College Algebra Essex County College LMS Project …faculty.essex.edu/~bannon/m100/mth.100.fall.2014.pdfCollege Algebra MTH-100 Essex County College Division of Mathematics LMS Project

College AlgebraMTH-100

Essex County CollegeDivision of Mathematics

LMS ProjectClass Notes

Contents

1 Introduction 5

2 Review for Assignment mth.100.02.01 62.1 Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Review for Assignment mth.100.02.02 133.1 Solving Simple Algebraic Word Problems . . . . . . . . . . . . . . . . . . . . 133.2 Solving Simple Consecutive Integers . . . . . . . . . . . . . . . . . . . . . . . 143.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4 Review for Assignment mth.100.02.03 174.1 Mixture Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 Uniform Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5 Review for Assignment mth.100.02.04 225.1 Properties of Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

6 Review for Assignment mth.100.05.01 306.1 Integer Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306.2 The Basic Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

7 Review for Assignment mth.100.05.02 367.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

8 Review for Assignment mth.100.05.03 388.1 Polynomial Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

9 Review for Assignment mth.100.05.04 439.1 Polynomial Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

10 Review for Assignment mth.100.05.05 4710.1 Polynomial Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4710.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

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11 Review for Assignment mth.100.05.06 5411.1 Special Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5411.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

12 Review for Assignment mth.100.05.07 5712.1 Zero-Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5712.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

13 Review for Exam #1 6413.1 Sample Questions for Review . . . . . . . . . . . . . . . . . . . . . . . . . . 6413.2 Answers to Sample Questions for Review . . . . . . . . . . . . . . . . . . . . 66

14 Review for Assignment mth.100.06.01 7314.1 Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7314.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

15 Review for Assignment mth.100.06.02 7915.1 Finding an LCM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7915.2 Using an LCM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7915.3 Addition and Subtraction of Rational Expressions . . . . . . . . . . . . . . . 80

15.3.1 Same Denominator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8015.3.2 Different Denominator . . . . . . . . . . . . . . . . . . . . . . . . . . 81

15.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

16 Review for Assignment mth.100.06.03 8816.1 Complex Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8816.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

17 Review for Assignment mth.100.06.05 9117.1 Solving of Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 91

17.1.1 Essential Rules and Procedures for Solving Equations . . . . . . . . . 9117.1.2 Work Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

17.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

18 Review for Assignment mth.100.07.01 9618.1 Rational Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

18.1.1 The Basic Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9618.1.2 Meaning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

18.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

19 Review for Assignment mth.100.07.02 10219.1 Simplifying Radical Expressions . . . . . . . . . . . . . . . . . . . . . . . . . 102

19.1.1 The Product Property . . . . . . . . . . . . . . . . . . . . . . . . . . 10219.1.2 Initial Radical Simplification . . . . . . . . . . . . . . . . . . . . . . . 10219.1.3 Adding/Subtracting Radicals . . . . . . . . . . . . . . . . . . . . . . 103

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19.1.4 Multiplying Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . 10319.1.5 Dividing Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10319.1.6 Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

19.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

20 Review for Midterm Exam 11320.1 Sample Questions for Review . . . . . . . . . . . . . . . . . . . . . . . . . . 11320.2 Answers to Sample Questions for Review . . . . . . . . . . . . . . . . . . . . 116

21 Review for Assignment mth.100.07.03 12821.1 Solving Radical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12821.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

22 Review for Assignment mth.100.07.04 13122.1 Introduction to Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . 131

22.1.1 Operations on Complex Numbers . . . . . . . . . . . . . . . . . . . . 13122.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

23 Review for Assignment mth.100.08.01 13723.1 Solving Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 13723.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

24 Review for Assignment mth.100.08.02 14224.1 Perfect Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14224.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

25 Review for Assignment mth.100.08.03 14525.1 The Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14525.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

26 Review for Assignment mth.100.08.04 14826.1 Related to Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . 14826.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

27 Review for Assignment mth.100.03.01 15027.1 Points, Midpoints and Distance . . . . . . . . . . . . . . . . . . . . . . . . . 15027.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

28 Review for Assignment mth.100.03.02 15328.1 Introductory Functions and Relations . . . . . . . . . . . . . . . . . . . . . . 15328.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

29 Review for Assignment mth.100.03.03 15729.1 Graphing Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15729.2 Special Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

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29.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

30 Review for Assignment mth.100.03.04 16530.1 Slope of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16530.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

31 Review for Assignment mth.100.03.05 16931.1 Equations of a Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16931.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

32 Review for Assignment mth.100.03.06 17232.1 Parallel and Perpendicular Lines . . . . . . . . . . . . . . . . . . . . . . . . . 17232.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

33 Review for Class Test #2 17533.1 Sample Questions for Review . . . . . . . . . . . . . . . . . . . . . . . . . . 17533.2 Answers to Sample Questions for Review . . . . . . . . . . . . . . . . . . . . 178

34 Review for Assignment mth.100.04.01 18834.1 System of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 18834.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

35 Review for Assignment mth.100.04.02 19235.1 Addition or Elimination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19235.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

36 Review for Assignment mth.100.04.04 19636.1 Word Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19636.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

37 Review for Assignment mth.100.09.01 19937.1 Graphing Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19937.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

38 Review for Assignment mth.100.11.02 20438.1 Graphing Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20438.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

39 Review for Comprehensive Final Exam 20939.1 Sample Questions for Review . . . . . . . . . . . . . . . . . . . . . . . . . . 20939.2 Answers to Sample Questions for Review . . . . . . . . . . . . . . . . . . . . 211

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1 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 follows willrefer to 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 setof notes (available on the course webpage); or please feel free to use a portable computer inclass to view these notes.

You may contact Ron Bannon if you have any questions or concerns regarding thisdocument and its content.

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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http://mth100.mathography.org

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2 Review for Assignment mth.100.02.01

2.1 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!

2.2 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

Solution: We’ll discuss this in class.

The final answer is: Yes.

2. Solve and check.

z + 6 = 11

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The final answer is: x = 5

3. Solve and check.

x +1

3=

2

3


The final answer is: x =1

3

4. Solve and check.

a− 2

5=

2

3


The final answer is: a =16

15

5. Solve and check.

−2

3x =

1

2


The final answer is: x = −3

4

6. Solve and check.

5

6+

n

12= −1

4


The final answer is: n = −13

7. Solve and check.

m + 1.32 = −2.39

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The final answer is: m = −3.71

8. Solve and check.x

5= −3



9. Solve and check.

3b− 3 = 5b


The final answer is: b = −3

2

10. Solve and check.

5

8− 2y

3=

5

4



16


7− 5z = 5z − 9


The final answer is: z =8

5


2x + 3− 5x = 9

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1

6− 5t = 6


The final answer is: t = −7

6


37

24=

7

8− 5x

6



5


5z − 4 (3− 2z) = 2 (3z − 2) + 6


The final answer is: z = 2


3y + 3 (y + 1) = 27


The final answer is: y = 4


−2 [3x− 5 (2x− 3)] = 3x− 8

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8− 5 (4− 5x) = 3 (4− x)− 8x



3


5 + 3 [1 + 2 (2x− 3)] = 6 (x + 5)



3


5 [x + 2 (3− x)] = 3 [2 (4− x)− 5]




5 [4− 2 (s− 7)] = 5 (2− 4s)


The final answer is: s = −8


5− 2x

2+

x− 3

8=

3

8

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0.5 (2x + 100)− 0.1 (x− 10) = 87



24. Solve for x.

y = mx + b


The final answer is: x =y − b

m

25. Solve for b.

D =a + b− c

3


The final answer is: b = 3D + c− a

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3.1 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’t thatalso unknown. Yes, the large number is also unknown, but it is related to the small number.Since we know that the small number and the large number, when added together, resultsin 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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3.2 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.

3.3 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.


The final answer is: The number is 3.

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.

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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.


The final answer is: The integers are 13 and 15.

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


The final answer is: The number is −50.

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


The final answer is: The integers are 9 and 17.

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.


The final answer is: The numbers are 4, 5, and 8.

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


The final answer is: The integers are 42, 43, and 44.

9. Five times the smallest of three consecutive odd integers is eight more than twice thelargest. Find the integers.1

1This problem has no solution, but you will still need to show all work in reaching this conclusion.

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The final answer is: Although tricky, the solution to the equation is not an integer, sothere is no solution to this problem. Solving this problem results in the smallest integerbeing 16

3, and that’s clearly not an integer!

10. Five times the smallest of three consecutive odd integers is seven more than twice thelargest. Find the integers.


The final answer is: The integers are 5, 7, and 9.

11. Five times the smallest of three consecutive even integers is seven more than twice thelargest. Find the integers.2


The final answer is: No answer because the only solution turns out to be odd!

12. The sum of two numbers is fourteen. One less than three times the smaller number isequal to the larger number. Find the numbers.


The final answer is: The numbers are 3.75 and 10.25.

13. 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.


The final answer is: The numbers are 8 and 10.

2This problem has no solution, but you will still need to show all work in reaching this conclusion.

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4.1 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 every mixtureproblem is essentially the same—you need to clearly understand what is being mixed 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) weighing 20 pounds that is40% silver. You should (we’ll discuss in class) be able to answer the following 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%

4.2 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.

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


The final answer is: 11 quarters.

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?


The final answer is: 7 quarters.

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.


The final answer is: The passenger train’s rate is 48 mph and the freight train’s rate is30 mph.

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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.


The final answer is: 9 and 18 mph.

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?


The final answer is: 14 pounds of the $5.50 per pound coffee and 16 pounds $4.00 perpound coffee.

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.


The final answer is: 76 adults tickets.

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?


The final answer is: 100 ounces.

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. Find thenumber of 15¢ stamps in the collection.


The final answer is: Four 15¢ stamps.

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?

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The final answer is: 9 L of 75% solution and 21 L of 25% 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?


The final answer is: 3 hours.

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 tenth ofa percent.)


The final answer is: 36%

12. At noon a train leaves Washington D.C., headed for Charleston, South Carolina, adistance of five-hundred miles. The train travels at sixty mph. At one pm a second trainleaves Charleston headed for Washington D.C., traveling at fifty mph. How long afterthe train leaves Charleston will the two trains pass each other?


The final answer is: 4 hours.

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


The final answer is: 7 quarts.

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.


The final answer is: The passenger train’s rate is 50 mph; and the freight train’s rate is30 mph.

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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?


The final answer is: 60 pounds of hamburger that is 15%; and 30 pounds of hamburgerthat is 21%.

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?


The final answer is: 36 miles.

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.


The final answer is: 420 and 530 mph.

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?


The final answer is: 45 miles.

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?


The final answer is: 2.7 ounces of 70% rubbing alcohol; and 1.5 ounces of water.

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5.1 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 whenever youmultiply or divide by a negative number (see last rule). In class we will do many examplesto illustrate what needs to be done.

5.2 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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5.3 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


Interval notation: [−5, ∞)

Set-builder notation: {x | x ≥ −5}Graph:

Figure 1: Graph of x + 8 ≥ 3.


−4x ≤ −12


Interval notation: [3, ∞)

Set-builder notation: {x | x ≥ 3}Graph:

Figure 2: Graph of −4x ≤ −12.

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4x + 4 ≥ 3x− 1




Figure 3: Graph of 4x + 4 ≥ 3x− 1.


