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# DrDelMath

## College Algebraby MillerSUMMARY

### Preliminary/Review:

#### Section 0.1: Tips for Success in Mathematics

Instruction is much more than presentation of information. Instruction may include events that are generated by a page of print, a picture, a television program, a combination of physical objects, potentially many other stimuli, as well as activities of a teacher. Instruction is a deliberately arranged set of external events designed to support internal learning processes.

Teaching refers to the activities of the teacher. Therefore teaching is only one part (I think it is an important part) of instruction.

The purpose of early college level algebra courses is to introduce the student to the use of abstraction, generalization, deductive reasoning and critical thinking while exploring the patterns and relationships of a variety of algebraic entities including, but not limited to, equations, inequalities, algebraic fractions, polynomials, and functions. [What Are We Studying]

A parallel concept that is important to the study of mathematics is the concept of binary relation. The most general definition of binary relation does not help understand algebra at this level, so the following definition is restricted to a particular kind of binary relation.

As you work your way through this very long chapter keep in mind that future chapters will not contain as much material. I have presented a lot of review material (supplemental to what is in the textbook) to insure that we are all starting out on a level playing field. It is very important that you work hard to master every detail of each of these concepts (review or new) so the future material can be learned with greater ease.

In addition to the suggestions in the textbook, you should also read the following essay on this website.
Studying Mathematics

A very important and an extremely effective Study Skill is provided in an Intermediate Algebra text (formerly used at STLCC) and is repeated here.
"Many of the terms used in this text may be new to you. It will be helpful to make a list of new mathematical terms and symbols as you encounter them and to review them frequently."

I suggest that a second list of important properties also be constructed. Whether you use flash cards or a notebook list, it is imperative that you regularly review these definitions and important properties. You must know them precisely -- your own words will not be sufficient.

To realize the importance of definitions, read the essay: Mathematics Without Definitions.

#### Section 0.3: Important Background Reading

Be sure that you have studied, have memorized, and understand all content in the Special Topic essay titled Introduction to Sets
Solutions are provided for those exercises.

Be sure that you have studied, have memorized, and understand all content in the Special Topic essay titled Sets of Numbers.

Also study this brief essay about graphs of certain sets of numbers.
Here is a Review of adding and subtracting on the number line.
Here are some Exercises for your practice.
Solutions are provided.

Be sure that you have studied, have memorized important definitions and properties, and understand all content in Chapter R of the Miller College Algebra text.

You are now ready to begin studying Chapter 1 which should be mostly enhanced review.

### Chapter 1:   Equations and Inequalities

#### Section 1.1: Linear Equations in One Variable

Equations: Identify and Classify

Equations: Solutions, Graphs, and Solving

THE FIRST TWO PROPERTIES OF EQUATIONS

Equations: Linear Equations in One Variable

At this point you should study Section 7 of this chapter. Stop when you get to the material about absolute value inequalities.
You should study Pages 1 - 3 of this Essay.

Here are some Exercises for your practice.
Solutions are forthcoming.

RATIONAL EQUATIONS in One Variable

Equations: Rational Equations in One Variable

The only reason a rational expression does not make sense is if the denominator is zero.
So the domain of a rational expression is all real numbers except those which cause a 0 in the denominator.

Recall that real numbers are polynomials, they are called constant polynomials.

In this section when we speak of solving rational equations, the reference is to rational equations which are not polynomial equations. We are concerned here with equations that have variables in the denominators. To solve that kind of equation we must be aware of the following VERY important facts.

Examples of working with rational expressions and rational equationsHERE
Examples of Solving Linear Equations and Solving Rational EquationsHERE

