INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith

Whether you are learning Algebra for the first time or brushing up on your math skills, Professor Eric Smith's comprehensive course on Algebra 1 will help you become an expert in no time. Professor Smith begins each lecture with common terminology and methods pertinent to the lesson. He explains concepts clearly, illustrates them with colorful slides, and reinforces them with plenty of sample problems and solutions. This fun, straightforward course in Algebra 1 will keep you engaged and covers everything from Order of Operations to Graphing Functions, Inequalities, Quadratics, and Complex Numbers. Professor Eric Smith has been teaching math at the high school and college level for 10+ years.

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I. Properties of Real Numbers
Basic Types of Numbers 30:41
Intro 0:00
Objectives 0:07
Basic Types of Numbers 0:36
Natural Numbers 1:02
Whole Numbers 1:29
Integers 2:04
Rational Numbers 2:38
Irrational Numbers 5:06
Imaginary Numbers 6:48
Basic Types of Numbers Cont. 8:09
The Big Picture 8:10
Real vs. Imaginary Numbers 8:30
Rational vs. Irrational Numbers 8:48
Basic Types of Numbers Cont. 10:55
Number Line 11:06
Absolute Value 11:44
Inequalities 12:39
Example 1 13:16
Example 2 17:30
Example 3 21:56
Example 4 24:27
Example 5 27:48
Operations on Numbers 19:26
Intro 0:00
Objectives 0:06
Operations on Numbers 0:25
Subtraction 1:33
Multiplication & Division 2:19
Exponents 3:24
Bases 4:04
Square Roots 4:59
Principle Square Roots 5:09
Perfect Squares 6:32
Simplifying and Combining Roots 6:52
Example 1 8:16
Example 2 12:30
Example 3 14:02
Example 4 16:27
Order of Operations 12:06
Intro 0:00
Objectives 0:06
The Order of Operations 0:25
Work Inside Parentheses 0:42
Simplify Exponents 0:52
Multiplication & Division from Left to Right 0:57
Addition & Subtraction from Left to Right 1:11
Remember PEMDAS 1:21
The Order of Operations Cont. 2:27
Example 2:43
Example 1 3:55
Example 2 5:36
Example 3 7:35
Example 4 8:56
Properties of Real Numbers 18:52
Intro 0:00
Objectives 0:07
The Properties of Real Numbers 0:23
Commutative Property of Addition and Multiplication 0:44
Associative Property of Addition and Multiplication 1:50
Distributive Property of Multiplication Over Addition 3:20
Division Property of Zero 4:46
Division Property of One 5:23
Multiplication Property of Zero 5:56
Multiplication Property of One 6:17
Why Are These Properties Important? 6:53
Example 1 9:16
Example 2 13:04
Example 3 14:30
Example 4 16:57
II. Linear Equations
The Vocabulary of Linear Equations 12:22
Intro 0:00
Objectives 0:09
The Vocabulary of Linear Equations 0:44
Variables 0:52
Terms 1:09
Coefficients 1:40
Like Terms 2:18
Examples of Like Terms 2:37
Expressions 4:01
Equations 4:26
Linear Equations 5:04
Solutions 5:55
Example 1 6:16
Example 2 7:16
Example 3 8:45
Example 4 10:20
Solving Linear Equations in One Variable 28:52
Intro 0:00
Objectives 0:08
Solving Linear Equations in One Variable 0:34
Conditional Cases 0:51
Identity Cases 1:09
Solving Linear Equations in One Variable Cont. 2:00
Multiplication Property of Equality 2:43
Steps to Solve Linear Equations 3:14
Example 1 4:22
Example 2 8:21
Example 3 12:32
Example 4 14:19
Example 5 17:25
Example 6 22:17
Solving Formulas 12:02
Intro 0:00
Objectives 0:06
Solving Formulas 0:18
Formulas 0:26
Use the Same Properties as Solving Linear Equations 1:36
Multiplication Property of Equality 1:58
Steps to Solve Formulas 2:43
Example 1 3:56
Example 2 6:09
Example 3 8:39
Applications of Linear Equations 28:41
Intro 0:00
Objectives 0:10
Applications of Linear Equations 0:43
The Six-Step Method to Solving Word Problems 0:55
Common Terms 3:12
Example 1 5:03
Example 2 9:40
Example 3 13:48
Example 4 17:58
Example 5 23:28
Applications of Linear Equations, Motion & Mixtures 24:26
Intro 0:00
Objectives 0:21
Motion and Mixtures 0:46
Motion Problems: Distance, Rate, and Time 1:06
Mixture Problems: Amount, Percent, and Total 1:27
The Table Method 1:58
The Beaker Method 3:38
Example 1 5:05
Example 2 9:44
Example 3 14:20
Example 4 19:13
III. Graphing
Rectangular Coordinate System 22:55
Intro 0:00
Objectives 0:11
The Rectangular Coordinate System 0:39
The Cartesian Coordinate System 0:40
X-Axis 0:54
Y-Axis 1:04
Origin 1:11
Ordered Pairs 2:10
Example 1 2:55
The Rectangular Coordinate System Cont. 6:09
X-Intercept 6:45
Y-Intercept 6:55
Relation of X-Values and Y-Values 7:30
Example 2 11:03
Example 3 12:13
Example 4 14:10
Example 5 18:38
Slope & Graphing 27:58
Intro 0:00
Objectives 0:11
Slope and Graphing 0:48
Standard Form 1:14
Example 1 2:24
Slope and Graphing Cont. 4:58
Slope, m 5:07
Slope is Rise over Run 6:11
Don't Mix Up the Coordinates 8:20
Example 2 9:39
Slope and Graphing Cont. 14:26
Slope-Intercept Form 14:34
Example 3 16:55
Example 4 18:00
Slope and Graphing Cont. 19:00
Rewriting an Equation in Slope-Intercept Form 19:39
Rewriting an Equation in Standard Form 20:09
Slopes of Vertical & Horizontal Lines 20:56
Example 5 22:49
Example 6 24:09
Example 7 25:59
Example 8 26:57
Linear Equations in Two Variables 20:36
Intro 0:00
Objectives 0:13
Linear Equations in Two Variables 0:36
Point-Slope Form 1:07
Substitute in the Point and the Slope 2:21
Parallel Lines: Two Lines with the Same Slope 4:05
Perpendicular Lines: Slopes are Negative Reciprocals of Each Other 4:39
Perpendicular Lines: Product of Slopes is -1 5:24
Example 1 6:02
Example 2 7:50
Example 3 10:49
Example 4 13:26
Example 5 15:30
Example 6 17:43
IV. Functions
Introduction to Functions 21:24
Intro 0:00
Objectives 0:07
Introduction to Functions 0:58
Relations 1:03
Functions 1:37
Independent Variables 2:00
Dependent Variables 2:11
Function Notation 2:21
Function 3:43
Input and Output 3:53
Introduction to Functions Cont. 4:45
Domain 4:46
Range 4:55
Functions Represented by a Diagram 6:41
Natural Domain 9:11
Evaluating Functions 12:02
Example 1 13:13
Example 2 15:03
Example 3 16:18
Example 4 19:54
Graphing Functions 16:12
Intro 0:00
Objectives 0:09
Graphing Functions 0:54
Using Slope-Intercept Form 1:56
Vertical Line Test 2:58
Determining the Domain 4:20
Determining the Range 5:43
Example 1 6:06
Example 2 7:18
Example 3 8:31
Example 4 11:04
V. Systems of Linear Equations
Systems of Linear Equations 25:54
Intro 0:00
Objectives 0:13
Systems of Linear Equations 0:46
System of Equations 0:51
System of Linear Equations 1:15
Solutions 1:35
Points as Solutions 1:53
Finding Solutions Graphically 5:13
Example 1 6:37
Example 2 12:07
Systems of Linear Equations Cont. 17:01
One Solution, No Solution, or Infinite Solutions 17:10
Example 3 18:31
Example 4 22:37
Solving a System Using Substitution 20:01
Intro 0:00
Objectives 0:09
Solving a System Using Substitution 0:32
Substitution Method 1:24
Substitution Example 2:35
One Solution, No Solution, or Infinite Solutions 7:50
Example 1 9:45
Example 2 12:48
Example 3 15:01
Example 4 17:30
Solving a System Using Elimination 19:40
Intro 0:00
Objectives 0:09
Solving a System Using Elimination 0:27
Elimination Method 0:42
Elimination Example 2:01
One Solution, No Solution, or Infinite Solutions 7:05
Example 1 8:53
Example 2 11:46
Example 3 15:37
Example 4 17:45
Applications of Systems of Equations 24:34
Intro 0:00
Objectives 0:12
Applications of Systems of Equations 0:30
Word Problems 1:31
Example 1 2:17
Example 2 7:55
Example 3 13:07
Example 4 17:15
VI. Inequalities
Solving Linear Inequalities in One Variable 17:13
Intro 0:00
Objectives 0:08
Solving Linear Inequalities in One Variable 0:37
Inequality Expressions 0:46
Linear Inequality Solution Notations 3:40
Inequalities 3:51
Interval Notation 4:04
Number Lines 4:43
Set Builder Notation 5:24
Use Same Techniques as Solving Equations 6:59
'Flip' the Sign when Multiplying or Dividing by a Negative Number 7:12
'Flip' Example 7:50
Example 1 8:54
Example 2 11:40
Example 3 14:01
Compound Inequalities 16:13
Intro 0:00
Objectives 0:07
Compound Inequalities 0:37
'And' vs. 'Or' 0:44
'And' 3:24
'Or' 3:35
'And' Symbol, or Intersection 3:51
'Or' Symbol, or Union 4:13
Inequalities 4:41
Example 1 6:22
Example 2 9:30
Example 3 11:27
Example 4 13:49
Solving Equations with Absolute Values 14:12
Intro 0:00
Objectives 0:08
Solve Equations with Absolute Values 0:18
Solve Equations with Absolute Values Cont. 1:11
Steps to Solving Equations with Absolute Values 2:21
Example 1 3:23
Example 2 6:34
Example 3 10:12
Inequalities with Absolute Values 17:07
Intro 0:00
Objectives 0:07
Inequalities with Absolute Values 0:23
Recall… 2:08
Example 1 3:39
Example 2 6:06
Example 3 8:14
Example 4 10:29
Example 5 13:29
Graphing Inequalities in Two Variables 15:33
Intro 0:00
Objectives 0:07
Graphing Inequalities in Two Variables 0:32
Split Graph into Two Regions 1:53
Graphing Inequalities 5:44
Test Points 6:20
Example 1 7:11
Example 2 10:17
Example 3 13:06
Systems of Inequalities 21:13
Intro 0:00
Objectives 0:08
Systems of Inequalities 0:24
Test Points 1:10
Steps to Solve Systems of Inequalities 1:25
Example 1 2:23
Example 2 7:28
Example 3 12:51
VII. Polynomials
Integer Exponents 44:51
Intro 0:00
Objectives 0:09
Integer Exponents 0:42
Exponents 'Package' Multiplication 1:25
Example 1 2:00
Example 2 3:13
Integer Exponents Cont. 4:50
Product Rule for Exponents 4:51
Example 3 7:16
Example 4 10:15
Integer Exponents Cont. 13:13
Power Rule for Exponents 13:14
Power Rule with Multiplication and Division 15:33
Example 5 16:18
Integer Exponents Cont. 20:04
Example 6 20:41
Integer Exponents Cont. 25:52
Zero Exponent Rule 25:53
Quotient Rule 28:24
Negative Exponents 30:14
Negative Exponent Rule 32:27
Example 7 34:05
Example 8 36:15
Example 9 39:33
Example 10 43:16
Intro 0:00
Objectives 0:07
Terms 0:33
Coefficients 0:51
Like Terms 1:29
Polynomials 2:21
Monomials, Binomials, Trinomials, and Polynomials 5:41
Degrees 7:00
Evaluating Polynomials 8:12
Adding and Subtracting Polynomials Cont. 9:25
Example 1 11:48
Example 2 13:00
Example 3 14:41
Example 4 16:15
Multiplying Polynomials 25:07
Intro 0:00
Objectives 0:06
Multiplying Polynomials 0:41
Distributive Property 1:00
Example 1 2:49
Multiplying Polynomials Cont. 8:22
Organize Terms with a Table 8:23
Example 2 13:40
Multiplying Polynomials Cont. 16:33
Multiplying Binomials with FOIL 16:48
Example 3 18:49
Example 4 20:04
Example 5 21:42
Dividing Polynomials 44:56
Intro 0:00
Objectives 0:07
Dividing Polynomials 0:29
Dividing Polynomials by Monomials 2:10
Dividing Polynomials by Polynomials 2:59
Dividing Numbers 4:09
Dividing Polynomials Example 8:39
Example 1 12:35
Example 2 14:40
Example 3 16:45
Example 4 21:13
Example 5 24:33
Example 6 29:02
Dividing Polynomials with Synthetic Division Method 33:36
Example 7 38:43
Example 8 42:24
VIII. Factoring Polynomials
Greatest Common Factor & Factor by Grouping 28:27
Intro 0:00
Objectives 0:09
Greatest Common Factor 0:31
Factoring 0:40
Greatest Common Factor (GCF) 1:48
GCF for Polynomials 3:28
Factoring Polynomials 6:45
Prime 8:21
Example 1 9:14
Factor by Grouping 14:30
Steps to Factor by Grouping 17:03
Example 2 17:43
Example 3 19:20
Example 4 20:41
Example 5 22:29
Example 6 26:11
Factoring Trinomials 21:44
Intro 0:00
Objectives 0:06
Factoring Trinomials 0:25
Recall FOIL 0:26
Factor a Trinomial by Reversing FOIL 1:52
Tips when Using Reverse FOIL 5:31
Example 1 7:04
Example 2 9:09
Example 3 11:15
Example 4 13:41
Factoring Trinomials Cont. 15:50
Example 5 18:42
Factoring Trinomials Using the AC Method 30:09
Intro 0:00
Objectives 0:08
Factoring Trinomials Using the AC Method 0:27
Factoring when Leading Term has Coefficient Other Than 1 1:07
Reversing FOIL 1:18
Example 1 1:46
Example 2 4:28
Factoring Trinomials Using the AC Method Cont. 7:45
The AC Method 8:03
Steps to Using the AC Method 8:19
Tips on Using the AC Method 9:29
Example 3 10:45
Example 4 16:50
Example 5 21:08
Example 6 24:58
Special Factoring Techniques 30:14
Intro 0:00
Objectives 0:07
Special Factoring Techniques 0:26
Difference of Squares 1:46
Perfect Square Trinomials 2:38
No Sum of Squares 3:32
Special Factoring Techniques Cont. 4:03
Difference of Squares Example 4:04
Perfect Square Trinomials Example 5:29
Example 1 7:31
Example 2 9:59
Example 3 11:47
Example 4 15:09
Special Factoring Techniques Cont. 19:07
Sum of Cubes and Difference of Cubes 19:08
Example 5 23:13
Example 6 26:12
Solving Quadratic Equations by Factoring 23:38
Intro 0:00
Objectives 0:08
Solving Quadratic Equations by Factoring 0:19
Zero Factor Property 1:39
Zero Factor Property Example 2:34
Example 1 4:00
Solving Quadratic Equations by Factoring Cont. 5:54
Example 2 7:28
Example 3 11:09
Example 4 14:22
Solving Quadratic Equations by Factoring Cont. 18:17
Higher Degree Polynomial Equations 18:18
Example 5 20:22
Intro 0:00
Objectives 0:12
Linear Factors 0:38
Not All Quadratics Factor Easily 1:22
Principle of Square Roots 3:36
Completing the Square 4:50
Steps for Using Completing the Square 5:15
Completing the Square Works on All Quadratic Equations 6:41
Discriminants 8:25
Solving Quadratic Equations - Summary 10:11
Example 1 11:54
Example 2 13:03
Example 3 16:30
Example 4 21:29
Example 5 25:07
Intro 0:00
Objectives 0:08
Using a Substitution 0:53
U-Substitution 1:26
Example 1 2:07
Example 2 5:36
Example 3 8:31
Example 4 11:14
Intro 0:00
Objectives 0:09
Squared Variable 0:40
Principle of Square Roots 0:51
Example 1 1:09
Example 2 2:04
Quadratic Formulas and Applications Cont. 3:34
Example 3 4:42
Example 4 13:33
Example 5 20:50
Intro 0:00
Objectives 0:06
Axis of Symmetry 1:46
Vertex 2:12
Transformations 2:57
Graphing in Quadratic Standard Form 3:23
Example 1 5:06
Example 2 6:02
Example 3 9:07
Completing the Square 12:02
Vertex Shortcut 12:16
Example 4 13:49
Example 5 17:25
Example 6 20:07
Example 7 23:43
Polynomial Inequalities 21:42
Intro 0:00
Objectives 0:07
Polynomial Inequalities 0:30
Solving Polynomial Inequalities 1:20
Example 1 2:45
Polynomial Inequalities Cont. 5:12
Larger Polynomials 5:13
Positive or Negative Intervals 7:16
Example 2 9:01
Example 3 13:53
X. Rational Equations
Multiply & Divide Rational Expressions 26:41
Intro 0:00
Objectives 0:09
Multiply and Divide Rational Expressions 0:44
Rational Numbers 0:55
Dividing by Zero 1:45
Canceling Extra Factors 2:43
Negative Signs in Fractions 4:52
Multiplying Fractions 6:26
Dividing Fractions 7:17
Example 1 8:04
Example 2 14:01
Example 3 16:23
Example 4 18:56
Example 5 22:43
Adding & Subtracting Rational Expressions 20:24
Intro 0:00
Objectives 0:07
Adding and Subtracting Rational Expressions 0:41
Common Denominators 0:52
Common Denominator Examples 1:14
Steps to Adding and Subtracting Rational Expressions 2:39
Example 1 3:34
Example 2 5:27
Adding and Subtracting Rational Expressions Cont. 6:57
Least Common Denominators 6:58
Transitioning from Fractions to Rational Expressions 9:08
Identifying Least Common Denominators for Rational Expressions 9:56
Example 3 11:19
Example 4 12:36
Example 5 15:08
Example 6 16:46
Complex Fractions 18:23
Intro 0:00
Objectives 0:09
Complex Fractions 00:37
Dividing to Simplify Complex Fractions 1:10
Example 1 2:03
Example 2 3:58
Complex Fractions Cont. 9:15
Using the Least Common Denominator to Simplify Complex Fractions 9:16
Example 3 10:42
Example 4 14:28
Solving Rational Equations 16:24
Intro 0:00
Objectives 0:07
Solving Rational Equations 0:23
Isolate the Specified Variable 1:23
Example 1 1:58
Example 2 5:00
Example 3 8:23
Example 4 13:25
Rational Inequalities 18:54
Intro 0:00
Objectives 0:06
Rational Inequalities 0:18
Testing Intervals for Rational Inequalities 0:38
Steps to Solving Rational Inequalities 1:05
Tips to Solving Rational Inequalities 2:27
Example 1 3:33
Example 2 12:21
Applications of Rational Expressions 20:20
Intro 0:00
Objectives 0:07
Applications of Rational Expressions 0:27
Work Problems 1:05
Example 1 2:58
Example 2 6:45
Example 3 13:17
Example 4 16:37
Variation & Proportion 27:04
Intro 0:00
Objectives 0:10
Variation and Proportion 0:34
Variation 0:35
Inverse Variation 1:01
Direct Variation 1:10
Setting Up Proportions 1:31
Example 1 2:27
Example 2 5:36
Variation and Proportion Cont. 8:29
Inverse Variation 8:30
Example 3 9:20
Variation and Proportion Cont. 12:41
Constant of Proportionality 12:42
Example 4 13:59
Variation and Proportion Cont. 16:17
Varies Directly as the nth Power 16:30
Varies Inversely as the nth Power 16:53
Varies Jointly 17:09
Combining Variation Models 17:36
Example 5 19:09
Example 6 22:10
Rational Exponents 14:32
Intro 0:00
Objectives 0:07
Rational Exponents 0:32
Power on Top, Root on Bottom 1:05
Example 1 1:37
Rational Exponents Cont. 4:04
Using Rules from Exponents for Radicals as Exponents 4:05
Combining Terms Under a Single Root 4:50
Example 2 5:21
Example 3 7:39
Example 4 11:23
Example 5 13:14
Simplify Rational Exponents 15:12
Intro 0:00
Objectives 0:07
Simplify Rational Exponents 0:25
Product Rule to Simplify Square Roots 1:11
Applications of Product and Quotient Rules 2:17
Higher Roots 2:48
Example 1 3:39
Example 2 6:35
Example 3 8:41
Example 4 11:09
Intro 0:00
Objectives 0:07
Like Terms 1:29
Bases and Exponents May be Different 2:02
Bases and Powers Must be Same when Adding and Subtracting 2:42
Example 1 4:47
Example 2 6:00
Simplify the Bases to Look the Same 7:25
Example 3 8:23
Example 4 11:45
Example 5 15:10
Intro 0:00
Objectives 0:08
Rules for Working With Radicals 0:26
Don’t Distribute Powers 2:54
Rationalizing Denominators 6:40
Example 1 7:22
Example 2 8:32
Multiply and Divide Radicals Cont. 9:23
Rationalizing Denominators with Higher Roots 9:25
Example 3 10:51
Example 4 11:53
Multiply and Divide Radicals Cont. 13:13
Rationalizing Denominators with Conjugates 13:14
Example 5 15:52
Example 6 17:25
Intro 0:00
Objectives 0:07
Isolate the Roots and Raise to Power 0:34
Example 1 1:13
Example 2 3:09
Example 3 7:54
Example 4 13:07
Complex Numbers 29:16
Intro 0:00
Objectives 0:06
Complex Numbers 1:05
Imaginary Numbers 1:08
Complex Numbers 2:27
Real Parts 2:48
Imaginary Parts 2:51
Commutative, Associative, and Distributive Properties 3:35
Adding and Subtracting Complex Numbers 4:04
Multiplying Complex Numbers 6:16
Dividing Complex Numbers 8:59
Complex Conjugate 9:07
Simplifying Powers of i 14:34
Shortcut for Simplifying Powers of i 18:33
Example 1 21:14
Example 2 22:15
Example 3 23:38
Example 4 26:33

Welcome to www.educator.com.0000

In this lesson we are to take a look at basic types of numbers.0002

We will see that there are many different ways that we can take numbers and start to classify them.0008

I will go over all these types of numbers and more in detail as we see how a number gets into each of the groups.0012

You will also see how you can represent these numbers on a number line.0020

Be handy for say comparing numbers and figure out what it means to take the absolute value of a number.0024

We will also see some symbols on how you can compare numbers meaning to our inequalities.0030

When it comes to numbers you can really break them down into various different groups and classify them according to their properties.0039

The most common types of groups that we can use to classify numbers are the natural numbers, whole numbers, 0046

integers, rational, irrational and imaginary numbers.0052

We will go over each of these groups in more detail.0057

In our first group will take a look at the natural numbers.0064

These are the numbers that do not contain any fractions or any decimals.0068

In fact they are sometimes called the counting numbers because they are some of the first number you learn when counting.0073

They contain the numbers 1, 2, 3, 4, 5 and it does go up from there so you know how we do not have a fractions and decimals and no negative numbers here.0078

In the next group we would start expanding on a lot of that last list just little bit and we also include the number 0.0092

Since we have all of the same numbers that we have before these natural numbers and we have that number 0,0100

you could say that all natural numbers are a type of whole number.0107

Watching a step in a few different times as we go through these groups of numbers, some numbers end up in more than 1 group.0112

An important part of this was that they contain a natural numbers and 0 to be a whole number.0118

Alright continuing around, we can also expand on those numbers by looking at the integers.0126

The integer is not only includes say the natural numbers but the negatives of all of our natural numbers and 0.0133

Again this makes all of our natural numbers and our whole numbers a type of integer.0141

You will see from the list that we got some nice numbers on here like -3, -2, - 1.0148

There is 0,1, 2, 3, 4 all the way up on that side.0154

The rational numbers are probably one of our most important groups.0161

These include all numbers that can be written as a fraction.0166

Now there is many different types of numbers that you can write as a fraction.0171

In fact all the numbers that we just covered previously can easily be turned into a fraction by putting them over 1.0176

A harder one says you determine whether you can write them as a fraction or not, or the one's that involve decimals.0183

Here is how you can tell if they are rational or not.0189

If that decimal terminates that means that stops, then you know you can write it as a fraction therefore it is rational.0193

If your decimal goes on and on forever and has a repeated block of numbers, then you may also write those as fractions, they are rational.0200

To help you figure out some of these, let us look at a few examples and see why they are all types of rational numbers.0209

The first one I'm looking at here is 3/17, we know how this one is already a fraction.0217

It is a pretty obvious choice that you can write as a fraction, it is rational.0224

This number 4 could have been one of our numbers on our natural number list and it is also one that we can write as a fraction fairly quickly by simply putting it over 1.0231

Since we can write as a fraction we know it is a type of rational number.0246

Some of the more difficult one, these are the ones that involve decimals.0251

In .161616 repeating of this one goes on and on forever and ever but it has about 16 to just keep repeating over and over again.0255

That is what I mean by repeated block of numbers.0266

Since it has a repeated block, it can be written as a fraction.0269

In fact, this one is written as 16/99, I know that it is a type of rational number.0272

The next one, 0.245 and then it stops, because it stops this is a type of terminating decimal.0279

It can be written as a fraction as well that we can count up the number of places in it and just put it over that number.0289

It is tens, hundreds, thousands, written as a fraction.0296

Look for these types of numbers when determining out your rational numbers. 0300

If we know what numbers can be written as a fraction, then we must also talk about the numbers that cannot be written as a fraction.0309

These types of numbers are irrational numbers. 0315

We saw many different types of numbers that could be written as a fraction.0319

They seem like they are might not be a whole lot that you can not write as a fraction0322

but it turns out there is many common numbers that simply cannot be written as a fraction.0327

More of the common ones are roots that cannot be reduced to any further.0331

If you have a decimal that goes on forever and does not have a repeated block of numbers in it, then that is a type of irrational number.0336

There are also many famous constants which happen to be irrational numbers.0345

They show up in many different areas.0349

Looking at my examples below to see why they are irrational numbers.0351

Here when looking at the square root of 57, I know that this does not reduce.0357

That gives you decide to punch this one into the calculator.0365

You will see that is has a decimal that just keep going on and on forever, if it does not have a repeated block of number.0368

That is how I know that that one is irrational.0374

That one is a little bit more clear to see because I can actually look at its decimal and see it has no repeated blocks and yet it goes on and on forever, it is irrational.0378

It is a very curious number and the variable it is one of those famous constants.0388

Pi is equal to 3.141592 and then it keeps going on and on forever.0393

And it does not have a repeated block of numbers in it, I know that it is irrational.0401

All the types of numbers we have cover those far are actually types of real numbers.0409

There is another group that is completely distinct from those real numbers.0414

Those are the imaginary numbers and you can usually recognize those ones because no contain an imaginary part with i in it.0418

The reasons why these will be so important is some equations might only have imaginary numbers as solutions.0426

As I said before, they are completely separate from our real numbers.0439

You would not have an imaginary number that also ends up on our list for real numbers, completely different things.0443

Here are some examples of some imaginary numbers.0449

I'm looking at 2i, I see that it has the (i) right next to it.0452

Definitely imaginary, 1/2 + 5/7i, I can see that (i) is in there.0456

This is one of our complex numbers but you know it is an imaginary number for sure.0464

At the end here I have the square root of -1, I do not see any (i) in there and why it could be an imaginary number.0470

We will learn that imaginary numbers come from taking the square root of negative quantities.0478

In fact, the square root of -1 is equal to (i), it is actually is an imaginary number. 0484

To understand why some numbers get to be on multiple groups, you have to take a step back and look at the big picture for this classification.0491

I'm trying out a nice diagram so you can see what numbers end up in which groups.0500

The most important distinction that you could make between numbers is probably whether they are real or imaginary.0506

Since those groups are completely separate.0512

Those in the real category we can go further and start breaking that down into many other different types of numbers.0515

Again, we do that according to the properties.0521

The most important distinction we make is whether we can write it as a fraction, we call these rational.0524

Or whether we can not write those as fractions, we call these irrationals.0529

That is how I'm connecting things with arrows here.0535

I'm doing that to show how these categories break down.0539

Rational numbers are types of real numbers and irrational numbers are types of real numbers.0543

Continue on with those numbers that can be written as fractions, those are the rational.0550

We move on to integers.0556

You will notice at that stage at we can drop with all our fractions, we do not have decimals anymore.0559

Now we have numbers like -2 , -1, 0, 1 and we go on from there.0563

As we continue classifying them, we get to our whole numbers.0571

In these ones now, we do not have any more negatives.0576

We start at 0, we have 1, 2, 3 and we go up from there.0579

On to our primal simplest list, those are the natural numbers.0585

They start at 1,2, 3 and they go up from there.0590

Remember, these ones are known as our counting numbers.0594

One way that you can use this diagram to help you classify numbers0599

is to know that if a number ends up in one of these categories it is also in all of the categories above it.0602

We can see this happen for some of our numbers.0609

Let us take the number 2, I see that it is definitely on my natural number lists, but it is also a type of whole number.0612

In addition, is a type of integer and I can take 2 and write it as a fraction.0621

It is a type of rational number which is of course a type of real number.0627

2 gets to be in all of those categories above it.0633

I will also take one that is not in quite as many groups.0638

For example let us just take the square root of 3, it is an irrational number.0640

But it is also in a category above it, it is a square root of 3 and it is a type of real number.0646

Now we know a little bit more about the different types of numbers.0657

We will show you how you can visualize a great way to compare them using what is known as a number line.0660

On a number line, we draw out a straight line and mark out some key values such as like -3, -2, all the way up from there.0668

We put the numbers that are smaller on the left and the larger numbers on the right.0679

In this way I can make good comparisons between numbers.0688

You can see that 0 is on the left side of 3, we could say that 0 is less than 3.0692

It is handy to be able to visualize numbers in this way when looking at their absolute value.0701

The absolute value of the number is its distance from 0 on a number line.0707

It is a quick example may be looking at the absolute value of 2.0713

Since 2 is exactly 2 away on a number line, I know that the absolute value of 2 is 2.0718

We will start another one, how about the absolute value of -3.0727

That one I can see is exactly 3 away on a number line, its absolute value is a +3.0733

We might develop some shortcuts and say wait a minute, the absolute value just takes the number and always makes a (+).0741

That is okay, that is the way it should work that is because our distances are always (+).0746

As long as we can go ahead and compare the numbers, we might as well pick up some new notation for doing this.0753

You can compare numbers using inequalities and use the following symbols.0758

You can use greater than, less then, greater than or equal to and less than or equal to.0763

The way these symbols work, is you want put the smaller number with the smaller end of the inequality sign.0770

And the larger end of the inequality sign with the larger number.0780

It could say something like -3 < 5, that would be a good comparison between the two.0787

We have seen a lot about classifying numbers and comparing them.0798

Let us go ahead and practice these ideas by classifying the following numbers.0801

Let us say from the list that all of the groups that the following numbers belong to.0806

Let me start with 2/3, first I think is 2/3 a real number or an imaginary number.0810

I do not see any (i) on it so I will call this a real number.0817

Now, I need to decide can I write it as a fraction or not.0824

This one is already a fraction I know that I can write as a fraction for sure.0829

I will call this a rational number.0833

Moving on from there, in my integers those containing numbers like -3, -2, -1, 0 end up from there.0837

That is how the integers, we do not have fractions, we do not have decimals.0846

This one does not get to be in the inter group or anything below that for that matter.0849

I could say 2/3 is a real number and I could say that 2/3 is a rational number.0854

Let us try another one of these, 2.666 repeating.0861

I do not see an imaginary part so I will say that this is definitely a real number.0865

We can not write it as a fraction, why do you see it has a repeated block of numbers that goes on and on forever, it is a type of rational number.0871

What else can I say? Is it an integer? No, it has the decimal part on it, it is not an integer.0883

I will leave that one as it is, moving on, the square root of 3.0892

This is a type of real number, it does not have any imaginary part on.0898

Can we write this one as a fraction or not? This one I can not.0903

In fact, when you look at the decimal, it goes on and on forever and it does not have that repeated block of numbers, irrational.0907

Since we do not have any more distinct groups of below irrational, we will go ahead and stop classifying that one.0918

Onto some other numbers, -5 that is a type of real number.0924

It looks good, can we write as a fraction?0931

You bet we will simply put it over 1, it is rational.0934

Is it an integer? it does not have any fractions, it does not have any decimals, I will say that it is an integer.0941

Is it a type of whole number? that is where I need to stop.0952

Whole numbers do not contain negative numbers.0956

-5 is real, it is rational and it is an integer.0959

On to the number 0, this one used to be in a lot of different groups.0965

0 is a type of real number.0970

You can write it as a fraction, we will say that it is rational.0975

It is a type of integer, since it is in between our negative numbers and our positive numbers.0983

It is definitely a whole number.0993

That is where this one stops getting classified because the natural numbers start at 1 and then go up from there.1002

One more, let us classify 9, this one is a type of real number.1009

We can write it as a fraction, I know that it is rational, it is definitely on our list of integers.1016

It is also a type of whole number and we can go just a little bit further with this one.1028

This is a type of natural number.1035

9 used to be in a lot of different groups.1039

It is a type of real number, a rational number, it is an integer, it is a whole number and is a type of natural number.1041

Let us try this in a slightly different way.1050

Here I have a giant group of numbers, we want to list out whether the numbers in some of our various different groups like imaginary, real, or irrational.1052

That way we can think of visualizing, classifying them in just a slightly different way.1060

Let us start out with the first one.1067

I want to figure out all the groups that -7 belongs to.1068

I know that it is a type of real number, let us go ahead and put it into that group.1072

Can we write this as a fraction or not, yes I can write it as a fraction.1078

Let us put it in our rational category.1081

Is it a type of an integer? Yes it is on my integer lists.1086

Is it a type of whole number? No, because our whole numbers do not contain negative.1091

We will stop classifying that number.1096

Let us try another one, negative the square root of 3, that is another type of real number.1098

However, that one I can not write as a fraction.1107

I better put it in the irrational category and then that one stop.1111

Moving on, -0.7 it is a type of real number.1117

This one can be written as a fraction, it is -7/10.1125

Let us go ahead and put it in our rational category.1129

Can we go any further from there?1134

Unfortunately not, because it contains those decimals and integers some contain decimals.1136

