## The Purpose of Homework and My Response The purpose of homework is to promote learning.  That’s it.  It’s not a way to earn a grade or something to keep kids busy.  It’s also not something that just must be completed in order to stay out of trouble.  Homework is a chance to try things independently, make mistakes and explore the nature of those mistakes in order to better learn the material at hand.

If students are not learning from the homework, it is a waste of time and effort.  There are a few things that could cause students not to learn from the homework.  Even if the assignments are of high quality, without the reflection and correction piece, students will not learn much from homework.

Reflection and correction go together.  It’s not about getting right answers, but thinking about what caused mistakes, identifying misconceptions or procedural inefficiencies and replacing those.  To reflect a student should NOT erase their incorrect working but instead should write on their homework, in pen, what went wrong and what would have been better.

It is quite possible more can be learned when reviewing homework than any other time.  It is certainly a powerful experience.

Textbooks and videos, tutors and peer help offer little appropriate support to help make homework, or practice, meaningful.  Textbooks only provide correct answers, YouTube videos usually do similar treatment to topics as textbooks offer.

I wish to help students learn and believe that reviewing work that has been done is too powerful of an opportunity to pass.  The trick is, how can I provide reflection and insight when to someone I am not sitting with and talking to?  I think I can help provide this reflection piece by doing all of the practice problems myself on a document camera and discussing pitfalls and mistakes, as well as sharing my thinking about the problems as I tackle them.  Further, I can share typical mistakes I see from students as they are learning topics.

So as I develop the Algebra 1 content I will be working on adding videos and short written responses to the assignments to help students think about what they’ve done, its appropriateness, correctness and their level of understanding.

## Adding and Subtracting Algebraic Fractions

part 2

Introduction to Algebraic Fractions
Part 2

In the last section we saw both how to reduce Algebraic Fractions, which if you recall, are also called Rational Expressions, but also how the math works. Because of the relationship between division and multiplication, and multiplication’s commutative property, we can reduce Algebraic Fractions like the one below:

$\frac{3a+6}{12{a}^{2}-9}$

When dealing with Algebraic Fractions, your work is not done until you’ve reduced completely. The example above is the unfinished answer to the first problem we will do to introduce addition and subtraction of Algebraic Fractions.

$\frac{2a-5}{12{a}^{2}-9}+\frac{a+11}{12{a}^{2}-9}$

Since these have a common denominator, we just add the like terms in the numerator. Note, in math when we say add, it means combine with addition and subtraction.

$\frac{2a-5+a+11}{12{a}^{2}-9}$

Combining like terms, we end up with:

$\frac{3a+6}{12{a}^{2}-9}$

But since each term has a factor of 3, we can reduce each term by 3:

$\frac{3a+6}{12{a}^{2}-9}\to \frac{\overline{)3}a+\overline{)3}\cdot 2}{\overline{)3}\cdot 4\cdot {a}^{2}-3\cdot \overline{)3}}=\frac{a+2}{4{a}^{2}-3}$.

Our answer has four total terms. While some share a common factor, not all four terms share a common factor, so we are finished.

If the denominators are the same, you just combine the numerators.

With subtraction it is slightly trickier. Let’s simplify a problem and see how this works.

5 $–$ 4 + 3

5 $–$ (4 + 3)

The two expressions above are not the same. The first equals four, while the second is -2. The parenthesis make a group of the four and three, which is being subtracted from the five. The four and the three are both being subtracted from the five. But, great care is in order here. It is easy to mess up these signs.

5 $–$ (4 $–$ 3) = 4

Because 5 $–$ 4 - -3 is 5 $–$ 4 + 3.

Remember that fraction bars also create groups in Algebra. So, instead of parenthesis, you will see:

$\frac{5}{x}-\frac{4+3}{x}$

This is the same as:

.

An example with variables would be:

$\frac{{x}^{3}-5{x}^{4}}{9}-\frac{5{x}^{4}-{x}^{3}}{9}$

Note: Exponents are repeated multiplication, they do not change from addition. Also, in order to be like terms, the variables and exponents must be the same. x3 and x5 are not like terms.

$\frac{{x}^{3}-5{x}^{4}-\left(5{x}^{4}-{x}^{3}\right)}{9}$

Caution and care are in order when dealing with subtraction and Algebraic Fractions!