4x + 2 < 10


Interval notation: (−∞, 2)

Set-builder notation: {x | x < 2}Graph:

Figure 4: Graph of 4x + 2 < 10.


3x + 2 ≤ −10


Interval notation: (−∞, −4]

Set-builder notation: {x | x ≤ −4}Graph:

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Figure 5: Graph of 3x + 2 ≤ −10.


5x + 2 < 3x− 6


Interval notation: (−∞, −4)

Set-builder notation: {x | x < −4}Graph:

Figure 6: Graph of 5x + 2 < 3x− 6.


2− 4x > 10


Interval notation: (−∞, −2)

Set-builder notation: {x | x < −2}Graph:

Figure 7: Graph of 2− 4x > 10.


5− 3x ≤ 20

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Figure 8: Graph of 5− 3x ≤ 20.


6x ≤ 19


Interval notation: (−∞, 19/6]

Set-builder notation: {x | x ≤ 19/6}Graph:

Figure 9: Graph of 6x ≤ 19.


2x + 7 > 3 (1− 2x)


Interval notation: (−1/2, ∞)

Set-builder notation: {x | x > −1/2}Graph:

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Figure 10: Graph of 2x + 7 > 3 (1− 2x).


3 +2

7x < 2x− 5


Interval notation: (14/3, ∞)

Set-builder notation: {x | x > 14/3}Graph:

Figure 11: Graph of 3 + 27x < 2x− 5.


−21 < 3x < −1


Interval notation: (−7, −1/3)

Set-builder notation: {x | − 7 < x < −1/3}Graph:

Figure 12: Graph of −21 < 3x < −1.


3

4x− 1

4< x− 5

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Interval notation: (19, ∞)

Set-builder notation: {x | x > 19}Graph:

Figure 13: Graph of 34x− 1

4< x− 5.


31

40x− 5

4<

4

5x +

19

20


Interval notation: (−88, ∞)

Set-builder notation: {x | x > −88}Graph:

Figure 14: Graph of 3140x− 5

4< 4

5x + 19

20.


8− 3 (x− 4) ≤ 2x + 10


Interval notation: [2, ∞)

Set-builder notation: {x | x ≥ 2}Graph:

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Figure 15: Graph of 8− 3 (x− 4) ≤ 2x + 10.


−5 ≤ 6x + 7 < 25


Interval notation: [−2, 3)

Set-builder notation: {x | − 2 ≤ x < 3}Graph:

Figure 16: Graph of −5 ≤ 6x + 7 < 25.

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6.1 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.”

6.2 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!

6.3 Examples


1. Simplify.(−2x0

)3 (3x2)2


Answer: −72x4

2. Simplify.

−18x3y5

27x4y6


Answer: − 2

3xy

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3. Simplify.(−3x−1y2

)2


Answer:9y4

x2

4. Simplify.(−3x2y

)2 (2x3y2

)3


Answer: 72x13y8

5. Simplify.(6x3y−3

9x4y−1

)2


Answer:4

9x2y4

6. Simplify.(6x3y−3

9x4y−1

)−1


Answer:3xy2

2

7. Simplify.(4x−2y3

6x4y−2

)−3

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Answer:27x18

8y15

8. Simplify.

(−2x)(3x−2

)2


Answer: −18

x3

9. Simplify.

25x4y7z2

20x5y9z8


Answer:5

4xy2z6

10. Simplify.(3x−1y−2

)2


Answer:9

x2y4

11. Simplify.(−2x−5

) (3x5)

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Answer: −6

12. Simplify.

(−2x)(−2xy−2

) (4x−2y

)−2


Answer:x6

4y4

13. Simplify.

3x−2y

9xy


Answer:1

3x3

14. Simplify.(−3xy−2

) (2x−1y

)−3


Answer: −3x4

8y5

15. Simplify.(6x−4yz−1

9xy−4z2

)−3

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Answer:27x15z9

8y15

16. Simplify.(−18x4y−2z3

12xy−3z2

)−2


Answer:4

9x6y2z2

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7.1 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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7.2 Examples


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


Answer: P (−2) = −2

2. Simplify. Use a vertical and horizontal format.3(4x2 − 5x + 6

)+(−4x2 + 5x− 13

)


Answer: −7

3. Simplify. Use a vertical and horizontal format.4(4x2 + 7x− 5

)−(2x2 − 6x− 7

)


Answer: 2x2 + 13x + 2

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)


Answer:

S (x) = P (x) + R (x) = 2x5 + 3x4 − 8x− 3

T (x) = P (x)−R (x) = −2x5 + 3x4 − 2x + 13

S (−1) = 6

T (2) = −7

3You’re free to use whatever format you like.4You’re free to use whatever format you like.

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8.1 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 multiplicationsof increasing difficulty. It’s really just long multiplication, but you may not see this at first.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 you wereactually taught in grammar school, in reverse order though. Don’t worry here, you can stillmultiply using columns, just as you did in grammar school. This example here is just topoint 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 same process.

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.

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8.2 Examples


1. Multiply and simplify.

4x2 − x [x− 2 (4x− 5)]


Answer: 11x2 − 10x


(x− 2)(3x2 − 5x + 1

)


Answer: 3x3 − 11x2 + 11x− 2

3. Multiply and simplify.(x2 − 3x + 5

)(x− 3)


Answer: x3 − 6x2 + 14x− 15


(b− 3) (2b− 7) (b− 6)


Answer: 2b3 − 25b2 + 99b− 126


(2x− 3)(x2 − 3x + 5

)

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Answer: 2x3 − 9x2 + 19x− 15


(2x− 1)(−3x2 − 5x + 1

)


Answer: −6x3 − 7x2 + 7x− 1

7. Multiply and simplify.(2 + 3x− x2

)(2x− 1)


Answer: −2x3 + 7x2 + x− 2

8. Multiply and simplify.(x2 − 2x

)(2x + 5)


Answer: 2x3 + x2 − 10x


(x + 1) (x− 1)


Answer: x2 − 1


(3x− 2) (3x + 2)

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Answer: 9x2 − 4


(3x− 1) (x + 4)


Answer: 3x2 + 11x− 4


(2x− 3) (4x− 7)


Answer: 8x2 − 26x + 21


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


Answer: 30x2 + 61x + 30


(11x− 2y) (3x + 5y)


Answer: 33x2 + 49xy − 10y2


(3x− 5)2

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Answer: 9x2 − 30x + 25


(5x + 2y)2


Answer: 25x2 + 20xy + 4y2


(a + b)2 + (a− b)2


Answer: 2a2 + 2b2


(2x− 3)2 − (2x + 3)2


Answer: −24x

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9.1 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.

9.2 Examples


1. Divide.

x6 − 3x4 − x2

x2


Answer: x4 − 3x2 − 1

2. Divide.

8x2y2 − 24xy

8xy

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Answer: xy − 3

3. Divide.

3x2 − 5x + 6

15x


Answer:x

5− 1

3+

2

5x

4. Divide.

9x2y + 6xy − 3xy2

3xy


Answer: 3x + 2− y

5. Divide.(x2 − x− 6

)÷ (x− 3)


Answer: x + 2

6. Divide.(2x2 − 13x + 21

)÷ (x− 3)


Answer: 2x− 7

7. Divide.(6x2 − x− 15

)÷ (3x− 5)

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Answer: 2x + 3

8. Divide.(15x2 + 17x− 4

)÷ (3x + 4)


Answer: 5x− 1

9. Divide.(12x2 − 17x + 5

)÷ (3x− 2)


Answer: 4x− 3− 1

3x− 2

10. Divide.(2x2 + 7

)÷ (x− 3)


Answer: 2x + 6 +25

x− 3

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

Q (x).


Answer: 9x2 − 6x + 4

12. Divide.(x4 − 3x3 + 2x2 − 5x− 10

)÷ (x− 3)

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Answer: x3 + 2x + 1− 7

x− 3

13. Divide.

4x3 − 14x2 + 2x− 7

x2 − 4x + 3


Answer: 4x + 2− 2x + 13

x2 − 4x + 3

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10.1 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, try5

x2 + 5x− 14.

10.2 Examples


5Answer: (x + 7) (x− 2)

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1. Factor.

4x2 − 12x


Answer: 4x (x− 3)

2. Factor.

5− 20x


Answer: 5 (1− 4x)

3. Factor.

12x3 − 10x2


Answer: 2x2 (6x− 5)

4. Factor.

2x2y − 6x2y2 + 12xy


Answer: 2xy (x− 3xy + 6)

5. Factor.

3x2y2 − 9xy2 + 15y2

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Answer: 3y2(x2 − 3x + 5

)6. Factor.

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


Answer: (x + y) (4 + A)

7. Factor.

x2 − 3x + 4ax− 12a


Answer: (x− 3) (x + 4a)

8. Factor.

x2 − 5x− 2xy + 10y


Answer: (x− 5) (x− 2y)

9. Factor.

2x2 − 10x + 7xy − 35y


Answer: (x− 5) (2x + 7y)

10. Factor.

x2 + 5x + 6

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Answer: (x + 3) (x + 2)

11. Factor.

x2 + x− 6


Answer: (x + 3) (x− 2)

12. Factor.

x2 − 2x− 35


Answer: (x− 7) (x + 5)

13. Factor.

x2 − 2x− 3


Answer: (x− 3) (x + 1)

14. Factor.

40− 3x− x2


Answer: (8 + x) (5− x)

15. Factor.

x2 − 6x + 8

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Answer: (x− 2) (x− 4)

16. Factor.

x2 + 16x + 39


Answer: (x + 13) (x + 3)

17. Factor.

x2 − 15x + 56


Answer: (x− 8) (x− 7)

18. Factor.

3x2 + 7x + 2


Answer: (3x + 1) (x + 2)

19. Factor.

x2 − 3xy − 28y2


Answer: (x− 7y) (x + 4y)

20. Factor.

2y2 − 21y + 52

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Answer: (2y − 13) (y − 4)

21. Factor.

x4 − 22x3 + 120x2


Answer: x2 (x− 12) (x− 10)

22. Factor.

11z2 − 57z + 10


Answer: (11z − 2) (z − 5)

23. Factor.

3y4 + 54y3 + 135y2


Answer: 3y2 (y + 3) (y + 15)

24. Factor.

18y3 + 24y2 − 64y


Answer: 2y (3y − 4) (3y + 8)

25. Factor.

4x2y + 20xy − 56y

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Answer: 4y (x− 2) (x + 7)

26. Factor.

21b + 22b2 − 8b3


Answer: b (3 + 4b) (7− 2b)

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11.1 Special Factoring

This section of the textbook deals with factoring polynomials where special evident patternsmay exists.6 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.

11.2 Examples


1. Factor.

12t3 − 75t


Answer: 3t (2t + 5) (2t− 5)

2. Factor.

s2 − 81

6Special patterns 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 54 of 219





Answer: (s− 9) (s + 9)

3. Factor.

9w2 − 16


Answer: (3w − 4) (3w + 4)

4. Factor.

y4 − 16x4


Answer: (y − 2x) (y + 2x)(y2 + 4x2

)5. Factor.

