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\fancyfoot[CO,CE]{\it Beginning Algebra Tutorial by A. A. Tovar, Ph. D., Created Jan. 2009, Amended Feb. 2009.}
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\begin{document}
\title {Beginning Algebra Tutorial}
\author{Anthony A. Tovar, Ph. D.
\\Eastern Oregon University
\\1 University Blvd.
\\La Grande, Oregon, 97850 }
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\maketitle
\dominitoc
\faketableofcontents
\chapter{The Ideas of Algebra}
\minitoc
\section{Introduction}
The goal of this tutorial is to show how algebra is merely a formal
mathematical notation for what most people, even children, already know.
\section{The Fundamental Rule of Algebra}
Imagine two people, call them Person A and Person B, have the same amount of money.\\ If I take a dollar
from each,\\
\hspace{2 cm} {\it they still have the same amount of money}.\\
If I double each of their cash amounts,\\
\hspace{2 cm} {\it they still have the same amount of money}.\\
If I steal 3 dollars from each of them,\\
\hspace{2 cm} {\it they still have the same amount of money}.\\
Regardless of what I do to them, as long as I do it to {\bf both} of them,\\
\hspace{2 cm} {\it they still have the same amount of money}!\\
This is not difficult to understand, even kids know this.
Let's do a very simple example:
\begin{EX}
A basketball team has a 3-point specialist who only shoots 3-pointers.
If he scored 12 points, how many 3-pointers did he make.
\end{EX}
\begin{SOL}
I've asked kids this type of question before, and they always answer
correctly: ``4 3-pointers". Then I've asked them, ``how did you do it"?
and they respond with something like ``easy, divide 12 by 3". ``How did you
know how that you should divide?" ``I don't know." What they're doing
is using algebra, even if they don't know that's what they're doing.
From an algebra point of view, i.e. a written point
of view, the problem is
\begin{equation}
3 \times (some\; number\; of \; baskets) = 12
\end{equation}
In algebra, we usually call an unknown quantity a ``variable". Often we
signify this variable as ``x". So, the equation is
\begin{equation}
3x = 12
\end{equation}
To find $x$, we need to isolate it. We do this by dividing both
sides of the equation by 3, which results in
\begin{equation}
x = 4
\end{equation}
Of course, we can only divide each side by 3 because of what I call
the Fundamental Rule of Algebra. (Note: There is a ``Fundamental Theorem
of Algebra" which is very different).
\end{SOL}
\begin{EX}
Jack wants to complete a 12 mile bike ride. He crashes after 8 miles.
He gets back on his bike to start riding. How many miles does he have
to go?
\end{EX}
\begin{SOL}
In equation form, it looks like
\begin{equation}
8 + (some \;number\;of\;miles) = 12
\end{equation}
or
\begin{equation}
8 + x= 12
\end{equation}
Subtracting 8 from both sides yields
\begin{equation}
x= 4
\end{equation}
\end{SOL}
From the Fundamental Rule of Algebra, we can add, subtract, multiply, or
divide both sides by any number\footnote{Actually, adding, subtracting,
multiplying, or dividing both sides by zero is not useful.}
Let's do a composite example with both addition and multiplication:
\begin{EX}
Solve the equation
\begin{equation}
\underbrace{3x+5}_{Person A} = \underbrace{14}_{Person B}
\end{equation}
\end{EX}
\begin{SOL}
Think of this as two people who have the same amount of money.
Person A has $3x+5$ dollars and Person B has 14 dollars. To solve
for the variable $x$, we need to ``isolate" it one side of the equation.
The simplest way to do this is to {\it steal} 5 dollars from each
person! If we do that, then the equation becomes
\begin{equation}
3x = 9
\end{equation}
Now, dividing both sides by 3 yields the desired result
\begin{equation}
x = 3
\end{equation}
\end{SOL}
\begin{quote}
{\bf
At this level, all algebra is is formally writing down a set of rules that we
already know!}
\end{quote}
\noindent
Let's look at some more rules of algebra:
\section{More Laws of Algebra}
\subsection{Commutative Law of Addition}
Tom has no money. If you give him 5 dollars and then give him 10 dollars,
then he has the same amount of money as if you would have given him 10 dollars
and then given him 5 dollars. Duuuuhhhhhhhhh!
Well, we actually have a name for this absolutely obvious result, and it's
called the ``Commutative Law of Addition". Mathematically, we write it
like this
\begin{equation}
5 + 10 = 10 + 5
\end{equation}
or more generally,
\begin{equation}
a + b = b + a
\end{equation}
\noindent
Maybe we shouldn't give this law a name because it's so obvious, or perhaps call
it the "Duh" Law, but in mathematics we choose to give it an intimidating name.
