Finding the "Hidden Wows"

I was working with some middle school teachers-to-be who are noticing some of the aspects of algebra we tend to take for granted after years of practice and application, and they are noticing how difficult it can be to anticipate the strategies and struggles of students who are just learning algebra.
 
One way I get around that is to try to hold myself still with a problem for a bit and look for the hidden connections. With kids, I might tell them we're looking for the Hidden Wows.

Suppose I am preparing a lesson for sixth graders to introduce problems of this form:

Solve: ax = b

 My planning might start out sounding a bit like this:

First of all: These are so easy, you just divide both sides by a. How will I make this interesting? And what in the world are they going to struggle with? The problems are all the same!

Well, how about (1/2)x = 5. I actually wouldn't divide by (1/2) in that one, I would you multiply by 2. Interesting. I wonder how else you could solve that.

Actually, you don't need to divide or multiply both sides by anything... you can just use your reasoning:

"Half of what number is 5? Well, 10 of course!!"

Wow! I'll try to remember to value that if anyone suggests that.

But the book tells us we are supposed to "divide by 1/2" on both sides.

Ok, so what if I do that? We get:

x = (5 div 1/2).

Uh oh, they'll need to understand what it means to divide 5 by 1/2? How would they think about that? Well last year they learned long division of decimals, I wonder if any of them will go there: I could write 1/2 as 0.5, and then do long division:

       ___           
0.5 | 5       -->  Yuck. No "Wows" yet. We need to "move the decimal points".
    ____ 
5 | 50     --> Well, that's easier, anyway. The answer is 10.

Oh, and that makes sense! The quotient of 5 / 0.5 should be 10 because there are ten 0.5's in 5.

Wow! That's a neat connection.

I wonder why we can move the decimal points like that. Hmm...

Let's see...  5 / 0.5 can be renamed 50 / 5 (multiply numerator and denominator by 10). Hey, that's neat! Moving the decimal point in a long division problem is just another way of renaming fractions. That's cool!

Wow! I wonder if there's an easier way for them to divide 5 by 0.5...


So moving the decimal point in long division is just like renaming the fraction. I wonder... what if instead of renaming using 10s (which moves the decimal), what if I renamed using 2s (double both the divisor and dividend). Then I could "rename the quotient" this way:

       ___
0.5 |  5
    ___
1 | 10

Wow! It's just like rewriting 5 / 0.5 as 10 / 1. My teachers never showed me that!

I wonder what else I can notice.

Well, the numerator and denominator of 5 / 0.5 are the numbers from the original problem, (1/2)x = 5. And if we multiply both sides of that equation by 2 (just like we did when renaming the fraction) the whole equation becomes much simpler: in fact, it becomes 1x = 10. Well that's easy!


Wow! I wonder if I could multiply both sides by 10, too, like we did to rename 5/0.5 as 50/5? Let's see:

(1/2)x = 5  ---->  (10)(1/2)x = (10)(5)  -----> 5x = 50.

Well, that's not much easier, but at least the fraction is gone now. I could think of solving 5x=50 as "5 times what equals 50," which is a little easier than "(1/2) times what equals 5". 

Hey, it's another ax = b problem, except without the fraction! The book tells us we need to solve ax=b by dividing both sides by a. In this case, 5x = 50, that's probably what I would do: divide both sides by 5. But.... I could also multiply both sides by the reciprocal of 5:   (1/5)5x = (1/5)50.

So solving (1/2)x = 5 can be done by:

• multiplying both sides by 10 and then dividing both sides by 5, or 
• multiplying both sides by 2, or 
• dividing both sides by (1/2). 

Wow, you know what? This 6th grade algebra is kind of fun...
Here's one for you to try: 

 Find the "Hidden Wows" in the problem

x - 5 = -3

Let us know what you find! Happy hunting!