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Archive for the ‘Elementary Math’ Category

How to Let Kids Use Calculators Without Ruining Their Ability to Think Numerically


Hi folks,

Now that summer has officially begun, I’m enjoying a certain distance from the heat of the school year, and that distance gives me a chance to reflect.

One set of ideas that my mind keeps poking around again and again is this:  a) the weakness in actual number sense among today’s elementary and secondary students,
b) the concomitant modern focus on teaching Number Sense during these school years, and c) the now-rampant overuse of calculators.

I find it interesting that Number Sense has become a “big important new topic” that math instructors are required to teach. I also find it interesting that the new focus on Number Sense has been growing steadily at the very same time that students in so many parts of our country have become more and more calculator dependent.

Could there be a connection?

Yes, undoubtedly! Back when I set up shop tutoring math, K-12, in 1990, Santa Fe (NM) Public School students were not permitted to use calculators willy-nilly. Because of that, our students were not calculator-dependent. Students were expected to know the truths of arithmetic forwards and backwards, and wouldn’t have dreamed of reaching for a calculator to find the value of something so simple as, say, 7 + 5, as happens routinely today. Yes, routinely! I should know; I’m a professional math tutor.

What’s more, I’d say that students in the 1990s generally understood concepts such as odd and even numbers, prime and composite numbers, how to prime factorize, how to find the GCF and the LCM, and the many other skills that are part of the “new area of math instruction we call Number Sense.

That’s because teachers used to require students to use their minds to work with numbers. Students used to grind out 7/18 + 5/12 by hand, not by pressing buttons. They used to figure out the LCM of 22 and 30 by using an algorithm rather than by tapping an app. They used to prime factorize numbers using the good old factor tree and simplify radicals by thinking rather than by pressing a sequence of buttons and scrolling through the numbers flashing across their LCDs.

You can probably see where I’m going with this. Today’s math students have become overly calculator dependent. That dependence on calculators, in turn, has made them deficient at the skills in the topic area we call Number Sense. And precisely because today’s students are so deficient at number sense, precisely because they have been allowed to become so dependent on their e-devices rather than on their mental devices, curriculum designers have devised this whole new area of math, Number Sense, that now gets taught as its own “thing” rather than being an integral thread of everyday math instruction. Number sense used to be something students developed naturally, by mentally working with numbers, day-in, day-out, using paper and pencil and mental math.

Lest I be called a Luddite, I’m not saying that calculators have no place in the math curriculum. But as a tutor who has helped students with math for some 27 years now, I can say with certainty that today’s students’ innate ability to work with numbers, play with numbers and calculate with numbers has been dulled and frankly allowed to atrophy because calculators have become an all-too-easy, all-too-available crutch.

In this way, math curricula and math educators who overly promote calculator usage have done a great disservice to students. The good news, though, is that  teachers could correct course without too much trouble.

Teachers could still allow students to use calculators, quite appropriately, for higher-order processes — such as graphing two functions to see where they intersect, and to see if the answer found that way comports with the answer attained by solving the systems simultaneously by hand — while at the same time disallowing calculator usage for arithmetic calculations.

I’d like to see teachers get their students back to basics in this way because, from my perspective, we’re raising a new generation of students, many of whom have little ability to calculate mentally and little understanding of how numbers work. As a result, these children (soon-to-be adults) are unnecessarily vulnerable.

They’re vulnerable because they cannot tell if they are receiving the correct change from a cashier. They’re vulnerable because they cannot tell if their car or home interest payment is correct. And they’re vulnerable in a larger sense because they lack the ability to easily think numerically, i.e., quantitatively. And when people lack the fundamental ability to think quantitatively, even having a calculator won’t save them in many situations. That’s because they might not even know what operation to do to find a solution in a real-world situation.

But in an even more direct and practical sense, the new calculator-dependent students are vulnerable because they have been set up to struggle mightily in their college math classes. That’s because nearly all U.S. colleges require students to take math tests without using calculators!

So I say let’s get back to basics, and let’s do it in a smart way. Let’s continue to let students use calculators for higher-order thinking skills, but let’s disallow calculators for ALL arithmetic so that students will be required to once again become strong in those critical fundamental skills and so that they will re-gain the natural form of Number Sense that is their right and their due.

 

 

 

 

 

“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.

 

 

 

 

 

 

 

Math Trick: Find the GCF for 3 or More Numbers


Find the GCF, your teacher says … not just for 2 numbers, but for 5 of them.

And yes, you need to do it by prime factorizing.

Can’t you just hear the students’ groans?!

But what if there were a way to do this without prime factorizing? Could it really be?

Yes!

What I’m about to teach you is a technique that lets you find the GCF of as many numbers as you wish, and with much greater ease than the old factoring technique. (by the way, I don’t really hate the factoring technique … it actually teaches you a lot about numbers … but it can get annoying!).