We can stop classifying that one.1142

Moving on to 0, 0 is a type of real number.1146

It is a type of rational, it is a type of integer and it is a type of whole number.1151

It gets to be in a lot of different groups.1160

Remember, it is not a natural number since that starts at 1 and goes up.1162

On the 2/3, that one is definitely a real number and since it is already a fraction, I know it is a rational number.1167

It is not an integer since it is a fraction, 2/3.1178

The square root of 11, it is a real number, it does not contain an imaginary parts.1187

This one cannot be written as a fraction and I will put it in the irrational category and then stop classifying that one.1194

On to our famous number here, pi.1202

Pi is a type of real number, even though it is a little unusual, it does go on and on forever.1207

It is a type of real number and it is irrational since I cannot write it as a fraction.1212

We will stop classifying them since there is no two groups below irrational.1220

On to the number 8, this one is going to be in a lot of different groups.1226

It is a type of real number, I can write as a fraction by putting it over 1.1231

It is on our integer lists, it is on our whole number list and it is a type of natural number, a lot of different things.1237

On to 15/2, I will say that that is a type of real number.1247

I can write it as a fraction, let us put it in our rational category.1254

Unfortunately it is not an integer so I will not put it in that one.1259

Then number12, 12 is a type of real number.1266

We can write it as a fraction by putting it over 1, let us put in rational.1271

It is a type of integer, it is a type of whole number and since the natural number starts at 1 and then goes 2, 3, 4.1276

All we have from there I know that it is a natural number.1285

Just one more to do, the number 3i.1291

I have to throw an imaginary number on my list so it will immediately drop that into the imaginary bin.1295

And that is all the more classifying we will do with that one.1301

Since again imaginary numbers and real numbers are completely distinct from one another.1305

What you will know is that most of these categories are all types of real numbers.1310

We have classified numbers, what gets better about comparing them on a number line or just being to plot them out.1318

The way we plot out a number on a number line is we find it.1325

Say using one of our markers below and put a big (dot) to where it is.1330

If I want to graph something out like 3 on a number line, I will find 3 and I will place a big old dot right at 3.1333

Once I applied it out, I can do some good comparisons.1342

We can see that since 3 is to the left of 4, that 3 is less than 4.1346

Since 3 is on the right side of -1, 9, 0, 3 is greater than -1.1355

Let us spot out a few more, -2 on our number line.1361

We would find -2 and go ahead and put up the big old dot there.1366

When it gets in to fractions and decimals it does get a little bit more difficult but you can still put these on the number line as well.1371

This one is 5/3 and I do not see any 5/3 in my markers here on the bottom.1378

What I can do is I can break down each little section into thirds and mark out the 5th one.1384

1/3 and more thirds and more thirds.1390

We are looking for 5/3, 1,2, 3, 4, 5, we put that big dot right here.1395

Now we can better compare where 5/3 is into other numbers.1403

5/3 < 2 but it is greater than 1.1407

Alright, -3.75 that would be the same as -3 and 75/100.1414

That can also be written as -3 and 3/4.1425

That tells me I need to break down my number line into quarters.1430

1/4, 1/4, 1/4 and 1/4.1436

I'm looking to mark out 3 whole sections and then 3 quarters.1446

And we are going the negative directions 3,1, 2, 3 and we will put up the old dot there.1451

We can see that -3.75 > -4 and it is also less than a -3.1458

Let us use our number lines so that we can actually line up various different numbers and see which ones are smaller than the other ones.1470

Be just a rough sketch of the number lines, I'm not going to be too accurate with my thick marks.1481

But I just used it so I know how they compare to one another.1487

Let us go ahead and start with our first number here and put -7 on a number line.1493

Since it is a negative number, I'm going to aim for somewhere on the left side here -7.1498

-3 is a little bit more than that, I will put it on the positive side over here.1508

Let me put a spot there for 3.1515

-0.7, that is not very big and is definitely larger than -7 and less than 3.1519

Let us go ahead and put it right here - 0.7.1527

0 is a good number and put it greater than -0.73.1539

2/3 is larger than 0, I will put on the right side.1550

Alright on to something little bit trickier, the square root of 15.1561

I know that that is less than 4, since the square root of that 16th is something a bit larger will be on the right side.1566

It is greater than 3, since the square root of 9 would be 3.1573

I'm going to put this one larger than 3, square root of 15.1577

It is a good one, definitely larger than square root of fifteenths.1587

-7/2, that one is about -7 1/2, I mean -3 1/2.1597

Let us put that one down here -7 1/2, -5 and one more number pi.1606

3.1415 a little bit larger than 3, put it a little bit larger than 3.1625

Now that we have used our number line, it gets some comparisons among all these.1636

We will simply list them from smallest all the way up to largest.1641

-7, -5, -7/2, -0.7, 0, 2/3, 3, pi, square root of 15 and 8, not bad.1646

For this last example, we will go ahead and use our inequality symbols like less than or greater than to go ahead and compare these 2 numbers.1670

If you want you can use a number line to plot them out before using these symbols.1676

Let us try the first one, comparing 6 and 2.1684

When I plot these out, 2 is on the left side of 6.1690

I know that 2<6, it is my smaller number, I will drop my inequality symbols so that I show that 2 is less than 6.1695

I can also say that 6>2.1705

Let us try another one, -7 and 5.1711

It is tempting to say that -7 is bigger but our negatives are on the left side and our positives are on the right side.1722

You can see that -7 is less than 5.1730

Let us write that out, -7<5.1735

-5 and -3, -5 is further down the -3, I know that it will be less than -3.1745

One more 2.3 and 5.7, 2.3, 5.7 will be much larger.1765

I know the 5.7 > 2.3 or in the order that they are in 2.3 <5.7.1780

These symbols are handy and in showing the comparison especially where they are on a number line.1790

One thing I did not use here is the or equals to symbol.1796

I could have put that in for all of these spots, 6 is greater than or equal to 2.1802

Or I could have said -7 is less than or equal to 5.1807

That is because it also takes into the possibility that the numbers could have been equal.1812

The reason why I did these is I can see that all of the numbers are not equal.1819

And it is a little bit more flexible when using this other one.1826

Watch for the or equal to symbol to show up when we are doing a lot of our inequalities, these ones are good.1831

Thank you for watching www.educator.com.1838

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at operations on numbers.0002

When you hear me use that word operations, I'm talking about ways that we can combine together numbers or do something to a number.0007

It gets very familiar things such as adding, subtracting, multiplying and dividing.0015

As well as exponents and square roots.0020

To combine numbers together, we use a lot of familiar operations in order to do so.0028

Again, adding, subtracting, stuff like that.0033

Note that depending on the types of numbers being used, certain rules applies that will help us put them together.0036

The rules that I will focus on are the ones that involve positive and negative numbers.0042

How should we deal with those negative signs? Let us see what we can do with addition.0048

When you add numbers that have the same sign then you are looking at adding their absolute values together.0055

The results if they do have the same sign or have the same sign as the original numbers.0062

In other words, if you are adding the other two positive numbers its result will be positive.0068

For adding together two negative numbers, then your result will be negative.0073

Now if you happen to have two numbers that you are adding together and they are different in sign,0079

then we will actually handle these using subtraction.0084

Wait for the rules on what to do with positive and negative on my next slide.0087

When dealing with subtraction and what we want to do is subtract the number that is smaller in absolute value0095

from the number that is larger in absolute value and looking at each of their absolute values taking the smaller one from the larger one.0102

When looking at the final result, the answer will have the same sign as whichever number was larger in absolute value.0111

If your larger number was or the larger number in absolute value is negative, then the result will be negative.0119

Remember that we will also think of adding negative numbers using subtraction in this way.0128

In different way we are writing it but instead we are really using subtraction.0134

The other two familiar operations that we have are multiplication and division.0142

The way we handle a lot of signs with these ones are by thinking of this.0147

If we multiply a negative × a negative then the result will be positive.0153

A negative times a negative results positive.0158

If we take two things that are different in sign such as negative times a positive, then the result will be negative.0164

Now these are the same two rules that we end up using for division.0175

If we divide a negative by a negative, you will get a positive.0180

Negative divided by negative result positive.0184

If they are different in sign such as negative divided by positive, then the result is negative.0189

Keep these in mind when working with that your signs, multiplication and division.0198

When we have an exponent, we can think of it as taking a number and multiplying it by itself in a certain number of times.0210

For example, maybe I have 37, 7 would be our exponent.0218

I can interpret that as multiplying that number by itself 7 times or some other key vocabulary you want to pick up on.0224

The exponent is that number raised next of the 3, exponent.0236

In the base of the number is the number that we are actually being repeatedly multiplied, base.0244

Once we see what number that is and how many times we need to multiply it, usually we can go ahead and simplify from there.0252

There are some common exponents that we usually give other names.0258

If I'm taking 5 and I'm raising it to the power of 2, 5 × 5, this is often said 52.0262

Another good common one would be something like 23, 2 × 2 × 2 that would be 8.0278

I could also say that this is 23.0290

Key on these special words for some of these other powers.0294

Another operation that we can do with numbers is taking the root of a number.0303

The principal square root of a number is the non negative number (n)0309

that you know if I were to say multiply it by itself, I would end up giving that (n).0313

This seems a little funny special worry but let us see if I can describe it using (n) as an example.0319

Let us say I wanted to find out the square root of 25.0326

What I'm looking for is what number would multiply it by itself in order to get a 25? That has to be 5.0330

It is what I'm talking about here, positive number such that when it is multiplied by itself you get that number underneath the root.0340

For this reason we have a little bit of a problem with our negatives underneath the root.0353

After all, what number would you multiply it by itself in order to get -16?0360

We learned from our rules of positive and negative numbers that -4 and +4 would work to get 16 0364

but unfortunately they are different in sign and we need them to be exactly the same, that is not going to work.0373

We are dealing with the principal square root because we are only interested in the positive numbers0382

that when multiplied by themselves would give us that number.0387

A perfect square is a number that is the square of a whole number and this one is usually reduced very nicely.0393

For example the square root of 9 would be the example of a perfect square, as it reduces down to a nice whole number, 3.0401

If you have a (a, b) being non negative real numbers, 0416

then there are a few different ways that you can say combine or rip apart those roots.0421

3 multiplied underneath the same root and you can apply the root to each of those pieces.0426

If you are dividing and you have a root then you can put it over each of its pieces in the numerator and in the denominator.0432

You can use these rules in two different ways to simplify or combine words together.0442

We will see that a lot in some future lessons.0446

I'm pointing this out now so you do not make a common mistake.0449

Do not try and split up your roots over addition or subtraction.0453

We do not have a rule to do that yet or a good way to handle it.0458

In fact, in this example I have written below you can see that the two are not equal by simply evaluating each side.0461

9 + 16 would be a 25 and the square root of 25 is 5.0469

Looking at the right side square root of 9 is 3, square root of 16 is 4.0475

I'm putting those together I will get 7 and you can see that these things are not the same.0482

Be very careful when working with your roots.0492

Now that we know a few things about combining these, let us go through some examples and just practice with them.0498

The first one I want to add together a -3 and a -6.0504

I'm adding together two numbers that have exactly the same sign.0510

I will look at their absolute value and add those together.0514

The absolute value of 3 is 3, the absolute value of -6 is 6, if I combine those together I will get 9.0519

Since I'm adding together numbers that have exactly the same sign, the result will also have the same sign.0531

Adding two negative numbers my result is negative, -3+ -6 is a -9.0537

Moving on, 19 + a -12, I want to think of this as a subtraction problem since their signs are different.0545

How do I handle subtraction? Again I will look at their absolute value.0558

The absolute value of 19 and the absolute value of a-12, 19, 12.0562

I will subtract the smaller number from the larger number 19-12, what will that give me? I will get7.0571

I have to determine what sign this should be.0584

In subtraction we take the same sign as the larger number.0588

19 was larger in absolute value, it was positive, I know my result is 7.0594

If we do get a positive result as our answer, we do not write that positive sign up there.0604

We are just having 7, we will assume that that is a positive 7.0609

-8-11, an interesting way we can look at the problem, we could look at this as adding a -11.0614

The reason why I have looked at it in that way is that I could use my rules for addition.0626

If I'm adding things that have the same sign, I have looked at the absolute value of each of them, 8, 11.0631

Then I can simply add up those two values and get 19.0640

Since I'm adding two negative numbers, I know my result will also be negative, -19.0645

Let us try another one, 8 - -13.0651

When you subtract the negative this is another good situation that you could end up rewriting in a much simpler form.0657

When you subtract the negative, you can change it into addition.0662

This is 8 + 13 and now I'm adding together two positive numbers.0667

You made a positive 21 and one more, negative the absolute value of a -4 + 9.0673

Let us start inside those absolute values and see what we can do.0683

I'm adding together two things but they have different signs.0687

Let us look at the absolute value of a -4 and the absolute value of 9, 4, 9.0692

We want to subtract the smaller value from the larger value, 9-4 and that result would be 5.0699

I know what sign should that have, or the number that is larger in absolute value is 9.0708

And that was a positive value over here, I'm looking at a 5.0716

There are a lot of other things I have left out here so far, 0721

those would be the absolute values and that leading negative sign.0724

Let us go ahead and put those in there now.0727

I want to do the absolute value of 5, all that is simply 5.0731

I still have that negative sign hanging out front and it is been there since the very beginning.0740

I can see that after done evaluating this one all the way, my answer is actually a 5.0743

Let us work on adding, multiplying and dividing the following numbers.0752

We only have one rule to take care of that is when we are multiplying together two negative numbers, we get a positive.0757

If we are multiplying together two numbers that are different in sign, the result should be negative.0766

4 × -7, I just want to think of 4 × 7 that would be 28.0775

Since they are different in sign, I know that this will be a -28.0784

Moving on, -6 × -5, 6 × 5 would give me 30 and now here I have a - × - I know the result will be positive.0789

But again we usually do not write that positive sign in there so just leave this as 30.0801

-12 ÷ 3, 12 ÷ 3 would be a 4, negative ÷ positive would be negative, our result is a -4.0806

One last one, -60 ÷ a -5, 60 ÷ 5 goes in there 12 times and negative ÷ negative is positive, my result is a 12.0820

You need to be very careful with these rules for multiplication, make sure you have these memorized.0835

Let us do a few involving our exponents.0843

Remember we can think of these as repeated multiplication.0846

2/34, it can be the same as 2/3 × 2/3 × 2/3 × 2/3, we are doing it 4 times.0850

I simply multiply it across the top and across the bottom 2 × 2 × 2 × 2 would be 16.0862

Then I have 3 × 3 × 3 × 3 = 81.0871

All the numbers here are positive so I know my result is positive.0876

Here is a tricky one, -5 ×-5 × -5, let us take this two other times so we can what is going on here.0881

Here I have a -5 × -5 the result of taking 5 × 5 would be 25.0896

Taking a negative × negative I would know that this would be 25.0904

That looks good, let us go ahead and work in this last value of -5.0910

We want to multiply 25 × -5, the result there would be 125.0918

Since I'm multiplying a positive × a negative result is -125, -125 would be my answer.0929

The next one looks very similar but it is actually very different.0939

That one is 23 and that negative sign is just out front of all that.0943

We recognize that the 2 is the base and that the negative is not included in that base0949

since there is no parenthesis given to group it in that way.0956

We want to figure this one as 2 × 2 × 2, I have multiplied it out three times.0960

And as for what to do with that negative sign, if we have not put it up front it is a long pretty ride.0966

2 × 2 × 2 that would give me an 8, all of those numbers are positive, 8.0973

Of course let us put our negative sign out front since it was out front at the very beginning.0979

We can see that our result is -8.0984

Now on to some square roots, these ones we simply just want to break them down and simplify them as much as possible.0990

The first one I have the square root of 64.0998

Think for yourself what number would you have to multiply it by itself in order to get 64?1001

I have a couple of options that could be 8 and 8, that would have given me 64.1008

Or it could be -8 × -8, that would also give me 64.1013

We are only interested in the positive values that do so, let us not worry about those -8.1018

We will say that our answer for the square of 64 is 8.1024

Moving on, the square root of 169 ÷ 81.1031

This one we can use one of our rules and break up the root over the top and over the bottom.1037

Now we can end up taking the root of each of these individually.1045

What number multiplied by itself would give us a 169? That have to be a 13.1048

A number that would be multiply by itself to get an 81? 9.1055

The answer in this one is 13/9.1060

Continuing on, this one has a negative sign out front but that is not underneath the root.1064

I would not worry about it just yet instead let us just focus on the square root of 36.1071

That would be 6, of course we will go ahead and put our negative sign and see that our final result is a -6.1079

One more, this last one involves the square root.1091

What two numbers when multiplied together would give us a -49 remember they must be the same.1095

We got a few problems, do not we? If I try and use 7 and 7 that would give me 49, that does not work.1102

If I try and use a couple of-7, that does not work, that still gives me 49.1109

I can not use one positive and one negative even though those give me a -49.1114

Those are not the same sign, one is positive and one is negative.1121

What is going on here, if you remember about your types of numbers, these are imaginary numbers.1125

I will leave that one just as it is until we learn about simplifying imaginary numbers in some later lessons. 1137

Some various different ways that you can go ahead and combine numbers using some very familiar operations.1144

Remember that most of the rules that I covered will give you some tips on what to do when they are different in signs.1150

Positive, negative, negative, positive and all those will be handy in figuring out the overall sign of your answer.1155

Thank you for watching www.educator.com.1166

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at the order of operations.0002

As we will see the order of operations is a great way that we can start combining numbers and figure out what we should do first.0009

This one involve things like what should we do at parentheses and exponents0015

and when should we do our multiplication, division, addition and subtraction. 0020

When trying to simplify much larger expression with many different types of operations present, we have to figure out what to do first.0027

Our order of operations gives us a nice run back on what we should be doing.0037

The very first thing that we should do is work inside our grouping symbols.0042

It means if you see parentheses or brackets work inside of those first.0047

Then move on to simplifying your exponents, things raised to a power.0052

Once you have both these in care, move on to your multiplication and division.0058

If you see lots of multiplication and division next to each other, remember to work these ones from left to right.0063

Now you have to do any remaining addition and subtraction.0072

And again when it comes to which of those is more important simply work those from left to right as well.0074

One handy way that you can remember of this entire list of that is the order of operations is to remember PEMDAS.0081

PEMDAS stands for parentheses, exponents, multiplication, division, addition and subtraction.0090

Let us try it out.0096

A great way that you can remember these is Please Excuse My Dear Aunt Sally.0105

I often heard a lot of my students use that one to make sure that they came out straight.0110

Be very careful if you are using this to memorize what to do first because sometimes when using it, it looks like multiplication is more important.0120

But work these ones from left to right.0129

The same thing applies to your addition and subtraction, work those from left to right.0135

Sometimes you will deal with a larger expression that has a fraction in it.0150

Even though you might not see some grouping symbols, think of the top and bottom as their own group.0157

That means work to simplify the numerator and get everything together up there.0163

And work to simplify your denominator, get everything together down there before we continue on with the simplification process.0166

As a quick example, let us look at this slide.0173

We have (-2 × 5) + (3 × -2) / (-5-3).0176

I'm going to work on the top part as its own group, and the bottom part as its own group.0183

Let us see what does this, -2 × 5 would give me -10, 3 × -2 =-6.0191

On the bottom in that group I have that -5 -3 =-8.0202

Okay -10 - 6=-16 and on the bottom I still have a -8.0212

We worked to look inside each of those groups and simplify them using our order of operations in there.0220

I simply have a -16 / -8 and that is a 2.0226

Watch for those large fractions to play a part.0232

Let us try some examples now that we know more about the order of operations and see how we can bring these into a much simpler expression.0237

This one is ((5 - 2)2 + 1))/ -5, we also write down PEMDAS.0245

This will act as our roadmap as we are going through the problem.0257

I want to look for grouping symbols or parentheses to see where I need to start.0261

That 5 - 2 looks like a good area, we will do that first, 5 - 2 is 3.0266

The only other grouping that I'm concerned with is the top and bottom of the fraction.0278

There is only one thing on the bottom so I'm just going to now focus on the numerator.0283

I can see that I have some exponents, I have a 32 in there.0290

And now let us do that, 32 is 9, it is getting better.0294

I want to move on to my multiplication and division.0302

Looking at the top and bottom of the fraction individually I do not see any multiplication or division, I can move on.0306

Addition and subtraction, why I do have some addition on the top, I put those together to get 10/-5.0314

We are looking at 10 ÷ -5 and now I can say that my result is a -2, this one is done.0322

You can see how we move through that order of operations as our road map.0330

In this next one we want to evaluate a (-12 × -4/3) - (5 × 6) ÷ 3, let us go over the map.0338

I do not see too much in terms of grouping but I do have this group of numbers over here.0354

Let us go ahead and take care of those.0361

Inside I have (5 × 6) ÷ 3, what should I do in there? I got multiplication and division.0363

Those ones remember we are working from left to right.0370

On the left side there I have multiplication then we actually do the division.0374

5 × 6 is a 30, now do the 30 ÷ 3 and get 10.0383

We have taken care of that grouping.0394

I'm just going to copy down some these other things and then we will continue on.0396

Our grouping is done, now on to exponents.0406

I do not see any exponents here so now on to multiplication and division.0410

We will do multiplication I got a -12 × -4/3.0416

A negative × a negative would give me a positive, multiplying on the top that would be 48/3.0421

Because of my fraction there, I do have some division I could take 48 and divided by 3 = 60.0434

On to addition and subtraction 16 – 10 = 6.0444

I have completely simplified this one and I can call it done.0451

This next one I have (12 ÷ 4) × (√5 - 1).0458

Starting with my grouping symbols and parentheses, I could consider everything underneath the square root as its own little group.0469

Let us work on simplifying that, I'm writing here 5-1 is a 4,12÷ 4 × √4, taking care of the square root entirely.0475

I'm looking at 12 ÷ 4 × 2, moving on do I see any exponents? No exponents.0499

On to multiplication and division, this is that tough one.0509

It is tempting to say that multiplication is more important but it is not.0512

Simply work these guys from left to right.0516

In this case, we are going to do the division first, 12 ÷ 4 is 3.0519

Then we are actually taking that and multiply it by 2 and get 6, this one is completely simplified.0527

Let us look at our example that involves lots and lots of different things.0538

I have (8 × 4) - (32 × 5) + (2 × the absolute value of -1) / (-3 × 2/3) +1 0542

With so many different things in here we have to be careful in what to do first.0562

I'm dealing with a fraction here I want the top as its own group and the bottom as its own0567

and work inside each of those and try to simplify them.0572

Let us look at the top a little bit.0575

Inside of that I do not see any additional grouping symbols so I will try and do any exponents on the top.0578

I do have a 32, let us change that into a 9.0587

I have the absolute value of -1, might as well we go ahead and take care of that as well.0594

We are doing a little bit of simplifying on the top, let us see if there is any exponents in the bottom.0602

83 × 23 and change out into -3 × 8 and of course we still have the + 1.0607

Continuing on, looking at the top I do not have any additional parentheses, I do not have any additional exponents, multiplication and division.0621

A lot of multiplication on the top, 4 × 8 would give me 32, 9 × 5 =45, 2 × 1=2.0629

On to the bottom, -3 × 8=-24 and then +1, multiplication and division done.0644

On to addition and subtraction and we are going to do this from left to right.0655

I will do 32 - 45, what do we got from there?0660

Let us imagine our technique for combining numbers that have different signs.0665

I'm just subtracting here, I get a result of 13.0673

The one that is larger in absolute value is the -45 so my result is a -13.0677

Looking at the bottom-23 almost done.0685

-11 at the top divided by -23, this one is completely simplified as 11/23.0693

When dealing with multiple operations it is important that we do have a roadmap in order to get through all of these.0705

Feel free to use PEMDAS also that you keep everything in order.0711

As you use PEMDAS, if you get down to your multiplication and division then use them from left to right.0715

If you get down to your addition and subtraction, again use those from left to right.0720

Thank you for watching www.educator.com0725

Welcome back to www.educator.com.0000

In this lesson we will take a look at the properties of real numbers.0002

Some of these properties involve the commutative, associative, distributive, identity, inverse and 0 properties.0008

It seems like that is quite a bit to keep track of but once we get into the nuts and bolts of it, you will see it is not so bad.0015

There are several properties that can help you put different numbers together.0028

Some of the very first properties are the commutative property and the associative property.0034

We will first deal with that commutative property first.0041

What the commutative property says is that the order of addition simply does not matter.0044

It also says that the order of multiplication does not matter.0050

It seems like you should, but a quick example shows that actually it does not make much of a difference.0054

For example, if I have 2 × 3 × 4, I'm just multiplying all those numbers together.0060

I actually do this from left to right, this would be 6 × 4, I will get 24.0070

Since the order does not matter, I can feel free to scramble up that order a little bit just like that.0079

3 × 4 × 2 and again do these from left to right, 3 × 4=12 × 2.0087

12 × 2 you will see that the answer is still the same.0095

When you have addition or multiplication, the order you have a little bit of freedom with, you can scramble them up just a little bit.0102

With the associative property, we are also changing something and we are saying that the grouping does not matter.0110

If you are dealing with the associative property of addition then changing the grouping of addition does not matter.0117

The associative property for multiplication then the grouping of multiplication does not matter.0124

Let me take a quick look at a nice quick example to see what I mean by the grouping.0129

Let us look at (2 + 3) + 4, here I'm adding a bunch of different numbers.0135

Up with parentheses to show that I'm grouping the first two together.0143

According to our order of operations, we would have 2 and 3 together first.0147

This would give us a result of 5 + 4, giving us a final result of 9.0153

If I take those same numbers and the same exact order and this time I decide to group together the 3 and the 4 together,0160

you will see that you will get exactly the same answer.0170

We will do what is inside parentheses first, 3 + 4 is 7 and then we will go ahead and we will put that 2 with the 7.0173

I'm sure enough you will notice that our answers are exactly the same.0184

The associative property says that the grouping for addition and multiplication does not matter.0188

The distributive property is a neat one, it deals with multiplication and addition.0203

It says that we can distribute our multiplication over addition.0207

We looked at several different examples of this but one of the classic things is something like this.0212

I have 2 × (3 + 4), according to the distributive property I will take this multiplication with the 2 and distribute it over addition.0221

I actually multiplied it by the 3 and then multiply it by 4.0233

This will be 2 × 3, this will be 2 × 4 and I can take care of each of those individually.0239

Get a 6 + 8, combine those together and get a 14.0248

Just to highlight that this property is valid, I will also do the same problem using my order of operations to take care of the part inside parentheses first.0256

3 + 4 is 7 and 2 × 7 is 14, I'm sure enough just like we should we get the same answer.0268

This is a unique one involves multiplication and addition.0281

Some other properties that you want to keep in mind have to deal with 0 and 1.0288

With the division of property of 0, we say that 0 divided by any number we get the results is 0.0295

If we are trying to divide by 0, that is something that we can not do.0305

For that one we say that it is undefined.0311

Be very careful for this one, many people will try and divide by 0 later on, simply can not do it.0315

When it comes to 1, 1 is a little bit easier to deal with.0323

When you divide a number by 1, the number remains unchanged, nothing happens.0326

Think of 5 ÷ 1, you will still get 5, or 7 ÷ 1 =7.0332

A number divided by itself is 1, that will be like 5 ÷ 5, number divided by itself results to 1.0338

All of these properties seem like they might be arbitrary, but all of them play a very important role.0348

On to the multiplication property of 0 and let us see what we can do with that.0360

With the multiplication property of 0, when you multiply it by 0 the result is simply 0.0366

Think of stuff like 3 × 0 result 0.0372

Multiplication property of 1 says that when you multiply a number by 1, the number remains unchanged.0377

3 × 1 now 3 stays exactly the same, it is still 3.0383

With the addition property of 0, when you add 0 to a number, the number remains unchanged, 3 + 0 is still 3.0389

These last few properties that deal with 0 and 1, they can be a little confusing to keep straight.0400

But again the most important is you can not divide it by 0, watch out for that one.0407

You may be thinking that all of these properties and say okay what is the big deal?0416

Why do I necessarily need to know these? 0419

Will they become extremely important when you are manipulating large expressions, especially when you get into a lot of the solving stuff later on.0421

What they do is they allow us to take an expression change in many different ways without changing what it is.0429

For example, if I want to look at an example of the students work,0436

what allows them to do many of these steps is some of those of properties that we covered earlier.0441

In fact, let us take a close look at this and see what properties we will use.0446

In the very first part of this I see (7 + 2) × (x + 9) and in the next part I know that the 2 is moved inside parentheses and now I have 18.0451

What happened there is they took the 2 and they distributed it over the x and the 9.0462

What allows a person to do that part of this work is our distributive property.0469

In the next little bit of work, I see that they have actually switch the order of the 2x and the 18.0478

They have moved around where things are.0483

Are they allowed to do this? The answer is yes.0487

We are only dealing with addition right there so they can change the order just fine.0489

This is our commutative property, let us see what is going on in the next step.0496

In the next step, everything is in the same exact order but now the parentheses have moved into a different spot.0504

Now they are actually around the 7 and 8.0510

Since only addition is being shown here, then I know that this is a valid step, this is our associative property.0514

It looks like they did some simplifying to combine the 7 and 18 together and they switched the location of the 25 and 2x.0527

It looks like in that one they have used the commutative property again.0536

The big thing to take away from this is that we will be manipulating expressions quite a bit.0541

These properties that work in the background that allows to do many of these manipulations.0546

Switching the order of things or switching our grouping as we go along.0551

It is time to get into some examples and see what we can do about identifying many of these great properties.0557

Here I have lots of different examples and we just simply want to identify the property that is being used.0564

The first one I have 6 + - 4 = -4 + 6, carefully look at that and see what is changed.0570

One thing that picked up in my mind is I'm looking at the order of things and the order has been switched.0579

I will call this my commutative property.0586

Since this is dealing with addition, I could take this a little bit further and say this is the commutative property of addition.0598

The next one let us see what we got, 7 × 11 × 18 = (11 × 8) × 7, it is tricky here.0617

Let us see what is changed, the grouping is actually exactly the same.0626

The 11 and 8 are being grouped together and over here the 11 and the 8 are still being grouped together.0632

It has nothing to do with grouping.0637

This is another one where the order has changed.0639

Since the order has changed, it is still our commutative property.0642

This one has nothing to do with addition.0651

It is actually multiplication so I will say commutative property of multiplication.0652

Nice, I like it, next problem (11 × 7) + 4 = (11 × 7) + (11 × 4) more interesting.0663

This one has multiplication and addition, in fact it looks like they took the 11 and multiplied it by the 7 and by the 4.0672

We can recognize that as our distributive property.0681

That is the unique one, it has both multiplication and addition.0694

Continuing on, pi × √2 =√2 × pi, someone has been messing around with the order of things, another one of my commutative properties.0699

Since I have multiplication here, commutative property of multiplication.0720

One more for this example 6 + 2 + 7 = 6 + 2 + 7.0728

We noticed with that one all the numbers are in exactly the same order.0736

The order is untouched but what is different here is that the parentheses are showing up in an entirely different spot.0741

They are changing the grouping, this is our associative property.0749

If you change the order of the numbers present, then you are dealing with your commutative property.0773

If you change the grouping, you have your associative property.0780

With these ones we have some sort of property present and we want to fill in the blank so that the property ring is true.0786

The first one it looks like they are trying to display the commutative property.0796

I have 3 + 4 + 5= 3 + 5+ _, the commutative property changes the order of things.0799

It looks like we are changing the order of the 4 and the 5 here.0808

I see I got the 5, we will put the 4 in there and that would be a good example of the commutative property.0812

I technically, the commutative property of addition.0818

The next one is the associative property that one changes grouping so the order should stay exactly the same.0821

Let us write the same exact numbers in the same exact order,4, 5, and 3, it looks good.0829

Now I can see that since the parentheses are in a different spot, the grouping has changed.0837

Alright distributive property, that one involves multiplication over division.0844

I can see that the two has been taken to the 7 and the 3, I'm just missing the 7.0856

We have a good example of all 3 properties.0866

Example 3 deals with our division by 0, remember that we can not divide 0.0873

It is important to recognize what numbers can we sometimes not use, those are known as restricted values.0878

With these ones we are going to try and come up with a number that our letter x in this first one can not be.0886

Let us say we do not want the bottom to be 0, if x was 4 then what we will have on the bottom is 4 - 4.0896

That would definitely give us a 0.0911

In terms of our restricted value, we would say that x can not equal 4, since 4 is restricted.0915

It can not be 4 when we get that 0, let us try the next one.0926

Looking at the bottom of that fraction, I have 5 + a, what would (a) have to be to give me a 0 on the bottom?0931

I'm adding to 5 and thinking about 0 but it is possible if we start thinking negative numbers.0942

In fact what happens if a =-5, based on what you have seen there on the bottom would be 5 - 5 and that would give you 0.0948

My restricted value, we would say that (a) can not be -5, I want that 0 on the bottom.0959

Let us try this one, it looks a little bit more difficult.0968

What can (y) not being what is restricted?0973

I will be subtracting 1 from it but I'm also multiplying by 4.0980

The one thing that is going to do it, I have to borrow my fractions but 1/4 is actually my restricted value.0988

To see this, I'm thinking of taking 1/4 and multiply it by 4, that gives you 1.0995

When we subtract 1 you get 0, you know that (y) can not equal a 1/4.0999

Keep in mind that you can not divide by 0 and sometimes there are simply values that are restricted, you can not use them.1009

One last example here, we will look at the distributive property and how we can use it to simplify some expressions.1020

You may have see me use these lines on top of the number here, actually show the multiplication of what I need to do.1028

Definitely a good practice you should adopt.1035

In the first one, I looked at 2 being multiplied by (a) and 2 being multiplied by (k).1039

This will give me (2 × a) + (2 × k), I will leave that one as it is.1046

Moving on to the next one, here is -5 × (4 - 2x), this one used multiplied by both parts inside those parentheses.1056

-5 × 4 =-20, we will do that first, -5 × -2x.1069

Negative x negative is a positive, we will say10x.1077

One more, you may be looking at that one the same way then why are we not using the distributive property.1085

