Contents · College Algebra MTH-100 Essex County College Division of Mathematics Sakai Web Project...

College AlgebraMTH-100

Essex County CollegeDivision of Mathematics

Sakai Web ProjectClass Notes

Contents

1 Syllabus 6

2 Introduction 11

3 Review for Assignment mth.100.02.01 123.1 Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4 Review for Assignment mth.100.02.02 164.1 Solving Simple Algebraic Word Problems . . . . . . . . . . . . . . . . . . . . 164.2 Solving Simple Consecutive Integers . . . . . . . . . . . . . . . . . . . . . . . 174.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

5 Review for Assignment mth.100.02.03 195.1 Mixture Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.2 Uniform Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

6 Review for Assignment mth.100.02.04 236.1 Properties of Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

7 Review for Assignment mth.100.03.01 267.1 Points, Midpoints and Distance . . . . . . . . . . . . . . . . . . . . . . . . . 267.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

8 Review for Assignment mth.100.03.02 278.1 Introductory Functions and Relations . . . . . . . . . . . . . . . . . . . . . . 278.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

9 Review for Assignment mth.100.03.03 299.1 Graphing Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299.2 Special Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

10 Review for Assignment mth.100.03.04 3210.1 Slope of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3210.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

11 Review for Assignment mth.100.03.05 3411.1 Equations of a Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3411.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

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12 Review for Assignment mth.100.03.06 3612.1 Parallel and Perpendicular Lines . . . . . . . . . . . . . . . . . . . . . . . . . 3612.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

13 Review for Exam #1 38

14 Review for Assignment mth.100.04.01 3914.1 System of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 3914.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

15 Review for Assignment mth.100.04.02 4215.1 Addition or Elimination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4215.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

16 Review for Assignment mth.100.04.04 4516.1 Word Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4516.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

17 Review for Assignment mth.100.05.01 4717.1 Integer Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4717.2 The Basic Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4717.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

18 Review for Assignment mth.100.05.02 5018.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5018.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

19 Review for Assignment mth.100.05.03 5219.1 Polynomial Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5219.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

20 Review for Assignment mth.100.05.04 5520.1 Polynomial Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5520.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

21 Review for Assignment mth.100.05.05 5721.1 Polynomial Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5721.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

22 Review for Assignment mth.100.05.06 6022.1 Special Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6022.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

23 Review for Assignment mth.100.05.07 6223.1 Zero-Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6223.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

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25 Review for Assignment mth.100.06.01 6625.1 Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6625.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

26 Review for Assignment mth.100.06.02 6926.1 Finding an LCM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6926.2 Using an LCM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6926.3 Addition and Subtraction of Rational Expressions . . . . . . . . . . . . . . . 70

26.3.1 Same Denominator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7026.3.2 Different Denominator . . . . . . . . . . . . . . . . . . . . . . . . . . 71

26.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

27 Review for Assignment mth.100.06.03 7427.1 Complex Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7427.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

28 Review for Assignment mth.100.06.05 7628.1 Solving of Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 76

28.1.1 Essential Rules and Procedures for Solving Equations . . . . . . . . . 7628.1.2 Work Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

28.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

29 Review for Assignment mth.100.07.01 7929.1 Rational Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

29.1.1 The Basic Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7929.1.2 Meaning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

29.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

30 Review for Assignment mth.100.07.02 8230.1 Simplifying Radical Expressions . . . . . . . . . . . . . . . . . . . . . . . . . 82

30.1.1 The Product Property . . . . . . . . . . . . . . . . . . . . . . . . . . 8230.1.2 Initial Radical Simplification . . . . . . . . . . . . . . . . . . . . . . . 8230.1.3 Adding/Subtracting Radicals . . . . . . . . . . . . . . . . . . . . . . 8330.1.4 Multiplying Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . 8330.1.5 Dividing Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8330.1.6 Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

30.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

31 Review for Assignment mth.100.07.03 8931.1 Solving Radical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8931.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

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32 Review for Assignment mth.100.07.04 9132.1 Introduction to Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . 91

32.1.1 Operations on Complex Numbers . . . . . . . . . . . . . . . . . . . . 9132.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92


34 Review for Assignment mth.100.08.01 9634.1 Solving Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 9634.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

35 Review for Assignment mth.100.08.02 9935.1 Perfect Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9935.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

36 Review for Assignment mth.100.08.03 10136.1 The Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10136.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

37 Review for Assignment mth.100.08.04 10337.1 Related to Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . 10337.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

38 Review for Assignment mth.100.09.01 10438.1 Graphing Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10438.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

39 Review for Assignment mth.100.11.02 10639.1 Graphing Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10639.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107


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Errors in this document should be reported to Ron Bannon.

[email protected]

This document may be shared with students and instructors of MTH 100 only.

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1 Syllabus

Essex County CollegeMathematics and Physics Division

Syllabus for MTH 100 §006 — Introductory College MathematicsSpring 2012 Class Syllabus

Lecturer: Ron Bannon

Office: 2210Phone: 973-877-1886Email: [email protected]

Course Website: http://mth100.mathography.org/

Regular Office Hours: Tuesday & Thursday 8:30 a.m.–10:55 a.m.Appointment Office Hours: Monday & Wednesday 11:30 a.m.–12:55 p.m.Class Meetings: Tuesday, Thursday & Friday 11:30 a.m.–12:50 p.m.Classroom: Main Building, Room 4142First Class §013: Tuesday, January 10, 2012Last Class §013: Friday, April 20, 2012

• Course Description: This course covers topics including special products, factoring,and other operations on polynomials, rational and radical expressions, integral and ra-tional exponents, and scientific notation. In addition, analytic and graphical methodsof solving linear equations, linear systems, literal equations, and elementary polyno-mial equations are covered. Students are also introduced to the analytic geometry offunctions, including lines, circles, and parabolas. Diverse applications are emphasizedthroughout the course.

• Required Materials: Textbook Introductory College Mathematics, 8th edition (cus-tom version of the textbook Intermediate Algebra, an Applied Approach, 8th edition,includes AIM Practice Sheets and Nolting Study Skills Workbook), by Aufmann &Lockwood; published by Brooks/Cole, Cengage Learning, 2011; Package (includestextbook, DVDs, Student Solutions Manual, and WebAssign access card) ISBN: 1-111-66227-4. Students should also have access to a simple calculator, notebook, andpencil or pen.

• Course Prerequisites Grade of C or better in MTH 092 or placement.

• Course Prerequisites & Course Co-requisites None.

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• General Education Goals

MTH 100 is affirmed in the following General Education Foundation Category: Quanti-tative Knowledge and Skills. The corresponding General Education Goal is as follows:Students will use appropriate mathematical and statistical concepts and operations tointerpret data and to solve problems.

• Course Goals Upon successful completion of this course, students should be able todo the following:

1. demonstrate knowledge of the fundamental concepts and theories from algebraand geometry;

2. utilize various problem-solving and critical-thinking techniques to set up and solvereal-world applications;

3. communicate accurate mathematical terminology and notation in written and/ororal form in order to explain strategies to solve problems as well as to interpretfound solutions; and

4. use calculators effectively as a tool to solve such problems as those describedabove.

• Measurable Course Performance Objectives (MPOs) Upon successful comple-tion of this course, students should specifically be able to do the following:

1. Demonstrate knowledge of the fundamental concepts and theories from algebraand geometry:

1.1 solve equations of various types (linear, quadratic, literal, rational, and rad-ical);

1.2 solve linear inequalities;

1.3 solve systems of linear equations;

1.4 factor a polynomial;

1.5 perform basic operations on polynomials, rational expressions, radicals, andcomplex numbers;

1.6 simplify exponential expressions;

1.7 find the equation of a line based on given geometric properties;

1.8 graph lines, parabolas and circles in the Rectangular Coordinate System; and

1.9 determine whether a given relation is a function, find its domain, and usefunction notation.

2. Utilize various problem-solving and critical-thinking techniques together with al-gebra to set up and solve application problems taken from a variety of disciplines:

2.1 apply algebraic methods to solve varied real-world applications (such as, con-secutive integer problems, coin/stamp problems, distance problems, invest-ment problems, area problems, and work problems) that can be modeled by alinear equation, quadratic equation, rational equation, or system of equations.

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3. Communicate accurate mathematical terminology and notation in written and/ororal form in order to explain strategies to solve problems as well as to interpretfound solutions:

3.1 write and explain solutions to application problems related to the coursematerial using appropriate mathematical terminology and notation.

4. Use calculators effectively as a tool to solve such problems as those describedabove:

4.1 use a calculator to perform basic arithmetic operations such as evaluatingpowers and roots.

• Methods of Instruction

Instruction will consist of a combination of lectures, class discussion, individual study,and computer lab work.

• Class Requirements All students are required to:

1. Read the textbook1 and do the suggested review problems in a timely manner.

2. Be an active participant in all classes.

3. Complete all written and/or electronic homework and adhere to assignment dead-lines.

4. Take exams/quizzes in class and adhere to the exam/quiz schedule.2

• Grading Criteria

1. Homework is worth 20%.

2. Three exams worth 20% each.

3. Cumulative final exam is worth 20%.

Tentative grade breakdown:

A if average ≥ 94%;

B+ if 88% ≤ average < 94%;

B if 82% ≤ average < 88%;

C+ if 76% ≤ average < 82%;

C if 70% ≤ average < 76%;

D if 64% ≤ average < 70%;

F if average < 64%;

1The e-book is available through WebAssign and students may wish to read the book while they are doingassignments online.

2Advanced notice will be given in class when an assessment is scheduled.

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• Academic Integrity Dishonesty disrupts the search for truth that is inherent in thelearning process and so devalues the purpose and the mission of the College. Academicdishonesty includes, but is not limited to, the following:

plagiarism — the failure to acknowledge another writers words or ideas or to giveproper credit to sources of information;

cheating — knowingly obtaining or giving unauthorized information on any test/examor any other academic assignment;

interference — any interruption of the academic process that prevents others fromthe proper engagement in learning or teaching; and

fraud — any act or instance of willful deceit or trickery.

Violations of academic integrity will be dealt with by imposing appropriate sanctions.Sanctions for acts of academic dishonesty could include the resubmission of an assign-ment, failure of the test/exam, failure in the course, probation, suspension from theCollege, and even expulsion from the College.

• Student Code of Conduct All students are expected to conduct themselves asresponsible and considerate adults who respect the rights of others. Disruptive behaviorwill not be tolerated. All students are also expected to attend and be on time allclass meetings. No cell phones or similar electronic devices are permitted in class.Please refer to the Essex County College student handbook, Lifeline, for more specificinformation about the Colleges Code of Conduct and attendance requirements.

• Tentative Class Schedule The course will follow, in order, the following sectionsand suggested problems.3 Prepared notes are available4 that will be used in class tohelp guide the student trough the material. We will move quickly, and students will betested on materials assigned. To remain current a student should come to each class.

– Chapter 2, §2.1: Solving First-Degree Equations

– Chapter 2, §2.2: Applications: Puzzle Problems

– Chapter 2, §2.3: Applications: Mixture and Uniform Motion Problems

– Chapter 2, §2.4: First-Degree Inequalities (Objective A)

– Chapter 3, §3.1: The Rectangular Coordinate System (Objectives A and B)

– Chapter 3, §3.2: Introduction to Functions

– Chapter 3, §3.3: Linear Functions (Objectives A, B, and C)

– Chapter 3, §3.4: Slope of a Straight Line

– Chapter 3, §3.5: Finding Equations of Lines (Objectives A and B)

3In the notes that follow you will see that assignments are in the format, chapter.section.exercise, matchingthe corresponding chapters, sections, and exercises from the textbook.

