Kiss those Math Headaches GOODBYE!

Archive for January, 2014

How to Remove (“Unpack”) Algebraic Terms from Parentheses


As you’re probably aware, I’m a big believer in using stories to bring math to life. Especially when you’re teaching tricky concepts, using a story can be the “magic switch” that flicks on the light of understanding. Armed with story-based understanding, students can recall how to perform difficult math processes. And since people naturally like stories and tend to recall them, skills based on story-based understanding really stick in the mind. I’ve seen this over and over in my tutoring.

Stories from My Tutoring Work

The kind of story I’m talking about uses an extended-metaphor, and this way of teaching  is particularly helpful when you’re teaching algebra. Ask yourself: what would you rather have? Students scratching their heads (or tearing out their hair) to grasp a process taught as a collection of abstract steps? Or students grasping  a story and quickly seeing how it guides them in doing the math? I think the answer is probably pretty clear. So with this benefit in mind, let’s explore another story that teaches a critical algebraic skill: the skill of  “unpacking” terms locked inside parentheses.

To get the picture, first imagine that each set of parentheses, weirdly or not, represents a corrugated cardboard box, the kind that moving companies use to pack up your possessions. Extending this concept, the terms inside parentheses represent the items you pack when you move your goodies from one house to another.  Finally, for every set of parentheses (the box), imagine that you’ve hired either a good moving company or a bad moving company. (You can use a good company for one box and a bad company for a different “box” — it changes.) How can you tell whether the moving company is good or bad? Just look at the sign to the left of the parentheses. If the moving company is GOOD, you’ll see a positive sign to the left of the parentheses. If the moving company is BAD, you’ll spot a negative sign there.

Here’s how this idea looks:

+ (    )     The + sign here means you’ve hired a GOOD moving company for this box of stuff.

– (    )     This – sign means that you’ve hired a BAD moving company to pack up this box of things.

Now let’s put a few “possessions” inside the boxes.

+ (2x – 4)  This means a GOOD moving company has packed up your treasured items: the 2x and the – 4.

– (2x – 4)  Au contraire! This means that a BAD moving company has packed up the 2x and the – 4.

[Remember, of course, that the term 2x is actually a + 2x. No sign visible means there’s an invisible + sign before the term.]

What difference does it make if the moving company is GOOD or BAD? A big difference! If it’s a GOOD company, it packs your things up WELL.  Result: when you unpack your items, they come out exactly the same way in which they went into the box. So since a good moving company packed up your things in the expression:  + (2x – 4), when you go to unpack your things, everything will come out exactly as it went in. Here’s a representation of this unpacking process:

+ (2x – 4)

=      + 2x – 4

Note that when we take terms out of parentheses, we call this “unpacking” the terms. This works because algebra teachers fairly often describe the process of taking terms out of (   ) as “unpacking” the terms. So here’s a story whose rhetoric  matches the rhetoric of the algebraic process. Convenient, is it not?

Now let’s take a look at the opposite situation — what happens when you work with a BAD (boo, hiss!) moving company. In this case, the company does such a bad job that when you unpack your items, each and every item comes out  “broken.” In math, we indicate that terms are “broken” by showing that when they come out of the (  ), their signs,  + or – signs, are the EXACT OPPOSITE of what they should be. So if a term was packed up as a + term, it would come out as a – term.  Vice-versa, if it was packed up as a – term, it would come out as a + term. We show the process of unpacking terms packed by a BAD moving company, as follows:

– (2x – 4)

=      – 2x + 4

And that pretty much sums up the entire process. Understanding this story, students will be able to “unpack” terms from parentheses, over and over, with accuracy and understanding.

But since Practice Makes Perfect, here are a few problems to help your kiddos perfect this skill.

PROBLEMS:

“Unpack” these terms by removing the parentheses and writing the terms’ signs correctly:

a)  – (5a + 3)

b)  + (5a – 3)

c)  – (– 3a + 2b – 7)

d)  + (– 3a + 2b – 7)

e)  6 + (3a – 2)

f)  6 – (3a – 2)

g)  4a + 6 + (– 9a – 5)

h)  4a + 6 – (– 9a – 5)

ANSWERS:

a)  – (5a + 3)   =   – 5a – 3

b)  + (5a – 3)  =  + 5a – 3

c)  – (– 3a + 2b – 7)  =  + 3a – 2b + 7

d)  + (– 3a + 2b – 7) = – 3a + 2b – 7

e)  6 + (3a – 2)  =  + 3a + 4

f)  6 – (3a – 2)  =  – 3a + 8

g)  4a + 6 + (– 9a – 5)  =  – 5a + 1

h)  4a + 6 – (– 9a – 5)  =  + 13a + 11


Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like how Josh explains these problems, you’ll certainly  like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 

Nigeria Algebra Workshop — Tuesday Morning!


