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DrDelMath

Intermediate Algebra
Elayn Martin-Gay

"One cannot apply what one knows in a practical manner if one does not know anything to apply."
-- Robert Sternberg

Section 0.2: Four Important Properties of the Real Numbers





Appendix B: Equation Review,
Chapter 9: Compound Inequalities, Absolute Value Equations and Inequalities

Section B.0: Review: Fundamentals of Equations









About Equations and Classifying Equations

Important:The Law of Trichotomy is a property of the Real numbers. It is true for any subset of the Real numbers. It is not true for any other sets of numbers. It is not true for sets of any other mathematics objects.

This Law of Trichotomy states that when comparing two real numbers, or expressions representing real numbers, there are ALWAYS three possibilities. These possibilities are represented by the normal symbols <, =, > for the three binary relations.
It is beneficial to consider all three any time you want to consider any one of them. Think of them as inseparable siblings - triplets.
Any time you are working with an equation think of the two siblings obtained by replacing the = symbol with < and >.
Any time you are considering an inequality think of the two siblings obtained by replacing the inequality with the other inequality and the equal symbol =.

Examples: When considering 3x + 5 = 9 you should also consider 3x + 5 < 9 and 3x + 5 > 9
Examples: When considering 2x3 + 5x < 9 you should also consider 2x3 + 5x > 9 and 2x3 + 5x = 9
Examples: When considering 2x2 + 5x - 4 > 0 you should also consider 2x2 + 5x - 4 < 0 and 2x2 + 5x - 4 = 0



Linear equations in one variable are equations of the form ax + b = 0.
Linear inequalities in one variable are inequalities of the form ax + b < 0 and ax + b > 0.
Note that in each of these three statements, an algebraic expression ax + b is compared to the number 0.
Clearly these three statements resemble the Law of Trichotomy.
In fact The Law of Trichotomy tells us that for each real number x:

As we examine linear equations in one variable and linear inequalities in one variable side by side we will uncover even stronger ties between the three expressions ax + b < 0, ax + b = 0 and ax + b > 0

When solving ax + b = 0, the other two siblings ax + b < 0 and ax + b > 0 cry out for attention.
When solving ax + b < 0, the other two siblings ax + b = 0 and ax + b > 0 cry out for attention.
When solving ax + b > 0, the other two siblings ax + b = 0 and ax + b < 0 cry out for attention.

You should recognize that when you solve any one of the three, solutions to the other two are close at hand and may be determined without another "solution process" or another bunch of calculations.

The relationship between solution sets for linear equations in one variable and linear inequalities in one variable is an example of relationships between equations and inequalites of any degree involving any number of variables.
It is an important principle to observe and learn!

Fundamentals of Equations:
Fundamentals of Inequalities:

About Solutions of Equations and Inequalities. Graphs of Equations and Inequalities. Equivalent Equations and Inequalities

The following properties of equations and inequalities are extremely important. These properties are the first things that should come to mind when you encounter an equation or inequality in any context.

These properties are the basic tools for:

Properties of Equations:
Properties of Inequalities:

About Basic Properties of Equations and Inequalities

Section B.1a: Linear Equations and Inequalities in One Variable

Linear Equations in One Variable:
Linear Inequalities in One Variable:

About Linear Equations and Inequalities in One Variable

Putting It Together

Summary Solving Linear Equations and Inequalities in One Variable

Exercises for Section B1:a
Solve Exercises 3, 4, 9, 15,and 17. Be sure you can state the mathematics principle which permits you to perform each step in the process.
Sketch the graph of each of the equations in Exercises 3, 4, 9, 15, and 17.
Write the corresponding inequalities for the equations in Exercises 3, 4, 9, 15, and 17.
Use the first three properties of inequalities to solve the inequalities you associated with the equations in Exercises 3 and 4.
Graph each inequality on its own line, then sketch the graph of the equation and both inequalities on the same line.
What observations can you make about this last graph?
Use the graph of the equation, results of testing one point, and deductive reasoning to solve the two inequalities you associated with the equations in Exercises 9, 15, and 17.
Graph each inequality on its own line, then sketch the graph of the equation and both inequalities on the same line.
What observations can you make about this last graph?
Answer Question 47. Be sure to write your conclusion/answer and all reasoning which produced that conclusion.
Answer Questions 51 and 52. Be sure you can state the mathematics principle which permits you to perform the steps in your work.

Because some may not yet own a textbook, I will reproduce exercises here.
Solve each of the following:
3) 3x - 4 - 5x = x + 4 + x
4) 13x - 15x + 8 = 4x + 2 - 24
9) 5(y + 4) = 4 (y + 5)
15) 6x - 2(x - 3) = 4 (x + 1) +4
17) 3/8 + b/3 = 5/12
47) Which solution strategies are incorrect and why?
a) Solve (y - 2)(y + 2) = 4 by setting each factor equal to 4.
b) Solve (x + 1)(x + 3) = 0 by setting each factor equal to 0.
c) Solve z2 + 5z + 6 = 0 by factoring z2 + 5z + 6 and setting each factor equal to 0.
d) Solve x2 + 6x + 8 = 10 by factoring x2 + 6x + 8 and setting each factor equal to 10.
51) Find the value of K which will make the two equations equivalent x/6 + 4 = x/3 and x + K = 2x
52) Find the value of K which will make the two equations equivalent 5x/4 + 1/2 = x/2 and 5x + K = 2x

Section B.1b: Quadratic Equations in One Variable: Solving with Factoring and Zero Factor Property

If the reader needs a quick review of basic factoring techniques it is available HERE
These factoring techniques are absolutely essential. If the reader is rusty -- study the review.

