ALGEBRA

ESSENTIALS

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ALGEBRA

ESSENTIALS

David A. Santos

Olgha B. Davis

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David A. Santos and Olgha B. Davis. Algebra Essentials.
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CONTENTS

Preface

Acknowledgments

Part I: First Impressions

Chapter 1: Why Study Algebra?

1.1

The Value in Studying Algebra

Supplementary Exercises

1.2

What is Elementary Algebra All About?

Supplementary Exercises

1.3

Puzzles

Supplementary Exercises

Part II: Arithmetic Review

Chapter 2: Arithmetic Operations

2.1

Symbolical Expression

Supplementary Exercises

2.2

The Natural Numbers

Supplementary Exercises

2.3

Fractions.

Supplementary Exercises

2.4

Operations with Fractions

Supplementary Exercises

2.5

The Integers

Supplementary Exercises

2.6

Rational, Irrational, and Real Numbers

Supplementary Exercises

Part III: Algebraic Operations

Chapter 3: Addition and Subtraction

3.1

Terms and Algebraic Expressions

Supplementary Exercises

3.2

More Suppression of Parentheses

Supplementary Exercises

Chapter 4: Multiplication

4.1

Laws of Exponents

Supplementary Exercises

4.2

Negative Exponents

Supplementary Exercises

4.3

Distributive Law

Supplementary Exercises

4.4

Square of a Sum

Supplementary Exercises

4.5

Difference of Squares

Supplementary Exercises

4.6

Cube of a Sum

Supplementary Exercises

4.7

Sum and Difference of Cubes

Supplementary Exercises

Chapter 5: Division

5.1

Term-by-Term Division

Supplementary Exercises

5.2

Long Division

Supplementary Exercises

5.3

Factoring I

Supplementary Exercises

5.4

Factoring II

Supplementary Exercises

5.5

Special Factorizations

Supplementary Exercises

5.6

Rational Expressions

Supplementary Exercises

Part IV: Equations

Chapter 6: Linear Equations in One-Variable Expressions

6.1

Simple Equations

Supplementary Exercises

6.2

Miscellaneous Linear Equations

Supplementary Exercises

6.3

Word Problems

Supplementary Exercises

Chapter 7: Quadratic Equations in One-Variable Expressions

7.1

Quadratic Equations

Supplementary Exercises

7.2

Complete the Square

Supplementary Exercises

7.3

Quadratic Formula

Supplementary Exercises

Part V: Inequalities

Chapter 8: Linear Inequalities

8.1

Intervals

8.2

One-Variable Linear Inequalities

Supplementary Exercises

8.3

Using One-Variable Linear Inequalities to Solve Word Problems

Supplementary Exercises

8.4

Compound Linear Inequalities

Supplementary Exercises

Appendix A: Review Exercises

Chapter 2 Review Exercises

Chapter 3 Review Exercises

Chapter 4 Review Exercises

Chapter 5 Review Exercises

Chapter 6 Review Exercises

Chapter 7 Review Exercises

Chapter 8 Review Exercises

Appendix B: Answer Keys for Chapter Supplementary Exercises

Chapter 1 Answers

Chapter 2 Answers

Chapter 3 Answers

Chapter 4 Answers

Chapter 5 Answers

Chapter 6 Answers

Chapter 7 Answers

Chapter 8 Answers

Appendix C: Answer Key for Odd-Numbered Review Exercises

Appendix D: Theorems, Corollaries, and Proofs

Appendix E: Keywords

Index

PREFACE

Welcome to Algebra Essentials! This book is designed for instructors to introduce algebra to students in approachable and meaningful ways. Algebra Essentials is intended to increase your knowledge and skills needed to succeed in collegiate mathematics courses. The eight chapters in this book present many examples and exercises—with complete solutions—that can be used to practice algebraic concepts. You will receive the most benefit from this book if you use all of the components included. For self-correcting exercises, figures from the text (including 4-color), theorem proofs, and additional information, check out the companion disc. Additional instructor supplements for use as a textbook are available from the publisher.

Exercises

This self-study text contains hundreds of exercises to supplement the numerous examples and aid in the drill and mastery of the concepts:

1) Supplementary Exercises (End of each section)

These are found at the end of each section of every chapter (e.g., after Sections 2.1, 2.2, 2.3, etc.)

Answers to ALL of these exercises may be found in Appendix B at the end of the book.

2) Review Exercises (End of the book)

Over 100 of these exercises (included to reinforce or review concepts) may be found in

Appendix A. Answers to the odd-numbered Review Exercises may be found in Appendix C.

3) Self-correcting Mini-Quizzes

The companion disc in the back of the book includes seven, multiple-choice, self-correcting mini-quizzes on arithmetic, equations, factoring, inequalities, like-terms, exponents, and word problems.

Companion files on the disc are also available by writing to the publisher at info@merclearning.com.

ACKNOWLEDGMENTS

Many people have contributed to this work in very significant ways. First, we would like to thank Margaret Hitczenko, John Majewicz, and José Mason for their numerous suggestions and ideas. We also would like to thank Iraj Kalantari, Lasse Skov, and Don Stalk for their editorial suggestions.

It is with great humility and gratitude that Olgha Davis thanks Tuere Bowles, Christine Grants, David Keim, and Roberta Mahatha, who were instrumental in providing support and insightful guidance throughout this project. And last, but far from least, Olgha Davis could not have completed this book without her two beautiful daughters, Dalya and Andreya Davis.

David A. Santos
Olgha B. Davis

PART 1

FIRST IMPRESSIONS

CHAPTER 1

WHY STUDY ALGEBRA?

This chapter is optional. However, it is necessary to answer two commonly asked questions about learning algebra: Why do adult learners need to learn algebra? And when will they use it in the real world? We attempt to answer these questions in the Value in Studying Algebra, What Is Elementary Algebra All About? and the Puzzles sections.

1.1

THE VALUE IN STUDYING ALGEBRA

Mathematicians like to boast about the multiple applications of their discipline to real-world problems, as commonly found in science, business, engineering, etc. For example, chemical compounds must obey certain geometric arrangements, which in turn specify how they behave. Financial firms use mathematical theory to explain long-term markets behavior. Biomedical engineering researchers use mathematics to model the way virus strands twist and better understand the behavior of viruses.

These applications are perhaps too complex for a new math learner to master and require many years of math studies beyond algebra to understand them. However, learning algebra is the beginning of a journey that gives you the tools and the necessary skills to solve more complex systems. You might still ask why a person who is not planning to become a scientist, business analyst, or biomedical engineer must learn algebra. It is important because:

Algbera is a gateway course, used by other disciplines as a hurdle for college admission.

Algebra is part of our cultural legacy, much like art and music, and its mastery is expected by all who want to be considered educated.

Algebra strengthens deductive reasoning.

Algebra provides a first example of an abstract system.

Despite the above reasons, what if you still claim that you cannot learn algebra? Is algebra not for you? Unlike foreign languages, algebra is a more universal language, much like music, and to fully appreciate its power you must be willing to learn it. Most teachers will claim that anyone can learn algebra. Most mathematicians and mathematics instructors will state that the major difficulties in learning algebra stem from previous difficulties with arithmetic. Therefore, if you are reasonably versed in arithmetic, you should not have any trouble with learning algebra. Keep in mind that everyone is capable of learning algebra, so take it one step at a time and learn the basic rules before moving on to the more advanced concepts.

Supplementary Exercises

Exercise 1.1.1 Comment on the following statement: There is no need to learn mathematics, because today all calculations can be carried out by computers.

Exercise 1.1.2 Comment on the following statement: I was never good at math. I will never pass this algebra class.

Exercise 1.1.3 Comment on the following statement: Only nutritionists should know about the basics of nutrition, since that is their profession. Only medical doctors should know about the basics of health, since that is their profession. Only mathematicians should know about algebra, since that is their profession.

1.2

WHAT IS ELEMENTARY ALGEBRA ALL ABOUT?

Elementary algebra generalizes arithmetic by treating quantities in the abstract. In other words, mathematical expressions use variables (symbols) instead of values (numbers) to represent an algebraic equation. For example, consider two arithmetic problems: 2 + 3 = 3 + 2 and 4 + 1 = 1 + 4. In elementary algebra, our arithmetic problem can be expressed for any two numbers a, b and rewritten as

a + b = b + a.

Algebra allows the formulation of problems and their resolution by treating an unknown quantity formally with a variable or a symbol, such as a or b. For example, we will learn later how to calculate

123456789^2 – (123456787)(123456791)

without using a calculator. But the power of algebra goes beyond these curiosities.

Supplementary Exercises

Exercise 1.2.1 You start with $100. You give 20% to your friend. But it turns out that you need the $100 after all in order to pay a debt. By what percent should you increase your current amount in order to restore the $100? (The answer is not 20%.)

Exercise 1.2.2 A bottle of wine and its cork cost $1. The bottle of wine costs 80¢ more than the cork. What is the price of the cork, in cents?

1.3

PUZZLES

The purpose of the puzzles in this section is to demonstrate some mathematical problem-solving techniques: working backwards, searching for patterns, and case-by-case analysis.

Example 1.1 A frog is in a 10 ft well. At the beginning of each day, it leaps 5 ft up, but at the end of the day it slides 4 ft down. After how many days, if at all, will the frog escape the well?

Solution: The frog will escape after seven days. At the end of the sixth day, the frog has leaped 6 feet. Then at the beginning of the seventh day, the frog leaps 5 more feet and is out of the well.

Example 1.2 Dale should have divided a number by 4, but instead he subtracted 4. He got the answer 48. What should his answer have been?

Solution: We work backwards. He obtained 48 from 48 + 4 = 52. This means that he should have performed 52 ÷ 4 = 13.

Example 1.3 When a number is multiplied by 3 and then increased by 16, the result obtained is 37. What is the original number?

Solution: We work backwards as follows. We obtained 37 by adding 16 to a number. Working backwards, 37 − 16 = 21. We obtained this 21 by multiplying a number by 3. Again working backwards, the number 21 ÷ 3 = 7. Thus the original number was 7.

Example 1.4 You and I play the following game. I tell you to write down three 2-digit integers between 10 and 89. Then I write down three 2-digit integers of my choice. The answer comes to 297, no matter which three integers you choose (my choice always depends on yours). For example, suppose you choose 12, 23, 48. Then I choose 87, 76, 51. You add

12 + 23 + 48 + 87 + 76 + 51 = 297.

Again, suppose you chose 33, 56, 89. I then choose 66, 43, 10. Observe that

33 + 56 + 89 + 66 + 43 + 10 = 297.

Explain how I choose my numbers so that the answer always comes up to be 297?

Solution: Notice that I always choose my number so that when I add it to your number I get 99. Therefore, I end up adding 99 three times — and 3 × 99 = 297.

Example 1.5 What is the sum of all the positive integers from 1 to 100? This problem can be written as the following expression:

1 + 2 + 3 + … + 99 + 100.

Solution: Pair up the numbers into the fifty pairs, as follows:

(100 + 1) = (99 + 2) = (98 + 3) = … = (50 + 51).

This provides 50 pairs that each add up to 101, so the desired sum is 101 × 50 = 5050. Another solution to this problem will be given in Example 6.23.

Supplementary Exercises

Exercise 1.3.1 What could St. Augustine mean by mathematicians making prophecies? Could he have meant a profession other than mathematician?

Exercise 1.3.2 Can you find five even integers whose sum is 25?

Exercise 1.3.3 Doris entered an elevator in a tall building. She went up 4 floors, down 6 floors, up 8 floors, and down 10 floors. She then found herself on the 23rd floor. In what floor did she enter the elevator?

Exercise 1.3.4 A natural number is called a palindrome if it reads the same both forwards and backwards. For example, 1221 and 100010001 are palindromes. The palindrome 10001 is strictly between two other palindromes. Which two?

Exercise 1.3.5 In the figure below, each square represents a digit. Find the value of each missing digit.

Exercise 1.3.6 Is it possible to replace the letter a in the square below so that every row has the same sum of every column?

Exercise 1.3.7 Fill each square in the figure below with exactly one number from

{1, 2, 3, 4, 5, 6, 7, 8, 9}

so that the square becomes a magic square, that is, a square where every row has the same sum as every column, and as every diagonal.

Is there more than one solution?

Exercise 1.3.8 A brother and a sister collected 24 coins. The brother collected twice as many coins as his sister. How many did each collect?

PART 2

ARITHMETIC REVIEW

CHAPTER 2

ARITHMETIC OPERATIONS

This chapter reviews the mathematical operations of addition, subtraction, multiplication, and division of numbers and introduces exponentiation and root extraction. It is expected that most of the material here will be familiar to you. Nevertheless, arithmetic operations are presented in this chapter in such a way so that algebraic generalization can be easily derived from them and applied into algebraic equations.

2.1

SYMBOLICAL EXPRESSION

The study of algebra begins by interpreting the meaning of its symbols. The use of symbols or letters denotes arbitrary numbers and frees us from the use of a long list of cases. For example, suppose that

Since numbers are infinite, it is an impossible task to list all cases. Instead, the use of an abstract system to represent the statement above saves space. Instead of listing all the possible cases for infinite numbers, you simply replace a number with an arbitrary symbol such as x. Where x is a number, then

x + 0 = x,

without knowing the exact value of the arbitrary number x.

Normally the words increase, increment, and augment, are associated with addition. Thus, if x is an unknown number, the expression “a number increased by seven” is translated into symbols as x + 7. Later in this chapter, this expression can also be equivallenty writtened as 7 + x.

Similarly, the words decrease, decrement, diminish, and difference are associated with subtraction. Thus if x is an unknown number, the expression “a number decreased by seven” is translated into symbols as x − 7. Note that this expression is different than “seven decreased by a certain number” or the expression 7 − x, which is discussed later in this chapter.

Multiplication is commonly associated with the word product. Thus if x is an unknown number, the expression “the product of a certain number and seven” is translated into symbols as 7x. Notice here the use of juxtaposition to denote the multiplication of a letter and a number. In other words, the use of the × (times) symbol or the · (central dot) symbol is omitted. The following are all equivalent:

7x, 7 ⋅ x, 7(x), (7)(x), 7 × x.

Notice that a reason for not using × when we use letters is to avoid confusion with the letter x. Also, the expression could have been written as x7, but this usage is uncommon.

Multiplication operation symbols are necessary in order to represent the product of two numbers. Thus, write the product as 5 ⋅ 6 = 5 × 6 = (5)(6) = 30 so that the reader is not confused with the number 56.

A few other words are used for multiplication by a specific factor. If the unknown quantity is a, then twice the unknown quantity is represented by 2a. Thrice, or to triple, the unknown quantity is represented by 3a. The square of an unknown quantity is that quantity multiplied by itself. For example, the square of a is aa, which is represented in short by a^2. Here a is the base and 2 is the exponent. The cube of an unknown quantity is that quantity multiplied by its square, so for example, the cube of a is aaa, which is represented in short by a^3. Here a is the base and 3 is the exponent.

The word quotient will generally be used to denote division. For example, the quotient of a number and 7 is denoted by x ÷ 7, or equivalently by x/7 or The reciprocal of a number x is 1/x or

Example 2.1 If a number x is tripled and its value is increased by five, we obtain 3x + 5.

Example 2.2 If x is larger than y, the difference between x and y is x − y. However, if y is larger than x, the difference between x and y is y − x.

Example 2.3 If a and b are two numbers, then their product is ab. Later we will see that this is the same as ba.

Example 2.4 The sum of the squares of x and y is x^2 + y^2. However, the square of the sum of x and y is (x + y)2.

Example 2.5 The expression can be translated as “twice a number is diminished by the reciprocal of its square.”

Example 2.6 If n is an integer, its predecessor is n − 1 and its successor is n + 1.

Example 2.7 You begin the day with E eggs. During the course of the day, you fry Y omelettes, each requiring A eggs. How many eggs are left?

Solution: E – YA, since YA eggs are used in frying Y omelettes. Thus, E –YA eggs are left.

Example 2.8 An even natural number has the form 2a, where a is a natural number. An odd natural number has the form 2a + 1, where a is a natural number.

Example 2.9 A natural number divisible by 3 has the form 3a, where a is a natural number. A natural number leaving remainder 1 upon division by 3 has the form 3a + 1, where a is a natural number. A natural number leaving remainder 2 upon division by 3 has the form 3a + 2, where a is a natural number.

Example 2.10 Find a formula for the nth term of the arithmetic progression

2, 7, 12, 17, ….

Solution: We start with 2, and then keep adding 5, thus

The general term is therefore of the form 2 + 5(n − 1), where n = 1, 2, 3, … is a natural number.

Example 2.11 Find a formula for the nth term of the geometric progression

6, 12, 24, 48, ….

Solution: We start with 6, and then keep multiplying by 2, thus

6 = 6 ⋅ 2^0,

12 = 6 ⋅ 2^1,

24 = 6 ⋅ 2^2,

48 = 6 ⋅ 2^3, ….

The general term is therefore of the form 3 · 2n-1, where n = 1, 2, 3, … is a natural number.

Example 2.12 The law of formation and conjecture in a general formula can be represented as follows:

Solution: Notice that the right-hand side consists of the last number on the left times its successor, which is then divided by 2. Thus we are asserting that

Supplementary Exercises

Exercise 2.1.1 If a person is currently N years old, what was his age 20 years ago?

Exercise 2.1.2 If a person is currently N years old, what will his age be in 20 years?

Exercise 2.1.3 You start with x dollars. Then you triple this amount, and finally you increase that resulting amount by 10 dollars. How many dollars do you now have?

Exercise 2.1.4 You start with x dollars. Then you add 10 dollars to this amount, and finally you triple the resulting amount. How many dollars do you now have?

Exercise 2.1.5 A crocheted baby blanket uses 5 balls of yarn. I start the day with b balls of yarn and crochet s blankets. How many balls of yarn do I have at the end of the day?

Exercise 2.1.6 Think of a number and then double it. Next, add 10. Half your result, and then subtract your original number. After these five steps, your answer is 5 regardless of your original number! If x is the original number, explain by means of algebraic formula each step.

Exercise 2.1.7 What is the general form for a natural number divisible by 4? Leaving remainder 1 upon division by 4? Leaving remainder 2 upon division by 4? Leaving remainder 3 upon division by 4?

Exercise 2.1.8 Find a general formula for the nth term of the arithmetic progression

1, 7, 13, 19, 25, ….

Exercise 2.1.9 Identify the law of formation and conjecture a general formula:

Exercise 2.1.10 Identify the law of formation and conjecture a general formula.

Exercise 2.1.11 You start the day with q quarters and d dimes. How much money do you have? Answer in cents. If by the end of the day you have lost a quarters and b dimes, how much money do you now have? Answer in cents.

Exercise 2.1.12 Let x be an unknown quantity. Translate into symbols the expression “the cube of a quantity is reduced by its square, and then what is left is divided by 8.”

Exercise 2.1.13 A man bought a hat for h dollars. He then bought a jacket and a pair of jeans. If the jacket is thrice as expensive as the hat, and the jeans are 8 dollars cheaper than the jacket, how much money did he spend in total for the three items?

Exercise 2.1.14 A farmer sold a cow for a dollars and gained b dollars as profit. What is the real cost of the cow?

Exercise 2.1.15 A salesman started the year with m dollars. The first month he gained x dollars, the next month he lost y dollars, the third month he gained b dollars, and the fourth month lost z dollars. How much did he have at the end of the fourth month?

2.2

THE NATURAL NUMBERS

This section gives an overview of the natural numbers. The natural numbers are typically used to count and to order objects. Start with two symbols, 0 and 1, and an operation +, adjoining the elements

Observe that this set is infinite and ordered, that is, you can compare any two elements and tell whether one is larger than the other. Define the symbols

Beyond 9, reuse these symbols by also attaching a meaning to their place. Thus

Definition 2.1 A positional notation or place-value notation system is a numeral system in which each position is related to the next by a constant multiplier of that numeral system. Each position is represented by a limited set of symbols. The resultant value of each position is the value of its symbol or symbols multiplied by a power of the base.

As you know, we use base-10 positional notation. For example, in

1234 = 1 ⋅ 1000 + 2 ⋅ 100 + 3 ⋅ 10 + 4 ⋅ 1,

1 does not mean “1,” but 1000; 2 does not mean “2,” but 200, etc.

Before positional notation became standard, simple additive systems (sign-value notation) were used, such as the value of the Hebrew letters, the value of the Greek letters, and Roman Numerals. Arithmetic with these systems was incredibly cumbersome.

Definition 2.2 The collection of all numbers defined by the recursion method above is called the set of natural numbers, and we represent them by the symbol N, that is,

ℕ = {0, 1, 2, 3, …}.

Natural numbers are used for two main reasons:

1.

Counting, for example, “There are 10 sheep in the herd.”

2.

Ordering, for example, “Los Angeles is the second largest city in the United States.”

FIGURE 2.1 The Natural Numbers ℕ.

The natural numbers are interpreted as a linearly ordered set of points, as shown in Figure 2.1, to induce an order relation as defined below.

Definition 2.3 Let a and b be two natural numbers. We say that a is (strictly) less than b, if a is to the left of b on the natural number line. We denote this by a < b.

The symbol ∈ is used to indicate that a certain element belongs to a certain set. The negation of ∈ is ∉. For example, 1 ∉ ℕ because 1 is a natural number, but ∉ ℕ because is not a natural number.

FIGURE 2.2 Addition in ℕ.

FIGURE 2.3 Multiplication in ℕ.

FIGURE 2.4 Multiplication in ℕ.

You may think of the addition of natural numbers as concatenation of segment lengths. For example, in Figure 2.2 a segment whose length is b units is added to a segment whose length is a units, resulting in a new segment whose length is a + b units.

Multiplication is somewhat harder to interpret. In Figure 2.3, consider form ΔOAC with segment OA = 1 and segment OC = b. Extend the segment [OA] to point B, with AB = a. Through point B draw a line parallel to segment [AC], meeting segment [OC] extended at point D. By the similarity of ΔOAC and ΔOBD, segment CD = ab. Another interpretation of multiplication occurs in Figure 2.4, where a rectangle of area ab (square units) is formed having sides of a units by b units.

Observe that if we add or multiply any two natural numbers, the result is a natural number. We encode this observation in the following axiom.

Axiom 2.1 (Closure) ℕ is closed under addition, that is, if a ∈ ℕ and b ∈ ℕ then also a + b ∈ ℕ. ℕ is closed under multiplication, that is, if x ∈ ℕ and y ∈ ℕ then also x y ∈ ℕ.

If 0 is added to any natural number, the result is unchanged. Similarly, if a natural number is multiplied by 1, the result is also unchanged. This is encoded in the following axiom.

Axiom 2.2 (Additive and Multiplicative Identity) 0 ∈ ℕ is the additive identity of ℕ, that is, it has the property that for all x ∈ ℕ it follows that

x = 0 + x = x + 0.

1 ∈ ℕ is the multiplicative identity of ℕ, that is, it has the property that for all a ∈ ℕ it follows that

a = 1a = a1.

Again, it is easy to see that when two natural numbers are added or multiplied, the result does not depend on the order. This is encoded in the following axiom.

Axiom 2.3 (Commutativity) Let a ∈ ℕ and b ∈ ℕ. Then a + b = b + a and ab = ba.

Two other important axioms for the natural numbers are now given.

Axiom 2.4 (Associativity) Let a, b, c be natural numbers. Then the order of parentheses when performing addition is irrelevant, that is,

a + (b + c) = (a + b) + c = a + b + c.

Similarly, the order of parentheses when performing multiplication is irrelevant,

a(bc) = (ab)c = abc.

Axiom 2.5 (Distributive Law) Let a, b, c be natural numbers. Then

a(b + c) = ab + ac,

and

(a + b)c = ac + bc.

The product mn is, simply put, shorthand for addition. That multiplication has the equivalent expressions

Thus (3)(4) means the following expression

(3)(4) = 3 + 3 + 3 + 3 = 4 + 4 + 4 = 12.

Therefore solving the expression

(3)(5) + (6)(4)

requires performing the multiplication first and then the addition, obtaining

(3)(5) + (6)(4) = 5 + 5 + 5 + 6 + 6 + 6 + 6 = 39,

or more concisely

(3)(5) + (6)(4) = 15 + 24 = 39.

In turn, a shorter form for multiplication by the same number is exponentiation, which is defined as follows:

Definition 2.4 (Exponentiation) If n is a natural number greater than or equal to 1, then the nth power of a is defined by

Here, a is the base and n is the exponent. If a is any number different from 0, then we define

a^0 = 1.

without attaching any meaning to 0^0.

Example 2.13 Powers of 2 is a very important concept in computers. A bit is a binary digit taking a value of either 0 (electricity does not pass through a circuit) or 1 (electricity passes through a circuit). Then,

Since 2^10 ≈ 1000, 2^10 is referred to as a kilobit.

Example 2.14 Notice that 2^3 = 8 and 3^2 = 9 are consecutive powers.

Notice that ab is not ab. Thus 2^3 = (2)(2)(2) = 8, and not (2)(3) = 6.

In any expression containing addition and exponentiation, the exponentiation is performed first, since it is really a shortcut for writing multiplication.

Example 2.15

The order of operations can be forced by grouping symbols, like parentheses ( ), brackets [ ], or braces { }.

Example 2.16

(3 + 2)(5 + 3) = (5)(8) = 40,
(3 + 2)2 = (5)2 = 25,
(5 + (3 + 2(4))2)3 = (5 + (3 + 8)2)3 = (5 + (11)2)3 = (5 + 121)3 = 126^3 = 2000376.

Observe that (3 + 2)2 = 25 but that 3^2 + 2^2 = 9 + 4 = 13. Thus exponentiation does not distribute over addition.