All I see is a negative sign but imagine that as -1 and you will see that the distributive property will come in handy just fine.1090

We will take -1 and distribute it among both parts on the insides of those parentheses.1098

-1 × -2 = 2 and -1 × 10q= -10q.1104

In all of these examples, we want to take the multiplication and distribute it out over addition.1117

It even works for some of these subtraction problems here because subtraction is just like adding a negative number.1122

Thanks for watching www.educator.com.1130

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at how you can multiply and divide rational expressions.0003

There is a lot to know when it comes to rational expressions.0011

We will first start off by looking at what values would be undefined or restricted in some of these rational expressions.0013

We will also take quite a bit of time to figure out to how simplify them and put them into lowest terms.0021

There will be a few situations where you have to recognize that there are equivalent forms of these different rational expressions.0028

They may look different but they are the same.0035

Finally we will get into that multiplication and division process so you can see it all put together.0038

A lot of things that we will do these rational expressions ties into rational numbers in general.0046

In terms of the techniques and how we simplify them.0052

Remember that a rational number is any number that can be written as a fraction.0056

The number 2/3 is a good example of one of these rational numbers. 0061

We are working with rational expressions you can consider those of the form p/q when p and q are both going to be polynomials.0066

Let me give you some examples of what these rational expressions look like.0076

2x + 3 ÷ 5x2 - 7.0080

Since both of those are polynomials and they are simply being divided that will be a good rational expression. 0086

How about x3 - 27 ÷ 4x + 3? That will also be a good example of another rational expression.0093

What this rational expression since we have some variables in them, we are always concerned about dividing by 0. 0107

We do not want use any values for x that would make the bottom 0.0113

Think of this expression up here x + 5 ÷ 2x - 4.0117

I can plug in many, many different values for my x as long as I do not get a 0 on the bottom.0123

In fact if I try and plug in a 2 into this that would give me 0 on the bottom.0130

2 × 2 - 4 and that is bit of a problem. 0136

Since it gives me that 0 the bottom this expression is undefined at 2 or I might simply make a note somewhere and say x can not equal 2.0141

For many of the expressions that you will see on the future slides here, 0151

we will assume that it can take on many different values, but not those restricted values.0156

When we are into the multiplying and dividing we will always be concerned with taking our answer down to its simplest form.0165

How is it that we go ahead and reduce a fraction to it simplest terms?0173

Remember how you do this process with some similar type of rational expressions.0178

We will try and cancel out a lot of the extra p’s that are present.0183

If I’m looking at 6x2 ÷ 2x2 I could look of this as 2 × 3 × x × x × x and in the bottom would be 2 × x × x.0188

As long as I have multiplication there I can simplify it by canceling out a lot of these extra factors and get something like 3x.0204

The good news is that type of simplifying is the same process we want to do for rational expressions.0215

We want to look for their common factors. 0222

These are polynomials we will probably have to factor first, so that we have the actual factors in their p’s that are multiplied.0226

Watch how that works for this guy 4y + 2 ÷ 6y + 3.0233

Let us factor the top and bottom.0238

On the top there is a 2 in common and I can take it out from both parts.0241

That would be 2y + 1 and on the bottom it looks like there is a 3 in common, we will take that out, 2y + 1.0246

Notice we have this common piece 2y + 1 that is multiplied on the top and bottom.0259

We can cancel that out and we would just be left with 2/3.0266

Remember that we can only cancel out factors, pieces that are multiplied.0272

If you attempted to do some canceling at the very beginning, you can not do that, not yet because you are dealing with addition.0277

Be very careful as we try and simplify your expressions.0287

Some answers may look a little different from other answers, but actually they may just be equivalent.0294

One thing that can make equivalent expression is where you put the negative sign. 0301

You could put it in the top, the bottom or out front of one of these rational expressions.0306

In all cases, they represent the same exact quantity.0313

You will say that they are all equivalent.0317

This also applies to your rational expressions, but sometimes it is not quite as obvious. 0320

For example, maybe I'm looking at x - 7 ÷ 2.0328

Let us go ahead and put a negative sign on the top.0334

If I do that, that would be -x and may be distribute it through with that negative + 7 ÷ 2.0339

I could have also given that negative sign to the bottom x – 7 ÷ 2 and give that to the bottom and that would give me x - 7 ÷ -2.0350

These two expressions are equivalent.0367

They do equal the same thing, even though they look a little different.0369

Be careful if you are working on your homework, working with other people and you do not get exactly the same answer, 0375

you might actually have equivalent expressions just inside a different form.0380

Now that we know about simplifying and some things to watch out for how do you get into the multiplication and division process.0389

I want to think back on how you do this with fractions.0396

Before multiplying fractions together and it is a nice process of multiplying across the top and multiplying across the bottom. 0400

The same thing applies to your rational expressions.0409

We will be dealing with polynomials for sure, but will just multiply across the top and across the bottom.0412

In order for this to work, we must know how to factor our polynomials. 0418

That way we can end up just multiplying their factors together.0423

When we are all done, we want to make sure that we have written it in lowest terms.0427

Try and cancel out any extra factors after you are done multiplying.0431

If you know how the multiplication process works, then you will also know a lot about the division process.0438

Think about how this works when you divide fractions.0445

From looking at something like 2/3 ÷ 5/7 and we have been taught to flip the second fraction and then multiply.0448

Which is of course, multiply across the top and multiply across the bottom.0456

There is some good news and this also applies to our rational expressions. 0460

If you want to multiply them together, flip your second rational expression and then multiply the two.0465

Factoring will definitely help in this process that way we have to keep track of the individual factors, and where they go.0472

Always write your answer in lowest terms when you are all done.0479

That is quite a bit of information just on simplifying and multiplying and dividing.0485

We will look at some quick examples and see how this works out.0491

We want to take all of these rational expressions and put them into lowest terms.0497

We will be going through a simplification process.0501

Notice how in a lot of these we are dealing with addition and subtraction, do not cancel out yet until you get it completely factored.0505

I have (x2 - y2) ÷ (x2 + 2xy + y2).0514

These look like some very special formulas that we had earlier.0520

I have the difference of squares on top. 0523

I have a perfect square trinomial on the bottom so I can definitely factor these.0525

I have x + y x – y on the top and on the bottom x + y, x + y.0533

Notice how we have a common factor the x + y.0549

We can go ahead and cancel that out. 0554

This will leave us with an x - y ÷ x + y.0558

I can be assured that this is in the lowest terms because there are no other common factors to get rid of.0565

The next ones are very tricky.0571

Notice how the top and bottom almost look like the same thing.0574

It is tempting to try and cancel out.0578

Be careful, we cannot cancel them out unless they are exactly the same thing.0581

One thing to notice here is a -5 and here is 5. 0586

Those are not the same thing, they are different in sign.0590

And same thing over here, this is w2 and this one is –w2.0593

Those are not the same in sign.0597

If you end up with a situation like this where they are almost the same, you are dealing with subtraction and the order is just reversed.0600

You can factor out a -1 from either the top or bottom.0607

If I factor out a -1 from the top then what is left over?0614

-1 × what will give me a w2, is a -w2 and let us see if I take out 5, that should do it.0620

-1 × -w = w, -1 × 5 = -5.0633

All of that is on top and I still have my 5 - w2 on the bottom.0641

We are getting a little bit closer and things are starting a matchup in sign a little bit better.0646

I’m just simply going to reverse the order of these and you will see that they are common factors.0652

-1 is still out front, (5 - w2) (5 + w2) and now I can go ahead and cancel these out.0657

The only thing left here is a -1.0677

That looks much nicer than what we started with.0680

In the next one I have 25q2 - 16/12 – 15q.0684

This one is going to take a little bit more work but I see I have one of those special cases on the top.0691

That is another difference of squares.0697

(5q + 4) (5q – 4) 0700

Let us see if we can do anything with the bottom.0712

Does anything go into 12 and 15?0714

These both have a 3 in common, let us take that out.0718

We are looking pretty good and we can see that this is getting pretty somewhere to the previous example.0725

These look almost the same we are dealing with subtraction, but the order is just reversed. 0732

We are going to take out a negative from the bottom so that they will be exactly the same.0736

(5q + 4) (5q – 4)0741

There you will take out the 3, let us take out -1 as well.0750

That will give us -4 + 5q.0753

It is better starting to match the top we just have to reverse the order.0758

Reading on the bottom 5q - 4 and now we can see we have a common factor to go ahead and get rid of.0773

The answer to this one would be 5q + 4 ÷ -3..0782

The last one involves 9 – t ÷ 9 + t.0791

In this one, another one that looks very close.0796

Unfortunately there is not a whole lot we can do to simplify it.0800

You might be wondering why cannot we just cancel out some 9 and call it good from there.0805

We can only cancel out common factors, things that are multiplied.0810

We cannot cancel out the 9 nor we can cancel out the t’s.0814

Unfortunately there is nothing to factor from the top or factor from the bottom. 0818

This one is simplified just as it is.0822

Be on the watch out for cases like this and know when you can cancel out those extra terms.0834

In this next few we are going to go through the multiplication process and then try and bring it down to lowest terms.0844

We just have to multiply across the top and then multiply across the bottom. 0850

That will make our lives a little bit easier.0854

Let us start off with this first one.0856

Multiplying across the top I will have 8x2 × 9 / 3xy2.0859

From here I can cancel a lot of my extra stuff.0873

I will cancel out extra 3 that it is the 9.0876

I can cancel out one of these x’s here and I can cancel out one of these y’s.0880

Let us see what is left over.0886

I still have 8x, 3, a single y on the bottom.0888

This is 24x ÷ y and that only multiply the two together, but I brought it down to lowest terms.0895

Onto the next one, multiplying the top together will give me 3t – (u × u) / (t × 2) × (t – u).0905

One obvious common term is that t – u, let us go ahead and get rid of that.0926

We would have left over 3u t × 2 or just make my brain feel better, 2 × t.0933

We have multiplied those together and reduce it to its lowest terms.0944

I want to point out something, back here it is tempting to go through the distribution process and put the 3 and t and the 3 and u, 0949

but actually you do not want to do that just yet. 0958

Go ahead and leave them into your factors because it will make it much easier to cancel them out.0961

If you do end up distributing them, you have to pull them back a part into their factors later on.0966

You are not saving yourself any work.0971

Leave the factors in there or if is not factored already go ahead and factor it so you can easier multiply and reduce.0974

Let us get into a much bigger one.0984

In this one we want to multiply together and then put into its lowest terms.0987

(x2 + 7x + 10 / 3x + 6) × (6x – 6 / x2 + 2x – 15)0992

This is a rather large one, but I'm not going to multiply together the tops first or the bottoms just yet.1005

I’m going to work on factoring just for little bit. 1011

How would my first polynomial here factor?1017

My first terms would need to be an x that will give me x2 and my other terms would have to be 2 and 5.1023

That will be the only way that I can get that 10 and they would add to be the 7 in the middle.1031

That would factor that and we will also factor the bottom.1037

It looks like there is a common 3, let us pull that out.1041

We have just factored that first rational expression.1047

Onto the next one I see it has a common 6, x – 1 and on the bottom I see a trinomial.1051

Let us go ahead and break that down into two binomials.1061

X2 + 2x – 15.1065

What would that factor into?1068

Now that I have all the individual factors, I can multiply them together much easier.1075

I will simply write all of the factors on the top and multiply that out.1079

All the factors in the bottom, 3x + 2 x + 5 and x – 3.1092

In that step they are all multiplied together.1102

Let us cancel out a lot of those extra terms.1105

x + 5 were gone, x + 2 those were gone.1109

Let us cancel out one of the 3 and 6.1116

The only thing we have left over is 2 and x – 1 / x -3.1120

Since there are no more common factors I know that this one is finally in lowest terms.1129

This takes care of a lot of multiplication let us get into dividing these and also putting them into lowest terms.1138

With this one needs one need one extra step, we have to flip the second rational expression and then multiply and reduce.1144

Let us start off with this first one.1154

I have 9x2 ÷ 3x + 4.1156

I will flip the second one, 3x + 4 ÷ 6x3.1162

If I’m going to multiply the two together, I need to write all of their factors, the top ones together.1173

We will write all the factors of the bottom one together.1184

Now they are multiplied.1189

We can go ahead and cancel a lot of those extra terms.1192

We have a common 3x + 4 in the top and bottom, let us get rid of that, we do not need that.1195

We cancel out an extra 3 in the top and bottom.1201

On top we have x2 and in the bottom we have x3 so an x2 will cancel out leaving me x in the bottom.1207

That is quite a bit of canceling.1216

Let us see if we can figure out what is left over.1217

I still have a 3 on top, we still have a 2 and x on the bottom.1219

3 / 2x is the only thing left when we reduce it to lowest terms.1225

The next one looks like a do- see, let us go ahead and take this one nice and easy.1231

We will first write the first rational expression, just as it is.1237

We will go ahead and flip the second rational expression.1247

4r -12 and -r2 r + 31250

If I'm going to multiply this together we will go ahead and write all of the factors on the top.1263

4r2 – 1 on the bottom 2r + 1 -r2 and r + 31272

If we multiply things together let us see what we can cancel out.1290

One initial thing, I see an r + 3, let us get rid of that.1294

I got an r on top and r2 so let us get rid of some r.1299

I think there is even a little bit more we can cancel out as long as we recognize that we have a very special polynomial on top.1305

Notice how we have the difference of squares on top the 4r2 -1.1318

We can write that as 2r + 1 and 2r - 1.1325

We can see that we do have an extra factor hiding in there that we can get rid of. 1333

We can get rid of that 2r + 1.1338

What is left over, I have 4 × 2r -1 on the top / -r in the bottom.1346

This one is completely factored.1354

I got one more example and this is another large one.1365

We will have to just walk through it very carefully.1369

I have xw - x2 / x2 – 1 then we are dividing that by x - w / x2 + 2x +1.1371

Let us see what we can do.1381

Let us go ahead and rewrite it and have the second one flipped over.1383

Multiply it by x2 + 2x + 1x – w.1395

We will go ahead and start putting things together. 1407

Let us multiply (xw - x2) (x2 + 2x + 1) ÷ (x2 – 1) (x – w)1410

This one looks like it is loaded with lots of things that we can go ahead and factor.1428

Let us go ahead and do that and we are going to do that bit by bit.1432

I’m looking at this very first factor here and notice how they both have terms in there that have x.1436

We can factor out an x from each of those.1443

That will leave us with w – x.1449

In this one over here we can factor it into an x + 1 and x + 1.1454

It is one of our perfect squares.1460

(x + 1)(x +1)1465

On to the bottom, this one right here is the difference of squares x + 1 x -1.1473

Unfortunately this guy we do not have anything special about it, I will just write it.1486

That will allow me to at least cancel out a few things.1493

I have an extra x + 1 that I will go ahead and get rid of.1496

It looks like I can almost get rid of something else.1500

Notice how we have this w – x and x – w.1503

It is almost the same thing, and it is involving subtraction. 1507

I need to factor out a -1 from one of these.1511

Let us do it to the top -w + x.1515

I still have an x + 1 on top / x – 1 and x – w.1522

-x will rearrange the order of these guys and switch them around.1537

That will give us (x – w) (x + 1) / (x – 1) (x – w).1544

We can see sure enough, there is another piece that we can go ahead and get rid of.1556

We only have one thing left, -x × x +1 / x – 1.1563

Since we have no other common factors, this is finally in its lowest terms. 1575

When it comes to working with these rational expressions, remember how all of these processes work, which is normal fractions.1580

Whether that means multiplying fractions by going across the top and bottom1587

or dividing fractions by flipping the second one.1591

Then cancel out your extra factors so you can be assured that is brought down to lowest terms.1593

Thank you for watching www.educator.com.1599

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at adding and subtracting rational expressions. 0003

In order for this process to workout correctly I have to spend a little bit of time working on finding a least common denominator. 0009

This will help us so we can rewrite our rational expressions and actually get them together. 0017

Then we will get into the addition process.0022

We will look at the some examples that have exactly the same denominator.0025

We will look at others which have different denominators.0029

You will see that it will involve many of the similar things.0036

So far we have covered a lot about multiplying and dividing rational expressions, 0044

but we also need to pick up how we can add and subtract them.0048

One key that will help us with this is you want to think of how you add and subtract simple rational number.0053

Think of those fractions, how do you add and subtract fractions?0059

One key component is that we need a common denominator before we can ever put those fractions together.0063

Think of how that will play a part with your rational expressions. 0070

Let us first look at some numbers okay.0076

Let us suppose I had 2/1173 and 5/782 and I was trying to add these or subtract them.0078

No matter what I'm trying to do, I have to get a common denominator. 0088

One thing that will make this process very difficult is that I chose some very large numbers in order to get that common denominator.0093

One thing that could help out when searching for this common denominator is not to take the numbers directly since they are so large.0101

You have to break them down into their individual factors.0108

Down here I have the same exact numbers, but I have broken them down into 3 × 17 × 23.0112

I have broken down the 782 into 2 × 17 × 23.0119

What this highlights is that the numbers may not be that different after all. 0123

After all, they both have 17 and 23 in common and the only thing that is different is this one has a 3 and this one has a 2.0128

When trying to find that common denominator, I want to make sure it has all the pieces necessary 0139

or I could have built either one of these denominators.0145

It must have the 2, 3 and also those common pieces of the 17 and the 23.0147

This is the number that you would have been interested in for getting these guys together.0153

When working with the rational expressions, we will be doing the same process.0161

We want them to have exactly the same denominator, but I would not be entirely obvious to look at the denominator and know what that is.0166

We will have to first factor those denominators, we can see that the pieces present in which ones are common and which ones are not.0176

We will go ahead and list out the different factors in the denominator first.0184

We will also list out the variables, when it appears the greatest number of times.0192

We will list the factors multiplied together to form what is known as the least common denominator.0200

Think of the least common denominator as having all the factors we need and we could have built either one of those original denominators.0206

Let us take a quick look at how this works with some actual rational expressions.0216

I want to first look at factoring the bottoms of each of these.0221

8 is the same as 2 × 2 × 2 and y4 we will have bunch of y all multiplied by each other.0225

For 12 that would be 3 × 2 × 2 and then a bunch of y, 6 of them.0236

When building the least common denominator, I will first gather up the pieces that are not the same.0245

Here I have the 3 and this one has an extra 2.0254

I need both of these in my least common denominator 2 and 3.0259

This one has a couple of extra y.0267

Let us put those in there as well. 0270

Once we have spotted all the differences between the two then we can go ahead and highlight everything that is the same. 0274

We have a couple more twos and 1, 2, 3, 4 y.0282

Let us package this altogether. 0290

2 × 2 × 2 = 8 × 3 = 24 and then I have 6y, y6.0292

Let us do some shortcuts here. 0303

One, you could have just figured out the least common denominator of the 8 and the 12 that will help you get the 24.0305

You can take the greatest value of y6 and gotten the y6.0312

We use techniques like that to help us out when looking for that least common denominator.0321

Some of our expressions may get a little bit more complicated than single monomials on the bottom. 0330

Let us see how this one would work.0335

This is 6 /x2 - 4x and 3x – 1/ x2 – 6.0336

In order to figure out what our common denominator needs to be, we are going to have to factor first.0343

Let us start with that one on the left and see if we can factor the bottom.0350

It looks like it has a common x in there.0353

We will take out and x from both of the parts.0356

For the other rational expression that looks like the difference of squares.0363

x + 4 and x – 40370

We just have to look with these individual factors. 0376

I can see that what is different is this x + 4 piece and the x piece.0381

Let us put both of those into our common denominator first.0388

I have x and x + 4 then we can go ahead and include the pieces that are common.0392

Any common pieces will only include once.0402

This down here represents what our least common denominator would be. 0406

We need an x, x + 4, and x -4. 0411

Finding the least common denominator is only half the battle. 0419

Once you find the least common denominator, you have to change both of your rational expressions 0422

so that they contain this least common denominator. 0428

Once you have identified it go one step farther and rewrite the expressions so that they have this least common denominator. 0433

Let us watch how this works with our numbers.0441

That way we could get a better understanding before we get into the rational stuff.0443

Here are these fractions that I had earlier and you will notice that the bottom is already factored.0447

Our LCD in this case was 2 × 3 × 17 × 23.0454

Now suppose I want them to both have this as their new denominator.0463

2 × 3 × 17 × 23, 2 × 3 × 17 × 23.0472

When looking at the fraction on the left here the only difference between this and the new LCD that I wanted to have is it is missing a 2.0483

I could give it a 2 on the bottom but just to balance things out I will also have to give it a 2 on the top. 0495

On the top of this one will be 2 × 2.0504

I better highlight that this 2 was the one we put in there.0510

For the other one, it needs to have that 3, I will give it a 3 on the top and there is where the 3 came from in the bottom.0515

You can see that we give the missing pieces to each of the other fractions. 0526

If I was looking to add or subtract these I will be in pretty good shape since they have exactly the same denominator.0532

We want to do the same process with our rational expressions, give to the other rational expressions its missing pieces 0538

so it can have that least common denominator.0545

We can find a least common denominator now, which means we can get to the process of adding and subtracting our rational expressions. 0551

Think of how this works with our fractions.0560

If I have two fractions and I have exactly the same denominator then I will leave that denominator exactly the same 0562

and I will only add the tops together.0570

This works as long as my bottom is not 0.0574

This will be the exact same thing that we will do for our rational expressions, the only difference is that this P, Q, and R 0578

that you see is my nice little example, all of those represent polynomials instead of individual numbers.0585

As soon as we get our common denominator we will just add the tops together. 0591

If they do not already have a common denominator, we have to do a little bit of work.0599

It means we want to find a common denominator and often times we will have to factor first before we can identify what that is. 0603

Then we will have to rewrite the expression so that both of them have this least common denominator. 0613

Once we have that then we will go ahead and add the numerators together and leave that common denominator in the bottom. 0619

Even after that we are not necessarily done.0627

Always factor at the very end to make sure that you are in the lowest terms.0629

Sometimes when we put these together we can do some additional canceling and make it even simpler. 0633

The subtraction process is similar to the addition process.0644

You will go through the process of finding the common denominator. 0650

Make sure they both have it, and then you will end up just subtracting the tops.0652

Remember though you want to subtract away the entire top of the second fraction.0660

Often to do this, it is a good idea to use parentheses and distribute through by your negative sign on the top part.0665

That way we would not forget any of your signs. 0671

It is usually a very common mistake when subtracting these rational expressions.0674

Let us look at this example and add the rational expressions.0682

I have (3x / x2 – 1) + (3 / x2 – 1). 0686

The good news is our denominators are already exactly the same.0691

I will simply keep that as my common denominator in the bottom and we will just add the tops.0697

Even though we have added this and put into a single rational expression, we are not done. 0707

We want to make sure that it is in lowest terms.0711

Let us go ahead and factor the top and bottom, see if there are any extra factors hiding in there.0716

As I factor the top, this will factor into 3 × x +1 and then we can factor the bottom, this will be x + 1 and x – 1.0722

I can definitely say yes there is a common piece in there, it is an x +1 and we can go ahead and cancel that out.0735

I'm left with a 3 / x - 1 and now I have not only added the rational expressions, this is definitely in lowest terms.0743

Let us try this on another one.0758

We want to add together the two rational expressions I have (-2/w + 1) + (4w/w2 -1).0760

This one is a little bit different.0769

The denominators are not the same. 0771

Let us see if we can figure out what the denominator should have in the bottom by factoring them out and seeing what pieces they have.0775

w2 - 1 is the difference of squares which would break down into w + 1 and w -1.0785

It looks like that first fraction is missing a w -1.0796

We have to give it that missing piece.0802

We will write it in blue.0808

I will give it an extra w -1 on the bottom and on the top just to make sure it stays the same.0809

Our second fraction already has our least common denominator, so no need to change that one.0819

Now that it has a common denominator we will keep it on the bottom and we will simply add the numerators together.0826

I got -2w -1 + 4w.0835

We cannot necessarily leave it like that, I’m going to go ahead and continue combining the top, 0841

maybe factor and see if there is anything else I can get rid of.0846

Let us distribute through by this -2.0851

-2w + 2 + 4w / (w +1) (w -1)0853

(-2w + 4w) (2w + 2)0868

I think I already see something that will be able to cancel out.0876

Let us factor out a 2 in the top.0880

Sure enough, we have a w + 1 in the top and bottom that we can get rid of.0887

That is gone. 0895

Our final expression here is 2 / w -1 and now we have added the two together and brought it down to lowest terms.0898

Let us do a little bit of subtraction.0910

This one involves (5u /u – 1 – 5) + (u /u -1).0912

This is one of our nice examples and that we are starting off and has exactly the same denominator. 0919

That is good so we can go ahead and just subtract the tops.0925

I will have 5u - the other top 5 + u.0931

Note what I did there, I still have the entire second top and I put it inside parentheses and I'm subtracting right here. 0940

One common mistake is not to put those parentheses in there and you will only end up subtracting the 5.0950

You do not want to do that.0956

You want to subtract away the entire second top.0956

To continue on, I want to see if there is anything that might cancel.0961

I’m going to try and crunch together the top a little bit and see if I can factor. 0964

Let us distribute through by that negative sign.0969

5u - 5 - u / u -10974

That will give me 5u – 1 = 4u – 5/ u – 1.0981

It looks like the top does not factor anymore.0995

This guy is in lowest terms.0997

Let us tackle one more last one and these ones involve denominators that are not the same.1008

We are going to have to do a lot of work on factoring and seeing what pieces they have before we even get into the subtraction process. 1019

Let us go ahead and factor the rational expression on the left.1026

On the bottom I can see that there is an (a) in common.1032

Over on the other side I can factor that into a - 5 and another a – 5.1042

They almost have the same denominator.1056

They both have that common a - 5 piece and the one on the left has an extra a.1059

And one on the right has an additional a – 5.1063

Let us give to the other one the missing pieces. 1066

Here is our rational expression on the left, we will give it an additional a – 5.1078

With this one it already has a - 5 twice, we will give it an additional a on the bottom and on the top.1091

Now that they have exactly the same denominator we can focus on the tops 3a a– 5 - 4a /(a – 5) (a- 5).1104

It looks like we can do just a little bit of combining on the top. 1123

I have (3a2 - 15a – 4a) / (a)(a – 5)(a-5)1127

When combined together the -15a and the -4a, 3a2 – 19a.1142

We have completely subtracted these we just need to worry about factoring and canceling out any extra terms.1156

On the top I can see that they both have an extra a in common, 3a-19.1168

Let us take that out of the top.1174

We will cancel out that a and now we brought it down into lowest terms. 1183

I have 3a - 19 / (a – 5) (a -5)1189

Whether you are adding or subtracting rational expressions make sure that you have your common denominator first.1202

Once you do, you just have to focus on the tops of those rational expressions by putting them together.1209

Once you do get them together remember you are not done yet, feel free to factor one more time and reduce it to lowest terms. 1215

Thank you for watching www.educator.com.1222

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at some complex fraction.0002

We will first have to do a little bit about explaining what a complex fraction is 0012

and then I’m going to show you two techniques and how you can take care of them.0015

In the first one we will write out these complex fractions as a division problem 0018

and then we will go ahead and use the method of this common denominator in order to simplify them.0023

Each of these methods has their own advantages but they both should work when dealing with simplifying complex fraction. 0029

What a complex fraction is, it is either the numerator or the denominator is also a fraction.0040

Here is a good example of a complex fraction using numbers. 0047

You will notice that the main division bars is actually sitting right here.0050

But in the numerator I have 2/5 and in the denominator I have 1/7. 0055

If I have fractions made up of other fractions. 0061

These are exactly the types of things that we are looking to simplify.0064

Since we are on a lot of rational expressions, then we will not only look at just numbers but we will look at more complex fractions like this. 0072

One of the first techniques that we can use to clean this up is to use division.0082

We will use that main division bar and write this as a division problem. 0089

Here I have (x + ½) ÷ (6x + 3)/4x.0095

It is the main division bar right there that will turn into division.0101

That means we will have to use all of our tools for simplifying the left and right, and eventually be able to get them together.0106

In a previous lesson we learn that we need to flip the second rational expression then multiply across the top and bottom.0115

That is exactly what you will see with these.0120

Let us grab on these rational expressions and give it a try.0125

You have (t2 u3) / r ÷ t4u/r2.0127

We want to identify what is on the top and what is on the bottom?0135

We want to write those again as a division problem.0140

(t2 u3) / r ÷ t4u/r20144

It looks pretty good.0157

We want to turn this into a multiplication process by flipping that second rational expression t2 u3/ r will now be multiplied by r2 ÷ t4u.0158

That looks much better.0177

We can go ahead and multiply across the top.0178

I will just put all of these on the top and multiply across the bottom.0182

I will put of all of these on the bottom.0187

Now that we have this we would simply go through and cancel out our common factors.0190

We will get rid of t2 on the top and t2 in the bottom.0198

Making a t2 in the bottom.0204

We can cancel out u, bring this down to u2 and we can cancel out on r.0207

We have r in the top.0217

Let us write down everything that is left over.0220

u2 r ÷ t2 and then we could consider this one simplified.0222

It changes it into a problem that we have seen before.0231

You just have to do a lot of work with simplifying.0234

When it gets to simplifying a complex fraction, even that process is not necessarily the easiest to go through.0240

In fact, you will find that this next problem is quite lengthy. 0248

We have (1/x + 1 + 2/y – 2) ÷ (2 /y – 2 – 1/x + 3).0251

Let us identify everything on the top and everything in the bottom.0262

That way we can simply rewrite this 1/x + 1 + 2/y – 2 all of this is being divided by everything on the bottom 2/y – 2 – 1/x + 3.0269

If I have any hope I'm doing this as a division problem that I need to normally flip that second fraction.0298

Notice how in this one I do not have a single fraction.0305

It is tempting to say, hey why we just flip both of them but that is not how division works.0309

We need to combine it into a single fraction before we can flip it and then do the multiplication.0313

Let us see if we can get these guys together with some common denominators. 0322

The common denominators on this side are the x + 1 and y – 2.0327

In order to get those together, I would have to give this fraction, y - 2 on the top and bottom.0343

To make it workout over here we will give the top and bottom of that one x +1 and the 2 is still up there.0353

It looks like that first piece will turn into y – 2 + let us go ahead and distribute this guy in there (2x + 2) ÷ (x + 1) (y – 2).0365

That just takes that and crunches it down a little bit.0384

Let us focus on this other one.0388

We need a common denominator and I see there is a y - 2 and x + 3.0390

I will need to give the fraction on the left an additional x + 3.0408

We still have the 2 in there.0412

2 × x + 30414

Over on the other side let us give the top and bottom of that one y -2.0417

When those are put together, we will do a little bit of distributing here.0424

We will have (2x + 6) – (y + 2) ÷ (y – 2) (x + 3).0431

Let us go ahead and write this again and see if we can do the actual division.0447

It looks like I can cancel out a few things in here.0455

Let us save ourselves a little bit of work.0458

y + 2x ÷ (x + 1) (y -2), we are dividing it by the second fraction.0461

Here is when I’m going to flip and multiply it, multiplied by 2x - y + 8.0476

I have combined the 6 and the 2 together, y - 2 x + 3.0487

I can just combine the tops and bottoms. 0498

Quite a lot of pieces in here that is okay, at least I see one piece will cancel out and that is the y – 2.0516

We are left with y + 2x and (x + 3) ÷ (x +1) and 2x - y + 8. 0528

Even though it can be a lengthy process, by rewriting it as a division process and using our tools from before0543

you will see is that it is possible to reduce and simplify this complex fractions. 0549

The other method which can often be a lot cleaner is using the least common denominator 0556

in order to clear out all the fractions in the top and bottom.0561

In order for this method to work, you must find the least common denominator of all the little fractions present in your complex fraction.0566

Go ahead and look at your numerator and denominator and think of all the least common denominator for all those fractions. 0573

Once you find that LCD, then you are going to multiply on the top and the bottom of the main fraction 0580

or on the top and bottom of the main division bar.0586

This will clear up things immensely but you have to be careful on canceling out.0591

You do not want to accidentally cancel out something that you should not.0596

You will see that you will clear out a bunch of stuff and then you end up simplifying just as you would normally.0598

No matter what method you use, you should get the same answer. 0608

Use the method that you are more comfortable with. 0614

I like to recommend a method two because it is usually much cleaner than using the first method.0620

However, anything that is cleaner use less opportunities for mistakes. 0627

One downside to the second method is usually happen so quickly it is hard to keep track of everything that was in there.0633

Let us try this second method with the following complex fraction. 0643

I have (2/s2 t + 3 /st2) ÷ (4/s2 t2 – 1/st).0647

Let us first see if we can identify the least common denominator.0656

Try and pick out all these little denominators here, see what the LCD would be.0661

They all have some s and the largest one I see there is an s2. 0668

They all have t’s but the largest one in there is a t2.0675

I’m going to take this and I’m going to multiply it on the top and the bottom of this expression so I can rewrite it.0680

There is our main division bar.0698

Let us get to multiplying.0708

On the top of this entire main division bar I will multiply it by this LCD.0709

I will do the same thing on the bottom to keep things nice and balanced. 0715

The top and bottom both have two terms, so we will definitely make sure that we distribute each of these parts.0719

We are going to write the result of this multiplication and then watch how many things will cancel out in the next step.0730

(2s2 t2 / s2 t) + (3s2 t2 / st2) ÷ (4s2 t / s2 t2 - s2 t/st).0737

It looks messy and it looks I have actually made things even more complicated.0768

Watch what is going to go take a vacation here soon.0772

I have these s2 they will go away. 0776

Then I have a single t and t up top, they will go away.0779

On the next one, there is my s one of those will be gone and both my t2 are gone.0783

Onto the bottom s2 is gone, s2 is gone.0790

I think I am missing some of my squares.0797

Our t2 are gone and we can get rid of one of these t’s.0805

The denominators of all those little fractions, these guys that we are so worried about at the very beginning, 0817

all of them have been cancel out in some way or another. 0823

This means as soon as we write down our leftover pieces, this one is simplified.0826

2t + 3s, those are the only things that survived up here ÷ 4 - st and those are the only things that survived on the bottom. 0831

This one is in its most simplified form.0847

It is nice, quick and easy method you just have to properly identify the LCD first.0852

You will know you are using the method right if all of these denominators end up going away.0858