1− 64x3


Answer: (1− 4x)(1 + 4x + 16x2

)6. Factor.

x2y2 − 7xy2 − 8y2


Answer: y2 (x + 1) (x− 8)

7. Factor.

8x3 + 27

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Answer: (2x + 3)(4x2 − 6x + 9

)8. Factor.

x2b2 + 6xb2 + 9b2


Answer: b2 (x + 3) (x + 3)

9. Factor.

2z3 − 8z2y + 8zy2


Answer: 2z (z − 2y) (z − 2y)

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12.1 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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12.2 Examples


1. Solve for x.

4x2 − 12x = 0


Answer: x = 0 or x = 3

2. Solve for y.

(y + 2) (y − 3) = 14


Answer: y = −4 or y = 5

3. Solve for z.

z (2z + 1) (3z − 2) = 0


Answer: x = 0 or x = −1

2or x =

2

3

4. Solve for a.

a2 + a = 6


Answer: a = −3 or a = 2

5. Solve for x.

x2 = 2x + 35

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Answer: x = −5 or x = 7

6. Solve for t.

15 = 8t− t2


Answer: t = 3 or t = 5

7. Solve for s.

s2 = 15s− 56


Answer: s = 7 or s = 8

8. Solve for y.

3y3 + 54y2 + 135y = 0


Answer: y = −15 or y = −3 or y = 0

9. Solve for x.

4x2 + 20x = 56



10. Solve for y.

3y2 + 7y − 6 = 0

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Answer: y = −3 or y =2

3

11. Solve for z.

2z2 − z = 3


Answer: z = −1 or z =3

2

12. Solve for t.

2t2 + 13t + 15 = 0


Answer: t = −5 or t = −3

2

13. Solve for x.

3x2 = 16x− 5


Answer: x =1

3or x = 5

14. Solve for v.

8v2 + 33v + 4 = 0

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Answer: v = −4 or v = −1

8

15. Solve for y.

2y3 − 5y2 − 32y + 80 = 0



2or y = 4

16. Solve for z.

24z2 − 24z = 18


Answer: z = −1

2or z =

3

2

17. Solve for c.

19c− 15c2 = 6


Answer: c =3

5or c =

2

3

18. Solve for b.

b2 + 8b− 15 = (2b + 3) (b− 5)

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Answer: b = 0 or b = 15

19. Solve for t.

12t3 = 75t


Answer: t = −5

2or t = 0 or t =

5

2

20. Solve for x.

(x− 1) (x + 2) = 4



21. Solve for n.

(5n− 1) (n + 2) = (2n− 1) (3n + 2)


Answer: n = 0 or n = 8

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


Answer: −12 or 11

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

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Answer: 6 ft.× 14 ft.

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13 Review for Exam #1

Test on Chapters 2 and 5. This exam will include materials covered up to this point. Forreview you may look back over your notes and the examples done in class. You should alsocomplete all homework assignments up to this point.

13.1 Sample Questions for Review

These problems will not be covered in class.

1. Solve for t.

8t− t2 = 15

2. A 24% acid solution is mixed with a 14% acid solution to form a 19% acid solution. Howmany liters of the 24% acid solution are used to make 10 liters of the 19% acid solution?

3. Simplify and leave your answer with positive exponents only.(4ab−6

a−2b−1

)2

4. Solve for y.

y2 − 10y = −16

5. Two small planes start from the same point and fly in opposite directions. The firstplane is flying 40 mph slower than the second plane. In two hours, the planes are 500miles apart. Find the rate of the slower plane.

6. Divide using long division.(10x2 + 9x− 5

)÷ (2x− 1)


)(x− 2)

8. Solve for y.

2y3 − 5y2 − 32y + 80 = 0

9. Solve for x.

2 [3− (5x− 4)] = −6

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10. Solve the inequality 7x + 4 ≤ 19x + 6. Write the solution in set-builder notation.

11. Solve the formula for the given variable.

V =1

3Ah; A

12. Simplify and leave with positive exponents only.(3a−4z5

)213. Factor completely.

36a3b2 − 9a2b2 − 81a2b4

14. The sum of four consecutive even integers is 108. Find the integers.

15. Solve the inequality 19− 3 (x− 6) ≤ 3x + 7. Write the solution in interval notation andgraph your solution on the number line.

16. How many pounds of candy worth 70 cents per pound must be mixed with 30 poundsof candy worth 90 cents per pound to produce a mixture which can be sold for 85 centsper pound?

17. Find the solution of −2 [3x− (4x + 6)] = 20.

18. Solve for y.

A =x + y − z

5

19. Factor completely.

2b6 + 16b3

Page 65 of 219




13.2 Answers to Sample Questions for Review

Your work should look similar to the work provided for each problem.

1. Solve for t.

8t− t2 = 15

Solution:

8t− t2 = 15

0 = t2 − 8t + 15

0 = (t− 3) (t− 5)

Answer: t = 3 or t = 5


Solution: Let x represent the number of liters of the 24% acid solution.

0.24x + 0.14 (10− x) = 0.19 · 10

0.24x + 1.4− .14x = 1.9

0.1x + 1.4 = 1.9

0.1x = 0.5

x = 5

Answer: 5

3. Simplify and leave your answer with positive exponents only.(4ab−6

a−2b−1

)2

Solution:(4ab−6

a−2b−1

)2

=

(4a3

b5

)2

=16a6

b10

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Answer:16a6

b10

4. Solve for y.

y2 − 10y = −16

Solution:

y2 − 10y = −16

y2 − 10y + 16 = 0

(y − 2) (y − 8) = 0

Answer: y = 2 or y = 8

5. Two small planes start from the same point and fly in opposite directions. The firstplane is flying 40 mph slower than the second plane. In two hours, the planes are 500miles apart. Find the rate of the slower plane.

Solution: Let x represent the rate of the slower plane.

2 (x + 40) + 2x = 500

2x + 80 + 2x = 500

4x + 80 = 500

4x = 420

x = 105

Answer: 105 mph.

6. Divide using long division.(10x2 + 9x− 5

)÷ (2x− 1)

Solution: 5x + 7

2x− 1)

10x2 + 9x− 5− 10x2 + 5x

14x− 5− 14x + 7

2

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Answer: 5x + 7 +2

2x− 1


)(x− 2)

Solution:(x2 − 7x + 4

)(x− 2) = (x− 2)

(x2 − 7x + 4

)= x3 − 7x2 + 4x− 2x2 + 14x− 8

= x3 − 9x2 + 18x− 8

Answer: x3 − 9x2 + 18x− 8

8. Solve for y.

2y3 − 5y2 − 32y + 80 = 0

Solution:

2y3 − 5y2 − 32y + 80 = 0

y2 (2y − 5)− 16 (2y − 5) = 0

(2y − 5)(y2 − 16

)= 0

(2y − 5) (y − 4) (y + 4) = 0


2or y = 4

9. Solve for x.

2 [3− (5x− 4)] = −6

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Solution:

2 [3− (5x− 4)] = −6

2 [3− 5x + 4] = −6

2 [7− 5x] = −6

14− 10x = −6

20 = 10x

2 = x

Answer: x = 2

10. Solve the inequality 7x + 4 ≤ 19x + 6. Write the solution in set-builder notation.

Solution:

7x + 4 ≤ 19x + 6

−2 ≤ 12x

−1

6≤ x

Answer:

{x | x ≥ −1

6

}

11. Solve the formula for the given variable.

V =1

3Ah; A

Solution:

V =1

3Ah

3V = Ah3V

h= A

Answer: A =3V

h

12. Simplify and leave with positive exponents only.(3a−4z5

)2Page 69 of 219




Solution:

(3a−4z5

)2=

(3z5

a4

)2

=9z10

a8

Answer:9z10

a8


36a3b2 − 9a2b2 − 81a2b4

Solution:

36a3b2 − 9a2b2 − 81a2b4 = 9a2b2(4a− 1− 9b2

)Answer: 9a2b2 (4a− 1− 9b2)

14. The sum of four consecutive even integers is 108. Find the integers.

Solution: Let x, x + 2, x + 4 and x + 6 represent the four consecutive even integers.

(x) + (x + 2) + (x + 4) + (x + 6) = 108

4x + 12 = 108

4x = 96

x = 24

Answer: 24, 26, 28 and 30.

15. Solve the inequality 19− 3 (x− 6) ≤ 3x + 7. Write the solution in interval notation andgraph your solution on the number line.

Page 70 of 219




Solution:

19− 3 (x− 6) ≤ 3x + 7

19− 3x + 18 ≤ 3x + 7

37− 3x ≤ 3x + 7

30 ≤ 6x

5 ≤ x

Answer: [5, ∞). Graph below.

Figure 17: 19− 3 (x− 6) ≤ 3x + 7

16. How many pounds of candy worth 70 cents per pound must be mixed with 30 poundsof candy worth 90 cents per pound to produce a mixture which can be sold for 85 centsper pound?

Solution: Let x represent the number of pounds of candy worth 70 cents per pound.

70x + 90 · 30 = 85 (x + 30)

70x + 2700 = 85x + 2250

150 = 15x

10 = x

Answer: 10

17. Find the solution of −2 [3x− (4x + 6)] = 20.

Solution:

−2 [3x− (4x + 6)] = 20

−2 [3x− 4x− 6] = 20

−2 [−x− 6] = 20

2x + 12 = 20

2x = 8

x = 4

Answer: x = 4

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18. Solve for y.

A =x + y − z

5

Solution:

A =x + y − z

55A = x + y − z

5A− x + z = y

Answer: y = 5A− x + z


2b6 + 16b3

Solution:

2b6 + 16b3 = 2b3(b3 + 8

)= 2b3 (b + 2)

(b2 − 2b + 4

)Answer: 2b3 (b + 2) (b2 − 2b + 4)

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14.1 Rational Expressions

You should already be familiar with functional notation and how to evaluate functions fora given value. In addition, you should have some understanding of domain of a function,especially the part about not dividing by 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.

14.2 Examples


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

f (x) =x− 3

2x− 1

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Answer: f (−2) = 1

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

f (x) = − 12

x2 − 4x + 6


Answer: f (−3) = −4

9

3. Find the domain of the function.

f (x) =x + 1

x + 5


Answer: All real numbers, x 6= −5.


f (x) =3x− 1

2x− 7


Answer: All real numbers, x 6= 7

2.


f (x) =9x + 4

(3x + 5) (x− 3)


Answer: All real numbers, x 6= −5

3, 3.

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6. Find the domain of the function;

f (x) =3− 2x

x2 − 3x− 18


Answer: All real numbers, x 6= −3, 6.

