\documentclass[12pt]{book} \usepackage{amsmath} \usepackage{latexsym} \newcounter{example} \newtheorem{thm}{Theorem}[section] \newtheorem{rem}[thm]{Remark} \newtheorem{exa}[example]{Example} \newtheorem{defn}[thm]{Definition} \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. \setlength{\unitlength}{.6cm} \begin{picture}(20,10) \put(8,2){\line(0,1){6}} \put(8,2){\line(1,0){6}} \put(8,8){\line(1,-1){6}} \put(6.5,4.3){{\Large 1}} \put(11.5,5){{\Large $\sqrt{2}$}} \put(11,.3){{\Large 1}} \end{picture} 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. \setlength{\unitlength}{.6cm} \begin{picture}(22,4)(-11,-2) \linethickness{0.3mm} \put(0,0){\vector(1,0){11}} \put(0,0){\vector(-1,0){11}} \linethickness{.2mm} \multiput(-10,-0.2)(1,0){21}{\line(0,1){0.4}} \linethickness{.3mm} \multiput(-10,-0.3)(5,0){5}{\line(0,1){0.6}} \linethickness{.4mm} \put(0,-0.4){\line(0,1){0.8}} \put(-10.7,-1){\small $-10$} \put(-9.4,-1){\small $-9$} \put(-8.4,-1){\small $-8$} \put(-7.4,-1){\small $-7$} \put(-6.4,-1){\small $-6$} \put(-5.4,-1){\small $-5$} \put(-4.4,-1){\small $-4$} \put(-3.4,-1){\small $-3$} \put(-2.4,-1){\small $-2$} \put(-1.4,-1){\small $-1$} \put(-0.1,-1){\small $0$} \put(.9,-1){\small $1$} \put(1.9,-1){\small $2$} \put(2.9,-1){\small $3$} \put(3.9,-1){\small $4$} \put(4.9,-1){\small $5$} \put(5.9,-1){\small $6$} \put(6.9,-1){\small $7$} \put(7.9,-1){\small $8$} \put(8.9,-1){\small $9$} \put(9.9,-1){\small $10$} \end{picture} 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. \setlength{\unitlength}{.6cm} \begin{picture}(22,4)(-11,-2) \linethickness{0.3mm} \put(0,0){\vector(1,0){11}} \put(0,0){\vector(-1,0){11}} \linethickness{.2mm} \multiput(-10,-0.2)(1,0){21}{\line(0,1){0.4}} \linethickness{.3mm} \multiput(-10,-0.3)(5,0){5}{\line(0,1){0.6}} \linethickness{.4mm} \put(0,-0.4){\line(0,1){0.8}} \put(-0.1,-1){\small $0$} \put(-5,0){\circle*{.3}} \put(2,0){\circle*{.3}} \put(3.6,0){\circle*{.3}} \put(7.3,0){\circle*{.3}} \put(7,.5){$7.3$} \put(3.2,.5){$\sqrt{13}$} \put(2,.5){$2$} \put(-5,.5){$-5$} \put(11,-2){$\Box$} \end{picture} 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. \setlength{\unitlength}{.6cm} \begin{picture}(22,4)(-11,-2) \linethickness{0.3mm} \put(0,0){\vector(1,0){11}} \put(0,0){\vector(-1,0){11}} \linethickness{.2mm} \multiput(-10,-0.2)(1,0){21}{\line(0,1){0.4}} \linethickness{.3mm} \multiput(-10,-0.3)(5,0){5}{\line(0,1){0.6}} \linethickness{.4mm} \put(0,-0.4){\line(0,1){0.8}} \put(-0.1,-1){\small $0$} \put(3,0){\circle*{1.5}} \linethickness{2mm} \put(3.0,0){\vector(1,0){9}} \put(11,-2){$\Box$} \end{picture} \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. \setlength{\unitlength}{.6cm} \begin{picture}(22,4)(-11,-2) \linethickness{0.3mm} \put(0,0){\vector(1,0){11}} \put(0,0){\vector(-1,0){11}} \linethickness{.2mm} \multiput(-10,-0.2)(1,0){21}{\line(0,1){0.4}} \linethickness{.3mm} \multiput(-10,-0.3)(5,0){5}{\line(0,1){0.6}} \linethickness{.4mm} \put(0,-0.4){\line(0,1){0.8}} \put(-0.1,-1){\small $0$} \put(3,0){\circle{1}} \linethickness{2mm} \put(3,0){\vector(1,0){9}} \put(11,-2){$\Box$} \end{picture} 