Example 2.17 Each element of the set

{10, 11, 12, …, 19, 20}

is multiplied by each element of the set

{21, 22, 23, …, 29, 30}.

If all these products are added, what is the resulting sum?

Solution: This is asking for the product (10 + 11 + · · · + 20) (21 + 22 + · · · + 30) after all the terms are multiplied. But 10 + 11 + · · · + 20 = 165 and 21 + 22 + · · · + 30 = 255. Therefore we want (165) (255) = 42,075.

Definition 2.5 To evaluate an expression with letters means to substitute the values of its letters by the equivalent values given.

Example 2.18 Evaluate a^3 + 3abc when a = 1, b = 2, c = 3.

Solution: Substituting the variables a, b, c with their given values, 1, 2, and 3,

1^3 + 2^3 + 3^3 + 3(1)(2)(3) = 1 + 8 + 27 + 18 = 54.

The root is the reverse mathematical operation of exponentiation and can be defined as follows:

Definition 2.6 (Roots) Let m be a natural number greater than or equal to 2, and let a and b be any natural numbers. We write that In this case we say that b is the mth root of a. The number m is called the index of the root.

In the special case when m = 2, we do not write the index. Thus we will write rather than The number is called the square root of a. The number is called the cubic root of a.

Example 2.19 Let

Example 2.20 Let

Having an idea of what it means to add and multiply natural numbers, let us define subtraction and division of natural numbers by means of those operations. This is often the case in mathematics: we define a new procedure in terms of old procedures.

Definition 2.7 (Definition of Subtraction) Let m, n, x be natural numbers. Then the statement m − n = x means that m = x + n.

Example 2.21 To compute 15 − 3 think of which number when added to 3 returns 15, clearly then 15 − 3 = 12 since 15 = 12 + 3.

Definition 2.8 (Definition of Division) Let m, n, x be natural numbers, with n ≠ 0. Then the statement m ÷ n = x means that m = xn.

Example 2.22 To compute 15 ÷ 3, think of which number when multiplied by 3 returns 15. Clearly then 15 ÷ 3 = 5 since 15 = 5 ⋅ 3.

Neither subtraction nor division are included in ℕ. For example, 3 − 5 is not a natural number, and neither is 3 ÷ 5. Again, the operations of subtraction and division misbehave in the natural numbers; they are not commutative. For example, 5 − 3 is not the same as 3 − 5 and 20 ÷ 4 is not the same as 4 ÷ 20.

Supplementary Exercises

Exercise 2.2.1 Find the numerical value of 1^12^23^3.

Exercise 2.2.2 Find the numerical value of

Exercise 2.2.3 Find the numerical value of 3 ⋅ 4 + 42.

Exercise 2.2.4 Evaluate (a + b)(a − b) when a = 5 and b = 2.

Exercise 2.2.5 Evaluate (a^2 + b^2)(a^2 − b^2) when a = 2 and b = 1.

Exercise 2.2.6 If today is Thursday, what day will it be 100 days from now?

Exercise 2.2.7 If the figure shown below is folded to form a cube, then three faces meet at every vertex. If for each vertex we take the product of the numbers on the three faces that meet there, what is the largest product we get?

Exercise 2.2.8 A freelance graphic designer must print 4500 cards. One contractor can print the cards in 30 days and another contractor can print them in 45 days. How many days would be needed if both contractors are working simultaneously?

Exercise 2.2.9 For a collegiate commencement ceremony, the students of a school are arranged in 9 rows of 28 students per row. How many rows could be made with 36 students per row?

Exercise 2.2.10 Oscar rides his bike, being able to cover 6 miles in 54 minutes. At that speed, how long does it take him to travel a mile?

Exercise 2.2.11 For which values of the natural number n is 36 ÷ n a natural number?

Exercise 2.2.12 Doing only one multiplication, prove that

(666)(222) + (1)(333) + (333)(222) + (666)(333) + (1)(445) + (333)(333)
+ (666)(445) + (333)(445) + (1)(222) = 1000000.

Exercise 2.2.13 A truck with 5 tires (four road tires and a spare tire) travelled 30,000 miles. If all five tires were used equally, how many miles ofwear did each tire receive?

Exercise 2.2.14 A quiz has 25 questions with 4 points awarded for each correct answer and 1 point deducted for each incorrect answer, with zero points for each question omitted. Anna scores 77 points. How many questions did she omit?

Exercise 2.2.15 A calculator displays the computed result of the product

987654 × 745321

as the number 7.36119E11, which means 736,119,000,000. Explain the process to find the last six missing digits.

Exercise 2.2.16 How many digits does 4^165^25 have?

Exercise 2.2.17 A farmer has decided to pose the following problem: “If one gathers all of my cows into groups of 4, 5, or 6, there will be no remainder. But if one gathers them into groups of 7 cows, there will be 1 cow left in one group.” The number of cows is the smallest positive integer satisfying these properties. How many cows are there?

Exercise 2.2.18 Create a new arithmetic operation ⊕ by letting a ⊕ b = 1 + ab.

1.

Compute 1 ⊕ (2⊕3).

2.

Compute (1 ⊕ 2) ⊕ 3.

3.

Is your operation associative? Explain.

4.

Is the operation commutative? Explain.

2.3

FRACTIONS

This section provides a review of fractions in terms of arithmetic operations.

Definition 2.9 A (positive numerical) fraction is a number of the form where m and n are natural numbers and n ≠ 0. Here m is the numerator of the fraction and n is the denominator of the fraction.

Given a natural number n ≠ 0, we divide the interval between consecutive natural numbers k and k + 1 into n equal pieces. Figures 2.5, 2.6, and 2.7 show examples with n = 2, n = 3, and n = 4, respectively. Notice that the larger n is, the finer the partition.

FIGURE 2.5 Fractions with denominator 2.

FIGURE 2.6 Fractions with denominator 3.

FIGURE 2.7 Fractions with denominator 4.

Notice that in many of the fractions written above, the numerator is larger than the denominator, commonly called improper fractions. In many instances, improper fractions are converted into to mixed numbers. For example, is an improper fraction that can also be rewritted into a mixed number as In this book, improper fractions are preferred since it is the common usage in Algebra and in modern computer algebra programs.

By doing more of the diagrams above you may have noticed that there are multiple names for, say, the natural number 2. For example

This observation is a particular case of the following theorem.

Theorem 2.1 (Cancellation Law) Let m, n, k be natural numbers with n ≠ 0 and k ≠ 0. Then

Thus given a fraction, if the numerator and the denominator have any common factors greater than 1, that is, any nontrivial factors, we may reduce the fraction and get an equal fraction.

Definition 2.10 Two fractions such that are said to be equivalent. If b < y, then is said to be a reduced form of

It is possible to prove that any fraction has a unique reduced form with minimal denominator, the so-called equivalent fraction in lowest terms. This depends on the fact that the natural numbers can be factored uniquely into primes. Although we will accept this result, we will not prove it here.

Example 2.23 To reduce to lowest terms, observe that

Since 13 and 15 do not have a nontrivial factor in common, is the desired reduction.

The reduction steps above are not unique. For example, if we had seen right away that 8 was a common factor, we would have obtained obtaining the same result. No matter how many steps you take, as long as there is valid cancellation to do and you perform them all, you will always obtain the correct result.

Example 2.24 Find a fraction with denominator 120 equivalent to

Solution: Observe that 120 ÷ 24 = 5. Thus

Supplementary Exercises

Exercise 2.3.1 Four comrades are racing side by side down a dusty staircase. Freddy goes down two steps at a time, Lorance three, George and Aragorn five. If the only steps with the footprints of all four comrades are at the top and the bottom, how many steps have just one footprint?

Exercise 2.3.2 What fraction of an hour is 36 minutes?

Exercise 2.3.3 Express in least terms.

Exercise 2.3.4 Find an equivalent fraction to with denominator 3990.

Exercise 2.3.5 Arrange in increasing order:

Exercise 2.3.6 Four singers take part in a musical round of 4 equal lines, each finishing after singing the round through 3 times. The second singer begins when the first singer begins the second line, the third singer begins when the first singer begins the third line, the fourth singer begins when the first singer begins the fourth line. What is the fraction of the total singing time when all the singers are singing simultaneously?

2.4

OPERATIONS WITH FRACTIONS

We now define addition of fractions. We would like this definition to agree with our definition of addition of natural numbers. Recall that we defined addition of natural numbers x and y as the concatenation of two segments of length x and y. We then define the addition of two fractions and as the concatenation of two segments of length and The properties of associativity and commutativity stem from this definition. The problem we now have is how to concretely apply this definition to find the desired sum. From this concatenation definition it follows that natural numbers a and b ≠ 0.

It follows from the above property that

We now determine a general formula for adding fractions of different denominators.

Theorem 2.2 (Sum of Fractions) Let a,b,c,d be natural numbers with b ≠ 0 and d ≠ 0. Then

Observe that the trick for adding the fractions in the preceding theorem was to convert them to fractions of the same denominator.

Definition 2.11 To express two fractions in a common denominator is to write them in the same denominator. The smallest possible common denominator is called the least common denominator.

Example 2.25 Add

Solution: A common denominator is 5 · 7 = 35. We thus find

In the preceding example, 35 is not the only denominator that we may have used. Observe that Adding,

This shows that it is not necessary to find the least common denominator in order to add fractions, simply a common denominator.

In fact, let us list the multiples of 5 and of 7 and let us underline the common multiples on these lists:

The multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, …,

The multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, ….

The sequence

35, 70, 105, 140, …,

is a sequence of common denominators for 5 and 7.

Example 2.26 To perform the addition observe that 7 · 5 · 2 = 70 is a common denominator. Thus

A few words about subtraction of fractions. Suppose Observe that this means that ad ≥ bc, which in turns means that ad − bc ≥ 0. Then from a segment of length we subtract one of In much the same manner of addition of fractions, then

Now, since ad − bc ≥ 0, ad − bc is a natural number and so is a fraction.

We now approach multiplication of fractions. We saw that a possible interpretation for the product xy of two natural numbers x and y is the area of a rectangle with sides of length x and y. We would like to extend this interpretation in the case when x and y are fractions. That is, given a rectangle of sides of length we would like to deduce that is the area of this rectangle.

Theorem 2.3 (Multiplication of Fractions) Let a, b, c, d be natural numbers with b ≠ 0 and d ≠ 0. Then

Example 2.27 We have

Alternatively, we could have cancelled the common factors, as follows,

Example 2.28 Find the exact value of the product

Solution: We have,

We now tackle division of fractions. Recall that we defined division of natural numbers as follows. If n ≠ 0 and m, x are natural numbers, then m ÷ n = x means that m = xn. We would like a definition of fraction division compatible with this definition of natural number division. Therefore, we give the following definition.

Definition 2.12 Let a, b, c, d be natural numbers with b ≠ 0, c ≠ 0, d ≠ 0. We define the fraction division

We would like to know what above is in terms of a, b, c, d. For this purpose we have the following theorem.

Theorem 2.4 (Division of Fractions) Let a, b, c, d be natural numbers with b ≠ 0, c ≠ 0, d ≠ 0. Then

that is, in Definition 2.12 is

Definition 2.13 Let c ≠ 0, d ≠ 0 be natural numbers. The reciprocal of the fraction is the fraction

Theorem 2.4 says that in order to divide two fractions we must simply multiply the first one by the reciprocal of the other.

Example 2.29 We have,

Supplementary Exercises

Exercise 2.4.1 Complete the “fraction puzzle” below.

Exercise 2.4.2 Find the exact numerical value of

Exercise 2.4.3 Find the exact numerical value of

Exercise 2.4.4 Find the value of

Exercise 2.4.5 Find the exact numerical value of

Exercise 2.4.6 If

where the fraction is in least terms, find a^2 + b^2.

Exercise 2.4.7 Find the exact numerical value of

Exercise 2.4.8 John takes 2 hours to paint a room, whereas Bill takes 3 hours to paint the same room. How long would it take if both of them start and work simultaneously?

Exercise 2.4.9 Naomi has 16 yards of gift-wrap in order to wrap the gifts for the Festival of Lights at the community center. Each gift requires yards of paper. How many gifts can she wrap?

Exercise 2.4.10 Evaluate

Exercise 2.4.11 Find the value of

Exercise 2.4.12 At a college 99% of the 100 students are female, but only 98% of the students living on campus are female. If some females live on campus, how many students live off campus?

Exercise 2.4.13 What would be the price of a -mile trip with the following taxi company?

2.5

THE INTEGERS

The introduction of fractions in the preceding section helped solve the problem that the natural numbers are not closed under division. We now solve the problem that the natural numbers are not closed under subtraction.

Definition 2.14 A natural number not equal to 0 is said to be positive. The set

{1, 2, 3, 4, 5, …}

is called the set of positive integers.

Definition 2.15 Given a natural number n, we define its opposite −n as the unique number −n such that

n + (−n) = (−n) + n = 0.

The collection

{−1, −2, −3, −4, −5, …}

of all the opposites of the natural numbers is called the set of negative integers. The collection of natural numbers together with the negative integers is the set of integers, which we denote by the symbol ℤ. A graphical representation of the integers is given in Figure 2.8.

FIGURE 2.8 The integers.

There seems to be no evidence of usage of negative numbers by the Babylonians, Pharaonic Egyptians, or the ancient Greeks. It seems that the earliest usage of them came from China and India. In the seventh century, negative numbers were used for bookkeeping in India. The Hindu astronomer Brahmagupta, writing around a.d. 630, shows a clear understanding of the usage of negative numbers.

Thus it took humans a few millennia to develop the idea of negative numbers. Perhaps because our lives are more complex now, it is not so difficult for us to accept their existence and understand the concept of negative numbers.

Let a ∈ ℤ and b ∈ ℤ If a > 0, then −a < 0. If b < 0, then −b > 0. Thus either the number, or its “mirror reflection” about 0, is positive, and in particular, for any a ∈ ℤ, –(−a) = a. This leads to the following definition.

Definition 2.16 Let a ∈ ℤ. The absolute value of a is defined and denoted by

Example 2.30 |5| = 5 since 5 > 0. | − 5| = −(−5) = 5, since −5 < 0.

Letters have no idea of the sign of the numbers they represent. Thus it is a mistake to think, say, that +x is always positive and −x is always negative.

We would like to define addition, subtraction, multiplication, and division in the integers in such a way that these operations are consistent with those operations over the natural numbers and so that they gain closure, commutativity, associativity, and distributivity under addition and multiplication.

We start with addition. Recall that we defined addition of two natural numbers and of two fractions as the concatenation of two segments. We would like this definition to extend to the integers, but we are confronted with the need to define what a “negative segment” is. This we will do as follows. If a < 0, then −a > 0. We associate with a a segment of length | − a|, but to the left of 0 on the line, as in Figure 2.9.

FIGURE 2.9 A negative segment. Here a < 0.

So we define the addition of integers a, b, as the concatenation of segments. Depending on the sign of a and b, we have four cases. (We exclude the cases when at least one of a or b is zero, these cases being trivial.)

Example 2.31 (Case a > 0, b > 0) To add b to a, we first locate a on the line. From there, we move b units right (since b > 0), landing at a + b. Notice that this case reduces to addition of natural numbers, and therefore, we should obtain the same result as for addition of natural numbers. This example is illustrated in Figure 2.10. For a numerical example (with a = 3, b = 2), see Figure 2.11.

FIGURE 2.10 a + b with a > 0, b > 0.

FIGURE 2.11 3 + 2.

Example 2.32 (Case a < 0, b < 0) To add b to a, we first locate a on the line. From there, we move b units left (since b < 0), landing at a + b. This example is illustrated in Figure 2.12. For a numerical example (with a = −3, b = −2), see Figure 2.13.

FIGURE 2.12 a + b with a > 0, b > 0.

FIGURE 2.13 −3 + (−2).

Examples 2.31 and 2.32 conform to the following intuitive idea. If we associate positive numbers to “gains” and negative numbers to “losses,” then a “gain” plus a “gain” is a “larger gain” and a “loss” plus a “loss” is a “larger loss.”

Example 2.33 We have,

(+1) + (+3) + (+5) = +9,

since we are adding three gains, and we thus obtain a larger gain.

Example 2.34 We have,

(−11) + (−13) + (−15) = −39,

since we are adding three losses, and we thus obtain a larger loss.

We now tackle the cases when the summands have opposite signs. In this case, borrowing from the preceding remark, we have a “gain” plus a “loss.” In such a case it is impossible to know beforehand whether the result is a gain or a loss. The only conclusion we could gather, again, intuitively, is that the result will be in a sense “smaller,” that is, we will have a “smaller gain” or a smaller loss.” If the “gain” is larger than the loss, then the result will be a “smaller gain.” And if the “loss” is larger than the “gain,” then the result will be a “smaller loss.”

Example 2.35 (Case a < 0, b > 0) To add b to a, we first locate a on the line. Since a < 0, it is located to the left of O. From there, we move b units right (since b > 0), landing at a + b. This example is illustrated in Figure 2.14. For a numerical example (with a = −3, b = 2), see Figure 2.15. Again, we emphasise, in the sum (−3) + (+2), the “loss” is larger than the “gain.” So when adding, we expect a “smaller loss,” fixing the sign of the result to be minus.

FIGURE 2.14 a + b with a < 0, b > 0.

FIGURE 2.15 −3 + 2.

Example 3.36 (Case a > 0, b < 0) To add b to a, we first locate a on the line. From there, we move b units left (since b < 0), landing at a + b. This example is illustrated in Figure 2.16. For a numerical example (with a = 3, b = −2), see Figure 2.17.

FIGURE 2.16 a + b with a > 0, b < 0.

FIGURE 2.17 3 + (–2).

Example 2.37 We have,

(+19) + (−21) = −2,

since the loss of 21 is larger than the gain of 19, and so we obtain a loss.

Example 2.38 We have,

(−100) + (+210) = +110,

since the loss of 100 is smaller than the gain of 210, and so we obtain a gain.

We now turn to subtraction. We define subtraction in terms of addition.

Definition 2.17 Subtraction is defined as

a − b = a + (−b).

Example 2.39 We have, (+8) − (+5) = (+8) + (−5) = 3.

Example 2.40 We have, (−8) − (−5) = (−8) + (+5) = −3.

Example 2.41 We have, (+8) − (−5) = (+8) + (+5) = 13.

Example 2.42 We have, (−8) − (+5) = (−8) + (−5) = −13.

We now explore multiplication. Again, we would like the multiplication rules to be consistent with those we have studied for natural numbers. This entails that if we multiply two positive integers, the result will be a positive integer. What happens in other cases? Suppose a > 0 and b < 0. We would like to prove that ab < 0. Observe that

a(b - b) = 0 ⇒ ab − ab = 0 ⇒ ab + a(−b) = 0 ⇒ ab = −a(−b).

Since −b > 0, a(−b) is the product of two positive integers, and therefore positive. Thus −a(−b) is negative, and so ab = −a(−b) < 0. We have proved that the product of a positive integer and a negative integer is negative. Using the same trick we can prove that

(−x)(−y) = xy.

If x < 0, y < 0, then both −x > 0, −y > 0, so the product of two negative integers is the same as the product of two positive integers, and so it is positive. We have thus proved the following rules:

(+)(+) = (−)(−) = + and (+)(−) = (−)(+) = −.

Intuitively, you may think of a negative sign as a reversal of direction on the real line. Thus the product or quotient of two integers of different signs is negative. Two negatives give two reversals, which is to say, no reversal at all—thus the product or quotient of two integers with the same sign is positive. The sign rules for division are obtained from and are identical to those of division.

Example 2.43 We have,

(−2)(5) = −10, (−2)(−5) = +10, (+2)(−5) = −10, (+2)(+5) = +10.

Example 2.44 We have,

(−20) ÷ (5) = −4, (−20) ÷ (−5) = +4, (+20) ÷ (−5) = −4, (+20) ÷ (+5) = +4.

The rules of operator precedence discussed in the section of natural numbers apply.

Example 2.45 We have,

Example 2.46 We have,

As a consequence of the rule of signs for multiplication, a product containing an odd number of minus signs will be negative and a product containing an even number of minus signs will be positive.

Example 2.47

(−2)2 = 4, (−2)3 = −8, (−2)10 = 1024.

Notice the difference between, say, (−a)2 and −a^2. (−a)2 is the square of −a, and so it is always nonnegative. On the other hand, −a^2 is the opposite of a^2, and therefore it is always nonpositive.

Example 2.48 We have,

Supplementary Exercises

Exercise 2.5.1 Perform the following operations mentally.

1.

(−9) − (−17)

2.

(−17) − (9)

3.

(−9) − (17)

4.

(−1) − (2) − (−3)

5.

(−100) − (101) + (−102)

6.

|−2| − |−2|

7.

|−2| − (−|2|)

8.

|−100| + (−100) − (−(−100))

Exercise 2.5.2 Place each of the following nine integers

{−4, −3, −2, −1, 0, 1, 2, 3, 4}

once in the diagram below so that every diagonal sum is the same.

Exercise 2.5.3 Place each of the nine integers {–2, –1, 0, 1, 2, 3, 4, 5, 6} once in the diagram below so that every diagonal sum is the same.

Exercise 2.5.4 Complete the “crossword” puzzle with 1s or –1s.

Exercise 2.5.5 Evaluate the expression

when a = 2, b = −3, and c = 5.

2.6

RATIONAL, IRRATIONAL, AND REAL NUMBERS

Definition 2.18 The set of negative fractions is the set

The set of positive fractions together with the set of negative fractions and the number 0 form the set of rational numbers, which we denote by ℚ.

The rules for operations with rational numbers derive from those of operations with fractions and with integers. Also, the rational numbers are closed under the four operations of addition, subtraction, multiplication, and division. A few examples follow.

Example 2.49 have

It can be proved that any rational number has a decimal expansion which is either periodic (repeats) or terminates, and that any number with either a periodic or a terminating expansion is a rational number. For example, has a terminating decimal expansion, and has a repeating one. By long division you may also obtain

and as you can see, the periods may be longer than what your calculator can handle.

What about the numbers whose decimal expansion is infinite and does not repeat? This leads us to the following definition.

Definition 2.19 A number whose decimal expansion is infinite and does not repeat is called an irrational number.

From the discussion above, an irrational number is one that cannot be expressed as a fraction of two integers.

Example 2.50 Consider the number

0.1010010001000010000010000001…,

where the number of 0s between consecutive 1s grows in sequence: 1, 2, 3, 4, 5, …. Since the number of 0s is progressively growing, this infinite decimal does not have a repeating period and therefore must be an irrational number.

Using my computer, when I enter I obtain as an answer

1.4142135623730950488016887242097

Is this answer exact? Does this decimal repeat? It can be proved that the number is irrational, so the above answer is only an approximation and the decimal does not repeat. The first proof of the irrationality of is attributed to Hippasus of Metapontum, one of the disciples of Pythagoras (c. 580 b.c.e.–c. 500 b.c.e.). The Greek world view at that time was that all numbers where rational, and therefore this discovery was anathema to the Pythagoreans who decided to drown Hipassus for his discovery.

It can be proved that if n is a natural number that is not a perfect square, then is irrational. Therefore, etc., are all irrational.

In 1760, Johnn Heinrich Lambert (1728 – 1777) proved that n is irrational.

In particular, then, it would be incorrect to write etc., since π is not rational. All of these are simply approximations, and therefore we must write π ≈ 3.14, or etc.

Definition 2.20 The set of real numbers, denoted by ℝ, is the collection of rational numbers together with the irrational numbers.

Supplementary Exercises

Exercise 2.6.1 Evaluate the expression

when x = − 1, y = 2, and z = −3.

Exercise 2.6.2 Evaluate the following expressions when

1.

2x + 3y

2.

xy − x − y

3.

x^2 + y^2

Exercise 2.6.3 In this exercise, you are allowed to use any of the operations +, −, +, ÷, ⋅,!, and exponentiation. You must use exactly four 4s. Among your fours you may also use .4. The n! (factorial) symbol means that you multiply all the integers up to n. For example, 1! = 1, 2! = 1 ⋅ 2 = 2, 3! = 1 ⋅ 2 ⋅ 3 = 6, 4! = 1 ⋅ 2 ⋅ 3 ⋅ 4 = 24. With these rules, write every integer, from 1 to 20 inclusive. For example,

Exercise 2.6.4 Without using any of the signs +, −, ÷, ·, but with exponentiation being allowed, what is the largest number that you can form using three 4s? Again, you must explain your reasoning.

Exercise 2.6.5 Find the value of

Exercise 2.6.6 Suppose you know that What should 0.1111… 0.1 be?

Exercise 2.6.7 Find the value of 121(0.09).

Exercise 2.6.8 Use a calculator to round 121(0.09). to two decimal places.

Exercise 2.6.9 Use a calculator to round to two decimal places.

Exercise 2.6.10 Let a, b be positive real numbers. Is it always true that

PART 3

ALGEBRAIC OPERATIONS

CHAPTER 3

ADDITION AND SUBTRACTION

This chapter illustrates the mathematical techniques of addition and subtraction of algebraic equations, with an emphasis on Terms and Algebraic Expressions and More Suppression of Parentheses.

3.1

TERMS AND ALGEBRAIC EXPRESSIONS

If one classroom has a group of 7 freshmen students, 8 sophomore students, and 3 junior students—and another classroom has 4 freshmen students, 4 sophomore students, and 1 junior student—we could write the following equation addressing the total number of students:

(7F + 8S + 3J) + (4F + 4S + 1J) = 11F + 12S + 4J.

The procedure of collecting like terms, which we will explain shortly, draws essentially from the concept used in this example.