7. Simplify.

x2 − 3x

2x− 6


Answer:x

2

8. Simplify.

5xy − 3y

9− 15x


Answer: −y

3

9. Simplify.

x2 + 7x− 8

x2 + 6x− 7


Answer:x + 8

x + 7

10. Simplify.

3x3 − 12x

6x3 − 24x2 + 24x

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Answer:x + 2

2 (x− 2)

11. Simplify.

8x− 12

14x + 7· 42x + 21

32x− 48


Answer:3

4

12. Simplify.

x2 − 2x− 24

x2 − 5x− 6· x

2 + 5x + 6

x2 + 6x + 8


Answer:x + 3

x + 1

13. Simplify.

x2 + 2x− 35

x2 + 4x− 21· x

2 + 3x− 18

x2 + 9x + 18


Answer:x− 5

x + 3

14. Simplify.

25− x2

x2 − 2x− 35· x

2 − 8x− 20

x2 − 3x− 10

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Answer:10− x

x− 7

15. Simplify.

4x2y3

15a2b3÷ 6xy

5a3b5


Answer:2ab2xy2

9

16. Simplify.

6x3 + 7x2

12x− 3÷ 6x2 + 7x

36x− 9


Answer: 3x

17. Simplify.

x2 − 49

x4y3÷ x2 − 14x + 49

x4y3


Answer:x + 7

x− 7

18. Simplify.

x2 − x− 2

x2 − 7x + 10÷ x2 − 3x− 4

40− 3x− x2

Page 77 of 219





Answer:8 + x

4− x

Page 78 of 219





15.1 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.

15.2 Using an LCM

The initial purpose7 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

7Will also use it to simplify and solve rational equations.

Page 79 of 219




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.

15.3 Addition and Subtraction of Rational Expressions

15.3.1 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 80 of 219




15.3.2 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

15.4 Examples


1. Find the LCM of the polynomials.

6x2y, 18xy2


Answer: 18x2y2


6x2, 4x + 12


Answer: 12x2 (x + 3)


8x2 (x− 1)2 , 10x3 (x− 1)

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Answer: 40x3 (x− 1)2


(2x− 1) (3− 5x) , (2x− 1)2 (x− 5)


Answer: (2x− 1)2 (3− 5x) (x− 5)


x2 − 2x− 24, x2 − 36


Answer: (x + 4) (x− 6) (x + 6)


2x2 − 7x + 3, 2x2 + x− 1


Answer: (2x− 1) (x− 3) (x + 1)


x2 + 3x− 18, x− 3, x− 2


Answer: (x− 3) (x + 6) (x− 2)

8. Find the LCD.

4

x,

3

x2

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Answer: x2

9. Rewrite the fractions in terms of the LCD.

4

x,

3

x2


Answer:4x

x2,

3

x2


a2

x (x + 7),

a

(x + 7)2


Answer:a2x + 7a2

x (x + 7)2,

ax

x (x + 7)2


3

x (x− 5),

2

(x− 5)2


Answer:3x− 15

x (x− 5)2,

2x

x (x− 5)2


7

x2 + x− 2,

x

x + 2

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Answer:7

(x + 2) (x− 1),

x2 − x

(x + 2) (x− 1)


3x

x− 4,

5

x2 − 16


Answer:3x2 + 12x

(x− 4) (x + 4),

5

(x− 4) (x + 4)


x− 1

x2 + 2x− 15,

x

x2 + 6x + 5


Answer:x2 − 1

(x + 1) (x + 5) (x− 3),

x2 − 3x

(x + 1) (x + 5) (x− 3)

15. Simplify.

7

4y+

11

6y− 8

3y


Answer:11

12y

16. Simplify.

4x− 3

6x+

2x + 3

4x

Page 84 of 219





Answer:14x + 3

12x

17. Simplify.

2x + 9

9x− x− 5

5x


Answer:x + 90

45x

18. Simplify.

x− 2

3x2− x + 4

x


Answer: −3x2 + 11x + 2

3x2

19. Simplify.

3 +x− 1

x + 1


Answer:4x + 2

x + 1

20. Simplify.

x− 2

8x2− x + 7

12xy

Page 85 of 219





Answer:3xy − 6y − 2x2 − 14x

24x2y

21. Simplify.

2

x− 3+

5

x− 4


Answer:7x− 23

(x− 3) (x− 4)

22. Simplify.

4

2x− 1− 5

x− 6


Answer: − 6x + 19

(2x− 1) (x− 6)

23. Simplify.

2x

x2 − x− 6− 3

x + 2


Answer:9− x

(x + 2) (x− 3)

24. Simplify.

6− 2x

2x2 − 15x + 28+

x− 2

x− 4− x + 6

2x− 7

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Answer:x2 − 15x + 44

(x− 4) (2x− 7)=

x− 11

2x− 7

Page 87 of 219





16.1 Complex Fractions

A complex fraction is a fraction where the numerator, denominator, or both contain a fraction.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.

16.2 Examples


1. Simplify.

4 + 32

9− 32


Answer:11

15

2. Simplify.

8x2 + 8

x5x2 + 5

x

Page 88 of 219





Answer:8

5

3. Simplify.

8 + 8x

3− 3x2


Answer:8x

3 (x− 1)

4. Simplify.

a− 749a− a


Answer: − a

a + 7

5. Simplify.

x− 1x

1x

+ x


Answer:x2 − 1

x2 + 1

6. Simplify.

5− 15x+3

7− 21x+3

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Answer:5

7

7. Simplify.

1 + 2x+3

1 + 9x−4


Answer:x− 4

x + 3

8. Simplify.

x− 2 + 92x+7

x + 3− 12x+7


Answer:x− 1

x + 4

9. Simplify.

1− 8x− 9

x2

1 + 6x

+ 5x2


Answer:x− 9

x + 5

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17.1 Solving of Rational Equations

17.1.1 Essential Rules and Procedures for Solving Equations

1. Simplify both sides of the equation. This often involves clearing fractions by multiplyingboth 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 91 of 219




17.1.2 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!

17.2 Examples


1. Solve.

3x + 4

12− 1

3=

5x + 2

12− 1

2


Answer: x = 2

2. Solve.

12

3x− 2=

6

5


Answer: x = 4

3. Solve.

3− 12

x= 7


Answer: x = −3

4. Solve.

3

x− 2=

4

x

Page 92 of 219





Answer: x = 8

5. Solve.

5

3x− 4= − 3

1− 2x


Answer: x = −7

6. Solve.

2x

x + 2− 5 =

7x

x + 2


Answer: x = −1

7. Solve.

x

x + 12=

1

x + 5



8. Solve.

x− 6

x− 3=

2x

x− 3



9. Solve.

x

x + 4= 3− 4

x + 4

Page 93 of 219





Answer: No solution.

10. Solve.

x + 2

x2 − x− 2+

x + 1

x2 − 4=

1

x + 1


Answer: x = −3

11. Solve.

5x

x + 2= 3− 10

x + 2


Answer: No solution.

12. Solve.

4

x− 6− 7

x + 6=

5

x2 − 36


Answer: x = 61/3

13. Solve.

x

2x− 9− 3x =

10

9− 2x


Answer: x = 5 or x = −1/3

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?

Page 94 of 219





Answer: 3 hours

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 apprenticerequired 8 more hours to finish the job. How long would it take the experienced bricklayer,working alone, to do the job?


Answer: 22 hours

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?


Answer: 20 hours

Page 95 of 219





18.1 Rational Exponents

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

18.1.1 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;

18.1.2 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 96 of 219




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.8

x1/n = n√x, xm/n = n

√xm =

(n√x)m

18.2 Examples


1. Simplify.

161/2


Answer: 4

2. Rewrite the exponential expression as a radical expression.

31/3


Answer:3√

3

3. Simplify.

82/3

8Here we are assuming that x1/n is a real number.

Page 97 of 219





Answer: 4


51/2


Answer:√

5

5. Simplify.

163/4


Answer: 8


23/5


Answer:5√

23 =5√

8

7. Simplify.

253/2


Answer: 125

8. Simplify.(64

27

)−2/3

Page 98 of 219





Answer:9

16

9. Simplify.

x1/2x7/2


Answer: x4

10. Simplify.

xx−1/3


Answer: x2/3

11. Simplify.

a−7/15a4/5a2/3


Answer: a

12. Simplify.

x1/4

x3/4


Answer:1

x1/2= x−1/2

13. Simplify.

x−4/7

x−2/7

Page 99 of 219





Answer: x−2/7 =1

x2/7

14. Simplify.(x6)−2/3


Answer: x−4 =1

x4

15. Simplify.(x−5/6

)−18


Answer: x15

16. Simplify.(x−4/3

)−3/8


Answer: x1/2

17. Simplify.(x4/5y3/10

)10


Answer: x8y3

18. Simplify.(x5y−10

)−2/5

Page 100 of 219





Answer: x−2y4 =y4

x2

19. Simplify.(x1/2y−7/6

y−5/6

)−6


Answer: x−3y2 =y2

x3

20. Simplify.

y1/3(y2/3 + y−1/3

)


Answer: y + 1

Page 101 of 219





19.1 Simplifying Radical Expressions

19.1.1 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.

19.1.2 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 102 of 219




19.1.3 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

19.1.4 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

19.1.5 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 exists inthe denominator of an expression. This also includes fractional radicands. Here’s an example,

√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

19.1.6 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 103 of 219




19.2 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!9

1. Simplify.√x3y4z9

Solution: We’ll discuss this is class.

Answer: xy2z4√xz

2. Simplify.√18x2y11z16


Answer: 3xy5z8√

2y

3. Simplify.

3√

x11y14z15


Answer: x3y4z5 3√

x2y2

4. Simplify.

3√

54− 3√

128


Answer: − 3√

2

5. Simplify.

3√

2x + 11√

2x

9Assume that variables represent non-negative real numbers.

Page 104 of 219





Answer: 14√

2x

6. Simplify.

2√

2x3 + 5x√

18x


Answer: 17x√

2x

7. Simplify.

2√

18x2y3 − xy√

50y


Answer: xy√

2y

8. Simplify.

24√

162x5 + 2x4√

32x


Answer: 10x4√

2x

9. Simplify.√9y3 −

√16y3 + y

√9y


Answer: 2y√y

10. Simplify.√

21 ·√

6

Page 105 of 219





Answer: 3√

14

11. Simplify.

3√

25 · 3√

5


Answer: 5

12. Simplify.√3x7y ·

√21x7y6


Answer: 3x7y3√

7y

13. Simplify.

3√

4x2y2 · 3√

8xy12


Answer: 2xy4 3√

4y2

14. Simplify.√

35(√

35−√

5)


Answer: 35− 5√

7

15. Simplify.(3− 2

√5)(

3 + 2√

5)

Page 106 of 219





Answer: −11


√2)2


Answer: 19− 6√

2

17. Simplify.(5 + 3

√2)(

3− 2√

2)


Answer: 3−√

2

18. Simplify.(1−√x) (

1 +√x)


Answer: 1− x


√x) (

2− 3√x)


Answer: 2 + 6x− 7√x

20. Simplify.√

7√3

Page 107 of 219





Answer:

√21

3

21. Simplify.

1√2


Answer:

√2

2

22. Simplify.

13√

2


Answer:3√

4

2

23. Simplify.

3√5x


Answer:3√

5x

5x

24. Simplify.

33√

3x2

Page 108 of 219





Answer:3√

9x

x

25. Simplify.

12x2

4√

27x


Answer: 4x4√

3x3

26. Simplify.

2√3x


Answer:2√

3x

3x

27. Simplify.

√48x2

√12x


Answer: 2√x

28. Simplify.√45x3y5√15x2y

Page 109 of 219





Answer: y2√

3x

29. Simplify.

153√

25


Answer: 33√

5

30. Simplify.

10x3√

25x2


Answer: 23√

5x

31. Simplify.√15x2y4√75x7y3


Answer:

√5xy

5x3

32. Simplify.

1

1−√

2


Answer: −1−√

2

Page 110 of 219




33. Simplify.

3

2−√

3


Answer: 6 + 3√

3

34. Simplify.

5

3−√

3


Answer:15 + 5

√3

6

35. Simplify.

3

2−√

5x


Answer:6 + 3

√5x

4− 5x

36. Simplify.

1 +√x

1−√x


Answer:1 + x + 2

√x

1− x

37. Simplify.√

3−√

5√2−√

3

Page 111 of 219





Answer:√

10 +√

15−√

6− 3

38. Simplify.

2√x− 4

√y

2√x− 5

√y


Answer:4x− 20y + 2

√xy

4x− 25y

Page 112 of 219




20 Review for Midterm Exam

Midterm Exam on Chapters 2, 5, 6, and parts10 of 7. This exam will include materialscovered up to this point. For review you may look back over your notes and the examplesdone in class. You should also complete all homework assignments up to this point.