If any of them are still in there double check to see what LCD you used.0862

Let us try this one more time but something a little bit more complicated. 0869

This one is (2y + 3 / y – 4) ÷ (4y2 – 9) + (y2 – 16).0873

Let us examine these denominators so we can find our LCD.0882

We are looking at this one this is the same as y - 4 and y + 4.0888

Over here it already has the y -4 in it.0897

The only piece that I am missing is the y + 4.0900

My LCD will contain both of these parts.0905

I have the y – 4 and y + 4, both of those in there.0908

I’m going to take that and we are going to multiply it on the top and bottom of the original.0917

Let me just quickly rewrite this and I’m going to rewrite it with the factored form on the bottom.0923

We will take our LCD and we will multiply it on the top and on the bottom of our main fraction here.0939

y – 4 y + 4.0952

Let us go ahead and put everything together and let us see how this looks.0959

I have (2y + 3 ) (y – 4) (y + 4) ÷ y – 4.0965

Then comes our main division bar right there.0980

On the bottom is (4y2 – 9) × (y – 4) × (y + 4) ÷ (y – 4) (y + 4).0985

Watch how many things will cancel in this next step.1005

y – 4 and y – 4 those are gone.1009

y – 4, y – 4, y + 4, y + 4, 4 those are gone.1013

All of these problems that we had at the very beginning, they are no longer problem.1017

They are gone.1022

We will simply write down all of the left over pieces.1023

2y + 3 y + 4 4y2 – 9.1028

Be careful, there is still some additional reducing that you can do even after using your LCD like this.1044

One thing that I can see is that I can actually continue factoring the bottom.1050

Let us write that out.1056

2y + 3/ y + 4 and this will be over I have different squares on the bottom so, 2y + 3, 2y – 3.1058

I’m sure enough now we can more easily that I have an extra 2y – 3 in the bottom and that is gone as well.1074

This one finally reduces down to y + 4 / 2y -3 and now we are finally done.1081

The second method is definitely handy and clears up a lot of fractions very quickly.1092

Use whatever method you are more comfortable with.1097

Thank you so much for watching www.educator.com. 1101

Welcome back to www.educator.com.0000

In this lesson we are going to work on solving rational equations.0003

Two things that we will look at is I can solve a equation that contains a rational expression.0009

I will throw in an example or we deal with a formula that also involves a rational expression.0016

Recall that when you are working with equations that have a lot of fractions in them we will often use the least common denominator 0025

to go ahead and clear out all of those fractions. 0032

These particular types of equations you definitely want to make sure that you check your solutions. 0036

This will be especially important for these types of equations this is because we are dealing with expressions that involve fractions,0041

we will have some restricted values, values that we cannot have.0051

Some of these values may try and show up as our solution.0056

We simply have to get rid of them.0059

They are not valid and they will make our denominator 0.0061

When working with these expressions that have a lot of fractions we like to keep the least common denominator throughout the entire process0068

or that way we can go back in and make sure that makes sense in the original.0076

When you see my example for formulas here again and know that the same process plays out. 0086

We like to use that least common denominator to go ahead and try and clear things out0092

Of course, our goal is to isolate which ever variable we are trying to solve for.0096

When we are all done with that formula we still have an equation of these words 0103

and for more variables still floating around in there.0107

That is okay as long as we get that variable we are looking for completely isolated. 0109

That is how we will know it is solved.0114

Let us take a look at one of these.0119

I have 1-2 / x + 1 = 2x / x +1 and we can see that this is a rational equation because looking at all these rational expressions I have, 0121

think of those fractions. 0131

To help out I’m going to work on getting a common denominator.0134

Already these 2 parts of it have an x + 1 in the bottom.0139

We are just going to focus on trying to give an x + 1 to the one over there on the left.0144

I can do that by taking 1 / 1 and then multiplying the top and bottom of that by x + 1.0151

(x + 1)(x +1) - 2 / (x + 1) = 2x /(x +1)0162

On the left side that allows us to finally combine things now that we have a common denominator. 0175

In doing so, we just put the top together.0182

x + 1 - 2 / x +10184

Doing a little bit of simplifying let us go ahead and subtract 2 from 1, x - 1 now x + 1 = 2x / x +1.0196

At this stage notice how we have two fractions that are set equal to one another and the bottoms are exactly the same.0211

Since the bottoms are the same then I know that the tops of these must also be the same.0219

Let us just focus on the top part for a little bit and see if we can find a solution out of that.0225

x - 1 = 2x this one is not so bad to solve.0230

Let us just go ahead and subtract an x from both sides and I think we will be able to get a possible solution.0236

It looks like a possible solution is that -1 = x.0245

Be careful this can sometimes happen with these rational equations. 0249

It looks like we have done all of our steps correctly and it looks like we have found a solution, but this guy does not work.0253

To see why it does not work, take this possible solution all the way back to one of our original denominators 0260

and you can see that if you try and plug it in and it makes the bottom 0.0267

We have to get rid of that possible solution.0272

Since we do not have any other possibilities left, we can say that this particular equation has no solution. 0275

With these types of equations, even when you find something that looks like it is a solution you have to go back 0284

and check to make sure that it make sense in the original.0291

Oftentimes you may end up getting rid of things that will simply make that denominator 0.0294

Let us try a different one.0301

This one is 2 /(p2 - 2p) = 3 /(p2 – p)0302

Before working with this, I need to see what is in the denominator.0311

To get a better idea we are going to go ahead and factor.0316

In this side I can see that they both have a p in common, we will go ahead and take that out.0321

I get p × p -2.0326

On the right side here, it looks like it has a p in common as well, p × p -1.0330

They almost have the same denominator, but I need to essentially give each one its missing factor.0339

Let us give the left side p -1 and we will give the right side p -2.0346

Just copy this down and see where the extra pieces are coming from.0353

On the left I will put in p -1 on the bottom and on the top.0365

The one on the right we will give p - 2 in the bottom and p -2 on top.0374

Distributing this will give us (2p -2) / and now we have this nice common denominator equals (3p – 6) / (p) (p -1) (p -2)0385

This is right back to that situation we had before.0406

The bottoms of each of these fractions are exactly the same I know that the tops of the fractions will also end up being the same. 0409

Let us extract out just the tops and focus on those.0418

If we have to solve this what can we do?0425

I will go ahead and add 2 to both sides 2p = 3p - 4 and let us go ahead and subtract 3p from both sides, -p = -4.0428

One last step, let us go ahead and multiply both sides by -1.0448

My possible solution is that p = 4.0455

We cannot necessarily just assume that that is the solution, not until we go back to the original 0459

and make sure that it is not going to give us 0 in the bottom.0464

Let us check to see what factors are in the bottom.0468

Some of our restricted values that we simply cannot have is 1p = 0, we cannot have that one. 0473

We also cannot have p = 2 and we cannot have p = 1.0480

Those three values are restricted.0487

Fortunately when we look down at what we got, it is not any of these restricted values.0490

We will go ahead and keep it as our solution.0497

The solution to this one is p = 4. 0499

Let us look at a little bit large one, one that has a few more things in the denominator.0505

This one is 1 / x -2 + 1/5 =2/5 × (x2 – 4)0511

Quite a bit going on the bottom and it looks like we definitely need to factor one of our pieces 0518

just so we can see what total factors we have in here.0523

1/x – 2 + 1/5 = 2/ I have 5 and then the x2 – 4 that happens to be the difference of squares.0528

(x + 2)(x – 2)0545

It looks like this will be our LCD and we are going to give it to these other fractions here with missing pieces.0550

The 1/x – 2 and see what we can put into that one. 0560

It needs to have a 5, we will put that down below and up top and it also needs to have an x + 2 and below and up top.0569

Onto the next fraction here, this is 1/5 and it looks like it is missing an x - 2 piece so we will throw that in there.0585

We are missing x + 2 missing both of those pieces.0597

On the other side of our equal sign it already has the LCD, so we just leave it just as it is.0607

We spread out quite a bit but now we are going to focus on the tops of everything since all the bottoms are now exactly the same.0616

We will also use our distribution property a little bit to help us out by multiplying 5 × 2 0626

and we will take the two binomials over here and we will foil them so we can put those together.0633

5x + 10 would be that first piece on the top of the first fraction.0641

We have our first terms x2, outside terms 2x, inside terms – 2x and last terms -4.0649

We want that all equal to 2.0658

You can notice this is just the tops of all of our fractions.0661

Let us go ahead and combine some things.0666

My + 2x and my - 2x will go ahead and cancel each other out and we will have x2, I will take care of that one.0669

5x took care of that one and combining the 10 and -4, +6 = 2.0680

We are almost there. Let us just go ahead and subtract 2 from both sides and see that we need to solve a quadratic.0689

We are going to solve x2 + 5x + 4 is all equal to 0.0698

This quadratic is not too bad I can end up factoring it using reverse foil without too much problem.0703

Our first terms x and x and two terms that need to multiply to give us 4 but add to give us 5.0710

I am thinking of 1 and 4, those will do it, both positive.0719

This will give us two possible solutions.0725

When I take x + 1 = 0 and x + 4 = 0, x could equal -1 or x could equal -4.0727

There are two possible things that I'm worried about.0739

Before we put our seal of approval on them, 0744

let us go back to some of those original fractions and see what some of our restricted values are.0746

Looking at this first one, the bottom would be 0 when x is equal to 2 so I know one of my restricted values x can not equal 2.0752

The bottom of this one is a number 5, so that one is not is not going to be 0.0762

Over here I can see that when x =-2 I will run into a 0 on the bottom, x cannot equal -2.0768

This factor here is exactly the same as the first one.0776

We already have that list as one of our restricted values. 0780

As long as I know my possible solutions are 2 and -2, it looks like we are okay.0783

Since I'm looking at -1 and -4, both of these are going to be valid solutions, none of them are restricted.0789

The process of solving these rational equations you just have to work on combining them and focus on the tops for a bit.0796

Let us look at that formula as I promised.0807

To see how you could solve the formula involving some these rational expressions.0809

To make things interesting, let us go ahead and solve for y.0814

In order to get rid of a lot of these things in the bottom we will identify our lowest common denominator.0822

One has an x another one has y and the other one has z.0830

It looks like we will need all three of those parts, x, y and z in our new denominator.0833

We will go through and give each of them their missing pieces.0840

Here we have our originals, let us throw in the missing.0846

The first one it already has an x in the bottom, we will give it y and z on the top and the bottom.0852

The next one already has a y, so give it the x and z.0860

For the last one we have z so xy.0868

This will ensure that all of the bottoms are exactly the same. 0878

We will go ahead and turn your attention to just the tops of all of these and we will continue moving forward.0882

2yz = 1 × x × z that will be just xz + xy.0889

We are trying to get these y isolated all by himself and looks like we have two of them, one on the left side and one on the right side.0899

Let us get them on the same side by subtracting an xy from both sides.0907

2yz – xy0914

We can factor out that common y that way we will have 1y to deal with.0922

Let us go ahead and factor out the common y out front.0928

y and then left over we will have 2z – x.0931

We are almost free with this one.0941

To get the y completely isolated, let us finally divide both sides by that 2z – x.0946

y = xz / 2z - x and this one is solved because we have completely isolated the variable we are looking for.0952

In a sense we have created a new type of formula.0970

Be careful in solving many of these different types of rational equations.0973

Make sure you always check your solutions and use that least common denominator to help yourself out. 0976

Thank you for watching www.educator.com.0982

Welcome back to www.educator.com.0000

In this lesson we are going to go ahead and take care of rational inequalities.0002

There are a few different techniques that you could use for solving rational inequalities,0007

but I’m just going to focus on looking at a table and keeping track of the sign for these.0011

Recall that when we are trying to solve an inequality that involves a polynomial, we want it in relation to 0. 0020

The reason why we are doing that is we just have to focus on whether it is positive or negative, a lot less to handle.0029

This will make looking at the intervals that represent our solution a little bit easier to find.0039

You will see that the process for solving these rational inequalities looks a lot like 0047

the process for solving our polynomials that we covered earlier.0051

We will still look at factoring it down, figuring out where each of those factors are equal to 0 0055

and looking at tables so we can test the values around each of those 0.0060

The actual process for solving a rational inequality looks like this. 0067

The very first thing that we are going to is we are going to set the inequality in relation to 0 on one side.0072

It means get everything shifted over, so you have that 0 sitting over there. 0078

Then we will go ahead and solve the related rational equation.0084

Solve the same thing and put an equal sign in there and see when it is equal to 0.0088

We will also figure out where the denominator can be equal to 0.0093

Those are some of our restricted values that we cannot use.0097

The reason why it is important to solve it and find out where the bottom is 0 is at those points it could change sign. 0100

We will call those particular points our critical values.0108

It is around those points that it could change in signs, we are interested in what happens.0113

To determine what it does around those we will use a few test points 0119

that will help us to determine which individual interval satisfy the over all inequality. 0124

Until we get to finding our intervals, we are not quite done yet. 0132

We also have to pay close attention to the end points to see which ones should be included and which ones should not be included.0135

I will give you a few tips on that, watch for that.0143

Here are my tips.0149

Remember that when you are working with these rational inequalities and get everything over onto one side, 0150

we might combine it into one large rational expression.0155

If we got 3 or 4 of them, put them all down into one.0159

Make sure you factor both the top and the bottom.0164

You need to see where each of the factors is equal to 0.0166

A lot of factoring.0169

You will never include any values that make the denominator 0.0172

We cannot divide by 0 even if it has an or equal in there, never include anything that makes the bottom 0.0176

What that said, if we are dealing with a strict inequality like greater than or less than 0184

then we will not include any points on the end points.0189

If we have greater than or equal to, less than or equal to, we will include where the numerator or the top will equal 0. 0196

Those are the only end points we have to worry about including.0209

That is a lot of information, let us jump in and look at some of these examples.0215

I want to solve when -1/x + 1 is greater than or equal to -2/x - 1. 0219

Rather than worrying about clearing up fractions or anything like that, 0226

let us get everything over onto one side and get it in relation to 0.0229

I'm going to add 2/x - 1 to both sides.0236

My goal here is to combine these fractions into one large fraction.0251

Do not attempt to clear out these fractions like you would with a rational equation. 0256

If you clear out the fractions you will lose information about the denominators 0261

and we want to check around points where the bottoms could be 0.0265

Do not clear those out.0269

My LCD that I will need to use will be (x + 1)( x – 1)0271

Let us give that to each of our fractions.0280

-1 will give this one x -1.0288

We will do that on the bottom and on the top then we will give the other one x +1.0292

Let us combine this into a single fraction here.0311

We would have to do a little bit of distributing so - x + 1 + 2x + 2 / x + 1 x - 1 greater than or equal to 0.0317

Just a little bit more combining on the top that is a - x + 2 would be a single x and then 1 + 2 would be 3.0337

It is quite a bit of work but we have condensed it down into a single rational expression on the left there and we also have it in relation to 0.0355

That is a good thing.0363

I want to figure out where would this thing be equal 0? 0365

It will equal 0 anywhere in the top would equal to 0. 0369

The top is equal to 0 at x = -3.0375

We also want to check where would the denominator be 0?0383

The bottom equal 0 at two spots when x =- 1, 1x =1.0389

All three of these values are what I call our critical values, 0399

and it is around these values that we need to check the sign of our rational expression here.0403

That way we can see whether it is positive or negative. 0408

Just like before when we are dealing with those polynomial inequalities, this is where our table come into play. 0412

First start out by drawing a number line and putting these values on a number line.0419

I want to start with the smallest ones so -3 and then we will work our way up from there - 1 and 1.0425

Along one side of this table we will go ahead and we will write down the factors that our rational inequality here.0435

I have x + 3 x +1 and x – 1.0442

Now comes the part where we simply grab a test values and see what it is doing around these.0450

Our first test value we need to pick something that is less than -3.0457

Let us choose -4, that is on the correct side and we will put it into all of our factors to see what sign they have.0461

-4 + 3 = -4 + 1 still - and -4 -1=-5, negative.0469

We will select a different test value between -1 and -3.0484

-2 will work out just great and we will put it into all of our factors.0490

Let us see what that does.0494

-2 + 3 that would give us something positive, -2 + 1 that will be negative and -2 – 1, negative.0497

We are doing pretty good and now we need a test value between -1 and 1.0512

I think 0 is a good candidate for this.0518

That will be a nice one to test out.0519

0 + 3 = 3, 0 + 1 =1, 0-1=-1.0522

One last test value we need something larger than 1.0532

I will put 2 into all of these.0536

2 + 3= 5, 2 + 1 =3, 2 -1 =10539

My chart here is keeping track of the individual sign of all the factors.0547

We will look at this column by column, so we can see how they all package up for our original rational expression here.0553

In this first column I have a negative and it is being divided by a couple of other negative parts.0561

What I’m thinking of visually in my mind here is that this will look a lot like the following.0572

I have a negative value on top and a couple of negative values in the bottom.0578

When those negative values in the bottom combine they will give us another positive value.0583

We have a negative divided by a positive.0591

That means the overall result of that first interval is going to be negative.0595

In the first interval it is a negative.0603

For the next one I will have a positive ÷ negative × negative and that will end up positive.0607

Then I will have a positive ÷ positive × negative = negative.0616

My last I will have positive ÷ couple of other positive values, everything in there is positive.0623

I know what my rational expression is doing on each of the intervals.0630

I know when it is positive, I know when it is negative and things are looking pretty good.0634

Looking at our rational inequality over here I can see that I'm interested in the values that are greater than or equal to 0. 0639

That means I want to figure out what are the positive intervals.0647

In this chart that we have been keeping track of all that so I can actually see where it is positive. 0652

I have between -3 and - 1. 0659

I have from 1, all the way up to infinity. 0662

Both of those would be some positive values.0665

Let us write down those intervals.0669

I'm looking at between -3 and -1 and from 1 all the way up to infinity.0670

There is one last thing that we need to be careful of what endpoints should I include, which endpoints should I not?0679

Notice how we are dealing with or equals to. 0688

I want to include places where the top of my rational expression could have been 0.0695

It is a good thing I highlighted it early on, it fact they are right here. 0700

I know that the top will be equals 0 at -3.0703

I'm going to include that in my intervals.0708

We never include spots where the bottom is 0. 0712

We have marked those out and I will use parentheses to show that those should not be included.0716

Let us go ahead and put our little union symbol so we can connect those two intervals.0723

This interval from -3 all the way up to -1 and -1 to infinity 0728

that would be our interval that represents the solution for the rational inequality.0734

Let us try one that looks fairly small, but actually has a lot involved in it.0743

This one is x -5 / x -10 is less than or equal to 3.0748

Like before let us get everything over on to one side first. 0754

That way it is in relation to 0.0757

x -5 / x -10 -3 is less than or equal to 0.0760

We need to worry about getting that common denominator.0773

I see I have x -10 in the denominator, let us give that x -10 to the top and bottom of our 3.0776

Looking much better from now, we will go ahead and combine these rational expressions right here.0799

I think we need to do a little bit of work, let us go ahead and distribute this -3, as long as we have it there.0807

x -5 - 3x + 30 / x – 10 you want to know where is that less than or equal to 0.0813

Just a few things that we can combine we will go ahead and do so.0827

-2x from combining our x and let us see a 25 will be from combining 5 and 30.0832

We want to know where is that less than or equal to 0.0850

We are in good shape so far.0855

Now that we have crunched it down into one rational expression, we are interested in where is it equal to 0. 0857

This will be equal to 0 wherever the top is equal to 0.0865

We are going to do a little bit of work with this one, but we can figure out where the top is equal to 0.0869

We will have to move the 25 to the other side and divide both sides by -2.0874

I have 25/2.0881

Let us make a little note is where the top is equal to 0. 0885

Okay, that looks pretty good.0890

Let us figure out where the bottom equal 0, that is not so bad that will be x = 10.0892

It is around those two values that we will go ahead and check to see what the sign is, around 10 and 25/2.0903

Let us go to our table here.0915

The smaller of these values would actually be our 10 and 25/2.0922

We will put that over there. 0929

Let us check the sign and see how these turnouts.0933

- 2x + 25 and the other one would be x – 10.0936

Let us grab some sort of test value that is less than 10. 0947

One thing I can see here is I will plug in a 0 that is less than 10.0952

If I plug it in the first one I will get 25.0958

If I plug in 0 for the other one I will get -10.0961

That is my sign for that one.0965

On to the middle interval I need something between 10 and 25/20969

I think 11 will work out just fine.0974

If I put 11 into that top one I will get -22 + 25 that will be positive.0979

If I put 11 into the bottom one, 11 - 10 would be 1. 0987

Okay, interesting.0994

I need to pick something larger than 25 / 2.0997

25/2 is about 12 ½ let us go ahead and test out something like 20.1002

If I put 20 into the top one I will have -2 × 20 that will be -40 + 25 that will give me a negative value. 1010

If I put 20 into the bottom one it will be 20 – 10 and that will be a 10.1019

These two things are being divided, to look at our over all sign we will take each of these signs and go ahead and divide them.1026

Positive ÷ negative = negative, positive ÷ positive = positive and negative ÷ positive = negative.1035

What am I interested in for this particular one I want to know where it will be less than or equal to 0. 1046

Let us look for those negative values. 1053

I can see one interval here everything less than 10 and I have another interval here where it is greater than 25/2.1057

Let us write down those intervals.1066

From negative infinity up to 10, from 25/2 all the way up to the other infinity.1067

One last thing, we need to figure out what end points we should include and which ones we should not include.1076

This one does have or equals 2 so I will include the places where the top is equal to 0.1082

We already marked those out, the top is equal to 0 at 25/2 so we will include it. 1089

We never include spots where the bottom is equal to 0, so I will not include the 10.1097

We will finish off by putting a little union symbol in between. 1104

The intervals that represent the solution are from negative infinity up to 10 and from 25/2 all the way up to infinity. 1108

And then we can consider this inequality solved.1118

It is a lengthy process, but we will use that table to help you organize your information 1121

and watch at what point you can and cannot include in the solution intervals. 1126

Thank you for watching www.educator.com.1132

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at applications of rational equations.0002

In some of the examples I have cooked up we will look at how some examples involves a just numbers.0009

There are ones that involve motion and some of my favorite ones that involve work.0015

This one is going to be a little bit trickier to think about, but once you get the process done 0020

you will see that the work problems are not so bad.0024

Depending on the unknowns in the problem and depending on how we should go ahead and package everything up 0031

there could be a few situations that actually lead to rational expressions. 0037

Think about those fractions or where an unknown ends up on the denominator. 0041

What we want to do is be able to solve these using many of the techniques that we have picked up for rational equations 0050

and see how we can recognize these in various different situations, such as numbers, motion, and especially work.0056

The work problems are kind of unique and that you want to know how to represent the situation. 0067

If you know how long it takes to complete an entire job then you know the rate of work is given by the following formula 1/T.0075

The way you can read this is by using that T for the amount it takes to do that one job.0084

Here is a quick example. 0090

Let us say it takes Betty 7 hours to paint her entire room. 0092

Well, that means that every single hour 1/7 of the room is going to be painted.0098

We are going to make a little chart for just keeping track of everything.0103

Maybe x will be my time and let us say how much of the room has been painted so far.0109

One hour, two hours, three hours and make sure you jump all the way to 7.0120

After she is in the room for 7 hours, she will have the entire room painted, the whole thing.0127

If you scale it back, what if she is only working for one hour then only 1/7 of the room is painted.0135

If she is in there for two hours, 2/7 of the room is painted and if she is in there for three 3/7.0144

You can see that we are just incrementing this thing by exactly 1/7 every time.0151

We can make some adjustments to our formula here and say that T would be the amount it takes to do that one for the person 0156

and may be multiply it by x, that would represent how long they have been doing that particular job.0165

Watch for that to play a key component with our work problem in just a bit.0173

Let us first see our example of numbers and just see something where our variable ends up in the denominator.0180

In this one we have a certain number and we are going to add it to the numerator and subtract it from the denominator of 7/3.0186

The result equals the reciprocal of 7/3 and we are interested in finding that number.0194

Let us first write down our unknown. 0202

x is the unknown number.0205

We construct a model situation here.0217

Take that unknown number and add it to the numerator, but subtract it from the denominator of 7/3.0219

Here is 7/3, so we are adding it to the numerator and subtract it from the denominator.0227

The result equals the reciprocal of 7/3, that is like 7/3 but we flipped it over.0238

This will be our rational inequality here.0245

What we have to do is work on solving it.0249

To solve many of our rational equalities we work on finding a least common denominator 0253

which I can see for this one will be 3 - x and 7.0261

Let us give that missing piece to each of the fractions.0268

Here is my original (x + x) (3 – x) = 3/7 let us use some extra space in here.0272

The one on the left, it could use 7, let us put that in there.0286

The one on the right it is missing the 3 – x.0294

At that point, the denominators will be exactly the same.0302

We will just go ahead and focus on the tops of each of these.0307

7 × 7 + x = 3, 3 – x.0312

Continuing on and solving for our x we will go ahead and distribute our 7 and 3.0321

That will give us 49 7x = 9 - 3x.0328

Moving along pretty good. 0336

Let us go ahead and add 3x to both sides giving us a 49 + 10x = 9.0338

We will go ahead and subtract 49 from both sides.0351

I have x = 10x is equal to -40.0359

Dividing both sides by 10, I have that x = -4.0368

Just like when we are working with equations, it has to make sense in our original.0374

Looking at the original rational expression I have a restricted value of 3 and I know that it is not 3 that will make the bottom 0.0379

I do not get any restricted values from the other faction because it is simply always 7 on the bottom.0388

Since the -4 is not 3, I'm going to keep it as a valid solution.0394

That one looks good.0402

Let us look at one that involves motion.0407

We will set these up using a table and also use that same idea to help us organize this information.0409

A boat can go 10 miles against a current in the same time it can go 30 miles with the current.0416

The current flows at 4 mph, find the speed of the boat with no current.0422

We have an interesting situation.0428

We have a boat that looks something like this and we have the flow of the river. 0430

Now in one situation, it is fighting against the current, and the way you want to think of that in relation to its speed 0439

is that the speed of the river is taking away some of the speed of the boat.0447

You will see a subtraction process.0452

If the boat is going in the same direction of the river, they are both helping each other out 0456

and you will see an addition problem with both of their speed.0461

You will know they are both helping each other out.0464

Let us see if we organize this information so I can see how to connect it.0468

We need to think of two different situations.0492

We are either going against, or we are going with the river.0495

We will look at the rate, the time, and the distance.0502

This will help us keep track of everything. 0511

And of course we are leaving unknown in here.0513

Since we are finding the speed of the boat with no current, let us set that as our unknown.0516

x is the speed of the boat with no current.0522

I think we have a good set up and we can start organizing our information.0540

In the first bit of this problem we know that it can go 10 miles where it is going against the current.0545

That is its distance.0553

It went 10 miles when it is going against the current.0554

It can do that in the same time it can go 30 miles with the current.0558

A little bit of different information, this one will be 30 when it is going that way.0564

The current of the river flows at 4 mph. 0570

If we are looking at the speed of the boat and it is going against that river, probably this will be the boat - the current.0575

If we are looking at it going with the river, that will be the speed of the boat + the speed of the current.0585

They are helping each other out.0590

The only thing we do not know in here is the time, but I do know that the time was exactly the same for both of these situations.0593

Let us see what do we got here.0603

I will go ahead and create an equation for each of these.0604

x -4 × time = 10 and x + 4 × time = 30.0607

I know the times are exactly the same for each of these, let us solve them both for time.0620

This one I will go ahead and divide both sides by x - 4.0628

In this one I will divide by x + 4.0631

And I'm ready to develop that rational equation. 0638

I will set each of these equal to each other since the times are equal to each other.0641

I have a rational equation then I can go ahead and try and solve.0651

10 ÷ x – 4 = 30 ÷ x + 40655

To get through solving process, we find our common denominator.0661

I'm going to give x + 4 on the left side here and I will give x - 4 to the other side.0668

We will note that we made the denominators exactly the same.0687

We just need to focus on the tops of each of these.0691

Continuing on, you will distribute 10 and we will distribute 30.0702

10x + 40 = 30x -1200710

Subtracting a 10x from both sides will give us 40 = 20x -120 and let us go ahead and add 120 to both sides.0720

160 = 200739

We can divide both sides by 20 and I have that x is equal to 8.0745

Let us make sure that it makes sense.0754

Some restricted values I have for my equations here I know that x cannot equal 4 and -4 0757

and fortunately both that we have found for a possible solution is neither of those.0766

We can say the speed of the boat in still water would be 8 mph and then this guy is done.0772

Do not be afraid to use those tables from earlier to organize your information.0790

Joe and Steve operated a small roofing company and Mario can roof an average house alone in 9 hours.0800

Al can roof a house alone in 8 hours.0809

We want to know how long will it take them to do their job if they work together.0812

First we need to figure out the rates of each of them individually.0818

Let us go ahead and focus on Joe.0821

Joe can roof an average house alone in 9 hours.0830

Looking at just Joe we know that every hour he will get 1/9 of that house done.0834

Steve over here can roof the house in 8 hours.0844

He is working every single hour 1/8 of that house will be done. 0853

We can put in that time on it.0858

Let x be the number of hours.0861

You have 1/9 × however many hours they work and 1/8 × qualified by however many hours Steve works.0871

We want to know how long it will take them to do the job if they work together.0881

We will take each of their work that they are doing and we will add them since they are working together, 0889

we want to know when they will complete one job.0895

We have all of our information here and we can go ahead and try and solve this.0899

Our LCD would be 72.0906

We will multiply that through on all parts.0911

Doing some reducing 72 and 9 = 8, 72 and 8 = 9 and now we have an equation 8x + 9x = 72.0923

Combining together what we have on the left side this would be 17x is equal to 72 0942

and we can divide both sides of that by 17 to get x = 72 ÷ 17.0951

It is looking pretty good.0959

That represents how long it will take them to work together.0961

If you want to represent that as a decimal, you could go ahead and take 72 ÷ 17.0964

When I did that I got about 4.24 hours, I did round it.0971

That gives me a better idea of how long it took them when working together.0980

Anytime when you are working with these types of problems, it should be less than any one of them working by themselves,0985

since they are helping each other out.0992

This one looks good.0995

In this last example we are going to look at a water tank that has two hoses connected to it.0998

Even though this is not a work problem you can see that we can set it up in much the same way.1003

The information that we have is that the first hose can fill the entire tank in 5 hours. 1011

The second hose connected to this tank it can empty it in 3 hours. 1018

If we start with a completely full tank and then we turned both of them open, 1023

the question is how long will it take it to empty the entire tank?1029

Let us have an accrued picture of what we are dealing with here.1034

This would be our water tank and we have one hose that is going in and one hose that is going out. 1039

The hose that is going in, it can fill the tank in 5 hours.1054

That means if we just leave it on every hour 1/5 of that tank will fill up.1058

I know its rate is 1/5.1063

If I look at emptying the tank it is a much smaller time to empty it, 1/3.1068

If I did have them both open I would know that the second one 1076

would be able to empty the tank since it empties faster than the first one can fill it.1080

Let us see what that we can do to set this one out.1086

The first thing I want to consider was how much the tank is going to be empty every single hour.1091

Starting with the 1/3 I know that every hour that passes by the 1/3 of it will be emptied out.1097

x is the number of hours.1106

The 1/5 coming in is not emptying the tank but it is actually filling it back up.1118

We will say it is the opposite of emptying the tank by 1/5 and it will do that for every hour.1125

We want to know is when will one tank be completely empty. 1132

We have a lot of similar components for this one and it often look like a lot of our work problems.1137

All we have to do is go ahead and solve it.1145

Our LCD here that would be 15.1148

Let us multiply all parts by 15 and see what that does.1153

15, 15 and 15.1159

Canceling out some extra stuff here I get 5x - 3x = 15.1163

Doing a little bit of combining on the left side I have 2x =15 or x = 15 ÷ 2 or 7 ½ hours.1173

Notice how in this case it is taking longer since both of them are open. 1190

The reason for this is that they are not working with each other.1194

They are working against each other.1198

One was trying to fill the tank, and one is trying to empty the tank.1199

Use those bits of information you can find along the word problem to give you a little bit of intuition about your final solution. 1205

In that way you can be assured that it does make sense in the context of the problem.1212

Thank you for watching www.educator.com.1217

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at a very interesting application of our rational.0002

We will look at variation and proportion.0006

We will have to do a little bit of work just to explain what I mean by the word variation.0012

We will break this down into few other things.0017

We will look at direct variation, inverse variation, and combined variation. 0019

This particular section is filled with lots of different word problems all of them involving variation.0025

Let us see what we can do.0032

When you hear that word variation we are talking about a connection between two variables, 0036

such that one is a constant multiple of the other.0041

You will see some nice handy formulas that will highlight how one is a constant multiple of the other.0044

If you are looking for little bit more intuition on the situation then you can grasp on to that. 0050

Intuitively a variation can be thought of as a special connection and the two basic types are direct and inverse.0055

In this connection when one quantity goes up then the other one would go down.0063

In that one I'm talking about an inverse variation.0069

If I’m thinking about those two quantities when one goes up and the other one goes up as well, that will be an example of a direct variation. 0075

It is how one affects the other one and they could be moving in the same direction our going in the opposite directions.0084

To model this type of situation that has variation in it, you can end up setting up a proportion.0094

Be very careful on where you put each of the individual components and make sure you line them up correctly.0100

To model a direct variation, you could use the following proportion x1 / y1 = x2 / y2.0107

The way want to interpret those subscripts that you are seeing on everything is that all of the values with a subscript of 1 involved one situation.0116

Everything with the 2 involves a second situation.0126

Since the x are both on the top that will be from one particular type of thing and y will be from another type of things.0133

We will line those up to make sure that they at least agree.0142

It seems a little vague but let us get into an example 0148

and you will see how I do set this up using a proportion and how everything does line up.0151

This one says that an objects weight on the moon varies directly as its weight on earth.0156

Neil Armstrong, the first person to step on the mood weighed 360 pounds on Earth, when he is on the moon he was only 60 pounds. 0163

The question is if we take just an average person who weighs 186 pounds, what will their weight be if they go to the moon?0174

We have that this is a direct variation because it pointed out and says it is a direct variation.0186

Let us just think a little bit and see if that makes sense.0192

Intuitively in a direct variation when one quantity goes up then should be the other one.0196