$\frac{{x}^{3}-5{x}^{4}-5{x}^{4}+{x}^{3}}{9}$

Combining like terms we get

$\frac{2{x}^{3}-10{x}^{4}}{9}$.

Since there is not a common factor between all terms, we are done.

Adding and subtracting Algebraic Fractions with unlike denominators involves finding the LCM (lowest common multiple) of each denominator. We will restrict our denominators to monomials for now, as to keep this appropriate for beginning Algebra students.

Let’s begin with the denominators a and b. All we know is that a and b are numbers that cannot be zero, but we don’t know their exact value. So, we assume they are relatively prime, making their LCM their product, a×b.

$\frac{3}{a}+\frac{2}{b}$

We arrive at common denominators through multiplication. Don’t get confused here, we are multiplying by a number that equals one if it were reduced, but we don’t want to reduce until we are finished. Also, note, that since e are multiplying by one, the expression will look different but will have the same value. It is not unlike the difference between a twenty dollar bill versus a ten and two five dollar bills.

Our denominator will be ab. To change a into ab, we multiply by b. We multiply b by a, writing the variables in alphabetical order will help to recognize that they are the same. (Sometimes students write ab and then ba. While they’re the same, the order in which they’re written can confuse you.)

$\frac{b}{b}\cdot \frac{3}{a}+\frac{2}{b}\cdot \frac{a}{a}$

In many respects, adding or subtracting Algebraic Fractions is easier because there is less calculation taking place.

Let’s walk through an uglier problem involving subtraction and negative signs.

$\frac{7{x}^{2}-2y}{5{x}^{2}y}-\frac{4x-2}{5{x}^{2}}$

We need to find the LCM of . The LCM will be 5x2y. So we need to multiply the second fraction by y over y.

$\frac{7{x}^{2}-2y}{5{x}^{2}y}-\frac{4x-2}{5{x}^{2}}\cdot \frac{y}{y}$

Now we are multiplying the entire group of 4x $–$ 2 by y, not just whatever term is written next to the y.

$\frac{7{x}^{2}-2y}{5{x}^{2}y}-\frac{4xy-2y}{5{x}^{2}y}$

Normally I would not write the step above, but did so to help make sure you understand why the fraction on the right is 4xy, not just 4x. We have to distribute the y to the entire group.

Now with care for that negative sign, let’s put it all together.

$\frac{7{x}^{2}-2y-\left(4xy-2y\right)}{5{x}^{2}y}$

which will become:

$\frac{7{x}^{2}-2y-4xy+2y}{5{x}^{2}y}$

Combining like terms, we get:

$\frac{7{x}^{2}-4xy}{5{x}^{2}y}$.

Before we can say we’re finished we need to check for a common factor (GCF) that could be divided out. These don’t always exist but if one does, and you had the answer correct up to that point, it would be a shame to mess up the last little step, so check.

$\frac{7{x}^{\overline{)2}}-4\overline{)x}y}{5{x}^{\overline{)2}}y}=\frac{7x-4y}{5xy}$.

In review, you need a common denominator which will be the LCM of the denominators. You must take care to both distribute property in the numerator, and watch for sign errors, especially with subtraction, when combining like terms. The last thing is to check for a common factor between all terms when you’ve finished combining like terms.

Practice Problems
Instructions: Add or Subtract as indicated

1. $\frac{3x}{5}+\frac{2}{a}$

2. $\frac{4}{9a}+7$

3. $\frac{4}{9a}+\frac{7}{9}$

4. $\frac{4}{9a}+\frac{7}{a}$

5. $\frac{2x-1}{3{x}^{2}}-\frac{2x-1}{{x}^{2}}$

## Combining Like Terms & The Order of Operations

Be fair-warned…this might seem pedantic, and I agree, it is.  But, with good cause. The point of this discussion is to cause students to pause and think about something they’ve always done in math without any understanding.  It’s true, you can combine like terms, but why?  To that point, and I do not say this to put anybody down, but I bet most math teachers have never wondered this on their own.

### Why Can’t We Combine Unlike Terms?

I pose this question to Algebra 1 students every year, for over a decade now.  The responses are always with an air of incredulity.  Their looks say to me:

Come on Mr. Brown, you should be smarter than to ask such a dumb question!