1. One member of a gardening team can mow and clean up a lawn in nine hours. Withboth members of the team working, the job can be done in six hours. How long would ittake the second member of the team, working alone, to do the job?

2. Simplify.

2√

32x2y3 − xy√

50y

3. Simplify and rationalize the denominator.

√5 +√

3√5−√

3

4. Simplify.

√44x−

√11x

5. Simplify.

(−27)−2/3

6. Solve for x. Check your answer.

4x− 3

8− 2− x

4=

5

8

7. Solve the inequality 3 (x + 5) > −9. Write the solution in set-builder notation.

8. Simplify.(2√

6 + 3)2

10Up to and including multiplication and division of radical expressions.

Page 113 of 219




9. Simplify.

1 +8

t

1− 64

t2

10. Solve for y.

y

y + 1=

4

y + 7

11. Simplify.(b−5/6

a−1/3

)12

12. Simplify.

3√−8x15y18

13. Simplify.

v − 5 +14

v + 4

4v + 12− 8

v + 4

14. Simplify.

x2 − 81

ax + 9a− bx− 9b

15. Divide.

x2 + 2x− 15

x2 − 7x + 10÷ x2 − 4x + 3

x2 − 3x + 2

16. Multiply.

x2 − 16

2x2 + 7x− 4· 2x2 − x

x2 − 4x

17. Simpify.√48x11y5√

3x5y

Page 114 of 219




18. Subtract and simplify.

x

x2 − 6x + 9− 3

x2 − 9

19. Solve for w.

3

w − 4+ 2 =

5

w − 4

20. Solve for z.

2

z − 2+

1

z + 1=

9

z2 − z − 2

21. The width of a rectangle is 2 inches less than twice its length. Its area is 40 square inches.Find the length and width of the rectangle.

22. One grocery clerk can stock a shelf in 60 minutes, whereas a second clerk requires 30minutes to stock the same shelf. How long would it take to stock the shelf if the twoclerks worked together?

23. Solve for n.

−3

5n + 2 = 7

24. A collection of fifty coins has a value of $4.05. The collection contains only nickels anddimes. Find the number of dimes in the collection.

25. Solve for y.

Ax + By = C

26. Simplify.

x2 − 5x− 14

2x− 14

27. An experienced bricklayer can work twice as fast as an apprentice bricklayer. If the twobricklayers take six hours to do a job when they work together, how long would it takeeach bricklayer to do the job alone?

Page 115 of 219






1. One member of a gardening team can mow and clean up a lawn in nine hours. Withboth members of the team working, the job can be done in six hours. How long would ittake the second member of the team, working alone, to do the job?

Solution: Let x present the numbers of hours it takes the second member of the team,working alone, to do the job.

6 · 1

9+ 6 · 1

x= 1

6x + 54 = 9x

54 = 3x

18 = x

Answer: The second member of the team, working alone, would take 18 hours to dothe job.

2. Simplify.

2√

32x2y3 − xy√

50y

Solution:

2√

32x2y3 − xy√

50y = 2√

16x2y2 · 2y − xy√

25 · 2y= 8xy

√2y − 5xy

√2y

= 3xy√

2y

Answer: 3xy√

2y

3. Simplify and rationalize the denominator.

√5 +√

3√5−√

3

Page 116 of 219




Solution:

√5 +√

3√5−√

3=

√5 +√

3√5−√

3·√

5 +√

3√5 +√

3

=5 +√

15 +√

15 + 3

5− 3

=8 + 2

√15

2

= 4 +√

15

Answer: 4 +√

15

4. Simplify.√

44x−√

11x

Solution:

√44x−

√11x = 2

√11x−

√11x

=√

11x

Answer:√

11x

5. Simplify.

(−27)−2/3

Solution:

(−27)−2/3 =1

(−27)2/3

=1(

3√−27

)2=

1

(−3)2

=1

9

Answer:1

9

Page 117 of 219




6. Solve for x. Check your answer.

4x− 3

8− 2− x

4=

5

8

Solution:

Solving:

4x− 3

8− 2− x

4=

5

8(4x− 3)− 2 (2− x) = 5

4x− 3− 4 + 2x = 5

6x− 7 = 5

6x = 12

x = 2

Checking:

4 · 2− 3

8− 2− 2

4=

5

88− 3

8− 0 =

5

85

8=

5

8

Answer: x = 2. This solution was also checked and shown to be valid.

7. Solve the inequality 3 (x + 5) > −9. Write the solution in set-builder notation.

Solution:

3 (x + 5) > −9

3x + 15 > −9

3x > −24

x > −8

Answer: {x | x > −8}.

8. Simplify.(2√

6 + 3)2

Page 118 of 219




Solution:(2√

6 + 3)2

=(

2√

6 + 3)(

2√

6 + 3)

= 24 + 6√

6 + 6√

6 + 9

= 33 + 12√

6

Answer: 33 + 12√

6

9. Simplify.

1 +8

t

1− 64

t2

Solution:

1 +8

t

1− 64

t2

=1 · t2 +

8

t· t2

1 · t2 − 64

t2· t2

=t2 + 8t

t2 − 64

=t (t + 8)

(t + 8) (t− 8)

=t��(t + 8)

��(t + 8) (t− 8)

=t

t− 8

Answer:t

t− 8

10. Solve for y.

y

y + 1=

4

y + 7

Page 119 of 219




Solution:

y

(y + 1)· (y + 7) (y + 1) =

4

(y + 7)· (y + 7) (y + 1)

y (y + 7) = 4 (y + 1)

y2 + 7y = 4y + 4

y2 + 3y − 4 = 0

(y − 1) (y + 4) = 0

Answer: y = 1 or y = −4

11. Simplify.(b−5/6

a−1/3

)12

Solution:(b−5/6

a−1/3

)12

=

(a1/3

b5/6

)12

=a4

b10

Answer:a4

b10

12. Simplify.

3√−8x15y18

Solution:

3√−8x15y18 = −2x5y6

Answer: −2x5y6

13. Simplify.

v − 5 +14

v + 4

4v + 12− 8

v + 4

Page 120 of 219




Solution:

v − 5 +14

v + 4

4v + 12− 8

v + 4

=(v − 5) · (v + 4) +

14

v + 4· (v + 4)

(4v + 12) · (v + 4)− 8

v + 4· (v + 4)

=v2 + 4v − 5v − 20 + 14

4v2 + 16v + 12v + 48− 8

=v2 − v − 6

4v2 + 28v + 40

=(v + 2) (v − 3)

4 (v + 2) (v + 5)

=��(v + 2) (v − 3)

4��(v + 2) (v + 5)

=v − 3

4v + 20

Answer:v − 3

4v + 20

14. Simplify.

x2 − 81

ax + 9a− bx− 9b

Solution:

x2 − 81

ax + 9a− bx− 9b=

(x− 9) (x + 9)

a (x + 9)− b (x + 9)

=(x− 9) (x + 9)

(x + 9) (a− b)

=(x− 9)��(x + 9)

��(x + 9) (a− b)

=x− 9

a− b

Answer:x− 9

a− b

15. Divide.

x2 + 2x− 15

x2 − 7x + 10÷ x2 − 4x + 3

x2 − 3x + 2

Page 121 of 219




Solution:

x2 + 2x− 15

x2 − 7x + 10÷ x2 − 4x + 3

x2 − 3x + 2=

(x− 3) (x + 5)

(x− 2) (x− 5)· (x− 1) (x− 2)

(x− 1) (x− 3)

=��(x− 3) (x + 5)

��(x− 2) (x− 5)·�

��(x− 1)��(x− 2)

��(x− 1)��(x− 3)

=x + 5

x− 5

Answer:x + 5

x− 5

16. Multiply.

x2 − 16

2x2 + 7x− 4· 2x2 − x

x2 − 4x

Solution:

x2 − 16

2x2 + 7x− 4· 2x2 − x

x2 − 4x=

(x + 4) (x− 4)

(2x− 1) (x + 4)· x (2x− 1)

x (x− 4)

=��(x + 4)��(x− 4)

��(2x− 1)��(x + 4)·�x��(2x− 1)

�x��(x− 4)= 1

Answer: 1

17. Simpify.√48x11y5√

3x5y

Solution:√48x11y5√

3x5y=

√48x11y5

3x5y

=√

16x6y4

= 4x3y2

Answer: 4x3y2

Page 122 of 219




18. Subtract and simplify.

x

x2 − 6x + 9− 3

x2 − 9

Solution:

x

x2 − 6x + 9− 3

x2 − 9=

x

(x− 3) (x− 3)− 3

(x + 3) (x− 3)

=x · (x + 3)− 3 (x− 3)

(x− 3) (x− 3) (x + 3)

=x2 + 3x− 3x + 9

(x− 3) (x− 3) (x + 3)

=x2 + 9

(x− 3) (x− 3) (x + 3)

Answer:x2 + 9

(x− 3) (x− 3) (x + 3)

19. Solve for w.

3

w − 4+ 2 =

5

w − 4

Solution:

3

(w − 4)· (w − 4) + 2 · (w − 4) =

5

(w − 4)· (w − 4)

3 + 2w − 8 = 5

2w − 5 = 5

2w = 10

w = 5

Answer: w = 5

20. Solve for z.

2

z − 2+

1

z + 1=

9

z2 − z − 2

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Solution:

2

z − 2+

1

z + 1=

9

z2 − z − 22

(z − 2)+

1

(z + 1)=

9

(z − 2) (z + 1)

2 · (z + 1) + 1 · (z − 2) = 9

2z + 2z − 2 = 9

3z = 9

z = 3

Answer: z = 3

21. The width of a rectangle is 2 inches less than twice its length. Its area is 40 square inches.Find the length and width of the rectangle.

Solution:

Let x represent the length.

x (2x− 2) = 40

2x2 − 2x = 40

2x2 − 2x− 40 = 0

2(x2 − x− 20

)= 0

2 (x− 5) (x + 4) = 0

So x = 5 or x = −4, but the length cannot be negative, so the length is 5 inches. Toget the width just note that it’s 2 inches less than twice its length, or 8 inches.

Answer: the length is 5 inches and width is 8 inches.


Solution: Let x represent the number of minutes it will take to stock the shelf if the

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two clerks worked together.