Again, consider the following way of adding 731 and 695. Since

731 = 7 · 10^2 + 3 · 10 + 1, 695 = 6 · 10^2 + 9 · 10 + 5,

we could add in the following fashion, without worrying about carrying:

(7 · 10^2 + 3 · 10 + 1) + (6 · 10^2 + 9 · 10 + 5) = 13 · 10^2 + 12 · 10 + 6 = 1300 + 120 + 6 = 1426.

Definition 3.1 An algebraic expression is a collection of symbols (letters and/or numbers). An algebraic expression may have one or more terms, which are separated from each other by the signs + (plus) or – (minus).

Example 3.1 The expression 18a + 3b − 5 consists of three terms.

Example 3.2 The expression 18ab^2c consists of one term. Notice in this case that since no sign precedes 18ab^2c, the + sign is tacitly understood. In other words, 18ab^2c = + 18ab^2c.

Example 3.3 The expression a + 3a^2 − 4a^3 − 8ab + 7 consists of five terms.

Definition 3.2 When one of the factors of a term is a number, it is called the numerical coefficient (or coefficient, for short) of the term. In the expression ab, a is the base and b is the exponent.

Example 3.4 In the expression 13a^2b^3, 13 is the numerical coefficient of the term, 2 is the exponent of a and 3 is the exponent of b.

Example 3.5 In the expression a^2bd, 1 is the numerical coefficient of the term, 2 is the exponent of a and d is the exponent of b.

Example 3.6 In the expression −ac, −1 is the numerical coefficient of the term and c is the exponent of a. Notice then that the expression −1ac is equivalent to the expression −ac.

Definition 3.3 If two terms

1.

agree in their letters, and

2.

for each letter appearing in them the exponent is the same,

then we say that the terms are like terms.

Example 3.7 The terms −a, 7a, 5a, are all like terms.

Example 3.8 The terms 4a^2b^3, 7b^3a^2, 5a^2b^3, are all like terms. Notice that by commutativity of multiplication, 7b^3a^2 = 7a^2b^3.

Example 3.9 The terms 4a and 4a^2 are unlike terms, since the exponent of a in 4a is 1 and the exponent of a in 4a^2 is 2.

Example 3.10 The terms 4ab^2 and 4ab^3 are unlike terms, since the exponent of b in 4ab^2 is 2 and the exponent of b in 4ab^3 is 3.

To add or collect like terms, we simply add the coefficients of the like terms.

Example 3.11 We have,

(a + b) + (a + b) + (a + b) = a + a + a + b + b + b = 3a + 3b

By the commutative law, we may also write this as 3b + 3a.

Example 3.12 We have,

Example 3.13 We have,

Example 3.14 We have,

Example 3.15 We have,

Example 3.16 We have,

By the commutative law, we may also write this as −13b + 6a.

Example 3.17 We have,

By the commutative law, we may also write this as −12x + 4.

Example 3.18 We have,

(x^2 + xy + y^2 + xy^2) + (−x^2 + 2yx − 3y^2 − x^2y) = 3xy − 2y^2 + xy^2 − x^2y.

This last expression is one of the many possible ways to write the final result, thanks to the commutative law.

Supplementary Exercises

Exercise 3.1.1 A boy buys a marbles, wins b marbles, and loses c marbles. How many marbles does he have? Write one or two sentences explaining your solution.

Exercise 3.1.2 Collect like terms mentally.

1.

(−9a) + (+17a) − 8a

2.

(−17a) + (−9a) + 20a

3.

− a − 2a − 3a − 4a

4.

10a − 8a + 6a − 4a + 2a

5.

−5a − 3b − 9a − 10b + 10a + 17b

Exercise 3.1.3 Collect like terms:

(a + b^2 + c + d^2) + (a^2 − b − c^2 − d) + (a^2 − b^2 + c − d) + (a − b − c^2 − d^2).

Exercise 3.1.4 Collect like terms.

1.

(a + 2b + 3c) + (3a + 2b + c)

2.

(a + 2b + 3c) + (3a − 2b − c)

3.

(a + 2b + 3c) + (3a + 2b + c) + (a + b + c)

4.

(a + b + c) + (a − b + c) + (−a − b − c)

5.

(x^2 − 2x + 1) + (x^2 + 2x + 1)

6.

(x^3 − x^2 + x − 1) + (2x^2 − x + 2)

7.

(2x^3 − 2x^2 − 2) + (2x^3 − 2x − 1) + (2x^3 − x^2 − x + 1)

8.

Exercise 3.1.5 Find the value of the expression

(2x^2 + x − 2) + (−x^2 − x + 2)

when x = 2.

Exercise 3.1.6 Is it always true that a + a = 2a?

Exercise 3.1.7 Is it always true that a + a = 2a^2?

Exercise 3.1.8 Is it always true that a + a^2 = 2a^3?

Exercise 3.1.9 If (8x − 5)5 = Ax^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F, find A + B + C + D + E + F.

Exercise 3.1.10 Give three different equivalent forms for the expression 3yx − x^2 y.

3.2

MORE SUPPRESSION OF PARENTHESES

A minus sign preceding an expression in parentheses negates every term of the expression. Thus

− (a + b) = −a − b.

Example 3.19 We have,

Example 3.20 We have,

Example 3.21 We have,

Example 3.22 We have,

Recall that the distributive law states that for real numbers a, b, c,

a(b + c) = ab + ac.

We now apply this to algebraic expressions.

Example 3.23 Multiply term by term: 2(a + 2b − 3c).

Solution: We have,

2(a + 2b − 3c) = 2a + 2(2b) + 2(−3c) = 2a + 4b − 6c.

Example 3.24 Multiply term by term: −2(a + 2b − 3c).

Solution: We have,

−2(a + 2b − 3c) = −2a + (−2) (2b) + (−2)(−3c) = −2a −4b + 6c.

Example 3.25 Divide term by term:

Solution: We have,

We prefer to use improper fractions and write rather than Also, the expressions and are identical, that is,

We may combine the distributive law and combining like terms.

Example 3.26 Combine like terms: 2(a − 2b) + 3(−a + 2b).

Solution: We have,

Example 3.27 Combine like terms: 2(a − 2b) −3(–a + 2b).

Solution: We have,

Example 3.28 Combine like terms:

Solution: We have,

Example 3.29 Combine like terms:

Solution: We have,

Supplementary Exercises

Exercise 3.2.1 A rod measuring a + b inches long was cut by b − c inches. How much of the rod remains? Write one or two sentences explaining your answer.

Exercise 3.2.2 A group of x people is going on an excursion and decides to charter a plane. The total price of renting the plane is D dollars, and everyone will pay an equal share. How much does each person pay for their share? On the day of departure, p people are no-shows. How much more will the people who show up have to pay in order to cover the original cost?

Exercise 3.2.3 Suppress parentheses and collect like terms:

(a + b^2 + c + d^2) − (a^2 − b − c^2− d) + (a^2 − b^2+ c − d) − (a − b − c^2− d^2).

Exercise 3.2.4 Expand and simplify:

2(a + b − 2c) − 3(a − b − c).

Exercise 3.2.5 Suppress parentheses and collect like terms.

1.

(a + 2b + 3c) − (3a + 2b + c)

2.

(a + 2b + 3c) + (3a + 2b + c) − (a + b + c)

3.

(a + b + c) − (a − b + c) − (−a − b − c)

4.

(x^3 − x^2 + x − 1) − (2x^2 − x + 2)

5.

(2x^3 − 2x^2 − 2) − (2x^3 − 2x − 1) − (2x^3 − x^2 − x + 1)

6.

(8.1a − 9.2b) − (−0.2a − 3.1b) − (−a + 3b)

7.

8.

9.

(5♣ − 7 ♠) −(−5♣ − 7 ♠)

Exercise 3.2.6 Collect like terms and write in a single fraction:

Exercise 3.2.7 Combine like terms.

1.

2(−x + 2y) + 3(x − 2y) − (x + y) + (y − 2x)

2.

3.

Exercise 3.2.8 Match each expression on the left with the equivalent expression on the right.

Exercise 3.2.9 Mary starts the day with A dollars. Her uncle Bob gives her enough money to double her amount. Her aunt Rita gives Mary 10 dollars. Mary has to pay B dollars in fines, and spent 12 dollars fueling her car with gas. How much money does she has left?

Exercise 3.2.10 Simplify each of the following expressions as much as possible.

1.

(2 − 4)x − 5(−6)x

2.

5 − 2(2x)

3.

(5 − 2)(1 − 2^2)3x

Exercise 3.2.11 Remove all redundant parentheses. If removing parentheses changes the expression, explain.

1.

(6t) + x

2.

6(t + x)

3.

(ab)2

4.

(ab − c)2

Exercise 3.2.12 Fill in the box to make the expression true.

Exercise 3.2.13 Fill in the box to make the expression true.

1.

1−t = t + 1

2.

3x+y−t = + 3x

Exercise 3.2.14 What is the opposite of 4 − y + 2?

Exercise 3.2.15 What is the additive inverse of

CHAPTER   4

MULTIPLICATION

This chapter illustrates the techniques used to perform multiplication operations in algebra. We introduce mathematical operations in the following sections: Laws of Exponents, Negative Exponents, Distributive Law, Square of a Sum, Difference of Squares, Cube of a Sum, and Sum and Difference of Cubes.

4.1

LAWS OF EXPONENTS

Recall that if a is a real number and if n is a natural number, then

with the interpretation that a^0 = 1 if a ≠ 0.

Suppose we wanted to compute a^2a^5. We would proceed as follows:

We know that a^2 = aa and a^5 = aaaaa, thus:

(a^2)(a^5) = (aa)(aaaaa) = aaaaaaa = a^7,

The above example can be generalized as the following theorem.

Theorem 4.1 (First Law of Exponents) Let a be a real number and m, n natural numbers. Then

am an = am + n

Example 4.1 (ab^2c^3)(a^3b^3c) = a^1 + 3b^2 + 3c^3 + 1 = a^4b^5c^4.

Example 4.2 (ax)(b2y)(a2x b2y) = ax + 2xb2y + 2y = a3xb4y.

Example 4.3 (ax − 2y + z)(a2x − y − z)(ax + y + z) = ax − 2y + z + 2x − y − z + x + y + z = a4x − 2y + z.

Example 4.4 5^5 + 5^5 + 5^5 + 5^5 + 5^5 = 5(5^5) = 5l + 5 = 5^6 = 15625.

We may use associativity and commutativity in the case when the numerical coefficients are different from 1.

Example 4.5 (5x^2)(6x^3) = 30x^5.

Example 4.6 (−2ab^2c^3)(3abc^4) = −6a^2b^3c^7.

We now tackle division. As a particular case, observe that

and, a^3 = a^5 −2. This generalizes as the following theorem.

Theorem 4.2 (Second Law of Exponents) Let a ≠ 0 be a real number and m, n natural numbers, such that m ≥ n. Then

Example 4.7

Example 4.8

Example 4.9

Example 4.10

Example 4.11

Example 4.12

Example 4.13

Example 4.14

Suppose we wanted to compute (a^2)5. Since a quantity raised to the fifth power is simply the quantity multiplied by itself five times, we have:

(a^2)5 = (a^2)(a^2)(a^2)(a^2)(a^2) = a^2 + 2 + 2 + 2 + 2 = a^10,

upon using the first law of exponents. In general we have the following law dealing with “exponents of exponents.”

Theorem 4.3 (Third Law of Exponents) Let a be a real number and m, n natural numbers. Then

(am)n = amn

Theorem 4.4 (Fourth Law of Exponents) Let a and b be a real numbers and let m be a natural number. Then

(ab)m = ambm.

Example 4.15 We have,

Example 4.16 (ax^3)2x = a(x^3)(2x) = a2x^4.

Example 4.17 25^34^3 = (25 · 4)3 = 100^3 = 1000000.

Supplementary Exercises

Exercise 4.1.1 Simplify:

Exercise 4.1.2 Simplify:

Exercise 4.1.3 Find the exact numerical value of:

Exercise 4.1.4 Find the exact numerical value of:

Exercise 4.1.5 If

find n.

Exercise 4.1.6 If

3^2001 + 3^2002 + 3^2003 = a3^2001,

find a.

Exercise 4.1.7 Demonstrate that:

Exercise 4.1.8 Divide:

Exercise 4.1.9 Multiply:

Exercise 4.1.10 Divide:

Exercise 4.1.11 Multiply:

Exercise 4.1.12 Divide:

Exercise 4.1.13 Multiply:

Exercise 4.1.14 Simplify and write as a single number:

Exercise 4.1.15 If then m = _____________ and n = _____________.

4.2

NEGATIVE EXPONENTS

Let a ≠ 0 be a real number and let n be a natural number. Then

This gives an interpretation to negative exponents. Observe also that taking reciprocals,

Example 4.18

Example 4.19

Example 4.20

Example 4.21

Example 4.22 Simplify, using positive exponents only:

Solution: ►

Example 4.23 Simplify, using positive exponents only:

Solution: ► We have,

Supplementary Exercises

Exercise 4.2.1 Simplify, using positive exponents only.

Exercise 4.2.2 Demonstrate that (ax+3by−4)(a2x−3b−2y+5) = a3xa1−y.

Exercise 4.2.3 Demonstrate that

Exercise 4.2.4 Simplify and express with positive exponents only:

Exercise 4.2.5 Calculate the exact numerical value:

Exercise 4.2.6 Calculate the exact numerical value:

Exercise 4.2.7 Calculate the exact numerical value:

4.3

DISTRIBUTIVE LAW

Recall that the distributive law says that if a, b, and c are real numbers, then

a(b + c) = ab + ac

(a + b)c = ac + bc.

We now apply the distributive law to multiplication of terms.

Example 4.24 We have,

Example 4.25 We have,

From the distributive law we deduce that

(a + b)(c + d) = (a + b)c + (a + b)d = ac + bc + ad + bd.

Example 4.26 We have,

Because of commutativity, we may also proceed as follows:

Example 4.27 We have,

Because of commutativity, we may also proceed as follows:

We also recall the meaning of exponentiation.

Example 4.28 We have,

Example 4.29 To expand (x + 1)3, we first find

(x + 1)2 = (x + 1)(x + 1) = x(x + 1) + 1(x + 1) = x^2 + x +x + 1 = x^2 + 2x + 1.

Then we use this result to find the following:

Supplementary Exercises

Exercise 4.3.1 Expand and simplify.

1. x(x^2 + x + 1)

2. 2x(x 2 − 2x + 3)

3. (2x + l)(2x + l)

4. (2x + l)(2x − l)

5. (2x − l)(2x − l)

Exercise 4.3.2 Expand and simplify:

(x 2 + 1) + 2x(x^2 + 2) − 3x^3(x^2 + 3).

Exercise 4.3.3 Expand and simplify:

4x 3(x^2 − 2x + 2) − 2x 2(x^2− x^2− l).

Exercise 4.3.4 Multiply and collect like terms.

1. −2x(x + 1) + 3x 2(x − 3x^2)

2. −2x(x + 1)2

3. a^2c(ab − ac) + ac 2(a^2 − 2b)

4. (x + 2y + 3z)(x − 2y)

Exercise 4.3.5 Expand the product: (2x + 2y + 1)(x − y + 1).

Exercise 4.3.6 Expand and collect like terms: (x + 2)(x + 3) − (x − 2)(x − 3).

Exercise 4.3.7 Expand and collect like terms: (x − 1)(x^2 + x + 1) − (x + 1)(x^2 − x + 1).

Exercise 4.3.8 Expand and simplify: (x + y − z)(x − y + 2z).

Exercise 4.3.9 Expand the product: (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca).

Exercise 4.3.10 Expand the product: (a +b +c)(x + y +z).

Exercise 4.3.11 Prove that the product of two even integers is even.

Exercise 4.3.12 Prove that the product of two odd integers is odd.

Exercise 4.3.13 Prove that the product of two integers that leave remainder 3 upon division by 4 leaves remainder 1 upon division by 4.

Exercise 4.3.14 Prove that the product of any two integers leaving remainder 2 upon division by 3 leaves remainder 1 upon division by 3.

Exercise 4.3.15 Prove that the product of two integers leaving remainder 2 upon division by 5 leaves remainder 4 upon division by 5.

Exercise 4.3.16 Prove that the product of two integers leaving remainder 3 upon division by 5 leaves remainder 4 upon division by 5.

Exercise 4.3.17 If x^2 + x − 1 = 0, find x^4 + 2x^3 + x^2.

Exercise 4.3.18 Expand and collect like terms:

2x(x + 1) − x 2 − x(x − 1).

Then use your result to evaluate

2 · (11112) · (11113) − 11112^2 − (11111) · (11112).

without a calculator.

4.4

SQUARE OF A SUM

In this section, we study some special products that often occur in algebra.

Theorem 4.5 (Square of a Sum) For all real numbers a, b the following identity holds

(a ± b)2 = a^2 ± 2ab + b^2.

FIGURE 4.1 (a + b)2 = a^2 + 2ab + b^2

The identity (a + b)2 = a^2 + 2ab + b^2 has a graphical justification in terms of areas of squares, as seen in Figure 4.1. The area of the large (a + b) × (a + b) square is (a + b)2. In terms, this area can be divided into two square regions, one of area a^2 and the other of area b^2, and two rectangular regions each of area ab.

Example 4.30 We have,

(2x + 3y)2 = (2x)2 + 2(2x) (3y) + (3y)2 = 4x^2 + 12xy + 9y^2.

Example 4.31 We have,

(2 − x 2)2 = (2)2 − 2(2) (x 2) + (x^2)2 = 4 − 4x 2 + x 4.

The principle illustrated here can be used to facilitate mental arithmetic.

Example 4.32 One may compute mentally 52^2 as follows:

52^2 = (50 + 2)2 = 50^2 + 2(50) (2) + 2^2 = 2500 + 200 + 4 = 2704.

The formula (a + b)2 = a^2 + 2ab + b^2 = a^2 + b^2 + 2ab gives the square of a sum in terms of the sum of squares and the product of the numbers. Thus knowing any two of them will determine the other.

Example 4.33 If the sum of two numbers is 3 and their product 4, find the sum of their squares.

Solution: ► Let the two numbers be x, y. Then x + y = 3 and xy = 4. Therefore

9 = (x + y)2 = x^2 + 2xy + y^2 = x^2 + 8 + y^2 ⇒ x^2 + y^2 = l.

Supplementary Exercises

Exercise 4.4.1 Expand and simplify: (3 − y)2.

Exercise 4.4.2 Expand and simplify: (5 − 2x^2)2.

Exercise 4.4.3 Expand and simplify: (2ab 2− 3c 3d^4)2.

Exercise 4.4.4 Expand and simplify: (x + 2y)2 + 4xy.

Exercise 4.4.5 Expand and simplify: (x + 2y)2− 4xy.

Exercise 4.4.6 Expand: (xy − 4y)2.

Exercise 4.4.7 Expand: (ax + by)2.

Exercise 4.4.8 Expand and simplify: (a + 2)2 + (a − 2)2.

Exercise 4.4.9 Expand and simplify: (a + 2)2 − (a − 2)2.

Exercise 4.4.10 Expand and simplify: (a + 2b + 3c)2.

Exercise 4.4.11 Expand and simplify: (a + 2b − c)2.

Exercise 4.4.12 If , find

Exercise 4.4.13 Find the value of n in (10^2002 + 25)2 − (10^2002 − 25)2 = 10n.

Exercise 4.4.14 Given that 2x + 3y = 3 and xy = 4, find 4x^2 + 9y^2.

Exercise 4.4.15 If the difference of two numbers is 3 and their product 4, find the sum of their squares.

Exercise 4.4.16 Given that x + y = 4 and xy = −3, find x^2 + y^2.

Exercise 4.4.17 Given that x + y = 4 and xy = −3, find x^4 + y^4.

Exercise 4.4.18 Expand: (x + y + z)2.

Exercise 4.4.19 Demonstrate that

(x + y + z + w)2 = x^2 + y^2 + z^2 + w^2+ 2xy + 2xz + 2xw + 2yz + 2yw + 2zw.

4.5

DIFFERENCE OF SQUARES

In this section, we continue our study of special products.

Theorem 4.6 (Difference of Squares) For all real numbers a, b the following identity holds

(a + b)(a − b) = a^2 − b^2.

Example 4.34 (4x − 5y) (4x + 5y) = 16x^2 − 25y^2.

Example 4.35 (xy − 5y) (xy + 5y) = x^2y^2 − 25y^2.

Example 4.36 Show that (a − b − c)(a + b + c) = a^2 − b^2 − c^2 − 2bc.

Solution: ► We have

Example 4.37

Example 4.38 We have

Example 4.39 We have

Example 4.40 Prove that

x^2 − (x + 2)(x − 2) = 4.

Solution: ► We have

Explain how to find the exact value of the following, mentally:

(987654321) (987654321) − (987654323) (987654319),

Since the first term is two values away from the third term, we can set x = 987654321. Then x + 2 = 987654323 and x − 2 = 987654319, from where (987654321)(987654321) − (987654323) (987654319) = 4

follows.

Supplementary Exercises

Exercise 4.5.1 Multiply and collect like terms: (a + 4)2 − (a + 2)2.

Exercise 4.5.2 Prove that

(x − y)(x + y)(x^2 + y^2) = x^4 − y^4.

Exercise 4.5.3 Multiply and collect like terms: (a + 1)4 − (a − 1)4.

Exercise 4.5.4 Show, without expressly computing any power of 2, that

(2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)(2^16 + 1) = 2^32 − 1.

Exercise 4.5.5 Find

1^2 − 2^2 + 3^2 − 4^2 + ··· + 99^2 − 100^2.

Exercise 4.5.6 Without using a calculator, determine the exact numerical value of

(123456789)2 − (123456787)(123456791).

Exercise 4.5.7 Without using a calculator, determine the exact numerical value of

(666 666 666)2 − (333 333 333)2.

4.6

CUBE OF A SUM

We introduce now a special product formula for the cube of a sum. This formula is valid for both real and complex numbers.

Theorem 4.7 Let a and b be real numbers. Then

(a + b)3 = a^3 + 3a^2b + 3ab^2 + b^3.

Corollary 4.1 Let a and b be real numbers. Then

(a − b)3 = a^3 − 3a^2b + 3ab^2 − b^3.

Example 4.41 We have,

(x + 2y)3 = x^3 + 3x^2(2y) + 3x(2y)2 + (2y)3 = x^3 + 6x^2y + 12xy^2 + 8y^3.

Example 4.42 We have,

(3x − 2y)3 = (3x)3 − 3(3x)2(2y) + 3(3x)(2y)2 − (2y)3 = 27x^3 − 54x^2y + 36xy^2 − 8y^3.

Observe that

3ab(a + b) = 3a^2b + 3ab^2,

therefore, we may write the cube of a sum identity as

(a + b)3 = a^3 + b^3 + 3ab(a + b).

Example 4.43 Given that a + b = 2 and ab = 3, find a^3 + b^3.

Solution: ► We have

a^3 + b^3 = (a + b)3 − 3ab (a + b) = 2^3 − 3(3) (2) = 8 − 18 = −10.

Supplementary Exercises

Exercise 4.6.1 Expand and simplify: (5x + l)3.

Exercise 4.6.2 Multiply and collect like terms: (x + 1)3 + (x + 1)2.

Exercise 4.6.3 Multiply and collect like terms: (x −1)3 − (x + 2)2.

Exercise 4.6.4 Multiply and collect like terms: (x − 2)3 − (x − 2)2.

Exercise 4.6.5 Given that a − b = 6 and ab = 3, find a^3 − b^3.

Exercise 4.6.6 Given that a + 2b = 6 and ab = 3, find a^3 + 8b^3.

4.7

SUM AND DIFFERENCE OF CUBES

In this section, we study our final special product.

Theorem 4.8 (Sum and Difference of Cubes) For all real numbers a, b the following identity holds

(a ± b)(a^2 ∓ ab + b^2) = a^3 ± b^3.

Example 4.44 (a − 2b)(a^2+ 2ab + 4b^2) = a^3− 8b^3.

Example 4.45 (5a + 2b) (25a^2 − 10ab + 4b^2) = 125a^3 − 8b^3.

Supplementary Exercises

Exercise 4.7.1 Expand and simplify: (x − 2)(x^2 + 2x + 4).

Exercise 4.7.2 Expand and simplify: (x + 8)(x^2 − 8x + 64).

Exercise 4.7.3 Given that a − b = 6 and ab = 3, find a^3 − b^3.

CHAPTER   5

DIVISION

This chapter reviews the mathematical operations of division in algebraic expressions. We introduce the concepts of Term-by-Term Division, Long Division, Factoring, Special Factorizations, and Rational Expressions.

5.1

TERM-BY-TERM DIVISION

From the distributive law we deduce

Example 5.1

Be careful not to confuse an expression of the form with one of the form Also note that an expression of the form can be evaluated in various ways. For example, one may evaluate first the product ab and then divide this by c, or one may divide a by c and then multiply this by b, etc.

Example 5.2 We have,

This may also be evaluated in the following manner,

as before.

Example 5.3

Example 5.4 We have,

We may also proceed as follows,

When a negative exponent arises in a term, we prefer to express the term with positive exponents.

Example 5.5

Supplementary Exercises

Exercise 5.1.1 Perform the division:

Exercise 5.1.2 Divide term by term:

Exercise 5.1.3 Divide:

Exercise 5.1.4 Divide term by term:

Exercise 5.1.5 Divide:

Exercise 5.1.6 Divide term by term:

Exercise 5.1.7 Divide term by term:

Exercise 5.1.8 Divide:

Exercise 5.1.9 Divide:

5.2

LONG DIVISION

We must now confront the problem when the divisor consists of more than one term. The algorithm for algebraic long division resembles that for long division of natural numbers.
Example 5.6 Find (−x + x^2 − 6) ÷ (x + 2).