### Interleaved Practice.

Textbook p.109:
Questions 1 - 12. (Try hard to remember what words are left out of these important statements. If, after some effort, you cannot fill in the blank, look it up and study it. In all cases check your answers by finding the statement in the text.
Questions 13 and 14: Classify each equation as linear or nonlinear. Don't bother solving these -- there will be time for that later.
Questions 15,18, 27: What kind of equations are these? Before you begin to solve these equations, write down a procedure that you intend to use to solve the equation.Solve the equation and state the reason for each step. Use the roster method in a complete sentence to state the solution set. How does the process you used compare with the procedure you wrote down before starting the solving process?
Questions 15 - 34 Each equation is a linear equation in one variable. Rewrite the equations in Questions 15, 17, 19, 27, and 30 in standard form as presented in the definition of a linear equation in one variable. Be sure to provide reasons for each step in the process.
Questions 43 - 48: Follow the directions in the text. Be sure to provide reasons for each step in the process.Use the roster method in a complete sentence to state the solution set.
Questions 70, 76, 79, 81, 88: Follow the directions in the book.Be sure to provide reasons for each step in the process.
Question 35 as directed.
Answer the questions posed in 109 - 114.
Answer questions posed in 115 and 117.
Questions 53 - 68: What kind of equations are these? How does the process of solving these equations differ from solving linear equations in one variable?
Questions 53, 57, 61, 58: Solve these equations . Use the roster method in a complete sentence to state the solution set. Be sure to provide reasons for each step in the process.
Not from p.109
For Questions 1 - 6 use the sets A, B, and C, as defined here.
A = {2, 3, 5, 7}, B = {2, 4, 5, 6}, C = { 12, 28, 35}.
Question 1. The set {3, 7} is a subset of A. Using the subset symbol we can write this same fact much more efficiently as $\left\{3,7\right\}\subset A$.
Use the subset symbol to make a similar statement for each of those subsets of A which have two elements.
Question 2. 12 is an element of C. Using the symbol for "is an element of" we can write this more efficiently by writing $12\in C$.
Use the symbol for "is an element of" to make a similar statement about each element of C.
Question 3. Determine the set $A\cap B$.
How is $A\cap B$ related to B?
How is $A\cap B$ related to A?
Question 4. Determine the set $B\cap C$.
How is $B\cap C$ related to B?
How is $B\cap C$ related to C?
Question 5.In the same manner consider the one other obvious intersection.
Question 6. Determine $A\cup B$.
How is $A\cup B$ related to A?
How is $A\cup B$ related to B?

For Questions 7 - 9 use the set $A=\left\{4,0,\sqrt{13},-\frac{5}{8},\sqrt{49},-14,83\right\}$.
In Questions 7 - 9, use the roster method to specify the set.
Question 7. $A\cap Q=$
Question 8. $A\cap Z=$
Question 9. $A\cap R=$

For Questions 10 - 12 insert a symbol in the blank to make the statement true.
Question 10.
Question 11.
Question 12.

Question 13. List all the subsets of {2, 7, b, 9, k} which have three elements. Use the roster method to write the sets.

#### Section 1.2: Applications and Modeling

Begin by reading this introduction to modeling

Equations: Important Formulas

As we progress through Intermediate Algebra and College Algebra we will review, relearn, or learn for the first time quite a number of formulas. Most of the formulas encountered in Intermediate and College Algebra are formulas that every adult should know. We begin with the following list of formulas for areas of some familiar geometric shapes. Two very useful geometric terms are polygon and quadrilateral. For interactive information about polygons.

Learn to set up models using known formulas: HERE
Learn to set up models using known geometry: HERE

Click HERE for some interesting animated derivations of area formulas. (Tom Richmond - Western Ky. Univ.)

About Variables, Algebraic Expressions, and Basic Formulas

Equations: Percent Formulas

Each percent problem is solved by referring to a basic formula that relates percentage, percent, and base of the percentage.

Use this verbal statement and the basic percent formula to help translate percent problems into an equation. Finally remember the meaning of percent is "per 100" and convert all percents to decimals.

Equations: Temperature Conversion

It is not required to memorize the following three formulas, but you should be able to work with them.

### Minimal List of Exercises Page 126.

If you understand the previous material you should be able to answer the following questions.
Each of these questions should be answered. Most of these questions are included in the MyMathLab homework requirement. If a particular concept is difficult for you, you should study the related text material and then try to answer some additional questions from the list provided by the department. In each case you are expected to make an honest adult evaluation of your understanding of the concept. Your ability to answer these questions is one tool to help you make that evaluation.
Section 1.3 Exercise Set: 33, 35, 37, 48, 57, 68, 74.