For some this makes mathematics in general intimidating, smug, or incomprehensible.
Though some naughty mathematicians enjoy you feeling this way, the real reason
the name is used is because it is fairly precise.
\subsection{Commutative Law of Multiplication}
This one is a little less obvious, unless you think a little
bit about what multiplication really is. It says that if you
have 2 dollars and tripled your money, you would have the same
if you would have 3 dollars and doubled your money. Mathematically,
\begin{equation}
2 (3) = 3(2)
\end{equation}
or more generally,
\begin{equation}
a (b) = b (a)
\end{equation}
To make this result sensible, let's consider multiplication more
closely.
To do that, think about sets. If I doubled 3 dollars, then
I'd have 2 sets of 3 dollars, as shown in the figure:
\begin{center}
\includegraphics[scale=0.5]{graphics/mult.eps}
\end{center}
Tripling 2 dollars is having 3 sets of 2 dollars, as in the second
part of the figure. Clearly, we have the same number of dollars
either way.
If we wanted to count our pennies, we could put them into piles (or sets).
If we had 6 piles of 8 pennies, we'd have 6(8) or 48 pennies.
Indeed, this is why multiplication was created.
\subsection{Associative Laws of Addition and Multiplication}
Another obvious one. If you have three pockets with money in them, it
wouldn't matter which order you counted them in.
You could count the money in your 2 front pockets, add the total and
then add it to the amount in your back pocket. Or, you could add the
money in your back pocket to the money in one of your front pockets,
then add that amount to your other front pocket. Either way, you
have the same amount of money!
Mathematically,
\begin{equation}
(a + b) + c = a + (b+c)
\end{equation}
Similarly, the Associative Law of Multiplication is
\begin{equation}
(a b) c = a (bc)
\end{equation}
\subsection{Laws of Algebra Example}
\begin{EX}
Solve
\begin{equation}
3x + 2 + 5x + 9 = 3x +4 -x -2
\end{equation}
\end{EX}
\begin{SOL}
First, combine like terms using the laws above
\begin{equation}
8x + 11 = 2x +2
\end{equation}
subtracting $2x$ from each side,
\begin{equation}
6x + 11 = 2
\end{equation}
subtracting $11$ from each side,
\begin{equation}
6x = -9
\end{equation}
dividing $6$ from each side,
\begin{equation}
x = -9/6
\end{equation}
Reducing the fraction (dividing top and bottom by 3) yields
\begin{equation}
x = -3/2
\end{equation}
\end{SOL}
\section{The Foil Method}
\begin{EX}
Suppose you have two pockets, 4 Dollars in one and 5 Dollars in the other.
Mystery Man A says he will triple the {\bf total} number of coins you have.
Mystery Man B says that he will triple the coins in the first pocket and then
triple the coins you have in the second. You can only choose one.
Which one should you choose?
\end{EX}
\begin{SOL}
After a little thought, I'm sure you will agree that it does not
matter, the total amount will be the same in either case. Mathematically,
we would write this result as
\begin{equation}
3(4+5) = 3(4) + 3(5)
\end{equation}
In general, we would write
\begin{equation}
a(b+c) = a(b) + a(c)
\end{equation}
\end{SOL}
We could actually generalize the last result even more:
\begin{equation}
(a+b)(c+d) = ac + ad + bc + bd
\end{equation}
I actually came close to inventing this myself in the $4^{th}$ grade.