When I press for an answer they’re suddenly tongue-tied.  The most common answer, often spit at me with some venom is:

Because they’re not like terms.

Okay then, why can you add things that ARE like terms?

### Are We in Violation of PEMDAS

The order of operations is the set of instructions by which math is to be performed.  It’s kind of a big deal!  Mess it up and you’re in a lot of trouble.  And yet, how is it that we can combine like terms if there’s multiplication occurring first?

Consider 3x + 5x – 2a.  Even the most average of 9th grade, Algebra 1 students will tell you that obviously equals 8x – 2.  My question is this:  How can you add 3and 5x if there’s multiplication taking place.  The three and the x are multiplying, right?  And in the order of operations, doesn’t multiplication come before addition?

Student Responses

When I ask students why like terms can be combined even though there’s multiplication taking place and they add before multiplying, they just say, “cuz.”  This is a classic example of something that’s been taught and accepted without understanding.  It’s not a complicated issue, but promoting mathematical literacy is my crusade of late.

Consider the product of four and five.  I remember being a young child and counting groups of five to find the product of five and a number.  4 x 5 = 5 + 5 + 5 + 5

Good to go on that point?

Then 3x = x + x  + x

And 5x = x + x + x + x + x

Then 3x + 5x =  x + x  + x  + x + x + x + x + x8x

Because multiplication is repeated addition, and we’re adding the same thing repeatedly, we can combine the coefficients of x.

Why Can’t You Combine Unlike Terms?

Consider our original expression: 3x + 5x – 2a.  Two times a is a + a.  Three and five makes 8x.  But x and a are probably not the same thing (we don’t REALLY know because they’re unknown), so we cannot combine them with addition.

8x + 2

x + x  + x  + x + x + x + x + x + a + a

This cannot be simplified further than 8x + 2a.

To combine unlike terms would be in violation of the order of operations.  Since and a are different, we would need to multiply 8 and 2 by them respectively, before adding them.

I hope this short discussion has caused you to pause and think about what you thought you knew about Algebra.  Let me know what you think in the comments.  Thank you, again, for reading.

Philip

## Promoting Mathematical Literacy through Non-Procedural Questions

Math was okay until they threw the alphabet in it.

I have heard, Math was okay until they threw the alphabet in it, so many times from so many adults, who I wish were trying to be witty, but are in fact completely serious.  And these are reasonable, intelligent people.  Why does the abstraction of a variable or an unknown become so confusing that it creates a huge disconnect?

I’d like to share with you an assignment that I believe will help students transition from the concrete application and properties of Real Numbers to the abstractions we deal with in Algebra and mathematics beyond.

The desired outcomes of this assignment are:

• Improve mathematical literacy by encouraging students to read the mathematical meanings created by the spatial arrangement of numbers and symbols
• Improve their understanding of the Order of Operations and to help student realize that the Order of Operations is not its own topic, compartmentalized, but rather an over-arching understanding of how math is performed.
• Promote abstract thinking about numbers and their properties
• Introduce some concepts that will come into play later in Algebra, like finding the x-intercepts of a quadratic once it is factored

How to Introduce the Assignment

Students must be aware that they will be dealing with abstract ideas and that there are sometimes more than one right answer.  Also, a student can be right, but not completely right, they could also be wrong, but not always.  By fostering a healthy discussion about these problems you can introduce the idea that in order for something to be mathematically true, it must always be true.  If a single circumstance is untrue, then the statement is untrue.

Consider the problem: Given that a and b are real numbers, and the following is true, what do you know about the numbers a and b?   a×b=0

A student might say, In this case a and b are both zero.

That is correct in one case, but there are many cases where that is not true.  It is true that one of them must be zero.

The First Prompt

Given that a and b are real numbers, and the following statement is true, what can you conclude at the numbers a and b?

a – b = 0

At first have them think and write on their own.  Make sure they’re all working, not avoiding this uncomfortable notion.

After a given amount of time (short, maybe one minute), instruct them to talk with two different people.  Be clear that the expectation is that they take turns, one person shares, the other listens and responds.

After the time is up, have a whole-class discussion, but avoid being the authority until the discussion is winding down.  Only be the authority on the subject to help summarize.

The Second Prompt

With the same properties of the numbers and the statement being true, provide them with the equation:

(a + 5)(b –7) = 0

Conduct conversations as you did with the first, maybe allowing more time for them to talk together as this is a more complicated situation.