Solution: ► First, we rewrite −x + x^2 − 6 so that the exponents of each of the terms are listed in decreasing order: x^2 − x − 6. Next, display the dividend and divisor as follows:

Now by what must x (the term with the largest degree in the divisor) be multiplied to return x^2 (the term with the largest degree in the dividend)? The answer is and so we write x in the quotient,

Multiply this x of the quotient by the divisor, obtaining x(x + 2) = x^2 + 2x and change all signs:

Add, and obtain

The new dividend is now −3x − 6. Again by what must x (the term with the largest degree in the divisor) be multiplied to return −3x (the term with the largest degree in the dividend)?

The answer is and so we write −3 in the quotient,

Multiply this −3 of the quotient by the divisor, obtaining −3(x + 2) = −3x − 6 and change all signs:

Add, and obtain

Since the remainder is 0, the division ends and we conclude that

Example 5.7 Find (3x^4 − 2x^3 − 9x^2 + 5x − 6) ÷ (x − 2).

Solution: ► The terms of 3x^4 − 2x^3 − 9x^2 + 5x − 6 are already written with exponents in decreasing order. Next, display the dividend and divisor as follows:

Now, what must x (the term with the largest degree in the divisor) be multiplied by to return the expression 3x^4 (the term with the largest degree in the dividend)? The answer is and so we write x in the quotient,

Multiply this 3x^3 of the quotient by the divisor, obtaining 3x^3(x − 2) = 3x^4 − 6x^3 and change all signs:

Add, and obtain

The new dividend is now 4x^3 − 9x^2. Again by what must x (the term with the largest degree in the divisor) be multiplied to return 4x^3 (the term with the largest degree in the dividend)? The answer is and so we write +4x2 in the quotient,

Multiply this +4x^2 of the quotient by the divisor, obtaining +4x^2(x − 2) = 4x^3 − 8x^2 and change all signs:

Add, and obtain

The new dividend is now −x^2 + 5x. Again by what must x (the term with the largest degree in the divisor) be multiplied to give −x^2 (the term with the largest degree in the dividend)? The answer is and so we write −x in the quotient,

Multiply this −x of the quotient by the divisor, obtaining −x(x − 2) = −x^2 + 2x and change all signs:

Add, and obtain

The new dividend is now 3x − 6. Again by what must x (the term with the largest degree in the divisor) be multiplied to give 3x (the term with the largest degree in the dividend)? The answer is and so we write 3 in the quotient,

Multiply this 3 of the quotient by the divisor, obtaining 3(x − 2) = 3x − 6 and change all signs:

Add, and obtain

Since the remainder is 0, the division ends and we conclude that

We now give an example where there is a remainder.

Example 5.8 Find (2x^2 + 1) ÷ (x + 1).

Solution: ► Display the dividend and divisor as follows:

Now by what must x (the term with the largest degree in the divisor) be multiplied to give 2x^2 (the term with the largest degree in the dividen)d? The answer is and so we write 2x in the quotient,

Multiply this 2x of the quotient by the divisor, obtaining 2x(x + 1) = 2x^2 + 2x and change all signs:

Add, and obtain

The new dividend is now −2x + 1. Again by what must x (the term with the largest degree in the divisor) be multiplied to give −2x (the term with the largest degree in the dividend)? The answer is and so we write −2 in the quotient,

Multiply this −2 of the quotient by the divisor, obtaining −2(x + 1) = −2x − 2 and change all signs:

Add, and obtain

Now, the remainder is 3, which has lower degree than the divisor x + 1, therefore the division ends. We write

We provide below additional examples, without explanations.

Example 5.9 We have,

Example 5.10 We have,

Example 5.11 We have,

and therefore we write

Supplementary Exercises

Exercise 5.2.1 Perform the division: (x^2− 2x + 1) ÷ (x − 1).

Exercise 5.2.2 Perform the division: (x^2 + 5x + 6) ÷ (x + 2).

Exercise 5.2.3 Perform the division: (x^3 + 1)÷(x + 1).

Exercise 5.2.4 Expand and collect like terms:

Exercise 5.2.5 Expand and collect like terms:

5.3

FACTORING I

Definition 5.1 To factor an algebraic expression is to express it as a product.

The idea behind factoring is essentially that of using the distributive law backwards,

ab + ac = a(b + c),

where the dextral quantity is a decomposition into factors of the sinistral quantity.

We will begin by giving examples where the greatest common divisor of the terms is different from 1, and therefore it may be removed. Thus an expression with two terms will be factored as

a + b = ■(□1 + □2),

an expression with three terms will be factored as

a + b + c = ■ (□1+ □2 + □3),

and so on.

Example 5.12 Factor 20a^2b^3 + 24a^3b^2.

Solution: ► The greatest common divisor of 20 and 24 is 4, therefore it can be removed. Also for the two expressions a^2 and a^3, their greatest common divisor is the one with the least exponent, that is a^2. Similarly, the greatest common divisor of b^2 and b^3 is b^2. Thus 4a^2b^2 is a common factor of both terms and the desired factorization has the form

20a^2b^3 + 24a^3b^2 = 4a^2b^2(□1 + □2).

We need to determine □1 and □2. We have

20a^2b^3= 4a^2b^2□1 ⇒ □1 = 5b,

and

The desired factorization is thus

20a^2b^3 + 24a^3b^2 = 4a^2b^2(5b + 6a).

Multiplying, 4a^2b^2(5b + 6a) = 20a^2b^3 + 24a^3b^2.

Example 5.13 Factor −30x^2yza^2b + 36xy^2za + 16x^3y^2z^2b.

Solution: ► The greatest common divisor of −30, 36, 16 is 2. The letters x, y, z appear in all three terms, but a and b do not appear in all three terms. The least power of x appearing in all three terms is 1, and same for y and z. Thus the greatest common divisor of the three terms is 2xyz and the desired factorization is

2xyz(−15xa^2b + 18ya + 8x^2yzb).

Example 5.14 Factor ab + a^2b + ab^2.

Solution: ► The desired factorization is clearly

ab + a^2b + ab^2 = ab(1 + a + b).

Example 5.15 Prove that the sum of two even integers is even.

Solution: ► Let 2a and 2b be two even integers. Then

2a + 2b = 2(a + b),

that is, twice the integer a + b, and therefore an even integer.

Example 5.16 Factor 14a^3b^2(x + y)2(x − y)3 + 20ab^3(x + y)3(x − y)4.

Solution: ► The desired factorization is clearly

14a^3b^2(x + y)2(x − y)3 +20ab^3(x + y)3 (x − y)4 = 2ab^2(x + y)2(x − y)3 (7a^2 + 10b(x + y)(x − y)).

Example 5.17 Resolve into factors: ax^2 + a + bx^2 + b.

Solution: ► In this case, no factor greater than 1 is common to all terms. We have, however,

ax^2 + a + bx^2 + b= a(x^2 + 1) + b(x^2 + 1).

Notice now that x^2 + 1 is a common factor of both summands, and finally,

ax^2 + a + bx^2 + b = a(x^2 + 1) + b(x^2 + 1) = (x^2 + 1)(a + b).

Example 5.18 Resolve into factors: x^3 + x^2 + x + 1.

Solution: ► Like in the preceding problem, we have no factor greater than 1 being common to all terms. Observe that

x^3 + x^2 + x + 1 = x^2(x + 1) + l(x + 1).

Notice now that x + 1 is a common factor of both summands, and finally,

x^3 + x^2 + x + l = x^2(x + l) + l (x + l) = (x + l)(x^2 + l).

Example 5.19 Resolve into factors: x^2 − ax + bx − ab.

Solution: ► We have,

x^2 − ax + bx − ab = x(x − a) + b(x − a) = (x − a)(x + b).

Sometimes it is necessary to rearrange the summands.

Example 5.20 Resolve into factors: 12a^2 − 4ab − 3ax^2 + bx^2.

Solution: ► We have,

12a^2 − 4ab − 3ax^2 + bx^2 = (12a^2 − 3ax^2) − (4ab − bx^2) = 3a(4a − x^2) − b(4a − x^2) = (4a − x^2)(3a − b).

Supplementary Exercises

Exercise 5.3.1 Factor −2a^2b^3 from −4a^6b^5 + 6a^3b^8 − 12a^5b^3 − 2a^2b^3.

Exercise 5.3.2 Factor 2a^3b^3 from 4a^3b^4 − 10a^4b^3.

Exercise 5.3.3 Factor from

Exercise 5.3.4 Factor −1 from −x + 2y − 3z.

Exercise 5.3.5 Prove that the sum of two odd integers is even.

Exercise 5.3.6 Resolve into factors: x^3 − x^2.

Exercise 5.3.7 Factor 125a^4b^5c^5− 45a^5b^3c^4 + 5a^3b^2c^4 − 300a^4b^2c^8 − 10a^3b^2c^5.

Exercise 5.3.8 Resolve into factors: 5x^5 − 10a^7x^3 − 15a^3x^3.

Exercise 5.3.9 Factor 3x^3y +4x^2y^3 − 6x^6 − 10x^4.

Exercise 5.3.10 Factor 3m^6p^4q^2 − 9m^5p^2qx + 3m^7p^3qx +3m^4p^2q − 6m^5p^4qx^2y.

Exercise 5.3.11 Resolve into factors: 38a^2x^5 + 57a^4x^2.

Exercise 5.3.12 Decompose into factors: a^2 + ab + ac + bc.

Exercise 5.3.13 Decompose into factors: a^2 − ab + ac − bc.

Exercise 5.3.14 Decompose into factors: y^3 − y^2 + y − 1.

Exercise 5.3.15 Decompose into factors: 2x^3 + 3x^2 + 2x + 3.

Exercise 5.3.16 Decompose into factors: a^2x +abx + ac + aby + b^2y + bc.

5.4

FACTORING II

In this section, we study factorizations of expressions of the form ax^2 + bx + c, the so-called quadratic trinomials. Let us start with expressions of the form x^2 + bx + c. Suppose that

x^2 + bx + c = (x + p) (x + q).

Then, upon multiplying,

x^2 + bx + c = (x + p) (x + q) = x^2 + (p + q) x + pq.

Thus c = pq and b = p + q, that is, the constant term is the product of two numbers, and the coefficient of x is the sum of these two numbers.

Example 5.21 To factor x^2 + 5x + 6, we look for two numbers whose product is 6 and whose sum is 5. Clearly 2 and 3 are these two numbers and so,

x^2 + 5x + 6 = (x + 2)(x + 3).

You should multiply the last product to verify the equality.

Example 5.22 To factor x^2 +7x + 6, we look for two numbers whose product is 6 and whose sum is 7. Clearly 1 and 6 are these two numbers and so,

x^2 + 7x + 6 = (x + 1)(x + 6).

You should multiply the last product to verify the equality.

Example 5.23 To factor x^2 − 5x + 6, we look for two numbers whose product is 6 and whose sum is −5. Clearly −2 and −3 are these two numbers and so,

x^2 − 5x + 6 = (x − 2)(x − 3).

Example 5.24 To factor x^2 − x − 6, we look for two numbers whose product is −6 and whose sum is −1. Clearly −3 and +2 are these two numbers and so,

x^2 − x − 6 = (x − 3)(x + 2).

Example 5.25 To factor x^2 + x − 6, we look for two numbers whose product is −6 and whose sum is +1. Clearly +3 and −2 are these two numbers and so,

x^2 + x − 6 = (x + 3)(x − 2).

Example 5.26 Following are additional examples to inspect:

Example 5.27 Following are additional examples to inspect:

Example 5.28 Following are additional examples to inspect:

In an instance when the coefficient of the x^2 is different from 1, more cases must be considered. The exercise is more difficult, but the method is essentially the same.

Example 5.29 By inspection,

Example 5.30 By inspection,

The inquiring reader may wonder whether all quadratic trinomials ax^2 + bx + c factor in the manner above, that is, into linear factors for which all coefficients are integers. The answer is no. For example, if x^2 − 2 does not factor into linear factors whose coefficients are integers because there are no two integers whose product is −2 and whose sum is 0. If we augment our choice for the coefficients for linear factors, then we may write

but we will not consider these cases in this book.

Supplementary Exercises

Exercise 5.4.1 Decompose into factors: a^2 − 11a + 30.

Exercise 5.4.2 Decompose into factors: a^2 − 38a + 361.

Exercise 5.4.3 Decompose into factors: a^4b^4 + 37a^2b^2 + 300.

Exercise 5.4.4 Decompose into factors: x^2 − 23x + 132.

Exercise 5.4.5 Decompose into factors: x^4 − 29x^2 + 204.

Exercise 5.4.6 Decompose into factors: x^2 + 35x + 216.

Exercise 5.4.7 Resolve into factors: 5x^2 + 17x + 6.

Exercise 5.4.8 Resolve into factors: 14x^2 + 29x − 15.

5.5

SPECIAL FACTORIZATIONS

Recall the special products we studied in the Chapter 4 on multiplcation. We list them here for easy reference.

Difference of Squares

x^2 − y^2 = (x − y)(x + y).

Difference of Cubes

x^3 − y^3 = (x − y)(x^2 + xy + y^2).

Sum of Cubes

x^3 + y^3 = (x + y)(x^2 − xy + y^2).

Square of a Sum

x^2 + 2xy + y^2 = (x + y)2.

Square of a Difference

x^2 − 2xy + y^2 = (x − y)2.

We will use these special factorizations and the methods of the preceding sections to treat more complicated factorization problems.

Example 5.31 To factor x^4 − 9y^2, observe that it is a difference of squares, and so

x^4 − 9y^2 = (x^2 − 3y)(x^2 + 3y).

Example 5.32 To factor x^3 − 1, observe that it is a difference of cubes, and so

x^3 − 1 = (x − 1)(x^2 + x + 1).

Sometimes we need to use more than one method.

Example 5.33 To factor x^3 − 4x, factor a common factor and then the difference of squares:

x^3 − 4x = x(x^2 − 4) = x(x − 2)(x + 2).

Example 5.34 To factor x^4 − 81, observe that there are two difference of squares:

x^4 − 81 = (x^2 − 9)(x^2 + 9) = (x − 3)(x + 3)(x^2 + 9).

Example 5.35 To factor x^6 − 1, observe that there is a difference of squares and then a sum and a difference of cubes:

x^6 − 1 = (x^3 − 1)(x^3 + 1) = (x − 1)(x^2 + x + 1)(x + 1)(x^2 −x + 1).

The following method is useful to convert some expressions into differences of squares.

Example 5.36 We have

Example 5.37 We have

This trick is often used in factoring quadratic trinomials, where it is often referred to as the technique of completing the square. We will give some examples of factorizations that may also be obtained using the trial− and -error method of the preceding section.

Example 5.38 We have

Example 5.39 We have

The techniques learned may be used to solve some purely arithmetic problems.

Example 5.40 Find the prime factor greater than 9000 of

1, 000, 002, 000, 001.

Solution: ► We have

Therefore, the prime factor greater than 9000 is 9901.

Supplementary Exercises

Exercise 5.5.1 Given that x +2y = 3 and x −2y = −1, find x^2 − 4y^2.

Exercise 5.5.2 Resolve into factors: x^4 − 16.

Exercise 5.5.3 Resolve into factors: (a + b)2 − c^2.

Exercise 5.5.4 Resolve into factors: x^3 − x.

Exercise 5.5.5 Find all positive primes of the form n^3− 8, where n is a positive integer.

5.6

RATIONAL EXPRESSIONS

To reduce an algebraic fraction, we must first factor it.

Example 5.41 Reduce the fraction

Solution: ► We have

Example 5.42 Express the fraction in lowest terms.

Solution: ► We have

To add or subtract algebraic fractions, we first write them in common denominators. We will prefer to expand all products and collect like terms.

Example 5.43 Add:

Solution: ► We have,

Example 5.44 Add:

Solution: ► We have,

Example 5.45 Gather all the fractions:

Solution: ► We have,

Multiplication and division of algebraic fractions is carried out in a manner similar to the operations of arithmetical fractions.

Example 5.46 Multiply:

Solution: ► We have,

Example 5.47 Divide:

Solution: ► We have,

Example 5.48 The sum of two numbers is 7 and their product 21. What is the sum of their reciprocals?

Solution: ► Let x, y be the numbers. One has x + y = 7, xy = 21, therefore

Example 5.49 Prove that

Solution: ► Proceeding from the innermost fraction

Supplementary Exercises

Exercise 5.6.1 Add:

Exercise 5.6.2 Subtract:

Exercise 5.6.3 Add:

Exercise 5.6.4 Subtract:

Exercise 5.6.5 Gather the fractions:

Exercise 5.6.6 Gather the fractions:

Exercise 5.6.7 Gather the fractions:

Exercise 5.6.8 Gather the fractions:

Exercise 5.6.9 Gather the fractions:

Exercise 5.6.10 Gather the fractions:

Exercise 5.6.11 The sum of two numbers is 7 and their product 21. What is the sum of the squares of their reciprocals?

PART 4

EQUATIONS

CHAPTER   6

LINEAR EQUATIONS IN ONE-VARIABLE EXPRESSIONS

This chapter applies the mathematical operations of addition, subtraction, multiplication, and division in algebraic linear equations. We introduce linear equations in sections on Simple Equations, Miscellaneous Linear Equations, and Word Problems.

6.1

SIMPLE EQUATIONS

Definition 6.1 An identity is an assertion that is true for all values of the variables involved.

Example 6.1 From the square of the sum formula we get (x + 1)2 = x^2 + 2x + 1. Since this is valid for all real values of x, this is an identity.

Example 6.2 The assertion 4 + x = 6 − x is not an identity. For example, if x = 0 we have 4 + 0 = 6 − 0, which is patently false. For x = 1 we have, however, 4 + 1 = 6 − 1, which is true. Thus the assertion is sometimes true and sometimes false.

Example 6.3 The assertion x = x + 1 is never true. How could a real number be equal to itself plus 1?

Definition 6.2 An equation of condition (or equation, for short), is an assertion that is true for only particular values of the variables employed.

Example 6.2 is an example of an equation.

Definition 6.3 The variable whose value must be found is called the unknown variable. The process of finding a value that satisfies the equation is called solving the equation.

In this section our main focus will be linear equations with one unknown variable.

Definition 6.4 A linear equation in the unknown x is an equation of the form ax + b = c, where a ≠ 0, and b, c are real numbers.

In order to solve simple equations, we will consider the following axioms:

Axiom 6.1 If equals are added to equals, the sums are equal.

Axiom 6.2 If equals are multiplied by equals, the products are equal.

In the following examples we will use the symbol ⇒, which is read “implies.”

Example 6.4 Solve for x:

x − 3 = −9

Solution: ►

Verification:

Example 6.5 Solve for x:

x + 3 = −9

Solution: ►

Verification:

Example 6.6 Solve for x:

x + 3a = −9a

Solution: ►

Verification:

Example 6.7 Solve for x:

x − a + b = 2a + 3b

Solution: ►

Verification:

Example 6.8 Solve for x:

2x = −88

Solution: ►

◄

Verification:

Example 6.9 Solve for x:

2ax = 4a

Solution: ►

Verification:

Example 6.10 Solve for x:

2ax = 4a^2

Solution: ►

Verification:

Example 6.11 Solve for x:

Solution: ►

Verification:

Example 6.12 Solve for x:

Solution: ►

Verification:

Example 6.13 Solve for x:

Solution: ►

Verification:

Example 6.14 Solve for x:

Solution: ►

Verification:

Example 6.15 Solve for x:

0 = 12x

Solution: ►

Verification:

Example 6.16 Solve for x:

3x − 5 = 28

Solution: ►

Verification:

Example 6.17 Solve for x:

ax − 2a = 2ab − a

Solution: ►

Verification:

Supplementary Exercises

Exercise 6.1.1 Solve for the variable z: z + 4 = −4.

Exercise 6.1.2 Solve for the variable z: z − 4 = −4.

Exercise 6.1.3 Solve for the variable z: 4z = −44.

Exercise 6.1.4 Solve for the variable z: 4z = −3.

Exercise 6.1.5 Solve for the variable

Exercise 6.1.6 Solve for the variable

Exercise 6.1.7 Solve for the variable

Exercise 6.1.8 Solve for the variable

Exercise 6.1.9 Solve for the variable z: 5z − 3 = 22.

Exercise 6.1.10 Solve for the variable

Exercise 6.1.11 Solve for the variable N: 3N = 3A.

Exercise 6.1.12 Solve for the variable N:

Exercise 6.1.13 Solve for the variable N:

Exercise 6.1.14 Solve for the variable N: N − 5A = 7A+A^2.

Exercise 6.1.15 Solve for the variable N: N + 8X = −9X.

Exercise 6.1.16 Solve for the variable N:

Exercise 6.1.17 Solve for the variable N:

Exercise 6.1.18 Solve for the variable N:

Exercise 6.1.19 Solve for the variable N: N + 2 = A + 14.

6.2

MISCELLANEOUS LINEAR EQUATIONS

Example 6.18 Solve for x:

Solution: ► The denominators are 3, 4 and 2, and their least common multiple is (3, 4, 2) = 12. We multiply both sides of the equation by this least common multiple, and so

Example 6.19 Solve for a:

Solution: ► The denominators are 6 and 15, and their least common multiple is (6, 15) = 30. We multiply both sides of the equation by this least common multiple, and so

Example 6.20 Solve for x:

Solution: ◄ Cross-multiplying,

Example 6.21 Write the infinitely repeating decimal 0.345 = 0.345454545 … as the quotient of two natural numbers.

Solution: ► The trick is to obtain multiples of x = 0.345454545 … so that they have the same infinite tail, and then subtract these tails, cancelling them out.[1] Observe that

Example 6.22 Find the fraction that represents the repeating decimal

4.321 = 4.32121212121…

Solution: ► Put x = 4.32121212121…. Then 10x = 43.21212121… and 1000x = 4321.2121212121…. Therefore

1000x − 10x = 4321.2121212121… − 43.212121212121 … 4278.

Finally

Example 6.23 Find the sum of the first hundred positive integers, that is, find

1 + 2 + 3 + · · · + 99 + 100.

Solution: ► Notice that this is Example 1.5. Put

x = 1 + 2 + 3 + · · · + 99 + 100.

The crucial observation is that adding the sum forwards is the same as adding the sum backwards, therefore

x = 100 + 99 + · · · + 3 + 2 + 1.

Adding,

therefore,

Example 6.24 Consider the arithmetic progression

2, 7, 12, 17, ….

Is 302 in this progression? Is 303 in this progression?

Solution: ► Observe that we start with 2 and keep adding 5. Thus

2 = 2 + 5 · 0, 7 = 2 + 5 · 1, 12 = 2 + 5 · 2, 17 = 2 + 5 · 3, …

so the general term has the form 5n + 2, which are the numbers leaving remainder 2 when divided by 5. Since

5n + 2 = 302 ⇒ 5n = 300 ⇒ n = 60,

is an integer, 302 is in this progression. Since

is an not integer, 303 is not in this progression.

Some equations are nonlinear, but may be reduced to a linear equation. Sometimes it is necessary to simplify the expressions involved before solving the equation, as seen in the following examples:

Example 6.25 Solve for x:

(3x + 1)2 + 6 + 18(x + 1)2 = 9x(3x − 2)+65.

Solution: ►

Example 6.26 Solve for x:

(2x + 1)(2x + 6) − 7(x − 2) = 4(x + 1)(x − 1)− 9x

Solution: ►

Example 6.27 Solve for x:

Solution: ► Cross-multiplying,

Supplementary Exercises

Exercise 6.2.1 Solve for x: 2(3x − 4) − 4(2 − 3x) = 1.

Exercise 6.2.2 Solve for x:

Exercise 6.2.3 Solve for x:

Exercise 6.2.4 Solve for x:

Exercise 6.2.5 Solve for x:

Exercise 6.2.6 Solve for x: ax + b = c.

Exercise 6.2.7 Solve for x:

Exercise 6.2.8 Solve for x:

Exercise 6.2.9 Solve for x:

Exercise 6.2.10 Solve for x:

Exercise 6.2.11 Solve for x:

Exercise 6.2.12 Solve for x:

Exercise 6.2.13 Solve for x: 2(3 − x) = 3x − 4

Exercise 6.2.14 Solve for x: (2 − x)(x + 3) = −x(x − 4)

Exercise 6.2.15 Solve for x: (x − a)b = (b − x)a.

Exercise 6.2.16 Solve for x:

Exercise 6.2.17 Write the infinitely repeating decimal

0.123 = 0.123123123… as the quotient of two positive integers.

6.3

WORD PROBLEMS

Example 6.28 Find two numbers whose sum is 28, and whose difference is 4.

Solution: ► Let x be one of the numbers, then the other number is 28 − x. Then we have

The numbers are x = 16 and 28 − x = 28 − 16 = 12.

Example 6.29 Divide $47 between Peter, Paul, and Mary, so that Peter may have $10 more than Paul, and Paul $8 more than Mary.

Solution: ► Let p be Paul’s amount in dollars. Then Peter has p + 10 dollars and Mary has p − 8 dollars. Then we have

Thus Paul has $15, Peter has $25, and Mary has $7.

Example 6.30 The sum of three consecutive odd integers is 609. Find the numbers.

Solution: ► Let the numbers be x − 2, x, x + 2. Then

The numbers are x − 2 = 201, x = 203, and x + 2 = 205.