4http://mth100.mathography.org/

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– Chapter 3, §3.6: Parallel and Perpendicular Lines

– Chapter 4, §4.1: Solving Systems of Equations in Two Variables by Graphingand Substitution

– Chapter 4, §4.2: Solving Systems of Linear Equations by the Addition Method(Objective A)

– Chapter 4, §4.4: Application Problems

– Chapter 5, §5.1: Exponential Expressions (Objectives A, B, and C)

– Chapter 5, §5.2: Introduction to Polynomial Functions

– Chapter 5, §5.3: Multiplication of Polynomials (Objectives A, B, and C)

– Chapter 5, §5.4: Division of Polynomials (Objectives A and B)

– Chapter 5, §5.5: Factoring Polynomials

– Chapter 5, §5.6: Special Factoring (Objectives A, B, and D)

– Chapter 5, §5.7: Solving Equations by Factoring

– Chapter 6, §6.1: Multiplication and Division of Rational Expressions

– Chapter 6, §6.2: Addition and Subtraction of Rational Expressions

– Chapter 6, §6.3: Complex Fractions

– Chapter 6, §6.5: Rational Equations (Objectives A and B)

– Chapter 7, §7.1: Rational Exponents and Radical Expressions

– Chapter 7, §7.2: Operations of Radical Expressions

– Chapter 7, §7.3: Solving Equations Containing Radical Expressions

– Chapter 7, §7.4: Complex Numbers

– Chapter 8, §8.1: Solving Quadratic Equations by Factoring and by TakingSquare Roots

– Chapter 8, §8.2: Solving Quadratic Equations by Completing the Square

– Chapter 8, §8.3: Solving Quadratic Equations by the Quadratic Formula (nodiscussion of discriminant)

– Chapter 8, §8.4: Solving Equations that are Reducible to Quadratic Equations(Objectives B and C)

– Chapter 9, §9.1: Properties of Quadratic Functions (Objectives A and B, butno discussion of the zeros of a function)

– Chapter 11, §11.2: The Circle

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2 Introduction

You should make every effort to attend all scheduled classes. The notes that follow are notself-contained and you will need to attend class in order to make sense out of the contentthat follows. The notes just give structure to the classroom lectures, and will keep you ontrack with your assignments. You should note that almost every section that follow will referto an assignment in WebAssign. It is incredibly important that you visit

http://mth100.mathography.org

and get started with WebAssign immediately! You should also print out the complete notes(available on the course webpage) or purchase a copy (link provided online).

If you have any questions or concerns about WebAssign or the notes, you need to contactRon Bannon as soon as possible.

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3 Review for Assignment mth.100.02.01

3.1 Solving Equations

• An equation expresses the equality of two mathematical expressions.

Example: The left and right side of the equal sign are mathematical expressions.

2x + 3 = x− 1

• A solution of an equation is a number that, when substituted for the variable, resultsin a true equation.

Example: x = −4 is a solution of

2x + 3 = x− 1,

because

2 (−4) + 3 = (−4)− 1 ⇒ −5 = −5.

• To solve an equation means to find a solution of the equation. The goal is to rewritethe equation in the form variable = constant, because the constant is the solution.

Example: Given

2x + 3 = x− 1,

the solution is

x = −4.

• Essential Rules and Procedures for Solving Linear Equations

1. Simplify both sides of the equation. This may mean clearing the equation offractions and decimals.

2. Get the variable term on one side by adding or subtracting the same variableexpression to each side of the equation.

3. Get the constant term on the other side (opposing side of the variable term) byadding or subtracting the same constant to each side of the equation.

4. Divide both sides of the equation by the variable term’s numerical coefficient.

5. Check your solution in the original equation.

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Example: Solve for x.

6x + 3 (2x− 1) = 5 (x− 2) + 1 + 5x

6x + 6x− 3 = 5x− 10 + 1 + 5x

12x− 3 = 10x− 9 Simplified.

2x− 3 = −9 Subtract 10x from both sides.

2x = −6 Add 3 to both sides.

x = −3 Divide both sides by 2.

Now check it.

6 (−3) + 3 (2 (−3)− 1) = 5 (−3− 2) + 1 + 5 (−3)

−18 + 3 (−7) = 5 (−5) + 1 + (−15)

−18 + (−21) = (−25) + 1 + (−15)

−39 = −39 Q.E.D.

So, x = −3 is a solution to the equation 6x + 3 (2x− 1) = 5 (x− 2) + 1 + 5x.

What’s incredibly important here, is that equality demands equal treatment, and sincethis is an algebra class, we need to understand that treatment invariably relates toaddition, subtraction, multiplication and division. In essence, whatever you do to oneside, you must also do it to the other. Be fair!

3.2 Examples

We’ll do the following examples in class. In your notebook you should show all work foreach and every problem. Furthermore, you should learn to identify your answer by clearlyboxing them!

1. Is 5 a solution of the following equation?

6y − 5 = 5y

2. Solve and check.

z + 6 = 11

3. Solve and check.

x +1

3=

2

3

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4. Solve and check.

a− 2

5=

2

3

5. Solve and check.

−2

3x =

1

2

6. Solve and check.

5

6+

n

12= −1

4

7. Solve and check.

m + 1.32 = −2.39

8. Solve and check.

x

5= −3

9. Solve and check.

3b− 3 = 5b

10. Solve and check.

5

8− 2y

3=

5

4


7− 5z = 5z − 9


2x + 3− 5x = 9


1

6− 5t = 6


37

24=

7

8− 5x

6

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5z − 4 (3− 2z) = 2 (3z − 2) + 6


3y + 3 (y + 1) = 27


−2 [3x− 5 (2x− 3)] = 3x− 8


8− 5 (4− 5x) = 3 (4− x)− 8x


5 + 3 [1 + 2 (2x− 3)] = 6 (x + 5)


5 [x + 2 (3− x)] = 3 [2 (4− x)− 5]


5 [4− 2 (s− 7)] = 5 (2− 4s)


5− 2x

2+

x− 3

8=

3

8


0.5 (2x + 100)− 0.1 (x− 10) = 4

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4.1 Solving Simple Algebraic Word Problems

I’ll be brief, and so will most of the word problems that follow. The basic idea behindsolving an algebraic word problem is to identify what is the unknown. For example, giventhe following word problem:

Twenty percent of what number is seventy-five?

We first need to identify that we don’t know the number, so we say,“let x represent thenumber.” Then we simply reread the English and translate it into an algebraic equation.

20

100· x = 75

Now, solving for x, we have.

20

100· x = 75

1

5· x = 75

5 · 1

5· x = 5 · 75

x = 375

So the answer to the original question is, the number is 375.

For most students, the biggest problem is understanding what they’ve read. For example,let’s say we’re given the following word problem:

The sum of two numbers is twelve. The total of three times the smaller numberand six amounts to seven less than the product of four and the larger number.Find the two numbers.

After reading the problem you should understand that there are two numbers, and thatthey have been named small and large. We all need to be clear that we’re just identifyingone unknown though, that is, our problems have only one variable. So for this particularproblem we say, “let s represent the small number.” But what about the large one, isn’tthat also unknown. Yes, the large number is also unknown, but it is related to the smallnumber. Since we know that the small number and the large number, when added together,results in 12, we have:

(larger number) + s = 12,

so the larger number must be 12− s. Now let’s reread the problem to find the equation.

3s + 6 = 4 (12− s)− 7

If you solve this equation you’ll see that s = 5, hence the larger number is 7. The answer tothis particular word problem is, the numbers are 5 and 7.

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4.2 Solving Simple Consecutive Integers

It is often re-stated that, “God created the integers and all else was the work of man” as areferential adage coined by a very famous mathematician named Kronecker. So I guess weneed to learn to deal with integer problems, after-all they’ve become deified.

You’ll be given three types:

Consecutive integers in abstract algebraic form look like this:

x ∈ Z, {x, x + 1, x + 2, . . .} .

Consecutive even integers in abstract algebraic form look like this:

x ∈ Z and is even, {x, x + 2, x + 4, . . .} .

Consecutive odd integers in abstract algebraic form look like this:

x ∈ Z and is odd, {x, x + 2, x + 4, . . .} .

You should realize that these integers are ordered and you need to familiarize yourself withwords that describe order. For example, you may say first, second, or third ; or possiblesmallest and largest. The textbook uses simple words to describe order, and they are prettyconsistent, for example, if they say middle integer we should immediately know that there’sone above (last) and one below (first). Consistency in naming is important.

4.3 Examples

We’ll do the following examples in class. In your notebook you should show all work foreach and every problem. Furthermore, you should learn to identify your answer by boxingit! Also, you will not be given any credit unless you use algebra to solve, that is, you needto identify an unknown, and write the correct equation.

1. Four more than three times a number is thirteen. Find the number.

2. What number must be added to the numerator of 3/10 to produce the fraction 4/5?

3. The difference between nine times a number and six is twelve. Find the number.

4. The sum of two integers is twenty-eight. Seven times the smaller integer is sixteen morethan five times the larger integer. Find the integers.

5. Twice the difference between a number and twenty-five is three times the number. Findthe number.

6. One integer is eight more than another integer. The sum of the integers is twenty-six.Find the integers.

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7. The sum of three numbers is seventeen. The second number is twice the first number,and the third number is three less than the second number. Find the three numbers.

8. The sum of three consecutive integers is one hundred twenty-nine. Find the integers.

9. Five times the smallest of three consecutive odd integers is eight more than twice thelargest. Find the integers.5

10. The sum of two numbers is fifteen. One less that three times the smaller number is equalto the larger number. Find the numbers.

11. The sum of two numbers is eighteen. The total of three times the smaller number andtwice the larger number is forty-four. Find the two numbers.

5This problem has no solution, but you will still need to show all work in reaching this conclusion.

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5.1 Mixture Problems

Mixture problems are word problems where items or quantities of different values are mixedtogether. I recommend using a table/diagram/chart to organize the information for eachmixture problem you attempt. Using a table/diagram/chart allows you to think of onenumber at a time instead of trying to handle the whole mixture problem at once.

We’ll do a variety of problems in class, and it should be clear that each and everymixture problem is essentially the same—you need to clearly understand what is beingmixed together, and what is of interest in the mixture. For example, if you have two alloys(metal mixtures), (A) one weighing 10 pounds that is 30% silver and the other (B) weighing20 pounds that is 40% silver. You should (we’ll discuss in class) be able to answer thefollowing questions:

• How many pounds of silver are there in alloy (A)?

10 · 30% = 3 pounds of silver.

• How many pounds of silver are there in alloy (B)?

20 · 40% = 8 pounds of silver.

• How many pounds would you have in total if you combined alloy (A) and (B)?

10 + 20 = 30 pounds of alloy.

• How many pounds of silver would you have in total if you combined alloy (A) and (B)?

3 + 8 = 11 pounds of silver.

• If you combined alloy (A) and (B), what would the percent silver be of this alloy?

11

30· 100% ≈ 36.7%

5.2 Uniform Motion

Uniform motion is motion at a constant rate. For example, if you’re driving a car usingcruise control you are moving at a constant rate. That is, if you set the cruise control at 50mph (rate) and you drive for three hours (time), it should be clear that you travel 150 miles(distance).

Here’s the relationship:

distance = rate · time,

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or in algebraic form:

d = r · t.

Now when we are doing word problems we need to describe these three variables in termsof constants or one unknown only. For example, if we’re going x miles per hour for 3 hourswe have:

t = 3 , r = x , d = rt = 3x .

You should already be familiar with word problems of this type, but we will nonetheless goover it again.

5.3 Examples

We’ll do the following examples in class. In your notebook you should show all work foreach and every problem. Furthermore, you should learn to identify your answer by boxingit! Also, you will not be given any credit unless you use algebra to solve, that is, you needto identify an unknown, and write the correct equation.

1. A collection of seventeen coins has a value of $3.35. The collection contains dimes andquarters. Find the number of quarters in the collection.

2. A coin collection contains nickels, dimes, and quarters. There are twice as many dimesas quarters and five more nickels than dimes. The total value of all the coins is $4.10.How many quarters are in the collection?

3. A passenger train leaves a depot 1.5 hours after a freight train leaves the same depot.The passenger train is traveling 18 miles per hour faster than the freight train. Find therate of each train if the passenger train overtakes the freight train in 2.5 hours.

4. Two cyclists start from the same point and ride in opposite directions. One cyclist ridestwice as fast as the other. In three hours, they are eighty-one miles apart. Find the rateof each cyclist.

5. A coffee merchant combines coffee costing $5.50 per pound with coffee costing $4.00 perpound. How many pounds of each should be used to make thirty pounds of a blendcosting $4.70 per pound?

6. Tickets for a school play sold for $7.50 for each adult and $3.00 for each child. The totalreceipts for 117 tickets sold were $693.00. Find the number of adult tickets sold.

7. A silversmith combined pure silver that cost $16.90 per ounce with 70 ounces of a silveralloy that cost $15.20 per ounce. How many ounces of pure silver were used to make analloy of silver costing $16.20 per ounce?