The Third Act

Switching gears from properties of variables to applying some properties of real numbers will promote their understanding of when the associative and commutative properties as well as challenge their understanding of how math is written.  We are really trying to promote their ability to read and write math and their fact that the spatial arrangement has meaning in math.

Have the students try and add parenthesis, as many as they like, to the following equation, so that it will be true.  Have them try it on their own first, then provide a short amount of time for peer discussion.

3•2 – 72 + 5 = 80

When you conduct a whole-class discussion, make sure it’s student lead, your role is as a mediator, not a disseminator of facts.

Fourth Act

Instruct students to create a similar problem by making a statement they know is true and removing the parenthesis.   For example, they might make up:

8(5 – 3) + 11 = 27

But would only write

8•5 – 3 + 11 = 27

When they’re completed their problem, have them show you.  Once everybody has a problem, hand out 3×5 cards.  On the front of the card the student will write their problem, without the parenthesis.  On the back, they’ll write their name.  Have them pass the cards forward and you can distribute them at to another class the following day.

Last Thing for the Lesson:

The idea here is the same as with the previous activity, but we are accessing their knowledge from a different angle.  They will be given an expression with parenthesis and be asked if the parenthesis can be removed without changing the value of the expression.

For example:  (5 + 4) – 2, or 11 – (4 – 9).

Introduce these problems in the same fashion, with quiet thinking first, then small group discussions, then whole-class discussion.

The Homework:

The homework is critical here because it will challenge students to think and examine how the way in which math is written changes the meaning.  It will also force them to think about numbers in a general fashion.

Timing

The last thing I’d like to mention is that this could easily be done over two separate days depending on the aptitude of the class you’re teaching.

I hope this is helpful and food for thought.

## Why Teaching Properties of Real Numbers is Important If you are going to do a fraction review, the lesson here might be of some help.  I believe things are best reviewed in context, but this is a decent set of information that also introduces the real numbers and some other basics of math. The PDF icon to the left has a lesson outline you can feel free to use with the PowerPoints of in any way you see fit.

The structure is all there in the lessons, but they're not over scripted.  Remember, I believe the majority of a lesson should be spontaneous.  It should be anticipated and prepared for, but how the lesson really unfolds depends on the audience.

Below you will find an overview of how and why I teach real numbers as well as two PowerPoint icons you can download and use as your own.  I only ask that you share where you found them.

Anything you purchase from Amazon.com through the banner below goes to producing more materials, and at no cost to you.

### What Good Is It?

The Real Number Line has always been one of the dullest lessons I have to teach.

Natural Numbers are the set of numbers you can count on your fingers, beginning with one.  The Whole Numbers are the Natural Numbers and Zero...Integers are ...

Blah Blah Blah

I have to teach it because it's in the curriculum.  And I always wonder, what use is it if a student knows the difference between a whole number and a natural number?

It is hypocritical of me to complain in such a fashion because I laud the virtues of education being greater than a set of skills or a body of knowledge.  Education is about learning to think, uncovering something previously unknown that ignites excitement and interest.  Education should change how you see yourself, how you think about the world.  It should enrich our lives.

Teaching the Real Number Line can be a huge first step in that direction, if done properly.

### Math is About Ideas, Not Just Computation

There are some rich, yet entirely approachable, mathematical ideas that can be introduced with the Real Number Line (RNL).  For example, a series of questions to be posed to students could be:

1.  The Natural Numbers are infinite, meaning, they cannot be counted entirely.  How do we know that?
2. The Integers are also infinite.  How do we know that?
3.  Is infinity a number?
4. Which are there more of, Natural Numbers of Integers?  How can you know, if they're both infinite?

The idea of an axiom can be introduced.  Most likely, students assume math is true, or entirely made up, but correct or incorrect, because it is written in a book and claimed to be such by a teacher.  The idea of how we know what we know and if math is an invention or a discovery can be introduced by talking about axioms.  For example:

1. Is it true that 5 + 4 = 4 + 5 ?
2. If a and are Real Numbers, would it always be true that b = b?  (What if they were negative?)
3. Is it also true that b = b - a?  How do we know that?
4. Is the following also true:  If a = b, and b = c, then a = c?  How do we know?