Hi everyone,

I have some exciting news.

I will be conducting a workshop this Tuesday morning, and the workshop is with a school in Nigeria.

Thanks to my good friend Ibraheem Dooba (aka Professor Brainy), and also thanks to the marvels of modern technology, this Skype workshop has been all set up and is ready to go.

The school whose teachers are receiving the workshop (not sure if that’s the right way to phrase it, or if there is a way to phrase that!) is the Esteem International School, in Nigeria’s capital city, Abuja. This is a wonderful school for elementary age children. Like students in the United States, students at the Esteem School are learning aspects of algebraic thinking in elementary school. And since teaching algebra to young students is a challenge, that will be the focus of my workshop.

If anyone here or anywhere in the world is interested in receiving a workshop on this topic, please let me know, and we’ll set one up.

What it’s like to tutor math


Today I realized something about being a tutor.

Stories from My Tutoring Work

Tales from the Tutoring Experience

A big part of it — maybe as much as half it — involves nothing more than  …   being nice.

By that I mean being kind.

By which I mean that if someone looks at you, as a young man did today, shaking his head and saying, “It’s crazy … I don’t know what 3 x 6 is,” I don’t laugh or chuckle or say anything remotely mean or mocking. Instead I just say, “It’s o.k. Look, I tutor people every day who don’t know what 3 x 6 is. Who cares, really? Let’s just try to figure it out … or use a calculator, as long as your teacher doesn’t mind.”

Really. That is a lot of what being a math tutor is about. Being nice. Really nice. Really understanding. And being there to be accepting of people no matter how much mental pain they may be in about math. Because there’s a lot of pain out there. Many people are carrying loads of pain about math. They feel dumb. They feel like it’s some huge reflection on whether or not they can make it in the world.

And so it is my job, as a tutor, to listen to their worries and to assure them that they will get better. And that even if they keep struggling, as they probably will, to some extent, it is ok. They can still live good lives, and math is not going to define or confine them,” to quote Bob Dylan a bit.

Don’t get me wrong in terms of what I said up above. A math tutor has to know the material … extremely well. And he needs to know how to teach the material and the skills of math. But once he gets that down, once he gains in competence, he can really open up his heart and help people with the emotional struggles they go through with math as well.

I generally like being a tutor. It feels satisfying. I love seeing people go through the gradual transition from hating to have to see me, to feeling somewhat ok about it, to starting to feel good because their grades are going up and they are starting to get it better. They start to walk taller, literally as well as figuratively. They come right into the office, after several weeks or months of working with me, and they tell me exactly what they need help with. They become their own best advocates. And they get over that horrible feeling that math is holding them back.For the most part, to be honest, the students I work with don’t end up loving math. For the most part, they go from hating it to feeling ok about it. And that is ok with me. I just want students to feel like they can understand many of the parts of math and to feel competent in relation to math. Seeing that transition occur is the best reward I can get, and it actually happens more often than not.

So if you or someone you know needs a math tutor, I suggest you find one. It can make a big difference in a person’s life. A good tutor can really help a young person grow.

How to Combine Positive & Negative Numbers — Quickly and Easily


If you or someone you know struggles when combining numbers with opposite signs — one positive, the other negative — this post is for you!

To be clear, I’m referring to problems like these:

 – 2 + 7 [first number negative, second number positive], or

+ 13 – 20 [first number positive, second number negative]

To work out the answers, turn each problem into a math-story. In this case, turn it into the story of a tug-of-war battle. Here’s how.

In the first problem, – 2 + 7, view the – 2 as meaning there are 2 people on the “negative” team; similarly, view the + 7 as meaning there are 7 people on the “positive” team.

There are just three things to keep in mind for this math-story:

1)  Every “person” participating in the tug-of-war is equally strong.

2)  The team with more people always wins; the team with fewer people always loses.