When one quantity goes down, then so should the other one.0201

That is what is going on with our moon weight.0204

If I decide to go on and get heavier and heavier on Earth, that same thing is going to happen on the moon.0207

I will get heavier and heavier on the moon. 0214

This is a type of direct proportion.0217

Let us see what we can set up.0223

x1 / y1 = x2 / y20224

To help out I’m going to highlight two things, we will go ahead and keep our Earth weights on top 0231

and we will go ahead and put our moon weights on the bottom. 0238

Let us look at our first situation with our Neil Armstrong.0245

He weighed 360 pounds on Earth, and only 60 pounds on the moon.0250

360 on Earth and 60 on the moon.0255

I will compare that with our average person there.0267

They weigh 186 pounds on the Earth, but we have no idea how much they weigh on the moon.0270

I'm going to put that as my unknown.0277

Everything on the tops of these fractions represents an earth weight and everything on the bottom represents a moon weight.0280

You can see that we have developed one of these rational equations and now we simply have to solve it.0287

This one is not too bad if you go ahead and reduce that fraction on the left, it goes in there 6 times.0292

We can multiply both sides by x.0301

I have 6x = 186. 0308

Finally dividing both sides by 6 I will get that x =31.0316

This reveals how much that average person, the 186 pounds would weigh when they are on the moon.0327

There are lots of other types of situations that could involve a direct proportion and they might not come out and say its direct proportion.0338

Look for clues along the way to help out.0344

This one says that a maintenance bill for a shopping center containing 270,000 ft2 is \$45,000.0348

What is the bill of the first store in the center that is only 4800 ft2?0356

Let us think of what type of a variation this should be.0361

If I have a large store as a part of the shopping mall and I will have to end up paying a much larger maintenance or bill because of my larger store. 0366

If I have a small store then my bill should be very small.0378

Those quantities are moving in the same direction. 0382

I’m going to say this is an example of direct variation.0385

We will set it up using our proportion.0392

We want to keep things straight. 0403

What do each of these quantities represent?0405

Let us say the top will be our square footage and we will make the bottom the bill.0409

How much it cost?0421

Let us go ahead and substitute some of this information we have. 0424

If I’m looking at the entire shopping center I know how big it is and I know the bill for the entire place.0427

270,000, 45,000 and now I can look at my much smaller store, it only has 4800 ft2 0435

and this one we have no idea what the bill is so I will leave that one as my unknown.0453

I think we can do a little bit of simplifying with this one as well. 0459

I will divide these two and we get 6 and I need to just solve this rational equation which is not too bad as long as we multiply both sides by x.0464

I have 6x = 4800. 0481

Divide both sides by 6 and then I think we will have our solution x =800.0488

Even if they do not come out and tell you what type of variation is going on here, 0497

look at the values present to see if things are moving in the same direction, or in opposite directions.0502

For inverse variation we want to make sure that if one quantity goes up, the other one goes down.0510

One way that we can model it now with our proportions is to set up like this.0516

We will go ahead and use us a subscript for one situation and we will put them on opposite sides of our equal sign.0521

We will get that inverse relationship. 0528

When one goes up, the other one will go down.0531

For our second situation we will put these on two opposite sides of the equal sign.0535

One thing that will not change is that both of these x will represent the same type of thing 0541

and both of these y will still represent the same type of thing.0546

We can definitely note that the things are on opposite sides of the equal sign for these given situations.0550

For my inverse variation problem I will look at the current in a circuit and we are told that it varies inversely with the resistance. 0562

If I have a current that is 30 amps when the resistance is 5 ohms, find the current for resistance of 25 ohms.0572

I have a lot of things in here and we just want to keep track of each of them.0581

Let us see if we can highlight each of these situations. 0586

The current is 30 amps when the resistance is 5.0589

We will be looking for the current when the resistance is 25.0596

Let us get our proportion out here.0605

I’m using these little subscripts and put them on opposite sides.0611

Onto the first situation, we will keep everything on the top. 0615

Let us say that is our resistance, we will put everything on the bottom as the current.0624

Our first situation, we know that the current is going to be 30 when the resistance is 5.0635

In that situation I will put my 5 and 30.0641

For the other situation I want to figure out what the current is.0647

I have no idea what that is but I know that the resistance is 25 ohms.0651

This is my initial proportion that I have set up.0657

All we have to do is go through the process of solving it.0661

This one is not too bad. 0664

Let us multiply both sides by 30x and see what is left over.0666

30 × 5, since the x will cancel out equals 25 × x since the 30 cancel out.0673

Okay looking pretty good. 0688

Also 5 × 30 =150 equals 25x and we can simply divide both sides by 25 to get our answer. 0690

The current is 6 amps.0708

When dealing with these variation problems always look at your solution and see if it makes sense in the context of the problem. 0722

One thing that I'm noticing in here is that I start off with a resistance of 5 and a current of 30.0729

One thing I'm doing with my current is that I'm lowering it, and I'm taking it from 30 down to 25.0736

Since I have an inverse relationship that is going to have an effect on our amps is the current.0744

That should bring it up.0751

Originally I started with 5 and I can see my answer is 6.0753

It did go up and things are working in the right direction.0757

Since earlier, we said that it variation is relationship when one is a constant multiple of the other.0764

We can model this using some nice formulas.0769

For direct variation you can use the formula y = k × x.0773

For your inverse variation, you could use y = k ÷ x.0779

We have two variables in each of these equations that we are looking at connecting are the x and y.0785

The k in each of these is known as our constant of proportionality.0791

It connects how they are related. 0795

A lot of problems that you will end up doing with these variations, if you want to use these formulas 0799

then you will end up going through three steps to figure out what is going on. 0806

You might first go ahead and identify what type of variation you are using so that you can take one of these formulas.0811

Then you will end up using a little bit of known information to go ahead and solve for that constant of proportionality, k.0818

Once you know what k is, then you can usually figure out some new information by substituting it into the formula.0825

I got a couple of examples where we are going through these 3 steps.0832

Let us first just see an example of using an inverse of variation using one of these formulas.0840

This one says that the speed of water through a hose is inversely proportional to a cross section of the area of the hose. 0846

If a person places their thumb on the end of the hose and decreases the area by 75%, what does this do the speed of the water?0855

Let us think about what we got here.0867

We are looking for a connection among the cross area section of the hose and the speed of the water.0868

We are told that they are set up inversely so when one goes down, the other one should go up.0876

For decreasing the area then we should expect it to increase the speed of the water.0879

In fact we can be a little bit more precise, but despite playing around with those formulas for a little bit. 0887

Our relationship is that the speed of the water is inversely proportional to a cross section of the area.0893

Maybe x = k/a.0900

I'm going to decrease the area by 75%, one way that I could model that is by simply multiplying our area by what is left over, 25%.0905

Let us relate this to the original before I manipulated it.0923

25 is the same as ¼ I'm dividing by ¼.0928

When you divide by a fraction that is the same as multiplying.0936

This is multiply by 4.0943

Now we can see the difference between the original situation.0947

The area decreased by 75%.0956

They are almost exactly the same but in the second situation, it is 4 times as much.0966

It will actually increase the speed by 4 times.0971

There are lots of other types of variations that you can go ahead and work into a problem.0978

Some of these will look similar to previous types that we covered.0982

Let us go through how each of these are connected. 0984

If I say that y varies directly to some power of x then there are some k that connects those two things. 0990

You will notice that it looks like our direct proportion only since we are dealing with an nth power, 0997

like a 3rd power 4th power that we put that exponent right on the other variable.1004

Our y varies inversely as the nth power looks much like our inverse relationship as well.1013

But since we are doing it to the nth power, notice how we put that exponent right on the other variable.1020

In this one is a new one that involves a few more variables, this is our joint variation.1029

y varies jointly as x and z so more than one variable now.1034

If there some sort of positive constant such that y = k × x × z.1039

In this one the k is our constant and the variables are being connected, the y, x, and z.1046

You can start mixing and matching these various different types of variation.1057

The reason why that we do not want to start mixing and matching them is we can get some more complicated models.1064

With these more complicated models we are hoping to be able to more accurately model what is happening in a real life situation.1071

Usually it does a better job than any one variation could do by itself. 1079

The key when putting all of those different types of variations together is to look for clues in the actual problem itself.1085

Here is a formula that I have it is T = k × x × z3 /√y .1092

Here is how I could describe that problem using the language of these variations. 1100

The k right here is my constant of proportionality and we are looking to get it connect together the t, x, z, and y.1107

Here is what I can say, the t varies jointly with x and z3.1116

It is telling me that I have my constant of proportionality but the x and z should end up in the top and the z should be cubed.1123

t also varies inversely with the √y.1133

Since it is an inverse relationship I will put that one in the bottom and make sure to put in that square root.1138

Watch for those keywords so we can figure out where we should put things.1145

Let us try some of these variation problems and watch as I walk through the three steps of 1151

starting with some template solving for our constant of proportionality, k and the last part figure out some new information. 1155

What do I know so far?1167

I know that y varies jointly as x and z.1168

Just on that little part, let us go ahead and make a nice formula. 1173

y varies jointly as x and z.1177

y = 12 when x = 9 and z = 3.1184

I can use that information right there to go ahead and solve for my constant of proportionality. 1191

y =12, x = 9, z= 31198

The only unknown I have in there is that k value.1206

We will combine things I will get 27 divide both sides by 27 and then maybe reduce that by dividing the top and bottom by 3.1210

Now that we know much more I can end up revising our formula. 1228

y = xz and now I can say that the k value is this 4/9.1236

Now that we know about our constant of proportionality, let us find z when y = 6 and x = 15.1244

We will substitute those directly in there and you will see the only unknown I have left will be that z.1253

Y= 6, 4/9 × 15 and we are not sure what z is now we have to solve for it.1259

A little bit of work.1270

I think I can cancel out an extra 3 here.1272

I have 6 = 20/3 × z we could multiply both sides by 3.1276

I will get rid of those 3, 18 = 20z.1292

Lastly let us go and divide both sides by 20. 1298

18 ÷ 20 =z1301

We will just go ahead and put that into lowest terms.1306

Divide the top and bottom by 2 and I know that 9/10 is equal to z.1310

If you follow along we start with some sort of template for our variation, we solved for our unknown constant of proportionality k.1316

We have used it to figure out some new information about the other variables.1325

For this last type of example let us look at one that is a little bit more of a word problem.1331

This one says that we have the maximum load of the beam and how much you can support varies jointly 1336

with the width of the beam and the square of its depth, but it varies inversely as the length of the beam.1343

Some of that initial information there is just giving us a formula for how everything is connected.1352

We will definitely set up the equation that will model that situation.1358

Let us continue on.1364

Assume that a beam that is 3 feet wide, 2 feet deep, 30 feet long and it can support 84 pounds.1365

What is the maximum load a similar beam can support if it is 2 feet wide, 5 feet deep and 100 feet long?1371

A lot of information is floating around there.1378

Let us start off by seeing if we can set up this formula.1381

The maximum load that a beam can support varies jointly as the width of the beam and the square of its depth.1385

Since this is a jointly, we are talking about k × width × the square of its depth or d2.1395

It varies inversely with the length, I will put the length on the bottom.1408

This part here is our formula that we will end up using.1413

Assuming that a beam is 3 feet wide, 2 feet deep and 30 feet long, and it can support 84 pounds, 1421

let us use that to figure out our unknown constant sitting right here that k.1426

I know it can support 84 pounds, we do not know k.1432

3 feet wide, I put that in for width, 2 feet deep and it is 30 long.1440

The only unknown is that k.1448

Let us go ahead and simplify it just little bit in here.1451

I have 3 × 22 and 22 is 4 = 121455

I can get that k a little bit more by itself if I multiply both sides by 30, 84 × 30.1463

Let us go ahead and put those together and see that would be 2520.1473

Divide both sides by 12.1483

That is where we have been able to figure out what that unknown constant is.1488

Now that I have that I can go back to the original formula.1494

M = 210w × d2 / l1498

We will use this new formula to figure out a little bit of known information.1506

We will look at it for the other beam where we are looking for its maximum load if we know its width, depth, and its length.1512

Let us give it a try.1520

That formula we developed was its maximum load = 210 × width × d2 / length.1525

Let us put in our information.1536

For this new beam, its width is 2 feet and its depth is 5. 1541

It also has a length of 100. 1549

As soon as we simplify all that we will be able to figure out what its maximum load should be.1551

210 × 2 = 420 × 25 ÷ 1001558

Let us do some simplifying here 420 × 25 = 10500 ÷ 100 and looks like this reduces to 105.1573

This new beam can support a maximum of 105 pounds.1593

Being able to recognize various different types of variation 1599

and combine them all together is essential for some more complicated problems.1602

Remember that if you have those nice formulas then you can go through 3 steps.1607

One, build a formula or equation that models the situation, solve for your unknown k 1611

Thank you for watching www.educator.com.1622

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at rational exponents. 0002

The neat part about rational exponents is you will see that we will develop a way and connect them to our radicals.0011

You will see that there are a few rules for working with these rational exponents and a lot of them come for just our rules for exponents.0019

We will look at a few ways that you can combine terms that have some these rational exponents on them.0027

We have seen many different types of radicals. 0034

We have seen exponents but there is actually a great connection between the two. 0039

If you have a radical of some index like a square root or third root and it is raised to a power, 0044

you can write this in one of the following two ways.0051

You can write it as the root of that expression ratio power or you can write it as the expression raised to a fractional power. 0054

One thing to notice were the location of everything has gone to.0066

The power in each of these problems I have marked that off as a that shows up on the top of the fraction. 0070

The root is going to be the bottom of the fraction.0083

You can take any type of radical and end up rewriting it as a fractional exponent or as a rational exponent.0089

To get some quick practice with this let us try some examples.0098

I have 36^½, -271/3, 13/2 and -93/2.0102

We are going to evaluate these up by turning them into a radicals. 0110

Okay, so this first one, the way we interpret that is that I'm going to put it under a root with an index of two.0116

This is like looking at the √ 36, which is of course just 6.0123

For the other one, if I see -271/3 that will be like taking -27 to the 3rd root so this one is a -3.0132

In the next one notice how we have a power and a root to deal with. 0146

813/20149

There are two ways you could look at this.0153

You could say this is 813 and we are going to take the square root or you can take the square root of 81, 0155

then raise the result to the third power.0163

Both of them would be correct, but I suggest going with the one that is a lot easier to evaluate.0166

I’m thinking of this one. 0171

If you take 81 and raise it to the third power we are going to get something very large 0173

and try to figure out what is the square of that is going to be a little difficult.0177

But look at the one on the right, I can figure out what the √ 81 is, I get 9.0180

We can go ahead and take 93 it would be 729.0189

One last one, I have 93/2 and a negative sign out front.0201

Let us first write that as the √93 as for this negative sign it is still going to be out front.0206

I have not touched it.0215

I'm not including that in everything because there is no parenthesis around the -9 in the original problem.0217

I’m starting to simplify this.0224

The √9 would be 3 then I will take 33 and get -27.0226

In all of these situations I'm looking at the top of that fraction make it a power, we get the bottom to see what the root needs to be. 0236

The good part about taking all of our radicals and writing them using these exponents, 0247

it means that we can use a lot of our tools that we have already developed for exponents and we have done quite a bit of them.0252

In fact a quick review, we have a product rule for exponents, a power rule 0258

and we have gotten different ways that we could go ahead and combine them. 0264

We also have rules on how to deal with fractions like adding subtractions, subtracting fractions. 0269

We have our zero exponent rule, our quotient rule and negative exponent rules. 0273

All of these rules will help us when working with our radicals.0278

Just watch on what our base is and what we need to do from there. 0284

Now using some of these rules if you do have radicals you might be able to combine them under a single root.0293

That involves using a common denominator most of the time.0305

You can use this tool for working with rational expressions.0311

Watch how I find the common denominator for some of my problems and actually get everything under one root.0315

Let us do these guys a try.0323

64/27-4/3 0325

Before I get to that 4/3 part I’m going to apply some of my other rules for exponents and specifically that negative exponent. 0329

One way I can treat a negative exponent is it will change the location of the things that it is attached to.0339

I’m going to write this as (27/64)4/3.0344

I’m going to give that 4/3 to the top and to the bottom that is using my quotient rule.0352

I will end up rewriting the top and bottom using my radicals. 0362

I’m looking at the 3rd root of 27 and we will take the 4th power of that 0369

and then we will take the third root of 64 and take the 4th power of that one.0376

Let us see what this gives us.0390

On top the 3rd root of 27 that will be 3 and 3rd of 64 =4 and both of these are still raised to the 4th power.0393

We will go ahead and take care of that multiplication.0403

81 / 2560407

Being able to take it and write it as radicals, it meshes well with all the rest of our rules.0413

Let us use the quotient rule on this next one.0421

47/4 ÷ 45/40423

I need to subtract my exponents.0429

Good thing both of these have the same denominator this will simply be 42/4.0434

That continues to simplify this would be 4^½ which written as a radical is the √4 which all simplifies down to 2.0442

Be very comfortable with switching back and forth between those rational exponents and your radicals.0451

On to ones that are a little bit more difficult. 0461

These will involve trying to simplify much larger expression and some of the terms will have those rational exponents. 0464

Okay, so here I have (r^¼ y5/7)28 ÷ r5.0474

I can apply my 28 to both of the parts on the insides since they are being multiplied.0486

Let us see what this looks like.0493

28 × 1/4 = 28/4 and I have y5 × 28 ÷ 7 and all of that is being divided by r5.0495

Let us see if we can simplify some of those fractions.0512

How may times this 4 go into 28? 7 times and 7 will go into 28 four times.0515

This is (r7 y4)/r50527

We can go ahead and reduce our y.0534

5 of them on the bottom and with 5 of them on top that will leave us with an r2 and y4.0536

Let us try another one.0548

This one has a lot of fractions and a lot of negative signs.0549

(P-1/5 q-5/2)/(4-1 p-2 q-1/5)-20553

I’m going to use my rule to apply this -2 to all of my exponents.0565

I’m sure that will help get rid of a bunch of different negatives.0572

-2 × -1/5 = 2/5, -2 × -5/2 = q50576

Then on to the bottom, -1 × -2 = 2 and p4 = 4, q2/5.0593

That does simplify it quite a bit.0610

At least I can see that this guy right here will be 16 but we will also have to reduce these a little bit more. 0611

I'm going to take care of these ones I want to think what is 2/5 – 4.0620

If we can find a common denominator it will helps out with that ones.0626

2/5 - 20/5 we will call that one -18/5, so I know that I will have 18/5 on the bottom.0629

Let us try it out.0646

I have a 16 on the bottom and now I discovered I have a p218/5 on the bottom as well.0648

These ones we can reduce. 0656

I want think of 5-2/5.0659

The common denominator there will be 5, 25/5 - 2/5 = 23/5 we will put that on top q23/5.0663

We have our final simplified expression. 0679

In this next two I have some radicals and we will go ahead and write them as our rational and see what we can do from there.0686

The top, this would be y2/3 ÷ y2/50696

If I’m going to end up simplifying using our quotient rule, we will look at this as 2/3 – 2/5.0705

We need a common denominator on those fractions to put them together.0713

Let us look at this as over 15.0721

10/15 – 6/150726

This would give us y4/15.0731

Then we can continue writing this as a radical if I want it to be the 15th root y4.0735

Let us try this other one.0752

I have z and I’m looking for the 5th root of it, I will write that as z1/5.0753

When I’m taking the 3rd root of all of that, that is like to the 1/3.0761

My rule for combining exponents in this way says I need to multiply the two together.0769

Z1/150774

If I will write this one as a radical them I’m looking at the 15th root of z.0779

There are many of the different examples that we can get more familiar with using these radicals and rational exponents.0786

Let us do one where we work on writing it using these radicals.0796

I’m looking at the 8th root of the entire 6z5 – 7th root 5m40805

Be careful when it comes to trying to combine things any further from there.0824

We have not covered yet on what to do with addition.0828

Most of our rules cover our rules for exponents you got to be careful on what you do with addition.0834

I’m going to leave this one just as it is and not work on combining.0843

One problem that I will have is that my bases are not the same in anyway.0849

This is good as it is.0855

Be very more familiar with taking the radicals and turning them into rational exponents.0858

Remember that the top will represent the power and the bottom of those fractions will represent the index of the radical.0864

Thank you for watching www.educator.com.0871

Welcome back to www.educator.com.0000

In this lesson we are going to look at simplifying rational exponents.0003

The things we will look at are our product rule, quotient rule and what to do when they involve things like radicals.0009

We will also look at a few radicals that involve variables and how can we get into simplifying much higher roots.0018

If you have two non negative real numbers call them A and B.0028

And what you can do with them when they are being multiplied underneath a radical is put them under the same radical.0033

If those are already underneath the same radical and you are also free to go and separate them back out. 0041

They are each underneath their own radical.0047

This rule right here is good for either simplifying or combining things.0050

You can use it in both direction, combine or break apart. 0055

One very important thing to note is that this rule only works for multiplication, 0060

do not try and use any type of splitting thing if you have addition or subtraction.0065

Where the simplifying part comes into play is by taking a very large number, large expression and breaking it up into those individual parts.0075

We can consider a radical simplified when there is no perfect square under the radical anymore.0085

Look at these different radicals, find the perfect squares that are in there and split up the roots over those perfect squares.0093

In a similar rule, we have that we can either combine things under radical when we are dealing with division 0106

or if they are both under division, we can split up over the top and over the bottom.0112

This works as long as the value on the bottom is not equal to 0. 0118

Of course that is because we cannot divide by 0. 0122

Feel free to use this rule in both directions either combine or separate them.0127

Note that this rule only works for division.0131

Again, do not try and use this for addition or subtraction. 0133

If we happen to have a variable underneath the root and it is raised to the power, 0140

we can assume that variable represents a non negative number.0143

That will make sure that we are dealing with real numbers and not imaginary numbers.0149

The product in the quotient rule that we have just covered apply when all the variables up here under a radical sign as well. 0155

We are dealing with non negative real numbers with these various different rules.0162

You can even use these rules for some much higher roots and when you are looking to simplify those it depends on the index. 0171

For example, if you are looking to simplify a cubed root, then what we are doing is 0183

we are trying to take out all the cubes that are underneath that root. 0187

Here is a quick example. 0190

Maybe we are looking at the 3√a3 without simplifying it to just an (a).0192

In cube roots this happens to be true whether (a) is positive or negative.0198

As long as you are dealing with square roots, the square root of the square number is always non negative. 0204

The product and quotient will be applied to both our higher roots as well as just square roots.0212

Let us practice a few of these rules with some example questions. 0221

If you want to simplify some numbers like the √4/49 feel free to split it up over the top and or the bottom.0225

And then you can take each of these separately. 0234

This would be 2/7.0237

Here us one where they already have a root over the top and the bottom but I cannot simplify them individually. 0245

I’m going to use that the quotient rule in the other direction and we will put them underneath 1 root.0251

From there I can figure out how many times 3 goes into 48.0259

I have that 3 goes in the 48 at least 10 times. 0269

I have 18 left over, that would be 3 × 6.0277

By combining them underneath one root it allowed to go ahead and simplify the numbers a little bit better.0291

I just have to look at √16 is just 4.0296

Whether you are putting things underneath the same root or separating them out sometimes they cannot be simplified.0302

In this next one I will look at the root of the top and the root of the bottom.0311

With the bottom I can take the √36 is just 6 but the √ 5 has to stay as it is because it is not perfect square.0316

Let us look at one more.0331

You want to look at the √3/8 × √7/20335

With this one I have two different radicals.0347

Do not let those fractions distract you we are looking to combine them underneath the same radical.0351

I will just combine them together using multiplication.0359

Once I do that I could go ahead and continue simplifying from here. 0365

This will be √21/60 and on the bottom 16 is a square number so I can even go further.0368

√21 / √16 will be √21 / 40379

All of these examples are designed to get you more familiar with either taking that root 0386

and putting over both the parts or combining them into a single root.0390

Let us try a few that have some variables in them and notice how the same process works out. 0397

What I want to do is simplify the √x6 and I could imagine splitting up my x6 into lots of x2.0404

If I use my rule to split up the root over each of those I will have √x2, √x2, I will have that three different times.0419

Each of these would simplify to just an x.0429

I can go ahead and package that altogether as x3.0433

Notice how that fits with some of our other rules such as rewriting this as x6 / 2 which is x3. 0436

All our rule stays nice and consistent with one another, they both agree.0448

Let us try another one.0454

We will slip this one up over 100 and over p8.0457

I could think of the √100 that is 10 and then for the p8 that could be (p4)2.0465

That square and square root can take care of each other 10p4.0475

That one will be good.0481

Let us try another one. This one is √7/y40485

I will put the square root on the top and in the bottom.0492

We want to look at that y4 as (y2)2.0498

In that way we can see that this square and square root will take care of each other.0505

I’m left with the √7/y20511

In this higher root the same rules apply.0523

When it gets down to breaking them up and simplifying them, nothing changes.0526

We just have to be worried about looking for our cubed numbers or in some of these other examples a number raised to the fourth power.0531

Starting with 108, I want to think of this one, I think about a cubed number.0539

108 is the same as 4 × 27.0553

That is important because 27 is one of my cubic numbers.0558

I will split it up over the 4 and 27.0565

We can go ahead and simplify that 3rd root of 27.0570

That one will be just 3.0577

We like to put our numbers first, let me write this one as 3 × 3rd root of 40581

Note how we do not simplify the 4 because it is not a cubic number, it is a square number.0589

Let us try the next one, the 4th root of 160.0597

I want to think of numbers raised to the fourth power.0603

If you want you can even make a list.0606

14 = 1, 24 = 16, this is 4 16 × 10, the 4th root of 16, 4th root of 10.0607

We will go ahead and simplify the 4th root of 16 is 2, the 4th root of 100627

That one is done.0633

Let us use the quotient rule on this last problem.0640

On top I will have the 4th root of 16 and in the bottom 4th root of 6250644

We saw earlier that the 4th root of 16 we will go ahead and simplify that will be just 2.0652

With 625 that is 54, 2/5.0658

Let us do the same thing with higher roots but we will deal with some variables underneath the roots.0671

The first one I have 3rd root of z9.0679

I could look at this as (z3)3.0685

What I want to do in this is highlight this root and that 3rd root end up getting rid of each other.0693

What is left over is z3.0700

Note that this meshes with some of our earlier work and I could write this as z9/3 and I will still get z3.0705

All of our rules are staying nice and consistent with one another.0715

In here the3rd root of 8x6.0719

I’m looking at 3rd root of 8 and 3rd root of x6.0724

That will reduce to 2 and this would be like (x2)3.0730

2 × x2 that will be the final reduced expression.0745

I have the 3rd root of 54 t50753

In this one I will try to break down as much as possible but remember if we have things that are still not cubic, 0757

we have to leave them underneath the root.0763

Let us see.0766

What cubic numbers can I find in 54?0767

That will be the same as 27 ×2 and what cubic numbers can I grab from t5?0772

That is a t3 and t2.0785

Notice here I am thinking of my product rule for exponents and how I have to add those exponents together to get 5.0790

I’m splitting them up just this way so that I would have one of them as cubed.0797

We could take the cubed of everything in here.0805

Cube of 27, 2 and t3, t2 0808

Some of these will simplify and some of them would not.0817

3rd root of 27 is 3, 3rd root of 2 has to stay, 3rd root of t3 is t, and the 3rd root of t2 has to stay.0821

Gathering up what was able to be taken out I will have 3t × 3rd root of 2t20837

Simplify and bring out as much as you can, and if you cannot go ahead and leave them under the root.0846

One more, I will first split up the root over the top and on the bottom.0853

A15 / 3rd root of 640860

On the top of this that is (a5)3 underneath the cube root.0868

For the 64 on the bottom that is something that I could have just take the cube root of.0879

This would simplify into a5 and 3rd root of 64 is 4.0888

That one is simplified.0895

Be familiar with your rules for these radicals especially when it gets to being able to split things up 0900

and simplify over each of the individual components.0907

Thank you for watching www.educator.com.0910

Welcome back to educator.com.0000

In this lesson we are going to take a look at adding and subtracting radicals.0003

You may notice in a lot of my other lessons I avoid trying to add or subtract radicals as much as possible.0009

It is because there is a few problems that you will run into when you try and these radical expressions.0015

Look at some of the problems that we want to avoid. 0020

Do not get into the actual rule for adding and subtracting that way you will know 0023

what situations that you can add and subtract and which situations you cannot.0027

A lot of the other rules that we have picked up for radicals so far we have been doing lots of multiplication and division.0035

You will notice that in those rules they follow pretty much exactly from the rules of exponents.0042

There is a nice product rule, quotient rule, and they seem to mimic one another.0048

To understand some of the difficult things that we run into with adding, subtracting we have to look at what happens 0053

when some of our rules for adding, subtracting when you have things with exponents. 0060

For example, I want to put together x2 + 3x what problems when I run into? can I put those two things together or not?0064

You will notice that you will run into quite a bit of a problem.0075

These ones do not have the same exponents, I cannot put them together.0079

Since the terms are not like terms and they have different exponents maybe we can change it around and try a different situation. 0084

Let us go ahead and try ab2 + r2.0100

Can we put those together? After all the exponents here are exactly the same. 0104

Can we combine those like terms?0110

These one are not like terms.0114

We ran into a very similar problem the bases are not the same with these two.0115

Those ones are not the same, they do not have the same base then we cannot put them together. 0124

We usually run into one of those two problems where we are dealing with radicals.0131

Either they do not have the same base or they do not have the same exponents. 0135

Here is a quick example involving radicals so we can see what I’m talking about.0139

Here I have 3rd root of x + √30142

If I write those as exponents, then it is like x1/3 + 3^½.0147

These have different bases and they have different exponents, there is no reason why you should be able to put these together.0153

After seeing many of these different examples, you might that be under the false assumption that we cannot add any radicals together.0163

In some instances you will be able to put these radicals together you have to be very careful on certain conditions. 0173

One, when putting radicals together you have to make sure that their powers, those would be the indexes of each 0182

of the radicals are the same and you have to make sure that the radicand or the bases are exactly the same.0187

I can put together the 3rd root of 5x with the other 3rd root of x.0194

It is completely valid because when written as exponents I have the bases the same and they are both raised to the power of 1/3.0200

How would I go about actually putting them together?0210

I will treat them just almost like an entire variable.0213

If I was adding u + u I will get two u.0216

Notice how this common pieces here.0221

Another way of saying that is, we simply add together our coefficients out front so one of those should equal two of them.0224

Make sure that you keep in mind that if you are going to add and subtract radicals 0237

you must have the same index on those radicals and you must have the same radicand.0242

That is the part underneath radicals.0246

Once you get to the addition or subtraction process, look at your coefficients out front, so (5 × √x) + (3 × √x) – (6 × √x) .0248

I’m looking at 5 + 3 - 6. 0260

That will give me a result of 2.0265

This is exactly the same process that you would go through if you are just adding like terms.0270

That will be u2 + 7u2 - u2 then you are just looking at these initial coefficients like 2 + 7 - 1, and that would give you the 8.0275

Let us see if we can take a look at some examples on when we can add and subtract these radicals.0288

In the first one I'm looking at 3 × 4th root of 17 – 4th root of 170294

Let us check, the indexes are exactly the same we are looking at the 4th root and our radicands. 0300

That is the part underneath, they are both 17.0308

We are going to look at the coefficients, 3 - 1 = 2.0312

I have 2 4th root of 17 and that one is good.0318

Let us look at this other one.0326

21√a + 4 3rd root of a0328

It is tempting to want to put these ones together but we cannot do it. 0335

This one is the square root and this one is a cubed root.0340

It must have the same index if you have any hope of getting those together.0353

Let us look at some others.0357

On this one we want to add or subtract if possible.0361

I have 3 + √xy + 2 × √xy0365

Both of these are dealing with square roots and that is good.0373

Both of these are with an xy underneath that root.0376

We will simply add together their coefficients.0381

This will give us a 5√xy.0386

Let us see how that works for the next one.0392

7 × 5th root of u3 - 3 × 5th root of y3.0393

That is so close.0400

Both of them have a 5th root and things are being raised to the third power, all of that is matching up but the variables are completely different. 0402

One is a u and one is a y.0412

There are a few instances where the indexes are the same, but it looks like that radicand on the part underneath is completely different.0431

It is tempting to write those off and say okay, I probably cannot put those together using addition or subtraction.0440

Sometimes if you can do a little bit of simplification and get them the same then you can go ahead and put those together.0446

Let me show you an example of numbers. 0453

Suppose I wanted to put together √2 + √8 and just looking at them I will say that wait 8 and 2 they are not the same, I cannot put them together.0455

The √8 over here that is the same as 4 × 2 and I can take √4 and that would leave me with 2 × √2 .0466

I can simply rewrite the next one as 2 × √2 .0480

In doing so now my radicals are exactly the same and I can simply focus on these coefficients out front.0485

I can see that 1 + 2 does equal my 3. 0493

Do not be afraid to try and simplify these a little bit before you get into the addition or subtraction process.0497

Let us try that and keep it in line with these ones.0504

We want to rewrite the expressions and then try and add or subtract them if possible.0507

The first one I have a -√ 5 + 2 × √125.0513

I have -√ 5, 2 and 125 if I want to end up rewriting that, that is a 5 × 25 so that one reduces.0520

I have the √5 × 5.0542

Let us write that as √ 5 + 10√ 5 and now that I have my radicals the same now just focus on this coefficients -1 + 10 would be 9√ 5.0548

The next one I chose a big number but no worries, we can take care of this one.0568

We are looking for the 4th root of 3888 + 7 × 4th root of 30575

If I have any hope of putting these together I want to match this 4th root of 3 over here.0584

As I go searching for ways to break down that very large number, the very first thing I'm going to try and break it down with is 3.0591

Let us see if I can.0600

It is the same as 1296 × 3 that is good because if I look at 1296 I can take the 4th root of that and I will get 6.0608

I simply have to add together these other radicals here by looking at their coefficients.0631

6 + 7 = 13 4th root of 30637

Let us try one more, √72x - √32x 0648

Let us try and simplify this as much as possible.0655

With the first one, looking at 72 is the same as 36 × 2 and with 32 that 16 × 2.0659

Notice how I have the square numbers underneath here, but I can go ahead and simplify.0674

√36 that will be 6 and I still have that 2 underneath there, √16 will be 4 and there is the 2x for that one.0679