The idea here is not to teach students the difference between the Associative Property and the Commutative Property, but to use these properties to introduce students to math as a topic that can be discussed, and that it is not about answer getting, but instead about ideas.

For more on this topic and a few other related items, visit this page.

### Why Are Some Rational Numbers Non-Terminating Decimals? If you had a particularly smart group of students, you could pose this question.  I mean, after all, 1/3 = 0.3333333333333333...  And yet, we are told rational numbers include decimals that can be written as a fraction (the ratio of two integers).

How it works is sometimes very clear and clean.  For example, 0.7 is said, "Seven tenths." And "Seven tenths," can also be written as the ratio of seven and ten.  And the number seven tenths is of course equal to itself, regardless of how it is written.  The number 0.27 is said, "twenty seven hundredths," which can easily be written as the ratio of twenty seven, for the numerator, and one hundred, as the denominator.  And this can continue so long as the decimal terminates.  But try the same thing with the a repeating decimal and you do not end up with things that are equal. The algorithm to convert a repeating, but non-terminator decimal into a fraction is pretty straight forward. But that does not address why a rational number would be a non-terminating decimal. Click the PPT Icon to the left to download a lesson on converting repeating decimals into fractions for honors students.  It includes a proof of why the square root of two is irrational.

The simple reason that some rational numbers cannot be expressed as terminating decimals has to do with our numbering system.  We use base 10 numbers.  Our decimal system provides us an easy way to write fractions with denominators that are powers of 10.

That means that we have ten numbers, including zero, that fill up one column, like a car’s odometer.  When you travel 9 miles the odometer will read 000009.  When you travel the tenth mile the odometer will read 000010.

Not all of our number systems are base ten. While the metric system is base 10, or at least translates into powers of ten, the Imperial system is base 12 from inches to feet, but 5,280 feet to the next unit of miles, and so on.

Time is another great example of bases other than ten.  Seconds and minutes are base sixty.  You need sixty seconds before you have an hour, not ten.  But hours are base 24 because 24 hours are needed to make one of the next category, which is days.

In time, 25 minutes of an hour is the ratio: But in base ten this is 0.4166666666666666... Our decimal system does math in base ten, not base sixty.  This is not 41 minutes!  A typical mistake would be two say 25 minutes is 0.25 of an hour.

Back to our original example of 1/3.  Not all numbers can be cleanly divided into groups of ten, like 3.  If we had a base 3 numbering system, where after the third number we moved, then 1/3 would just be 0.1.  But in our numbering system, 0.1 is one tenth.

Other numbers, like four, translate into ten more easily.  Consider the following: The only issue remaining is that 2.5/10 is not a rational number because 2.5 is not an integer and rational numbers are ratios of two integers.  This can be resolved as follows: Let's try the same process with 1/3. As you can see, we will keep getting ten divided by three, forever.

This is a great example of how exploring a question can uncover many topics within the scope of the course being taught.

I hope this has caused you to pause and think of how exploring questions, relationships and properties in mathematics can lead to greater understanding than just teaching process and answer getting.

The video below is a fun way to explore some of the attributes of prime numbers in a way that provides insight into the nature of infinity.   All of the math involved is approachable to your average HS math student.

Here is a link to the blog post that goes into a little more detail than offered in the video:  Click Here.

## Algebraic Fractions Part 1

Algebraic Fractions Pt 1

Algebraic Fractions $–$ Rational Expressions

In this section we will learn how Algebraic Fractions can be multiplied, reduced and added or subtracted. This particular entry will cover reducing and how reducing uses the greatest common factor of all terms.

It is often the case that students that once struggled with fractions gain insight and confidence with the rational numbers.

An algebraic fraction, or rational expression, is just a ratio of two algebraic expressions. The difference between an algebraic expression and a number is the variable, or unknown value. (Note that they’re called expressions and not equations because they’re not equal to anything.)

For example, 5x, is the product of five and x. Since we do not know what x equals, we cannot carry out the multiplication. So, we just leave it written 5x.

Another algebraic expression would be 15x2. This is the product of 15, x, and x.

An algebraic fraction, or rational expression of these two could be $\frac{5x}{15{x}^{2}}$ .

This expression can be reduced and below we will see two ways to approach reducing algebraic fractions.

$\frac{5x}{15{x}^{2}}$ = $\frac{5x}{5x}\cdot \frac{1}{3x}$

These are equal because when multiplying rational exponents you multiply the numerators together and then multiply the denominators together.