The Zero Factor Property may be generalized to the following very useful statement.
If each of two expressions represent real numbers and their product is zero then one (or both) of the expressions is zero.

If the quadratic in a quadratic equation can be factored, then the Zero Factor Property (generalized form) may be used to conclude that the original quadratic equation is equivalent to the compound equality (obtained by setting each factor equal to zero) with the conjunction OR.

Example: The quadratic polynomial in the equation x2 - 4x - 21 = 0 may be factored so that x2 - 4x - 21 = 0 is equivalent to (x + 3)(x - 7) = 0.
As a result of The Zero Factor Property (x + 3)(x - 7) = 0 is equivalent to the compound equality x + 3 = 0 OR x - 7 = 0.
The solution set for x + 3 is {-3} and the solution set for x - 7 = 0 is {7}.
The solution set for the compound equality is the union {-3} ∪ {7} = {-3, 7}.
Therefore the solution set for the original quadratic equation is {-3, 7}.

The role of equivalance and The Zero Factor Property in the above process for solving a quadratic equation was probably neglected in previous courses. Because we want to generalize this process to many other situations, it is important now to pay attention to the importance of these two properties.

Exercises for Section B1b:
Solve Exercises 5, 7, 14, 16, 26, 29, 30, and 38. Be sure you can state the mathematics principle which permits you to perform each step in the process.
Sketch the graph of each of the equations in Exercises 5, 7, 14, 16, 26, 29, 30, and 38.
Write the corresponding inequalities for the equations in Exercises 5, 7, 14, 16, 26, 29, 30, and 38.
Use the first three properties of inequalities to solve the inequalities you associated with the equations in Exercises 5, 7, 26 and 30.
Graph each inequality on its own line, then sketch the graph of the equation and both inequalities on the same line.
What observations can you make about this last graph?
Use the graph of the equation, results of testing one point, and deductive reasoning to solve the two inequalities you associated with the equations in Exercises 14, 16, 29, and 38.
Graph each inequality on its own line, then sketch the graph of the equation and both inequalities on the same line.
What observations can you make about this last graph?
Answer Question 47. Be sure to write your conclusion/answer and all reasoning which produced that conclusion.
Was the answer to this question more obvious after studying Section B1b?
Answer Question 48. Be sure you can state the mathematics principle which permits you to perform the steps in your work.
NOTE: In Question 48 the reference is to equations IN ONE VARIABLE.

Because some may not yet own a textbook, I will reproduce exercises here.
Solve each of the following:
5) 12x2 + 5x - 2 = 0
7) z2 + 9 = 10z
14) n(2n - 3) = 2
16) 10x -2(x + 4) = 8(x - 2) + 6
26) y2/30 = y/15 + 1/2
29)(3t + 1)/8 = (5 + 2t)/7 + 2
30)4 - (2z + 7)/9 = (7 - z)/12
38)7c - 2(3c + 1) = 5(4 - 2c)
47) Which solution strategies are incorrect and why?
a) Solve (y - 2)(y + 2) = 4 by setting each factor equal to 4.
b) Solve (x + 1)(x + 3) = 0 by setting each factor equal to 0.
c) Solve z2 + 5z + 6 = 0 by factoring z2 + 5z + 6 and setting each factor equal to 0.
d) Solve x2 + 6x + 8 = 10 by factoring x2 + 6x + 8 and setting each factor equal to 10.
48)Describe two ways a linear equation in one variable differs from a quadratic equation in one variable.

Section 9.1: Compound Inequalities

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About Compound Inequalities

Assignment for Section 9.1:
Study Section 9.1 in the textbook.
Solve Exercises 1, 6, 13, 21, 26, 37, 41. Be sure you can state the mathematics principle which permits you to perform each step in the process.
As you consider these inequalities, please observe that the examples the author references and the work you do should be examples of how to use The First Three Properties of Inequalities. Don't try to learn how to solve each phony textbook problem type by example. Rather learn how to use fundamental principles so that you have tools with which to attack new problem types after you leave the classroom. The textbook problems should be considered self-test devices to check your understanding of the principles.