So why don’t they teach this new way in school? No idea. But let’s just focus on the technique because once you do, you’ll be so much faster at finding the GCF …  you’ll be amazing your friends and your teacher, too!

So just kick back, watch the video — and learn …. then do the practice problems at the end of the video, to become a whiz! And remember, if you ever want extra help in the form of tutoring, I’m available — worldwide — thanks to the power of online videoconferencing.

Enjoy!

— Josh

 

 

 

Fun Math Trick: Find the GCF for Any Two Numbers


“Get the GCF for these two numbers,” your teacher says.

“How?” you ask.

“Easy,” your teacher replies. “Do what I told you yesterday.”

But perhaps you don’t remember. Or maybe you don’t like the technique your teacher used, telling you to prime factorize the numbers …

Even though that felt awful, now you’re in luck because I’m going to show you another way to find the GCF, one that’s more intuitive and easier than the prime factoring technique. So just sit back and watch the following video. Then do the practice problems at the end of the video. And you’ll know something that even your teacher probably doesn’t know … a thoroughly original way to find the GCF of two numbers.

 

 

 

Fraction Hack #2: The Size of the Smaller Number


I received an interesting question from alert reader Ivasallay a couple of days ago … about fractions.

Responding to my post about the fraction “hack” of using the gap between fraction numbers, Ivasallay wrote: “What if the numerator is smaller than the gap?”

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Good question, and thanks for sharing it. My answer: Yes, the numerator could be smaller than the gap, and if it is, that can help us simplify fractions, too.

Now we could have a fraction like 15/6, in which the lesser of the two numbers is the denominator, so to keep our discussion general I’m going to talk, not about the numerator, but rather about the “smaller fraction number,” whether numerator or denominator.

The way this matters is as follows: like the gap, the smaller fraction number provides an upper limit, a greatest possible value, for the GCF of the fraction’s two numbers. So if the fraction is 12/90 (smaller number being 12), that means that the GCF can be no larger than 12. If the fraction is 3/1011, with lesser number 3, the GCF can be no larger than 3.

The reason should be obvious, and when I say this I really mean it. Take the fraction 6/792, for example. Could a number larger than 6 go into both 6 and 792? Well there may be a number larger than 6 that goes into 792 evenly, but nothing larger than 6 can go into 6 itself, right? A large peg can’t go through a tiny hole, right? So there you go. Nothing larger than 6 can go into both 6 AND 792. QED.

So what does this mean for you, the math student, or parent of a math student, or the teacher of math students? … I means you want to keep in mind that in actuality two different numbers will help you nail down the size of the GCF. One is the gap between the fraction numbers, and the other is our “new friend,” the smaller of the two fraction numbers.

And here’s another … hack fact. (Whenever I say that, you know we’re heading into ‘nerd-land,’ right?) For both limiting numbers, the gap and the smaller fraction number, the only numbers that can possibly go into both fraction numbers are the FACTORS of those limiting numbers. So for example, if your fraction is 6/50, with the smaller number of 6, the only numbers that can possibly go into 6 and 50 are the factors of 6: i.e., 6, 3, or 2.

A nice rule of thumb:  see which is smaller, the gap or the smaller fraction number. Then use that smaller number as your largest possible GCF. To nail this down, let’s do two example problems.

Example 1:  8/44. What’s smaller? 8 or the gap, 36. Obviously 8! So use 8. Test the factors of 8, which are 8, 4, 2. Notice that 8 doesn’t go into both 8 and 44. But 4 does, so 4 is the GCF, and using 4, the fraction simplifies down to 2/11.

Example 2:  22/36. What’s smaller? 22 or the gap, 14. Here the gap is smaller. So test the gap’s factors: 14, 7, 2. 14 doesn’t go into 22 and 36; nor does 7. But 2 does. So 2 is the GCF, and using 2, the fraction simplifies to 11/18.

Time for you all to try your hands at this fun practice, which catapults your “number sense” to new heights.

For each problem, 1) identify the fraction’s smaller number and the gap. 2) Say which of those two numbers is smaller. 3) Using that number’s factors, find the GCF. 4) Finally, using the GCF, simplify the fraction. Answers follow.