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8. A stamp collection consists of 3¢, 12¢, and 15¢ stamps. The number of 3¢ stamps isfive times the number of 12¢ stamps. The number of 15¢ stamps is three less than thenumber of 12¢ stamps. The total value of the stamps in the collection is $2.49. Findthe number of 15¢ stamps in the collection.

9. A hospital staff mixed a 75% disinfectant solution with a 25% disinfectant solution. Howmany liters of each were used to make 30 L of a 40% disinfectant solution?

10. A long distance runner started on a course running at an average speed of six mph.One-half hour later, a second runner began the same course at an average speed of 7mph. How long after the second runner started will the second runner overtake the firstrunner?

11. A silversmith mixed 15 grams of a 80% silver alloy with 60 grams of a 25% silver alloy.What is the percent concentration of the resulting alloy? (Round to the nearest tenthof a percent.)

12. At noon a train leaves Wasington D.C., headed for Charleston, South Carolina, a distanceof five-hundred miles. The train travels at sixty mph. At one pm a second train leavesCharleston headed for Wasington D.C., traveling at fifty mph. How long after the thetrain leaves Charleston will the two trains pass each other?

13. How many quarts of water must be added to 9 quarts of an 80% antifreeze solution tomake a 45% antifreeze solution?

14. A passenger train leaves a train depot two hours after a freight train leaves the samedepot. The freight train is traveling twenty mph slower than the passenger train. Findthe rate of each train if the passenger train overtakes the freight train in three hours.

15. A butcher has some hamburger that is 21% fat and some that is 15% fat. How manypounds of each should be mixed to make 90 pounds of hamburger that is 17% fat?

16. After a sailboat had been on the water for three hours, a change in wind direction reducedthe average speed of the boat by five mph. The entire distance sailed was fifty-sevenmiles. The total time spent sailing was six hours. How far did the sailboat travel in thefirst three hours?

17. Two planes are 1425 miles apart and are traveling toward each other. One plane istraveling 110 miles per hour faster than the other plane. The planes meet in 1.5 hours.Find the speed of each plane.

18. A ferry leaves a harbor and travels to a resort island at an average speed of 18 milesper hour. On the return trip, the ferry travels at an average speed of 10 miles per hourbecause of fog. The total time for the trip is 7 hours. How far is the island from theharbor?

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19. Rubbing alcohol is typically diluted with water to 70% strength. If you need 4.2 oz of45% rubbing alcohol, how many ounces of 70% rubbing alcohol and how much watershould you combine?

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6.1 Properties of Inequalities

Let a,b, c, and d be real numbers.

1. Transitive Property.

a < b and b < c ⇒ a < c

2. Addition of Inequalities.

a < b and c < d ⇒ a + c < b + d

3. Addition of Constants.

a < b ⇒ a + c < b + c

4. Multiplying by Constants.

(a) If c > 0.

a < b ⇒ ac < bc

(b) If c < 0.

a < b ⇒ ac > bc

Okay, the above properties will be discussed, but for the most part they do not need tobe memorized. You’ll see that solving a linear inequality is nearly identical to solving alinear equation—you’re isolating the variable—but you need to be very careful wheneveryou multiply or divide by a negative number (see last rule). In class we will do manyexamples to illustrate what needs to be done.

6.2 Notation

Once an inequality is solved—the variable is isolated—you’ll need to get used to describingyour answers. Always keep in mind that your solutions should always be read from thevariable first. For example,

x > 2 and 2 < x,

are both read “x is greater than two.” That’s the English, but you’ll also need to describethe solution using notation, and we have a variety of ways of doing this. Here goes:

Graphical: We’ll do this for almost every problem. You’ll need to first solve for the variable,read the solution in English, then graph what you’ve read on a simple number line.

Set Builder: This is rather simplistic and results from the actual solution. Again, this willbe demonstrated using examples.

Interval: Once you’ve got the graph, this will be a natural extension of what you’re seeingin the graph. Again, this will be demonstrated using examples.

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6.3 Examples

We’ll do the following examples in class. In your notebook you should show all work for eachand every problem. Furthermore, you should learn to identify your answer by boxing it!

1. Solve, graph, and express the solution set using interval and set-builder notation.

x + 8 ≥ 3


−4x ≤ −12


4x + 4 ≥ 3x− 1


4x + 2 < 10


3x + 2 ≤ −10


5x + 2 < 3x− 6


2− 4x > 10


5− 3x ≤ 20


6x ≤ 19


2x + 7 > 3 (1− 2x)

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3 +2

7x < 2x− 5


0 < 2− 3 (x + 1) < 20


3

4x− 1

4< x− 5


31

40x− 5

4<

4

5x +

19

20


8− 3 (x− 4) ≤ 2x + 10


−5 ≤ 6x + 7 < 25

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7.1 Points, Midpoints and Distance

We’ll review plotting points (x, y) in the Cartesian6 plane (abscissa or x-coordinate, ordi-nate or y-coordinate). Visualizing points in a plane should eventually become natural, andconcepts such as distance and midpoint between points should become a natural extensions.

In class we will develop two formulas related to two points in a plane. It is typical tolabel these two points P1 or (x1, y1) and P2 or (x2, y2).

Distance: The distance between P1 and P2 is given by:

d =

√(x2 − x1)

2 + (y2 − y1)2

Midpoint: The midpoint between P1 and P2 is given by:(x1 + x2

2,y1 + y2

2

)Again, these formulas will be thoroughly discussed in class.

7.2 Examples


1. Graph the ordered pairs (−5, 0) and (0, −3), determine the distance between thesepoints, and the midpoint between these two points.

2. Find the distance, to the nearest hundredth, between the given points. Then find thecoordinates of the midpoint of the line segment connecting the points.

P1 (4, 5) and P2 (6, 3)

3. Graph the ordered-pair solutions of

y = −x2 + 5

when x = −3 and 1. Find the distance, to the nearest hundredth, between those twopoints. Then find the coordinates of the midpoint of the line segment connecting thepoints.

4. Find the distance, to the nearest hundredth, between the given points. Then find thecoordinates of the midpoint of the line segment connecting the points.

P1 (5.3, −6.4) and P2 (−2.7, 3.1)

6Cartesian refers to Rene Descartes and his ideas. Rene Descartes, (1596–1650), was a French philosopher,mathematician, and man of science. He concluded that everything was open to doubt except consciousexperience and existence as a necessary condition of this: “Cogito, ergo sum” (I think, therefore I am). Inmathematics, he developed the use of coordinates to locate a point in two or three dimensions.

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8.1 Introductory Functions and Relations

Definition of a Function: A function7 f from a set A to a set B is a relation that assignsto each element x in the set A exactly one element y in the set B. The set A is the domain(the x values) of the function f , and the set B contains the range (the values of y that arepaired with x). You should note that the set B is not necessarily the range, it just containsthe range.

Characteristics of a function from set A to set B

• Each element of A must be matched with an element from B.

• An element from A cannot be matched with two different elements of B.

Not all relationships are functional and you’ll need to evaluate when a relationship isfunctional. We’ll keep it very simple!

One last point: never divide by zero! So if you asked to find the domain of function

f (x) =5

x− 3,

you’re basically being asked what values of x are allowed? I hope you can see that x = 3will result in a division by zero—so the domain of this function is all real numbers exceptthree. This is often written: R, x 6= 3.

Don’t get your brain into a bunch, we’ll do this material by example—there’s very littleto remember.

8.2 Examples


1. State whether the relation is a function.

{(−9, −7) , (−7, −2) , (−2, 0) , (0, 4)}


{(4, −3) , (0, 0) , (6, 0) , (7, −3)}


{(4, −3) , (2, 0) , (6, −4) , (7, −2) , (2, 7)}7Often denoted f (x) and read f of x.

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4. Given f (x) = 4x− 3, evaluate f (−3)

5. Given g (x) = x2 − 3x + 2, evaluate g (−2)

6. Given

k (y) =y − 2

y + 2,

evaluate k (−1)

7. Find the domain and range of the function.

{(8, −5) , (10, 7) , (6, −1) , (5, −3)}

8. What values are excluded from the domain of the function?

g (x) = − 9

x− 2


f (x) =x− 2

9


h (x) = 3x4 − 2x3 + 4x2 − 9x + 17

11. Find the range of the function defined by the equation and the given domain.

h (x) = 7x− 2; domain = {−9, −6, −3, 0, 3}


f (x) = 2x2 − 3x + 1; domain = {−1, −0, 1, 2, 3}


k (x) =3− x

x2 + 1; domain = {−2, −1, −0, 1, 2}

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9.1 Graphing Lines

Now we will plot the solutions we obtained from simple two-variable linear relationships.For example, here’s a simple two variable linear relationship.

2x + 3y = 8

You should be able to verify that x = 4 and y = 0 is a solution of this equation; however,there’s an infinite number more. For example, you should also be able to verify that x = −2and y = 4 is another solution of this equation. Here’s partial unordered list.

{x = 4, y = 0; x = −2, y = 4; x = −8, y = 8; x = 10, y = −4}

Now let’s plot these points to see a simple linear pattern.

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

Figure 1: {x = 4, y = 0; x = −2, y = 4; x = −8, y = 8; x = 10, y = −4}

These point are collinear. When we draw a line through these dots we are actuallydescribing the solution set visually.

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

Figure 2: 2x + 3y = 8

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9.2 Special Points

x-intercept: The point at which the graph (line) crosses the x-axis. At the x-interceptthe y coordinate is zero. To find the x-intercept, just set y = 0 and solve for x. Forexample:

2x− 3y = 6 set y = 0

2x− 3 · 0 = 6 solve for x

2x = 6

x = 3

So the x-intercept is (3, 0).

y-intercept: The point at which the graph (line) crosses the y-axis. At the y-intercept the xcoordinate is zero. To find the y-intercept, just set x = 0 and solve for y. For example:

2x− 3y = 6 set x = 0

2 · 0− 3y = 6 solve for y

−3y = 6

y = −2

So the y-intercept is (0, −2).

Sometimes you’ll be asked to find one, or possibly both of these points. Be aware,that occasionally these points will be difficult to plot, and in those cases I suggest you findeasy points to plot—don’t worry, the graph will include these difficult points anyway. Forexample, find the x and y-intercept for:8

3x + 5y = −1.

To plot this however, I do not suggest that you use the x and y-intercept, but rather findsome simpler points. For example:

{x = −2, y = 1; x = 3, y = −2; x = 8, y = −5}

9.3 Examples


1. Graph.

y = −3x + 2

8The y-intercept is (0, −1/5), and the x-intercept is (−1/3, 0).

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2. Graph.

f (x) =7

6x

3. Graph.

f (x) =3

5x− 1

4. Graph.

f (x) = −3

2x− 2

5. Graph.

2x− y = 5

6. Graph.

5x + 2y = 7

7. Graph.

y = x

8. Graph.

y = 2

9. Graph.

x = −1

10. Find the x and y-intercepts and graph the function.

2x + y = 4

11. Find the x and y-intercepts and graph the function.

3x− 2y = 6

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10.1 Slope of a Line

We have already graphed many lines, and we have repeatedly seen a pattern in how thepoints have changed. Clearly we were able to follow a pattern in both the x and y. As wefollowed any graph from point-to-point, you should notice that ratio of this changing patterndoes not change. Here we define this ratio as slope, where

Slope = m =change in y

change in x=

∆y

∆x=

y2 − y1x2 − x1

.

Taking the prior example,

3x + 5y = −1,

and looking at its graph you should be able to determine the slope of the line. Just followfrom one point to another, for example, suppose we take (−2, 1) and (3, −2) as our twopoints, we have:

Slope = m = −3

5=−2− 1

3 + 2.

Another remarkable feature is that if you can solve the equation for y, then the slope is thecoefficient of x. Taking the above example.

3x + 5y = −1

5y = −3x− 1

y = −3

5x− 1

5

Here you should notice that the coefficient on the x is identical to the slope. Furthermore,the constant is identical to the y coordinate of the y-intercept.

10.2 Examples


1. Find the slope of the line containing the points

P1 (4, −5) and P2 (−6, 3) .

2. Find the slope of the line containing the points

P1 (4, 8) and P2 (4, −7) .

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3. Find the slope, b, x-intercept, and y-intercept of the line whose equation is

3y = 2x− 6.

