and see what happens.
We see that this solution, (-6, -2) does work for our equation, and we have now found a point on the line (that is not a fraction). From here we can easily complete a table of solutions using the common difference from the original equation:
It may be appropriate to now do an exercise that will ask us to determine the degree of difficulty (easy, middle easy, or hard), determine 2 or more points on the line formed by the equation, and graph the line based on the points you determined. Click here from Worksheet 4.
Other concepts to be covered: