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\begin{document}
\chapter{Beginning Algebra Review}
\section{The Real Number Line}
To begin your adventure into algebra try an
experiment. Open to a random page in one of the later sections of
this book and read a page. Most likely you did not understand some
of what you read. \footnote[1]{If you did understand it all you
might want to consider transferring into a different course!} One
enjoyable consequence of this experiment is that by the end of the
semester you will be able to look back and say with a hearty
chuckle, ``Was there really a time when the logarithmic function
seemed mysterious?'' The main goal of the experiment, however,
is to help us think about the process of learning math.
There are two difficulties in understanding the language of math,
and the same two problems that can make life difficult for you as
an algebra student probably made life difficult for your
instructor as a graduate student studying advanced math. First,
there is the use of mathematical notation which will naturally
seem foreign to the unfamiliar reader. On the page you looked at
did you see any symbols that you could not identify? Perhaps
there were symbols you had seen before but they were used in an
unfamiliar manner? Second, the subject of conversation in
mathematics is not simply facts but {\it concepts}. While a fact
might be understood in a few minutes, to understand a mathematical
concept well requires much practice and study.
Why do mathematicians use these concepts and notation? For the
{\it exact same reasons that you do!} While it might be more
difficult at first, in the long run it is the easiest and most
natural way to discuss the problems we wish to understand.
Given a strong enough
desire to understand these problems and enough time to work out
the solutions, you would probably come up with a system not
too different than the algebra in this book. ``Hold on!'' you object,
``I would never have come up with this in a hundred years!'' The
fact is, it took mathematicians {\it more} than a hundred
years to come up with all the ideas and notations you will learn
in this course. Because you don't have that long you will have to
sometimes take things on faith, jump in, and start trying out a concept
before you entirely see where it fits in the grand scheme of
things.
Consider the set of numbers. Without knowing it you have already
employed abstract notation and expanded your concept of numbers,
both of the techniques mentioned above that make math
powerful though sometimes difficult. When you were first
introduced to numbers as a child they were probably associated with
the idea of a number of objects: two blocks or three apples. A
handy way to keep track of numbers was to use your fingers.
Later you learned that numbers could
be abstractly represented using the curious symbols 1, 2, 3, 4,
... It required a little work making the transition from the
reliable fingers to this new notation, but in the long run it was
well worth it. How would you like to go back to the old finger
system to do your taxes?
Similarly, as you used numbers you were naturally forced to
introduce new concepts to describe what a number is. To begin with
there were the {\bf counting numbers} $\{1,2,3,...\}$. After
learning the counting numbers you soon realized that a number is
needed to represent ``nothing at all.'' This suggests that the set
should be expanded to include 0 to create the {\bf whole numbers}
$\{0,1,2,3,4...\}$. Next, you came to realize that as wonderful as
they are the whole numbers are not enough. They are fine for
counting students and cars, but when counting apples or dollars
you might have two and a half apples or three dollars and fifty
cents. Also, once you opened a bank account you saw that it could
be very useful to distinguish between withdrawals (negative
numbers) and deposits (positive numbers). This leads us to define
the {\bf integers}, the set consisting of all the counting numbers
and their negatives. To include fractions, we define the {\bf
rational numbers}, the set that consists of all numbers that can
be written $\frac{p}{q}$ where p and q are integers. Notice that p
or q could be negative, so the rational numbers include negative
fractions. Also, q could be the integer 1 so the rational numbers
include integers like $\frac{7}{1}$, though we would generally
write simply 7.
Do we need to expand the set of numbers any further? Are there any
numbers that are not rational? The Greeks thought of numbers
geometrically, as representing the length of the side of a square
or triangle, and asked the question can a triangle have a length
that ia not a rational number? The answer is yes! Numbers which
are not rational, called {\bf irrational numbers} do in fact
exist. The right triangle with two sides of length one
has a third side with length $\sqrt{2}$, which is turns out is not
rational.
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Using a calculator one finds that $\sqrt{2} \approx
1.41421356231$. This is merely an approximation; the decimal
representation of an irrational number goes on forever and never
falls into a repeating pattern.
The irrational and rational numbers together make up the {\bf real
numbers}, the set we will be using most often. Do we need to
expand the set of numbers any further? \footnote[2]{We will come back to this
question in Chapter 7. In the meantime be thinking.}
Formally, a {\bf set} is a collection of objects, called {\bf
elements}. The most common sets encountered in algebra will be
sets of numbers, either large sets like the set of all real
numbers or smaller sets consisting of a few numbers that are
solutions to an equation. More generally a set might consist
of students in a class or days of the week.
Sets with only a few elements can be described in {\bf roster
notation} which lists all the elements, as in the set $\{ 1,4,5
\}$ or the set $\{$Sunday, Monday, Tuesday, Wednesday, Thursday,
Friday, Saturday$\}$. How can we describe very large or even
infinite sets? One option is to list enough elements to establish
a pattern for determining the entire set. For example, the set of
integers can be written $\{ ...-5,-4,-3,-2,-1,0,1,2,3,4,5...\}$. A
second option is {\bf set builder} notation which gives the
conditions that an element must satisfy to qualify for membership
in the set, as in the following examples.
\begin{exa}
Use set builder notation to describe the set of rational numbers.
\end{exa}
\noindent {\bf SOLUTION:} The set of rational numbers is $$\bigg\{x
\quad | \quad x=\frac{p}{q} \quad where \quad p \quad and \quad q
\quad are \quad integers \quad and \quad q \neq 0 \bigg\}. \Box $$
\begin{exa}
Use set builder notation to describe the set of even numbers.
\end{exa}
\noindent {\bf SOLUTION:} The even numbers are given by the set
$$\{x \quad | \quad x=2k \quad where \quad k \quad is
\quad an \quad integer \}. \Box
$$
It will often be useful to visualize the real numbers as measuring
off distance along the {\bf real number line}, with the numbers
growing larger as we go from left to right.
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Notice that the number
line contains a continuous spectrum of numbers, not just the
integers.
\begin{exa}
Plot the points $2$, $-5$, $7.3$, and $\sqrt{13}$ on the
real number line.
\end{exa}
{\bf SOLUTION:} To graph numbers like $7.3$ it may be necessary
to approximate. Using a calculator one finds that
$\sqrt{13}$ is about equal to 3.6.
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In addition to graphing single points on the number line we may wish
to graph a large set of numbers, as in the examples below.
\begin{exa}
Graph the set $\{ (x,y) \quad | \quad x\geq 3 \}$.
\end{exa}
{\bf SOLUTION:} This set contains three and all the real numbers
larger than three, so we shade in $3$ and everything to the right.
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\begin{exa}
Graph the set $\{ (x,y) \quad | \quad x > 3 \}$.
\end{exa}
{\bf SOLUTION:} This example is almost identical to the
previous example, but to indicate that $3$ is {\bf not}
included in the set we use an open circle.
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In the next example there are two conditions on the items in the
set.
\begin{exa}
Graph the set $\{ (x,y) \quad | \quad x \leq -2 \quad or
\quad x>5 \}$.
\end{exa}
{\bf SOLUTION:} First consider what the graph would look like
for the two inequalities separately.
$ x \leq -2 \quad$
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$ x>5 \qquad$
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To be including a point only needs to be in one set {\it or}
the other, so the graph including all the points from {\it both}
sets.
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The distance a number $x$ is from $0$ on the real number line is
called the {\bf absolute value} of $x$, written $|x|$. Notice that
for a positive number taking the absolute value does not change the
number, while taking the absolute value of a negative number makes
it positive. In light of that we could use the following alternative
definition of absolute value.
\begin{displaymath}
|x| =\left\{ \begin{array}{ll}
x & \textrm{if $x \geq 0$}\\
-x & \textrm{if $x<0$}\\
\end{array} \right.
\end{displaymath}
\begin{exa}
Evaluate the following expressions involving absolute values.
\end{exa}
a) $|-10|=10$
b) $|3|=3$
c) $|-\frac{1}{3}|= \frac{1}{3}$
d) $-|-7|=-7$
e) $-|4|=-4$
\vspace{.5 in}
{\bf \LARGE Exercises 1.1}
\vspace{.25 in}
List the described set of numbers then graph the numbers on the real number line.
\begin{enumerate}
\item The whole numbers less than 5
\item The counting numbers less than 10
\item The integers between -5 and 3
\item The whole numbers between -5 and 7
\item The whole numbers smaller than $\frac{7}{2}$
\vspace{.25 in}
Write the interval corresponding to the given set of real numbers then graph the interval on the real number line.
\item The set of real numbers between 1 and 2
\item The set of real numbers between -3 and 5
\item The set of real numbers between -1 and 1 inclusive
\item The set of real numbers between 0 and 10 inclusive
\item The set of real numbers greater than 1 and less than or equal to 2
\item The set of real numbers greater than or equal to 4 and less then 7
\item The set of numbers greater than 1
\item The set of numbers less than 10
Evaluate the following absolute values.
\item $|-6|$
\item $|6|$
\item $|0|$
\item $|-\frac{4}{3}|$
\item $|-5.02|$
What number should be placed in the absolute value to make the statement true?
\item $|?|=4$
\item $|?|=0$
\item $|?|=-7$
Write the interval corresponding to the given graph of the real number line.
\item
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\item
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\end{enumerate}
\newpage
\section{Graphing in the Cartesian Plane}
\setcounter{example}{0}
In this section the Cartesian plane is introduced allowing algebra
to be translated into a two-dimensional picture. By understanding
the relation between an algebraic problem and its graph you will
have two ways approach most problems. In some cases, looking at
the graph will give the key insight to understanding what is going
on in a problem.
The {\bf Cartesian plane} consists of the {\bf x-axis} which is
identical to the real number line and a new vertical axis, the
{\bf y-axis}. The ordered pair {\bf (x,y)} is identified with the
point with is right x units and up y units. For example, (5,3) is
over 5 and up 3. The numbers $5$ and $3$ are the {\bf coordinates}.
If the coordinates are negative then the point lies down or to the
left. The central starting point $(0,0)$ is called the
{\bf origin}.
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Notice that the x and y axes divide the plane into four
quadrants, sometimes numbered I, II, III, and IV.
\begin{exa}
Graph the points $(6,2)$, $(-3, -4)$, and $(-2.5, 4)$ in the
Cartesian plane.
\end{exa}
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$\Box$
\newline
Instead of graphing a few isolated points it is usually more interesting
to graph a large set of points satisfying a given condition. As a starting
point, try a few points and consider whether or not the x and y values meet
the condition for being in the set. In Chapter Two we will refine our
approach to graphing.
\begin{exa}
Graph the set $\{ (x,y) \quad | \quad x=2 \}$.
\end{exa}
{\bf SOLUTION:} For a point to be in the set its x coordinate must be 2 but
the y coordinate can be anything. Hence, points in the set include $(2,0)$,
$(2,2)$, $(2,-3)$, and $(2,7)$.
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In fact, any point that is two to the right of the y-axis will have an
x value of 2, so the graph is a vertical line. $\Box$
The {\bf graph of an equation} is the graph containing all points whose x
and y values make the equation true. In the above example, then, we could have
simply said ``Graph the equation x=2.''
\begin{exa}
Graph the set $\{ (x,y) \quad | \quad x>0 \qquad and \qquad
y<0 \}$.
\end{exa}
{\bf SOLUTION:} The first requirement that $x>0$ implies
that only points to the {\it right} of the x-axis are in the set.
The requirement that $y<0$ implies that we want only those points
with negative y values, those points {\it below} the y axis.
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These are precisely the points that are in the fourth quadrant. $\Box$
The next problem we consider is determining the distance between
two given points in the Cartesian plane. Consider the points A, B,
and C below.
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\put(5.7,5){$C(x_2,y_2)$}
\linethickness{.3mm}
\put(-2.3,-1.4){\line(4,3){8}}
\put(-2.3,-1.4){\line(1,0){8}}
\put(5.7,-1.4){\line(0,1){6}}
\end{picture}
To allow us to answer the question for any pair of points the
coordinates of A and C are labeled $(x_1,y_1)$ and $(x_2,y_2)$
where $x_1,x_2,y_1$ and $y_2$ can be any real numbers. (Do
we need to label point C with $(x_3, y_3)$?)
It is easy to find the distance between A and B since they are
on the same horizontal line. Geometrically, we could measure off
the distance along the x-axis. Algebraically, the distance is
just the difference between the two x-coordinates, $x_2-x_1$.
Similarly, the distance between B and C is $y_2-y_1$. Finding the
length $d$ of the edge from A to C, however, is more difficult.
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\put(5.7,-1.4){\circle*{.3}}
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\put(-3,-2.5){$A$}
\put(5.9,-1.4){$B$}
\put(5.7,5){$C$}
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\put(-2.3,-1.4){\line(1,0){8}}
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\put(.5,-2.4){$x_2-x_1$}
\put(5.9, 1.5){$y_2-y_1$}
\put(1.7,2.5){$d$}
\end{picture}
If we can think of a way to find the length of the longest side using
the length of the other two then our goal will be achieved. Fortunately,
long ago the curious natives of the Isle of Pythagorea spent many hours
pondering this question and left us with their theorem.