It is useful to separate 5x and 15x2 in this fashion because a number divided by itself equals one.

$\frac{5x}{15{x}^{2}}$ = $\frac{5x}{5x}\cdot \frac{1}{3x}$

$\frac{5x}{15{x}^{2}}$ = $1\cdot \frac{1}{3x}$

So this would be: $\frac{5x}{15{x}^{2}}$ = $\frac{1}{3x}$.

To reduce you find what factors the numerator and denominator share and recognize that those shared factors are being divided by themselves, resulting in the number one.

The greatest common factor is what gets divided out of both the numerator and denominator. Another way to see this is below:

$\frac{5x÷5x}{15{x}^{2}÷5x}=\frac{\frac{5x}{5x}}{\frac{15{x}^{2}}{5x}}=\frac{1}{3x}$

This method is less clear to see, but the math is the same.

Regardless of the method, the key piece of information required to reduce is the greatest common factor. The greatest common factor of two expressions is the largest expression that divides into the expressions in question.

For example: has a greatest common factor of 3xy, because 3xy is the largest thing that divides into all three terms.

Let’s look at another example and use a table for factoring.

15a4b3, 20a7b, 30a7b2

 15a4b3 20a7b 30a7b2 3$•$5$•$a$•$a$•$a$•$a$•$b$•$b$•$b 2$•$2$•$5$•$a$•$a$•$a$•$a$•$a$•$a$•$a$•$b 2$•$3$•$5$•$a$•$a$•$a$•$a$•$a$•$a$•$a$•$b$•$b

These are all of the factors of each of these expressions. To find the GCF we can make a list of the repeated factors, factors that are in common between all three expressions.

5$•$ a$•$a$•$a$•$a$•$b

If we divide this GCF out of each term, we would be left with:

 15a4b3 3$•$5$•$a$•$a$•$a$•$a$•$b$•$b$•$b = 3b2

 20a7b 2$•$2$•$5$•$a$•$a$•$a$•$a$•$a$•$a$•$a$•$b = 4a3

 30a7b2 2$•$3$•$5$•$a$•$a$•$a$•$a$•$a$•$a$•$a$•$b$•$b = 6a3b

Where this will come into play is with reducing something like:

$\frac{15{a}^{4}{b}^{3}-20{a}^{7}b}{30{a}^{7}{b}^{2}}$

By dividing out the GCF of all three terms we are left with:

$\frac{15{a}^{4}{b}^{3}-20{a}^{7}b}{30{a}^{7}{b}^{2}}=\frac{3{b}^{2}-4{a}^{3}}{6{a}^{3}b}$

Reducing:

To reduce an algebraic fraction all terms must have a common factor. Terms can be separated by the fraction bar or by addition or subtraction. The expression below has three terms, two in the numerator and one in the denominator. In order to reduce, all terms must share a common factor. What students will often do when reducing this expression is reduce the a’s, just leaving the expressions of $–$b. Sometimes, they will realize that a divides into itself one time, so they will write
1
$–$ b.

If we assign some relatively prime numbers for a and b, and evaluate each of these figures we will see that they are not all equal. If the reducing was correct, each expression would be equal.

Let a = 5, b = 3

Only Figure 1 is correct. The others are incorrect because in order to reduce all terms must have a common factor. Here is why.

The order of operations governs the process in which we perform mathematical calculations. The fraction bar groups together the terms in the numerator, even though there are not any parenthesis. Operations grouped together must be carried out before any other operations. Division is reducing, which takes places after the group’s operations are completed.

So why can we divide (reduce) before carrying out the group’s operations? Consider the following example for an idea of why this is before reading the why this works.

$\frac{15x+5}{10x}$

The GCF of all the terms is five. This expression could be written as it is below.

$\frac{15x+5}{10x}=\frac{5\left(3x+1\right)}{5\cdot 2x}$

And we could write that as follows:

$\frac{5\left(3x+1\right)}{5\cdot 2x}=\frac{5}{5}\cdot \frac{3x+1}{2x}$

And five divided by five is one. The product of one and anything is, well, that anything. Multiplying by one does not change the value (that is why one is called Identity).