4. Graph the line using slope and y-intercept.

y =3

4x− 2

5. Graph the line using slope and y-intercept.

2y − 3x = 7

6. Graph the line that passes through the point (−2, −1) and has slope 2/3.

7. Graph the line that passes through the point (9, −6) and has slope −1/8.

8. Determine the value of k such that the points whose coordinates are given lie on thesame line.

(k, −2) , (0, −3) , (9, −6)

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11.1 Equations of a Lines

We will basically focus on two forms when asked to write an equation of a line. I prefer thepoint-slope form, but we will still do both forms in class. The fact remains though, if youknow the point-slope form of the line it is usually trivial to get the slope-intercept form.This will be illustrated in class.

Slope-intercept form: The graph of the equation

y = mx + b

is a line whose slope is m and y-intercept is (0, b).

Point-slope form: The point-slope form of the equation of a line that passes through thepoint (x1, y1) and has slope m is

y − y1 = m (x− x1) .

It is essential that you know slope, and at least one point if you are going to use the point-slope form. If however, you decide on the slope-intercept form, you will certainly need slope,but you also need to know the y-intercept.

11.2 Examples


1. Find the equation of the line that contains the given point and has the given slope. (Lety be the dependent variable and let x be the independent variable.)9

(0, 9) , m = 3

2. Find the equation of the line that contains the given point and has the given slope. (Lety be the dependent variable and let x be the independent variable.)

(4, 5) , m =4

5

3. Find the equation of the line that contains the given point and has the given slope. (Lety be the dependent variable and let x be the independent variable.)

(3, 0) , m = −4

5

9That is, you should write your final answer as y = mx + b.

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4. Find the equation of the line that contains the given points. (Let y be the dependentvariable and let x be the independent variable.)

(0, 6) , (1, 9)


(−3, 4) , (1, 9)


(−5, 3) , (1, 3)


(−7, −4) , (−7, 4)

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12.1 Parallel and Perpendicular Lines

If two (non-vertical) lines are parallel then their slopes are numerically the same. If two lines(one of which is non-vertical) are perpendicular then their slopes are negative reciprocals ofone another.

In other words/notation, given two non-vertical lines with slopes m1 and m2 respectively.

Parallel: These lines are parallel if and only if m1 = m2.

Perpendicular: These lines are perpendicular if and only if m1 · m2 = −1. By writingm1 ·m2 = −1, you should be able see it as the same as writing

m1 = − 1

m2

or m2 = − 1

m1

,

which is the same as saying the slopes are negative reciprocals of one another.

Graphing may help, especially in those cases where computing slopes is impossible (x = 1for example). However, you should keep in mind that slopes are generally easy to computeand compare.

12.2 Examples


1. Find the slope of the line whose equation is 3y + 6x = −7.

2. Find the slope of the line that is perpendicular to the line whose equation is 3x+6y = 1.

3. Find the slope of the line that is parallel to the line whose equation is y−1 = −7 (x + 2).

4. Find the slope of the line that is perpendicular to the line whose equation is x = −1.

5. Is the line y = −4x + 3/7 parallel to the line y = 4x + 7/3?

6. Is the line that contains the points (2, −3) and (0, 9) parallel to the line that containsthe points (−5, −4) and (−7, 2)?

7. Find the equation of the line containing the point whose coordinates are (3, 5) andparallel to the graph of 2x + y = −3.

8. Find the equation of the line containing the point whose coordinates are (4, 3) andperpendicular to the graph of 3x− 4y = 20.

9. Find the equation of the line containing the point whose coordinates are (−1, 2) andperpendicular to the graph of 3y + 6x = −7.

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10. Find the equation of the line containing the point whose coordinates are (−2, −3) andperpendicular to the graph of y = 1.

11. Find the equation of the line containing the point whose coordinates are (5, 7) andperpendicular to the graph of x = 1.

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13 Review for Exam #1

This exam will include materials covered up to this point. For review you may look backover your notes and the examples done in class. You should also complete all homeworkassignments up to this point.

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14.1 System of Linear Equations

We will now solve linear systems with two variables—usually x and y, but other variablesmay have been used—of the following form.10{

Ax + By = CDx + Ey = F

The methods that we will use to solve these linear systems: elimination, substitution, andgraphical will be reviewed by example. You should be able to do all three methods, but it’sokay to prefer one method over the others.

Here’s a graph of this linear system to start us off. Can you visually determine the pointof intersection? It should be relatively easy to see where the two lines cross one another(the point), and you should be able to check that the point satisfies both equations. Thistechnique may overwhelm some, but it’s a good place to start.

-4 -3 -2 -1 0 1 2 3 4 5

-2

-1

1

2

3

4

5

6

Figure 3: Linear system, 2x + 3y = 5 and 5x− 2y = −16.

In general it is not so easy to graph two lines accurately. So we’ll need other methods.For example suppose we have the following two lines

y = 2x− 1 and x + 3y = 4,

and you’re ask for the point of intersection. Graphing is certainly okay, but here I want tosuggest substitution! that is use the fact that y = 2x − 1 in the second equation, Here’s

10The upper case letters represent constants.

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what I mean:

y = 2x− 1 Equation 1

x + 3y = 4 Equation 2

x + 3 (2x− 1) = 4 Making the substitution!

x + 6x− 3 = 4

7x = 7

x = 1

Now that you know that x = 1, you can substitute this value into Equation 1 to get y = 1

14.2 Examples


1. Determine whether the ordered pair (0, 1) is a solution of the system of equations.{2x − 5y = −3x + 5y = 8

2. Solve by graphing. (If equations are independent, give answer as a point; If the equationsare dependent, enter x for the value of x and y in terms of x; if the system is inconsistent,just state inconsistent.){

x + y = −3x − y = 5


x + 5 = 1y − 1 = 2


2x + 3y = 6y = −2

3x + 2


x − 2y = 12y − x = 1

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2x − 4y = 0y = x + 1

7. Solve by substitution. (If equations are independent, give answer as a point; If theequations are dependent, enter x for the value of x and y in terms of x; if the system isinconsistent, just state inconsistent.){

2x − 4y = 0y = x + 1


8x − 5y = 35y = 4x − 19


6x + 8y = −12y = 3x − 4

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15.1 Addition or Elimination

A typical system of equations looks like this:{Ax + By = CDx + Ey = F

.

Having everything aligned certainly makes things more orderly. What we’ll be doing nowis adding (addition method) these two equations together (equal signs must line up) in thehope that one variable will be eliminated (elimination method). This may involve additionalsteps, such as:

• Rearranging the equations in standard form.

• Multiplying one or both equation to create coefficients of one of the variables to beopposites.

As an example, let’s try:{2x − 3y = 12y − x = 1

.

First rearrange so everything lines up. Although the only real requirement is that the equalsigns must be alligned!{

2x − 3y = 1−x + 2y = 1

.

If we add these equations together we do not get an elimination. So, instead let’s make thecoefficients of the x variable opposite by multiplying the second equation be 2.{

2x − 3y = 1−2x + 4y = 2

.

Now add these two equations together to get:

y = 3.

Now you can take this value (y = 3) and use substitution to get the value for x.

x = 5

Okay, you should probably/definitely check your answer in the original two equations. Yes,it works fine. By the way, if substitution is too difficult, you may decide to use eliminationtwice. However, you should still make some attempt to check your final answers in theoriginal equations.

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15.2 Examples


1. Solve by the addition method. (If equations are independent, give answer as a point; Ifthe equations are dependent, enter x for the value of x and y in terms of x; if the systemis inconsistent, just state inconsistent.){

x − y = 2x + y = 8


x + y = 34x − y = 17


x − 2y = −2−3x + 6y = 6


6x − 4y = 296x + 2y = −1


6x + 3y = 64x + 2y = 5


13x − 2

5y = −1

23x + 3

5y = 5

Page 43 of 108





4x − 2y = 7x − 22x + 8y = 5y + 8


3 − 2x = 4 − 3y3x + 2y = 5

Page 44 of 108





16.1 Word Problems

Not much new here, except now we’ll be using systems of linear equations to solve theseword problems. However, please feel free to solve these word problems in the old algebraicway. Systems of linear equations, once mastered, can often make word problems easier toset-up and complete.

16.2 Examples


1. A pharmacist has two vitamin-supplement powders. The first powder is 25% vitaminB1 and 5% vitamin B2. The second is 25% vitamin B1 and 15% vitamin B2. How manymilligrams of each powder should the pharmacist use to make a mixture that contains165 milligrams of vitamin B1 and 43 milligrams of vitamin B2?

2. A motorboat traveling with the current went 15 miles in 1 hour. Against the current, ittook 3 hours to travel the same distance. Find the rate of the boat in calm water andthe rate of the current.

3. The total value of the quarters and dimes in a coin bank is $6.55. If the quarters weredimes and the dimes were quarters, the total value of the coins would be $7.45. Findthe number of quarters in the bank.

4. A chemist has two alloys, one of which is 10% gold and 15% lead and the other whichis 30% gold and 50% lead. How many grams of each of the two alloys should be used tomake an alloy that contains 70 grams of gold and 109.5 grams of lead?

5. A rowing team rowing with the current traveled 36 kilometers in 3 hours. Rowing againstthe current, the team rowed 12 kilometers in the same amount of time. Find the rate ofthe rowing team in calm water and the rate of the current.

6. During one month, a homeowner used 300 units of electricity and 130 units of gas for atotal cost of $88.00. The next month, 230 units of electricity and 250 units of gas wereused for a total cost of $82.50. Find the cost per unit of gas.

7. Flying with the wind, a plane flew 1,200 miles in 6 hours. Against the wind, the planerequired 10 hours to fly the same distance. Find the rate of the plane in calm air andthe rate of the wind.

8. A contractor buys 15 yards of nylon carpet and 19 yyardsd of wool carpet for $1,750. Asecond purchase, at the same prices, includes 20 yards of nylon carpet and 22 yards ofwool carpet for $2,150. Find the cost per yard of the wool carpet.

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9. A merchant mixed 10 pounds of a cinnamon tea with 2 pounds of spice tea. The 12-pound mixture cost $20. A second mixture included 12 pounds of the cinnamon tea and8 pounds of the spice tea. The 20-pound mixture cost $38. Find the cost per pound ofthe cinnamon tea and of the spice tea.

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17.1 Integer Exponents

This section of the textbook usually proves difficult for most students. However, please tryto focus on the basics, and when we’re dealing with exponents, this is what you need tounderstand.

x−n = 1 · 1

xn=

1

n factors of x.... =

...

x−4 = 1 · 1

x4=

1

xxxx

x−3 = 1 · 1

x3=

1

xxx

x−2 = 1 · 1

x2=

1

xx

x−1 = 1 · 1

x=

1

xx0 = 1 · x0 = 1

x1 = 1 · x1 = x

x2 = 1 · x2 = xx

x3 = 1 · x3 = xxx

x4 = 1 · x4 = xxxx... =

...

xn = 1 · xn = n factors of x.

Your textbook, of course, will provide a plethora of rules, but I will only box a few that

we will absolutely need. All boxed rules basically will help us rewrite negative exponentsas positive exponents. That is the direction we need to go in, and your instructions willgenerally read, “Simplify and rewrite so there are no negative exponents.”

17.2 The Basic Rules

If m, n, and p are integers, then:

1. x0 = 1, x 6= 0;

2. xm · xn = xm+n;

3. (xm)n = xmn;

4. (xmyn)p = xmpynp;

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5.xm

xn= xm−n, x 6= 0;

6.

(xm

yn

)p

=xmp

ynp, y 6= 0;

7. x−n =1

xn, x 6= 0;

8. xn =1

x−n, x 6= 0;

9.

(x

y

)−n

=(yx

)n, y 6= 0, x 6= 0;

Yes, lots of rules, but note that only three are boxed, and for the most part, it is all you’llever need. Be patient though, this will require a bit of getting used to. Remember thisthough, negative exponents need to be looked at and taken care of quickly. One at a time!

17.3 Examples


1. Simplify.(−2x0

)3 (3x2)2

2. Simplify.

−18x3y5

27x4y6

3. Simplify.(−3x−1y2

)24. Simplify.(

−3x2y)2 (

2x3y2)3

5. Simplify.(6x3y−3

9x4y−1

)2

Page 48 of 108




6. Simplify.(6x3y−3

9x4y−1

)−1

7. Simplify.(4x−2y3

6x4y−2

)−3

8. Simplify.

(−2x)(3x−2

)29. Simplify.