One day I had an odd thought - that $7 \times 7$ must be close to
$6 \times 8$. Sure enough,
\begin{equation}
7 \times 7 = 49 \;\;\;\;6 \times 8 =48\;\;\;\; \mbox{off by 1}
\end{equation}
So, I continued in this way:
\begin{equation}
6 \times 6 = 36 \;\;\;\;5 \times 7 =35\;\;\;\; \mbox{off by 1}
\end{equation}
\begin{equation}
5 \times 5 = 25 \;\;\;\;4 \times 6 =24\;\;\;\; \mbox{off by 1}
\end{equation}
\begin{equation}
4 \times 4 = 16 \;\;\;\;3 \times 5 =15\;\;\;\; \mbox{off by 1}
\end{equation}
\begin{equation}
3 \times 3 = 9 \;\;\;\;2 \times 4 =8\;\;\;\; \mbox{off by 1}
\end{equation}
\begin{equation}
2 \times 2 = 4 \;\;\;\;1 \times 3 =3\;\;\;\; \mbox{off by 1}
\end{equation}
\begin{equation}
1 \times 1 = 1 \;\;\;\;0 \times 2 =0\;\;\;\; \mbox{off by 1}
\end{equation}
I knew that this was some new form of fancy math, bigger than
what we were being taught. However, I couldn't proceed because
I had never dreamed of the idea of representing numbers with
letters. But, we can! In general, we would write
\begin{equation}
a \times a \;\;and\;\;(a-1) \times (a+1) \;\;\;\; \mbox{must be off by 1}
\end{equation}
so that
\begin{equation}
a^2 =(a-1)(a+1) +1
\end{equation}
or more commonly
\begin{equation}
(a-1)(a+1) =a^2 -1
\end{equation}
The reason math people don't talk about this much is because
with the rules
of algebra and the foil method, it is an absolutely obvious result:
\begin{equation}
(a-1)(a+1) =a^2 + a -a - 1 = a^2 -1
\end{equation}
However, they should think about it! It has important ramifications for
understanding geometry, physics, biology, chemistry, and the whole world
around us! Here's how:
\begin {equation}
a^2 > (a+1)(a-1)
\end{equation}
for any $a$. O.K., but you can also show that
\begin {equation}
a^2 > (a+2)(a-2)
\end{equation}
for any $a$, and that
\begin {equation}
a^2 > (a+3)(a-3)
\end{equation}
and furthermore
\begin {equation}
a^2 > (a+b)(a-b)
\end{equation}
for any $b$ and any $a$. What this means is that if we have a
piece of string (of fixed length) and want to make a rectangle with the
largest area, we should make it a square.
We can see this by drawing a picture:
\begin{center}
\includegraphics[scale=0.5]{graphics/square.eps}
\end{center}
The first is a square with four sides, each of length $a$. The
area of the square is $a\times a$. The second is a rectangle, two sides
of which are $a+1$ and two side of which are $a-1$. They both have
the same perimeter, but the area of the square is larger. Put another
way,
\begin{quote}
{\bf The Optimum Rectangle is the one with the most Symmetric Shape -
the Square.}
\end{quote}
What's the most optimum shape? If we had a fixed perimeter and
wanted to maximize the area and the shape didn't have to be a rectangle,
what shape would it be? Congratulations, you guessed it:
\begin{quote}
{\bf The Most Optimum Shape is the Most Symmetric Shape - the Circle.}
\end{quote}
I saw an interesting film in a high school biology class. Some scientists
wanted to determine, based on function, what the optimum shape would
be for a Red Blood Cell. They programmed a computer to determine this,
and they let it run for several weeks (would take just hours, now).
What did the computer find? The optimum shape for a Red Blood Cell is
the shape that Red Blood Cells actually have. They are Round!
\begin{quote}
{\bf Nature Likes Symmetry because Symmetry is Optimal.}
\end{quote}
The most common theory physicists have for how the universe works? ``Super-Symmetry."
These results are not surprising for someone who has thought about
the ramifications of basic algebra!
\section{Simultaneous Equations}
Consider the equation
\begin{equation}
ax+b = c
\end{equation}
This seems like a very hard equation, it looks like it has 4 unknowns. However,
it is a mathematics convention that letters at the beginning of the alphabet
(a,b,c) are ``constants" which are considered to be known, and letters at
the end of the alphabet (x, y, z) are considered to be unknowns which are called
``variables". So, a mathematician would assume that only $x$ in the equation is
unknown, and immediately solve for it:
\begin{equation}
x = \frac{c-b}{a}
\end{equation}
Of course, to do this one must assume that $a$ is not zero.
What if you have two variables? Well, you use the following rule:
\begin{quote}
{\bf
You must have the same number of INDEPENDENT equations as unknowns to obtain a solution.}
\end{quote}
\noindent
So, if you have two variables you must have two equations.
\begin{EX}
Solve
\begin{equation}
3x + 4y = 3 \;\; , \;\;
3x - 4y = 4
\end{equation}
The basic procedure, called the ``Substitution Method,"
to solve these equations would be to solve one of the equations
for $y$, and substitute it into the other equation, which would then only
have $x$'s in it. Then isolate the $x$, and you have the $x$ solution. You can
plug you $x$ solution into either of the original equations to get the $y$ solution.