1

60· x +

1

30· x = 1

x + 2x = 60

3x = 60

x = 20

Answer: 20 minutes

23. Solve for n.

−3

5n + 2 = 7

Solution:

−3

5n + 2 = 7

−3n + 10 = 35

−3n = 25

n = −25

3

Answer: n = −25

3

24. A collection of fifty coins has a value of $4.05. The collection contains only nickels anddimes. Find the number of dimes in the collection.

Solution: Let x represent the number of dimes.

10x + 5 (50− x) = 405

10x + 250− 5x = 405

5x + 250 = 405

5x = 155

x = 31

Answer: 31

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25. Solve for y.

Ax + By = C

Solution:

Ax + By = C

By = C − Ax

y =C − Ax

B

Answer: y =C − Ax

B

26. Simplify.

x2 − 5x− 14

2x− 14

Solution:

x2 − 5x− 14

2x− 14=

(x− 7) (x + 2)

2 (x− 7)

=��(x− 7) (x + 2)

2��(x− 7)

=x + 2

2

Answer:x + 2

2

27. An experienced bricklayer can work twice as fast as an apprentice bricklayer. If the twobricklayers take six hours to do a job when they work together, how long would it takeeach bricklayer to do the job alone?

Solution: Let x represent the number of hours it takes the experienced bricklayer tocomplete the job alone.

6 · 1

x+ 6 · 1

2x= 1

12 + 6 = 2x

18 = 2x

9 = x

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Answer: An experienced bricklayer would take 9 hours to do the job alone, and theapprentice bricklayer would take 18 hours to do the job alone.

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21.1 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.

21.2 Examples


1. Solve.√x− 5− 5 = 0


Answer: x = 30

2. Solve.√

2x + 88− 12 = 0

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Answer: x = 28

3. Solve.

3√

6x = −3


Answer: x = −9

2

4. Solve.

3√x− 5 + 2 = 0


Answer: x = −3

5. Solve.

4√

4x + 5 = 3


Answer: x = 19

6. Solve.

5√

3− 7x = −2


Answer: x = 5

7. Solve.√x + 1 +

√x− 3 = 2

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Answer: x = 3

8. Solve.

√4x−

√4x− 7 = 1


Answer: x = 4

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22.1 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 that ifi =√−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

22.1.1 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

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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

22.2 Examples


1. Simplify.

√−16


Answer: 4i

2. Simplify.

√−75


Answer: 5i√

3

3. Simplify.

7−√−49

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Answer: 7− 7i

4. Evaluate

−B +√B2 − 4AC

for A = −2, B = 6 and C = −29.


Answer: −6 + 14i

5. Simplify.

(3− 2i) + (4 + 5i)


Answer: 7 + 3i

6. Simplify.

(5− 7i)− (6− 19i)


Answer: −1 + 12i

7. Simplify.

(−4i) (5i)


Answer: 20

8. Simplify.

(1− i) (1 + i)

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Answer: 2

9. Simplify.√−5 ·√−80


Answer: −20

10. Simplify.

(5− 3i) (5 + 3i)


Answer: 34

11. Simplify.

(3− 4i)2


Answer: −7− 24i

12. Simplify.

i (3− 2i)


Answer: 2 + 3i

13. Simplify.

(2− 3i) (5 + 2i)

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Answer: 16− 11i

14. Simplify.

2

i


Answer: −2i

15. Simplify.

2

3i


Answer: −2

3i

16. Simplify.

2 + 5i

−3i


Answer: −5

3+

2

3i

17. Simplify.

3

1 + i

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Answer:3

2− 3

2i

18. Simplify.

6

5 + 2i


Answer:30

29− 12

29i

19. Simplify.

5

3− i


Answer:3

2+

1

2i

20. Simplify.

3− 4i

3 + 4i


Answer: − 7

25− 24

25i

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23.1 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.

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23.2 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


Answer: x2 − 6x− 5 = 0, A = 1, B = −6, C = −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)


Answer: x2 − 3x + 9 = 0, A = 1, B = −3, C = 9

3. Solve by factoring.

3x2 − 9x = 0




x2 + 4 = 5x




2x2 − 12x = 54

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2x2 + 5x = 42


Answer: x = −6 or x =7

2


x + 27 = x2 − 5x




(3x− 2) (2x + 1) = 3x2 − 15x− 10


Answer: x = −4 or x = −2

3

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


Answer: (x− 2) (x + 5) = x2 + 3x− 10 = 0

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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


Answer: (3x− 2) (7x + 5) = 21x2 + x− 10 = 0

11. Solve by using the square root method.

x2 = 9


Answer: x = ±3


16x2 − 9 = 0


Answer: x = ±3

4


(x− 2)2 = 25


Answer: x = 2± 5, it is preferred that you write x = 7 or x = −3 instead.


(x− 2)2 = 25

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(x + 2)2 = 3


Answer: x = −2±√

3

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24.1 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 perfect 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 quadraticequations. Only simple ones though. Here’s an example, solve x2 + 30x + 200 = 0 bycompleting the 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.

24.2 Examples


1. Solve by completing the square.

x2 + 8x− 9 = 0




x2 = 2x + 3




x2 + 2x− 1 = 0

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2


x2 + 4x− 1 = 0



5


x2 = 7− 2x


Answer: x = −1± 2√

2


x2 = 18x− 17




x2 − 14x + 113 = 0


Answer: x = 7± 8i

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25.1 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.

25.2 Examples


1. Solve using the quadratic formula.

x2 − 6x− 5 = 0

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Answer: x = 3±√

14


x2 − x− 3 = 0


Answer: x =1±√

13

2


3x2 − 7x + 3 = 0


Answer: x =7±√

13

6


5x2 = 1− 2x


Answer: x =−1±

√6

5


12x− 9 = x2

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Answer: x = 6± 3√

3


x2 + 18x + 82 = 0


Answer: x = −9± i


x2 + 8x + 25 = 0


Answer: x = −4± 3i

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26.1 Related to Quadratic Equations

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

26.2 Examples


1. Solve.√x− 5 + x = 7


Answer: x = 6

2. Solve.√2y − 1 = y − 8


Answer: y = 13

3. Solve.

2√x + 1 =

√5x + 4


Answer: x = 1 x = 9

4. Solve.

3

x= 1 +

2

2x− 1

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Answer: x = 1 or x =3

2

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27.1 Points, Midpoints and Distance

We’ll review plotting points (x, y) in the Cartesian11 plane (abscissa or x-coordinate, ordinateor y-coordinate). Visualizing points in a plane should eventually become natural, and conceptssuch 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.

27.2 Examples


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


Distance:√

34 ≈ 5.83

Midpoint: (−5/2, −3/2)

Graph: Be sure you can visualize the above two concepts. Use the graph paper provided.

11Cartesian 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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0

Figure 18: Graph Paper

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)


Distance:√

8 ≈ 2.83

Midpoint: (5, 4)


0


3. Graph the ordered-pair solutions of

y = −x2 + 5

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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.


Distance:√

80 ≈ 8.94

Midpoint: (−1, 0)


0


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)


Distance:√

154.25 ≈ 12.42

Midpoint: (1.3, −1.65)

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28.1 Introductory Functions and Relations

Definition of a Function: A function12 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 = 3 willresult in a division by zero—so the domain of this function is all real numbers except three.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.

28.2 Examples


1. State whether the relation is a function.

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


Answer: Yes.

12Often denoted f (x) and read f of x.

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{(4, −3) , (0, 0) , (6, 0) , (7, −3)}


Answer: Yes.


{(4, −3) , (2, 0) , (6, −4) , (7, −2) , (2, 7)}


Answer: No, because an element in the domain (2) maps to two different elements (0and 7) in the range.

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


Answer: f (−3) = −15

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


Answer: g (−2) = 12

6. Given

k (y) =y − 2

y + 2,

evaluate k (−1).


Answer: k (−1) = −3

7. Find the domain and range of the function.

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

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Domain: {8, 10, 6, 5, }Range: {−5, 7, −1, −3}

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

g (x) = − 9

x− 2


Answer: x = 2


f (x) =x− 2

9


Answer: None.


h (x) = 3x4 − 2x3 + 4x2 − 9x + 17


Answer: None.

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}


Answer: {−65, −44, −23, −2, 19}


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

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Answer: {6, 1, 0, 3, 10}


k (x) =3− x

x2 + 1; domain = {−2, −1, 0, 1, 2}


Answer: {1, 2, 3, 1, 1/5}

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29.1 Graphing Lines

Now we will plot the solutions we obtained from simple two-variable linear relationships. Forexample, 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 a 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 21: {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 22: 2x + 3y = 8

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29.2 Special Points

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

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, thatoccasionally these points will be difficult to plot, and in those cases I suggest you find easypoints to plot—don’t worry, the graph will include these difficult points anyway. For example,find the x and y-intercept for:13

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}

29.3 Examples


1. Graph.

y = −3x + 2

13The y-intercept is (0, −1/5), and the x-intercept is (−1/3, 0).

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Solution: We’ll discuss this in class. Use the graph paper to graph.

0


2. Graph.

f (x) =7

6x


0


3. Graph.

f (x) =3

5x− 1

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0


4. Graph.

f (x) = −3

2x− 2


0


5. Graph.

2x− y = 5

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0


6. Graph.

5x + 2y = 7


0


7. Graph.

y = x

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0


8. Graph.

y = 2


0


9. Graph.

x = −1

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0


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

2x + y = 4


0


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

3x− 2y = 6

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0


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30.1 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.

30.2 Examples


1. Find the slope of the line containing the points

P1 (4, −5) and P2 (−6, 3) .

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Answer: m = −4/5

2. Find the slope of the line containing the points

P1 (4, 8) and P2 (4, −7) .


Answer: Undefined.

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

3y = 2x− 6.


Answer: m = 2/3, b = −2, x-intercept is (3, 0), y-intercept is (0, −2).

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

2x + 3y = 9.


Answer: m = −2/3, b = 3, x-intercept is (9/2, 0), y-intercept is (0, 3).

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

(k, −2) , (0, −3) , (9, −6)


Answer: k = −3

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

y =3

4x− 2

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0


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

2y − 3x = 7


0


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

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0


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


0


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31.1 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. Thiswill 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.

31.2 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.)14

(0, 9) , m = 3


Point-slope: y − 9 = 3 (x− 0)

Slope-intercept: y = 3x + 9

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

14That is, you should write your final answer as y = mx + b.

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Point-slope: y − 5 = 45

(x− 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


Point-slope: y − 0 = −45

(x− 3)

Slope-intercept: y = −45x + 12

5

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)


Point-slope: y − 6 = 3 (x− 0) or y − 9 = 3 (x− 1)



(−3, 4) , (1, 9)


Point-slope: y − 4 = 54

(x + 3) or y − 9 = 54

(x− 1)


4


(−5, 3) , (1, 3)

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Point-slope: y − 3 = 0 · (x + 5) or y − 3 = 0 · (x− 1)

Slope-intercept: y = 0 · x + 3 or y = 3


(−7, −4) , (−7, 4)


Vertical line: x = −7

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32.1 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.

32.2 Examples


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


Answer: m = −2

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


Answer: m = 2

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

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Answer: m = −7

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


Answer: m = 0

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


Answer: No.

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)?


Answer: No.