Example 6.31 A glass of milk costs 40 cents more than a loaf of bread, but 50 cents less than a glass of juice. If the cost of the three items, in cents, is 730, what is the price of each item, in cents?

Solution: ► Let m be the price of a glass of milk in cents. Then bread costs m − 40 cents and juice costs m + 50 cents. This gives

m+(m − 40) + (m + 50) = 730 ⇒ 3m + 10 = 730 ⇒ m = 240.

Thus milk costs 240 cents, bread costs 200 cents and juice costs 290 cents.

Example 6.32 Currently, the age of a father is four times the age of his son, but in 24 years from now it will only be double. Find their ages.

Solution: ► Let s be the current age of the son. Then the current age of the father is 4s. In 24 years the son will be s + 24 and the father will be 4s + 24 and we will have

4s + 24 = 2(s + 24).

This gives

4s + 24 = 2s + 48 ⇒ 4s − 2s = 48 − 24 ⇒ s = 12.

Thus the son is currently 12 years old and the father is currently 48 years old.

Supplementary Exercises

Exercise 6.3.1 Six times a number increased by 11 is equal to 65. Find the number.

Exercise 6.3.2 Find a number which when multiplied by 11 and then diminished by 18 is equal to 15.

Exercise 6.3.3 If 3 is added to a number, and the sum multiplied by 12, the result is 84. Find the number.

Exercise 6.3.4 The sum of eleven consecutive integers is 2002. Find them.

Exercise 6.3.5 One number exceeds another by 3, and their sum is 27. Find them.

Exercise 6.3.6 Find two numbers whose sum is 19, such that one shall exceed twice the other by 1.

Exercise 6.3.7 Split $380 among Peter, Paul and Mary, so that Paul has $30 more than Peter, and Mary has $20 more than Paul.

Exercise 6.3.8 How much pure water should be added to 100 grams of 60% acid solution to make a 20% acid solution?

Exercise 6.3.9 Jane’s age is twice Bob’s age increased by 3. Bill’s age is Bob’s age decreased by 4. If the sum of their ages is 27, how old is Bill?

Exercise 6.3.10 The average of six numbers is 4. A seventh number is added and the new average increases to 5. What was the seventh number?

Exercise 6.3.11 Bill currently has five times as much money as Bob. If he gives $20 to Bob, then Bill will only have four times as much. Find their current amounts.

Exercise 6.3.12 Find a number so that six sevenths of it exceed four fifths of it by 2.

Exercise 6.3.13 The difference between two numbers is 8. If we add 2 to the larger number we obtain 3 times the smaller one. Find the numbers.

Exercise 6.3.14 Find two numbers whose difference is 10, and whose sum equals twice their difference.

Exercise 6.3.15 I bought a certain amount of avocados at four for $2; I kept a fifth of them, and then sold the rest at three for $2. If I made a profit of $2, how many avocados did I originally buy?

Exercise 6.3.16 Find a number whose fourth, sixth, and eighth of its value add up to 13.

Exercise 6.3.17 A fifth of the larger of two consecutive integers exceeds a seventh of the smaller by 3. Find the integers.

Exercise 6.3.18 I bought a certain number of oranges at three for a dollar and five sixths of that number at four for a dollar. If I sold all my oranges at sixteen for six dollars, I would make a profit of three and a half dollars. How many oranges did I buy?

Exercise 6.3.19 A piece of equipment is bought by a factory. During its first year, the equipment depreciates a fifth of its original price. During its second year, it depreciates a sixth of its new value. If its current value is $56,000, what was its original price?

Exercise 6.3.20 The difference of the squares of two consecutive integers is 665. Find the integers.

Exercise 6.3.21 In how many different ways can one change 50 cents using nickels, dimes or quarters?

Exercise 6.3.22 Five burglars stole a purse with gold coins. The burglars each took different amounts according to their meanness, with the meanest among the five taking the largest amount of coins and the meekest of the five taking the least amount of coins. Unfair sharing caused a fight that was brought to an end by an arbitrator. He ordered that the meanest burglar should double the shares of the other four burglars below him. Once this was accomplished, the second meanest burglar should double the shares of the three burglars below him. Once this was accomplished, the third meanest burglar should double the shares of the two burglars below him. Once this was accomplished, the fourth meanest burglar should double the shares of the meekest burglar. After this procedure was terminated, each burglar received the same amount of money. How many coins were in a purse if the meanest of the burglars took 240 coins initially?

1 That this cancellation is meaningful depends on the concept of convergence.

CHAPTER   7

QUADRATIC EQUATIONS IN ONE-VARIABLE EXPRESSIONS

This chapter applies the mathematical operations of addition, subtraction, multiplication, and division in algebraic quadratic equations. We introduce this concept through these sections: Quadratic Equations, Complete the Square, and Quadratic Formula.

7.1

QUADRATIC EQUATIONS

Definition 7.1 A quadratic equation in the unknown x is an equation in which a variable is squared or written to the second power of the form

ax^2 + bx + c = 0

where a ≠ 0, b, c are real numbers.

We will circumscribe ourselves to the study of quadratic trinomials ax^2 + bx + c that are amenable to the factorizations studied earlier. When a variable is squared in an equation, and if the product of two real numbers is zero, then at least one of them must be zero. In symbols,

ab = 0 ⇒ a = 0

or

b = 0.

Quadratic equations are solved by factoring their terms. Two expressions when multiplied will equal the quadratic equation.

Example 7.1 Solve the quadratic equation x^2 − x = 0 for x.

Solution: ► We first factor the expression x^2 − x = x(x − 1). Now, x(x − 1) = 0 is the product of two real numbers giving 0, therefore either x = 0 or x − 1 = 0. Thus either x = 0 or x = 1. One can easily verify now that 0^2 − 0 = 0 and l^2 −1 = 0.

Example 7.2 Solve the quadratic equation x^2 + 2x − 3 = 0 for x.

Solution: ► Factoring,

x^2 + 2x − 3 = 0 ⇒ (x + 3)(x − 1) = 0

When a quadratic equation is set to equal zero and the factors are known, you can find the solutions by appling the zero-product rule, where any number when multiplied by 0 is equal to 0.

Check: Substitute each solution (x = −3 and x = 4) into the original equation. Both solutions equal zero (−3 + 3 = 0, and 1 − 1 = 0).

To multiply expressions, multiply each term in the second expression by each term in the first expression. The acronym to remember is FOIL, where you multiply the terms in the following order: First, Outside, Inside, Last.

(x + 3)(x − 1) = x(x) + x(−1) + 3(x) + 3(−1) = x^2 − x + 3x − x + 3x − 3 = x^2 + 2x − 3

Example 7.3 Solve the quadratic equation 8x^2 − 2x = 15 for x.

Solution: ► First, transform the equation into 8x^2 − 2x −15 = 0. Now, factoring,

Some equations can be reduced to quadratic equations after proper massaging

Example 7.4 Solve for x:

Solution: ► Cross-multiplying,

The argument is that a product of real numbers is zero if and only if at least one of its factors can be used with equations of higher degree.

Example 7.5 Solve for x: x^3 − 4x = 0.

Solution: ► Factoring a common factor and then a difference of squares,

Supplementary Exercises

Exercise 7.1.1 Solve for x: x^2 − 4 = 0.

Exercise 7.1.2 Solve for x: x^2 − x − 6 = 0.

Exercise 7.1.3 Solve for x: x^2 +x − 6 = 0.

Exercise 7.1.4 Solve for x: x^2 − 4x = 5.

Exercise 7.1.5 Solve for x: x^2 = 1.

Exercise 7.1.6 If eggs had cost x cents less per dozen, it would have cost 3 cents less for x + 3 eggs than if they had cost x cents more per dozen. What is x?

Exercise 7.1.7 A group of people rents a bus for an excursion for $2,300, with each person paying an equal share. The day of the excursion, it is found that six participants do not show up, which increases the share of those present by $7.50. What was the original number of people renting the bus?

Exercise 7.1.8 Let

where there are an infinite number of fractions. Prove that x^2 − 6x − 1 = 0.

Exercise 7.1.9 Assume that there is a positive real number x such that

where there is an infinite number of x’s. What is the value of x?

Exercise 7.1.10 Factor:

x^2 + 20x + 19,

Exercise 7.1.11 Factor:

x^2 + 14x + 13,

Exercise 7.1.12 Factor:

z^2 + 14z + 49,

Exercise 7.1.13 Factor:

x^2 + 2x + 1,

Exercise 7.1.14 Factor:

x^2 + 6x + 9,

Exercise 7.1.15 Factor:

x^2 − 10x + 25,

Exercise 7.1.16 Factor:

x^2 − 6x + 9,

Exercise 7.1.17 Factor:

x^2 − 12x + 36,

Exercise 7.1.18 Solve for z. Write your answers as whole numbers and in the simplest form.

z(z − 2) = 0,

Exercise 7.1.19 Solve for n. Write your answers in the simplest form.

(3n + 4)(n − 2) = 0,

Exercise 7.1.20 Solve for t. Write your answers in the simplest form.

(5t − 9)(t + 2) = 0,

Exercise 7.1.21 Solve for x. Write your answers in the simplest form.

(5x + 9)(x − 2) = 0,

Exercise 7.1.22 Solve for n. Write your answers in the simplest form.

(2n + 3)(n − 13) = 0,

7.2

COMPLETE THE SQUARE

With quadratic equations like x^2 + bx, you can complete the square by adding

Example 7.6 Complete the square by filling in the number that makes the polynomial a perfect square.

x^2 + 44x + _____

Solution: ► Add to complete the square.

This quadratic equation can be written in a complete square, (x + 22)2 The perfect square was complete with the number 484.

Supplementary Exercises

Exercise 7.2.1 Complete the square by filling in the number that makes the polynomial a perfect square.

x^2 + 30x + _____

Exercise 7.2.2 Complete the square by filling in the number that makes the polynomial a perfect square.

x^2 − 6x + _____

Exercise 7.2.3 Complete the square by filling in the number that makes the polynomial a perfect square.

h^2 + 10h + _____

Exercise 7.2.4 Complete the square by filling in the number that makes the polynomial a perfect square.

x^2 − 16x + _____

Exercise 7.2.5 Complete the square by filling in the number that makes the polynomial a perfect square.

p^2 + 36p + _____

Exercise 7.2.6 Complete the square by filling in the number that makes the polynomial a perfect square.

x^2 + 28x + _____

Exercise 7.2.7 Complete the square by filling in the number that makes the polynomial a perfect square.

x^2 − 26x + _____

Exercise 7.2.8 Complete the square by filling in the number that makes the polynomial a perfect square.

h^2 − 10h + _____

Exercise 7.2.9 Complete the square by filling in the number that makes the polynomial a perfect square.

x^2 + 2x + _____

Exercise 7.2.10 Complete the square by filling in the number that makes the polynomial a perfect square.

x^2 − 2x + _____

7.3

QUADRATIC FORMULA

A quadratic equation ax^2 + bx + c = 0, where a ≠ 0, can be solved using the following quadratic formula:

Example 7.7 Use the quadratic formula to solve 4x^2 − 9x + 3 = 0,

Simplify and round to nearest hundredth

Example 7.8 Use the quadratic formula to solve p^2 − 8p + 6 = 0,

Simplify and round to nearest hundredth

Example 7.9 Use the quadratic formula to solve 6n^2 − 2n − 8 = 0,

Supplementary Exercises

Exercise 7.3.1 Find the solution for the following quadratic equation using the quadratic formula.

p^2 + 7p − 9 = 0,

Exercise 7.3.2 Find the solution for the following quadratic equation using the quadratic formula.

4x^2 − 3x − 8 = 0,

Exercise 7.3.3 Find the solution for the following quadratic equation using the quadratic formula.

3f^2 − 7f − 7 = 0,

Exercise 7.3.4 Find the solution for the following quadratic equation using the quadratic formula.

2z + 7z + 5 = 0,

Exercise 7.3.5 Find the solution for the following quadratic equation using the quadratic formula.

2q^2 + 4q + 2 = 0,

Exercise 7.3.6 Find the solution for the following quadratic equation using the quadratic formula.

2n^2 − 9n − 4 = 0,

Exercise 7.3.7 Find the solution for the following quadratic equation using the quadratic formula.

4x^2 − 3x − 9 = 0,

Exercise 7.3.8 Find the solution for the following quadratic equation using the quadratic formula.

5u^2 + 9u − 7 = 0,

Exercise 7.3.9 Find the solution for the following quadratic equation using the quadratic formula.

2w^3 − 9w^2 + 8w = 0,

Exercise 7.3.10 Find the solution for the following quadratic equation using the quadratic formula.

2m^3 + 8m^2 + 2m = 0,

PART 5

INEQUALITIES

CHAPTER   8

LINEAR INEQUALITIES

This chapter applies the mathematical operations of addition, subtraction, multiplication, and division in algebraic linear inequality equations. We introduce linear inequalities in sections covering Intervals, One-Variable Linear Inequalities, and Using One-Variable Linear Inequalities to Solve Word Problems.

8.1

INTERVALS

Linear inequalities are statements that mathematically express that two expressions or numbers are not equal in value. Solving linear inequalities is very similar to solving linear equations discussed earlier. The symbols below are used to show the relationship between the two sides of a linear inequality statement.

We will simply state the useful properties for the symbol <, as the same properties hold for the other three symbols (>, ≤, and ≥).

Axiom 8.1 Let a, b, c be real numbers. Then

1.

If a < b and b < c, then a < c, that is, inequalities are transitive.

2.

If a < b and c < d, then a + c < b + d, that is, addition preserves the inequality sense.

3.

If a < b and c > 0, then ac < bc, that is, multiplication by a positive factor preserves the inequality sense.

4.

If a < b and c < 0, then ac > bc, that is, multiplication by a negative factor reverses the inequality sense.

Definition 8.1 An interval is a set I of real numbers with the following property: for any two different elements x, y in I, any number a in between x and y also belongs to I.

From the definition we see intervals have infinitely many elements. For example, consider the interval in Figure 8.11. Every number between −2 and 1 is in this interval, so in particular, −1.5 belongs to it, so does −1.51, −1.501, −1.5001, etc.

Some examples of intervals are shown in Figures 8.1 through 8.12. If we define the length of the interval to be the difference between its final point and its initial point, then intervals can be finite, like those in Figures 8.9 through 8.12 or infinite, like those in Figures 8.1 through 8.8. In either case, intervals contain an infinite number of elements.

FIGURE 8.1 x < −1.

FIGURE 8.2 x > −1.

FIGURE 8.3 x < 1.

FIGURE 8.4 x > 1.

FIGURE 8.5 x ≤ −1.

FIGURE 8.6 x ≥ −1.

FIGURE 8.7 x ≤ 1.

FIGURE 8.8 x ≥ 1.

FIGURE 8.9 −1 ≤ x ≤ 2.

FIGURE 8.10 −1 < x <2.

FIGURE 8.11 −2 < x ≤ 1.

FIGURE 8.12 −1 ≤ x < 1.

8.2

ONE-VARIABLE LINEAR INEQUALITIES

We are now interested in solving linear inequalities in one-variable. The process will closely resemble that for solving linear equations in one-variable, with the caveat that if we ever divide or multiply by a negative quantity, then the sense of the inequality changes. This can be done by simply isolating the unknown variable on the left side of the inequality and moving the rest of the expresson to the right side of the inequality. To do this, make sure you perform the same mathematical operation on both sides of the inequality

Example 8.1 Solve the inequality and graph its solution set.

2x − 3 < −13

Solution: ► Add 3 to both sides of the inequality

►

Divide by 2, which is a positive quantity. The sense of the inequality is retained, and we gather,

x < − 5

FIGURE 8.13 Graphical solution for example 267: x < −5.

Example 8.2 Solve the inequality and graph its solution set.

− 2x − 3 ≤ − 13

Solution: ► Add 3 to both sides of the inequality

►

Divide by 2, which is a positive quantity. The sense of the inequality is reversed, and we gather,

FIGURE 8.14 Graphical solution for example 8.2 x ≥ −5.

Supplementary Exercises

Exercise 8.2.1 Find the set of all real numbers x satisfying the inequality and graph its solution set.

Exercise 8.2.2 Find the set of all real numbers x satisfying the inequality and graph its solution set.

4(x + 1) ≤ x − 5,

Exercise 8.2.3 Find the set of all real numbers x satisfying the inequality and graph its solution set.

8.3

USING ONE-VARIABLE LINEAR INEQUALITIES TO SOLVE WORD PROBLEMS

Inequalities can be used to solve real-life problems. The procedure to solve any word problems that have a range of answers is to set the problems up with inequalities, as follows:

1.

Assign a variable to the unknown numbers in the word problem.

2.

Set a relationship between the variable and the other numbers in the word problem.

3.

Use the appropriate inequality expression to define the relationship between the unknown variable and the other numbers.

Example 8.3 Mark wants to save at least $1,200 for a down payment on a used vehicle. What is the smallest amount of money he can set aside each month so that he can reach his goal in 6 months?

Solution: ► Let x be the amount of money that Mark will save each month, then 6x represents the amount he will save in 6 months. His savings must be greater than or equal to $1,200.

Solution: ► Mark should set a goal of saving a minimum of $200 per month.

Check: ► Replace x with 200 or 6(200) = 1200. If Mark saves $200 for 6 months, he will have $1,200, which is enough for the down payment on a used vehicle.

It is not necessary to use linear inequality statement to solve this type of problem. Most people will solve the problem using equations or their common sense. You know that $200 times 6 equals $1,200. Using common sense, you know that Mark can save even more if he puts away even a larger amount in savings each month. Thus, his savings must be greater than or equal to $200.

There are times when you need to set up word problems in the best way possible using linear inequality statements. The best setup for the following example is to use a distance formula.

Example 8.4 Andreya computes to work from Raleigh to Chapel Hill daily, a distance of 30 miles. If she drives no faster than 60 miles per hour, what is the total time in hours that the trip will take her each day?

Solution: ► To solve this problem, use the distance formula d = rt, where d is the distance, r is the rate of travel (speed), and t is the time in hours. The distance Andreya travels is 30 miles (d = 30) and the rate of travel is 60 miles per hours (r = 60). Therefore 60t represents distance. Her traveling speed must be less than or equal to 60.

In other words, if Andreya drives any slower than 60 miles per hour, her commute will take longer than hour or 30 minute, 60t ≥ 30.

Supplementary Exercises

Exercise 8.3.1 An unknown number is doubled and decreased in its value by 50 is less than 100. Which of the following is the correct statement?

(A)

x < 75

(B)

x > 75

(C)

x < 25

(D)

x > 25

Exercise 8.3.2 Dalya works at least 40 hours a week as a cashier at a department store, making $10.75 per hour. Write the inequality statement that expresses her biweekly take home pay.

Exercise 8.3.3 The perimeter of a square is larger than 72 inches. Set up the inequality expression to solve for the size of the square (s) in inches?

Exercise 8.3.4 A tennis ball is projected upward in the air with an intial velocity of V = 75 − 30t, where t is the time it travels in seconds. When will the velocity of the ball reach between 30 and 60 feet per second?

Exercise 8.3.5 Robert invests $10,000 into a financial portfolio, where 3% of his investment is placed into a 3% simple interest account. The remainder of his investment is placed in mutual funds that return an average of 11% interest. What is the least amount of investment that Robert can invest in mutual funds to receive $2,700 interest in a year?

8.4

COMPOUND LINEAR INEQUALITIES

Compound linear inequalities are simply inequality expressions that have more than one set of constraints. The process will closely resemble that for solving linear inequality equations in one-variable, with the caveat that if we divide or multiply by a negative quantity, then the sense of the inequality changes. This can be done by simply isolating the unknown variable on the left side of the inequality and moving the rest of the expresson to the right side of the inequality. To do this, make sure you perform the same mathematical operation on both sides of the inequality.

Example 8.5 Solve the compound inequalities and graph their solution set.

x − 15 < − 1

or

x − 13 ≥ 6

Solution: ► Treat each inequality expression separately. For example, add 15 to both sides of the inequality for the first expression and 13 to the other expression.

FIGURE 8.15 Graphical solution for Example 8.5: x < 14 or x ≥ 19.

Example 8.6 Solve the compound inequalities and graph their solution set.

x + 11 ≤ 15

or

x + 1 ≥ 13

Solution: ► Treat each inequality expression separately. For example, subtract 11 from both sides of the inequality for the first expression and subtract 1 from the other expression.

FIGURE 8.16 Graphical solution for Example 8.6: x ≤ 4 or x ≥ 12.

Example 8.7 Solve for n and graph the solution set.

n + 15 ≤ 15

or

n − 8 ≥ − 3

Solution: ► Treat each inequality expression separately. For example, subtract 15 from both sides of the inequality for the first expression and add 8 to the other expression.

FIGURE 8.17 Graphical solution for Example 8.7: x ≤ 4 or x ≥ 12.

Supplementary Exercises

Exercise 8.4.1 Find the set of all real numbers x satisfying the inequalities and graph solution set.

x − 10 > − 15

or

x + 1 < − 14,

Exercise 8.4.2 Find the set of all real numbers x satisfying the inequalities and graph solution set.

x − 15 > − 7

or

x + 19 < 13,

Exercise 8.4.3 Find the set of all real numbers x satisfying the inequalities and graph solution set.

16 ≥ x + 10 ≥ 13,

Exercise 8.4.4 Find the set of all real numbers x satisfying the inequalities and graph solution set.

12 ≥ x + 17 ≥ 4,

Exercise 8.4.5 Find the set of all real numbers y satisfying the inequalities and graph solution set.

−7 > y − 17 > −9,

Exercise 8.4.6 Find the set of all real numbers m satisfying the inequalities and graph solution set.

−14 > m − 19 > −17,

Exercise 8.4.7 Find the set of all real numbers b satisfying the inequalities and graph solution set.

b − 4 ≤ −5

or

b + 1 > 9,

Exercise 8.4.8 Find the set of all real numbers a satisfying the inequalities and graph solution set.

a − 5 ≥ 7

or

a > 10,

Exercise 8.4.9 Find the set of all real numbers z satisfying the inequalities and graph solution set.

z − 8 > −8

or

z − 12 < −22,

Exercise 8.4.10 Find the set of all real numbers c satisfying the inequalities and graph solution set.

−15 > c − 13 > −17,

Exercise 8.4.11 Find the set of all real numbers x satisfying the inequalities and graph solution set.

x − 1 ≥ 15

or

x + 8 ≤ 12,

Exercise 8.4.12 Find the set of all real numbers y satisfying the inequalities and graph solution set.

y + 9 ≥ 15

or

y + 4 < −2,

Exercise 8.4.13 Find the set of all real numbers x satisfying the inequalities and graph solution set.

−5 ≤ 2x + 1 ≤ 11,

Exercise 8.4.14 Find the set of all real numbers n satisfying the inequalities and graph solution set.

−3 ≤ n + 7 ≤ 17,

Exercise 8.4.15 Find the set of all real numbers p satisfying the inequalities and graph solution set.

17 > p + 16 > 11,

Exercise 8.4.16 Find the set of all real numbers x satisfying the inequalities and graph solution set.

x − 2 < 1

or

x − 12 ≥ −8,

Exercise 8.4.17 Find the set of all real numbers z satisfying the inequalities and graph solution set.

11 < z + 2 ≤ 17,

Exercise 8.4.18 Find the set of all real numbers p satisfying the inequalities and graph solution set.

−5 > p > −8.

Exercise 8.4.19 Find the set of all real numbers n satisfying the inequalities and graph solution set.

n − 9 > 7

or

v − 20 < −6,

Exercise 8.4.20 Find the set of all real numbers υ satisfying the inequalities and graph solution set.

v + 6 ≤ 7

or

v + 6 > 14,

APPENDIX A: REVIEW EXERCISES

CHAPTER 2 REVIEW EXERCISES

1.

Which of the following expressions best represents: “the excess of a number over 3”?

(a)

(b)

x − 3

(c)

3x

(d)

(e)

none of these

2.

Which of the following expressions best represents: “the sum of a number plus thrice another being diminished by 3”?

(a)

(b)

x + 3y − 3

(c)

x + 3(y − 3)

(d)

x + 2y − 3

(e)

none of these

3.

The algebraic expression can be translated as:

(a)

“The square of a number is increased by the reciprocal of the number.”

(b)

“The square of a number is reduced by the reciprocal of the number.”

(c)

“Twice a number is increased by the reciprocal of the number.”

(d)

“Twice a number is decreased by the reciprocal of the number.”

(e)

none of these

4.

(a)

(b)

(c)

6

(d)

(e)

none of these

5.

(−100) ÷ (10) ÷ (−10) =

(a)

1

(b)

10

(c)

100

(d)

−1

(e)

none of these

6.

(a)

(b)

(c)

−1

(d)

(e)

none of these

7.

(a)

(b)

(c)

(d)

(e)

none of these

8.

(a)

1

(b)

−1

(c)

(d)

(e)

none of these

9.

1024 ÷ 512 × 2 =

(a)

1

(b)

2

(c)

4

(d)

8

(e)

none of these

10.

(a)

1

(b)

(c)

2

(d)

(e)

none of these

11.

1 − (−2)3 =

(a)

−7

(b)

9

(c)

−9

(d)

7

(e)

none of these

12.

You have t $10 and w $20 bills. How much money, in dollars, do you have?

(a)

t + w

(b)

10t + 20w

(c)

200tw

(d)

30 + t + w

(e)

none of these

13.

You have t $10 and w $ 20 bills. How many “paper bills” do you have?

(a)

t + w

(b)

10t + 20w

(c)

200tw

(d)

30 + t + w

(e)

none of these

14.

