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Teaching |ax + b| + |x| = |x – c|

January 3, 2007 am31 10:27 am

Obviously, there are different variants as well. But how would you approach this with high school students? I realized that I was winging it, probably hybridizing what I learned in 1979 with some pedagogy I picked up over the last 10 years, and while my ‘method’ seemed to work perfectly well, maybe there is better out there.

First, I think this stuff is worth doing. Not that the kiddies are going to have to solve this to be allowed on the boat across the river, or to answer the wizard’s question to allow entry to the secret castle or anything that exciting (though, please do remember Samuel L Jackson and Bruce Willis struggling with “As I was going to St Ives… in one of the Die Hard’s)… Nah, this is worth doing for cases and as a foreshadowing of piecewise functions.

Cases

We started by proving that 2|x| = |2x| (I did that) and then they proved a|x| = |ax| for a > 0

(For diehard – haha – math teacher-types, click) ——–>

The proofs by cases are short, easy, clear.

Case I: x > 0, |x| = x and 2|x| = 2x. Also 2x is positive (+ * + = +), so |2x| = 2x.
Case II: x is 0, 2|x| = 0, |2x| = 0, done
Case III: x < 0, |x| = -x, so 2|x| = 2(-x), Also 2x is negative, so |2x| = -(2x), show 2(-x) = -(2x), which we can show is true (or just say, “done”)
All cases work, we are done.

Now, we move to something like |x + 11| + |x| = |2x – 5|
Those absolute values turn on/off at -11, 0 and 5/2 respectively, dividing the real numbers into 4 chunks. So now we have four cases, and we solve for each
I can’t do the graphic thing on the screen, but I set up a number line, and use vertical lines to divide it. I write one equation in each section, and solve beneath:

Case I, x < -11:  -(x + 11) + -(x) = -(2x – 5)  —> No solution
Case II, -11 < x < 0:   (x + 11) + -(x) = -(2x – 5)  —> x = -3 (check, consistent with case? yes. Good solution)
Case III, 0 < x < 5/2:  (x + 11) + (x) = -(2x – 5)  —> x = -3/2 (check, consistent with case? no. not a solution)
Case IV, x >5/2: (x + 11) + (x) = (2x – 5) —-> No solution

This makes sense to me, and except for a few verbal flourishes, is really what I was taught. I think it works. Any improvements?

5 Comments leave one →
  1. January 4, 2007 am31 8:10 am 8:10 am

    Number lines!! As I have babbled about in the past, I am enamored of using them as many ways as possible. Yay!

  2. January 4, 2007 am31 9:07 am 9:07 am

    Interesting…but what are the diehard math type supposed to click? I can’t find any clickable material near or on the arrow, unless you’re referring to one of the sites in the sidebar?

  3. January 4, 2007 am31 9:12 am 9:12 am

    From the main page, it is the link to see the rest of the post. I guess the words out – wordpress puts in a clickable “more” anyhow.

    As far as neat math diagrams, I started learning LaTeX; I should get back to it for using in here. That, or I should clear up my image overflow, and start taking and posting pics of hand-drawn work.

  4. Lsquared permalink
    January 4, 2007 pm31 7:11 pm 7:11 pm

    Ooo-nice problem. I wonder if I can find a use for it. I get a lot of college students who don’t know how to prove theorems using cases.

  5. January 5, 2007 am31 7:01 am 7:01 am

    Ah, I see…I spent at least a minute searching for something to click before deciding I had failed the test for diehard math types!

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