How many elements ar in the intersections which you computed?
In each of Exercises 13, 21, 26, 37, 41, state how many solutions are in the solution set.
In each of Exercises 1, 6, 13, 21, 26, 37, 41, sketch the graph of the sets (in #1 & #6) and solution sets.
For each of the solution sets in each of Exercises 13, 21, 26, 37, 41, explain how you know, without testing, that each number is a solution.
For each of the solution sets in each of Exercises 13, 21, 26, 37, 41, explain how you know, that no solution exists outside the set you calculated.
Sketch and label a Venn Diagram which illustrates the intersection of two sets.
Sketch and label a Venn Diagram which illustrates the union of two sets.
Use factoring and The Zero Factor Property to solve the equation 3y2 - y - 14 = 0. Explain every step.
Write the two sibling inequalities that correspond to 3y2 - y - 14 = 0.
Solve both of these inequalities.
Sketch the graph of all three on the same graph.
What observations can you make from this picture?
Write the first two properties of equations.
Write the first three properties of inequalities.
Write The Zero Factor Property.
When considering an equation and it two corresponding inequalities, the equation is frequently called the boundary equation. Why?
Write The Law of Trichotomy.
Is The Law of Trichotomy a property of quadratic equations or of the Real Numbers?

 

Section 9.2: Absolute Value Equations and
Section 9.3: Absolute Value Inequalities

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.

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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. That is what is done in most elementary textbooks. HOWEVER, there is an easier method. We will explain and demonstrate the easier method. This easier method depends on deductive reasoning as opposed to computations.

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.
If k is negative, we easily observe:

If k is 0, then we observe that the absolute value of an expression is 0 if and only if the expression is 0. Therefore if k is 0, the equations and inequalities are equivalent to ax + b < 0, ax + b = 0, and ax + b > 0. These are simple linear equations and linear inequalities which were previously mastered.

The only remaining case is when k is positive. Therefore in the remaining discussion we will assume that the constant k on the right is positive.

Fact:When k is positive:

The word equivalent as used in the above fact means | X | < k has the same solution set as -k < X < k.

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.

interval all three

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 structure.

The Law of Trichotomy informs us that when we consider any one of |ax + b| = k, |ax + b| < k, or |ax + b| > k 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 |ax + b| = k, |ax + b| < k, or |ax + b| > k.
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.

interval all three

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.

interval all three

The Process

  1. Whether solving |ax + b| = k, |ax + b| < k, or |ax + b| > k always solve |ax + b| < k computationally and then solve the other two by using deductive reasoning without any additional computation.

  2. Convert |ax + b| < k to its compact compound inequality form - k < ax + b < 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 to conclude the solution set for the equality is the set containing the endpoints of S.

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

 

About Absolute Value Equations and Inequalities

Page 563: Exercises: 1, 2, 9, 10. These should be solved by inspection using nothing but the definition of absolute value.
Page 563: Exercises: For each of Exercises 5, 15, 39, and 59 follow each of the following instructions in the order presented. Think about what you are doing. How does The Law of Trichotomy enter these considerations?
Write the two sibling/corresponding inequalities.
Observe that the "less than" inequality is "the easy one".
Write the less than inequality as a compact compound inequality which does not involve abasolute value.
How is this compact compound inequality related to the original less than inequality?
Solve the compact compound inequality.
Write that solution set in interval notation. Write that solution set with set builder notation.
Sketch the graph of the less than inequality.
Without computation determine the solution set of the original absolute value equation.
Write the solution set of the original equation using the roster method.
Graph the solution set of the equation on the graph of the less than inequality.
Without computation determine the solution set of the greater than inequality.
Write the solution set of the greater than inequality using interval notation.
Write the solution set of the greater than inequality using set builder notation.
Graph the solution set of the greater than inequality on the graph of the less than inequality and the equation.
What is the answer to the original question?
Look at the graph and determine the union of the three solution sets.
Look at the graph and determine the intersection of each pair of solution sets.

Page 568: Exercises: 1, 11, 22, 26. These should be solved by inspection using nothing but the definition of absolute value.
Page 568: Exercises: For each of Exercises 6, 15, 24, 34, and 63 follow each of the following instructions in the order presented. Think about what you are doing. How does The Law of Trichotomy enter these considerations?
Write the two sibling/corresponding equation and inequality.
Observe that the "less than" inequality is "the easy one".
Write the less than inequality as a compact compound inequality which does not involve abasolute value.
How is this compact compound inequality related to the original less than inequality?
Solve the compact compound inequality.
Write that solution set in interval notation. Write that solution set with set builder notation.
Sketch the graph of the less than inequality.
Without computation determine the solution set of the original absolute value equation.
Write the solution set of the original equation using the roster method.
Graph the solution set of the equation on the graph of the less than inequality.
Without computation determine the solution set of the greater than inequality.
Write the solution set of the greater than inequality using interval notation.
Write the solution set of the greater than inequality using set builder notation.
Graph the solution set of the greater than inequality on the graph of the less than inequality and the equation.
What is the answer to the original question?
Look at the graph and determine the union of the three solution sets.
Look at the graph and determine the intersection of each pair of solution sets.

Write the first two properties of equations.
Write the first three properties of inequalities.
Compute the exact area of a circle whose radius is 7.
Write the standard formula for the circumference of a circle.
What is the definition of π.
How many irrational numbers are in { x|3<x<5 }?
How many irrational numbers are in { x|xQ  and 3<x<5 }?