I’m looking at 6√2x -4 × √2x or 2 √2x0691

These ones are a little bit larger involving some much higher roots, but the same process applies.0708

We must get the part underneath the roots the same if we are going to be able to put these two together.0714

10 × 4th root of m3 is already broken down as far as it will go.0724

The next one I could look at the 6561 and try and take its 4th root and break it apart from its m3.0736

The good news is that one does break down, you will get 9.0751

4th root of m30763

10 4th root of m3 + 9d4th root of m3 and we could put those together 10 and 90 is 1004th root of m3 .0767

Let us try this next one here, this one is the 3rd root of 63xy2 – 3rd root of 125x4y50789

In this one we will not only need to simplify those numbers but also take care of the variables like the x and y.0799

First, the numbers the 3rd root of 64 what does that break down into?0811

That will go in there 4 times and I have a 3rd root of xy2, 0820

that one does not break down any further because both of those powers are smaller than the index of 3. 0827

The 3rd root of 125 would be 5 and let us see what we can do with those variables.0834

X3 × x that will be x4 and y3 × y2 that would give me y5.0843

This right here I can go ahead and take out of radicals. 0855

Okay, taking out the x, taking out the y, and I still have xy2.0865

Things are looking good and the part underneath the radical is now exactly the same, we will worry about our coefficients. 0873

Notice how our coefficients are not like terms, I will be able to write them simply as 4 - 5xy package them together and then write my radical.0880

It is like we are just factoring out this common piece and writing it outside here.0895

In all cases, make sure you get those radicals exactly the same and combine their coefficients.0904

One last example that we can see many of our different rules in action, we will try and combine 3rd root of 2 / x12 - 3 × 3rd root of 3 / x15.0912

Starting off, I'm going to use my quotient rule to break that up over the top and over the bottom.0927

We will break this one up over the top and over the bottom as well, looks pretty good.0938

I will go ahead and simplify the roots on the bottom.0948

23rd root of 2 / x4 - 3 × 3rd root of 3 / x5.0953

Since we are looking to combine things as much as possible, 0967

I will get a common denominator by putting an x on the bottom and on the top for the left fraction.0969

2x 3rd root of 2/x5 - 3 3rd root of 3 /x5.0979

As I continue trying to put them together here is one of those situations where we are stuck, we cannot move any farther from there.0993

Since this is 3rd root of 2 and this is the 3rd root of 3 and those are different.1000

You will know it is tempting but we cannot put them together anymore. 1006

We will leave this as 2x3rd root of 2 - 33rd root of 3/x5.1011

A lot of different rules to keep track of our radicals but as long as you remember the rules follow directly from the rules for exponents you should be okay. 1021

Be very careful in adding and subtracting those radicals and make sure everything is satisfied before you even attempt to put them together. 1031

Thank you for watching educator.com1040

Welcome back to www.educator.com.0000

In this lesson let us take a look at how you can multiply and divide radicals.0003

We will first cover some rules for multiplying and dividing radicals and then get into that division process 0011

and see how we want to rationalize the denominator that involves a radical expression.0018

There are only a few rules that you have to remember when working with radicals and the good news is we have seen many of these already. 0027

For example, you can change radicals into fractional exponents.0033

If you want to combine them for division you can separate those out if you want to separate them out over multiplication.0039

You can also add and subtract radicals as long as we make sure that the radicands and the indexes are exactly the same.0050

Radicals follow our properties for all the other types of numbers so we can also use 0059

the commutative property, associative property, even on these radicals.0064

The reason why that is important is because there is a few situations where you often try to apply some rules that do not work.0072

Treat these radicals like they are any other numbers. 0080

Let us see this as we walked through the following problem. 0084

I want to multiply the √6 + 2 × √6-30087

For this one treat it like binomials and use foil to multiply everything out.0093

Every terms multiplied by every other term. 0099

Let us see our first terms would have √6 × √6, outside terms -3 √6, inside terms 2√6 and our last terms 2 × -3 = 6.0103

Then we can use our other rules to go ahead and simplify this further. 0119

I see I have two roots here, I can put those underneath the same root and that is the same as the √36 or 6.0125

-3 × √6 + 2 × √6 can I add them together or not?0135

Yes I can because they have the same radical I will just do the coefficients.0142

-3 + 2 - √6 0147

- 6 is still there. 0157

You can continue simplifying your like terms by combining the 6 and -6 giving -√6.0159

We are using many of our tools that we have learned up to this point, in order to handle these radicals.0168

Watch out for lots of situations where you need to use foil.0176

In this example I have 4 + √7 2 it is tempting to try and take that 2 and distribute it over the parts in between.0181

However, do not do that. 0191

Do not even attempt any type of distribution with this one. 0194

What you should do with it is foil because as we learned in our exponent section, this stands for 4 + √7 and 4 + √7 .0197

Those two things are being multiplied by each other.0208

I can see where foil comes in to play.0212

I will take my first terms 16, outside terms, inside terms, and last terms.0216

Unlike before I could use some other rules to go ahead and combine things.0227

I can go ahead and add these two since they have the same radical and get 8√7.0232

I can combine these under the same radical and get the √49 and that is simply 7.0239

We can go ahead and finish off this problem, 16 + 7 would be 23 or 23 + 8 × √7.0249

Let us take a close look at division.0266

When dividing a radical expression we go ahead and rewrite it by rationalizing the denominator. 0269

If you have never heard that term before rationalizing the denominator, 0278

it is a way of rewriting it so that there is no longer a radical in the bottom.0281

That seems a little weird. 0286

I mean, if we are interested in dividing by radical why are we writing it that there is no radical in the bottom. 0287

It seems like we are side stepping the problem like we are not ending up dividing by the radical.0294

It is just a simpler way of looking at the whole division process and it is something that you have done before with fractions. 0299

For example, when we have 1/3 ÷ 2/5 you have been taught that you should always flip the second and then multiply.0307

Why are we doing that? Why do not we just go ahead and divide by 2/5?0314

What is the big deal with a turn it into multiplication problem?0317

One, we will show you how accomplish the same thing, but by flipping the second one multiplying it does it in a much simpler way.0321

Also supposed I write this problem as 1/3 ÷ 2/5, you recognize that this is one of our complex fractions. 0330

I can simplify a complex fraction by multiplying the top and bottom by the same thing.0341

I will multiply the top and bottom by 5/2, that should be able to get rid of our common denominator.0348

On top I would have 1/3 × 5/2 and on the bottoms 2/5 × 5/2 would be 1.0359

What we are sitting on the top there is the 1/3 and there is the 5/2 which comes from that rule we learned.0369

That we should take the second one, flip it and multiply. 0379

But notice how we are doing that by simplifying the bottom now we are dividing by 1.0382

It is a great way to end up rewriting the problem, and taking care of it in a much simpler way.0390

That is exactly what we want to do when we are rationalizing.0396

The actual steps for rationalizing the denominator look a lot like this. 0402

First, we will end up rewriting the rational expression, so that we will end up with no root in the bottom.0407

When we rationalize we try to get rid of that root.0413

We will do this by multiplying the top and bottom by the smallest number that gets rid of that radical expression.0416

That sometimes you can use some larger numbers, but it is best to use the smallest thing that will get rid of that radical. 0422

It saves you from doing some extra simplifying in the end.0429

If we are dealing with square roots I recommend try and make that perfect square 0434

in the denominator and that should be able to rationalize it just fine.0437

Let us see one of this division by radical in process and this is also known as rationalizing the denominator.0444

I have 2 ÷ √(2 )0451

I’m going to end up rewriting this so that there is no longer √2 in the bottom.0456

We will do this by multiplying the top and bottom by another root so that we will have a squared number in the root for the bottom.0462

On top I have 2√(2 ) and in the bottom I can go ahead and put these together and get √2 × 2, which is the √4.0471

I have created that square number on the bottom and now we can go ahead and simplify it.0485

That will be 2.0493

If you can simplify from there go ahead and do so, you will see that this one turns into the √2.0496

When I take 2 ÷√2 like the original problem says the √2 is my answer.0503

Let us try another one.0513

This one is 12 ÷ √50515

We are looking to multiply the top and bottom by something to get through that √5 in the bottom we will use another√5.0519

12 √5 for the top and bottom √5 × √5 would be the same as √25.0529

The good news is that one reduces and becomes just 5.0541

If you are doing a division, you are dividing by the √5 even though it looks like you are changing into multiplication problem.0549

This is also known as rationalizing the denominator.0558

You can use these tools like rationalizing the denominator for some much higher roots as well. 0565

The thing to remember by is that you want to multiply by the smallest root actually complete set root.0571

When we are dealing with the square roots it look like always multiplying by the same thing on the bottom.0577

With high roots, sometimes that might not be the case.0583

Let us look at this one.0586

1 / 3rd root of 2 if I try and multiply the top and bottom by 3rd root of 2 it is not going to get rid of that radical.0587

We will be left with the 3rd root of 2 / 3rd root of 4 and 4 is not a cubic number.0598

It is like it did not have enough of the number on the bottom to go ahead and simplify it completely.0604

Which should we multiply?0612

If I want a cubic number in the bottom but I'm going to use 3rd root of 4 .0615

When I take that onto the top and bottom you will see that indeed we do get that cubic number that we need.0625

From the bottom this would be 3rd root of 8 and on the top I will just have 3rd root of 4 . 0632

The bottom simplifies becomes 2 now my answer is 3rd root of 4 ÷ 2.0640

Let us try another one of those higher roots. 0652

This one is 3 ÷ 4th root of 90654

Let us see, what would I have to use with a 4th root of 9?0659

9 is the same as 3 and 3, it will be nice if I had even more 3’s underneath there.0664

Let us say a couple of more.0671

I will accomplish that by doing the top and bottom by 4th root of 9 .0674

3 × 4th root of 9 for the top and 4th root of 81 on the bottom.0682

Now it is looking much better. 0693

34th root of 9 on the top, the bottom simplifies to just the 3 and now we will cancel out these extra 3s. 0696

I have 4 4th root of 9 .0706

This one we want to rationalize the denominator and if you look at it you must think what denominator are we trying to rationalize?0715

Is there a root in the bottom?0722

Because of our rules that allow us to break up the root over the top and bottom, there is.0725

In this one we have the 3rd root of 2y / 3rd root of z.0731

We can see we have 3rd root of z and it definitely needs to be simplified.0744

How are we going to do that?0750

Since I'm dealing with a cubed root I will need an additional 2 z's for that bottom.0752

Let us use the cube root of z2.0758

Watch what that would give us here on the bottom.0765

3rd root of z3 on top, the 3rd root of 2y z2.0768

The bottom simplifies and there will no longer be any roots in the bottom.0779

3rd root of 2y z2 / z and I know that this one is done.0784

One thing that can make the rationalization a more difficult process since we have more than one term in the bottom.0797

Our main goal is to end up rewriting the bottom so that there is no longer a root. 0805

If we have more than one term we are going to end up using something known as 0812

the conjugate of the expression to go ahead and get rid of it. 0815

What the conjugate is, it is the same as our original expression, but it has a different sign connecting them.0820

That will allow us to get rid of that root.0827

To show you why we get through the root we will use an example. 0830

Here I have 4 + √3, the conjugate of this one would be the same I have a 4√3 it will have a different sign connecting them 4 - √3.0835

Watch what happens when I foil these two together.0852

The 4 + √3 and its conjugate.0855

Starting with the first terms I have 4 × 4, which would be 16. 0863

My outside terms would be -4√3, my inside terms will be 4√3.0868

I can move on to my last terms -√3 × √3.0877

A lot of things are happening when you multiply by its conjugate.0882

One, notice how our outside and inside terms where the same but one was positive and one was negative.0886

When you are dealing with conjugates that should always happen.0892

Those two things are gone.0897

We will focus was going on down here on the end.0899

-√3 × √3, -3 × 3 which be 16 - √9 which would be 9. 0903

The numbers may change to make it different but at this point, there is no more radical.0918

Because those two radicals multiply and I get that perfect square number, there is no more radicals to deal with.0925

I can just take 16 - 9 and get a result of 7.0932

By multiplying by that conjugate, I got rid of all instances of all radicals.0940

This is why it will be important to use it when getting rid of our radical on the bottom.0946

Let us see for these examples.0953

Notice we need that conjugate because we have two terms in the denominator. 0955

I'm going to multiply the top and bottom of this one by the conjugate.0962

√5 – 2 and √5 - 2 0968

When dealing with more than one term, remember that you will have to foil out the bottom.0976

Also remember that though you will have to possibly foil or even distribute the top.0984

The top 9√5 – 18 and working with the bottom and foiling that out my first terms would be √5 × √5 =5.0991

Outside terms and inside terms they are going to cancel out I know I’m on the right track.1006

2 × -2 -41015

We will go ahead and do some canceling and let us see what we have left over.1020

9√5 - 18 / 5 – 4 = 1 and this reduces to 9√5 - 18. 1024

Notice how we have divided by that radical because we have gotten rid of that radical in the bottom.1037

In this last example, let us look at rationalizing the following denominator.1046

We have 7 ÷ 3 - √x1050

With this one I’m going to have to multiply the top and the bottom by its conjugate.1055

You know what is next in there, this will be 3 + √x we will do that on the bottom and on the top. 1064

Let us see what this does to the top as we distribute and remember that on the bottom we will foil.1073

I get 21 + 7√x for the top.1082

On the bottom we have 9 + 3 × √x - 3 × √x and then - √x × √x.1089

If we do things correctly we should get rid of all those radicals in the bottom.1108

+ square root - square root those two will take care of each other and then my √x× √x = x.1114

This will leave us with 21 + 7 √x / 9 – x.1123

We have got rid of all those radicals in the bottom you can say that our denominator is rationalized.1133

You can see that when you are working with dividing radicals you always have to keep in mind 1139

what you would put on the bottom in order to get rid of all those radicals. 1144

If you only have a single term feel free to multiply by what would complete whatever that radical is.1148

Complete the square or complete the cube.1154

If you have more than one term use the conjugate to go ahead and rationalize the denominator.1157

Thank you for watching www.educator.com.1162

Welcome back to www.educator.com.0000

In this lesson we are going to go ahead and solve some radical equations. 0003

We only have one thing to pickup in this lesson it is just all the nuts and bolts on how you solve on these radical equations.0009

A radical equation is an equation in which we have variables present in our radical.0018

That will be something like this, I have √x-5 - √x-3 = 1.0026

To solve these we essentially want to end up isolating these roots then raise each side of our equation to a power in order to get rid of them. 0033

It has to be the appropriate power. 0044

For dealing with square roots we will square both sides.0045

If we have say 4th root then we will raise both sides to the 4th power.0048

With these ones it is extremely important that you go ahead and you check your solutions when you are done.0054

In the solving process and you raise both sides to a power we may introduce solutions or possible solutions that do not work in the original.0060

Always check your solutions for these types of equations.0069

Let us get an idea of what I’m talking about with √3x + 1 - 4 = 0. 0075

The very first thing you want to do when solving something like this is get that root all by itself. 0082

Let us go to both sides of the equation so we can do that. 0088

√3x+ 1 = 40092

We have isolated the root we are going to get rid of it by raising both sides to the power of two.0097

The reason why we are doing this is because we have a square root.0104

3x + 1 = 42 which is 16 0109

Now they got rid of your roots, we simply solve directly.0117

The type of equation that we might end up with could be quadratic, could be linear, but we use all of our other tools from here on out.0122

-1 from both sides we will get 15 and divide both sides by 3 will give us 5.0129

You will realize that this looks like a possible solution always go back to your original for these ones and check to see if it does work. 0136

I’m going to write this out here -4 does that equals 0 or not.0146

Let us put in 5, 3 × 5 would be 15 and all of that is underneath the root and then 15 + 1 would be 16.0156

√16=4, 4 -4 does equal 0, I know that this checks out and x equals 5 is our solution.0172

This one, we will go ahead and try and solve it is 5 + √x+ 7 = x.0191

Isolate our radical by subtracting 5 from both sides of the equation √ x + 7 = x - 5.0200

To get rid of that radical let us go ahead and raise both sides to the power of 2.0213

On the left side we will just have x + 7.0222

Be very careful on what is going on over here, keep your eyes peeled because notice how we have a binomial and we are squaring it.0226

In fact, you should look at it like this x - 5 × x – 5.0238

That way you can remember that this is how it should be foiled instead of trying to do some weird distribution with the two.0244

It does not work like that.0250

Let us see what we have when we foil.0253

Our first terms are x2, outside terms - 5x, inside terms - 5x, and our last terms 25.0256

We can go ahead and combine to see what we will get x2 - 10x + 25.0270

You can see that this one is turning into a quadratic equation because we have x2 term.0285

Since it is quadratic we will get all of our x on to the same side and see if we can use some of our quadratic tools in order to solve it.0291

0 = x2 - 11x + 180301

How shall we solve this quadratic? 0312

It is not too bad, it looks like we can use reverse foil to go ahead and break it up and see what our parts are.0315

x × x = x2 then 2 × 9 = 180322

I know that 2 and 9 are good candidate because If I add 2 and 9 I will get that 11.0330

Let us make these both negative.0335

Have it x = 2 or x = 9.0337

Two possible solutions and they are possible because they might not work.0342

Let us check in the original 5 + √ 2 + 7 does it equal 2?0347

Let us find out, 2 + √7 = √ 9, √9 is 3 so I get 8.0359

Unfortunately that looks like it is not the same as the other side.0372

This one does not work and we can mark it off our list.0377

Let us try the other one 5 + the square roots and we are testing out a 9 so we will put that in 9 and 9.0383

Underneath the square root, 9 + 7 would give us a 16 and √16 =4, I will get 9 which is the same thing as the other side. 0396

That one looks good. 0410

You will keep that one as your solution. 0413

Be very careful and always check these types of ones back in the original, you will know which ones you should throw out.0416

If there happens to be more than one radical in your equation then we can solve these by isolating the radicals one at a time.0427

Do not try and take care of them both at once. 0434

Just focus on one, get rid of that radical and focus on the other one and then get rid of that radical. 0436

It does not matter which radical you choose first.0442

If you have a bunch of them just go ahead choose one and go after it. 0446

Be very careful as you raise both sides to a power since you have more than one radical in there, 0450

when you raise both sides to power it will often end having to get foiled.0456

Let us see exactly what I'm talking about with this step, but will have to be extremely careful to make sure we multiply correctly.0460

Always make sure you check your solutions to see that they work in the original.0468

But this one has two radicals in it. 0474

I have √x - 3 + √x + 5 = 40477

It does not matter which radical you choose at the very beginning and I'm going to choose this one.0483

I’m going to work to isolate it and get all by itself x + 5 = 4 – and I have subtracted the other radical to the other side.0490

Since √x+ 5 is the one I’m trying to get rid of, I’m going to square both sides to get rid of it. 0505

That will leave me with just x + 5 on that left side.0514

Be very careful what happens over here on the other side, it is tempting to try and distribute the 2 but that is not how those work.0520

In fact if we are going to take look at this as two binomial so we can go ahead and distribute.0528

Let us go ahead and take care of this very carefully. 0542

Our first terms 4 × 4 = 16, our outside terms we are taking 4 × -√x – 3, -4 × √x - 3.0545

Inside terms would be the same thing -4 √x - 3 and now we have our last terms.0560

negative × negative = positive and then we have √x -3 × √x -3.0570

That will give us √x - 32 since it will be multiplied by themselves. 0581

When you do that step the first time, it usually looks like you have made things way more complicated. 0590

I mean, we are trying to get rid of our root but now it looks like I have 3 of them running around the page. 0596

It is okay we will be able to simplify and in the end we will end up with just one root, 0600

which is good because originally we started with two and if I get it down to one we are moving forward with this problem.0605

Let us see what we can do.0612

On that left side I have 16 these radicals are exactly the same, I will just put their coefficients together -8 √x -3 0614

The last one square of square root would be x -3.0626

Now you can start to combine a few other things. 0634

I will drop these parentheses here, if I subtract x from both sides that will take care of both of those x.0639

Let us see I can go ahead and subtract my 3 from the 16, 5 =13 - 8√x -3.0650

Let us go ahead and subtract a 13 from both sides, -13 – 13, -8 = -8√x -3.0665

It looks like we can divide both sides by -8, 1 =√x -3.0682

That is quite a bit of work, but all of that work was just to get rid of one of those radicals and we have accomplished that.0689

We got rid of that one. 0695

But we still have the other radical here to get rid of.0696

We go through the same process of getting it all alone on one side, isolating it, squaring both sides get rid of it, and solving the resulting equation.0700

This one is already isolated we are good there.0709

I will simply move forward by squaring both sides. 0711

1 = x -3 let us get some space. 0716

If I add 3 both sides this will be 4 = x.0724

Even after going through all of that work and just when you think you have found a solution, we got to check these things. 0729

We have to make sure that it will work in the original. 0735

4 - 3 + √4 + 5 we are checking does that equal 4.0741

Let us see what we have.0756

4 - 3 would give us √1 , 4 + 5 would be 9, so√4 =1 and √9 = 3 and fortunately 4 does equal 4.0758

We know that this solution checks out, 4 =x.0774

If you have more than one of those radicals just try and get rid of them one at a time.0781

You might be faced with some higher roots and that is okay.0788

You will end up simply using a higher power on both sides of your equation in order to get rid of them.0791

In this one I have the 3rd root of 7x - 8 = 3rd root of 8x+ 2.0796

If I'm going to get rid of these cube roots I will raise both sides of it to the power of 3.0802

Leaving me with 7x - 8 = 8x + 20810

I can work on just giving my x together.0818

-7x from both sides, -8 = x + 2.0822

We will subtract 2 from both sides and this will give us -10 = x.0833

Let us quickly jump back up here to the top and see if that checks out.0841

3rd root of 7 × putting that -10 - 8 and 8 × -10 + 20845

Let us see if they are equal.0866

This time I’m dealing with 3rd root of (-70-8) to be the 3rd root of -78.0868

The other side I have -80 + 2 which is the 3rd root of -78.0880

It looks like the two agree. 0887

I know that the -10 is my solution.0890

With those radicals try and isolate them, and then get rid of them by raising them to a power.0894

Always check your solutions with these ones to make sure they work in the original.0899

Thanks for watching www.educator.com.0904

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at complex number. 0002

I held off on this very special type of number so you could build a lot of things about them that you will see in the section 0007

such as multiplying them together which will look like multiplying polynomials and rationalizing denominator will look a lot like the division process.0013

Here are some other things that we will cover.0022

First we will get into a little bit about that the vocabulary from the complex numbers.0024

You have heard about imaginary numbers and you will see how that works with complex numbers. 0028

We will learn about the real and the imaginary part of complex numbers. 0033

Many of the properties involving these complex numbers are the same properties that you have seen before for real numbers. 0039

Then we will get into the nuts and bolts on how you can start combining these things. 0046

That means adding, subtracting, multiplying and dividing our complex numbers.0051

Near the very end we will also see some nice handy ways that we can simplify powers of (i) 0056

so you can always bring them down to the most simplest form.0061

An imaginary number (i) is defined as the √-1.0067

It seems like a small definition and it does not seem to be very useful, especially if we want talk about lots of different numbers.0075

By defining it in this way, the √-1 it can actually express the root of any negative number using this imaginary number (i).0082

Here is a quick example to show how it works. 0091

Suppose I’m looking at the √-7 using some of our properties we can go ahead and break it up into it -1 × 7.0093

And then break up the root over each of those parts.0102

You will see the definition shows up right there. 0105

I'm taking the √ -1.0108

It is that piece that gets turned into an (i) and now I'm representing this number as (i) × √7.0111

Think of doing this for many other numbers where you have a negative underneath the square root.0119

If I had -23 underneath the square root that will be (i)√23 and a negative comes out as the (i).0125

√ -16 would be (i) × 4 or just 4i.0137

How those imaginary numbers relate to our complex numbers?0150

Complex number is a number of the form A + B(i).0154

You can see that the imaginary number is sitting right there, right next to that B.0158

Those other values A and B are both real numbers.0164

We call A the real part of the complex number and B the imaginary part, since it is right next to the imaginary number.0168

You can represent a lot of different numbers using this nice complex form.0175

Maybe I will have a number like 3 – 5i that would be in a nice complex form.0180

I can read off the real and imaginary part.0186

We could throw in some fractions and in here as well, so maybe ½ × 2/3(i).0189

We can also have complex numbers like 0 + 3(i).0196

We would probably would not leave it like that, something like this would probably we will write as 3(i).0202

Notice how we can still say it is in a complex form where the real part is 0.0207

The good part about these complex numbers is they will obey most of the usual laws that you are familiar with.0217

Our commutative, associative, and distribution, all of those will work with our imaginary numbers as well. 0223

The things that we could do with real numbers, we can do with imaginary numbers as well. 0230

Some of the things that we will have to watch out for is when we get down in the simplification process, so keep a close eye on that.0236

Let us get into how you can start combining these numbers together.0247

To add or subtract complex numbers this is a lot like having like terms.0251

Be very careful to watch the signs when combining these numbers together.0257

To show you how this works I have 2 – 5(i) that complex number + 1 + 6(i).0262

Since I'm adding I'm going to go ahead and drop my parentheses here and just figure out which terms I should add together.0269

Notice how all of our real parts I can go ahead and put those together and I can go ahead and collect together the imaginary parts.0280

That is what I'm talking about when I say adding like terms. 0290

The real ones 2 + 1 would be 3 and the imaginary I have -5(i) -6(i) that will give me -11(i).0294

I will write those as 3 -11(i).0309

When you are working with subtraction we have to be very careful with your signs.0314

Let me show you, suppose I have 3 + 2(i) and now I'm subtracting a second number 4 + 7(i).0319

Parentheses will come in handy because I need to distribute this negative sign to both parts of that second complex number.0334

I will look at this as 3 + 2(i) – 4 – 7(i) and now I can see the parts and the like terms that I need to go ahead and put together.0343

3 – 4 = -1 and 2 – 7(i) = -5(i)0357

This would be written as -1 – 5(i)0369

Again, combine those like terms.0373

To multiply complex numbers think of multiplying together two binomials.0379

It looks a lot like the same process. 0384

When we are multiplying together two binomials we have great way of remembering that we can use the method of foil to accomplish that.0387

There is one additional thing you have to remember, when you have (i)2, we can go ahead and reduce that to -1,0394

that seems a little odd but let us see why that works.0401

If I'm looking at the √-1 and that is our (i) and suppose I take that and I square it.0404

I want to square that square root I will get -1 on the left.0413

That reveals that (i)2 = -1.0419

Watch for that to show up in the simplification process.0424

Let us go ahead and foil out these two complex numbers.0428

2 – 5(i) multiplied by 1 – 6(i)0432

We will begin by multiplying the first terms 2 × 1 = 2.0436

We will move on to those outside terms 2 × -6(i) = -12(i).0441

On the inside terms -5(i) and now let us go ahead and do the last terms -5(i) × -6(i), 0453

minus × minus would be +, 5 × 6 is 30 and now there is my (i)2.0465

We will go along and we will start crunching things down like we normally would do if they were a couple binomials.0473

2 – 17(i) now comes this interesting part this (i)2 here is the same as -1.0480

We will go ahead and rewrite it as a -1. 0491

You can see there is more than we can do to simplify this.0498

I’m not only multiplying the 30 × -1, but then I can go ahead and combine it with the two out front.0501

2 – 17(i) - 30 and then let us combine these guys that will give us -28 – 17(i).0508

This one is good.0524

With many of these problems where combining complex numbers you know that you are done 0527

when we finally get into that nice complex form.0532

You can easily see the real and imaginary parts.0535

To go ahead and divide complex numbers, we want to multiply the top and the bottom by the complex conjugate.0542

The complex conjugate of a number will look pretty much the same but it will be different in sign.0550

If I'm looking at the complex conjugate of A + B(i) that will be –B(i).0557

Let us just do some quick example. 0563

Let us see if I have 3 + 2(i) its complex conjugate would be 3 – 2(i). 0564

If I had -7 – 3(i), the complex conjugate would be -3 + 3(i).0573

The only thing that is changing is that negative sign or that sign on the imaginary part.0583

This will have the feel of rationalizing the denominator that is why we had to cover a lot of that information before working on these complex numbers.0591

I’m going to look at the bottom of this particular division problem 4 + 2(i) ÷ 3 + 5(i).0599

I'm going to find the complex conjugate of 3 + 5(i), that would be 3 – 5(i).0606

Once we find that complex conjugate we will multiply on the top and on the bottom of our complex fraction.0616

Multiply on top, multiply on the bottom.0628

This one involves quite a bit of work.0631

I have two terms in my complex number, I will have to foil the top and we will also have to foil out the bottom. 0634

This will make things look a lot more complicated at first but if you just go through the problem carefully you should do fine.0644

Taking our first terms I have 4 × 3 and that would give me 12.0656

The outside terms would be 4 × -5 (i) = -20(i).0662

On the inside 2(i) + 3(i) then would be 6(i) and our last terms 2 × -5 = -10(i)2.0668

Do not forget to multiply those (i) together as well.0680

On the button 3 × 3 = 9, outside will be -15(i), inside would be 15(i).0684

And then we have 5(i) × -5(i) = -25(i)2.0695

Some things you want to notice after you go through that forming process.0703

We are using that complex conjugate on the bottom, the outside and inside terms will end up canceling. 0707

If they do not cancel, make sure you have chosen the correct complex conjugate.0715

Notice that we have a couple of (i)2 showing up on the top and bottom. 0720

That comes from the (i) being multiplied together. 0725

Each of these will be need to change into -1.0728

Let us go through and do those two things and clean this problem a little bit.0733

I have 12 – 20 + 6(i) = -14(i), -10 × -1 ÷ 9 my outside and inside terms on the bottom cancel -25 × -1.0737

We can see the beauty of using that complex conjugate.0759

On the bottom we no longer have any more imaginary numbers. 0762

That one I was talking about how I have the feel of rationalizing the denominator and we are getting rid of those roots in the bottom.0766

Here we got rid of the imaginary numbers in the bottom.0772

Let us continue to simplify and see where we can take this problem.0776

12 – 14(i) + 10 ÷ (9 + 25) 0781

12 + 10 = 22 – 14(i) ÷ 34 0790

It is tempting to try and stop right there but do not do it, we want to continue doing this entire problem 0801

and try to get it in that nice complex form where I can see the real and imaginary part.0806

To get in that form I’m going to use the part where I breakup the 34 and 22 and under the 14.0811

It looks a lot like when we are taking polynomials when we are dividing by a monomial.0818

This will be 22 ÷ 34 -14(i) ÷ 34.0824

This particular one, both of those fractions can be reduced.0833

I will divide the top and the bottom by 2, 11/17 – 7(i) ÷ 17.0837

This one is actually completely done.0846

I can easily see my real part over here and my imaginary part -7/17.0849

That is I know that this one is done.0855

Division is quite a tricky process remember to multiply by the complex conjugate of the bottom 0859

and then work very carefully to simplify the problem from there.0864

It should be in its complex form when you are all done so, you can see the real and imaginary parts.0868

After working with doing some combining things with these various complex imaginary numbers, 0876

you may have noticed that you rarely see any higher powers of (i).0881

In fact the largest part of (i) that we came across was i2 and we turned it immediately into a -1. 0885

The interesting part is you can usually take much higher powers of (i) and end up simplifying them down.0892

To see out why this works for some much higher powers, let us just take a bunch and start picking them apart.0898

We see many instances while we are working with (i) and (i) was just equal √1 which of course we call (i).0905

When we have i2 and then we thought about squaring the √-1, so we got -1 as its most simplified result.0912

We will see what will happen with i3.0922

One way that you could interpret this would be i2 × (i).0927

The reason for doing that is because you can then take the i2 sitting right here and simplify that.0934

That will give us -1 × (i) the final result of that would be –i as it is simplest form.0942

Let us choose that same idea and see if we can go ahead and simplify i4.0955

That will be i2 × i2.0959

Both of the i2 = -10964

I have -1 × -1 = 10970

Take note how all of these are simplifying to something else that does not have any more powers on them.0975

Let us do a few more that hopefully we can develop a much easier or a shortcut way of doing this process.0983

Let us go on to i5.0990

This can be thought of as i4 × (i).0994

We have already done i4, it is way back here, that is equal to just 1.0999

1 × (i) I know this simplifies to (i).1003

On the next one i6, i4 × i2.1010

We have already done i4 that is just 1.1020

I will put in i2, I will just write across from it -1 so 1 × -1 = -1, very interesting.1024

I think we are starting to see a pattern of some of these (i).1035

Notice how they are looking exactly the same as the ones on the other side here.1038

Let us see if that continues.1044

I7 will be i4 × i3, 1 × i3 = -i.1046

Maybe the last one, i4 × i4, both of those are equal to one, this is equal to 1 as well.1058

Notice how we did starting to see same numbers.1069

There are some good patterns we can see, I mean we got past i4 and all of them contained an i4 1072

so we are getting this similar pieces right here.1079

If I was even to continue on to i9 then I will have even more groups of i4 and I could see that it would simplified down.1083

All of those much higher powers will end up simplifying down to something which does not have any more powers whatsoever.1097

We can take as much farther and develop a nice shortcut formula so we do not always have to string out into a bunch (i).1104

Let us see how to do that.1111

Looking at all of these examples you can see that there is a pattern on how they simplify.1115

In fact they repeats in blocks of 4 and you start with (i) then it goes -1, then –I, then 1, it just repeats over and over again.1120

To figure out where in that pattern you are at, you can take the exponent and simply divide it by 4 and observe what the remainder is.1129

From the remainder you will know exactly what it will simplify to.1138

I have made a handy table to figure out what that looks like 1142

If I’m looking at (i) to some large power, I will take the power and divide it by 4 1146

and the remainder happens to be 3, according to my chart here it will simplify to –i.1152

Let us do a quick example to see how this works and why it works.1160

Let us take i13.1166

If I wanted to I could imagine a whole bunch of i4 in here.1173

i4, i4, i4 and then i1.1178

If we add up all those exponents they add up to 13.1186

You can see that most of them are gone since i4 = 1.1191

1 × 1 × 1 × (i) =(i)1196

Let us use our shortcut formula to figure out the same thing. 1207

In the shortcut we take the exponent and divide that by 4.1210

13 ÷ 4, 4 goes in the 13 exactly 3 times.1216

Notice why that is important, that is exactly how many bunches of i4 I have.1227

When you are going through that division by 4 process you are counting up all of the i4 that will simply break down and simplify into 1.1233