#### Section 1.3: Complex Numbers (page 125).

A more detailed presentation of Complex Numbers is found HERE

Complex Numbers

### Minimal List of Exercises Page 135.

If you understand the previous material you should be able to answer the following questions.
Each of these questions should be answered. Most of these questions are included in the MyMathLab homework requirement. If a particular concept is difficult for you, you should study the related text material and then try to answer some additional questions from the list provided by the department. In each case you are expected to make an honest adult evaluation of your understanding of the concept. Your ability to answer these questions is one tool to help you make that evaluation.
Section 1.4 Exercise Set: 1, 5, 9, 13, 19, 21, 25, 27, 29, 41.

#### Section 1.5A: Quadratic Equations in One Variable(page 136)

Equations: Quadratic Equations in One Variable

Observe that a quadratic equation is an equation which can be written as a second degree polynomial equal to 0.

Quadratic equations may be solved by factoring in conjunction with the Zero Factor Property or by using the Quadratic formula.

Here are some Exercises for your practice.

Quadratic Equations: Two Less Used Methods for Solving

The Square Root Property

Solving a quadratic equation using the Square Root Property depends on the definition of square root of a nonnegative number.

Suppose we want to solve the equation x2 = 5. To solve that equation means to find a number x whose square is 5. According to the above definition the square root of 5 is a number whose square is 5. Note that the square of both the Principal Square Root and the negative square root make the equation true. Therefore the solutions are $±\sqrt{5}$ and the solution set is $\left\{±5\right\}$.

A slight variation of the above problem might be the solution of ${\left(x-3\right)}^{2}=16$.

Solving a quadratic equation with the Square Root Property is an application of the definition of Square Root and is useful in those very special occasions when the equation consists of a square equal to a number.

Completing The Square

The process of completing the square is a very useful tool which is used in many situations in mathematics. In fact it is the best approach to derive the Quadratic Formula.
Completing the square is not a sensible method for solving quadratic equations.
One could say completing the square was used to develop the quadratic formula to avoid continued use of completing the square to solve quadratic equations.

The first page of this essay summarizes everything about quadratic equations in one variable.

### Minimal List of Exercises Page 152.

If you understand the previous material you should be able to answer the following questions.
Each of these questions should be answered. Most of these questions are included in the MyMathLab homework requirement. If a particular concept is difficult for you, you should study the related text material and then try to answer some additional questions from the list provided by the department. In each case you are expected to make an honest adult evaluation of your understanding of the concept. Your ability to answer these questions is one tool to help you make that evaluation.
Section 1.5 Exercise Set: 3, 13, 17, 29, 65, 69, 71, 77, 80.

#### Section 1.5B: Quadratic Inequalities in One Variable

Inequalitiess: Quadratic Inequalities in One Variable

Note that these definitions involve a comparison of . According to The Law of Trichotomy, each real number makes exactly one of the comparisons true. Therefore the union of the solution sets for these three comparisons is equal to the set of Real Numbers. Furthermore no number is in more than one solution set.

As always the equation is called the boundary equation because its graph forms a boundary between the graphs of the two inequalities.
The exact configuration of the three graphs depends on the number of real solutions of the equation and can be determined with one or two simple tests.

At this point it is helpful to recall the following fact and to extend it to a consideration of the three graphs. These three cases arise as a result of The Law of Trichotomy applied to a comparison of the discriminant and 0.

We examine each case seperately.
Case 1: The discriminant is positive, the equation has two real solutions. The graph of the boundary equation consists of two points on the number line.

The graph of the boundary equation divides the real line into four sets (see Fig. 1):
• the graph of the boundary equation --- points h and k
• the interval $\left(h,k\right)$ between the two points h and k
• the ray $\left(-\infty ,h\right)$
• the ray $\left(k,\infty \right)$
The interval $\left(h,k\right)$ between the points h and k will be the solution set for either the less than inequality or the greater than inequality.
The union of the two rays will be the solution set for the other inequality.
Testing one number in either of the inequalities will determine which set is the graph of which inequality.

Here is an easier method. Just look at the leading coefficient.
If the leading coefficient is positive then the graph for the less-than inequality is the interval $\left(h,k\right)$
and the union $\left(-\infty ,h\right)\cup \left(k,\infty \right)$ of the two rays is the graph of the greater-than inequality.
If the leading coefficient is negative then the graph for the greater-than inequality is the interval $\left(h,k\right)$
and the union $\left(-\infty ,h\right)\cup \left(k,\infty \right)$ of the two rays is the graph of the less-than inequality.