$\frac{15x+5}{10x}=\frac{5\left(3x+1\right)}{5\cdot 2x}=\frac{3x+1}{2x}$

The reason we can reduce before completing the operations in the group (numerator in this case), is because of the nature of multiplication and division being interchangeable. For example:

$5\cdot 3÷5=5÷5\cdot 3$

You may object here because in a previous section we showed how the order of division cannot be changed without changing the value. For example:

$\begin{array}{l}8÷4=2\\ 4÷8=\frac{1}{2}\end{array}$

$\frac{15x+5}{10x}=\frac{5}{5}\cdot \frac{\left(3x+1\right)}{2x}=\frac{3x+1}{2x}$

Without going into too much detail, division is multiplication by the reciprocal, and there is multiplication by the same factor taking place in both numerator and denominator. So when reducing, you are simply dividing out that common factor before multiplying it, which is mathematically sound.

Regardless, it must be understood that to reduce an algebraic expression each term must contain a common factor. In the expression remaining from the example above, two of the terms contain a factor of x, but not the third. To reduce (divide), before adding the numerator together, would be in violation of the order of operations.

Practice Problems:

Reduce the following:

1.     $\frac{32{x}^{2}{y}^{4}z}{14{x}^{5}y}$

2.     $\frac{3a}{9{a}^{2}}$

3.     $\frac{5{x}^{m}{y}^{3}}{15{x}^{m}y}$

4.     $\frac{27{a}^{2}b+3{a}^{2}b}{99{a}^{5}{b}^{3}}$

5.     $\frac{7xy-2}{4{y}^{2}z}$

## Wrongful Punishment was the Best Thing

When I was in 1st grade I suffered punished from a wrongful accusation, well, kind of.  And the punishment would land people in jail today.  In front of the entire 1st grade class I was spanked, but not spontaneously.  I was paraded to the front of the room and quite a spectacle was made of the ordeal.  Then I was sent to the principal’s office where I suffered a similar fate.  And at home, once again.

And while it is true, I did knock all of the lunch boxes off of the stand where they were stored, and they did spill open and they did make a big mess, it was an honest accident, no intent involved.  And I was very willing to clean all of it up, all of it!  The opportunity was not provided.  Instead, I bent over, knees straight, palms placed flat on a chair at the front of the room while the teacher drew back her arm, equipped with a wooden paddle, and brought it forward squarely on my six-year-old hind end, with all of my peers in audience.  Thrice over it happened.

It was spring and I had been spanked at school so frequently that my parents made a bargain for me, a bribe really.  If I could manage to not get spanked for an entire week then I could go out for ice cream on Friday.  And, being that it was spring, the local ice cream shop had my favorite ice cream in the whole world … peppermint ice cream!

Somehow I mustered the will-power to keep the tornado of energy that sprung forth from my body under wraps.  I sat properly, as instructed by the teacher.  And the teacher’s name, Mrs. Fortenberry, was pronounced clearly and accurately, all week.  I raised my hand at appropriate times, passed papers forward neatly and returned from recess with punctuality.  There was no cutting in line at lunch, no teasing other kids during class and the shenanigans that transpired when the teacher would turn around ceased to occur.

The staying power of six-year-olds is questionable, but I held this superb behavior through all of Monday, Tuesday, Wednesday, Thursday, and half of Friday.  Upon returning from lunch on Friday I went to place my lunch box on the stand and it slipped.  I had not been goofing off, running around, just an honest accident.  But it was as if I threw a basketball at them, they toppled like dominos … to this day I marvel at the physics involved in knocking over all of those lunch boxes in such a fashion.  Entropy is sometimes spontaneous in the presence of six year old boys!

My protests of innocence might as well have been mute.

That evening my family went out for ice cream.  I’m sure there was plenty of the delicious peppermint ice cream eaten.  I wouldn’t know, though, because I was left home by myself.

And while this may sound tragic, I realized something from this experience. I had my first epiphany.  I realized that nobody believed me with just cause.  They should not have believed me, my reputation was well earned.

From this unjust punishment I realized the power of reputation.  I realized that if I wanted to be believed in uncertain circumstances, when light spilled over me in a questionable fashion, that I needed to have a reputation of being honest.  The only way to establish such a reputation was through being honest.

My behavior did improve after this.  Perhaps not getting the ice cream was the best thing that ever happened to me.

Sometimes the best lessons are hard earned.