It goes in there 3 times, 3 × 4 = 12 and I will subtract that away getting my remainder of 1.1243

That remainder tells me how many extra (i) I still have left over.1253

From there I can see I have only one (i) and I know it simplifies down to (i).1259

I get the same exact answer as the 4, i13 = (i).1267

Now that we know a lot more about these complex numbers let us go through some various examples 1275

of putting them together and see how we can simplify powers of (i).1279

This first one we want to do a subtraction problem 4 + 5(i) – 6 – 3(i).1284

With the addition and subtraction we are looking to add like terms.1290

Subtraction is a tricky one, you have to remember to distribute our sign onto both parts of that second complex number. 1295

I’m looking at 4 + 5(i) -6 + 3(i)1302

I can see my like terms, let us put together 4 and -6, and 5 and 3.1311

4 - 6 = -2, 5(i) + 3(i) = 8(i)1320

We can see that it is in that final complex form and we know that this one is done.1329

In this next example, we want to multiply two complex numbers. 1337

Think foil.1342

Be on the watch out for additional things that we will need to simplify such as the i2. 1346

Okay, starting off, multiplying our first terms together we will get 1.1351

Our outside terms we will multiply that will be 3(i) then we can take our inside terms 2(i) and then finally take our last terms 2(i) × 3(i) + 6i2.1357

We go through and start combining everything we can.1377

1 + I have my like terms here, so 3 + 2 = 5(i) + 6 and take the i2 and turn it into -1.1380

Let us continue 1 + 5(i) - 6 now we can see we have just a couple of more things that we can go ahead and combine.1397

This will be -5 + 5(i) and now I can easily see my real and imaginary part, so I know that one is done.1408

Let us divide these two complex numbers. 1420

I have 2 – 5(i) ÷ 1 – 6(i)1422

This is the lengthy process where we find the complex conjugate of the bottom of if we multiply the top and bottom of our fraction by that.1426

Let us first find that complex conjugate.1434

That will be 1 + 6(i).1437

We are going to use that on the top and on the bottom.1440

We have to remember that since we are multiplying we will have to foil out the top and the bottom.1445

Foil top and foil bottom.1452

Let us see what the result of that one looks like.1458

I’m looking at the top, first terms 2 × 1 =2, outside terms would give me 12(i), inside terms -5(i), and last terms -30(i)2.1461

Lots of terms in there, be very careful and keep track of them all.1479

On the bottom, first terms 1, outside 6i, inside -6i, and our last terms -36i2.1483

It is looking much better, looking pretty good. 1496

If we do this correctly, we should not have any more powers of (i) in the bottom.1498

Let us see what else we can simplify.1502

We will go ahead and combine these 2i on the top and we will combine these 2i of the bottom 1505

and then we will put these i2 to make them both -1.1510

2 and then we will have 7(i) - 30 × -1.1518

On the bottom 1 – 36 -1.1527

It looks like we have indeed accomplished our goal there is no longer imaginary numbers on the bottom. 1533

We continue putting things together 2 + 7(i) + 30 since the negative × negative is a positive and 1 + 36.1538

Just a few more things to put together 32 + 7(i) ÷ 37.1553

This point it is tempting to stop it but keep going until you get it into its nice complex form, we can see that real and imaginary part.1562

I’m going to give 3732 and 377.1570

7 /37(i) this one looks much better.1579

Sometimes you can go ahead and reduce those fractions but this one the way it is.1587

In this last problem we will go ahead and see if we can simplify some very large powers of (i).1595

Since they are very large powers we will go ahead and use our shortcut to take care of that process.1602

In the first one I have i107. 1607

Let us start off by taking that exponent, 107 and we will divide it by 4.1611

This will figure out all the bunches of i4 that we have hiding in there.1618

4 of those in the 10, twice and we get 8.1623

Subtract them away I will end up with 27.1629

That goes in there 6 times and 6 × 4 = 24 and then we can subtract that away. 1635

I'm down to 3 and that is smaller than 4 so I know that 3 is my remainder.1645

Think back, if my remainder is 3 then what does this entire thing simplify to?1653

It simplifies to –i.1660

If you ever forget the way you use that table, you can always build it very quickly by doing just the first few values of (i).1664

(i) = (i), i2 = -1, and we have –I and 1.1674

You quickly have built your table.1682

On to the next one i2013 something very large.1686

We will grab the exponent and we will divide it by 4.1692

4 goes into 20 5 times and I will get 20 exactly.1701

Let us put in our 0 placeholder over the 1 and then bring down the 13.1707

4 goes into 13 3 times and we will get 12.1713

This one shows that my remainder is 1.1720

What does this entire thing simplify down to?1725

If I get a remainder of 1, 2, 3 or 0, now I know what it will simplify down into.1729

It simplifies down just to (i).1736

There are many things you can do to work with these complex numbers but they are often mere things that you have learned before.1740

Keep track of this handy method for simplifying powers of (i) that way you can take those much higher powers 1745

and bring them down to something nice and compact.1751

Thank you for watching www.educator.com.1754

Welcome back to www.educator.com.0000

In this lesson we are going to start looking more at linear equation starting off with the vocabulary of linear equations.0003

There will be lots of new terms in here, it will definitely take some time to look at them all and what they mean and play around with them a little bit.0010

Some of the terms that will definitely get more familiar with are variable, term.0018

We will look at coefficients and we will definitely see how we can combine like terms.0026

We will be able to tell the difference between equations and expressions and get into a linear equation.0032

What we want to solve later on and solutions.0038

When looking at an equation, we often see lots of letters in there, those are our variables.0046

What they do is they represent our unknowns.0054

One favorite thing to use in a lot of equations is (x), but potentially we could use any letter.0058

You could use a, b, it does not really matter but most the time our unknown is (x).0064

A term is a little bit more than just that variable.0070

It is a number of variables or sometimes the product or quotient of those things put together.0074

To make it a little bit more clear, I have different examples of what I mean by a term.0079

All of these things down here are types of terms.0085

The first one is the product of an actual number and a variable.0088

Down here with the (k) it is simply just a variable all by itself.0093

A coefficient of a term is a number associated with that term.0102

If I'm looking at a term say 2m, the coefficient is the number right out front that is associated with that term.0109

I'm looking at another one like 5mq, but again the 5 would be the coefficient of that term.0121

Terms with the exactly the same variables that have the same exponents those are known as like terms.0133

There are two conditions in there you want to be familiar with.0140

It must have exactly the same variables and it must also have exactly the same exponents.0143

Let us say I have both of those and you can not consider them like terms.0150

Let us take a look at some real quick.0156

I'm looking at 5x and I'm looking at 5y, these are not like terms.0158

Why, you ask? They do not have the same variable.0169

One has (x) and the other one has (y).0172

How about 3x2 and 4x2, these are like terms.0176

These ones are definitely good because notice they have exactly the same variable and they have the same exponent.0188

They have both of those conditions.0203

This one is little bit trickier so be careful, they both have an (x), that looks good.0218

They both have a (y), that seems good but they have different exponents.0224

This one has the y2 and this one has nothing on its y.0230

I would say that these are not like terms.0238

An expression is the statement written using a combination of these numbers, operations, and variables.0244

This is when we start stringing things together so I might have a term 2x and then I decide use may be addition and put together a 4xy.0251

That entire thing would be my expression.0262

In the equation, we take a statement that two algebraic expressions are actually equal.0265

I can even borrow my previous expression to make an equation.0271

I simply have to set it equal to another expression, maybe 2x2.0276

What a lot of students like to recognize in these two cases is that in an equation there better be an equal sign somewhere in there.0285

With your expression there is not an equal sign because it is just a whole bunch of string of terms and coefficients, numbers, operations.0292

Since we are interested in linear equations and eventually getting solved for those, what exactly is a linear equation?0306

It is any equation that can be written in the form, ax + b = c.0314

There is some conditions on those a, b, and c.0321

Here we want a, b, and c to be real numbers, we would not have to deal with any of those imaginary guys.0324

We want to make sure that (a) is not 0.0330

The reason why we are throwing that condition in there is we do not want to get rid of our variable.0335

If (a) was 0, you would have 0 × x and then we would have a rare variable whatsoever.0340

It is an equation, not necessarily look like that but it almost can be written in that form, it is a linear equation.0347

A number is a solution of that equation if after substituting it in for the value the statement is true.0354

That means if you actually take out your variable and replace it with a number that is the solution.0361

Then it is going to balance out, it is going to be true with that number in there.0367

This first part we just want to identify the different parts of the equation so we can better feel of what we are looking at.0379

First of all I know that this is an equation because notice how we have an equal sign right there.0386

I have the expression 7x + 8 that is one expression one side and expression 15 on the other.0394

What else do I have here? I have my 7x and I have the 8 both of these are terms.0400

If I pick apart that one term on the left, I can say that the 7 is a coefficient and that the x is my variable.0410

There are many different parts of the equation and you want be able to keep track of it.0428

And probably terms are one of the most important for now.0432

Let us look at this one, let us see 30x = (4 × X) - 3 + (3 × 3) + (x + 2).0438

Again identify the parts of the equation, let us see.0444

It has the equal sign, I know it is an equation, it is important to recognize.0449

Let us see, over on this side I have a term 30x, I have this term, I have this term.0454

I have a bunch of different terms.0461

Terms are always combined using addition or subtraction.0465

That is how we can usually recognize them.0468

In my terms I have some coefficients but you know it might be easier to use my distributive properties to see even more of those coefficients.0471

Let us use that distributive property, let us take the 4 multiplied by the x and 3 and do the same thing with 3.0481

4x -12 + 3x + 6, not bad.0489

Looking at that I have even more terms, I got my 30x, I get 4x, 12, 3x, 6 and lots of different terms now.0497

Into those terms I can identify what its coefficient is and I can identify the variable, the x.0508

This one says simplify it by combining like terms.0528

Remember our like terms are terms that have exactly the same variable and they have exactly the same exponent.0532

We have to be careful on which things we can actually combine here.0539

Let us see I have 12w and 10w those are like terms, they both have a w to the first.0543

Over here is 8- 2w, all three of those are like terms, we only write them next to each other.0550

I know that I need to combine them.0558

The 9 and the 3 I would also consider those like terms because both of them do not have a variable associated with them.0567

I will combine those together.0574

Let us take care of everything with the (w), 12w + 10 would be 22w.0576

That would give me a 20w when combining those, now we will take care of 3 and 9, -3 + 9 = - 6.0591

It is important to recognize that you should not go any further than here because the 20w and 6, those are not like terms.0609

We are not going to put those together.0616

On to example 4, this one we want to simplify by combining like terms.0622

I have lots of grouping symbols in here so it is hard to pick out what my like terms are.0630

I think we can do it though but we might have to borrow our distributive property first.0637

I'm going to take this negative sign and distribute it inside my parentheses here.0641

Now that would give me (2y - 3y) - 4 + (y2 + 6y2), not bad.0648

I can see there is a few things I can combine, let us see.0665

Specifically I can put these y's together since both of them are single (y).0670

I can combine these ones together over here because both of them are y2.0675

Let us put those together, 1y - 3y would be -2y, I have a 4 hanging by itself, -4.0683

y2 + 6y2 would be a 7y2, that looks good.0695

Remember, I have not distributed my 2, it is not going to help me combine anymore like terms.0703

It will definitely help me see my final results, -4y - 8 + 14y2.0708

My final answer would be -4y - 8 + 14y2.0722

I would not combine those anymore together because none of those are like terms.0726

I have a single (y), I have an 8 that does not have any variables whatsoever.0731

I have that y2, definitely not like terms.0735

Alright, thanks for watching www.educator.com.0739

Welcome back to www.educator.com.0000

In this lesson we will look at more of the nuts and bolts of solving linear equations.0002

Specifically some of the things that we will actually do is we will look at types of solutions.0010

What does it mean to actually have a solution versus when it is consistent or inconsistent?0015

The things that will make this handy is how you can also deal with fractions when they are in the solving process.0023

It is something that my students always ask me.0031

What a solution means is a value that when you substitute it in for the variable, it makes the equation true.0039

We are looking for that magic value that would make the whole thing true.0046

In some cases, only one value would end up making the whole thing true.0051

If you only have one value that will do that or sometimes only two or three, we call this conditional.0057

The condition is that this variable must be that value.0064

There are other cases where we actually end up where any value will do.0069

It does not really matter what the value is as soon as you substitute it in for the variable it will make it a true statement.0075

If any value will do we say that the equation is an identity.0083

Lastly, there are very few types of equations that no matter what you try and substitute in for that variable, nothing seems to work.0089

No matter what that value is if nothing works then we say that that equation is a contradiction.0100

I have hidden a couple of those into the examples later on so watch for the one that is an identity and one that is a contradiction0106

and how we actually pick that out of just one that actually have a solution.0116

How do you exactly solve a linear equation?0123

You will use a few properties to help you out in the solving process.0126

It is a handy property but what it says is that you can add the same amount to both sides of the equation.0133

In fact, many people like to look at this as almost like a balance scale.0140

Whatever you add to one side of your equation then you better be sure to add it to the other side as well to make sure that it all balances out.0146

If you are going to add + 4 over here, make sure you also add + 4 to the other side.0155

There is also the multiplication property of equality.0163

For this one you are allowed to multiply both sides by the same value and it will keep the equation the same.0168

You do have to be a little bit more careful with that property because it works as long as you do not multiply both sides by 0.0175

But you can use any other number you want.0184

You can multiply both sides by 3, multiply both sides by (x) as long as (x) is not 0, it will keep that equation the same.0186

One thing that I like to do when solving equation is to think of this process, 0197

it often help with fractions as well as to get the variable that we are looking for all alone.0203

The very first thing that I like to do is scan it over and see if it does contain any fractions0209

and immediately clear them out if I can using some sort of common denominator.0213

After I have done that, I like to simplify each side of the equation as much as possible before shifting things to one side or the next.0217

We will simplify each side as much as possible first.0225

After we simplify both sides, we will try and isolate the variable on one side of the equation.0230

It may be difficult to do especially if there is more than one copy of that variable but if we can get them together usually we can isolate it just fine.0236

Lastly, it is always a good idea to check the solution just to make sure it actually works in the original problem.0244

That is always a good thing to do in case we make a mistake in one of these earlier steps.0251

We will definitely be able to find that out by the four steps when we check it by substituting it back into the equation.0256

Let us get into the examples and see how the solving process actually works.0264

I want solve (2 × x) +3 = 4x – 8, our goal is to figure out what the value for x is that would make this entire thing true.0270

In order to help this process out, I can see that it does not have any fractions, I’m going to try and simplify both sides and make it as compact as possible.0281

This means I'm going to use my distributive property on the left so they do not have my x inside parentheses 2x + 6.0291

Looking at both sides of the equation now, there is not much I can simplify if I was looking at just left or if I was looking at just the right.0310

What I want to do is to try and isolate my variable, in other words try and get it all alone.0320

Since I have the next on the left and on the right, I have to work on getting these together first.0326

In order to get them together, I will use my addition property of equality to add the same thing to both sides.0333

What I will add is -2x, we will put that on both sides and see the results.0340

2x and -2x, those two would cancel each other out and get rid of each other and just be left with 6.0351

On the right side, I have 4x - 2x so only 2x left – 8.0358

That is good, I worked out that way because now I only have one x to deal with and I can work on isolating that and getting it all by itself.0368

How am I going to do that? I better move this into the other side by adding 8 to both sides.0377

6 + 8 that will give me 14 and -8 + 8 will cancel each other out and be gone.0384

I have 14 = 2x, we just divide both sides by 2, I have 7 = x.0400

It looks like we have found a solution something that will make our equation true.0415

My question is does it actually work or not?0420

On to that last step, the one we actually see if this is the solution and we do that by substituting it back into the original problem.0424

I’m going to write the original problem, I’m going to write it all except for those x’s.0431

I’m going to take that value and put it in wherever I saw an x, let us see.0443

I had an x + 3 put 7 there and 4 × x now 4 × 7.0451

I'm doing here is simplify both sides of our equation now, let us see if it balances out.0458

2 × 7 + 3 is 10, 2 ×10 that would be 20, I’m not sure if these are equal, if we continue we find out.0465

4 × 7 is 28, 28 - 8 is 20, sure enough it looks like the value of 7 does make this equation true.0479

I know that 7 = x is my solution.0494

Let us try another this one, 3/4x – 1 = 7/5.0498

This one contains fractions so remember how I suggested taking care of them and you do not have to worry about quite as much.0507

We are going to try and find a common denominator that we can just multiply it by and get rid of our fractions.0515

Looking at our denominator, I have 4 and 5.0522

A common denominator in this case would be 20.0529

I’m going to multiply both sides by that 20, 20 and 20.0541

Let us write down the rest of our equation as well, 3/4x -1 and 7/5.0555

Sometimes it is no fun to deal with numbers like 20 but watch what it does when I start distributing it on the left side here,0567

and we will multiply the 20 and 7/5 together.0573

20 × 3/4 would be a 15x and 20 × 1 being – 20.0578

We do not have to deal with fractions on the outside of the equation anymore.0591

Perfect, I like it.0593

20 × 7/5 if we want to cancel out 5 from there, when you are looking at 4 × 7 = 28.0595

I just have to solve 15x – 20=28, let us work on getting an x all by itself by isolating it.0608

I will add 20 to both sides giving us 15x equal to 48 and then we will divide both sides by 15.0622

This will give us that x is equal to 48/15 but if we want we can even reduce that a little bit further.0650

I think 3 goes into the top and into the bottom.0657

This will use 16/5, not bad. 0665

You have x equals 16/5, now get in front of that last step.0672

Does it equal to 16/5? Does it not? Let us find out by playing it back into the original equation.0676

I’m just testing it out and see if it holds true.0682

The original equation is ¾ and I have that x, that is where we will put our number -1 and we will see if it is equals 7/5.0686

Let us see our number was 16/5, check and see what happens.0700

I can take out a 4 from the 16 so this would be 12/5 – 1, it will give me a common denominator of 5.0707

I would have 12/5 – 5/5 and how do we know? That is equal to 7/5.0734

This one definitely checks out, I do know that 16/5 is my solution, it looks good.0743

We will have something like this one, more fractions here and that is good.0755

4 + 5x = (5 x 3) + x and see what we can do this one.0758

Let us simplify both sides of the equation first and then go from there.0765

I'm going to distribute my 5 into the parentheses.0769

4 + 5x = 15 + 5x, it looks pretty good.0775

I want to get my x’s together and hopefully isolate it.0786

We will subtract 5x from both sides, 4 = 15.0791

Look at what happened there, when I subtract 5x from both sides, I lost all of my x’s completely.0805

Even worse than that than I thought I have left over this 4 -15 that is not true.0812

That does not make any sense, 4 does not equal 15.0819

What I was left here is known as a false statement.0823

What this is telling me is that since I did not make any mistakes that you do not know the value for x is going to work.0828

This is an example of one of those situations where we have a contradiction.0837

No matter what value for x you are trying use, it simply not going to work in this equation.0849

In fact, no solution exists for it.0854

Let us look at another one and see if we have any better luck.0860

In this one is -x + 3 = (1 + 4) - 2x/2.0863

We have a fraction, I'm going to try and get rid of those fractions first by multiplying everything through by a 2.0869

I will multiply it on the left side and I will multiply it on the right side, both by 2.0877

Just to keep things nice and balanced.0886

-x + 3, 1 + 4 - 2x / 2.0890

We multiplied both sides by 2, we can go ahead and distribute it.0901

Let us see what the result will be.0907

-2x + 6 equals, I have 2 + and when it distributes on this second part here, it is going to get rid of that 2 in the bottom, 4 - 2x.0911

Looking good, let us continue trying to combine our terms specifically those x’s and the numbers.0933

I will add 2x to both sides and you will get 6 = 2 + 4.0943

You will notice in this one that we lost our x’s but this one is a little different.0958

Instead of being left with other nonsense, this one we are left with 6 = 6 which is actually something that is true.0963

Before you get too worried, of course check all the steps and make sure they are all correct, which they are.0977

What this is trying to indicate here is that we have an identity.0982

It does not matter the value of x, you can use any value and the equation is still going to ring true.0985

Let us mark this one as an identity.0992

In short, that is a great way that we can identify both of those situations.0999

If it has an actual solution, it is a conditional equation.1003

You usually go through the solving process and you will be able to figure out what that value is and able to test it and see if it actually works.1007

If it is a contradiction, so nothing works, then you will go through the solving process and usually your variable drops away entirely,1014

and what you are left with is a false statement, it does not make sense.1022

That is even after all of your steps are completely correct.1026

If you have an identity like this one then you will go through that solving process and you will be left with a true statement1029

which indicates that you could use any value want for x.1036

Let us try some more examples.1042

In this next one I have 1/3x – 5/12 = ¾ +1/2 x.1046

There is a lot of fractions in here, let us see if we can take care of them all by getting some sort of common denominator.1053

What would 3/12, 4, 2 go into?1062

I think our common denominator would have to be a 12.1067

We are going to multiply both sides by 12.1072

Let us write everything we got here, 1/3x – 5/12 and I have ¾ + 1/2x.1084

Then we are going to distribute through on both sides and see what we get.1100

1/3 of 12 would be 4x, 12 × 5/12 would give us 5, then I will have 12 × 3/4.1108

That goes in there 9 times, then (12 × ½) + 6x.1124

Notice that common denominator, we clear out all those fractions then you do not have to worry about them.1134

What we have left here is 4x - 5 = 9 + 6x.1139

If you want to continue trying to simplify this by getting the x’s together and getting them isolated.1144

Our x over here and x over here let us get them together by subtracting 4x from both sides, -5 = 9.1153

6x – 4x that would be 2x, that looks good.1170

I only have one x to deal with.1177

Let us go ahead and subtract 9 from both sides, -9 -9 -14 = 2x.1180

There is only one last thing to do, divide both sides by 2.1194

It looks like our solution is -7 or we are not too sure until we check it.1200

Let us substitute it back into the equation and see what happens.1211

I have 1/3 - 5/12 = ¾ + ½ and into all of those blank spots where we are used to have the x, let us go ahead and put that -7 in there.1215

Let us see if these things are true or not, also -7 × 1/3, that is -7/3 – 5/12, is that equal to ¾ - 87/2?1237

I do not know, we got to do a lot more simplifying before we can figure that out.1259

We will use our common denominator to help us out, the common denominator of 12.1263

Let us see if I need a common denominator over there that would be -28/12.1270

A common denominator of 12 on the other side would be 9/12 minus, top and bottom by 640, 2/12.1279

Let us see what we can do, -28 - 5 would be -33/12.1294

What is going on the other side? 9/12 – 42.1306

It looks like I have another -33/12.1310

They are exactly the same and what that means is that x does equal in -7.1314

You can see when you go through that process of checking it, 1322

it can be a lot of work especially if you have lots of fractions and stuff that you are dealing with.1324

But it is a good idea just to make sure that the solution you come up with is the solution.1329

Let us do one more.1338

Here I have (3x -15)/4 + (x + 47)/7 = 11.1339

This one involves fractions, let us take care of them out of the gate by multiplying it by a common denominator, let us use 28.1347

A lot to keep track of in here, (3x -15)/4 + (x + 47)/7 = 28 × 11.1364

We will take that 28, we will actually distribute it to both parts on this left side.1379

We will take it to this giant fraction here and I will take it to this giant fraction over here.1384

What this is going to do is, it is going to allow us to cancel out those fractions in the bottom.1388

28/4 would give us 7, that is still going to be multiplied by that 3x -15 part.1394

When it distributes on the second piece, 28/7 reduces and that goes in there four times. 1404

We still have a 4 multiplied by x + 47.1411

Over on the side we have 28 × 11 that is 308, some fairly big numbers.1417

Let us continue distributing and see what we can do.1426

Let us take this 3 × 7, we will also take the 15 × 7 and do the same thing with 4 on this side.1429

Hopefully free of those x’s and get them out of parentheses.1440

21x - (7 × 15) = 105 + 4x.1444

Now I have 4 × 47 = 188, again, some big numbers but just have to push on through.1461

I have more than one copy of x here let us get those guys together.1476

I also have some things that do not have an x, let us get those together as well.1480

21 + 4x would be 25x - 105 and 188, I think that would give us 83.1487

I only have a single x to deal with so I will simply work on getting that isolated.1508

Let us subtract 83 from both sides so that the 25x is the only thing on the left side.1516

308 – 83 will give us 225, almost done.1529

Let us go ahead and divide both sides by 25 now.1539

This will give us x = 9.1549

This one is a lengthy process to get all the way down to just x = 9.1554

Even when you get that far, go ahead and double check it just to make sure that it actually works out.1558

I have my (3 × x -15)/4 on one side and both in those blank spots, let us go ahead and put that 9.1571

Let us start to simplify and see what we will get.1589

3 × 9 = 27/4, I see a 9 + 47 = 56.1592

We are hoping that this will equal 11.1610

27 - 15 that is 12/4 that is 56/7, we are getting closer.1614

12/4 is 3, 56/7 let us see that goes in their 8 ×, 3 + 8 =11.1625

Sure enough this one checks out.1639

I know that x does equal 9.1642

When going through the solving process, you do have a lot to keep track of.1648

The most important thing is truly working getting those x’s all by themselves and isolate it so that they are the only thing on one side of the equation.1652

If you have to deal with fractions, use a common denominator to clear them all out.1660

And also definitely go back and check your solution to make sure it is a solution.1664

Thank you for watching www.educator.com1671

Welcome back to www.educator.com.0000

In this lesson we will take a look at solving formulas and that is our only objective for our list of things to do.0002

But one thing I want you to know while going through this one is how similar formulas to solve the equations.0011

If you are curious what exactly is a formula and how they differ from those linear equations that we saw earlier.0020

A formula is a type of the equation and usually conveys some sort of fundamental principle.0025

Below, I have examples of all kinds of different formulas, D = RT and I = P × R × T.0031

What these represents are usually something else.0039

For example, D = RT we will look at that a little bit later on because it stands for distance = rate × time.0042

I = PRT, that is an interesting formula and it stands for interest = principal, rate, time.0058

It is not that these equations are too unusual but they actually have lots of good applications behind them.0075

One thing that is a little bit scarier with these formulas is you will notice how they have a lot more variables than what we saw earlier.0081

They have a lot of L, W and other could be multiple variables for just a single formula.0089

You know we have those multiple variables present in there and usually we are only interested in solving for a single variable in the entire thing.0097

The way we go about solving for that variable is we use exactly the same tools that we used for solving several linear equations.0105

You have the addition property for equality and you have the multiplication property for equality.0114

Both of these say exactly the same thing that they did before.0120

One, you can add the same amount to both sides of an equation, keep it nice and balanced.0124

And that you can multiply both sides of an equation by the same value and again keep it nice and balanced.0129

The only difference here is that we will be usually adding and subtracting or multiplying, dividing by a variable of some sort.0135

One thing that we usually thrown in as an assumption is that when we do multiply by something then we make sure that it is not 0.0142

If I do multiply it by a variable, multiply by both sides by an X, then I will make the assumption that X is not 0.0151

Just to make sure I do not violate the multiplication property of 0.0158

You have lots of variables and the very first thing is probably identify what variable you are solving for.0167

I usually like to underline it, highlight it in some sort of way just to keep my eye on it.0174

If you have multiple copies of this variable, try to work on getting them together so that you can eventually isolate it.0180

Some of these formulas do involve some fractions.0186

Use the common denominator techniques so you can clear them out and not have to worry about them.0189

Simplify each side of your equation as much as possible then isolate that variable that you are looking for.0196

Try and get rid of all the rest of the things that are around you by using the addition and multiplication property.0203

You can check to make sure that your solution works out by taking your solution for that variable and playing it back into the original.0212

When you do this for your formula, what you often see is that a lot of stuff will cancel out and you will be left with a very simple statement.0221

Usually we do not worry about that one too much unless we have some more numbers present in there but you can do it.0229

Let us actually look at a formula and watch the solving process.0238

You will see that it actually flows pretty easily.0242

In this first one I have P=2L + 2W.0245

This formula stands for the perimeter of a rectangle.0248

It goes through and adds up both of the lengths and both of the widths.0251

What we want to do with this one is we want to solve it for L.0257

Let us identify where L is in our formula right there.0261

What I’m going to do is I’m trying to get rid of everything else around it.0265

The 2W is not part of what I'm interested in so I will subtract that from both sides.0271

P and W are not like terms so even though it will end up on the other side of the equation, I will not be able to combine them any further.0280

Now you have 2 - 2 or P - 2W = 2L, continuing on.0291

I almost have that L completely isolated, let us get rid of that 2 by dividing.0298

Notice how we are dividing the entire left side of this equation by 2 so that we can get that L isolated.0310

What I have here is (P -2W)/2 = L.0322

One weird part about solving a formula is sometimes it is tough to tell when you are done 0328

because usually with an equation when you are done you will have a number like L = 5 or L = 2.0334

Since you have many variables in these formulas that when we get to the end what we developed is another formula right here.0340

It is just in a different order or different way of looking at things.0348

This formula that we have created could be useful if we were looking for L many times in a row and we had information about the perimeter and W.0351

It is solved, it is a good the way that it is and the way we know it is solved is because L is completely isolated.0362

Let us look at another one, in this one we want to solve for R.0371

I have Q= 76R + 37, a lots of odd ball numbers but let us go ahead and underline what we would be looking for.0375

We want to know about that R.0382

I think we can get rid of a few fractions and we would have to multiply everything through by 6.0387

Let me do that first, I will multiply the left side by 6, multiply the right side by 6.0395

Q = 76R + 37, on the right side we better distribute that 6 and then we will see that it actually does take care of our fraction like it should.0403

6Q = 7R, I know I have to take care of 37 × 6.0422

I guess I better do some scratch work.0434

37 × 6 I have 42, 3 × 6 is 18 + 4 = 22, a lot of 2.0436

6Q = 7R + 222, I do not have to deal with any more fractions at this point so let us continue isolating the R and get it all by itself.0452

I will subtract 222 from both sides, awesome.0463

One final step to get R all by itself, we will divide it by 7.0482

(6Q – 222)/7 = R and I will consider this one as solved.0490

The reason why we can consider this one done is because we have isolated R completely.0503

Even though we do have a Q still floating around in there, it is solved.0508

Do not get too comfortable with that, it is not equal to just a single number, very nice.0513

This one is a little different, we want to solve the following for V.0521

The reason why this one is a little bit different is I have a V over here but I also have another V sitting over there.0525

When you have more than one copy of the variable like this, you have to work on getting them together before you can get into the isolating process.0534

Let us take care of our fractions and see if we can actually get those V’s together and work on isolating it.0543

We only isolate one of them then it is not solved, we still have a V in there.0551

To take care of our fraction I will multiply both sides of this one by W, it is my common denominator.0556

(RV + Q) / W = W × 5V, on the left side those W’s would take care of each other.0569

I will be left with RV + Q = W × 5V.0585

In this point I do not have to deal with the fractions but notice we have not got those V’s any closer together so let us keep working on that.0594

If I'm going to get them together, I at least better get them on the same side of the equation.0602

I'm going to subtract say an RV from both sides right.0609

Now comes the fun part, I still have two V’s, I’m going to make them into one.0627

The way I’m going to do this is I’m going to think of my distributive property but I’m going to think of it in the other direction.0633

If both of these have a V then I will pull it out front and I will be left with W5 – R.0641

This step is a little bit tough when you see it at first but notice how it does work is that I'm taking both of these and moving them out front into a single V.0653

I'm sure if it is valid, go ahead and take the V and put it back in using our distributive property and you will see that you be right back at this step.0664

It is valid.0673

The important part of why we are using it though is now we only have a single V and we can work further by isolating it.0675

How do we get it all by itself, these V’s be multiplied by 5W – R, that entire thing inside the parentheses.0682

We will divide both sides by that and then it should be all alone 5W – R = V.0690

I literally just took this entire thing right here and divided it on both sides.0702

To your left, now I can call this one done because V is completely isolated, it is all alone.0709

There is no other V’s running around in there.0717

I have worked hard to get them together and I know it is completely solved.0719

Welcome back to www.educator.com.0000

In this lesson we will look at applications of linear equations.0002

Think of those linear equations but more of a word problem sense.0006

Some of the things we will cover is the six step method that you can use to approach some of these word problems and to use that method.0012

We will have to look at how you can start translating these math things into actual equations.0020

We will also loot at very specific types of word problems that you might encounter such as when you just have some unknown quantities involving numbers.0028

Some of that involve some trigonometry and ones that might just involve some consecutive integers.0036

How should you approach some of these word problems?0046

In the six step method, we like to try and understand as much about the problem as we can.0049

That is why in the first step we are looking at reading the problem as much as possible and gathering up information and trying to understand it.0055

Once we think we have enough information it is a good idea to assign some sort of variable to the unknown.0063

In some of these problems, it might look like there is more than one unknown 0069

and in cases like that you should assign a variable to the one that you know the least amount about.0072

From there we will be able to actually start writing equation from the information of the problem.0080

This is probably the most difficult step and sometimes you might get stuck on that one.0084

Once we have an equation when you go ahead and move on to solving that equation and see if we can actually get a solution.0091

We are not done even when we do get a solution it is a good idea to state the solution in the context of the actual problem.0098

That way we know if we get done we will say x = 3.0104

What exactly does x represent? Is it the number of miles driven?0107

Is it the number of beans in our basket? What exactly is the variable?0111

It is also important that we identify it early because it will make step five much easier.0117

When we are all done, we want to make sure that we check our solution to make sure that it is reasonable.0125

Especially with a lot of word problem, sometimes you might get a solution that just does not make sense in the context of the problem.0131

For example maybe I'm going through on solving something and I actually get time being -3 seconds.0137

It might be a problem because we can have negative seconds being time.0143

It is definitely a good idea to check your problem, check your solution and make sure that they make sense.0147

One thing I will note is that these steps often make it look like a nice linear process like you have simply moved through 1, 2, 3, 4, 5, 6. 0154

In practice that may not always happen.0163

Do not be too surprised if you end up going through step 1, step 2, and you try to build that equation in step 30165

and you realize you might not have all the information you need.0171

In cases like that, it is good to go right back to step 1 and see if we can gather up even more information.0175