Consider the arithmetic progression 3, 14, 25, 36, …, where 3 is on the first position, 14 is on the second, etc. Which number occupies the 101st position?

(a)

101

(b)

1092

(c)

1103

(d)

1114

(e)

none of these

15.

(a)

(b)

−3

(c)

(d)

−

(e)

none of these

16.

−5^2 + 4 − 9 − (−7) + (−9 + 5 · 3)2 =

(a)

−13

(b)

−1

(c)

63

(d)

13

(e)

none of these

17.

Perform the calculation: 1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + 9 − 10.

(a)

5

(b)

0

(c)

−5

(d)

1

(e)

none of these

18.

Perform the calculation:

(a)

(b)

(c)

(d)

(e)

none of these

19.

Perform the calculation:

(a)

(b)

(c)

(d)

(e)

none of these

CHAPTER 3 REVIEW EXERCISES

20.

Collect like terms: (x + y + 2z) − (x − 2y + z).

(a)

3y + z

(b)

2x − y + 3z

(c)

3y + 3z

(d)

3y − z

(e)

none of these

21.

Which of the following are like terms?

(a)

x^6 and 6x

(b)

2x^2 and −x^2

(c)

j^2kl and jk^2l

(d)

c and c^2

(e)

none of these

22.

2(a − 2b) −3(b − 2a) =

(a)

3a − b

(b)

−5a − 10b

(c)

8a − 7b

(d)

8a + 7b

(e)

none of these

23.

(a)

7x

(b)

7x − 4

(c)

x

(d)

x−4

(e)

none of these

24.

Collect like terms: 2(a − 2b) − (2a − b)

(a)

−3b

(b)

−b

(c)

4a − 5b

(d)

4a − 3b

(e)

none of these

25.

Collect like terms: −2(x^2 − 1) + 4(2 − x − x^2)

(a)

−6x^2 − 4x + 10

(b)

6x^2 + 4x + 10

(c)

−6x^2 + 4x + 10

(d)

−2x^2 − 4x +10

(e)

none of these

26.

Collect like terms: (−2a + 3b − c) + (5a −8b + c) =

(a)

7a − 11b − 2c

(b)

3a − 5b − 2c

(c)

−7a + 11b − 2c

(d)

3a − 5b

(e)

none of these

27.

−a + 2b + 3a − 4b =

(a)

−2a + 2b

(b)

2a − 2b

(c)

−4a − 6b

(d)

2a + 2b

(e)

none of these

28.

(a)

(b)

(c)

(d)

(e)

none of these

29.

(a)

(b)

(c)

(d)

(e)

none of these

30.

(−a + 2b − 3c) − (−a + 2b + 3c) =

(a)

−6c

(b)

4b − 6c

(c)

2a + 4b + 6c

(d)

2a − 6c

(e)

none of these

31.

Collect like terms:

(a)

2x − 2

(b)

2x − 3

(c)

x − 3

(d)

2A − B −2

(e)

none of these

32.

How many of the following expressions are equivalent to 5 + x^2y?

I: x^2y + 5,

II: 5 + yx^2,

III: 5 + xy^2,

IV: yx^2 + 5.

(a)

exactly one

(b)

exactly two

(c)

exactly three

(d)

all four

(e)

none

33.

Collect like terms:

(a)

−11x + 6

(b)

3x + 2

(c)

3x + 6

(d)

−3x + 2

(e)

none of these

CHAPTER 4 REVIEW EXERCISES

34.

(a)

3

(b)

3^12

(c)

(d)

3^7

(e)

none of these

35.

(a)

3

(b)

3^12

(c)

(d)

3^7

(e)

none of these

36.

(3^3)4 =

(a)

3

(b)

3^12

(c)

(d)

3^7

(e)

none of these

37.

3^3 ⋅ 3^4 =

(a)

3

(b)

3^12

(c)

(d)

3^7

(e)

none of these

38.

(a)

3

(b)

3^12

(c)

(d)

3^7

(e)

none of these

39.

(a)

1111

(b)

2222

(c)

4444

(d)

(e)

none of these

40.

Simplify and write with positive exponents only: (ab^2)2(a^3b−4).

(a)

a^5

(b)

a^5b^8

(c)

(d)

a^6

(e)

none of these

41.

Simplify and write with positive exponents only: (ab^2)2 ÷ (a^3b−4).

(a)

a^5

(b)

a^5b^8

(c)

(d)

a^6

(e)

none of these

42.

Perform the calculation:

(a)

(b)

(c)

(d)

(e)

none of these

43.

Perform the calculation:

(a)

0

(b)

20

(c)

(d)

(e)

none of these

44.

Multiply and write with positive exponents only: (x^4 y^7)(x−2 y−8.)

(a)

x^6 y^15

(b)

x^2 y

(c)

(d)

(e)

none of these

45.

Multiply and collect like terms: (x − 2) (x^2 + 2x + 4).

(a)

x^3 − 8

(b)

x^3 − 4x^2 − 4x − 8

(c)

x^3 + 8

(d)

x^3 − 4x^2 − 8

(e)

none of these

46.

Divide and write with positive exponents only

(a)

x^6 y^15

(b)

x^2 y

(c)

(d)

(e)

none of these

47.

(a)

(b)

9

(c)

(d)

(e)

none of these

48.

(x−1 y−2 z^3)−3 =

(a)

x^4 y^5 z^0

(b)

(c)

(d)

(e)

none of these

49.

(x^4 y^3 z^2) (x^2 y^4 z^2) =

(a)

(b)

x^2y

(c)

(d)

x^6y^7 z^4

(e)

none of these

50.

Multiply and collect like terms: (x + 2)(x−3) =

(a)

x^2 − 6

(b)

x^2 − x − 6

(c)

x^2 + x − 6

(d)

x^2 − 5x − 6

(e)

none of these

51.

Multiply and collect like terms: (2x − 1)(2x + 1) − 2(2x − 1) =

(a)

4x^2 − 4x

(b)

4x^2 − 4x + 1

(c)

4x^2 − 4x − l

(d)

1

(e)

none of these

52.

Multiply and collect like terms: (a + 2) (a − 2) − (a − 1) (a + 1) =

(a)

2a^2 − 3

(b)

−3

(c)

−5

(d)

2a^2 − 5

(e)

none of these

53.

Multiply and collect like terms: (a + 2)2 + (a − 2)2 =

(a)

2a^2 + 8

(b)

8a

(c)

2a^2 + 4

(d)

2a^2 − 4a + 8

(e)

none of these

54.

Multiply and collect like terms: (a + 2)2 − (a − 2)2 =

(a)

2a^2 + 8

(b)

8a

(c)

2a^2 + 4

(d)

2a^2− 4a + 8

(e)

none of these

55.

Expand and collect like terms: (x + 2)(x^2 − 4x+ 1)

(a)

x^3− 6x^2 + 9x − 2

(b)

x^3− 2x^2 −7x + 2

(c)

x^3 + 2x^2 − 9x + 2

(d)

x^4 − 1

(e)

none of these

CHAPTER 5 REVIEW EXERCISES

56.

Perform the division: (x^3 − 8x^2) ÷ (−x).

(a)

x^2 − 8x

(b)

x^2 + 8x

(c)

−x^2 + 8x

(d)

−x^2 − 8x

(e)

none of these

57.

Factor: 2x^2 − 3x.

(a)

x(2x − 3)

(b)

2x(x − 3)

(c)

x^2(2x − 3)

(d)

x(2x + 3)

(e)

none of these

58.

Factor: x^2 + 3x + 2.

(a)

(x + 1)(x − 2)

(b)

(x + 1)(x + 2)

(c)

(x − 1)(x + 2)

(d)

(x − 1)(x − 2)

(e)

none of these

59.

Factor: x^2 − 3x + 2.

(a)

(x + 1)(x − 2)

(b)

(x + 1)(x + 2)

(c)

(x − 1)(x + 2)

(d)

(x − 1)(x − 2)

(e)

none of these

60.

Factor: x^2 − x − 2.

(a)

(x + 1)(x− 2)

(b)

(x + 1)(x + 2)

(c)

(x − 1)(x + 2)

(d)

(x − 1)(x − 2)

(e)

none of these

61.

(a)

x^2 + 2x +1

(b)

x^2 − 2x − 1

(c)

x^2 − 2x + 1

(d)

x^2 + 2x − 1

(e)

none of these

62.

(a)

3x + 1

(b)

3x

(c)

3x + 2

(d)

6x^2

(e)

none of these

63.

(a)

3x + 1

(b)

3x

(c)

3x + 2

(d)

6x^2

(e)

none of these

64.

(9a^3b^3− 6a^2b^4 + 3a^2b^3) ÷ (3a^2b^3)

(a)

3a − 2b

(b)

3a − 2b + l

(c)

−6ab

(d)

3a^5b^6 − 2a^4b^7 + a^4b^6

(e)

none of these

65.

(x^2 + 5x − 14) ÷ (x − 2) =

(a)

x − 7

(b)

x + 7

(c)

x + 2

(d)

x + 12

(e)

none of these

66.

(6x^3 − 5x^2 + 7x − 2) ÷ (3x − 1) =

(a)

2x^2 + x + 2

(b)

2x^2 − x − 2

(c)

2x^2 − x + 2

(d)

2x^2 + x − 2

(e)

none of these

67.

(x^3 + 27) ÷ (x + 3) =

(a)

x^2 − 9x + 9

(b)

x^2 + 9x + 9

(c)

x^2 + 3x + 9

(d)

x^2 − 3x + 9

(e)

none of these

68.

(a)

x^3 + x^2

(b)

3x^3 + 2x^2

(c)

x^6 + x^3

(d)

x^12

(e)

none of these

69.

(a)

x^3 + x^2

(b)

3x^3 + 2x^2

(c)

x^6 + x^3

(d)

x^12

(e)

none of these

70.

Factor: xy^2 − x^2y.

(a)

xy(x − y)

(b)

xy(y − x)

(c)

x^2y^2(x − y)

(d)

xy(x + y)

(e)

none of these

71.

Factor: x^2− x − 12.

(a)

(x−3)(x + 4)

(b)

(x−6)(x + 2)

(c)

(x + 3)(x−4)

(d)

(x− 12)(x + 1)

(e)

none of these

72.

Factor: x^2− 12x + 27.

(a)

(x − 3)(x − 9)

(b)

(x + 3)(x + 9)

(c)

(x−3)(x + 9)

(d)

(x− 1)(x−27)

(e)

none of these

73.

Factor: x^3− 12x^2 + 27x.

(a)

x(x − 3)(x − 9)

(b)

x(x + 3)(x + 9)

(c)

x(x − 3)(x + 9)

(d)

x(x − 1)(x − 27)

(e)

none of these

74.

Factor: a^2 + 16a + 64.

(a)

(a + 8)2

(b)

(a − 8)(a + 8)

(c)

(a − 8)2

(d)

(a + 16)(a + 4)

(e)

none of these

75.

Factor: (a^3 + 16a^2 + 64a).

(a)

a(a + 8)2

(b)

a(a − 8)(a + 8)

(c)

a(a − 8)2

(d)

a(a + 16)(a + 4)

(e)

none of these

76.

Factor: a^2 − 64.

(a)

(a + 8)2

(b)

(a − 8)(a + 8)

(c)

(a − 8)2

(d)

(a +16)(a + 4)

(e)

none of these

77.

Divide: (6x^3 + x^2 − 11x − 6) ÷ (3x + 2)

(a)

2x^2 − x + 3

(b)

2x^2 − x − 3

(c)

2x^2 + x − 3

(d)

2x^2 + x + 3

(e)

none of these

78.

Reduce the fraction:

(a)

x

(b)

(c)

x + 1

(d)

x − 1

(e)

none of these

79.

Add:

(a)

(b)

(c)

(d)

(e)

none of these

80.

Reduce the fraction:

(a)

(b)

−1

(c)

(d)

(e)

none of these

81.

Reduce the fraction:

(a)

(b)

1

(c)

(d)

(e)

none of these

82.

Add the fractions:

(a)

(b)

(c)

(d)

(e)

none of these

83.

Perform the subtraction:

(a)

(b)

(c)

(d)

(e)

none of these

CHAPTER 6 REVIEW EXERCISES

84.

Solve the following equation for x: ax = 2.

(a)

(b)

x = 2 − a

(c)

(d)

x = a

(e)

none of these

85.

Solve the following equation for

(a)

x = b

(b)

(c)

x = 2b

(d)

(e)

none of these

86.

Solve the following equation for x: ax + 2b = 5b.

(a)

x =7ab

(b)

(c)

x = 3ab

(d)

(e)

none of these

87.

Solve the following equation for x: 2x − 1 = x.

(a)

x = 1

(b)

x = −1

(c)

(d)

(e)

none of these

88.

Solve the following equation for x: 1 − x = 2.

(a)

x =3

(b)

x = −1

(c)

x = −3

(d)

(e)

none of these

89.

Solve the following equation for x: 2x − 1 = 1 − 3x.

(a)

(b)

(c)

x = 0

(d)

(e)

none of these

90.

Solve the following equation for x: 2(x −1) = 3(1 − 3x).

(a)

(b)

(c)

(d)

(e)

none of these

91.

Solve the following equation for

(a)

(b)

(c)

(d)

(e)

none of these

92.

Solve the following equation for x: ax − 2b = 5b.

(a)

x = 7ab

(b)

(c)

x = 3ab

(d)

(e)

none of these

93.

Solve the following equation for

(a)

(b)

(c)

(d)

(e)

none of these

94.

Adam, Eve, Snake, and Jake share $240 between them. Adam has twice as much as Eve, Eve has twice as much as Snake, and Snake has twice as much as Jake. How much money does Eve have?

(a)

$15

(b)

$60

(c)

$64

(d)

$128

(e)

none of these

95.

Peter, Paul, and Mary share $153 between them. Peter has twice as much as Paul, and Mary has thrice as much as Peter. How much money does Peter have?

(a)

$17

(b)

$34

(c)

$51

(d)

$102

(e)

none of these

96.

The sum of three consecutive odd integers is 909. Which one is the last number?

(a)

301

(b)

302

(c)

303

(d)

305

(e)

none of these

97.

Jane’s age is twice Bob’s age increased by 3. Bill’s age is Bob’s age decreased by 4. If the sum of their ages is 27, how old is Bill?

(a)

3

(b)

7

(c)

4

(d)

17

(e)

none of these

98.

A jar of change contains nickels, dimes, and quarters. There are twice as many dimes as nickels, and twice as many quarters as dimes. If the jar has a total of $8.75, how many quarters are there?

(a)

5

(b)

7

(c)

14

(d)

28

(e)

none of these

99.

Currently, the age of a father is four times the age of his son, but in 24 years from now the father’s age will only be double his son’s age. Find the son’s current age.

(a)

12

(b)

36

(c)

48

(d)

6

(e)

none of these

CHAPTER 7 REVIEW EXERCISES

100.

Solve for x: 2x^2 −3x = 0.

(a)

x = 0 or

(b)

x = 0 or

(c)

x = 0 or

(d)

x = 0 or

(e)

none of these

101.

Solve for x: x^2 + 3x + 2 = 0.

(a)

x = −1 or x = 2

(b)

x = -1 or x = -2

(c)

x = 1 or x = −2

(d)

x = 1 or x = 2

(e)

none of these

102.

Solve for x: x^2 −3x + 2 = 0.

(a)

x = −1 or x = 2

(b)

x = −1 or x = −2

(c)

x = 1 or x = −2

(d)

x = 1 or x = 2

(e)

none of these

103.

Solve for x: x^2 − x − 2 = 0.

(a)

x = −1 or x = 2

(b)

x = −1 or x = −2

(c)

x = 1 or x = −2

(d)

x = 1 or x = 2

(e)

none of these

104.

If 2(x −1) ≥ x + 3 then

(a)

x ≥ 1

(b)

x ≥ 5

(c)

x ≤ 1

(d)

x ≤ 5

(e)

none of these

105.

If −2x > x + 3 then

(a)

x > −1

(b)

x > 1

(c)

x < −1

(d)

x < 1

(e)

none of these

106.

Which graph gives the correct solution to 2 − 5x ≥ − 3?

(a)

(b)

(c)

(d)

(e)

none of these

107.

Which graph gives the correct solution to 2 + 5x ≥ −3?

(a)

(b)

(c)

(d)

(e)

none of these

True or False Exercises.

108.

(2+3)10 = 2^10 + 3^10.

109.

(2 ⋅ 3)10 = 2^10 ⋅ 3^10.

110.

(2^5)2 = 2^7.

111.

It is always true that x(2x + 3) = x(2x)(3).

112.

(1^1 + 2^2 + 3^3 + 4^4 −5^5)0 + 4 = 5.

In each item below, select the choice(s) that indicates equivalence to the given expression. Denominators are presumed not to vanish.

113.

−a =

(a)

a(−1)

(b)

−1 · a

(c)

(d)

(e)

1 − 2a

114.

x y − z =

(a)

z − x y

(b)

yx − z

(c)

x y + (−z)

(d)

x(y − z)

(e)

−z + x y

115.

(a)

a −b ÷ 2

(b)

(a − b) ÷ 2

(c)

(d)

(e)

116.

(a)

(b)

2xy + 10x

(c)

(d)

(e)

2xy + 10x^2

117.

(a −2b)2 =

(a)

(a − 2b)(a − 2b)

(b)

a^2 − 4b^2

(c)

a^2 + 4b^2

(d)

a^2 − 4ab + 4b^2

(e)

−(2b − a)(a − 2b)

118.

(a)

1

(b)

−1

(c)

−2

(d)

(x − y)2

(e)

119.

(2x^2y^2)(3x y^4) =

(a)

6x^3y^6

(b)

6(x y^4)2

(c)

6(x y^2)3

(d)

(e)

120.

x(x^3 − 1) + 2(x^3 − 1)=

(a)

(x^3 − 1)x −2(−x^3 + 1)

(b)

2(x^3 − 1)+x(−x^3 + 1)

(c)

(x + 2)(x^3 − 1)

(d)

x^4 − x + 2x^3 − 2

(e)

x^4 − x +1(x^3 − 2)

CHAPTER 8 REVIEW EXERCISES

121.

3x − 4 ≤ 2 =

(a)

x ≤ 2

(b)

x ≤ 3

(c)

x ≤ 2

(d)

(e)

none of these

122.

(a)

x < 25

(b)

x > 5

(c)

x > 25

(d)

x > 1

(e)

none of these

123.

(a)

x ≥ 130

(b)

x ≥ 65

(c)

x > 25

(d)

x > 1

(e)

none of these

124.

3(m − 2) < m =

(a)

m > 3

(b)

m < 3

(c)

m < 6

(d)

m > 6

(e)

none of these

125.

−28 <3x − 4 < −10=

(a)

−8 < x < −2

(b)

8 < x < 2

(c)

x > −8

(d)

x < −2

(e)

none of these

126.

25 ≤ 4y − 3 ≤ 9

(a)

7 < y < 3

(b)

7 ≤ y ≤ 3

(c)

3 ≤ y ≤ 7

(d)

−7 ≥ y ≥ −3

(e)

none of these

127.

8(n − 1) + 5 ≥ 3(5n − 4)=

(a)

(b)

(c)

n ≤ 7

(d)

n ≤ 9

(e)

none of these

128.

(n − 1)(n − 3) ≤ 4n =

(a)

(b)

(c)

(d)

(e)

none of these

APPENDIX B:
ANSWER KEYS FOR CHAPTER SUPPLEMENTARY EXERCISES

CHAPTER 1 ANSWERS

1.2.1 The Alexandrian mathematician Diophantus, for the introduction of syncopation (symbols to express abstract quantities and procedures on them) in algebra.

1.2.2 The word algebra is Arabic, but algebra was studied long before the Arabs had any prominent role in history.

1.2.3 The word ossifrage is Latin for “breaking bones.” The word algebra is Arabic for “bone setting.”

1.3.1 You need to increase $20, and $20 is 25% of $80, and so you need to increase by 25%.

1.3.2 10¢. The cork costs 10¢ and the bottle 90¢.

1.4.1 St. Augustine is using the word mathematician in lieu of the word astrologer.

1.4.2 No. When we add an even integer to another even integer the result is an even integer. Thus the sum of five even integers is even, but 25 is odd.

1.4.3 The 27th floor.

1.4.4 Between 9999 and 10101.

1.4.5 2 75 3 6 − 2 56 1 = 24 9 75.

1.4.6 Observe that the first and second rows, and the second and third columns, add up to 8. Thus a = 4 works.

1.4.7 There are multiple solutions. They can be obtained by permuting the entries of one another. Here are two:

1.4.8 We can represent the brother’s and sister’s amounts by boxes, each box having an equal amount of coins, where two boxes are for the brother and one for the sister:

Then clearly, each box must have 8 coins. The brother has 16 and the sister has 8.

CHAPTER 2 ANSWERS

2.1.1 N − 20.

2.1.2 N + 20.

2.1.3 3 x + 10.

2.1.4 3 (x + 10).

2.1.5 In crocheting s baby blankets, I need 5s balls of yarn. Therefore, at the end of the day I have b − 5s balls of yarn.

2.1.6 On the first step you have x. On the second step you have 2x. On the third step you have 2x + 10. On the fourth step you have On the fifth step you have You are asserting that is identically equal to 5.

2.1.7 4n; 4n + 1; 4n + 2; 4n + 3, where n is a natural number

2.1.8 6n + 1, where n = 0, 1, 2, 3, ….

2.1.9 First observe that

etc. Then it becomes clear that if the last number on the left is n, then the right-hand side is n(n + 1)(2n + 1) divided by 6. Therefore we are asserting that

2.1.10 The general formula is

1 + 3 + … + (2n − 1) = n^2.

2.1.11 You have 25q + 10d cents at the beginning of the day, and then you lose 25a +10b, which leaves you with

(25q + 10d) − (25a + 10b)

cents.

2.1.12 The cube is x^3, the square is x^2. The cube reduced by the square is x^3 − x^2. In conclusion, what is left after division by 8 is

2.1.13 The jacket costs 3h and the jeans cost 3h − 8. Therefore he paid in total

h + 3h + 3h − 8 = 7h − 8.

2.1.14 The profit is that amount over the real price. If the real price is Γ, then Γ b is the real price plus the profit, that is, the farmer’s selling price for the cow. In other words, Γ + b = a, there-fore Γ = a − b.

2.1.15 The salesman starts with s and gains x for a total of s + x. He then loses y, for a total of s + x − y. After he gains b more dollars he has a running total of s + x − y + b, and finally, after losing z dollars he has a grand total of s + x − y + b − z.

2.2.1 108.

2.2.2 1.

2.2.3 28.

2.2.4 21.

2.2.5 15.

2.2.6 Since 100 = 7 × 14 + 2, 98 days from today will be a Thursday, and 100 days from today will be a Saturday.

2.2.7 Observe that 4 and 5 are on opposite sides, so they never meet. The largest product is 3 × 5 × 6 = 90.

2.2.8 The first contractor prints 4500 ÷ 30 = 150 cards per day, and the second, 4500 ÷ 45 = 100 cards per day. Therefore, if they worked simultaneously, they could print 150 + 100 = 250 cards per day, and it would take them 4500 ÷ 250 = 18 days to print all cards.

2.2.9 Seven.

2.2.10 9 minutes.

2.2.11 If 36 ÷ n is a natural number, then n must evenly divide 36, which means that n is a divisor of 36. Thus n can be any one of the values in {1, 2, 3, 4, 6, 9, 12, 18, 36}.

2.2.12 From the distributive law,

2.2.13 Travelling 30,000 miles with 4 tires is equivalent to travelling 120,000 miles on one tire. The average wear of each of the 5 tires is thus 120000 ÷ 5 = 24000 miles.

2.2.14 Anna answered 20 questions correctly. (She could not answer less than 20, because then her score would have been less than 19 × 4 = 76 < 77, and she could not answer more than 20, because her score would have been at least 21 × 4 − 3 = 81.77). To get exactly 77 points, Anna had to answer exactly 3 questions wrong, which means she omitted 2 questions.

2.2.15 The trick is to use a technique analogous to the one for multiplying, but this time using three digits at a time:

Thus

2.2.16 There are 28 digits, since

4^165^25 = 2^325^25 = 2^72^255^25 = 128 × 10^25,

which is the 3 digits of 128 followed by 25 zeroes.

2.2.17 The least common multiple of 4, 5, and 6 is 60, therefore the smallest positive multiple of 60 leaving remainder 1 upon division by 7. This is 120.

2.2.18 Using the definition of ⊕,

1 ⊕ (2 ⊕ 3) = 1 ⊕ (1 + 2(3)) = 1 ⊕ 7 = 1 + 1(7) = 8,

but

(1 ⊕ 2) ⊕ 3 = (1 + 1(2)) ⊕ 3 = 3 ⊕ 3 = 1 + 3(3) = 10.

Since these two answers differ, the operation is not associative.

Now, for the operation to be commutative, we must have, in every instance, a ⊕ b = b ⊕ a. Using the fact that multiplication of numbers is commutative, we deduce

a ⊕ b = 1 + ab = 1 + ba = b ⊕ a,

so our operation is indeed commutative.

2.3.1 Number the steps from 0 to N, where N is the last step, and therefore there are N + 1 steps. Notice that the top and the bottom of the stairs are counted as steps. Let us determine first the number of steps. Freddy steps on steps 0, 2, 4, …, N; Lorance steps on steps 0, 3, 6, …, N; George steps on steps 0, 4, 8, …, N; and Aragorn steps on steps 0, 5, 10, …, N. For this last step to be common, it must be a common multiple of 2, 3, 4, and 5, and therefore

N = lcm(2, 3, 4, 5) = 60.