25x4y7z2

20x5y9z8

10. Simplify.(3x−1y−2

)211. Simplify.(

−2x−5) (

3x5)

12. Simplify.

(−2x)(−2xy−2

) (4x−2y

)−2

13. Simplify.

3x−2y

9xy

14. Simplify.(−3xy−2

) (2x−1y

)−3

15. Simplify.(6x−4yz−1

9xy−4z2

)−3

16. Simplify.(−18x4y−2z3

12xy−3z2

)−2

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18.1 Polynomials

• A monomial is a number, a variable, or a product of numbers and variables. If thevariable has an exponent, it must be non-negative. Examples:

5, x, 3x2, −12x2y.

• A polynomial is a variable expression in which the terms are monomials.

3x− 2, 2x2 − 3x + 4, 4xy2 − 7x2y + 9xy, x3 + 2x− 4.

• The degree of a polynomial in one variable is the greatest exponent on a variable. Forexample, here’s a degree 3 polynomial.

−11x3 + 4x2 + 15x− 9

• A polynomial in one variable is usually written so the terms are listed in decreasingdegree order, for example, 2x2 − 3x3 − 9 should be re-written as:

−3x3 + 2x2 − 9

• Although not appropriate in this class, many write polynomials in increasing degreeorder, for example, 2x2 − 3x3 − 9 would be re-written as:

−9 + 2x2 − 3x3

• Simplifying still follows the same order. And like terms should be well understood atthis point. As an example, if we’re asked to simplify

2(3x2 − 2x + 4 + 3x− 5

)− 3

(x2 + 5x− 7− 2x2 − 9x + 11

),

we’d do the following:

2(3x2 + x− 1

)− 3

(−x2 − 4x + 4

)= 6x2 + 2x− 2 + 3x2 + 12x− 12

= 9x2 + 14x− 14

• Evaluating a polynomial for a given value of the variable, where we will typically usefunction notation,

f (x) = 3x2 − 4x + 2.

So if you’re asked to evaluate f (−2), you should do this:

f (x) = 3x2 − 4x + 2

f (−2) = 3 (−2)2 − 4 (−2) + 2

f (−2) = 12 + 8 + 2

f (−2) = 22

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18.2 Examples


1. Given P (x) = −3x2 − 2x + 6, evaluate P (−2).

2. Simplify. Use a vertical and horizontal format.11(4x2 − 5x + 6

)+(−4x2 + 5x− 13

)3. Simplify. Use a vertical and horizontal format.12(

4x2 + 7x− 5)−(2x2 − 6x− 7

)4. Given P (x) = 3x4 − 5x + 5 and R (x) = 2x5 − 3x − 8, find S (x) = P (x) + R (x),

T (x) = P (x)−R (x), S (−1), and T (2)

11You’re free to use whatever format you like.12You’re free to use whatever format you like.

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19.1 Polynomial Multiplication

I am sure you all recall the distributive property of multiplication.

a (b + c) = ab + ac or a (b− c) = ab− ac

This property can be extending to any number of terms.

Here we will apply the distributive property of multiplication to polynomial multiplica-tions of increasing difficulty. It’s really just long multiplication, but you may not see this atfirst. For example, if you were asked to evaluate,

553× 297,

you’d use the distributive property. Although it may not look like this, this is what youwere actually taught in grammar school, in reverse order though. Don’t worry here, you canstill multiply using columns, just as you did in grammar school. This example here is justto point out the process as it relates to the distributive property.

553× 297 = (500 + 50 + 3) (200 + 90 + 7)

= (500) (200 + 90 + 7) + (50) (200 + 90 + 7) + (3) (200 + 90 + 7)

= (100000 + 45000 + 3500) + (10000 + 4500 + 350) + (600 + 270 + 21)

= (148500) + (14850) + (891)

= 164241

Please don’t be overwhelmed by this, and try to do this multiplication the way you weretaught to do it. However, if you give it some thought, you’ll see that they’re the sameprocess.

Again, the book pretty much sticks with very simple numbers. Here’s an example.

(2x + 3y)(2x2 − 3xy + 5y2

)= (2x)

(2x2 − 3xy + 5y2

)+ (3y)

(2x2 − 3xy + 5y2

)=

(4x3 − 6x2y + 10xy2

)+(6x2y − 9xy2 + 15y3

)= 4x3 + xy2 + 15y3

You should also be aware that multiplications don’t always produce a simpler expression.However, let’s try this one to see what happens.

(3x− 2)(9x2 + 6x + 4

)= (3x)

(9x2 + 6x + 4

)+ (−2)

(9x2 + 6x + 4

)=

(27x3 + 18x2 + 12x

)+(−18x2 − 12x− 8

)= 27x3 + 18x2 + 12x− 18x2 − 12x− 8

= 27x3 − 8

Certainly a lot simpler than what we started with. You should note that multiplications,once done, should be simplified—if possible—by collecting like terms.

Page 52 of 108




19.2 Examples


1. Multiple and simplify.

4x2 − x [x− 2 (4x− 5)]


(x− 2)(3x2 − 5x + 1

)3. Multiple and simplify.(

x2 − 3x + 5)

(x− 3)


(b− 3) (2b− 7) (b− 6)


(2x− 3)(x2 − 3x + 5

)6. Multiple and simplify.

(2x− 1)(−3x2 − 5x + 1

)7. Multiple and simplify.(

2 + 3x− x2)

(2x− 1)

8. Multiple and simplify.(x2 − 2x

)(2x + 5)


(x + 1) (x− 1)


(3x− 2) (3x + 2)

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(3x− 1) (x + 4)


(2x− 3) (4x− 7)


(5x + 6) (6x + 5)


(11x− 2y) (3x + 5y)


(3x− 5)2


(5x + 2y)2


(a + b)2 + (a− b)2


(2x− 3)2 − (2x + 3)2

Page 54 of 108





20.1 Polynomial Division

This section of the textbook deals with polynomial division. Basically these problems willbe divided into two distinct types. One type, usually referred to a simple division involves amonomial divisor. It is suggested that when you have a monomial divisor, that you simplydo term wise simplifications, and do it carefully, one term at a time. For example:

16x2y − 20xy + 24xy2

4xy=

16x2y

4xy− 20xy

4xy+

24xy2

4xy= 4x− 5 + 6y

The other type, usually referred to a long division involves a binomial divisor. It is suggestedthat you carefully review numerical long division that you learned years earlier, because theprocess is actually identical for polynomials. For example, verify,

6258

59= 106 +

4

59

using long division.

The book has many examples, and I will also carefully review the process of long divisionas we move through these problems. Don’t despair, at first you might be totally confused.It takes time, and more importantly, it will take a good deal of practice on your part.

20.2 Examples


1. Divide.

x6 − 3x4 − x2

x2

2. Divide.

8x2y2 − 24xy

8xy

3. Divide.

3x2 − 5x + 6

15x

4. Divide.

9x2y + 6xy − 3xy2

3xy

Page 55 of 108




5. Divide.(x2 − x− 6

)÷ (x− 3)

6. Divide.(2x2 − 13x + 21

)÷ (x− 3)

7. Divide.(2x2 + 7

)÷ (x− 3)

8. Given Q (x) = 3x− 2 and P (x) = 27x3 + 8, findP (x)

Q (x).

9. Divide.(x4 + 3x2 − 10

)÷(x2 − 2

)10. Divide.

4x3 − 14x2 + 2x− 7

x2 − 4x + 3

Page 56 of 108





21.1 Polynomial Factoring

This section of the textbook deals with factoring polynomials, which simply means that wewill rewrite the polynomial as a product of simpler polynomials. For example x2 + 2x − 3can be rewritten as a product of (x + 3) and (x− 1). That is:

x2 + 2x− 3 = (x + 3) (x− 1) .

You can of course check this, and this is exactly what you did in the prior chapter. Inshort, to factor polynomials you should be very clear on how to multiply polynomials, andto recognize that we’re going in reverse. Your understanding polynomial multiplication iscritical to this process of factoring, and we’ll always check our factorings with multiplication.

Factoring, in general, is very difficult. However, our text will stick with simple examples,and will following these basic steps:

1. Greatest common factors (GCF). For example:

3x2 − 6x = 3x (x− 2) .

2. When you have four terms with no GCF, you will divide the problem into two groups,and then return to step one. If your initial groups are not working, just regroup. Forexample:

x2 − 3x + 4ax− 12a =(x2 − 3x

)+ (4ax− 12a) two separate groups

= x (x− 3) + 4a (x− 3) factor each group

= (x− 3) (x + 4a) GCF factoring

3. Quadratic factors of the form Ax2 +Bx+C, where we will basically use trail and errorto find the factors. Again, your ability to multiply will largely determine your abilityto factor.

Examples abound here, and we actually started this worksheet with one. For anotherexample, try13

x2 + 5x− 14.

21.2 Examples


13Answer: (x + 7) (x− 2)

Page 57 of 108




1. Factor.

4x2 − 12x

2. Factor.

5− 20x

3. Factor.

12x3 − 10x2

4. Factor.

2x2y − 6x2y2 + 12xy

5. Factor.

3x2y2 − 9xy2 + 15y2

6. Factor.

x2 − 3x + 4ax− 12a

7. Factor.

4 (x + y) + A (x + y)

8. Factor.

x2 − 5x− 2xy + 10y

9. Factor.

2x2 − 10x + 7xy − 35y

10. Factor.

x2 + 5x + 6

11. Factor.

x2 + x− 6

12. Factor.

x2 − 2x− 35

Page 58 of 108




13. Factor.

x2 − 2x− 3

14. Factor.

40− 3x− x2

15. Factor.

x2 − 6x + 8

16. Factor.

x2 + 16x + 39

17. Factor.

x2 − 15x + 56

18. Factor.

3x2 + 7x + 2

19. Factor.

x2 − 3xy − 28y2

20. Factor.

2y2 − 21y + 52

21. Factor.

x4 − 22x3 + 120x2

22. Factor.

11z2 − 57z + 10

23. Factor.

3y4 + 54y3 + 135y2

24. Factor.

18y3 + 24y2 − 64y

25. Factor.

4x2y + 20xy − 56y

26. Factor.

21b + 22b2 − 8b3

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22.1 Special Factoring

This section of the textbook deals with factoring polynomials where special evident patternsmay exists.14 Basically, I still want to stress trial-and-error as a legitimate technique, exceptfor the difference and sum of perfect cubes where you’ll need to memorize the two specialcases of cubic factoring. With practice you should be able to get most factorizations fairlyquickly. Please do not try to memorize the problems! Furthermore, do not let initial failureto factor prevent you from trying again . . . if it’s factorable, I’d like to believe that you’ll getit within three tries if you use your head and think a little. You will get stuck on occasion(after three tries), and then it’s time to move on.

22.2 Examples


1. Factor.

12t3 − 75t

2. Factor.

s2 − 81

3. Factor.

9w2 − 16

4. Factor.

y4 − 16x4

5. Factor.

1− 64x3

14Special patters are:

1. Difference of perfect squares: a2 − b2 = (a− b) (a + b)

2. Perfect squares, type I: a2 − 2ab + b2 = (a− b)2

3. Perfect squares, type II: a2 + 2ab + b2 = (a + b)2

4. Difference of perfect cubes: a3 − b3 = (a− b)(a2 + ab + b2

)memorize this formula

5. Sum of perfect cubes: a3 + b3 = (a + b)(a2 − ab + b2

)memorize this formula

Page 60 of 108




6. Factor.

x2y2 − 7xy2 − 8y2

7. Factor.

8x3 + 27

8. Factor.

x2b2 + 6ab2 + 9b2

9. Factor.

2z3 − 8z2y + 8zy2

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23.1 Zero-Product Rule

This section of the textbook deals with solving quadratic (or higher degree) equations. Allprior equations in this class have been linear, that is, degree one, but now we are confrontedwith a product of linear factors. There are many methods to solve equations that involve aproducts of linear factors, but it should be painfully obvious that if

a · b = 0,

then a = 0 or b = 0. This will be our only method, and it solely relies on having a productof linear factors equaling zero. For example, if we are asked to solve:

(x− 1) (x + 1) = 0,

we just need to determine when the linear factors equal zero, that is

x− 1 = 0 or x + 1 = 0.

Clearly, this occurs when x = 1 or x = −1. Many other examples will follow, but you needto be clear that we need a product of linear factors equaling zero if our method is to work!