Case 2: The discriminant is zero, the equation has one real solution. The graph of the boundary equation consists of one point on the number line.
The graph of the boundary equation divides the real line into three sets (see Fig. 3):
• the graph of the boundary equation --- a single point h
• the ray $\left(-\infty ,h\right)$ --- to the left of h
• the ray $\left(h,\infty \right)$ --- to the right of h
If the leading coefficient is positive, the union $\left(-\infty ,h\right)\cup \left(h,\infty \right)$ of the two rays is the graph of the greater-than inequality and the less-than inequality has no graph because its solution set is the empty set.
If the leading coefficient is negative, the union $\left(-\infty ,h\right)\cup \left(h,\infty \right)$ of the two rays is the graph of the less-than inequality and the greater-than inequality has no graph because its solution set is the empty set.

Case 3: The discriminant is negative, the equation has no real solutions. There is no graph of the boundary equation on the number line.
The solution set for the equation and one of the inequalities is the empty set. If the leading coefficient is positive, the graph of the greater-than inequality is the real number line because its solution set is R. The less-than inequality has no graph because its solution set is the empty set.
If the leading coefficient is negative, the graph of the less-than inequality is the real number line because its solution set is R. The greater-than inequality has no graph because its solution set is the empty set.

#### Section 1.6: Other Types of Equations in One Variable(Page 158).

RATIONAL EQUATIONS in One Variable

Equations: Rational Equations in One Variable

The only reason a rational expression does not make sense is if the denominator is zero.
So the domain of a rational expression is all real numbers except those which cause a 0 in the denominator.

Recall that real numbers are polynomials, they are called constant polynomials.

In this section when we speak of solving rational equations, the reference is to rational equations which are not polynomial equations. We are concerned here with equations that have variables in the denominators. To solve that kind of equation we must be aware of the following VERY important facts.

Examples of working with rational expressions and rational equationsHERE
Examples of Solving Linear Equations and Solving Rational EquationsHERE

### Minimal List of Exercises Page 168.

If you understand the previous material you should be able to answer the following questions.
Each of these questions should be answered. Most of these questions are included in the MyMathLab homework requirement. If a particular concept is difficult for you, you should study the related text material and then try to answer some additional questions from the list provided by the department. In each case you are expected to make an honest adult evaluation of your understanding of the concept. Your ability to answer these questions is one tool to help you make that evaluation.
Section 1.6 Exercise Set: 1, 3, 11, 14, 15, 19, 25, 33, 45, 53, 63, 65, 67, 75, 115.
Section 1.2 Exercise Set: Page 112: 31, 34, 40, 45, 60, 77.

Equations: Radical Equations in One Variable

#### Section 1.7a: Inequalitites in One Variable (page 172).

Inequalities: Fundamentals of Inequalities in One Variable

Inequalities: Solving Inequalities in One Variable

Properties of Inequalities:

Absolute Value Equations and Inequalities in One Variable

#### Section 1.7c: Equations and Inequalities Involving Absolute Values of Linear Expressions

Solving equations or inequalities involving absolute values is based on the definition of absolute value.

Notice the precise definition of absolute value has two cases:
Case 1: The expression inside the absolute value symbol is positive or zero.
Case 2: The expression inside the absolute value symbol is negative.

Procedure: To solve an equation or an inequality involving absolute values of variable expressions, it is necessary to consider the two equations which naturally result from the definition of absolute value.

Every equation or inequality involving absolute value is solved by considering the two cases as in this general example:
Generic Example: To solve an equation or inequality involving |GLOB|, two cases must be considered. The two cases arise from the definition of absolute value. Therefore to solve any equation or inequality involving |GLOB|, we consider:
Case 1: The equation that results from relacing |GLOB| with GLOB
Case 2: The equation that results from relacing |GLOB| with the opposite of GLOB.

The following two properties of equations and three properties of inequalities are also important tools when solving equations involving absolute value.

generating equivalent inequalities generating equivalent equations

These five properties might be used to:
Simplify the equation or inequality before considering the two cases.
Solve the equations or inequalities in Case 1 and Case 2.