3)  In the story we figure out by how many people the winning team “outnumbers” the other team. That’s simple; it just means how many more people are on that team than are on the other team. Example: if the negative team has 2 people and the positive team has 7 people, we say the positive team “outnumbers” the negative team by 5 people, since 7 is 5 more than 2.

Now to simplify such a problem, just answer three simple questions: 

1)  How many people are on each team?
In our first problem, – 2 + 7, there are 2 people on the negative team and 7 people on the positive team.

2)  Which team WINS?
Since there are more people on the positive team, the positive team wins.

3) By how many people does the winning team OUTNUMBER the losing team?
Since the positives have 7 while the negatives have only 2, the positives outnumber the negatives by 5.

Now ignore the answer to the intro question, Question 1, but put together your answers to Questions 2 and 3.

ANSWER TO QUESTION 2:  +

ANSWER TO QUESTION 3:  5

ANSWERS TOGETHER:  + 5

All in all, this tells us that:  – 2 + 7 = + 5

For those of you who’ve torn your hair out over such problems, I have good news …

… THEY REALLY ARE THIS SIMPLE!

But to believe this, it will help to work out one more problem:  + 13 – 20.

Here, again, are the common-sense questions, along with their answers.

1)  How many people are on each team?
In this problem, + 13 – 20, there are 13 people on the positive team and 20 people on the negative team.

2)  Which team WINS?
Since there are more people on the negative team in this problem, the negative team wins.

3) By how many people does the winning team OUTNUMBER the losing team?
Since the negatives have 20 while the positives have only 13, the negatives outnumber the positives by 7.

Just as you did in the first problem, put together your answers to Questions 2 and 3.

ANSWER TO QUESTION 2:  

ANSWER TO QUESTION 3:  7

ANSWERS TOGETHER:  – 7

All in all, this tells us that:  + 13 – 20  = – 7

Now try these for practice:

a)  – 3 + 9

b) + 1 – 4

c)  –  9 + 23

d)  – 37 + 19

e) + 49 – 82

Answer to Practice Problems:

a)  – 3 + 9 = + 6

b) + 1 – 4 = – 3

c)  –  9 + 23 = + 14

d)  – 37 + 19 = – 18

e) + 49 – 82 = – 33

Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like the way Josh explains these problems, you will very likely like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 

The Log Blog


Tales from the Tutoring Experience

Tales from the Tutoring Experience

It’s that time of year again when Algebra 2 students are all studying logs — not the lumberjack things, but ‘logarithms’ — so I’d like to present a concept that helps students work with logs.

I call it the “Whole-Part-Equality Principle” (as I’ve never seen it named by anyone else … is there a name for it? Anyone know?) But I prefer to call it the “Peter-Paul-Pennies-in-the-Pocket Principle.”

Here’s how it works. There’s Peter, and there’s Paul. We are told that Peter and Paul have no money except pennies, and they transport their pennies only in their right and left pants pockets (if anyone can think of a way to pack this story with even more p’s, please let me know).

Anyhow, we know three additional facts:

1)  The number of pennies that Peter is transporting equals the number of pennies that Paul is transporting.

2) Peter and Paul each have three pennies in their right pants pockets.

3)  Peter and Paul transport their pennies NOWHERE but in their pants pockets.

QUESTION:  What can we conclude about the number of pennies that Peter and Paul have in their left pants pockets?

ANSWER:  It’s obvious, right? While we don’t know how many pennies Peter and Paul could be transporting in their left pants pockets (it could be any number, right?), it is nevertheless clear that they must have the same number of pennies in their left pants pockets.

WHY?  View it like this … If the wholes are the same (the total number of pennies that Peter and Paul each has), and if one of two key parts are the same (the number of pennies that Peter and Paul have in their right pants pockets), then the other parts must also be equal (the number of pennies they have in their left pants pockets).

Why am I bringing this up? To point out an important principle.

This same principle — if the wholes are equal, and if one of their two parts are equal, then the other parts must also be equal — can be used to solve many log and exponent problems.

EXAMPLE 1:  Suppose you have this equation:  log x = log 7.2. What can we conclude? Well, the wholes are equal (meaning the left and right sides of this equation are equal), and the bases of the logs are equal (logs are always base 10 unless another base is given), therefore the remaining parts, the ‘arguments,’ also must be equal. The ‘argument’ is the term after the word ‘log,’ so for this equation the arguments are x and 7.2, and they must be equal … meaning that  x = 7.2.