\newline
\begin{tabular}{|c|}\hline
\\
{\bf The Pythagorean Theorem} \\
In any right triangle, the square of the longest side \\ is equal to
the sum of the squares of the other two sides.\\
\\
\hline \end{tabular}
\newline
\newline
Applying it to our triangle gives the equation
$$d^2=(x_2-x_1)^2+(y_2-y_1)^2.$$
Notice that we have virtually achieved our goal: once we are given the
coordinates for the x's and y's we will know the value of $d^2$. To
get the value of $d$ instead of $d^2$ we simply take the square root.
\newline
\begin{tabular}{|c|}\hline
\\
{\bf The Distance Formula} \\
\\The distance $d$ between the points $(x_1,y_1)$ and $(x_2,y_2)$
is given by \\ \\ $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$\\
\\
\hline \end{tabular}
\newline
\newline
\begin{exa}
Find the distance between (-2,-1) and (6,5).
\end{exa}
{\bf SOLUTION:} In this case $x_1=-2$, $y_1=-1$,
$x_2=6$, and $y_2=5$. Applying the distance formula
$$d=\sqrt{(6-(-2))^2+(5-(-1))^2}.$$
$$d=\sqrt{8^2+6^2}$$
$$d=\sqrt{64+100}$$
$$\qquad \qquad \qquad d=\sqrt{100}=10 \qquad \qquad \qquad \Box$$
Notice that it does not matter which point you choose for the first
point $(x_1,y_1)$. Since the distance from point $A$ to point $B$ is
the same as from $B$ to $A$ we can switch the order and get the
same distance. On the other hand, be careful not to switch an $x$
with a $y$ in the formula.
In the example above it happened that the distance was an integer value
so we had a perfect square. In general, if you pick two points at random
the distance will probably not be an integer value.
\begin{exa}
Find the distance between (3,2) and (14,8).
\end{exa}
{\bf SOLUTION:} $$d=\sqrt{(14-3)^2+(8-2)^2}.$$
$$d=\sqrt{11^2+6^2}$$
$$d=\sqrt{121+36}$$
$$\qquad \qquad \qquad d=\sqrt{157}\approx 12.52996 \qquad \qquad \qquad \Box$$
The wavy equals sign means ``is approximately equal to'' and is used to show
that the value has been rounded.
\vspace{.5 in}
{\bf \LARGE Exercises 1.2}
\vspace{.25 in}
Plot the following points in the xy-plane.
\begin{enumerate}
\item (3,6)
\item (6,3)
\item (0,0)
\item (-4,2)
\item (0, -3)
\item (-6,2)
Shade in the region of the xy-plane containing all the points in the sets described below.
\item $\{(x,y)|x=4\}$
\item $\{(x,y)|x \geq 4\}$
\item $\{(x,y)|x>4\}$
\item $\{(x,y)|y>4\}$
\item $\{(x,y)|x>y\}$
Find the distance between the pair of points.
\item (1,1) and (4,5)
\item (-4,2) and (8,7)
\item (0,0) and (-3, 7)
\item (-2,4) and (2, -3)
\end{enumerate}
\newpage
\section{Linear Equations in One Variable}
\setcounter{example}{0}
One of the main goals of algebra is to be able to solve equations.
To {\bf solve} an equation we must find the {\bf solution set},
that is, the set of all values that will make the equation true.
\begin{exa}
Find the solution sets for the follow equations.
\end{exa}
\noindent a) $x+4=9$
{\bf SOLUTION:} If x=5 then x+4=9 becomes the true equation 5+4=9,
hence the solution set is $\{5\}$.
\newline
\newline
b) $x+1=x+2$
{\bf SOLUTION:}There is no number that is the same whether it is
added to one or two, so in this case the solution set is the {\bf
empty set}, written as $\{ \, \, \}$ or $\emptyset$.
\newline
\newline
c) $x^2=4$
{\bf SOLUTION:}The value 2 is clearly in the solution set, since
$2^2=4$, and -2 is also a solution since $(-2)^2=4$, so the
solution set is $\{ -2,2 \}$.
\newline
\newline
d) $x+3=x+3$
{\bf SOLUTION:} Here {\it any real value} will make this equation
true, so the solution set is all real numbers. $\Box$
\newline
These examples demonstrate that the number of solutions might be
anywhere from zero to infinity. As we focus on different types of
equations we will see that we may be able to guess the number of
solutions based on the type of equation. In this section we will
look at linear equations, the most basic type of equation.
\begin{defn}
A {\bf linear equation} in one variable is an equation that can be
written in the form $$Ax+B=0$$ where A and B are real numbers and
$A\neq 0$.