In this section of the textbook the expression inside the absolute value symbol is always a linear expresion in one variable. Each of these equations or inequalities can be solved by directly using the two cases that arise from the definition of absolute value together with the two properties for generating equivalent equations and the three properties for generating equivalent inequalities. HOWEVER, there is an easier method. We will explain and demonstrate the easier method.

##### Solving Equations and Inequalities of the form |ax + b| = k,|ax + b| < k, and |ax + b| > k
Begin to understand this topic by recalling the Law of Trichotomy.

The Law of Trichotomy informs us that the constant k on the right side is negative, zero, or positive.
Case 1: If k is negative, we easily observe:

• $|ax+b|=k$ has no solutions because the absolute value is never negative.
• $|ax+b| has no solutions because the absolute value is never negative.
• The solution set for $|ax+b|>k$ is R because the absolute value is always positive.

Case 2: If k is 0, then we observe that the absolute value of an expression is 0 if and only if the expression is 0 to arrive at the following:
• $|ax+b|=0$ is equivalent to $ax+b=0$ which is easily solved.
• The solution set for $|ax+b|<0$ is the empty set $\varnothing$ because the absolute value of an expression is non negative.
• From The Law of Trichotomy we then conclude that the solution set for $|ax+b|>0$ is all real numbers except $-\frac{b}{a}$ (the solution of $|ax+b|=0$).

Case 3:If k is positive, the situation is a bit more involved and requires more reasoning.
In the remaining discussion we will assume that the constant k on the right is positive.

When we examine equations in two variables the following fact will be more obvious and natural. For the present it is best to simply remember the three situations.

Fact:When k is positive:

• the solution set for $|ax+b| is the set of numbers in an interval on the number line
• the solution set for $|ax+b|=k$ is the set containing only the endpoints of that interval
• the solution set for $|ax+b|>k$ is the set of numbers outside of that interval

The word equivalent as used in the above fact means

Fact: Neither the corresponding "greater than" inequality nor the corresponding equality can be written in such a compact form.

The equation is called the boundary equation because its graph forms a boundary between the graphs of the two inequalities.

The fact that the "less than" inequality is equivalent to a compact compound inequality makes it the easiest to solve because the compact form incorporates both cases into one computational process.

The Law of Trichotomy informs us that when we consider any one of we should, in fact, consider all three of them.

The Law of Trichotomy informs us that each real number is a solution to one of
Therefore the union of the three solution sets is R.
Therefore when all three are graphed on the same number line, the entire number line is used.

The Law of Trichotomy informs us that each real number is a solution to only one of the three.
Therefore the intersection of any two of the solution sets is the empty set.
Therefore when all three are graphed on the same number line, none of the graphs overlap.

### The Process

1. Whether solving always solve $|ax+b| computationally and then solve the other two by using deductive reasoning without any additional computation.

2. Convert $|ax+b| to the equivalent compact compound inequality without absolute value $-k

3. Solve the compact compound inequality by using the three methods to generate equivalent inequalities until simplest inequalities are obtained. The solution set will be an interval, call it S.
4. Use Deductive Reasoning and the fact that the equation is the boundary equation to conclude the solution set for the equality is the set whose elements are the two endpoints of S.

5. Use Deductive Reasoning and The Law of Trichotomy to conclude the solution set for the "greater than" inequality is everything else. So the solution set for the "greater than" inequality must be the two rays to the left of and to the right of S.

Here are some Exercises to assist your learning.
Solutions are provided.

About Absolute Value Equations and Inequalities

You should study Pages 6 - 13 of this Essay.

Pages 14 - 17 of this essay review a topic that many of you have seen before. Please review it now.

### Minimal List of Exercises Page 185.

If you understand the previous material you should be able to answer the following questions.
Each of these questions should be answered. Most of these questions are included in the MyMathLab homework requirement. If a particular concept is difficult for you, you should study the related text material and then try to answer some additional questions from the list provided by the department. In each case you are expected to make an honest adult evaluation of your understanding of the concept. Your ability to answer these questions is one tool to help you make that evaluation.
Section 1.7 Exercise Set: 1, 3, 5, 7, 9, 11, 13, 15, 17, 21, 25, 27, 31, 33, 35, 37, 41, 45, 51, 55, 57, 59, 61, 63, 65, 67, 73, 77, 81, 83, 87, 103, 104 .