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$ \setlength{\unitlength}{.3cm} \begin{picture}(22,4)(-11,-2) \linethickness{0.3mm} \put(0,0){\vector(1,0){11}} \put(0,0){\vector(-1,0){11}} \linethickness{.2mm} \multiput(-10,-0.2)(1,0){21}{\line(0,1){0.4}} \linethickness{.3mm} \multiput(-10,-0.3)(5,0){5}{\line(0,1){0.6}} \linethickness{.4mm} \put(0,-0.4){\line(0,1){0.8}} \put(-0.1,-1.5){\small $0$} \put(-2,0){\circle*{1}} \linethickness{1mm} \put(-2,0){\vector(-1,0){9}} \end{picture} $ x>5 \qquad$ \setlength{\unitlength}{.3cm} \begin{picture}(22,4)(-11,-2) \linethickness{0.3mm} \put(0,0){\vector(1,0){11}} \put(0,0){\vector(-1,0){11}} \linethickness{.2mm} \multiput(-10,-0.2)(1,0){21}{\line(0,1){0.4}} \linethickness{.3mm} \multiput(-10,-0.3)(5,0){5}{\line(0,1){0.6}} \linethickness{.4mm} \put(0,-0.4){\line(0,1){0.8}} \put(-0.1,-1.5){\small $0$} \put(5,0){\circle{1}} \linethickness{1mm} \put(5,0){\vector(1,0){6}} \end{picture} 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. \setlength{\unitlength}{.6cm} \begin{picture}(22,4)(-11,-2) \linethickness{0.3mm} \put(0,0){\vector(1,0){11}} \put(0,0){\vector(-1,0){11}} \linethickness{.2mm} \multiput(-10,-0.2)(1,0){21}{\line(0,1){0.4}} \linethickness{.3mm} \multiput(-10,-0.3)(5,0){5}{\line(0,1){0.6}} \linethickness{.4mm} \put(0,-0.4){\line(0,1){0.8}} \put(-0.1,-1){\small $0$} \put(5,0){\circle{.6}} \put(-2,0){\circle*{.6}} \linethickness{1mm} \put(5,0){\vector(1,0){6}} \put(-2,0){\vector(-1,0){9}} \put(11,-2){$\Box$} \end{picture} 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 \setlength{\unitlength}{.6cm} \begin{picture}(22,4)(-11,-2) \linethickness{0.3mm} \put(0,0){\vector(1,0){11}} \put(0,0){\vector(-1,0){11}} \linethickness{.2mm} \multiput(-10,-0.2)(1,0){21}{\line(0,1){0.4}} \linethickness{.3mm} \multiput(-10,-0.3)(5,0){5}{\line(0,1){0.6}} \linethickness{.4mm} \put(0,-0.4){\line(0,1){0.8}} \put(-0.1,-1){\small $0$} \put(1,0){\circle*{1}} \linethickness{2mm} \put(1.5,0){\vector(1,0){9}} \end{picture} \item \setlength{\unitlength}{.6cm} \begin{picture}(22,4)(-11,-2) \linethickness{0.3mm} \put(0,0){\vector(1,0){11}} \put(0,0){\vector(-1,0){11}} \linethickness{.2mm} \multiput(-10,-0.2)(1,0){21}{\line(0,1){0.4}} \linethickness{.3mm} \multiput(-10,-0.3)(5,0){5}{\line(0,1){0.6}} \linethickness{.4mm} \put(0,-0.4){\line(0,1){0.8}} \put(-0.1,-1){\small $0$} \put(4,0){\circle*{1}} \linethickness{2mm} \put(4.5,0){\vector(1,0){9}} \end{picture} \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}. \setlength{\unitlength}{.8cm} \begin{picture}(14,14)(-7,-7) \linethickness{0.3mm} \put(0,0){\vector(1,0){7}} \put(0,0){\vector(0,-1){7}} \put(0,0){\vector(0,1){7}} \put(0,0){\vector(-1,0){7}} \linethickness{0.1mm} \multiput(-6,-6)(1,0){13}{\line(0,1){12}} \multiput(-6,-6)(0,1){13}{\line(1,0){12}} \linethickness{.2mm} \multiput(-6,-0.2)(1,0){13}{\line(0,1){0.4}} \multiput(-.2,-6)(0,1){13}{\line(1,0){0.4}} \put(5,3){\circle*{.4}} \put(0,.5){\vector(1,0){5}} \put(1,.6){{\Large Over five units}} \put(5.5,0){\vector(0,1){3}} \put(5.6,1.5){{\Large Up three units}} \put(5.3,5){{\Huge I}} \put(-5.7,5){{\Huge II}} \put(-5.5,-5.5){{\Huge