Here, however, it's just easier to add the equations, so that
\begin{equation}
6x = 7 \;\; or \;\; x=7/6
\end{equation}
Plugging our $x$ solution back into the first equation results in
\begin{equation}
3(7/6) + 4y = 3 \;\; or \;\; 7/2 + 4y = 3 \;\; or \;\; 4y = -1/2 \;\;or \;\; y=-1/8
\end{equation}
\noindent
The reader should plug these $x$ and $y$ values into the each of the original
equations to prove that they are satisfied.
\end{EX}
\noindent
OK, you get the idea, but what is an $independent$ equation, or what are
$dependent$ equations? It's a slightly more advanced topic, and it's harder
to prove the answer, but I'll give you an example of two dependent equations:
\begin{equation}
2x + 3y = 3 \;\; , \;\;
2x + 3y = 4
\end{equation}
If you pick $x$ and $y$ so that $2x + 3y = 3$, then obviously the second equation
is not satisfied. There are {\bf no solutions} to this problem.
Here's another example:
\begin{equation}
2x + 3y = 3 \;\; , \;\;
4x + 6y = 6
\end{equation}
If you multiply the first equation by 2, it is identical to the second equation.
So, the second equation does not have any new information. There are {\bf infinite
solutions} to this set of equations.
\section{Algebra's Dirty Little Secret - Transcendental Equations}
I won't sugar coat it, {\bf You've Been Lied To!} However, the lies are
those of omission. It's not what you were told was wrong, it's that by not
telling you about transcendental equations, you got the wrong impression.
In particular,
\begin{quote}
{\bf
Not all algebra problems are solvable.}
\end{quote}
Generally, you are given only ones that are. {\bf People around the country (and the
world) have this strange idea that math is just a series of procedures, like
recipes - that it doesn't take any creativity. This is VERY WRONG}. In
algebra it's fairly simple to see whether a problem is solvable or not. It's
worse it calculus, because you can't just look at a problem and know whether
it can be solved. You can look at a problem, waste a lot of time not being able
to solve the problem, and conclude that the problem can't be done. (However, you
could still be wrong!) In a sense, then, Integral Calculus is as much an art as it is
a science. Back to algebra, consider the following equation:
\begin{equation}
\frac{2*x}{1+x*x} +3x*x*x(1-x) = 7
\end{equation}
There is no procedure to give the exact answer to that problem.\footnote{However,
one can come up with an approximate answer by making a series of guesses. Usually,
a computer is employed to make guesses. However, there are procedures to improve
upon a previous guess. So, the more guesses the computer makes, the closer to
the solution it gets. This is extremely important in physics and engineering
because the equation models a physical problem which they must have an answer.}
It should be noted that there is an alternate notation for the above type
of equation. In particular, $x *x*x$ is called $x^3$, so that the above equation
is
\begin{equation}
\frac{2x}{1+x^2} +3x^3(1-x) = 7
\end{equation}
Actually, $x^n$ is a type of ``function" called a polynomial....
\section{What's Next}
The next thing to do is to study different types of functions - get to know
them one at a time. In particular, you should get to know the elementary functions:
Sinusoidal functions ($sin(x)$, $cos(x)$, $tan(x)$), polynomial functions ($x^3 +2x^2$),
exponential functions ($2^x$, $3^x$). You should know that any combinations of
these elementary functions is an elementary function. You should know the graphs
of each of these functions. You should know what inverse functions are, when they
exist, and what they are for each of the elementary functions.
You should also learn basic methods in graphing. This certainly includes graphing
terminology such as horizontal, vertical, and other asymptotes, concavity,
monotonic, positive-definite, even/odd, bounded/unbounded, inflection points,
local and global maxima and minima. You should get a clear picture of what
$f(x+a)$ is, and how it differs from $f(x) + a$.
After you have attained some comfort level with algebra, then next subject
to tackle is basic geometry, following by a more advanced form of geometry
called trigonometry (which is just the geometry of triangles).
%%\chapter{Functions and their Graphs}
%%\section{Graphing Terminology}
%%Monotonic\\
%%Asymptotes (Horizontal and Vertical)\\
%%Positive-Definite\\
%%Concave Up\\
%%Inflection Point\\
%%Even/Odd\\
%%Bounded/Unbounded\\
%%Maxima/Minima\\
%%\section{Two Quick Graphing Rules}
%%f(x) + a\\
%%f(x-a)\\
%%\section{Common Functions}
%%\subsection{Linear Functions}
%%\subsection{Exponential Functions}
%%\subsection{Polynomials}
%%\subsection{Rational Functions}
%%\section{Inverse Functions}
%%\section{Solving Pseudo-Linear Equations}
\end{document}