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Archive for the ‘Algebra Survival Guide’ Category

Algebra Mistake #4: Combining Negative Numbers


Here’s a common mistake, and a very understandable one, too. Students need to combine two negative numbers, and they, of course, wind up with an answer that’s positive. Why? Because, they’ll say — pointing out that you yourself have told them this —  “Two negatives make a positive!”

This video gets to the root of this common misunderstanding by helping students understand exactly when two negatives make a positive, and when they don’t.

 

Make sure you watch the whole video, as there are practice problems at the end, along with their answers.

 

 

 

 

 

 

 

 

 

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“Algebra Survival” Program, v. 2.0, has just arrived!


The Second Edition of both the Algebra Survival Guide and its companion Workbook are officially here!

Check out this video for a full run-down on the new books, and see how — for a limited time — you can get them for a great discount at the Singing Turtle website.

 

Here’s the PDF with sample pages from the books: SAMPLER ASG2, ASW2.

And here’s the website where you can check out the books more fully and purchase the books.

 

 

 

 

 

 

 

“Unpacking” Terms from Parentheses


How do you get math terms out of parentheses? And what happens to those terms when you remove the parentheses?

It seems like the process should be simple. But this issue often plagues students; they keep getting points off on tests, quizzes, homework assignments.  What’s the deal?

The deal is that there’s a specific process you need to follow when taking terms out of parentheses, and what you do hinges on whether there’s a positive sign (+) or a negative sign (–) in front of the parentheses.

But not to worry. This video on this page settles the question once and for all. Not only that, but the video provides a story-based approach that you can teach (if you’re an instructor) or learn (if you’re a student) and remember (no matter who you are). Why? Because stories are FUN and MEMORABLE.

So kick back and relax (yes, it’s math, but you have a right to relax) and let the video show you how this process is done.

And in customary style, I present practice problems (along with the answers, too) at the end of the video so you can be sure you understand what you believe you understand.

 

 

 

 

 

How to Transform from Standard Form to Slope-Intercept Form


Is there any point to doing something the long way when you can just as well do it in a shorter, much more efficient way? I say, Heck no! We can do things in the “Triple-F” way:  Fast, Fun & Friendly, and with deep understanding, to boot.

High-Octane Boost for Math

High-Octane Boost for Math Ed

So in that spirit, today I’ll get us started on quickly and effortlessly converting a linear equation from what’s called standard form (Ax + By = C) to what we know as the good-old slope-intercept form (y = mx + b).

To better grasp standard form, let’s replace its mysterious A, B, and C with actual numbers:  4 for A, 2 for B and 8 for C. That gives us the more typical looking equation of an actual line:  4x + 2y = 8. Do you recall seeing this kind of equation in your algebra text and class? Sure you do. You get this kind of equation in the chapter(s) on the coordinate plane and in other spots, too.

Now usually when books teach us how to convert from this “standard” form to slope-intercept form, they tell us to solve the equation for y. Of course that works, but it takes too darn long.

To understand the quicker way, let’s have a little fun with the standard form of the equation: Ax + By = C

We’re going to start with this standard form and solve that for y. And as you’ll see, we’ll learn some useful things from the result.

To kick things off, we start with Ax + By = C, and we subtract the Ax term from both sides. That leaves us with this equation:

By = – Ax + C

Now take this new equation and divide both sides by B. That gives us this little gem of an equation:

y = (– A/B) x + C/B

I’m going to call this the magic equation both to give us a way to refer to it and to show us what’s so useful about it.

The big insight is that this magic equation is actually, believe-it-or-not, in slope-intercept form; we just need to SEE it that way. Here’s how.

In slope-intercept form (y = mx + b), notice that the y variable is all by its lonesome on the left side. Do we have that in the magic equation? Yes we do. So … CHECK!

In slope-intercept form, there’s a value called m (aka, the slope) that is multiplying the x variable. Do we have something in the magic equation that’s multiplying the x variable? Why yes, and it happens to be
(–A/B). So do we have the slope showing in the magic equation? Yes, the slope is:  (–A/B). So … CHECK!

Finally, in slope-intercept form, there’s a constant (i.e., a number term, not a variable term) that appears after the mx term. So do we have a constant after the mx term in the magic equation? Yes, indeed. We have C/B. Note that in any actual linear equation, B and C will be actual numbers, not variables. So the value you get when you divide B by C (the quotient B/C), also must be a real number, just as surely as the real numbers 8 and 2 gives us the real number 4 when we divide 8 by 2.

So, to address the final question, do we have a b-value in the magic equation? Yes, it’s (C/B). C/B is the y-intercept, the real number we call b in slope-intercept form. So once again … CHECK.

So all in all, do we now have the equation in slope-intercept form? Yes, indeed. You just need to realize that
(–A/B) is the slope, and (C/B) is the y-intercept.

In my next post I’ll show you how you use these results to quickly transform the equation from standard form to slope-intercept form. It will be amazing.

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