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


Answer: y − 5 = −2 (x− 3) or y = −2x + 11

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


Answer: y − 3 = −43

(x− 4) or y = −43x + 25

3

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


Answer: y − 2 = 12

(x + 1) or y = 12x + 5

2

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

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Answer: x = −2

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


Answer: y = 7

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33 Review for Class Test #2

Chapters 8, 3, and parts of chapter 7 (radical equations and complex numbers). For reviewyou may look back over your notes and the examples done in class. You should also completeall homework assignments up to this point.



1. Find the slope and y-intercept, then use this information to graph: y = −3

2x− 2.

2. Solve.

3

x= 1 +

2

2x− 1

3. Simplify.

3− 4i

3 + 4i

4. Solve.

√x− 5− 5 = 0


3x2 + 3 = 7x

6. Determine whether the relation is a function and explain your answer.

{(4, 1) , (8, 2) , (10, 5) , (4, −1) , (7, −3)}

7. Find the equation of the line that contains the point (0, 5) and has slope −4. Writefinal equation in slope intercept form.15

8. What value(s) are excluded from the domain of the function f (x) =x + 2

x− 5? Explain

your answer.

9. Find the slope of the line containing the points (9, −4) and (5, 8).


f (x) = 2x− 3; D = {0, 1, 2, 3}15y = mx + b

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11. Find the equation of the line containing the point (3, −3) and perpendicular to the line

y =1

3x− 2.

12. Given the function f (x) = 2x3 − 5x− 2, find f (−2).

13. Find the domain of the relation. Then, determine whether the relation is a function andexplain your answer.

{(−2, 2) , (−1, 5) , (3, 7) , (−2, −2)}

14. Given

f (x) =x2 − 3x

x3 − x + 6,

find f (−1).

15. Find the x-intercept and y-intercept, then use this information to graph the equation:3x + 4y = 12.

16. Use the distance formula16 to find the distance between the points (−4, 2) and (4, −5).Round your answer to two decimal places.

17. Are the lines y − 5x = 3 and 5x − y = 4 perpendicular, parallel, or neither? Explainyour answer.

18. Graph: y = −3

2x− 2.

19. Solve for y.

3√

5y = −4

20. Find the equation of the line that contains the point (−3, 7) and has slope −3.

21. Write a quadratic equation that has integer coefficients and has solutions 11 and 3.

22. Find the equation of the line that contains the points (−5, 13) and (2, −1).

23. Solve for z

z2 + 64 = 0

16In class we developed a formula for the distance between two points in the plane. We labeled the twopoints P1 or (x1, y1) and P2 or (x2, y2).


d =

√(x2 − x1)

2+ (y2 − y1)

2

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24. Are the lines 4x− y = 2 and y +1

4x = −7 parallel, perpendicular, or neither? Explain

your answer.

25. Find the domain of the following function.

h (x) =2x− 1

x2 + 4x− 12

26. Simplify.

√−5 ·√−80

27. Solve for x

√6x + 13 = 1 +

√6x

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1. Find the slope and y-intercept, then use this information to graph: y = −3

2x− 2.

Solution:

Answer: m = −3

2and the y-intercept is (0, −2). Graph below.

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

-3

-2

-1

1

2

3

Figure 38: y = −3

2x− 2

2. Solve.

3

x= 1 +

2

2x− 1

Solution:

3

x= 1 +

2

2x− 1(3

x

)· x (2x− 1) =

(1 +

2

2x− 1

)· x (2x− 1)

6x− 3 = 2x2 − x + 2x

0 = 2x2 − 5x + 3

0 = (2x− 3) (x− 1)

Answer: x = 1 or x =3

2. You should check both answers!

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3. Simplify.

3− 4i

3 + 4i

Solution:

3− 4i

3 + 4i=

3− 4i

3 + 4i· 3− 4i

3− 4i

=9− 24i + 16i2

9− 16i2

=9− 24i− 16

9 + 16

=−7− 24i

25

= − 7

25− 24

25i

Answer: − 7

25− 24

25i

4. Solve.√x− 5− 5 = 0

Solution:

√x− 5− 5 = 0√

x− 5 = 5

x− 5 = 25

x = 30

You should also check! Answer: x = 30


3x2 + 3 = 7x

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Solution: First rewrite 3x2 + 3 = 7x as 3x2− 7x+ 3 = 0, and identify A = 3, B = −7,and C = 3. Now using the quadratic formula we get:

x =− (−7)±

√(−7)2 − 4 (3) (3)

2 · 3=

7±√

13

6

Answer: x =7±√

13

6

6. Determine whether the relation is a function and explain your answer.

{(4, 1) , (8, 2) , (10, 5) , (4, −1) , (7, −3)}

Solution:

Answer: This relationship is not functional because 4 maps to 1 and −1. For arelationship to be functional an element from the domain cannot be matched with twodifferent elements of the range.

7. Find the equation of the line that contains the point (0, 5) and has slope −4. Writefinal equation in slope intercept form.17

Solution:

y − 5 = −4 (x− 0)

y = −4x + 5

Answer: y = −4x + 5

8. What value(s) are excluded from the domain of the function f (x) =x + 2

x− 5? Explain

your answer.

Solution: Division by zero is not allow, and x− 5 = 0 when x = 5.

Answer: 5. The reason that 5 is excluded is that it would result in a division by zero.

9. Find the slope of the line containing the points (9, −4) and (5, 8).

17y = mx + b

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Solution:

m =8 + 4

5− 9= −3

Answer: m = −3


f (x) = 2x− 3; D = {0, 1, 2, 3}

Solution:

f (0) = 2 · 0− 3 = −3

f (1) = 2 · 1− 3 = −1

f (2) = 2 · 2− 3 = 1

f (3) = 2 · 3− 3 = 3

R = {−3, −1, 1, 3}

Answer: R = {−3, −1, 1, 3}

11. Find the equation of the line containing the point (3, −3) and perpendicular to the line

y =1

3x− 2.

Solution: The slope of the given line is1

3, so the slope of the line perpendicular to

this line is −3.

y + 3 = −3 (x− 3)

y + 3 = −3x + 9

y = −3x + 6


12. Given the function f (x) = 2x3 − 5x− 2, find f (−2).

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Solution:

f (−2) = 2 (−2)3 − 5 (−2)− 2

= −16 + 10− 2

= −8

Answer: f (−2) = −8

13. Find the domain of the relation. Then, determine whether the relation is a function andexplain your answer.

{(−2, 2) , (−1, 5) , (3, 7) , (−2, −2)}

Solution: Answer: D = {−2, −1, 3}. This relationship is not functional because −2maps to 2 and −2. For a relationship to be functional an element from the domaincannot be matched with two different elements of the range.

14. Given

f (x) =x2 − 3x

x3 − x + 6,

find f (−1).

Solution:

f (−1) =(−1)2 − 3 (−1)

(−1)3 − (−1) + 6

=1 + 3

−1 + 1 + 6

=2

3

Answer:2

3

15. Find the x-intercept and y-intercept, then use this information to graph the equation:3x + 4y = 12.

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Solution:

Answer: The x-intercept is (4, 0) and y-intercept is (0, 3). Graph below.

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

-3

-2

-1

1

2

3

Figure 39: 3x + 4y = 12


Solution:

d =

√(4 + 4)2 + (−5− 2)2

d =√

64 + 49

d =√

113

d ≈ 10.63

Answer: 10.63

17. Are the lines y − 5x = 3 and 5x − y = 4 perpendicular, parallel, or neither? Explainyour answer.



d =

√(x2 − x1)

2+ (y2 − y1)

2

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Solution:

y − 5x = 3

y = 5x + 3

m1 = 5

5x− y = 4

5x− 4 = y

m2 = 5

Answer: The lines are parallel because m1 = m2.

18. Graph: y = −3

2x− 2.

Solution:

Answer: Two point method used: (−2, 1) (0, −2). Graph below.

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

-3

-2

-1

1

2

3

Figure 40: y = −3

2x− 2

19. Solve for y.

3√

5y = −4

Solution:

3√

5y = −4

5y = −64

y = −64

5

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Answer: y = −64

5.

20. Find the equation of the line that contains the point (−3, 7) and has slope −3.

Solution:

y − 7 = −3 (x + 3)

y − 7 = −3x− 9

y = −3x− 2

Answer: y = −3x− 2

21. Write a quadratic equation that has integer coefficients and has solutions 11 and 3.

Solution:

(x− 11) (x− 3) = 0

x2 − 14x + 33 = = 0

Answer: x2 − 14x + 33 = 0

22. Find the equation of the line that contains the points (−5, 13) and (2, −1).

Solution:

Using the slope formula we get m = −2. Using (2, −1) we get:

y + 1 = −2 (x− 2)

y + 1 = −2x + 4

y = −2x + 3


23. Solve for z

z2 + 64 = 0

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Solution: Answer: z = ±8i

24. Are the lines 4x− y = 2 and y +1

4x = −7 parallel, perpendicular, or neither? Explain

your answer.

Solution:

4x− y = 2

4x− 2 = y

m1 = 4

y +1

4x = −7

y = −1

4x− 7

m2 = −1

4

Answer: The lines are perpendicular because m1 ·m2 = −1.

25. Find the domain of the following function.

h (x) =2x− 1

x2 + 4x− 12

Solution: Division by zero us undefined so we need to solve x2 + 4x − 12 = 0 andexclude these solutions from the domain.

x2 + 4x− 12 = 0

(x− 2) (x + 6) = 0

Answer: R, x 6= −6, 2

26. Simplify.

√−5 ·√−80

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Solution:

√−5 ·√−80 =

√−1 ·√

5 ·√−1 ·√

80

= i · i ·√

400

= i2 · 20

= −20

Answer: −20

27. Solve for x

√6x + 13 = 1 +

√6x

Solution:

√6x + 13 = 1 +

√6x(√

6x + 13)2

=(

1 +√

6x)2

6x + 13 = 1 + 2√

6x + 6x

12 = 2√

6x

6 =√

6x

36 = 6x

6 = 6

Answer: x = 6

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34.1 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.19{

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 41: 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 what

19The upper case letters represent constants.

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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

34.2 Examples


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


Answer: No.

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


Answer: (1, −4)


x + 5 = 1y − 1 = 2

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Answer: (−4, 3)


2x + 3y = 6y = −2

3x + 2


Answer: Same line.(x, −2

3x + 2

)5. Solve by graphing. (If equations are independent, give answer as a point; If the equations

are dependent, enter x for the value of x and y in terms of x; if the system is inconsistent,just state inconsistent.){

x − 2y = 12y − x = 1


Answer: No solution. These lines never intersect.


2x − 4y = 0y = x + 1


Answer: (−2, −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

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Answer: (−2, −1)


8x − 5y = 35y = 4x − 19


Answer: (5, 1)


6x + 8y = −12y = 3x − 4


Answer: (2/3, −2)

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35.1 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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35.2 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


Answer: (5, 3)


x + y = 34x − y = 17


Answer: (4, −1)


2x + 3y = 45x + 2y = −1


Answer: (−1, 2)


x − 2y = −2−3x + 6y = 6

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Answer: (x, 1 + x/2)


6x − 4y = 65x + 2y = 5


Answer: (1, 0)


13x − 2

5y = −1

23x + 3

5y = 5


Answer: (3, 5)


4x − 2y = 7x − 22x + 8y = 5y + 8


Answer: (−2, 4)


3 − 2x = 4 − 3y3x + 2y = 5

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Answer: (1, 1)

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36.1 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.