SIMPLIFY THE FRACTIONS:

a)   8/42

b)  12/20

c)  36/60

d)  18/96

e)  21/91

ANSWERS:

a)   8/42:  1)  smaller # = 8; gap = 34.  2)  8 < 34. 3)  GCF = 2. 4)  4/21

b)  12/20:  1)  smaller # = 12; gap = 8.  2)  8 < 12. 3)  GCF = 4. 4)  3/5

c)  36/60:  1)  smaller # = 36; gap = 24.  2)  24 < 36. 3)  GCF = 12. 4)  3/5

d)  18/96:  1)  smaller # = 18; gap = 78.  2)  18 < 78. 3)  GCF = 6. 4)  3/16

e)  21/91:  1)  smaller # = 21; gap = 70.  2)  21 < 70. 3)  GCF = 7. 4)  3/13

Josh Rappaport is the author of five math books, including the wildly popular Algebra Survival Guide and its trusty sidekick, the Algebra Survival Workbook. And FYI:  the 2nd Edition of the Survival Guide was just released in March, so get it while it’s hot off the press! If you’d like to get tutored by Josh, you can. Josh and his remarkably helpful wife, Kathy, use Skype to tutor students in the U.S. and Canada in a wide range of subjects. They also prep students for the “semi-evil” ACT and SAT college entrance tests. If you’d be interested in seeing your ACT or SAT scores soar, shoot an email to Josh, sending it to: josh@SingingTurtle.com  We’ll keep an eye out for your email, and in our office, our tutoring is always ON … except on Saturdays.

How to find the GCF of 3+ Numbers — FAST … no prime factorizing


Suppose you need to find the GCF of three or more numbers, and you’d really prefer to avoid prime factorizing. Is there a way? Sure there is … here’s how.

 

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Example:  Find the GCF for  18, 42 and 96

Step 1)  Write the numbers down from left to right, like this:

………. 18     42     96

[FYI, the periods: …. are there just to indent the numbers. They have no mathematical meaning.]

Step 2)  Find any number that goes into all three numbers. You don’t need to choose the largest such number. Suppose we use the number 2. Write that number to the left of the three numbers. Then divide all three numbers by 2 and write the results below the numbers like this:

2    |  18     42     96
……..  9     21     48

Step 3)  Find another number that goes into all three remaining numbers. It could be the same number. If it is, use that. If not, use any other number that goes into the remaining numbers. In this example, 3 goes into all of them. So write down the 3 to the left and once again show the results of dividing, like this:

2    |  18     42     96
3    |    9     21     48
……… 3      7      16

Step 4)  You’ll eventually reach a stage at which there’s no other number that goes into all of the remaining numbers. Once at that stage, just multiply the numbers in the far-left column, the numbers you pulled out. In this case, those are the numbers:  2 and 3. Just multiply those numbers together, and that’s the GCF. So in this example, the GCF is 2 x 3 = 6, and that’s all there is to it.

Now try this yourself by doing these problems. Answers are below.

a)   18, 45, 108
b)   48, 80, 112
c)   32, 72, 112
d)   24, 60, 84, 132
e)   28,  42, 70, 126, 154

Answers:
a)   GCF =  9
b)   GCF =  16
c)   GCF =  8
d)   GCF =  12
e)   GCF =  14

Everyday Life Sparks Mathematical Puzzles


So here’s the situation: you’re at the breakfast table, enjoying a bowl of steaming-hot steel-cut oats and maple syrup, and you just poured yourself a mug of black coffee. But then you realize you want to pour some milk in the coffee (sorry, purists). But the milk is in the frig, six feet away. So of course you walk to the frig, grab the milk, bring it to the table, pour some in your coffee, return the milk to the frig and sit back down. Question: could you have done this more efficiently?

Yes, of course. You could have brought your cup of coffee with you as you walked to the frig, poured the milk right there at the frig, returned the milk, and then walked back to the table.

Being Smart?

“Morning Joe”

When I realized this this morning, I thought … hmmm. Had I used a bit of forethought, I would save myself an entire round trip from the table to the frig. And while I have no problem making that extra trip (hey, just burned 1.3 calories, right?), the experience made me wonder if anyone has ever developed a mathematics of efficiency for running errands.

I could imagine someone taking initial steps for this. One would create symbols for the various aspects of errands. There would be a general symbol for an errand, and there would be a special ways of denoting: 1) an errand station (like the frig), 2)  an errand that requires transporting an item (like carrying the mug), 3) an errand that requires doing an activity (pouring milk) with two items (mug and milk) at an errand station, 4) an errand that involves picking something up (picking up the mug), and so on. Then one could schematize the process and use it to code various kinds of errands. Eventually, perhaps, one could use such a system to analyze the most efficient way to, say, carry out 15 errands of which 3 involve transporting items, 7 involve picking things up, and 5 involve doing tasks at errand stations. Don’t get me wrong! I have not even begun to try this, but I’ve studied enough math that I can imagine it being done, and that’s one thing I love about math; it allows us to create general systems for analyzing real-world situations and thereby to do those activities more intelligently.

Of course, one reason I’m bringing this up is to encourage people to think more deeply about things that occur in their everyday lives. Activities that appear mundane can become mathematically intriguing when investigated. A wonderful example is the famous problem of the “Bridges of Konigsberg,” explored by the prolific mathematician Leonhard Euler nearly 300 years ago.