If our quadratic (or higher degree) equation is not equal to zero, we will need to do thisfirst. Furthermore, we will always factor once we get the zero. For example, suppose you’reasked to solve for x?

x2 + 2 = 3x

First you will need to solve for zero, and I strongly encourage that you put your polynomialsin order.

x2 − 3x + 2 = 0

Now factor.

(x− 2) (x− 1) = 0

Now solve.

x− 2 = 0 or x− 1 = 0.

So the solution to x2 +2 = 3x is x = 2 or x = 1 . You should, of course, check you solutionsin the original equation—they, indeed, work!

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23.2 Examples


1. Solve for x.

4x2 − 12x = 0

2. Solve for y.

(x + 2) (x− 3) = 14

3. Solve for z.

x (2x + 1) (3x− 2) = 0

4. Solve for a.

a2 + a = 6

5. Solve for x.

x2 = 2x + 35

6. Solve for t.

15 = 8t− t2

7. Solve for s.

s2 = 15s− 56

8. Solve for y.

3y3 + 54y2 + 135y = 0

9. Solve for x.

4x2 + 20x = 56

10. Solve for y.

3y2 + 7y − 6 = 0

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11. Solve for z.

2z2 − z = 3

12. Solve for t.

2t2 + 13t + 15 = 0

13. Solve for x.

3x2 = 16x− 5

14. Solve for v.

8v2 + 33v + 4 = 0

15. Solve for y.

2y3 − 5y2 − 32y + 80 = 0

16. Solve for z.

24z2 − 24z = 18

17. Solve for c.

19c− 15c2 = 6

18. Solve for b.

b2 + 8b− 15 = (2b + 3) (b− 5)

19. Solve for t.

12t3 = 75t

20. Solve for x.

(x− 1) (x + 2) = 4

21. Solve for n.

(5n− 1) (n + 2) = (2n− 1) (3n + 2)

22. The sum of a number and its square is 132. Find the number.

23. The length of a rectangle is 2 feet more than twice the width. The area of the rectangleis 84 square feet. Find the length and width of the rectangle.

Page 64 of 108






Page 65 of 108





25.1 Rational Expressions

You should already be familiar with functional notation and how to evaluate functions fora give value. In addition, you should have some understanding of domain of a function,especially the part about not dividing be zero. This is particularly important when dealingwith rationals.

We will also be reducing a fraction, which essentially boils down to factoring and thencanceling. For example, here’s the proper method.

14

21=

2 · 73 · 7

=2

3· 7

7=

2

3· 1 =

2

3

Of course you probably wouldn’t do these steps for this particular example. However, youwould for this example.

x2 + 8x + 16

x2 − 2x− 24=

(x + 4) · (x + 4)

(x− 6) · (x + 4)=

x + 4

x− 6· x + 4

x + 4=

x + 4

x− 6· 1 =

x + 4

x− 6

Some will even shorten this further to:

x2 + 8x + 16

x2 − 2x− 24=

(x + 4)��(x + 4)

(x− 6)��(x + 4)=

x + 4

x− 6.

And I encourage everyone to follow this format on all problems that follow,

The process is to factor the numerator and denominator, then cancel common factors.Even when the problems become more involved the same steps will be followed.

When you’re confronted with a product to begin with, please do not multiply, mainlybecause you want factors anyway. In addition, when you’re presented with a division, youwill need to rewrite it as a product. That is,

a

b÷ c

d=

a

b· dc.

25.2 Examples


1. Given the following function, find f (−2).

f (x) =x− 3

2x− 1

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2. Given the following function, find f (−3).

f (x) = − 7

x2 − 4x + 2

3. Find the domain of the function.

f (x) =x + 1

x + 5


f (x) =3x− 1

2x− 7


f (x) =9x + 4

(3x + 5) (x− 3)

6. Find the domain of the function;

f (x) =3− 2x

x2 − 3x− 18

7. Simplify.

x2 − 3x

2x− 6

8. Simplify.

5xy − 3y

9− 15x

9. Simplify.

x2 + 7x− 8

x2 + 6x− 7

10. Simplify.

3x3 − 12x

6x3 − 24x2 + 24x

11. Simplify.

8x− 12

14x + 7· 42x + 21

32x− 48

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12. Simplify.

x2 − 2x− 24

x2 − 5x− 6· x

2 + 5x + 6

x2 + 6x + 8

13. Simplify.

x2 + 2x− 35

x2 + 4x− 21· x

2 + 3x− 18

x2 + 9x + 18

14. Simplify.

25− x2

x2 − 2x− 35· x

2 − 8x− 20

x2 − 3x− 10

15. Simplify.

4x2y3

15a2b3÷ 6xy

5a3b5

16. Simplify.

6x3 + 7x2

12x− 3÷ 6x2 + 7x

36x− 9

17. Simplify.

x2 − 49

x4y3÷ x2 − 14x + 49

x4y3

18. Simplify.

x2 − x− 2

x2 − 7x + 10÷ x2 − 3x− 4

40− 3x− x2

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26.1 Finding an LCM

The least common multiple (LCM) of two or more polynomials is the polynomial of leastdegree that contains all the factors of each polynomial. To find the LCM, first factor eachpolynomial completely. Then, the LCM is the product of each factor the greatest number oftimes it occurs in any one factorization.

Let’s first take a simple numerical example. Find the LCM of 50 and 60.

First factor.

50 = 2 · 5 · 560 = 2 · 2 · 3 · 5

The LCM is:

22 · 3 · 52 = 300.

Now let’s take a more complicated example. Find the LCM of x2−6x+9 and x2−2x−3.

First factor.

x2 − 6x + 9 = (x− 3) · (x− 3)

x2 − 2x− 3 = (x− 3) · (x + 1)

The LCM is:

(x− 3)2 · (x + 1) .

No need to multiple it out.

26.2 Using an LCM

The initial purpose15 of finding an LCM is so that we can rewrite fractions in terms of leastcommon denominator (LCD). The basic idea here is that

a

b=

a

b· 1 =

a

b· cc

=ac

bc.

So if you have a fraction

9

5

15Will also use it to simplify and solve rational equations.

Page 69 of 108




and you want to rewrite the fraction with another denominator, say 25, just do this:

9

5=

9

5· 1 =

9

5· 5

5=

9 · 55 · 5

=45

25.

Now let’s take a more complicated example. So if you have a fraction

15x

13y

and you want to rewrite the fraction with another denominator, say 26xy2, just do this:

15x

13y=

15x

13y· 1 =

15x

13y· 2xy

2xy=

15x · 2xy13y · 2xy

=30x2y

26xy2.

26.3 Addition and Subtraction of Rational Expressions

26.3.1 Same Denominator

If the denominators are the same.

Addition Don’t forget to look for reductions!

a

b+

c

b=

a + c

b

For example:

x

x2 − x− 12+

3

x2 − x− 12=

x + 3

x2 − x− 12=

(x + 3)

(x + 3) (x− 4)=

1

x− 4

Subtraction Don’t forget the parenthesis! Again, don’t forget to look for reductions!

a

b− c

b=

a− (c)

b

For example:

2x2

x2 − x− 12− 7x + 4

x2 − x− 12=

2x2 − (7x + 4)

x2 − x− 12

=2x2 − 7x− 4

x2 − x− 12

=(2x + 1) (x− 4)

(x + 3) (x− 4)

=2x + 1

x + 3

Page 70 of 108




26.3.2 Different Denominator

If the denominators are different you’ll need to find the least common multiple (LCM)of denominators and then rewrite each fraction. For example:

z

8y+

4z

3y=

z

8y· 3

3+

4z

3y· 8

8

=3z

24y+

32z

24y

=3z + 32z

24y

=35z

24y

26.4 Examples


1. Find the LCM of the polynomials.

6x2y, 18xy2


6x2, 4x + 12


8x2 (x− 1)2 , 10x3 (x− 1)


(2x− 1) (3− 5x) , (2x− 1)2 (x− 5)


x2 − 2x− 24, x2 − 36


2x2 − 7x + 3, 2x2 + x− 1


x2 + 3x− 18, x− 3, x− 2

Page 71 of 108




8. Find the LCD.

4

x,

3

x2

9. Rewrite the fractions in terms of the LCD.

4

x,

3

x2


a2

x (x + 7),

a

(x + 7)2


3

x (x− 5),

2

(x− 5)2


7

x2 + x− 2,

x

x + 2


3x

x− 4,

5

x2 − 16


x− 1

x2 + 2x− 15,

x

x2 + 6x + 5

15. Simplify.

7

4y+

11

6y− 8

3y

16. Simplify.

4x− 3

6x+

2x + 3

4x

17. Simplify.

2x + 9

9x− x− 5

5x

Page 72 of 108




18. Simplify.

x− 2

3x2− x + 4

x

19. Simplify.

3 +x− 1

x + 1

20. Simplify.

x− 2

8x2− x + 7

12xy

21. Simplify.

2

x− 3+

5

x− 4

22. Simplify.

4

2x− 1− 5

x− 6

23. Simplify.

2x

x2 − x− 6− 3

x + 2

24. Simplify.

6− 2x

2x2 − 15x + 28+

x− 2

x− 4− x + 6

2x− 7

Page 73 of 108





27.1 Complex Fractions

A complex fraction is a fraction where the numerator, denominator, or both contain a frac-tion. Our basic approach would be as follows:

a

bc

d

=a

b÷ c

d=

a

b· dc,

however, this only works if we’re dealing with very simple cases. In general it is best to find aminor LCD (the LCD of all fractions in the complex fraction’s numerator and denominator),and then use a builder to clear the minor fractions. For example:

34

+ 12

23− 1

4

=34

+ 12

23− 1

4

·(

12

12

)=

9 + 6

8− 3

=15

5= 3

Problems will vary in difficulty, but the process will remain the same, and generally theproblems will support following this process.

27.2 Examples


1. Simplify.

4 + 32

9− 32

2. Simplify.

8 + 8x

3− 3x2

3. Simplify.

a− 749a− a

Page 74 of 108




4. Simplify.

x− 1x

1x

+ x

5. Simplify.

4 + 32

9− 32

6. Simplify.

8x2 − 8

x5x2 + 5

a

7. Simplify.

5− 18x+3

7− 21x+3

8. Simplify.

1 + 2x+3

1 + 9x−4

9. Simplify.

x− 2 + 92x+7

x + 3− 12x+7

10. Simplify.

1− 8x− 9

x2

1 + 6x

+ 5x2

Page 75 of 108





28.1 Solving of Rational Equations

28.1.1 Essential Rules and Procedures for Solving Equations

1. Simplify both sides of the equation. This often involves clearing fractions by multiply-ing both sides of the equation by a non-zero factor.

2. Get the variable term on one side by adding or subtracting the same variable expressionto each side of the equation.

3. Get the constant term on the other side (opposing side of the variable term) by addingor subtracting the same constant to each side of the equation.

4. Divide both sides of the equation by the variable term’s numerical coefficient.

5. Check your solution in the original equation.

Well the procedure for solving rational equations will involve these steps. However, thesimplification process (as we did prior) will involve multiplying both sides (every term) bythe LCD (least common denominator).

Example: Solve for x.

5x

x + 2= 3 +

10

x + 2Multiply both sides by (x + 2).

5x

(x + 2)· (x + 2) = 3 · (x + 2) +

10

(x + 2)· (x + 2)

5x = 3x + 6 + 10

5x = 3x + 16

2x = 16

x = 8

Now check it. Yes, you must check it!

5x

x + 2= 3 +

10

x + 25 · 88 + 2

= 3 +10

8 + 240

10= 3 +

10

104 = 3 + 1

4 = 4

So the solution to the equation

5x

x + 2= 3 +

10

x + 2,

is x = 8 .

Page 76 of 108




28.1.2 Work Problems

Here the textbook will introduce work word problems. The simple formula is:

W = r · t

Where W represents work done, r represents the rate at which work is being done, andt represents time. These problems are not much different than uniform motion problems.Again, it boils down to reading, and understanding what you’ve read. Yes, we’ll do examplesin class!