EXAMPLE 2:   Suppose you have the equation:  log 2^x = log 16. Again, the wholes are equal, and the logs have the same base, so the arguments must be equal. That means that 2^x = 16. Since 2^4 = 16, x = 4, and that’s the answer.

EXAMPLE 3:  Suppose you have the equation:   a^log x = a^log 12.9. Since the wholes (the left and right sides of the equation) are equal, and since the bases are equal as they are both ‘a,’ therefore the only remaining parts, the exponents, must also be equal. So this means that log x = log 12.9. Following the same logic as we used in Examples 1 and 2, this means that x = 12.9.

Any questions? If so, please post as a comment. If not, please use this principle, and enjoy its profound practicality. (OK, I’m done.)

Josh Rappaport is the author of the Algebra Survival Guide and Workbook, which together comprise an award-winning program that makes algebra do-able! Josh also is the author of PreAlgebra Blastoff!, an engaging, hands-on approach to working with integers. All of Josh’s books, published by Singing Turtle Press, are available on Amazon.com

The “Unknown” Order of Operations


Talk about a major point that’s usually unspoken …

We make such a big deal out of the Order of Operations in Algebra, and yet there’s a second order of operations, equally important but seldom mentioned.

First, to clarify, the standard Order of Operations (caps on the two O’s to indicate this one) helps us simplify mathematical expressions. It tells us how to take a group of math terms and boil them down to a simpler expression. And it works great for that, as it should, as that’s what it’s designed for.

EXAMPLE:  this Order of Operations tells us that, given an expression like:  – 2 – 3(4 – 10), we’d first do the operations inside PARENTHESES to get – 6, then we’d MULTIPLY the 3 by that – 6 to get – 18. Then we would SUBTRACT the – 18 from the – 2, to get 16. You know, PEMDAS.

But it turns out that there’s another order of operations, the one used for solving equations. And students need to know this order as well.

In fact, a confusing thing is that the PEMDAS order is in a sense the very opposite of the order for solving equations. And yet, FEW people hear about this. In fact, I have yet to see any textbook make this critical point.  That’s why I’m making it here and now: so none of you  suffer the confusion.

In the Order of Operations, we learn that we work the operations of multiplication and division before the operations of addition and subtraction. But when solving equations we do the exact opposite: we work with terms connected by addition and subtraction before we work with the terms connected by multiplication and division.

Example: Suppose we need to solve the equation,
4x – 10 = 22

What to do first? Recalling that our goal is to get the ‘x’ term alone, we see that two numbers stand in the way: the 4 and the 10. We might  think of them as x’s bodyguards, and our job is to get x alone so we can have a private chat with him.

To do this, we need to ask how each of those numbers is connected to the equation’s left side. The 4 is connected by multiplication, and the 10 is connected by subtraction. A key rule comes into play here. To undo a number from an equation, we use the opposite operation to how it’s connected.

So to undo the 4 — connected by multiplication — we do division since division is the opposite of multiplication. And to undo the 10 — connected by subtraction — we do addition since addition is the  opposite of subtraction.

So far, so good. But here’s “the rub.” If we were relying on the PEMDAS Order of Operations, it would be logical to undo the 4 by division BEFORE we undo the 10 with addition … because that Order of Operations says you do division before addition.

But the polar opposite is the truth when solving equations!

WHEN SOLVING EQUATIONS, WE UNDO TERMS CONNECTED BY ADDITION AND SUBTRACTION BEFORE WE UNDO TERMS CONNECTED BY MULTIPLICATION OR DIVISION.

Just take a look at how crazy things would get if we followed PEMDAS here.

We have:  4x – 10 = 22

Undoing the 4 by division, we would have to divide all of the equation’s terms by 4, getting this:

x – 10/4 = 22/4

What a mess! In fact, now we can no longer even see the 10 we were going to deal with. The mess this creates impels us to undo the terms connected by addition or subtraction before we undo those connected by multiplication or division.

For many, the “Aunt Sally” memory trick works for PEMDAS. I suggest that for solving equations order of operations, we use a different memory trick.

I just remind students that in elementary school, they learned how to do addition and subtraction before multiplication and division. So I tell them that when solving equations, they go back to the elementary school order and UNDO terms connected by addition/subtraction BEFORE they UNDO terms connected by multiplication/division.