III}} \put(4.5,-5.5){{\Huge IV}} \end{picture} 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} \setlength{\unitlength}{.4cm} \begin{picture}(16,16)(-8,-8) \linethickness{0.3mm} \put(0,0){\vector(1,0){8}} \put(-10, 8){{\bf SOLUTION:}} \put(0,0){\vector(0,-1){8}} \put(0,0){\vector(0,1){8}} \put(0,0){\vector(-1,0){8}} \linethickness{0.1mm} \multiput(-7,-7)(1,0){15}{\line(0,1){14}} \multiput(-7,-7)(0,1){15}{\line(1,0){14}} \linethickness{.2mm} \multiput(-7,-0.2)(1,0){15}{\line(0,1){0.4}} \multiput(-.2,-7)(0,1){15}{\line(1,0){0.4}} \put(6,2){\circle*{.3}} \put(-3,-4){\circle*{.3}} \put(-2.5,4){\circle*{.3}} \put(6.2,2.3){$(6,2)$} \put(-2.8,-3.7){$(-3,-4)$} \put(-2.7,4.3){$(-2.5,4)$} \end{picture} $\Box$ \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 $(2,7)$. \setlength{\unitlength}{.3cm} \begin{picture}(16,16)(-8,-8) \linethickness{0.3mm} \put(0,0){\vector(1,0){8}} \put(0,0){\vector(0,-1){8}} \put(0,0){\vector(0,1){8}} \put(0,0){\vector(-1,0){8}} \linethickness{0.1mm} \multiput(-7,-7)(1,0){15}{\line(0,1){14}} \multiput(-7,-7)(0,1){15}{\line(1,0){14}} \linethickness{.2mm} \multiput(-7,-0.2)(1,0){15}{\line(0,1){0.4}} \multiput(-.2,-7)(0,1){15}{\line(1,0){0.4}} \put(2,0){\circle*{.4}} \put(2,2){\circle*{.4}} \put(2,-3){\circle*{.4}} \put(2,7){\circle*{.4}} \put(2,0){$(2,0)$} \put(2,2){$(2,2)$} \put(2,-3){$(2,-3)$} \put(2,7){$(2,7)$} \linethickness{1mm} \put(2,0){\vector(0,1){8}} \put(2,0){\vector(0,-1){8}} \end{picture} 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. \setlength{\unitlength}{.3cm} \begin{picture}(16,16)(-8,-8) \linethickness{0.3mm} \put(0,0){\vector(1,0){8}} \put(0,0){\vector(0,-1){8}} \put(0,0){\vector(0,1){8}} \put(0,0){\vector(-1,0){8}} \linethickness{0.1mm} \multiput(-7,-7)(1,0){15}{\line(0,1){14}} \multiput(-7,-7)(0,1){15}{\line(1,0){14}} \linethickness{.2mm} \multiput(-7,-0.2)(1,0){15}{\line(0,1){0.4}} \multiput(-.2,-7)(0,1){15}{\line(1,0){0.4}} \multiput(0,0)(.25,0){33}{\line(0,-1){8.5}} \end{picture} 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. \setlength{\unitlength}{.4cm} \begin{picture}(14,14)(-7,-7) \linethickness{0.3mm} \put(0,0){\vector(1,0){8}} \put(0,0){\vector(0,-1){7}} \put(0,0){\vector(0,1){7}} \put(0,0){\vector(-1,0){7}} \linethickness{.2mm} \multiput(-6,-0.2)(1,0){13}{\line(0,1){0.4}} \multiput(-.2,-6)(0,1){13}{\line(1,0){0.4}} \put(-2.3,-1.4){\circle*{.3}} \put(5.7,-1.4){\circle*{.3}} \put(5.7,4.6){\circle*{.3}} \put(-3,-2.5){$A(x_1,y_1)$} \put(5.9,-1.4){$B(x_2,y_1)$} \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. \setlength{\unitlength}{.4cm} \begin{picture}(14,14)(-7,-7) \linethickness{0.3mm} \put(0,0){\vector(1,0){8}} \put(0,0){\vector(0,-1){7}} \put(0,0){\vector(0,1){7}} \put(0,0){\vector(-1,0){7}} \linethickness{.2mm} \multiput(-6,-0.2)(1,0){13}{\line(0,1){0.4}} \multiput(-.2,-6)(0,1){13}{\line(1,0){0.4}} \put(-2.3,-1.4){\circle*{.3}} \put(5.7,-1.4){\circle*{.3}} \put(5.7,4.6){\circle*{.3}} \put(-3,-2.5){$A$} \put(5.9,-1.4){$B$} \put(5.7,5){$C$} \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}} \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}