28.2 Examples


1. Solve.

3x + 4

12− 1

3=

5x + 2

12− 1

2

2. Solve.

12

3x− 2= 3

3. Solve.

3− 12

x= 7

4. Solve.

3

x− 2=

4

x

5. Solve.

5

3x− 4= − 3

1− 2x

6. Solve.

2x

x + 2− 5 =

7x

x + 2

7. Solve.

x

x + 12=

1

x + 5

Page 77 of 108




8. Solve.

x− 6

x− 3=

2x

x− 3

9. Solve.

x

x + 4= 3− 4

x + 4

10. Solve.

x + 2

x2 − x− 2+

x + 1

x2 − 4=

1

x + 1

11. Solve.

5x

x + 2= 3− 10

x + 2

12. Solve.

4

x− 6− 7

x + 6=

5

x2 − 36

13. Solve.

x

2x− 9− 3x =

10

9− 2x

14. A ski resort can manufacture enough machine-made snow in 4 hours to open its steepestrun, whereas it would take naturally falling snow 12 hours to provide enough snow. Ifthe resort makes snow at the same time that it is snowing naturally, how long will ittake until the run can open?

15. An experienced bricklayer can work twice as fast as an apprentice bricklayer. After theyworked together on a job for 12 hours, the experienced bricklayer quit. The appren-tice required 8 more hours to finish the job. How long would it take the experiencedbricklayer, working alone, to do the job?

16. A roofer requires 10 hours to shingle a roof. After the roofer and an apprentice work ona roof for 2 hours, the roofer moves on to another job. The apprentice requires 14 morehours to finish the job. How long would it take the apprentice, working alone, to do thejob?

17. A cyclist and a jogger start from a town at the same time and head for a destination 19mi away. The rate of the cyclist is twice the rate of the jogger. The cyclist arrives 1.9 hbefore the jogger. Find the rate of the cyclist.

Page 78 of 108





29.1 Rational Exponents

The exponent rules we used in the past were applied to integers, now we will apply the samerule to rational exponents. Here, as a review, are the basic rules.

29.1.1 The Basic Rules

If m, n, and p are rational numbers, then:

1. x0 = 1, x 6= 0;

2. xm · xn = xm+n;

3. (xm)n = xmn;

4. (xmyn)p = xmpynp;

5.xm

xn= xm−n, x 6= 0;

6.

(xm

yn

)p

=xmp

ynp, y 6= 0;

7. x−n =1

xn, x 6= 0;

8. xn =1

x−n, x 6= 0;

9.

(x

y

)−n

=(yx

)n, y 6= 0, x 6= 0;

29.1.2 Meaning

By now you should have a good grasp of integer exponents, and this understanding shouldnow extend nicely to rational exponents. For example, x3 means three factors of x, so whatdoes x1/2 mean then? I like to think of rational exponents in the same way as I do integerexponents, so x1/2 must mean one or two equal factors of x then. Here’s some concreteexamples:

41/2 = (2 · 2)1/2 = 2

91/2 = (3 · 3)1/2 = 3

81/3 = (2 · 2 · 2)1/3 = 2

641/3 = (4 · 4 · 4)1/3 = 4

Page 79 of 108




Certainly these are simple cases, but they are nonetheless the first steps in working throughmore complex problems. Let’s try these using the same line of reasoning.

82/3 = (2 · 2 · 2)2/3 = 2 · 2 = 4

642/3 = (4 · 4 · 4)2/3 = 4 · 4 = 16

323/5 = (2 · 2 · 2 · 2 · 2)3/5 = 2 · 2 · 2 = 8

813/4 = (3 · 3 · 3 · 3)3/4 = 3 · 3 · 3 = 27

Notationally these rational exponents are often written using the root symbol.16

x1/n = n√x, xm/n = n

√xm =

(n√x)m

29.2 Examples


1. Simplify.

161/2

2. Rewrite the exponential expression as a radical expression.

31/3

3. Simplify.

82/3


51/2

5. Simplify.

163/4


23/5

7. Simplify.

253/2

16Here we are assuming that x1/n is a real number.

Page 80 of 108




8. Simplify.(64

27

)−2/3

9. Simplify.

x1/2x7/2

10. Simplify.

xx−1/3

11. Simplify.

a−7/15a4/5a2/3

12. Simplify.

x1/4

x3/4

13. Simplify.

x−4/7

x−2/7

14. Simplify.(x6)−2/3

15. Simplify.(x−5/6

)−18

16. Simplify.(x−4/3

)−3/8

17. Simplify.(x4/5y3/10

)1018. Simplify.(

x5y−10)−2/5

19. Simplify.(x1/2y−7/6

y−5/6

)−6

20. Simplify.

y1/3(y2/3 + y−1/3

)

Page 81 of 108





30.1 Simplifying Radical Expressions

30.1.1 The Product Property

If n√a and n

√b are positive real numbers, then n

√ab = n

√a · n√b and n

√a · n√b = n

√ab. For

example, let’s say we’re given√

75, we could rewrite this as follows:

√75 =

√25 ·√

3 = 5 ·√

3.

You may wonder what is the value of√

3, and you may even try it on your calculator—it’scertainly not nice, so we generally just write

√3. However, in the real world we are often

more concerned with meaning than symbol, so you should be able to use a calculator to getan approximate.

√75 ≈ 8.660254037844387

For the most part though, do not use a calculator unless you are asked to approximate theanswer! And please follow instructions when asked to approximate, for example, if you’reasked to find

√7 to the nearest hundredth, you would write

√7 ≈ 2.65,

but do not believe that you know the square root of 7, you’re just writing down an approxi-mate! In fact the

√7 can not be written in a simpler way.

30.1.2 Initial Radical Simplification

A radical expression (contains the root symbol) is in simplest form when the radicand (thething inside the root symbol) contains no factor that is a perfect power. For example the√

27 is not in simplest form because the radicand contains a perfect square (9).

√27 =

√9 ·√

3 = 3 ·√

3

The 3√

48 is not in simplest form because it contains a perfect cube (8).

3√

48 =3√

8 · 3√

6 = 2 · 3√

6

Numerically you should be able to recognize perfect squares, cubes, etc.. At least to amoderate level of numeracy. For variables it is a lot easier, for example

5√x6 is not in

simplest form because it contains a perfect fifth (x5).

5√x6 =

5√x5 · 5√x = x · 5

√x

Page 82 of 108




30.1.3 Adding/Subtracting Radicals

The distributive property is used to add/subtract radicals together.

5√

7 + 3√

7 = (5 + 3)√

7 = 8√

7

You’ll certainly need to simplify your radicals first in order to do this.

3√

8− 5√

2 = 6√

2− 5√

2 =√

2

30.1.4 Multiplying Radicals

The product property of radicals is used to multiply radical expressions with the same index.

n√a · n√b =

n√ab

Again, it is in your best interest to simplify first.√

24 ·√

27 = 2√

6 · 3√

3 = 6√

18 = 18√

2

30.1.5 Dividing Radicals

The quotient properties of radicals is helpful when dealing with fractional radicands orradicals as divsiors, it says that if n

√a and n

√b are real numbers and b 6= 0, then

n

√a

b=

n√a

n√b.

The convention here is that a radical expression is in simplest form when no radical existsin the denominator of an expression. This also includes fractional radicands. Here’s anexample,

√125√50

=

√125

50=

√5

2,

although these are equivalent, they’re all in violation of our convention. Here’s what we needto do

√125√50

=

√125

50=

√5

2=

√5√2·√

2√2

=

√10

2

30.1.6 Conventions

To get simplest radical form, your radical expressions must meet these conditions.

• The radicand contains no factor that is a perfect power.

• The radicand cannot be a fraction.

• A fraction cannot contain a radical as a divisor.

Tough stuff, I know, but we’ll do plenty of examples to illustrate simplest radical form.

Page 83 of 108




30.2 Examples

We’ll do the following examples in class. In your notebook you should show all work for eachand every problem. Furthermore, you should learn to identify your answer by boxing it!17

1. Simplify.√x3y4z9

2. Simplify.√18x2y11z16

3. Simplify.

3√

x11y14z15

4. Simplify.

3√

2x + 11√

2x

5. Simplify.

2√

2x3 + 5x√

18x

6. Simplify.

2√

18x2y3 − xy√

50y

7. Simplify.

3√

54− 3√

128

8. Simplify.

24√

162x5 + 2x4√

32x

9. Simplify.

4x 4√

81xy5 + 2y 4√

256x5y

10. Simplify.√9y3 −

√16y3 + y

√9y

17Assume that variables represent non-negative real numbers.

Page 84 of 108




11. Simplify.√16x7y5 + 7x2 −

√4x3y5 + 3xy2

√9x5y

12. Simplify.

√21 ·√

6

13. Simplify.

3√

25 · 3√

5

14. Simplify.√3x7y ·

√21x7y6

15. Simplify.

3√

4x2y2 · 3√

8xy12

16. Simplify.

√35(√

35−√

5)

17. Simplify.(3− 2

√5)(

3 + 2√

5)


√2)2

19. Simplify.(5 + 3

√2)(

3− 2√

2)


√5)(

6− 3√

5)


√x) (

2− 3√x)

Page 85 of 108




22. Simplify.(1−√x) (

1 +√x)

23. Simplify.

1√2

24. Simplify.

13√

2

25. Simplify.

3√5x

26. Simplify.

33√

3x2

27. Simplify.

8x4√

27x

28. Simplify.

1

1−√

2

29. Simplify.

3

2−√

5x

30. Simplify.

1 +√x

1−√x

31. Simplify.√

48x2

√12x

Page 86 of 108




32. Simplify.

√36x2

√4x

33. Simplify.√45x3y5√15x2y

34. Simplify.

√7√3

35. Simplify.

2√3x

36. Simplify.√x

5

37. Simplify.

33√

25

38. Simplify.

43√

25x2

39. Simplify.√15x2y4√75x7y3

40. Simplify.

3

2−√

3

Page 87 of 108




41. Simplify.

5

3−√

3

42. Simplify.

2√x− 3

43. Simplify.

√3−√

5√2−√

3

44. Simplify.

5− 3√

2

7− 2√

3

45. Simplify.

√x− 7

√x−√

2

46. Simplify.

2√x− 4

√y

2√x− 5

√y

Page 88 of 108





31.1 Solving Radical Equations

If two real numbers are equal, then the same powers of the numbers are equal. That is, ifa = b, then an = bn. This is also useful for equations involving radicals, but it may alsobe misleading, so I strongly suggest that you check your solutions in the original problem.Here’s a simple example.

√x = 5(√

x)2

= (5)2

x = 25

This certainly works. Now let’s try this one.

√x + 1 = x− 1(√

x + 1)2

= (x− 1)2

x + 1 = x2 − 2x + 1

0 = x2 − 3x

0 = x (x− 3)

The final line leads us to believe that x = 0 or x = 3. However, if you bother to check youwill see that only x = 3 works. Please check your solutions to see if they work.

31.2 Examples


1. Solve.

√2x + 88− 12 = 0

2. Solve.

3√

6x = −3

3. Solve.

√x− 5− 5 = 0

4. Solve.

3√x− 5 + 2 = 0

Page 89 of 108




5. Solve.

4√

4x + 5 = 3

6. Solve.

5√

3− 2x = −2

7. Solve.

√x + 1 +

√x− 3 = 2

8. Solve.

√4x−

√4x− 7 = 1

9. Solve.

√x + 5 +

√3x + 4 = x + 3

Page 90 of 108





32.1 Introduction to Complex Numbers

Fact is, you’ve probably been told that not all numbers are real and that some are imaginary.You may have further been told that they were invented and don’t really exists. That’shogwash—they not only exists, but were a necessary consequence of solving a certain classof problems—that is, they were discovered and not invented. I won’t dwell on this, and we’llstick to some very basic concepts.

The unit imaginary is denoted by i, where

i =√−1.

A complex number is written as a linear combination of a real and imaginary part, and isusually denoted by the letter z, where

z = a + bi.

Both a and b are real numbers, and i is just the imaginary unit. You should note thatif i =

√−1 then i2 = −1. That’s about it! However, since i is really a root you need to

follow the same conventions that were used in radical simplifications. And you should alwayssimplify your radicals first, that is, if you’re asked to multiple

√−8 by the

√−25, you’d do

this:

√−25 ·

√−8 = 5i · 2i

√2

= 10i2√

2

= −10√

2

32.1.1 Operations on Complex Numbers

Addition/Subtraction: just as you did with any algebraic expression. For example ifyou’re asked to add 3 + 2i to 7− 3i you’ll get 10− i. And if you’re asked to subtract2− 7i from 5− 3i you’ll get 3 + 4i. Really simple!