And this works quite well for most students. Try it and see if it works for you as well.

Josh Rappaport is the author of the Algebra Survival Guide and Workbook, which together comprise an award-winning program that makes algebra do-able! Josh also is the author of PreAlgebra Blastoff!, an engaging, hands-on approach to working with integers. All of Josh’s books, published by Singing Turtle Press, are available on Amazon.com

The Clouds Part, and a Log Rule MAKES SENSE!


Have you ever been befuddled by the rules for logs?

More specifically, have you ever looked at this rule:

log (v w) = log v + log w

and thought: Now why in the world is that true?! What exactly is this saying? I know that I, myself, have had that thought. And for me the desire to understand this rule never went away. Till I got it some time ago.

[By the way, keep in mind that the v and the w in the parentheses are multiplying each other, so that v w actually means: v times w]

And the good news is: I think I can explain this rule in a way so that pretty much everyone who knows basic algebra can grasp it.

O.K., first, I knew that this log rule was related to another rule, the  exponent rule that says:

(a^b) x (a^c) = a^(b + c)

Remember: this is the rule that says if you have two exponential terms  with the same base, and those two terms are multiplying each other, you just keep that base and add the exponents. For example:
(3^2)  x  (3^5) =  3^(2 + 5) = 3^7

But how exactly does this exponent rule relate to the more confusing-looking log rule?

To get ready to see this, one preliminary concept must be clear. The concept is that whenever you see a log term, you’re basically seeing an exponent. Why? Because every log represents an exponent. For example:  log 2 of 8 is the exponent of 3 since 2^3 = 8. 

Put another way, the term log 2 of 8 is asking a question. It’s asking: what exponent would you plunk on the right shoulder of the smaller number, 2, to get the much bigger number 8? The answer is 3, since 2^3 = 8.

Now you try this.

What question is log 3 of 81 asking? Answer: What exponent would we put on 3 to get 81?
What is the answer to this question? Answer:  4, since 3^4 = 81.
So based on all of that, log 3 of 81 = 4.

Now that we’ve got this concept straight, let’s look at the log rule again.

log (v w) = log v + log w

If we substitute in some numbers, this rule will be easier to think about. So let’s substitute 4 for v and 8 for w. After doing that we get:

log (4 x 8) = log 4 + log 8

Next, keep in mind that we can insert a base, and we can actually use any base we wish, as long as we use the same base for all three terms. A handy base would be 2 since 4 and 8 are both powers of 2. So when we use 2 as our base, the equation now reads:

log 2 of (4 x 8) = log 2 of 4 + log 2 of 8

One more thing before we tackle this sucker. Let’s  express the product inside parentheses as 32, which is ok since 4 x 8 equals 32, right? So now the equation reads:

log 2 of (32) = log 2 of 4 + log 2 of 8

Now, after all of that work, let’s finally have some fun. “Having fun,” of course, is relative, but if you’re a math person, “having fun” probably means: let’s  figure out what this crazy equation is saying. So here goes …

Based on what we’ve been saying, the left side of the equation asks the question: what exponent would we put on 2 to get the number 32. So what about that … ? What exponent would we stick on the left shoulder of 2 to get 32? The answer, of course, is 5, since 2^5 = 32. O.K., so far so good: the left side of this equation is clearly equal to 5.

Now how about the right side? While the left side asked one question, the right side asks two questions because it has two log terms. First, the term, log 2 of 4, asks: what exponent do we put on 2 to get the number 4? That, of course, is 2, since 2^2 = 4. And the next term, log 2 of 8, asks: what exponent do we put on 2 to get the number 8? That, of course, is 3, since 2^3 = 8.

So the two log terms on the right side are 2 and 3. And we are supposed to add those terms because the equation says to add them. And what is 2 + 3? It is 5, the same number we just got for the left side of the equation. So that is that. The rule works. We can see it working!

And all it is really saying (for this example) is this:

The exponent you put on 2 to get 32 [which is 5] is the sum of the exponents you put on 2 to get the factors of 32, 4 and 8. Or, stated more succinctly and more generally:  the exponent you put on a base to get a certain number is the sum of the exponents you put on that same base to get the factors of that certain number.

That is all that this formula is saying; nothing more, nothing less. So if you understand what I’ve explained here, you understand this rule more deeply. And that is a cool thing. So pat yourself on the back, and go  enjoy the rest of your day!