36.2 Examples


1. A pharmacist has two vitamin-supplement powders. The first powder is 25% vitamin B1and 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?


Answer: 560 milligrams of the first powder, and 100 milligrams of the second powder.

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.


Answer: The rate of the boat in calm water is 10 mph and the rate of the current is 5mph.

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.


Answer: 17 quarters.

4. A chemist has two alloys, one of which is 10% gold and 15% lead and the other which is30% 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?

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Answer: 430 grams of the 10% gold and 15% lead alloy; 90 grams of the 30% gold and50% lead alloy.

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.


Answer: The rate of the rowing team in calm water is 8 mph and the rate of the currentis 4 mph.

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.


Answer: The cost per unit of gas is $0.10.

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.


Answer: The rate of the plane in calm air is 160 mph and the rate of the wind is 40mph.

8. A contractor buys 15 yards of nylon carpet and 19 yards 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.


Answer: The cost per yard of the wool carpet is $55.

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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Answer: The cost per pound of the cinnamon tea is $1.50 and the cost per pound forspice tea is $2.50.

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37.1 Graphing Parabolas

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

-4

-3

-2

-1

1

2

Figure 42: 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 is the dashed-line in our graph above. It should be noted thatthis 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.

37.2 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


Vertex: (2, −5); AOS: x = 2

Use the graph paper provided.

0



y = x2 − 2x− 3

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Vertex: (1, −4); AOS: x = 1


0



y = −x2 + 2x− 3


Vertex: (1, −2); AOS: x = 1


0


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y = 2x2 − 12x + 17


Vertex: (3, −1); AOS: x = 3


0


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

f (x) = 2x2 − 16x + 34


Domain: (−∞, ∞); Range: [2, ∞)

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

f (x) = 4x2 − 9


Domain: (−∞, ∞); Range: [−9, ∞)

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

y = x2 − 5x + 6


x intercepts: (2, 0) and (3, 0); y-intercept: (0, 6)

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

f (x) = 12 + 7x + x2


Zeros: −3 and −4

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38.1 Graphing Circles

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

-4

-3

-2

-1

1

Figure 47: 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 a setof points that is equidistant 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 an arbitrary point on the circle we get20√

(x− h)2 + (y − k)2 = r.

Squaring both sides we get the standard form 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 circle you should alsobe 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. For

20We’re using the distance formula.

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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 circle you should alsobe able to find four easy points along the circle’s circumference:

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

38.2 Examples


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

(x− 7)2 + (y + 4)2 = 9


Center: (7, −4); Radius: r = 3


0


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

(x + 5)2 + (y − 3)2 = 4

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Center: (−5, 3); Radius: r = 2


0


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


Answer: (x + 2)2 + (y − 1)2 = 9 or x2 + y2 + 4x− 2y − 4 = 0

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


Answer: (x− 1)2 + (y + 1)2 = 1 or x2 + y2 − 2x + 2y + 1 = 0

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


Answer: (x− 1)2 + (y + 3)2 = 17 or x2 + y2 − 2x + 6y − 7 = 0

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

x2 + y2 − 2x + 12y + 12 = 0

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Answer: (x− 1)2 + (y + 6)2 = 52

0



x2 + y2 − 10x + 6y + 33 = 0


Answer: (x− 5)2 + (y + 3)2 = 12

0



x2 + y2 + 6x + 2y − 6 = 0

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Answer: (x + 3)2 + (y + 1)2 = 42

0


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39 Review for Comprehensive Final Exam

Comprehensive Final Exam on all course material covered. This exam will include materialscovered up to this point. For review you may look back over your notes and the examplesdone in class. You should also complete all homework assignments up to this point.




2. Find the equation of the line that contains the points (4, −4) and (2, 0).

3. Simplify.

2√

32x2y3 − xy√

98y


5. Simplify.

5

x2 + x− 12− 4

x2 + 5x + 4

6. Find the vertex and axis-of-symmetry (AOS), and then graph the parabola given by:

y = x2 − 2x− 4.

7. Solve{3x + 7y = 164x − 3y = 9


2b6 − 16b3



d =

√(x2 − x1)

2+ (y2 − y1)

2

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9. An office has 20¢ stamps and 28¢ stamps. All together the office has 40 stamps for atotal value of $8.80. How many of each type of stamp does the office have?

10. Sketch the graph of the circle given by the equation:

(x + 3)2 + (y − 3)2 = 4.

11. Solve for x.

5 [6− 4 (x− 4)] = 5 (4− 6x)

12. Graph.

2x + y = 4


x2 − 3x

x2 − 9· x

2 + 6x + 9

2x2 + 9x + 9

14. The length of a rectangle is 2 inches less than twice its width. Its area is 40 square inches.Find the length and width of the rectangle.

15. Simplify.(6a3b−5

a−2b−3

)2

16. Solve for c.

7

c− 5+ 3 =

13

c− 5

17. Solve for z.

3√

4z + 5 + 5 = 3

18. Solve for x.

x2 − 2x− 7 = 0

19. Simplify.

(3− 2i) (2 + 3i)


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Solution: Let x represent the time in minutes that it takes to stock the shelf if thetwo clerks worked together

x · 1

24+ x · 1

40= 1

5x + 3x = 120

8x = 120

x = 15

Answer: 15 minutes

2. Find the equation of the line that contains the points (4, −4) and (2, 0).

Solution: The slope is −2.

y − 0 = −2 (x− 2)

y = −2x + 4


3. Simplify.

2√

32x2y3 − xy√

98y

Solution:

2√

32x2y3 − xy√

98y = 8xy√

2y − 7xy√

2y

= xy√

2y

Answer: xy√

2y

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Solution:

d =

√(4 + 4)2 + (−5− 2)2

d =√

64 + 49

d =√

113

d ≈ 10.63

Answer: 10.63

5. Simplify.

5

x2 + x− 12− 4

x2 + 5x + 4

Solution:

5

x2 + x− 12− 4

x2 + 5x + 4=

5

(x + 4) (x− 3)− 4

(x + 4) (x + 1)

=5 · (x + 1)− 4 · (x− 3)

(x + 4) (x− 3) (x + 1)

=x + 17

(x + 4) (x− 3) (x + 1)

Answer:x + 17

(x + 4) (x− 3) (x + 1)

6. Find the vertex and axis-of-symmetry (AOS), and then graph the parabola given by:

y = x2 − 2x− 4.



d =

√(x2 − x1)

2+ (y2 − y1)

2

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Solution:

Answer: The vertex is (1, −5) and the AOS is x = 1. Graph below.

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

-5

-4

-3

-2

-1

1

Figure 53: y = x2 − 2x− 4

7. Solve{3x + 7y = 164x − 3y = 9

Solution: Using elimination (addition) by multiplying the top equation by 4 and thebottom by −3.{

12x + 28y = 64−12x + 9y = −27

Adding these equations together gives 37y = 37 or y = 1. Using y = 1 in the firstequation gives 3x + 7 = 16 or x = 3.

Answer: x = 3, y = 1


2b6 − 16b3

Solution:

2b6 − 16b3 = 2b3(b3 − 8

)= 2b3 (b− 2)

(b2 + 2b + 4

)Answer: 2b3 (b− 2)

(b2 + 2b + 4

)

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9. An office has 20¢ stamps and 28¢ stamps. All together the office has 40 stamps for atotal value of $8.80. How many of each type of stamp does the office have?

Solution: Let x represent the number of 20¢ stamps.

20x + 28 (40− x) = 880

20x + 1120− 28x = 880

−8x + 1120 = 880

−8x = −240

x = 30

Answer: Thirty 20¢ stamps and ten 28¢ stamps.

10. Sketch the graph of the circle given by the equation:

(x + 3)2 + (y − 3)2 = 4.

Solution: The center is (−3, 3) and the radius is 2. Graph below.

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

1

2

3

4

5

6

Figure 54: (x + 3)2 + (y − 3)2 = 4

11. Solve for x.

5 [6− 4 (x− 4)] = 5 (4− 6x)

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Solution:

5 [6− 4 (x− 4)] = 5 (4− 6x)

5 [6− 4x + 16] = 20− 30x

5 [22− 4x] = 20− 30x

110− 20x = 20− 30x

10x = −90

x = −9

Answer: x = −9

12. Graph.

2x + y = 4

Solution: Using two points (0, 4) and(2, 0) to graph.

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

-1

1

2

3

4

5

Figure 55: 2x + y = 4


x2 − 3x

x2 − 9· x

2 + 6x + 9

2x2 + 9x + 9

Solution:

x2 − 3x

x2 − 9· x

2 + 6x + 9

2x2 + 9x + 9=

x (x− 3)

(x + 3) (x− 3)· (x + 3) (x + 3)

(2x + 3) (x + 3)

=x��(x− 3)

��(x + 3)��(x− 3)· ��(x + 3)��(x + 3)

(2x + 3)��(x + 3)

=x

2x + 3

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Answer:x

2x + 3

14. The length of a rectangle is 2 inches less than twice its width. Its area is 40 square inches.Find the length and width of the rectangle.

Solution: Let x represent the width of the rectangle.

x (2x− 2) = 40

2x2 − 2x− 40 = 0

2 (x− 5) (x + 4)

The width must be 5 inches and the length 8 inches. Answer: The width must be 5inches and the length 8 inches.

15. Simplify.(6a3b−5

a−2b−3

)2

Solution:(6a3b−5

a−2b−3

)2

=

(6a5

b2

)2

=36a10

b4

Answer:36a10

b4

16. Solve for c.

7

c− 5+ 3 =

13

c− 5

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Solution:(7

c− 5+ 3

)· (c− 5) =

(13

c− 5

)· (c− 5)

7 + 3 · (c− 5) = 13

7 + 3c− 15 = 13

3c− 8 = 13

3c = 21

c = 7

Answer: c = 7

17. Solve for z.

3√

4z + 5 + 5 = 3

Solution:

3√

4z + 5 + 5 = 3(3√

4z + 5)3

= (−2)3

4z + 5 = −8

4z = −13

z = −13

4

Answer: −13

4

18. Solve for x.

x2 − 2x− 7 = 0

Solution: This quadratic is not factorable using rational numbers so the quadraticformula is being used. A = 1, B = −2 and C = −7

x =2±√

4 + 28

2

=2±√

32

2

=2± 4

√2

2

= 1± 2√

2

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Answer: x = 1± 2√

2

19. Simplify.

(3− 2i) (2 + 3i)

Solution:

(3− 2i) (2 + 3i) = 6 + 9i− 4i− 6i2

= 12 + 5i

Answer: 12 + 5i


Solution: Let x represent the number of liters of the 14% acid solution used.

0.14x + 0.24 (10− x) = 0.19 (10)

0.14x + 2.4− 0.24x = 1.9

2.4− 0.1x = 1.9

−0.1x = −0.5

x = 5

Answer: 5 liters of the 14% acid solution.

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END OF DOCUMENT!

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College Algebra Essex County College LMS Project …faculty.essex.edu/~bannon/m100/mth.100.fall.2014.pdfCollege Algebra MTH-100 Essex County College Division of Mathematics LMS Project