Multiplication: again, just as you did with any algebraic expression. However, if you geti2 in your multiplications you need to rewrite this as −1. Here goes:

(2− 3i) (3 + 2i) = 6 + 4i− 9i− 6i2

= 6− 5i + 6

= 12− 5i

Page 91 of 108




Division: just as you did with radical divisions! For example:

1

3 + 2i=

1

3 + 2i· 3− 2i

3− 2i

=3− 2i

9− 4i2

=3− 2i

13

=3

13− 2

13i

32.2 Examples


1. Simplify.

√−16

2. Simplify.

√−75

3. Simplify.

7−√−49

4. Evaluate

−B +√B2 − 4AC

for A = −2, B = 6 and C = −29.

5. Simplify.

(3− 2i) + (4 + 5i)

6. Simplify.

(5− 7i)− (6− 19i)

7. Simplify.

(−4i) (5i)

Page 92 of 108




8. Simplify.

(1− i) (1 + i)

9. Simplify.

√−5 ·√−80

10. Simplify.

(5− 3i) (5 + 3i)

11. Simplify.

(3− 4i)2

12. Simplify.

i (3− 2i)

13. Simplify.

(2− 3i) (5 + 2i)

14. Simplify.

2

i

15. Simplify.

2

3i

16. Simplify.

2 + 5i

−3i

17. Simplify.

3

1 + i

18. Simplify.

6

5 + 2i

Page 93 of 108




19. Simplify.

5

3− i

20. Simplify.

3− 4i

3 + 4i

21. Simplify.

2 + 13i

4 + i

22. Simplify.

3 + 9i

1− i

Page 94 of 108





This exam will include materials from chapters 6 and and only. For review you may lookback over your notes and the examples done in class. You should also complete all homeworkassignments up to this point.

Page 95 of 108





34.1 Solving Quadratic Equations

The general form of a quadratic equation is

Ax2 + Bx + C = 0,

where A, B and C are constants with A 6= 0. This equation is easy to solve and generallywe have the following methods available:

• Factoring. For example if 2x2 + x− 6 = 0, we can easily find the factored form

(2x− 3) (x + 2) = 0.

Using the zero-product rule we get x = 3/2, or x = −2.

• Square-root method. If we have the form (variable expression)2 = k, it follows that(variable expression) = ±

√k. For example if (2x− 3)2 = 25, we can easily do the

following:

(2x− 3)2 = 25

2x− 3 = ±5

Certainly the two simple equations here are easy to solve.

2x− 3 = −5 or 2x− 3 = 5

After solving these two simple equations we get x = −1, or x = 4.

• Quadratic formula. If we have the form Ax2 + Bx + C = 0 where A, B and C areconstants with A 6= 0, we can quickly solve for x by using this formula:

x =−B ±

√B2 − 4AC

2A.

For example, if 3x2 + 2x− 8 = 0 and use the quadratic formula we get:

x =−2±

√22 − 4 (3) (−8)

2 · 3=−2±

√100

6=−2± 10

6.

Simplifying this we get

x = −2 or x =4

3.

For now we will only concentrate on using factoring or square-root method. The quadraticformula will be discussed later, and we will also derive it using the square root method.

Page 96 of 108




34.2 Examples


1. Write the quadratic equation in standard form with the coefficient of x2 positive. Namethe values of A, B, and C.

x2 = 6x + 5

2. Write the quadratic equation in standard form with the coefficient of x2 positive. Namethe values of A, B, and C.

3 (x− 1)2 = 2x2 − 3 (x + 2)

3. Solve by factoring.

3x2 − 9x = 0


x2 + 4 = 5x


2x2 − 12x = 54


2x2 + 5x = 42


x + 27 = x2 − 5x


(3x− 2) (2x + 1) = 3x2 − 15x− 10

9. Write a quadratic equation that has integer coefficients and has as solutions the givenpair of numbers. (Use x as the variable.)

2 and − 5

Page 97 of 108




10. Write a quadratic equation that has integer coefficients and has as solutions the givenpair of numbers. (Use x as the variable.)

2

3and − 5

7

11. Solve by using the square root method.

x2 = 9


16x2 − 9 = 0


16x2 − 9 = 0


(x− 2)2 = 25


(x− 2)2 = 25


(x + 2)2 = 3

Page 98 of 108





35.1 Perfect Squares

When we square something the result is a perfect square. The following list of perfect squaresshould be easy to recognize:

{1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, . . . } .

You’ll also need to recognize the following list of prefect squares too.

(x + 1)2 = x2 + 2x + 1

(x− 1)2 = x2 − 2x + 1

(x + 2)2 = x2 + 4x + 4

(x− 2)2 = x2 − 4x + 4

(x + 3)2 = x2 + 6x + 9

(x− 3)2 = x2 − 6x + 9

(x + 4)2 = x2 + 8x + 16

(x− 4)2 = x2 − 8x + 16

(x + 5)2 = x2 + 10x + 25

(x− 5)2 = x2 − 10x + 25

(x + 6)2 = x2 + 12x + 36

(x− 6)2 = x2 − 12x + 36

... =...

After reviewing this list in class you should easily be able to see a pattern, and then be ableto complete the square on similar problems. For example, suppose you are given

x2 − 14x

and asked to add a number to this expression to make it a perfect square, you’d add 72 or49 to get:

x2 − 14x + 49 = (x− 7)2 .

We’ll use completing the square along with the square root method to solve quadratic equa-tions. Only simple ones though. Here’s an example, solve x2 + 30x+ 200 = 0 by completingthe square. The method will be as follow.

1. Isolate the constant: x2 + 30x = −200.

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2. Find the constant that will complete the square and add it to both sides.

x2 + 30x = −200

x2 + 30x + 152 = −200 + 152

(x + 15)2 = 25

3. Use the square root method to solve.

(x + 15)2 = 25

x + 15 = ±5

x = −15± 5

So we have x = −20 or x = −10.

I’m not claiming that this is the most straight forward method, but it is nonetheless animportant method to learn.

35.2 Examples


1. Solve by completing the square.

x2 + 8x− 9 = 0


x2 = 2x + 3


x2 + 2x− 1 = 0


x2 + 4x− 1 = 0


x2 = 7− 2x


x2 = 18x− 17


x2 − 14x + 113 = 0

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36.1 The Quadratic Formula

Now we’ll complete the square on:

Ax2 + Bx + C = 0, A 6= 0.

Here goes:

Ax2 + Bx + C = 0

x2 +B

Ax +

C

A= 0

x2 +B

Ax = −C

A

x2 +B

Ax +

(B

2A

)2

=

(B

2A

)2

− C

A(x +

B

2A

)2

=B2

4A2− 4AC

4A2(x +

B

2A

)2

=B2 − 4AC

4A2

x +B

2A= ±

√B2 − 4AC

4A2

x = − B

2A±√

B2 − 4AC

4A2

This last step is equivalent to:

x =−B ±

√B2 − 4AC

2A

This formula works on any quadratic equation, but I still think you should try to factor first,and if you can’t you should move onto this formula.

36.2 Examples


1. Solve using the quadratic formula.

x2 − 6x− 5 = 0


x2 − x− 3 = 0

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3x2 − 7x + 3 = 0


5x2 = 1− 2x


12x− 9 = x2


x2 + 18x + 82 = 0


x2 + 8x + 25 = 0

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37.1 Related to Quadratic Equations

The following equations will eventually reduce themselves to a simple quadratic, or if you’rereally lucky, a simple linear equations. Don’t despair—you’ll get it once you realize that weonly give you problems that work ! Oh, also, don’t forget to check.

37.2 Examples


1. Solve.

y4 − 34y2 − 225 = 0

2. Solve.

x4 + 24y2 − 25 = 0

3. Solve.√x− 5 + x = 7

4. Solve.√2y − 1 = y − 8

5. Solve.√

3z + 4 + 2z = 12

6. Solve.

b− 4 =√b− 2

7. Solve.√x + 5 =

√x + 65

8. Solve.√

19− 2x−√

9− x = 1

9. Solve.6y − 42

y − 6= 11− y

10. Solve.3

x= 1 +

2

2x− 1

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38.1 Graphing Parabolas

-6 -5 -4 -3 -2 -1 0 1 2 3

-4

-3

-2

-1

1

2

Figure 4: Partial graph of f (x) = y = x2 + 2x− 3, with important features indicated.

You should be able to graph a simple case of the general form of a parabola,

y = Ax2 + Bx + C.

I strongly suggest you start by creating a table with simple points, at least six. I do expectthat you are capable to finding the following key features though:

• x-intercepts by setting y = 0. They’re not always there, but in general they’re worthfinding. In our example above we would do the following.

y = x2 + 2x− 3

0 = x2 + 2x− 3

0 = (x + 3) (x− 1)

So the x-intercepts are: (−3, 0) and (1, 0).

• y-intercept by setting x = 0. That’s just too easy! The y-intercept in the exampleabove is: (0, −3).

• The vertex, which is the point (at least in our examples) that will either be the highest(maximum) or lowest (minimum) point on our graph. This point is dead-center of thex values of the x-intercepts. If you use the quadratic formula and take the averageyou’ll get a nice formula for the vertex.(

− B

2A, f

(− B

2A

))So in our example above we have (−1, −4) as the vertex.

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• The axis-of-symmetry which is the dashed-line in our graph above. It should be notedthat this line is a folding line of symmetry. The equation of this line is x = −1.

The examples that follow are really not difficult are are representative of what you’ll see inthe homework and on exams.

38.2 Examples


1. Find the coordinates of the vertex and the equation of the axis of symmetry for theparabola given by the equation. Then graph the equation.

y = x2 − 4x− 1


y = x2 + 2x− 3


y = −x2 + 2x− 3


y = 2x2 − 12x + 17

5. Graph the function. State the domain and range in interval, or set-builder notation.

f (x) = 2x2 − 16x + 34

6. Graph the function. State the domain and range in interval, or set-builder notation.

f (x) = 4x2 − 9

7. Find the x and y-intercepts of the graph of the parabola given by the equation.

y = x2 − 5x + 6

8. Find the zeros (x coordinate of the x-intercepts) of f .

f (x) = 12 + 7x + x2

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39.1 Graphing Circles

-2 -1 0 1 2 3 4 5 6 7

-4

-3

-2

-1

1

Figure 5: Graph of x2 + y2 − 2x + 2y − 2 = 0, with important features indicated.

On the graph above you should be able to label the center, indicated radius, and the fourpoints along the circumference of the circle. You should also understand that a circle is aset of points that is eqidistant from the center, so we can generalize the equations of a circleusing the distance formula. Here we will let r represent the radius, and (h, k) the center. Ifwe let (x, y) represent a arbitrary point on the circle we get18√

(x− h)2 + (y − k)2 = r.

Squaring both sides we get the standard from of a circle

(x− h)2 + (y − k)2 = r2

Again, the radius is r; the center is (h, k), and the point on the circle is (x, y). For example,if we have

(x− 3)2 + (y + 4)2 = 25,

the center is (3, −4), and the radius is 5. If you’re asked to graph the problem you shouldalso be able to find four easy points along the circle’s circumference:

(3, −9) , (3, 1) , (−2, −4) , (8, −4) .

If the equation given is not in standard form you will need to use completing the square.

18We’re using the distance formula.

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For example:

x2 + y2 + 10x− 6y + 25 = 0

x2 + 10x + y2 − 6y = −25

x2 + 10x + 25 + y2 − 6y + 9 = −25 + 25 + 9

(x + 5)2 + (y − 3)2 = 9

The center is (−5, 3), and the radius is 3. If you’re asked to graph the problem you shouldalso be able to find four easy points along the circle’s circumference:

(−5, 0) , (−5, 6) , (−8, 3) , (−2, 3) .

39.2 Examples


1. Sketch a graph of the circle given by the equation.

(x− 7)2 + (y + 4))2 = 9

2. Sketch a graph of the circle given by the equation.

(x + 5)2 + (y − 3))2 = 4

3. Find the equation of the circle with radius 3 and center (−2, 1).

4. Find the equation of the circle with radius 1 and center (1, −1).

5. Find the equation of the circle whose center is (1, −3) and that passes through the pointwhose coordinates are (2, 1).

6. Find the equation of the circle whose center is (−1, 3) and that passes through the pointwhose coordinates are (−2, −1).

7. Write the equation of the circle in standard form. Then sketch its graph.

x2 + y2 − 2x + 12y + 14 = 0


x2 + y2 − 10x + 6y + 27 = 0


x2 + y2 + 6x + 2y − 6 = 0

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