\end{defn}
Linear equations are sometimes called {\bf first order} equations
because the variable is not squared or cubed but only taken to the
first power. It is important to notice what {\it doesn't} happen
in a linear equation.
\begin{exa}
Which of the following equations are linear equations in one
variable?
\end{exa}
a) $3x-12=0$
b) $10y-50=2y+30$
c) $2w+14$
d) $8y=80$
e) $2a+4b=120$
f) $\sqrt{x}+4=0$
g) $\frac{1}{x+1}+\frac{1}{x}=\frac{1}{12}$
h)$\frac{1}{3}x+\frac{2}{5}=13$
\newline
\newline
\noindent {\bf SOLUTION:} Equations a, b, d, and h are the linear
equations. Notice that c is not an equation since there is no
equal sign. Equation e has two variables, a and b. Equation f is
disqualified because it has a square root, and g is not linear
because there is an x in the denominator. Notice, however, that we
{\it can} have fractions in a linear equation, since $\frac{1}{3}$
and $\frac{2}{5}$ are real numbers. $\Box$
\newline
Notice that equation b and equation d have the exact same solution
set, namely $\{10 \}$. Two such equations are called {\bf
equivalent}. Probably it was not immediately clear that 10 was a
solution for equation b, but it is easier to see that it is a
solution to equation d. In general our strategy for solving
equations is to use the rules of algebra to change an equation
like d to an equivalent but simpler equation like b, or better yet
the even simpler (but still equivalent) equation $y=10$. The
following two rules provide valuable tools for simplifying linear
equations.
\newline
\newline
\begin{tabular}{|l|}\hline
\\
{\bf {\large The Addition and Multiplication Properties for
Equations}}
\\
\\For any real numbers A, B, and C
\\
$A=B$ is equivalent to $A+C=B+C$ \\
$A=B$ is equivalent to $A
\cdot C=B \cdot C$ as long as $C \neq 0$
\\ \hline \end{tabular}
\\
\\
Informally, these properties say that we can add the same number
to both sides of the equation or multiply both sides of the
equation by the same number. Notice that if C happens to be a
negative number then we are adding a negative number to both
sides, which is the same as subtracting. For that reason we could
state a similar ``Subtraction Property.'' Similarly, the
Multiplication Property implies we can also {\it divide} both
sides by any number except zero.
\newline
\begin{tabular}{|l|}\hline
\\
{\bf {\large Strategy for Solving Linear Equations}}\\
\\
1. Combine like terms. (You may have to combine \\ terms again later.)\\
\\
2. Use the Addition Property to gather the terms with the
variable x on \\ one side of the equation and the numbers on the
other side.\\
\\
3. Use the Multiplication Property to eliminate the coefficient
\\ in
front of x.\\
\\
4. Check your answer.\\ \hline \end{tabular}
\begin{exa}
Solve: $2x-19=6-5x+2x$
\end{exa}
\noindent {\bf SOLUTION:} $2x-19=6-3x \qquad Combine\; like\;
terms.$
$2x-19+3x=6-3x+3x \qquad \quad Apply \; the\; Addition\;
Property.$
$5x-19=6 \qquad \qquad \qquad \qquad Combine\; like\; terms.$
$5x-19+19=6+19 \qquad \qquad \quad Apply \; the\; Addition\;
Property,\; again.$
$5x=25 \qquad \qquad \qquad \qquad \qquad Combine\; like\; terms.$
$\frac{5x}{5}=\frac{25}{5} \qquad \qquad \qquad Apply \; the\;
Multiplication\; Property.$
$x=5 \qquad \qquad \qquad \qquad \qquad \qquad Divide.$
{\it Check: } $$ 2 \cdot (5) -19 = ? \; 6-5 \cdot (5)+2 \cdot
(5)$$
$$10 - 19 = ? \; 6-25+10$$
$$ -9 = ? \; -9 \qquad \qquad True! \qquad \Box$$
\newline
Notice that the addition property effectively says that if there
is a $3x$ subtracted from the right side of the equation, as in
Example 3 above it is equivalent to adding $3x$ to the left side.
In other words, when moving a term to the other side of the equation
it is necessary to change the sign.