Since every step that George covers is also covered by Freddy, we don’t consider George in our count. Freddy steps alone on the steps which are multiples of 2, but not multiples of 3 or 5. Therefore he steps on the 8 steps

2, 14, 22, 26, 34, 38, 46, 58.

Lorance steps alone on the 8 steps that are multiples of 3, but not multiples of 2 nor 5:

3, 9, 21, 27, 33, 39, 51, 57.

Finally, Steve steps alone on the 4 steps that are multiples of 5 but not multiples of 2 or 3:

5, 25, 35, 55.

Our final count is thus 8 + 8 + 4 = 20.

2.3.2 of an hour.

2.3.3

2.3.4 Observe that 3990 = 210 · 19. Thus

2.3.5 Observe that 3 · 5 · 13 = 195 is a common denominator. Now

therefore is the smallest, is in the middle, and is the largest.

2.3.6 Three lines have been sung before the fourth singer starts, and after that he sings 12 more lines. So the total span of lines is 15. They start singing simultaneously from line 4, and the first singer is the first to end, in line 12. Thus 12 − 4 + 1 = 9 lines out of 15 are sung simultaneously and the fraction sought is

2.4.1 Here are the solutions.

2.4.2

2.4.3

2.4.4 We have,

2.4.5 1.

2.4.6 We have,

therefore a^2 + b^2 = 5^2 + 6^2 = 61.

2.4.7

2.4.8 In one hour John does of the job. In one hour Bill does Thus in one hour they together accomplish of the job. Thus in of an hour they accomplish the job. Since it takes them 72 minutes to finish the job.

2.4.9 We have and

so she is able to wrap 8 gifts.

2.4.10 Proceeding from the innermost fraction one easily sees that

2.4.11

2.4.12 Of the 100 students, only one is male. He is 2% of the on-campus population. Thus the whole on-campus population consists of 50 students, so there are 100 − 50 = 50 off-campus students

2.4.13 Observe that Therefore miles are at the rate of 40¢. The trip costs $0.85 + $.40 · 21 = $.85 + $8.40 = $9.25.

2.5.1

1. +8

2. −26

3. −26

4. 0

5. −303

6. 0

7. +4

8. −β100

2.5.2 Here is one possible answer.

2.5.3 Here is one possible answer.

2.5.4 Here are two answers.

2.5.5 We have,

2.6.1 We have,

2.6.2

2.6.3 Here are some possible answers. In some cases, there might be more than one answer, as for example, etc.

2.6.4 To do this problem correctly, you must compare all combinations of numbers formed with three fours.

Observe that

(4^4)4 = 256^4 = 4294967296,

44^4 = 3748096, 4^44 = 309485009821345068724781056,

and that

4(4^4) = 4^256

= 1340780792994259709957402499820584612747936582059239337772 35614437217640300735 →

→ 469768018742981669034276900318581864860508537538828119465699 46433649006084096

where we have broken the number so that it fits the paper.

Note: This last calculation was performed with a computer program. How does one actually check that the computer is right? At any rate, since each time we multiply a given quantity by 4 the quantity increases, we must have 4^44 < 4^256 = 44^4.

2.6.5

2.6.6

2.6.7 From the text, we know that and therefore

2.6.8

2.6.9

2.6.10 No, this is not always true. For example, but But, is it ever true? For and so it is true in this occasion.

CHAPTER 3 ANSWERS

3.1.1 After buying the marbles, he has +a marbles. After winning b marbles, he now has a + b marbles. After losing c marbles, he ends up with a + b − c. In the end, he has a + b − c marbles.

3.1.2

1. 0

2. −6a

3. −10a

4. 6a

5. −4a +4b

3.1.3 2a −2b +2c −2d +2a^2 −2c^2.

3.1.4

1. 4a + 4b + 4c

2. 4a + 2c

3. 5a +5b + 5c

4. a −b + c

5. 2x^2 + 2

6. x^3 + x^2 + 1

7. 6x^3 − 3x^2 − 3x − 2

8.

3.1.5 Since (2x^2 + x −2) + (−x^2 − x + 2) = x^2, the value is 2^2 = 4.

3.1.6 Yes, this is always true.

3.1.7 No, this is not always true. For example, if a = 10, then a + a = 10 + 10 = 20, but 2a^2 = 2(10)2 = 2 · 100 = 200. Observe, however, that 0 + 0 = 2 · 0^2 and that 1 + 1 = 2 · 1^2, so the assertion is true in these two cases. That it is only true in these two cases is more difficult to prove, and requires use of quadratic equations.

3.1.8 No, this is not always true. For example, if a = 10, then a + a^2 = 10 + 100 = 110, but 2a^3 = 2(10)3 = 2·1000 = 2000. Observe, however, that 0 + 0^2 = 2· 0^3, that 1 + 1^2 = 2 · 1^3, and that so the assertion is true in these three cases. That it is only true in these three cases is more difficult to prove, and requires use of quadratic equations.

3.1.9 Letting x = 1, one gathers that

243 = 3^5 = (8(1) − 5)5 = A + B + C + D + E + F.

3.1.10 Here are a few. Many more are possible.

1. 3x y − x^2 y

2. −x^2 y + 3yx

3. 3yx − yx^2

4. 3x y − yx^2

5. −yx^2 + 3x y

3.2.1 Since b −c inches are cut from a + b, we must subtract b − c from a + b, giving

(a + b) − (b − c) = a + b − b + c = a + c

therefore a + c inches remain.

3.2.2 Since there are x people to share the cost of D, the share of each person is If p persons disappear, there remain x − p persons, and the share of each would increase to This means they will have to pay

more.

3.2.3 2b + 2c + 2c^2 + 2d^2.

3.2.4 −a + 5b −c.

3.2.5

1. −2a + 2c

2. 3a + 3b + 3c

3. a + 3b + c

4. x3 − 3x2 + 2x − 3

5. −2x^3 − x^2 + 3x − 2

6. 9.3a − 3.1b

7.

8.

9. 10♣

3.2.6 We have

3.2.7

1. −2x − 2y

2. 3x − 5

3. 4x2 + 2x − 5

3.2.8 x − y by commutativity is −y + x, and so 1 matches with B. By the subtraction rule, x − (−y) = x + (+ y) = x + y, and so 2 matches with C. Finally, using the distributive law and the commutative law, −(x − y) = −x + y = y − x, and so 3 matches with A.

3.2.9 Mary starts with A, and her uncle Bob gives her A more so that she now has A + A = 2A. Her aunt Rita gives her 10 more, so that now she has 2A + 10. She pays B dollars in fines, therefore she is left with 2A + 10 − B. She spent 12 in fuel, so now she has the following total:

2A + 10 − B − 12 = 2A − B − 2,

3.2.10

1. (2 − 4)x − 5(−6)x = −2x + 30x = 28x

2. 5 − 2(2x) = 5−4x

3. (5 − 2)(1 − 2^2)3x = (3)(1 − 4)x = (3)(−3)3x = (3)(−27)x = −81x

3.2.11

1. (6t) + x = 6t + x, so the parentheses are redundant.

2. In 6(t + x), the parentheses are needed.

3. In (ab)2, the parentheses are needed.

4. In (ab − c)2, the parentheses are needed.

3.2.12

1. In

2. In

3.2.13

1. In is −1 or simply, −.

2. In 3x + y − t = + 3x the is y − t.

3.2.14 The opposite sought is

−(4 − y + 2) = −4 + y − 2.

3.2.15 The additive inverse sought is

CHAPTER 4 ANSWERS

4.1.1 −8x yz.

4.1.2 24abc.

4.1.3

4.1.4

4.1.5 The product equals

so n = 12.

4.1.6 We have

3^2001 + 3^2002 + 3^2003 = 3^2001(1 + 3 + 3^2) = (13)3^2001,

therefore a = 13.

4.1.7

4.1.8 x.

4.1.9

4.1.10 s −1.

4.1.11

4.1.12

4.1.13 1.

4.1.14 We have,

4.1.15 We have,

So m = 2 and n = 11.

4.2.1

6. a^2 b^5

4.2.2 We have,

(ax + 3by − 4)(a2x − 3b− 2y + 5) = ax + 3 + 2x − 3by − 4 − 2y + 5 = a3xb1 − y.

4.2 We have

4.2.4

4.2.5 2^8 = 256.

4.2.6 4.

4.2.7

4.3.1

1. x^3 + x^2 + x

2. 2x^3 − 4x^2 + 6x

3. 4x^2 + 4x + 1

4. 4x^2 −1

5. 4x^2 − 4x + 1

4.3.2 x^2 + 1 − 7x^3 + 4x − 3x^5.

4.3.3 4x^5 − 8x^4 + 8x^3 + 2x^2.

4.3.4

1. −9x^4 + 3x^3 − 2x^2 − 2x

2. −2x^3 − 4x^2 − 2x

3. a^3cb − 2ac^2b

4. x^2 − 4y^2 + 3zx − 6zy

4.3.5 We have,

4.3.6 10x.

4.3.7 − 2.

4.3.8 x^2 + zx − y^2 + 3z y − 2z^2.

4.3.9 We have,

4.3.10 From the distributive law we deduce that

(a + b + c)(x + y + z) = ax + ay + az + bx + by + bz + cx + c y + cz.

4.3.11 Let 2a be one of the integers and let 2b be the other integer. Then (2a)(2b) = 2(2ab), which is twice the integer 2ab and therefore it is even.

4.3.12 Let 2a + 1 be one of the integers and let 2b + 1 be the other integer. Then

where m = 2ab + a + b. Since m is an integer, the equality (2a + 1)(2b + 1) = 2m + 1 shows that the product (2a + 1)(2b + 1) leaves remainder 1 upon division by 2, that is, the product is odd.

4.3.13 Let 4a + 3 be one of the integers and let 4b + 3 be the other integer. Then

where m = 4ab + 3a + 3b + 1. Since m is an integer, the equality (4a + 3)(4b + 3) = 4m + 1 shows that the product (4a + 3)(4b + 3) leaves remainder 1 upon division by 4.

4.3.14 Let 3a +2 and 3b + 2 be two numbers leaving remainder 2 upon division by 3. Then

(3a + 2)(3b + 2) = 9ab + 6a + 6b + 4 = 3(3ab + 2a + 2b + 1) + 1,

which is a number leaving remainder 1 upon division by 3.

4.3.15 Let 5a + 2 and 5b + 2 be two numbers leaving remainder 2 upon division by 5. Then

(5a + 2)(5b + 2) = 25ab + 10a + 10b + 4 = 5(5ab + 2a +2b) + 4,

which is a number leaving remainder 4 upon division by 5.

4.3.16 Let 5a + 3 and 5b + 3 be two numbers leaving remainder 3 upon division by 5. Then

(5a + 3)(5b + 3) = 25ab + 15a + 15b + 9 = 5(5ab + 2a + 2b + 1) + 4,

which is a number leaving remainder 4 upon division by 5.

4.3.17 Observe that x^2 + x = 1. Therefore x^4 +x^3 = x^2 and x^3 +x^2 = x. Thus

x^4 + 2x^3 + x^2 = x^4 + x^3 + x^3 + x^2 = x^2 + x = 1.

4.3.18 We have

Putting x = 11112, one obtains from the preceding part that

2 · (11112) · (11113)−11112^2 − (11111) · (11112) = 3 · 11112 = 33336.

4.4.1 9 −6y + y^2.

4.4.2 25 − 20x^2 + 4x^4.

4.4.3 4a^2b^4 − 12ab^2c^3d^4 + 9c^6d^8.

4.4.4 x^2 + 8x y + 4y^2.

4.4.5 x^2 +4y^2.

4.4.6 We have,

(xy − 4y)2 = (xy)2 − 2(xy)(4y) + (4y)2 = x^2 y^2 − 8xy^2 + 16y^2.

4.4.7 We have,

(ax + by)2 = (ax)2 + 2(ax)(by) + (by)2 = a^2x^2 + 2abxy + b^2 y^2.

4.4.8 2a^2 +8.

4.4.9 8a.

4.4.10 a^2 + 4ab + 6ac + 4b^2 + 12bc + 9c^2.

4.4.11 a^2 + 4ab − 2ac + 4b^2 − 4bc + c^2.

4.4.12 Observe that

4.4.13 We have

therefore n = 2004.

4.4.14 We have

4.4.15 Let the two numbers be x, y. Then x − y = 3 and xy = 4. Therefore

9 = (x − y)2 = x^2 − 2xy + y^2 = x^2 − 8 + y^2 ⇒ x^2 + y^2 = 17.

4.4.16 We have

16 = (x + y)2 = x^2 +2xy + y^2 = x^2 − 6 + y^2 ⇒ x^2 + y^2 = 22.

4.4.17 From Exercise 4.4.16, x^2 + y^2 = 22. Observe that 2x^2 y^2 = 2(xy)2 = 2(−3)2 = 18. Therefore

484 = (x^2 + y^2)2 = x^4 + 2x^2 y^2 + y^4 = x^4 + 18 + y^4 ⇒ x^4 + y^4 = 466.

4.4.18 We have,

4.4.19 We have

4.5.1 We have

We may also use the square of a sum identity:

like before.

4.5.2 We have,

4.5.3 We have

4.5.4 Multiplying by 1 = 2 − 1 does not alter the product, and so,

4.5.5 We have,

using the result of Example 1.5.

4.5.6 Put x = 123456789. Then

(123456789)2 − (123456787)(123456791) = x^2 − (x −2)(x + 2) = x^2 −(x^2 − 4) = 4.

4.5.7 Using x^2 − y^2 = (x − y)(x + y),

4.6.1 125x^3 + 75x^2 +15x + 1.

4.6.2 x^3 + 4x^2 + 5x +2.

4.6.3 x^3 − 4x^2 − x −5.

4.6.4 x^3 − 7x^2 + 16x − 12.

4.6.5 We have,

216 = (a − b)3 = a^3 −3a^2b + 3ab^2 −b^3 = a^3 − b^3 −3ab(a − b) = a^3 − b^3 −3(3)(6) ⇒ a^3 − b^3 = 270.

4.6.6 We have,

4.7.1 x^3 −8.

4.7.2 x^3 + 512.

4.7.3 Notice that this is Exercise 4.6.5. We will solve it here in a different way. We have

a^3 − b^3 = (a − b)(a^2 +ab + b^2) = 6(a^2 + 3 + b^2),

so the problem boils down to finding a^2 +b^2. Now,

36 = (a − b)2 = a^2 −2ab + b^2 = a^2 − 6 + b^2 ⇒ a^2 + b^2 = 42,

therefore

a^3 − b^3 = 6(a^2 + 3 + b^2) = 6(42 + 3) = 270

as was obtained in the solution of Exercise 4.6.5.

CHAPTER 5 ANSWERS

5.1.1

5.1.2 2x^2 y + 3xy^2.

5.1.3 102x^4 y^4.

5.1.4 3a^2x + 2ax.

5.1.5 12a^5x^4.

5.1.6 −x + y +z.

5.1.7 −5a + 7b^2.

5.1.8 35a^2b^3.

5.1.9 − x^3 yz.

5.2.1 x − 1.

5.2.2 x + 3.

5.2.3 x^2 − x +1.

5.2.4 2x + 4.

5.2.5 2x^2 − 2x − 2.

5.3.1 We have

−4a^6b^5 +6a^3b^8 −12a^5b^3 −2a^2b^3 = −2a^2b^3(2a^4b^2 −3ab^5 +6a^3 +1).

5.3.2 4a^3b^4 −10a^4b^3 = 2a^3b^3(2b −5a).

5.3.3

5.3.4 −x + 2y −3z = −1(x − 2y + 3z).

5.3.5 Let 2a + 1 and 2b + 1 be odd integers. Then

2a + 1 + 2b + 1 = 2a + 2b + 2 = 2(a + b + 1),

that is, twice the integer a + b + 1, and therefore even.

5.3.6 x^2(x −1).

5.3.7 5a^3b^2c^4(25cb^3a − 9ba^2 + 1 − 60c^4a − 2c).

5.3.8 5x^3(x^2 −2a^7 −3a^3).

5.3.9 x^2(3xy + 4y^3 −6x^4 −10x^2).

5.3.10 3m^4p^2q(m^2p^2q −2p^2x^2 ym + m^3px − 3mx + 1).

5.3.11 19a^2x^2(2x^3 +3a^2).

5.3.12 (a + c)(a + b).

5.3.13 (a + c)(a − b).

5.3.14 (y − 1)(1 + y^2).

5.3.15 (2x + 3)(x^2 + 1).

5.3.16 (a + b)(xa + c + by).

5.4.1 (a − 5)(a − 6).

5.4.2 (a − 19)2.

5.4.3 (a^2b^2 + 25)(a^2b^2 + 12).

5.4.4 (x − 11)(x − 12).

5.4.5 (x^2 − 12)(x^2 − 17).

5.4.6 (x + 27)(x + 8).

5.4.7 (5x + 2)(x + 3).

5.4.8 (7x − 3)(2x + 5).

5.5.1 We have

x^2 −4y^2 = (x +2y)(x −2y) = (3)(−1) = −3.

5.5.2 (x −2)(x + 2)(x^2 + 4).

5.5.3 (a + b − c)(a + b + c).

5.5.4 x(x −1)(x + 1).

5.5.5 Since n^3 − 8 = (n −2)(n^2 + 2n + 2) and n − 2 < n^2 + 2n + 4, we must have n − 2 = 1 ⇒ n = 3 and n^3 − 8 = 19.

5.6.1

5.6.2

5.6.3

5.6.4

5.6.5

5.6.6

5.6.7

5.6.8

5.6.9

5.6.10

5.6.11 We have,

x + y = 7 ⇒ x^2 + 2xy + y^2 = 49 ⇒ x^2 + y^2 = 49 − 2xy = 49 − 2(21) = 7.

Observe also that x^2 y^2 = (xy)2 = 212. Therefore

CHAPTER 6 ANSWERS

6.1.1 z = −8.

6.1.2 z = 0.

6.1.3 z = −11.

6.1.4 z =

6.1.5 z = −16.

6.1.6 z = −33.

6.1.7 z = −36.

6.1.8 z = 3a + 3.

6.1.9 z = 5.

6.1.10 z = (b^2 + a)(a + 1) = b^2a + b^2 + a^2 + a.

6.1.11 N = A.

6.1.12 N = A^3 + A.

6.1.13

6.1.14 N = 12A + A^2.

6.1.15 N = −17X.

6.1.16 N = A^2 + 3A + 2.

6.1.17 N = B^7.

6.1.18

6.1.19 N = A + 12.

6.2.1

6.2.2 x = 6.

6.2.3

6.2.4 x = 3a.

6.2.5 x = b.

6.2.6

6.2.7

6.2.8 x = 1.

6.2.9

6.2.10

6.2.11 x = 5.

6.2.12 x = −13.

6.2.13 x = 2.

6.2.14

6.2.15 We have,

(x − a)b = (b − x)a ⇒ bx − ab = ab − ax ⇒ bx = −ax ⇒ (a + b)x = 0 ⇒ x = 0.

6.2.17 If x = 0.123123123…, then 1000x = 0.123123123… and giving 1000x −x = 123, since the tails cancel out. This results in

6.3.1 If x is the number, then

6x + 11 = 65 ⇒ 6x = 54 ⇒ x = 9,

so the number is 9.

6.3.2 If x is the number, then

11x − 18 = 15 ⇒ 11x = 33 ⇒ x = 3,

so the number is 3.

6.3.3 If x is the number, then

12(x + 3) = 84 ⇒ 12x + 36 = 84 ⇒ 12x = 48 ⇒ x = 4,

so the number is 4.

6.3.4 Let

x − 5, x − 4, x − 3, x −2, x − 1, x, x + 1, x + 2, x + 3, x + 4, x + 5,

be the eleven consecutive integers. Then

x − 5 + x − 4 + x − 3 + x − 2 + x − 1 + x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 2002 ⇒ 11x = 2002 ⇒ x = 182.

The numbers are

177,178,179,180,181,182,183,184,185,186,187.

6.3.5 If the two numbers are x and x +3, then

x + x + 3 = 27 ⇒ x = 12,

so the numbers are 12 and 15.

6.3.6 If x is one number, the other is 19−x. If x exceeds twice 19 − x by 1, then

x − 2(19 − x) = 1 ⇒ 3x − 38 = 1 ⇒ 3x = 39 ⇒ x = 13.

Thus one number is 13 and the other is 6.

6.3.7 If Paul has x dollars, then Mary has x + 20 and Peter has x − 30. Therefore

x + x + 20 + x −30 = 380 ⇒ 3x −10 = 380 ⇒ 3x = 390 ⇒ x = 130

Therefore Paul has $130, Mary has $150, and Peter has $100.

6.3.8 The amount of acid does not change. Therefore

0.60 = 0.20(100 + x) ⇒ x = 200,

thus 200 grams should be added.

6.3.9 If Bob’s age is x, then Bill’s age is x − 4 and Jane’s is 2x + 3. Therefore

2x + 3 + x − 4 + x = 27 ⇒ 4x −1 = 27 ⇒ 4x = 28 ⇒ x = 7.

Thus Bob is 7, Bill is 3, and Jane is 17.

6.3.10 The sum of the original 6 numbers is S = 6 · 4 = 24. If the 7th number is x, then

therefore x = 11.

6.3.11 If Bob currently has b dollars, then Bill has 5b dollars. After giving $20 to Bob, Bill now has 5b − 20 and Bob now has b + 20. We are given that

5b −20 = 4(b + 20) ⇒ 5b − 20 = 4b + 80 ⇒ b = 100.

So currently, Bill has $500 and Bob has $100.

6.3.12 If x is the number, then

Multiplying both sides of this equation by 35 we obtain

6.3.13 If one number is x, the larger is x +8. Thus

x + 8 + 2 = 3x ⇒ x + 10 = 3x ⇒ 10 = 2x ⇒ 5 = x,

so the smallest number is 5 and the larger is 13.

6.3.14 If one of the numbers is x, the other is x + 10. Thus

x + x + 10 = 2(10) ⇒ 2x + 10 = 20 ⇒ 2x = 10 ⇒ x = 5.

The smaller number is 5 and the larger is 15.

6.3.15 If x is the amount originally bought, then I spent dollars. Since I kept I must have sold of them, making dollars on this sale. My net gain is thus

Multiplying both sides by 30, we have

I originally bought sixty avocados.

6.3.16 If x is the number, then

Multiplying both sides by 24, we have

therefore the number is 24.

6.3.17 Let x and x + 1 be the integers. Then

The integers are 51 and 52.

6.3.18 Let x be the total amount of oranges bought at three for a dollar. On these I spent dollars. I bought oranges at four for a dollar, thus spending dollars. Notice that I have bought a total of oranges. If I sell all of them at sixteen for six dollars, I make 6 · dollars. Thus

Therefore I bought oranges.

6.3.19 Let x be the original price. After a year, the new price will After another year, the new price will be Therefore we must have

Thus the original price was $84,000.

6.3.20 Let the integers be n and n +1. Then

(n + 1)2 − n^2 = 665 ⇒ 2n + 1 = 665 ⇒ n = 332.

Thus the integers are 332 and 333.

6.3.21 There are ten different ways. We want the number of solutions of

5x + 10y + 25z = 50,

that is, of

x + 2y + 5z = 10,

with integer 0 ≤ x ≤ 10, 0 ≤ y ≤ 5, 0 ≤ z ≤ 2. The table below exhausts all ten possibilities.

6.3.22 Let there be a coins in the purse. There are five stages. The fifth stage is when all the burglars have the same amount of money. Let

ak, bk, ck, dk, ek

be the amount of money that each burglar has, decreasing lexicographically, with the a’s denoting the amount of the meanest burglar and ek denoting the amount of the meekest burglar. Observe that for all k we have

ak +bk +ck +dk +ek = a

On stage five we are given that

On stage four

On stage three we have

On stage two we have

On stage one we have

Since a_1 = 240, we deduce

Check: On the first stage the distribution is (from meanest to meekest):

240, 100, 40, 15, 5.

On the second stage we have

80, 200, 80 30, 10.

On the third stage we have

80, 80, 160, 60, 20.

On the fourth stage we have

80, 80, 80, 120, 40.

On the fifth stage we have

80, 80, 80, 80, 80.

CHAPTER 7 ANSWERS

7.1.1 x = 2 or x = −2.

7.1.2 x = 3 or x = −2.

7.1.3 x = −3 or x = 2.

7.1.4 x = 5 or x = −1.

7.1.5 x = 1 or x = −1.

7.1.6 If eggs had cost x cents less per dozen, I would have saved cents per egg. If eggs had cost x cents more per dozen, I would have lost cents per egg. The difference between these two prices is per egg. Thus if I buy x + 3 eggs and the total difference is 3 cents, I can write

which is to say x^2 − 3x − 18 = (x − 3)(x + 6) = 0. Since x has to be a positive number, x = 3.

7.1.7 Let x be the original number of people renting the bus. Each person must originally pay dollars. After six people do not show up there are x − 6 people, and each must pay dollars. We need

Rearranging,

(x − 6)(2300 + 7.5x) = 2300x ⇒ 7.5x^2 − 45x − 13800 = 0 ⇒ 75x^2 − 450x − 138000 = 0 ⇒ 75 (x + 40)(x − 46) = 0 ⇒ x ∈ {−40, 46}.

Since x must be positive, there were 46 people on the bus originally.

7.1.8 One has Later on you will learn how to solve this equation and see that There is some deep mathematics going on here. That the infinite expression makes sense depends on the concept of convergence, which is addressed in calculus courses.

7.1.9 We have x^2 = 2, and since x is positive, Like Exercise 7.1.8, this exercise is deep and depends on the concept of convergence.

