## Kiss those Math Headaches GOODBYE!

### How to Understand and then Forever Memorize the Midpoint Formula

In algebra we have many formulas to learn. But one problem is that those formulas are often hard to memorize. They are written with variables, and the variables frequently have subscripts, and the truth is that a lot of us don’t really understand what the formulas are saying or how they work. So of course that makes formulas difficult to memorize.

Enter the concept of “friendly formulas.” Friendly formulas are the very same formulas but written in a way that you can understand and therefore memorize much more easily. It’s an idea I have come up with through my many years of algebra tutoring, and idea is included in my Algebra Survival Guide, available through Amazon.com

In this post I describe the “friendly formula” for the midpoint formula.

So as a refresher, what is the midpoint formula all about?

Basically, it lets you find the midpoint of any line segment on the coordinate plane. Think of it this way. There’s some line segment on the coordinate plane called segment AB. That means that it has an endpoint at point A, another at point B. We are given the coordinate of points A and B. We want to find the coordinates of the point right in the middle of points A and B.

Now let’s make this idea easy. Suppose we focus only on the x-coordinates. Suppose the x-coordinate of point A is 2, and the x-coordinate of point B is 6. Ask yourself: what x-coordinate is perfectly in the middle of coordinates 2 and 6? It’s just like asking: what number is right in the middle of 2 and 6 on the number line? Well, wouldn’t that be 4, since 4 is two more than 2 and two less than 6? And indeed it is 4.

But notice that there’s another way to get 4, given the coordinates 2 and 6. We also could have just added 2 and 6 to get 8, and then divided 8 by 2, since 8 ÷ 2 = 4. In other words, we could have TAKEN the AVERAGE of the two x-coordinates, since taking an average of two numbers is adding them and dividing by two.

Could the midpoint formula actually be as easy as taking averages?!

Before we say yes, let’s test this idea for more complicated situations. We just saw that it works when both coordinates are positive. But suppose one coordinate is positive, the other negative. Let’s let one coordinate be
– 2, while the other is + 4. What number is right between those two coordinates on the number line? Well, the numbers are 6 apart, right? And half of 6 is 3, so we could just add 3 to – 2, and get + 1 as the point in between them. And we see that + 1 is three away from both – 2 and 4. But could we also get + 1 by averaging -2 and 4? Let’s try:
(- 2 + 4) / 2 = 2 / 2 = + 1. Averaging works again.

And finally, what about the case where both coordinates are negative? Suppose one coordinate is – 2, the other – 8. What number is right between those two numbers on the number line? Well, these numbers are also 6 apart, right? And half of 6 is 3, so we could just add 3 to – 8, and get – 5 as the middle. And we see that – 5 is three away from both – 8 and – 2. But can we also get – 5 by averaging – 8 and – 2? Let’s try: (- 8 + – 2) / 2 = – 10 / 2 = -5. Averaging worked here too!

Since the averaging process works for all three cases, this approach does works always, and in fact it is how the midpoint formula works.

The midpoint formula basically just averages the x-coordinates to get the x-coordinate of the midpoint. Then it averages the y-coordinates to get the y-coordinate of the midpoint.

So here is the “friendly formula” for the midpoint of any segment on the coordinate plane: Given a segment whose x- and y-coordinates are known,

MIDPOINT = (AVERAGE of x-coordinates, AVERAGE of y-coordinates)

And that’s all you have to memorize!

### Algebra Mistake #2: How to Understand the Difference between A x A and 2 x A without Confusion

Now that you’ve gotten a taste for the benefits of analyzing algebraic mistakes, it’s time to explore a second common mistake. This one is so common that nearly every student commits it at least once on the road to algebra success.

As you watch the video, notice how by thinking hard about two expressions, we can think this mistake through to its very root, thus discovering the core difference between two similar-looking algebraic expressions.

And along the road, we’ll learn a general strategy for decoding the meaning of algebraic expressions. What I like about this strategy is that you can use it to understand the meaning of pretty much any algebraic expression, and you’ll see that it’s not a hard thing to do. In fact, it just involves using numbers in a nifty way.

Best of all, students usually find this approach interesting, convincing and even a bit fun. So here goes, Common Algebra Mistake #2 …

### How to Solve Word Problem Using Easy Numbers

Recently I’ve been using a new technique to help students solve word problems, and nearly every day I am amazed at how helpful it is.

The technique helps students overcome their confusion with word problems.

The approach involves giving students permission to replace the numbers in a word problem with what I call easy, or “friendly numbers.” Essentially “friendly numbers” are just numbers that are easy to think about because they are simple, round numbers.

Here’s an example of the replacement process.

Word Problem as written:  Of the people who voted, 90 percent of them voted for Sammy. If 1930 people voted, how many of them voted for Sammy.

I was tutoring a student. Her response after reading this:  Huh?

Then I told her that it’s ok to temporarily replace the numbers in the problem  with “friendly numbers,” just to make the problem easier to grasp. I helped her see that in this problem she could temporarily replace the 90% with 50% and replace the 1,930 figure with a nice round number, like 600.

The student then picked up her pencil and wrote the problem like this:

Of the people who voted, 50 percent of them voted for Sammy. If 600 people voted, how many of them voted for Sammy.

Then I asked the student if she could figure out this problem. She said it now made sense. She went on to say that if 50 percent of the people voted for Sammy, that meant that half of the 600 people voted for Sammy. So that means that 300 people voted for Sammy.

Then I asked the student if she could come up with an equation to solve this problem. With a bit of help, she came up with:

.5 x 600 = # voting for Sammy

She solved this using decimal multiplication and got the right answer: 300 voted for Sammy.

Then I asked her if she could make a similar equation for the original problem, using the following questions as prompts:

What number in the original problem corresponds to your 50%? Answer:  90%

What number in the original problem corresponds to 600? Answer:   1,930

Once she saw these correspondences, I had the student write her equation for the “friendly numbers” problem. Then, just below that I had her write the corresponding equation for the original problem. Her work looked like this:

.5 x 600  =  # voting for Sammy

.9 x 1,930   =  # voting for Sammy

I asked her to now solve this using decimal multiplication, and she got the correct answer, 1,737

After going through this process I often ask students what made the original problem seem so hard . Usually they will say they don’t know, or they will sometimes say that they just couldn’t understand it.

From my work with “friendly numbers” I’ve come up with a theory. I believe that for many students, merely looking at “unfriendly numbers” has a “psych-out” factor. When kids get “psyched out” by those numbers, they go into a mental panic. And in that panic they lose their intuitive sense of what they need to do.

While this is a problem, it is not insurmountable. All we educators need to do is help the student re-cast the problem with “friendly numbers.” When they do, the “psych-out” factor vanishes, and students see what needs to be done. And generally students can transfer their sense of what needs to be done from the easier problem to the original problem. At that point they are on their way to solving it.

So I encourage you to teach students how to use “friendly numbers” when solving word problems. Perhaps you will also find that students can succeed once they first make the original problem easy to grasp.