Euler in 1736 was living in the town of Konigsberg, now part of Russia. The Pregel River, which flows through Konigsberg, weaves around two islands that are part of the town, and a set of seven lovely bridges connect the islands to each other and to the town’s two river banks. For centuries Konigsberg’s residents wondered if there was a way to take a walk, starting at Point A, crossing each bridge exactly once, and return to Point A. But no one had found a way to do this.

One of the famous Seven Bridges of Konigsberg

One of the famous Seven Bridges of Konigsberg

Enter Euler. The great mathematician sat down and simplified the problem, turning the bridges into abstract line segments and transforming the bridge entrance and exits into points. Eventually Euler rigorously proved that there is no way to take the walk that people had wondered about. This would be just an interesting little tale, but it has a remarkable offshoot. After Euler published his proof, mathematicians took his way of simplifying the situation and, by exploring it, developed two new branches of math:  topology and graph theory. The graph theory ideas that Euler first explored when thinking about the seven bridges sparked a branch of math that’s used today to determine the most efficient ways of connecting servers that form the backbone of the internet!

Of course, there’s also the classic example of Archimedes shouting “Eureka!” and running through the streets naked after seeing water rise in his bathtub. In that moment, Archimedes, who had been trying to help his king figure out if the crown that was just made for him had been created with pure gold, or with an alloy, saw that the water displacement would help him solve the problem. In the end, Archimedes determined that the crown was not pure gold, and the king rewarded the great thinker for his efforts.

As I write this, I find myself wondering if any of you readers can think of other situations in which everyday life experiences led mathematicians or scientists to major discoveries. It would be enlightening to hear more of these stories.

And, if no such stories spring to mind, check out this site, which lists several such stories.  http://www.sciencechannel.com/famous-scientists-discoveries/10-eureka-moments.htm

In any case, the way that such discoveries occur shows that you never know where a seemingly trivial idea might lead … so it’s good to keep your eyes and mind open.

How to Find the GCF for Three or More Numbers


To find the GCF for three or more numbers,  follow these steps:

1)  Determine which of the given numbers is smallest, then find the smallest difference between any pair of numbers.

2)  See what is smaller:  the smallest number, or the smallest difference. Whichever one  is smallest, that number is the GPGCF (Greatest Possible GCF). That means that this is the biggest number that the GCF could possibly be. Or, more formally we would say:  The GCF, if it exists, must be less than or equal to the GPGCF.

3)  Check if the GPGCF itself goes into all of the given numbers. If so, then it is the GCF. If not, list the factors of the GPGCF from  largest to the smallest and test them until you find the largest one that does divide evenly into the given numbers. The first factor (i.e., the largest factor) that divides evenly into the given numbers is, by definition, the GCF.

EXAMPLE:

Problem:  Find the GCF for 18, 30,  54.

1)  Note that the smallest number is 18, and  the smallest difference between the pairs is 12 [54 – 30 = 24;  54 – 18 = 36;  30 – 18 = 12] .

2)  Of those four quantities (the smallest number and the three differences), 12 is the least. This means that the
GPGCF = 12.

3) Check if 12 divides evenly into the three given numbers: 18, 30 and 54. In fact, 12 doesn’t divide evenly into ANY of these  numbers. Next we check the factors of 12, in order from largest to smallest. Those factors are: 6, 4, 3 and 2. The first of those that divides evenly into all three numbers is 6. [18 ÷ 6 = 3;  30 ÷ 6 = 5;  54 ÷ 6 = 9]. So the GCF = 6. And we are done.
MORE CHALLENGING PROBLEM:

Find the GCF for 24, 148, 200.

1)  Note that the smallest number is 24, and that the smallest difference between the pairs is 52 [200 – 148 = 52;  200 – 24 = 176;  148 – 24 = 124] .

2)  Of those four quantities (the smallest number and the three differences), 24 is the least. This means that for this problem, the GPGCF = 24.

3) Check if 24 divides evenly into the three given numbers: 24, 148 and 200. While 24 does divide evenly into 24, it does not divide evenly into 148 or 200. So next we check the factors of 24, in order from largest to smallest. Those factors are: 12, 8, 6, 4, 3 and 2. The first of those that divides evenly into the three given numbers is 4. [24 ÷ 4 = 6;  148 ÷ 4 = 37;  200 ÷ 4 = 50]. So the GCF = 4. And, once again, we are done.

The process may seem a bit long, but once you get used to it and start doing it in your mind, not on paper, you should find that it actually is quite fast. And you’ll find yourself figuring out the GCF for three or more numbers all in your mind — with no need for pencil and paper — while everyone around you will be making prime factor trees or using calculators. And surely that is a good feeling.

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! 

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! 

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!