7.1.10 (x + 1)(x + 19).

7.1.11 (x + 1)(x + 13).

7.1.12 (z + 7)2.

7.1.13 (x + 1)2.

7.1.14 (x + 3)2.

7.1.15 (x + 5)2.

7.1.16 (x + 3)2.

7.1.17 (x + 6)2.

7.1.18 z = 0 and z = 2. Remember the Zero Product property states that for all real numbers x and y:if xy = 0, then x = 0 or y = 0. Therefore (z − 2) = 0 and z = 0.

7.1.18 or n = 2.

7.1.20 and t = −2.

7.1.21. and x = 2.

7.1.22. and n = 13.

7.2.1. x^2 + 30x + 225.

Add to complete the square, and substitute in b = 30. This quadratic equation can be written in a complete square, (x + 15)2. The perfect square was complete with the number 225.

7.2.2. x^2 − 6x − 36.

Add to complete the square, and substitute in This quadratic equation can be written in a complete square, (x − 3)2. The perfect square was complete with the number 9.

7.2.3. h^2 + 10h + 25.

Add to complete the square, and substitute in b = 10. This quadratic equation can be written in a complete square, (h + 5)2. The perfect square was complete with the number 25.

7.2.4. x^2 − 16x + 64.

Add to complete the square, and substitute in b = 16. This quadratic equation can be written in a complete square, (x − 8)2. The perfect square was complete with the number 64.

7.2.5. p^2 + 36p + 324.

Add to complete the square, and substitute in b = 36. This quadratic equation can be written in a complete square, (x + 18). The perfect square was complete with the number 324.

7.2.6. x^2 + 28x 196.

Add to complete the square, substitute in b = 28. This quadratic equation can be written in a complete square, (x + 14)2. The perfect square was complete with the number 196.

7.2.7. x^2 − 26x + 169.

Add to complete the square, and substitute in b = −26. This quadratic equation can be written in a complete square, (x − 13)2. The perfect square was complete with the number 169.

7.2.8. h^2 − 10h + 25.

Add to complete the square, and substitute in b = −10. This quadratic equation can be written in a complete square, (h − 5)2. The perfect square was complete with the number 25.

7.2.9. x^2 + 2x + 1

Add to complete the square, and substitute in b = 2. This quadratic equation can be written in a complete square, (x + 1)2. The perfect square was complete with the number 1.

7.2.10. x^2 − 2x + 1.

Add to complete the square, and substitute in b = −2. This quadratic equation can be written in a complete square, (x − 1)2. The perfect square was complete with the number 1.

7.3.1 p ≈ 1.11 or p = −8.11.

Use the quadratic formulat to solve p^2 + 7p − 9 = 0 and substitute in a = 1, b = 7, and c = −9.

7.3.2 x ≈ 1.84 or x ≈ −1.09.

Use the quadratic formulat to solve 4x^2 − 3x − 8 = 0 and substitute in a = 4, b = −3, and c = −8.

7.3.3 f ≈ 3.09 or f = −0.76.

Use the quadratic formulat to solve 3f^2 − 7f − 7 = 0 and substitute in a = 3, b = −7, and c = −7.

7.3.4 z = −1 or

Use the quadratic formulat to solve 2z^2 + 7z + 5 = 0 and substitute in a = 2, b = 7, and c = 5.

7.3.5 q = −1.

Use the quadratic formulat to solve 2q^2 + 4q + 2 = 0 and substitute in a = 2, b = 4, and c = 2.

7.3.6 n ≈ 4.91 or n ≈ 0.41.

Use the quadratic formulat to solve 2n^2 − 9n − 4 = 0, and substitute in a = 2, b = −9, and c = −4.

7.3.7 x ≈ 1.92 or x ≈ 1.17.

Use the quadratic formulat to solve 4x^2 − 3x − 9 = 0, and substitute in a = 4, b = −3, and c = −9.

7.3.8 u ≈ 0.59 or x ≈ −2.39.

Use the quadratic formula to solve 5u^2 + 9u − 7 = 0, and substitute in a = 5, b = 9, and c = −7.

7.3.9 w = 0 or w ≈ 3.28 or w ≈ 1.22.

Notice that this equation can be rewritten as w(2w^2 − 9w + 8) = 0, so one solution is w = 0. Let’s find the rest of possible solutions by using the quadratic formulat to solve 2w^2 9w + 8 = 0, and substitute in a = 2, b = −9, and c = 8.

7.3.10 m = 0 or w ≈ −0.27 or w ≈ −3.73.

Notice that 2m^3 + 8m^2 + 2m = 0, can be rewritten as m(2m^2 + 8m + 2) = 0, so one solution is m = 0. Let’s find the rest of possible solutions by using the quadratic formulat to solve 2m^2 + 8m + 2 = 0, and substitute in a = 2, b = 8, and c = 2.

CHAPTER 8 ANSWERS

8.2.1 x > − 2.

In order to solve this expression, first collect like terms on one side of the expression then manipulate the expression such that the unknown variable x is on one side and the rest of the expression is grouped to the other side of the inequality. The following are the steps to solve this problem.

8.2.2 x ≤ − 3.

In order to solve this expression, first multiply the right-hand expression. Collect like terms on one side of the expression, then manipulate the expression such that the unknown variable x is on one side and the rest of the expression is grouped to the other side of the inequality. The following are the steps to solve this problem.

8.2.3 x > 1.

In order to solve this expression, first multiply the right-hand expression. Collect like terms on one side of the expression then manipulate the expression such that the unknown variable x is on one side and the rest of the expression is grouped to the other side of the inequality. The following are the steps to solve this problem.

8.3.1 (A) x < 75 Let x be the unknown number, therefore the expression can be written as:

8.3.2 $10.75 x ≤ $860 Let x equal the number of hours Dalya works. If she works the maximum number of hours, or 40, per pay will be 40($10.75), which equals $430 per week. Therefore, the inequality expression is set up so that the wage times the number of hours is equal to or less than 2($430), for two weeks (biweekly).

8.3.3 s < 18 Set up the perimeter of a square formula p = 4s with an inequality or 72 > 4s or 18 > s, by dividing both sides by 4, which can be rewritten as s < 18.

8.3.4 The velocity of the ball will be between 30 and 60 feet per second between 0.5 and 1.5 seconds after initial launch. Set up the compound inequality, and then solve for t:

Remember that since we divided by a negative number, we flipped

the inequality signs. Also, we rearrange the answer such that the

time range is listed between 0.5 and 1.5 seconds.

8.3.5 Robert needs to invest $30,000 at 11% to obtain a $2,700 interest income from his mutual funds account. Set up the simple interest formula equation as I = Prt, where I is the interest, P is the initial investment or principal, and r is the rate of interest expressed in decimal, and t is the time in years. Let x be the amount to invest in mutual funds. Then, we will have 10,000 − x left to invest in the simple interest account. The interest on the mutual funds investment will be x(0.11)(1) = 0.11x. The interest on the simple interest account will be (10000 − x)(0.03)(1) = 300 − 0.03x. Then the total interest is 0.11x(300−0.03x) = 0.08x + 300. The minimum amount to invest in mutual funds is $2,700, so 0.08x + 300 ≥ 2700 → 0.08x ≥ 2400 → x ≥ 30000. In other words, Robert will need to invest at least $30,000 in mutual funds in order to get $2,700 in interest income.

8.4.1 x > −5 or x > −15.

The following are the steps to solve the compound inequalities:

x − 10 > −15 or x + 1 < − 14

x − 10 + 10 > − 15 + 10 or x + 1 − 1 < − 14 − 1

x − 15 or x < −15.

8.4.2 >x > 8 or x < −6.

The following are the steps to solve the compound inequalities:

x − 15 > −7 or x + 19 < 13

x − 15 + 15 > − 7 + 15 or x + 19 − 19

x > 8 or x < −6.

8.4.3 3 ≤ x ≤ 6.

The following are the steps to solve the compound inequalities:

8.4.4 −6 ≤ x ≤ −14.

The following are the steps to solve the compound inequalities:

8.4.5 8 < y < 10.

The following are the steps to solve the compound inequalities:

8.4.6 2 < m < 5.

The following are the steps to solve the compound inequalities:

8.4.7 b ≤ −1 or b > 8.

The following are the steps to solve the compound inequalities:

b − 4 ≤ −5 or b + 1 > 9

b − 4 + 4 ≤ −5 + 4 or b + 1 − 1 > 9 − 1

b ≤ −1 or b > 8.

8.4.8 a ≥ 12 or a < 10.

The following are the steps to solve the compound inequalities:

a − 5 ≥ 7 or a < 10

a − 5 + 5 ≥ 7 + 5 or a < 10

a ≥ 12 or a < 10

8.4.9 z > 0 or z > −10.

The following are the steps to solve the compound inequalities:

z − 8 < −8 or z − 12 < −22,

z − 8 + 8 ≥ −8 + 8 or z − 12 + 2 < −22 + 12

z > 0 or z > −10.

8.4.10 −4 < c < 8.

The following are the steps to solve the compound inequalities:

−5 > c − 13 < −17

−5 + 13 > c − 13 + 13 < −17 + 13

8 > c > −4 or −4 < c < 8.

8.4.11 x ≥ 16 or x ≤ 4.

The following are the steps to solve the compound inequalities:

x − 1 ≥ 15 or x + 18 ≤ 12

x − 1 + 1 ≥ 15 + 1 or x + 8 − 8 ≤ 12 − 8

x ≥ 16 or x ≤ 4.

8.4.12 y ≥ 6 or y < −6.

The following are the steps to solve the compound inequalities:

y + 9 ≥ 15 or y + 4 < −2

y + 9 − 9 ≥ 15 + 9 or y + 4 − 4 < −2 −4

y ≥ 6 or y < −4.

8.4.13 −3 ≤ x ≤ 5.

The following are the steps to solve the compound inequalities:

8.4.14 −10 ≤ n ≤ 10.

The following are the steps to solve the compound inequalities:

−3 ≤ n + 7 ≤ 17 → −3 −7 ≤ n + 7 − 7 ≤ 17 − 7

−10 ≤ n ≤ 10.

8.4.15 −5 < p < 1.

The following are the steps to solve the compound inequalities:

17 > p + 16 > 11 → 17 − 16 > p + 16 − 16 > 11 − 16

1 > p > −5 or −5 < p < 1.

8.4.16 x < 3 or ≈x ≥ 4.

The following are the steps to solve the compound inequalities:

x − 2 < 1 or x − 12 ≤ −8

x − 2 + 2 < 1 + 2 or x − 12 + 12 ≥ −8 + 12

x < 3 or x ≥ 4.

8.4.17 9 < z ≤ 15.

The following are the steps to solve the compound inequalities:

11 < z + 2 ≤ 17 → 11 − 2 < z + 2 − 2 ≤ 17 − 2

9 < z ≤ 15.

8.4.18 −5 > p > −8.

The following are the steps to solve the compound inequalities:

−7 > p − 3 > − 11 → −7 + 3 > p − 3 + 3 > − 11 + 3

−5 > p > −8.

8.4.19 n > 16 or v < 14.

The following are the steps to solve the compound inequalities

n − 9 > 7 or v − 20 < −6

n − 9 + 9 > 7 + 9 or v − 20 + 20 < −6 + 20

n > 16 or v < 14

8.4.20 v ≤ 1 or >v 8.

The following are the steps to solve the compound inequalities:

v + 6 ≤ 7 or v + 6 > 14

v + 6 − 6 ≤ 7 − 6 or v + 6 − 6 > 14 − 6

v ≤ 1 or v > 8.

APPENDIX C: ANSWER KEY FOR ODD-NUMBERED REVIEW EXERCISES

1.

(B) x − 3. The excess of a number (x) over another number (3) is another way of saying the difference between two numbers or x and 3.

3.

(B) “The square of a number is reduced by the reciprocal of the number.”

5.

(A) 1 (−100) ÷ (10) ÷ (−10) = (−10) ÷ (−10) = 1.

7.

(B) Remember that division is the same as multiplying a number by its reciprocal:

9.

(C) 4. Remember the order of operations or the acrynom PEMDAS: Parentheses first, Exponents, Multiplication and Division (from left to right), the Addition and Subtraction (also, from left to right). To answer this question, first divide 1024 by 512 then multiply by 2 or (1024 ÷ 512) × 2 = 2 × 2 = 4.

11.

(B) 9 1 − (−2)3 = 1 − (−8) = 1 + 8 = 9.

13.

(A) t + w.

15.

(A) Find the common denominator for all three terms and solve the expression:

17.

(C) −5.

19.

(C) Find the common denominator (12) and solve as follows:

21.

(E) none of these.

23.

(B) 7x − 4. First, reduce the terms inside of the parentheses, then collect like terms to reduce the expression:

25.

(A) −6x^2 − 4x + 10. Multiply through each expression and collect like terms: −2(x^2 − 1) + 4(2 − x − x^2) = − 2x^2 + 2 + 8 − 4x − 4x^2 = (−2x^2 − 4x^2) − 4x + (2 + 8) = −6x^2 − 4x + 10

27.

(B) 2a − 2b. −a + 2b + 3a − 4b = (−a + 3a) + (2b − 4b) = 2a − 2b.

29.

(B) First, find a common denominator of 12 then solve the expression:

31.

(B) 2x − 3.

(x + x) + (−2 −1) = 2x − 3.

33.

(A) −11x + 6.

(−7x + 4x) + (4 + 2) = −11x + 6.

35.

(C)

37.

(D) 3^7. 3^3. 3^4 = 3(3+4) = 3^7.

39.

(B)

2(1111)(5−4) = 2(1111)1 = 2(1111) = 2222.

41.

(C)

43.

(C)

45.

(A) x^3 − 8. (x − 2)(x^2 + 2x + 4) = x^3 + 2x^2 + 4x − 2x^2 − 4x − 8 =

x^3 + (2x^2 − 2x^2) + (4x − 4x) − 8 = x^3 − 8.

47.

(A)

49.

(D) x^6 y^7 z^4. (x^4 y^3 z^2)(x^2 y^4 z^2) = x(4+2)y(3+4)z(2+2) = x^6 y^7 z^4.

51.

(B) 4x^2 − 4x + 1. (2x − 1)(2x + 1)−2(2x − 1) = (4x^2 − 2x + 2x − 1) − 4x + 2 = 4x^2 − 4x + 1.

53.

(A) 2a^2 + 8. (a + 2)2 + (a − 2)2 = (a^2 + 4a + 4) + (a^2 − 4a + 4) = 2a^2 + 8.

55.

(B) x^3 − 2x^2 − 7x + 2. (x + 2)(x^2 − 4x + 1) = x^3 − 4x^2 + x + 2x^2 − 8x + 2 = x^3 − 2x^2 − 7x + 2.

57.

(A) x(2x − 3).

59.

(D) (x − 1)(x − 2).

61.

(B) x^2 − 2x − 1.

63.

(D)

65.

(B) x + 7.

67.

(D) x^2 − 3x + 9.

69.

(D) x^12.

71.

(C) (x + 3)(x − 4).

73.

(C) x(x + 3)(x − 9). x^3 − 12x^2 + 27x = x(x^2 − 12x + 27) = x(x − 3)(x − 9).

75.

(A) a(a + 8)2. a^3 + 16a^2 + 64a = a(a^2 + 16a + 16a + 64) = a(a + 8)2.

77.

(B) 2x^2 − x − 3.

79.

(B)

81.

(A)

83.

(B)

85.

(C) x = 2b.

87.

(A) x = 1. 2x − 1 = x → 2x − 1 + 1 = x + 1 → 2x − x = 1 → x = 1.

89.

(B) 2x − 1 = 1 − 3x → 2x − 1 + 1 = 1 − 3x + 1 → 2x = 2 − 3x →

91.

(A)

93.

(D)

95.

(B) $34. Let P represent Peter’s share, A represent Paul’s share, and M represent Mary’s share. Then P + A + M = 153, P = 2A, M = 3P Rewrite each expression in terms of Peter’s share: P + A + M = 153 →

97.

(B) 7. Let J represent Jane’s age, B represent Bob’s age, and I represent
Bill’s age. Then J + B + I = 27, J = 2B + 3, I = B − 4. J + B + I = 27 → (2B + 3) + B + B − 4 = 27 →

4B − 1 = 27 → 4B = 28 → B = 7.

99.

(B) 36. Let F represent a father’s current age and S his son’s current age, where F = 4S. In 24 years the father’s age will be F = 24 and his son’s age will be S + 24, then F + 24 = 2(S + 24) → 4S + 24 = 2S + 48 → 2S = 24 → S = 12. Therefore the son’s current age is 12 years of age, and the son’s age in 24 years is 12 + 24 = 36.

101.

(B) x = −1 or x = −2. x^2 + 3x + 2 = 0 → (x + 1)(x + 2) = 0 → x = −1, x = −2.

103.

(A) x = −1 or x = 2. x^2 − x − 2 = 0 → (x + 1)(x − 2) = 0 → x = −1, x = 2.

105.

(C) x < −1.

107.

(A) x ≤ −1. 2 + 5x ≤ −3 → 2 + 5x - 2 ≤ −3 − 2 → 5x ≤ −5 →

109.

True.

111.

False. x(2x + 3) = 2x^2 + 3x.

113.

(A) a(−1), (B) −1 · a, (C) and (D)

115.

(B) (a − b) ÷ 2, (C) (D) and (E)

117.

(A) (a − 2b)(a − 2b), (D) a^2 − 4ab + 4b^2, and (E) −2(2b−a)(a−2b).

119.

(A) 6x^3y^6, (B) 6(xy^4)2, and (E)

121.

(A) x ≤ 2.

123.

(B) x ≤ 65.

125.

(A) −8 < x < − 2.

127.

(B) This problem can be solved using the following steps:

APPENDIX D: THEOREMS, COROLLARIES, AND PROOFS

Theorem 2.1 (Cancellation Law) Let m, n, k be natural numbers with n ≠ 0 and k ≠ 0. Then

Proof: Prove this for m ≤ n. For m > n the argument is similar. Divide the interval [0; 1] into n k pieces. Consider the k-th, 2k-th, 3k-th, …, nk-th markers. Since the nk-th marker has to be 1. Thus the n markers k-th, 2k-th, 3k-th, …, nk-th, form a division of [0; 1] into n equal spaces. It follows that the k-th marker is that is, the 2k-th marker is that is, etc., and so the mk-th marker is that is, as we wanted to prove.

Thus given a fraction, if the numerator and the denominator have any common factors greater than 1, that is, any nontrivial factors, we may reduce the fraction and get an equal fraction.

Theorem 2.2 (Sum of Fractions) Let a, b, c, d be natural numbers with b ≠ 0 and d ≠ 0. Then

Proof: From the Cancellation Law (Theorem 42),

proving the theorem.

The formula obtained in the preceding theorem agrees with that of (2.2) when the denominators are equal. For, using the theorem,

where we have used the distributive law.

Observe that the trick for adding the fractions in the preceding theorem was to convert them to fractions of the same denominator.

Theorem 2.3 (Multiplication of Fractions) Let a, b, c, d be natural numbers with b ≠ 0 and d ≠ 0. Then

Proof: First consider the case when a = c = 1. Start with a unit square and cut it horizontally into b equal segments. Then cut it vertically into d equal segments. We have now bd equal pieces, each one having an area of Since each piece is in dimension by , we have shown that

Now construct a rectangle of length and width Such a rectangle is obtained by concatenating along its length a segments of length and along its width c segments of length . This partitions the large rectangle into ac subrectangles, each of area Hence the area of the by rectangle is from where

proving the theorem.

Theorem 2.4 (Division of Fractions) Let a, b, c, d be natural numbers with b ≠ 0, c ≠ 0, d ≠ 0. Then

that is, in definition 53 is

Proof: Let us prove that satisfies the definition of division of fractions. Observe that

and hence is the right result for the division of fractions. Could there be another fraction, say that satisfies definition 53? Suppose

Then

Hence is the only fraction that satisfies the definition of fraction division.

Theorem 4.1 (First Law of Exponents) Let a be a real number and m, n natural numbers. Then

am an = am + n

Proof: We have

Theorem 4.2 (Second Law of Exponents) Let a ≠ 0 be a real number and m, n natural numbers, such that m ≥ n. Then

Proof: We have

Theorem 4.3 (Third Law of Exponents) Let a be a real number and m, n natural numbers. Then

(am)n = amn

Proof: We have

Theorem 4.4 (Fourth Law of Exponents) Let a and b be real numbers and let m be a natural number. Then

(ab)m = ambm

Proof: We have

Theorem 4.5 (Square of a Sum) For all real numbers a, b the following identity holds:

(a ± b)2 = a^2 ± 2ab + b^2

Proof: Using the distributive law,

and so (a + b)2 = a^2 + 2ab + b^2. Putting −b instead of b in this last identity,

(a − b)2 = (a + (−b))2 = a^2 + 2a(−b) + (−b)2 = a^2 − 2ab + b^2

proving both identities.

Theorem 4.6 (Difference of Squares) For all real numbers a, b the following identity holds:

(a + b)(a − b) = a^2 − b^2

Proof: Using the distributive law,

proving the identity.

Theorem 4.7 Let a and b be real numbers. Then

(a + b)3 = a^3 + 3a^2b + 3ab^2 + b^3

Proof: Using the distributive law and the square of a sum identity,

as claimed.

Corollary 4.1 Let a and b be real numbers. Then

(a − b)3 = a^3 − 3a^2b + 3ab^2 − b^3

Proof: Replace b by −b in Theorem 163, obtaining

as claimed.

Theorem 4.8 (Sum and Difference of Cubes) For all real numbers a, b the following identity holds:

(a ± b)(a^2 ∓ ab + b^2) = a^3 ± b^3

Proof: Using the distributive law,

from where the sum of cubes identity is deduced. We obtain

(a − b)(a^2 + ab + b^2) = a^3− b^3

upon replacing b by −b in the above sum of cubes identity.

Verification:

APPENDIX E: KEYWORDS

Addition

Additive identity

Algebra

Algebraic equation

Algebraic formula

Algebraic fraction

Algebraic generalization

Arithmetic operation

Arithmetic problem

Arithmetic progression

Axiom

Coefficient

Cumulative

Commutativity

Complete the square

Compound linear equality

Concatenation

Cube

Cubic root

Difference of cube

Difference of square

Distributive Law

Dividend

Division

Divisor

Elementary algebra

Equality

Equation

Equation of Condition

Exponential

Exponent

Exponentiation

Factor

Factoring

Factorization

Formula

Geometric progression

Greatest common divisor

Index of root

Interval

Irrational expression

Irrational number

Law of formation and conjecture

Length of interval

Linear Equation One-Variable

Magic square

Simple Equation

Multiplication

Multiplicative identity

Natural number

nth power

nth term

One–variable linear equality

One–variable linear inequality

Place-value notation system

Polynomial

Positional notation

Positive integer

Power

Product

Quadratic equation

Quadratic trinomials

Quadratic formula

Quotient

Rational expression

Rational number

Real number

Reciprocal

Recursion method

Remainder

Root

Root extraction

Special factorization

Square

Square of difference

Square of sum

Square root

Subtraction

Sum of cube

Symbol

Symbolic expression

Term-by-term division

Trice

Unknown variable

Variable

Vertex

Word Problem

INDEX

Abstract system, 3

Additive identity, 14

Algebra

division. see Division

elementary, 4

equation, 4

important, 3

learning algebra, 3

linear equations. see Linear equations

linear inequalities. see Linear inequalities

multiplication operations. see Multiplication

puzzles, 4–5

quadratic equations. see Quadratic equations

Algebraic expression, 37

Arithmetic operations

fractions, 19

addition of, 20–21

division, 23

multiplication of, 22

subtraction of, 22

integers

negative, 26, 29

positive, 25, 29

irrational number, 32

natural numbers

additive and multiplicative identity, 14

defined, 13

definition of division, 17

definition of subtraction, 17

distributive law, 15

positional notation/place-value notation system, 13

square root, 16–17

rational numbers, 31

real numbers, 33

Cancellation law, 19

Closure, 14

Common denominator, 21

Commutative law, 14, 38, 39

Compound linear inequalities, 105–106

Distributive law, 15, 40, 41

Division

factor

factoring I, 65–67

factoring II, 68–69

rational expressions, 72–74

special factorizations, 70–71

long division, 58–65

of fractions, 23

term-by-term division, 57–58

Elementary algebra, 4

Fractions, 19

addition of, 20–21

division of, 23

multiplication of, 22

subtraction of, 22

Integers

negative, 26, 29

positive, 25, 29

Irrational number, 32

Least common denominator, 21

Linear equations, 79

miscellaneous linear equations, 83–86

simple equations, 79–82

word problems, 87–88

Linear inequalities

compound linear inequalities, 105–106

intervals, 101–102

one-variable, 102–103

Multiplication

cube

difference of, 56

sum of 55, 56

distributive law, 49–52

laws of exponents, 45–48

negative exponents, 48

of fractions, 22

square

difference of, 53

sum, 52–53

Multiplicative identity, 14

Natural numbers

additive and multiplicative identity, 14

defined, 13

definition of division, 17

definition of subtraction, 17

distributive law, 15

long division of, 58

positional notation/place-value notation system, 13

square root, 16–17

Negative integers, 26

Numerical coefficient, 37

One-variable linear inequalities, 102–103

word problems, 103–104

Positional notation/place-value notation system, 13

Positive integers, 25

Puzzles, 4–5

Quadratic equations, 91–92

complete square, 94

formula, 95–96

Quadratic trinomials, 68

Rational numbers, 31

Real numbers, 33

Reciprocal, 23