In the next example we must first apply the distributive property.
\begin{exa}
Solve: $10(3t-1)=6t-4$
\end{exa}
\noindent {\bf SOLUTION:} $$30t-10=6t-4$$
$$30t-10-6t=-4$$
$$24t-10=-4$$
$$24t=-4+10$$
$$24t=6$$
$$t=\frac{6}{24}$$
$$\qquad \qquad \qquad \qquad t=\frac{1}{4} \qquad \qquad \qquad \Box$$
\newline
When solving a linear equation involving fractions, it is
generally easier if we first multiply each side of the equation by
the least common denominator (LCD) to eliminate the fractions.
When we multiply by the LCD we was making use of the
Multiplication Property.
\begin{exa}
Solve: $\frac{1}{6}z-\frac{1}{4}=\frac{2}{3}z-\frac{33}{12}$
\end{exa}
\noindent {\bf SOLUTION:} Here the least common denominator
is $12$. Remember to multiply {\it all} terms by the LCD.
$$12 \cdot \frac{1}{6}z-12\cdot \frac{1}{4}= 12 \cdot \frac{2}{3}z-
12 \cdot \frac{33}{12}$$
$$2z-3=8z-33$$
$$-3=8z-33-2z$$
$$-3=6z-33$$
$$-3+33=6z$$
$$30=6z$$
$$5=z$$
$$Check: \qquad \qquad \qquad \qquad$$ $$\frac{1}{6}\cdot 5-\frac{1}{4}\quad=? \quad \frac{2}{3}\cdot 5-\frac{33}{12}$$
$$\frac{5}{6}-\frac{1}{4}\quad=? \quad \frac{10}{3}-\frac{33}{12}$$
$$\frac{10}{12}-\frac{3}{12}\quad=? \quad \frac{40}{12}-\frac{33}{12}$$
$$\qquad \qquad \qquad \qquad \frac{7}{12}=\frac{7}{12} \qquad TRUE! \qquad \qquad \Box$$
\newline
When we solve problems involving interest rates or percentages we
will often work with linear equations with decimal coefficients.
It is often easier in this case to multiply both sides of the
equation by 100.
\begin{exa}
Solve: $.07x+.12(5-x)=.5$
\end{exa}
\noindent {\bf SOLUTION:} Multiplying by $100$ we get
$$7x+12(5-x)=50$$
$$7x+60-12x=50$$
$$-5x+60=50$$
$$-5x=-10$$
$$x=\frac{-10}{-5}$$
$$x=2$$
$$\qquad \qquad \qquad \qquad5=z \qquad \qquad \qquad \qquad \Box$$
\newline
For all of the examples so far the solution set has contained
exactly one value. Such equations are called {\bf conditional
equations} due to the fact that whether or not they are true
depends on the condition that x must equal that one value in the
solution set. While this is the most common case, it turns out
there are two other possibilities. It may happen that the equation
is actually a {\bf contradiction} so that there are no values
which will make it true. In the other extreme the equation may be
an {\bf indentity}, an equation that is {\it always} true
regardless of which real value x is assigned.
\begin{exa}
a) Solve: $3(2x-12)+4 = 6x-32 $
b) Solve: $3(2x-12)+4 = 6x-31 $
\end{exa}
\noindent {\bf SOLUTION:}We proceed as we did with the conditional equations
above. $$6x-36+4=6x-32$$
$$6x-32=6x-32$$
$$6x=6x$$
Since this is true regardless of what value we use for x, the equation is
an identity. The solution set is all real numbers.
For the second equation we get $$6x-32=6x-31$$
$$6x=6x-31+32$$
$$6x=6x+1$$
Subtracting $6x$ from both sides leaves us with $$0=1$$ which is clearly
never true. The solution set for a contradiction is the empty set. $\Box$
\vspace{.5 in}
{\bf \LARGE Exercises 1.3}
\vspace{.25 in}
For exercises 1 through 5, determine if the equation is a linear or nonlinear equation.
\begin{enumerate}
\item $x=1$
\item $x^2+9=x$
\item $2a+9=a-4$
\item $\sqrt{m+1}=9$
\item $5(x+1)+5(x-3)+9x=4$
Solve the linear equations below. Check your answer by substituting back into the original equation.
\item $3x+1=10$
\item $2x=6-4x$
\item $3(x+1)=9(x-4)+2$
\item $\frac{1}{2}x+\frac{1}{3}x=10$
\item $5.4-0.3x=2.1x$
\end{enumerate}
\end{document}
(
geocities.com/perryand)