When you are completely done with solving the problem, you will eventually move through all 6 of these steps0180

but you may end up bouncing back a few times if it looks like you do not have quite enough information.0185

If I’m taking a lot of word problems and being able to create equations then we want to be on the lookout for some keywords that are in those word problems.0196

We will use these keywords, these nice common terms to help us translate some of what we see in the problem into the actual formula.0204

Here are some terms that you should be familiar with.0212

When we are looking at addition, you want to look for words such as sum, together, total, more than or added to, 0215

Think of a number and 3, that would say I want x + 3.0227

For subtraction we are looking for words like difference, minus, less than and decreased by,0234

all of those usually signify subtraction.0239

You have to be a little bit careful on the order of subtraction, for example you might see something like 3 less than a number0242

and that would say you have like x – 3 because that would be 3 less than that number x.0250

The order is important and sometimes it seems a little backwards.0257

For multiplication, look for product, times, twice or three times they might say something like that and percent of.0261

Those all indicate that something should be multiplied together.0270

Division has some good terms, you are looking for quotient, divided by, or ratio.0276

In all of our equations, we will need an equal sign there somewhere so we will be looking for equals is or will be.0283

That would be a good way to signify where the equal will go in the equation.0292

We will definitely come over our problems and look for some of these key terms so we can build our equation and solve from there.0296

Let us go ahead and jump into some of the examples and we will see what we have.0305

Let us first just start off by reading it and seeing what information we can dry out.0309

This one says the quotient of a number and 6 is added to twice the number and the result is 8 less than the number, find the number.0314

It looks like I see lots of things packaged up in here and just coming over.0324

I'm dealing with the quotient, remember that is division.0330

Twice the number means we have some multiplication.0337

The result is there is our equal sign.0340

8 less than the number, this will represent subtraction.0344

There are lots of good pieces that are flying around in there.0349

I think the very thing I need to do now that I got some good information is.0352

Let us go ahead and identify our unknown.0356

If I’m going to be using an x, also that x is the number and that is the one we are looking for.0360

We will use that in our equation to package up all the rest of this information.0371

We have identified are variable, let us see if we can come over, and start putting it together.0376

I have the quotient of a number and 6.0382

The quotient of our number and 6 would be division or x ÷ 6.0386

That entire piece is added to twice the number, notice how I’m using multiplication.0393

The result is equals 8 less than the number, this is the one that seems a little backwards.0406

But we will take our number and subtract 8 so that we can get 8 less than our actual number.0415

We have read it over, identified our variable, we set up our equation, now we have to move through the actual solving process.0420

In this process we use a lot of our tools as before.0428

We try and clear out some of our fractions with a common denominator and get those x’s isolated onto one side.0431

With this one, I’m going to multiply everything through by 6.0437

This will definitely help us take care of that fraction that x ÷ 6.0453

We will use our distributive property, that will give us x + 12x = 6x - 48. 0461

Now we do not have to deal with our fractions and we can just work on simplifying each side of the equation and getting our x’s alone.0477

I see some like terms on the left, 1x and 12 x = 13x, 6x – 48.0484

I simplified each side now let us go ahead and move the 6x to the other side.0495

It looks like we are almost done.0514

Finally, we will divide by 7 the x will be completely isolated.0517

It looks like our number, the one we are looking for, is a -48/7.0532

In terms of does this fit the actual context of the problem?0539

This one is less of a real word problem that is why I do not have to deal with time or distance. 0542

It seems a perfect reason if that this is the number we are looking for.0547

We can always check the number to make sure that it works by taking it, and putting it back into the original.0552

If I want to take this and put it all the way back into all of the x here it should work out just fine.0559

That is the process in a nutshell to get your solution.0568

Let us look at some ones that are little bit more real world-ish and see how we can move through those.0573

This one says the perimeter of a rectangle is 16 times the width and length is 12 cm more than the width.0583

Find the length and width of the rectangle.0590

It seems like a pretty good problem I want to get as much information on this as I can.0594

I’m going to start off by trying to draw a picture of what we are dealing with here.0599

It sounds like we have some sort of rectangle and the perimeter of the rectangle is 16 times the width.0604

It may be my unknown here should be the width.0613

Let us mark that w is the width.0618

The length is 12 cm more than the width and I can mark out and then find the length and width of the rectangle.0630

I draw a little picture to give myself a sense of what is going on.0645

I have labeled my unknown w being the width and now I think we have enough information to start packaging this together into an equation.0649

Let us see what we can do.0657

The perimeter would be the sum of all the sides.0659

w + w that would take care of these two sides right here, + w + 12 + w +12 and that will take care of both of my lengths.0664

That entire thing is the perimeter right there and it is equal to 16 times the width.0687

It looks like a pretty good equation.0696

We just have to work on solving to make it all makes sense.0698

Let us see on the left side I have lots of w's and we can go ahead and combine all of them together into just 4w since all of those are like terms.0703

Then I have a 12 + 12 so we will say that is 24, all equal to 16w.0715

Simplify each side, let us go ahead and put our w's together, 24 = 12w.0727

One last step, divide both sides by 12 and looks like we get that 2=w.0740

It starts the work where we interpreted in the context of the problem.0753

Since we identify this earlier that w is the width I know that this represents the width of the rectangle 2.0756

Since our units in here are in centimeters and let us say that our width is 2 cm.0764

That is all we wrote we would actually be in a lot of trouble because we have not answered the entire problem.0778

In the entire problem we want to know the length and the width of this rectangle.0785

We only have the width so let us use this width to see if we can find the other part.0789

Since it says that the length is 12 cm more than the width then we can simply add 12 to this 2 and we will get the other.0796

Length is 14 cm.0807

We have answered both parts of this question I know that the width is 2 cm and the length is 14 cm.0819

Let us look at another one, this one says during a sporting match the US team won 6 more medals than Norway and the two countries won a total of 44 medals.0831

How many did each country win?0840

This is one of those examples where I said it looks like there is more than one variable in here.0842

After all, we have no idea how many medals that US won and I have no idea how many medals Norway won.0847

Let us see if we can pick apart and see which one we know the least about.0853

During a sporting match the US team won 6 more medals than Norway, I know a little bit about the US, 0856

they won 6 more than Norway and the two countries won a total of 44.0862

It looks like the one I know the least amount about is actually Norway.0867

I have no idea how many medals they won.0871

Let us set that up as our variable so x is the number of medals won by Norway.0872

If the US team won 6 more medals than Norway, I could represent that using addition, x + 6 that would represent the number of medals that the US won.0898

I want to look at their total wins and it should be 44.0913

I will add Norway's medals to that should equal 44.0918

Let me highlight the parts of this equation so you can see exactly what I'm doing here.0923

This x right here is the number of medals from Norway and this over here is the number of medals from the US.0927

You can see I'm taking both of those quantities and put it together and look at the total number of medals from both one.0938

Now that we have our equation, again let us work to solve it.0946

My two like terms I got x and x that would give me a 2x then I can work to get my x’s all by themselves.0950

Let us go ahead and subtract 6 from both sides 2x = 38.0964

Now we will divide by 2, this will give us x = 19.0972

We have got to the problem and we figured out what x is but again interpret it in the context of the problem.0993

x = 19 but what is x?1000

Earlier we said that x is the number of medals won by Norway so I know that Norway won 19 medals.1003

Let us write that down.1011

We are not done yet, we still have to say how much each country won.1025

I also need to know how many the US won.1030

Since we know that the US won exactly 6 more than Norway and I can just take 6 and add it to the 19, the US won 25 medals.1033

I just have to double check this quickly, add these together and make sure we get 44.1050

19 + 25 = 44 total, it looks like we are doing okay.1054

Let us look at another example and see how well that one turns out.1074

In this example, we are going to look at some angles and a little bit of trigonometry.1080

It is okay if you do not know anything about trigonometry, I will give you all the background information you need for this one1084

You can see how you need to put it together.1089

This one says find the measure of an angle such that its complementary angle and its supplementary angle both added to be 174°.1092

After looking at these terms in here and saying wait a minute what is supplementary and what is complementary?1101

Let me tell you.1107

Two angles are said to be complementary if you can add them together and they add to be 90°.1108

90° is complementary.1114

Two angles are said to be supplementary if you can add those together and you get 180°.1117

180° is supplementary.1122

What we are looking to do is you have some unknown angle and it has a complement and a supplement and we are looking at those two to get that 174°.1126

Let us first identify our unknown in this problem and that would be our angle, x is the unknown angle.1137

We have to use this unknown angle in order to build what its complementary angle would be and what its supplementary angle would be.1154

That is going to be a little bit tricky, because we want to know what would you have to add to this unknown one in order to get an x? 1162

In order to get 90 and since we do not know, how could we figured that out.1168

Let us do a quick example.1174

What if we knew that our angle was 70°, what we have to add to that in order to get 90?1175

It would not take that long to figure out, you should just take another 20° and sure enough you get 90°.1183

If you think about how you came up with that, you will be amazed that all you have to do is take 90 and subtract the angle that you knew about.1189

We will say that 90 - x that right there will be our complementary angle.1199

In a similar way you can also build your supplementary angle.1216

You can say that it will be 180° minus whatever angle you started with.1220

We have our complementary angle and we have our supplementary angle.1239

In the context of the problem you want to be able to look at these two added together so that would be 174°.1242

Let us just take both of these expressions here, put them together.1249

(90 – x) + (180 – x) there is our complementary and supplementary must be equal to 174°.1253

Let us work on combining our like terms and see what we can get.1266

90 + 180 = 270, -x, -x, -2x = 174.1270

We will subtract 270 from both sides, -2x = -96 and then let us go ahead and divide by -2, x = 48.1286

It looks like our unknown angle is 48°.1314

We can check to make sure that this is our angle by using it back into the context of the problem.1332

If this angle is 48 then what do I have to add in order to get 90°?1338

In other words, what would be its complementary angle?1343

If I take 48 and I add 42 that will give me 90, so 42 is the complementary angle.1353

Let us do the similar process, what would I have to add to 48 in order to get 180° or even if you take 180 - 48 to see what its supplementary angle would be.1365

Looks like 132.1388

If we take both of these and add them together, sure enough we get 174° like we should, so we know that our angle is 48°.1392

One last example and in this one we will deal with looking at a lot of different numbers but there is only one unknown1409

and we will look at it as being the one we know the least amount about.1416

This one says we have a piece of wire and it is 80 feet long.1421

We are going to cut this into 3 different pieces with the longest piece being 10 feet more than the middle sized piece 1425

and a short piece being 5 feet less than the middle sized piece.1431

Find the length of the three pieces.1434

It looks like we do not know anything about the short, the middle or the long one, but we will write them all just using one variable.1438

I know a little bit about all of them put together since I know that my total is 80 feet long.1447

I'm going to come in three pieces and the longest piece is 10 feet more than the middle.1454

I know a lot about the long one.1460

The middle sized piece and the shorter piece is 5 feet less than the middle sized piece.1463

I know a little bit about the middle one, its 5 feet more than the shorter one since the shorter one was 5 feet less.1471

I think what I need to set up as my unknown and we will say that x is the length of the short piece.1479

I will try and describe all the other pieces using just that short length right there.1499

The short piece is 5 feet less than the middle piece.1505

If I take x and I add 5 to it that should give me my middle piece just fine.1514

The long one is 10 feet more than the middle, then I can start with the middle and add another 10 and this would represent the long piece.1528

I need to take my short piece, my middle piece, and my long piece to be able to put all those together and represent the 80 total feet of wire.1543

Let us set up this equation.1552

short piece = x + our middle piece x + 5 + our long piece x + 15 all of this should equally total of 80 feet.1554

We got our equation so let us work to solve it.1572

Adding together our like pieces, I have 3 x’s but I can put together the 5 + 15 = 20.1575

I have 3x + 20 = 80.1588

Let us subtract 20 from both sides, giving us 3x = 60.1593

We will divide both sides by 3 and we will see what x needs to be, x = 20.1604

It looks like our shorter piece of wire is going to be 20 feet long.1613

We also want to identify what all the other ones need to be.1636

We can use these smaller expressions over here to figure out what they need to be.1639

Since the shorter piece is 5 feet less than the middle piece, you can simply add 5 to this and get that middle.1646

The middle piece is 25 feet long.1654

Our long piece since it is 10 more than middle one, we will add 10 + 25 = 35 feet long.1668

Now I have the information of all three bits of wire.1685

If we add all of these up, we should get a total of 80.1691

20 + 25= 45 + 35 sure enough adds up to a total of 80 feet.1695

You can see that the process of interpreting these word problems can be a little tricky 1703

But if you look for those keywords and try and hunt down your unknowns as much as possible, it can work out pretty well.1708

You can always look at the context of the problem to make sure your answers make sense.1714

Thanks for watching www.educator.com.1719

Welcome back to www.educator.com.0000

In this lesson we are going to continue on with our examples of applications of linear equations.0003

In this we are going to look at some trickier applications, the ones involving lot of motion and mixtures.0008

It usually gives students a lot of headaches so pay close attention on how we approach this with some nice methods that will organizer our information.0013

Our two big goals, motion and mixture problems is what we want work at.0026

And the two methods we will use to attack these will be using the table method and the beaker method.0031

You will see that the table method usually works pretty good for a lot of motion problems 0037

and the beaker method works pretty good when you are dealing with those mixtures.0040

As I said earlier, motion and mixture problems tend to be a weak area for many people.0048

The problem comes from being able to take all the information and organize it in some sort of way so they can actually build your equation.0053

In terms of equation, there is only two equations that work in the background.0061

For a lot of motion problems, it is the distance, the rate at the time, that all need to be related together.0066

The equation for that is distance = rate × time.0073

Let us go ahead and put it in there.0077

We have our distance, we have our rate, and we have our time.0078

With our mixture problems, there is a very similar formula that we are after and that is the one that deals with the pure substance in the mixture.0086

For that one we are looking at the amount of pure substance, multiply it by some sort of percent and the amount in the total mixture.0094

This guy over here is just the pure amount.0112

To help us organize these we will begin by using what is known as the table method.0120

In the table method, you build a small table usually has about three rows and about four columns in it0125

and we end up putting into this table is information about your distance, rate, and time.0132

The information about different things that is moving in the problem.0137

Every row in this table here usually represents a different object.0142

If I’m talking about to bicyclists, I might talk about bike number 1 and bike number 2.0147

That way I can keep track of each of their motions separately.0157

In each of the columns, we will put information about the rate, distance, and the time.0160

The way we usually do that is we have information about the rate here, the time here, and the distance over here.0166

What we are looking at is that equation of distance = rate × time, in fact it is hidden in the background.0179

rate × time = distance.0186

For that reason it will be important on how we pull the information out of this table.0191

We want to remember that we want to multiply across.0196

In this direction we will multiply the rate and time in order to get our distance and what we want to do from there is add down to get our total.0200

It will help us build that equation and even if it does involve a couple of objects moving.0211

The other method which is good for mixtures is known as the beaker method.0219

The way this method works is you end up creating a beaker for each of the mixtures you are putting together.0224

Maybe this one over here is my 30% solution, this guy over here is my 90% solution and I'm looking to create a new solution over here on this side.0230

You will notice I have already put in the addition and equals signs so I can see that I'm mixing two beakers together to get a third one over here.0241

There are two things that we want to record in each of these beakers.0250

First we will record the amount, how much is in this beaker.0254

Is it 50 L? Is it 40 L?0260

We want to know what the exact amount in each of these will be.0263

We will also want to know what percentage is each of the solution. 30%, 20%, we will definitely keep track of that for each of these beakers.0269

In order to create the equation from these we will multiply the amount and the percentage together from each of the beakers.0279

Watch for me to use this method when we get to those mixture problems.0300

Let us go ahead and see this method in action as we are looking at example 1.0307

This says that building an equation to represent the following problem.0313

We have two airplanes and they are leaving a city at the same exact time and they are flying in opposite directions.0317

If one flies at 410 mph and the other one 105 mph faster, how long will it take them to be a total of 3290 miles apart.0323

You can see what I'm saying, there is a lot of information in this problem that is very confusing to try and keep it all straight.0333

It will help us out, let us keep track of both of these planes.0339

We have plane 1 and plane 2, what do we want to know about these planes?0343

We want to know how fast they are going so we will keep track of their rate.0357

We want to know how long they have been traveling, their time.0361

Of course we want to know how far they have traveled so we can get their distance.0364

Let us see what we can now figure out about these planes and put it into the table.0372

They are leaving a city at the same time and fly in opposite direction and one flies at 410 mph.0378

Let us say that is plane 1, they are traveling at 410 mph and the other one is 120 mph faster, if I was to add 120 to 410 that will give me 530 mph.0384

I know how fast each of these planes are going, that is definitely good.0401

How long have they been traveling now?0404

I’m not sure I got a lot of information about that because that is what I'm looking for in the problem.0408

It says how long will it take them to be 3290 miles apart?0413

My time here is completely unknown.0418

I do not how you know long each of them have been travelling.0424

To think about how the distance is working to all of this, remember the distance is equal to the rate multiplied by time.0428

I have multiplied across to get my distance is 410x for plane 1.0436

My distance for the other one is 530x.0445

I get a sense of how far each of them had gone and I want to know when they will reach a total of 3290 miles.0452

I’m looking at their total distance.0460

We will add this last column down.0462

The distance from plane 1+ the distance from plane 2 must equal a total distance of 3290 miles.0466

You can see that the table helps out so we can build this equation.0479

It organizes that information and we can crunch it down.0485

Now that we have this equation, we can not necessarily stop.0490

We have to move forward and figure out what the solution is.0492

Let us see if we can solve this.0495

3290 miles and we also have the 410x + 530x = 3290.0503

The first thing one I want to do is get my x’s together.0518

Let us see what that will be, 940x = 3290.0521

We will divide both sides by 940 and this will give me x = 3 ½.0533

In the context of the problem, what was that mean?0548

x was our time so this tells us that the planes traveled for 3 hours and ½ of an hour or 30 minutes.0551

When we use this table again to set up another problem and you will see how it helps us organize that information.0577

In this example, we are dealing with two trains and they are leaving the downtown station at the same time and they are traveling in opposite directions.0587

One had an average speed of 10 mph more than the other one and at exactly 1/5 of an hour they were 12 miles apart.0595

What are their two speeds?0603

We are going to start this one often much the same way we did the last one.0605

We have two trains, we will label our rows.0607

I have train 1 and train 2.0611

I want to keep track of how fast they are going, their rates.0618

How long they have been traveling, their time?0623

How far they have gone, their distance?0626

Let us go to the problem over here to even pull out the information and put in the right spot in the table.0631

One have an average speed of 10 mph more than the other one.0636

That does not give me a definite speed for any of these trains.0642

I just know that one is traveling faster than the other one.0644

Let us say we have no idea how fast train 1 is going we will use our unknown for that.0650

Since the other one was traveling 10 mph faster, I can take its rate and add 10.0655

Onto time, it says at exactly 1/5 of an hour and that is pretty good.0666

At least I know how long each of them have been traveling so 1/5 of an hour for each of them in terms of the time.0671

Let us put these together and see if we can get our distance.0683

Multiply rate and time, I will get 1/5x, multiplying rate and time here 1/5x + 10.0686

After 1/5 an hour they were a total of 12 miles apart.0702

I'm looking at both of these distances added together so 1/5x + 1/5 x + 10 so when will their distances be a total of 12 miles apart?0706

There is my equation and I'm going to use and try and solve from here 1/5x + 1/5x + 10 = 12.0727

We will use our techniques for solving linear equations and see if we can pick this apart.0748

The one is coming to mine right now since I'm staring at those fractions is we will multiply by a common denominator to clear at those fractions.0753

Let us go ahead and multiply everything through by 5.0760

I will do that on the left side and I will do it on the right side x + x + 10 and 5 × 12 = 60.0764

We can start combining our like terms 2x + 10 = 60.0781

I will go ahead and subtract 10 from both sides and finally we will divide both sides by 2.0791

I will get that x = 25 but again what exactly does this represent?0810

If you remember when we hunted down our variables and we said what x was, this is actually the speed of our first train.0815

Let us write that down, train 1 is moving at 25 mph and since the other train is going 10 mph faster, we will say that train 2 is moving at 35 mph.0822

Now we have the speed of both trains.0856

Now that we have seen a couple of those the table methods, let us get into using the beaker method with some good mixture problems.0861

In this problem we want to know how many mL of a 40% acid solution must be mixed with 80 mL of a 70% acid solution0868

in order to get a new acid solution that is only 50%.0875

A lot of information and it is tough to visualize this one unlike the motion problem where you can actually see two trains moving.0881

This is why I use the beaker method so I have something a little bit more visual that I can grab onto each of our beakers.0887

Let us go ahead and start recording the amount and the percentage for each of these things.0894

Let us start with the amount so it says how many mL of a 40% acid solution must be mixed with 80 mL of a 70% acid solution?0902

It looks like it does not tell me how much of the 40% I have, so we will leave that as an unknown amount.0912

We will be mixing it with 80 mL of this other one.0922

The percentage of this beaker is our 40% and the other beaker is our 70%.0928

This new beaker on the, we are trying to mix the two together, how much will it have in terms of their amount?0942

I’m mixing some unknown amount + some 80 mL so this new beaker at the end will have a total of x + 80 mL.0949

It is important that the last beaker should have a total of both the amounts.0960

What is its percentage? This will have a 50% solution.0965

We have recorded the amounts, we have recorded the percentages, and we will multiply those two quantities together and actually get our equation.0971

(40% is the same as .4 multiplied by x) + (70% is the same as .7 multiplied by 80) all of this must be equal to .5 multiplied by x + 80.0981

By multiplying those two together and putting in the appropriate addition and equals I have now set up my equation and I can solve from here.1000

This one is like our equation that has fractions only this one has in decimals.1008

I’m going to multiply through by 10 to get rid of a lot of those decimals.1013

.4 multiplied by 10 = 4x, .7 multiplied by 10 will be 7 and .5 multiplied by 10 would be 5.1024

This is the equation that I need to solve in order to figure out the amount of that 40% solution.1041

Let me just copy this onto the next page and we will go from there.1049

I’m going to start combining together each side of this equation using my distributive property and then I can work on getting those x’s isolated.1072

4x + 560 = 5 × 8 another 0.1082

We will subtract 4x from both sides 560 = 400 + x subtract the 400 from both sides, so 160 = x.1100

What this is telling me is since x represents the amount of 40% solution, I know that we need 160 mL of the 40% solution.1127

It is always a good idea to keep track of what your unknown represents.1148

I have one last example and this one is using the beaker method again but this one involves a lot of coins.1154

This is to demonstrate that these two methods are flexible and you do not always have to use them 1162

with just a pure mixture or even just with distance, motion and travel.1167

They still can apply to a lot of other types of problems.1171

This one says a man has \$2.55 in quarters and nickels.1176

He has 9 more nickels than he does quarters.1180

How many nickels and quarters does he have?1183

We have to figure out what the two are mixed together.1185

It is like a mixture problem that is why we are going to use the beaker method to attack it.1190

The two things that I will try to keep track of is how much of quarters and nickels do I have and what would the amounts of each of these be.1194

I better label my beakers to make it a little bit easier.1205

We will say that this first one contains just nickels and the second one contains quarters.1208

Let us see what we know about the nickels.1219

He has 9 more nickels than quarters.1221

We do not know a lot about those quarters, do we?1223

The amount of quarters we will leave that as x and the amount of nickels there is 9 more than however many quarters he had, x + 9.1226

I can write down what each of them represent.1239

A quarter is the .25 and nickel.05.1245

What we are looking to do is to combine these together, the amount and how much each of them are to get a total amount in that last beaker.1253

He has a total of \$2.55 and the reason why I’m not writing down a specific amount or specific amount per coin in there1267

is because this one represents the two already put together just the \$2.05.1277

Let us go ahead and build our equation.1283

.05 multiplied by the amounts x + 9.1286

.25x = \$2.55 this one is similar to the previous one.1292

It has lots of decimals running around and I want to get rid of a lot of those decimals.1302

I’m going to multiply this one by hundreds and then we can do it just some nice numbers and no decimals.1307

This would make it 5 × x + 9 and 25x = 2.55 that is the equation that we will solve and figure out how many quarters and nickels we actually have.1316

5x + 9 + 25x = 2.55 working to simplify this equation I will distribute through on the left side of the equation giving me 5x + 45.1342

I will go ahead and combine my like terms 5x + 25x that will give 30x.1376

Then subtract 45 from both sides and you can get that x a little bit isolated.1390

Finally let us go head and divide it by 30 to see if we can figure out what that x is, so x = 7.1404

In the context of this problem, x represents the number of quarters we have.1416

We have 7/4 if you remember we have exactly 9 more nickels than quarters so we can add 9 to this number and figure out the total nickels.1423

We have 16 nickels.1440

We have figured out the total amount of both coins.1447

In each of these types of methods, they are great ways that you can get your information organized and into a good spot.1453

It will be handy with motion and with mixtures.1461

Thanks for watching www.educator.com.1465

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at the rectangular coordinates system.0002

It will help us out as we get into more of the graphing process with some of equations later.0006

Some of the specific things that we will look at as we are looking at our Cartesian coordinate system are just some of the parts of it.0012

We will learn all about the x axis, y axis, origin, quadrants and how we can plot an ordered pair.0018

When we get into plotting more things such as the graph we will look at looking at a table to do that graph.0026

And again how you can find the x and y intercepts of that line once on the graph.0032

The Cartesian coordinate system is formed by taking 2 number lines and putting them together after 0.0041

You will notice that one is in the up and down direction and one is completely horizontal.0047

Now the horizontal one, the one that goes completely flat, this guy right here, we will call this one our x axis.0053

The one that go straight up and down that will be known as our y axis.0064

The 0s of each of them when it both connects, we will also give that one a special name, we will call this point right here the origin.0071

The little tick marks that you see usually represent how far you have to go on each of these number lines.0087

You can see that it breaks down into these number lines into so much larger parts of this graph.0095

We will give each of these squares a name, we will call them quadrants.0103

We usually number these quadrants starting in the upper right corner, say quadrant 1 and move in a counterclockwise direction.0110

Now in each of these quadrants we can plot a point.0126

A point is an ordered pair and we look at it as a pair of the x and y put together.0131

If I'm looking at a point say out here, I have identify what it is x and y value are 0138

by looking at how may tick marks I have to go over on the x axis and this would be 2.0148

How may tick marks I have to go in the y direction, 1, 2, 3, 4, I can plot out that point.0154

Keep in mind that with all of these points that you see, the first one is the x and the second one is y.0163

Let us just do a real quick example of how we know little bit more about the parts of the graph and see if we can plot out some of these ordered pairs.0177

In all of these, the first one will represent an x value and the second will represent our y value.0185

We will start at the origin with each of these and moving the x and then the y direction.0191

Starting with first one, we need to plot 2, 4.0198

I’m going to start at the origin and I'm going to move 2 units to the right and 4 units up.0201

This point right here will be the point 2, 4.0213

The next point has a negative value in it and it is a negative in the x.0222

I actually move left 1, 2, 3 and then I will go up 7, that point is located right here, 4, -3, 7.0227

As long as you can keep track of which one was x and which ones what is y, not a bad task at all.0244

The next one is -1 and -5, at we will move in the x direction -1 and then down 5.0251

1, 2, 3, 4, 5, so there is our point -1, -5.0258

Just a couple of more to do, the next one is 3, -2, that is in the positive direction 1, 2, 3 and then down 1, 2.0269

And one last one to go, this one is 0, 4.0286

What should you do if 0, should you go right or should you go left?0293

Since it is at 0, we do not go right or left in the x direction, we actually stay at the origin and then we will move in the y direction 4.0297

Up 1, 2, 3, 4 that point is right here 0, 4.0304

Now that we have identified some points on here, we can actually say which quadrants each of these are in.0313

If we look at our first point up here, our 2, 4 it is in quadrant number 1 because it is in that first square.0319

The next point -3, 7 is in quadrant number 2.0331

3, -2 that one is in quadrant number 4.0345

The last point 0, 4 as the tricky one is not in quadrant 1, it is not in quadrant 2, it is actually right on the y axis.0351

Some more things about this rectangular coordinate system, when we get into actually trying to graph an equation on here0370

we can look at it as a special relation between its x values and its y values.0377

There are a lot of good things we can say about that equation.0383

To actually visually see what that relationship is and of course put it on the graph, 0386

we will go ahead and create a table of values and see how they are related.0390

Usually this is done by picking a lot of different values for one of the variables and see what the other corresponding other values need to be.0395

Here are very important terms you want to keep in mind.0402

Where the graph of that equation crosses the x axis that will be known as its x intercept.0404

In a similar fashion, if it crosses the y axis, we will call that the y intercept.0415

Just draw a general graph so this is not a line since it is all curvy but it still crosses the x axis in some spot so I will call this the x intercept over there.0422

It still crosses the y axis so it has a y intercept.0440

You want to be able to identify both of these points.0446

A little bit more about generating those points from the table.0453

I said that a line is a special relationship between its x values and its y values and to see that relationship, 0457

we want to be able to create a little chart and put in some values for either x or y.0463

Let us look at one I have 5x - 3y =12.0476

If I want to go ahead and try and visualize what this equation looks like, I need to know what values go on my graph.0486

What point should be there?0493

A good way to figure out if the point is on the graph or not, is to see if it simply satisfies the equation.0495

If I’m just picking some point like 3, 1, I can substitute the 3 in for x and 1 for y and see if it does satisfy it.0502

And see if 3, 1 is on the graph or not.0513

(5 × 3) – (3 × 1) does that equal 12, let us find out.0516

That will be 15 - 3 = 12 or 12 = 12.0524

What that tells me is that 3, 1 is one of the points on my graph, 1, 2, 3 on the x axis up one.0532

Let us pick another point, 0, -4.0546

But when I plugged this one in, I get 5 × 0 which is 0 - 3 × - 4 = 12.0554

I know that one is on the graph as well, 0, -1, 2, 3, 4.0570

In practice, we usually do not pick up points out of thin air and then test to see if they work or not.0583

Usually we end up picking a lot of different things for one variable then see what it has to be for the other variable.0588

Let us get a little bit of space in here and see how we would do this more in practice rather than just picking things out of thin air.0595

Let us pick out some more values for x.0603

For example what if x was 2, what with that 4y to be, we can find out by putting in the 2 for x and actually solving for y.0606

That will be 10 - 3y = 12, I will subtract the 10 from both sides and get y = -2/3.0620

I know sure enough that is another point that I can put on my graph, 2 in the x direction then down 2/3.0632

What you may notice is I start to do more and more points over here on the graph.0641

You will get a better sense of what the entire graph looks like.0645

I have done about 3 points here but I’m going to draw lines for all of them to say that if I had even more points they would all line up along there.0650

Let us get into some more examples and see what other parts of the graph we can identify0665

and get a little bit more into the nuts and bolts of using that chart to graph out an entire line.0669

In this example we just simply want to find the x intercepts and the y intercept of the line.0676

Remember, this is where it crosses the x axis and where crosses the y axis.0681

My x axis is horizontal and I can see it crosses right at that point.0686

That would be at -1, 2, 3, 4, 5 in the x direction and 0 in the y direction.0691

x intercepts is at -5, 0 it crosses the y axis as well right up here at 4.0701

For that one we would not go in the x direction whatsoever but we would go out 4, so its y intercept is at 0, 4.0717

With these ones we want to determine whether the point is on the line or not.0735

We can figure this out by substituting the values in for x and y.0739

The first one will always be an x and the next one a y.0746

-2 + 3y = 21, let us put in the -3 for x and 5 for y.0753

-2 × -3 = 6, 3 × 5 =15, for a total of 21, 6 + 15 = 21.0768

It looks like this one checks out since it makes the equation true, we can say that -3, 5 is on the line.0783

Let us do the same thing for the next one, my x is 1 and my y is 7.0794

Let see if this one rings true.0811

Plugging in 1 for x, 7 for y and simplify 3 + 4 is 7 and 5 × 7 is 35, these things are definitely not equal.0815

What is that tells us that 1, 7 is not on the line.0836

You can take a point, plug it into the equation and see if it satisfies it and makes it true.0845

Let us get into more of the graphing process.0852

Here is an entire equation and I want to know what its entire graph looks like.0854

Rather than just picking points out of thin air and testing them, we are going to try and create a few values of our own by generating them.0858

We will generate them using a nice little table.0867

I will pick some values for x and we will see what makes y.0870

To make this processing a little bit easier, we will take our equation and we will solve for y first.0875

Our equation is 2x + 3y =12, to solve this for y I would move the 2x to the other side by subtracting a 2x and then divide it by 3, I will get -2/3x + 4.0881

This is the same equation, I just manipulated it a little bit so I can work with it a little bit easier.0905

We are going to pick some values for x, plug them in and see what they make y.0912

It does not matter what values you pick but I do recommend choosing some negative values or may be throw in 0 and also choose some positive values.0918

Let us choose -3, -2, -1, 0, 1, 2 and 3.0927

With each of these, imagine what happens if you take -3 plug it in into the equation, what result do you get.0938

If you do some scratch work that is okay, y = -2/3 we multiply that by -3, add 4 to it and see what the result is.0944

-2/3 × -3 that would give us 2, 2 + 4 =6.0955

You put that on the other side, I know that -3, 6 is one of the points on our graph.0966

-3, 1, 2, 3, 4, 5, 6 there is a point right up there.0972

Let us take up the -3 and try this again, this time we will plug in -2.0982

This would give us 4/3 + 4 which should be the same as 5 and 1/3.0991

It is okay if you get fractions with these just the way it turns out since you are choosing all kinds of different values for x.1001

-2 up 1, 2, 3, 4, 5 and just a third more there is that point right there.1009

Let us take out the -2 and put in -1 and let us see if this works out.1019

negative × negative would be a positive, 2/3 + 4 = 4 2/3.1032

-1 up 1, 2, 3, 4 and 2/3 right above there.1043

I have not even graphed all of the points yet and you can see that they all seem to be following along in a straight line.1049

We will do one more then will go ahead and connect the dots.1058

Let us see what happens if x =0, 0 × -2/3 would be 0, 0 + 4 = 4.1066

We will put that point on there as well right up here at 4 and now we will just connect all of our dots and this would be our graph of the equation.1078

Using this table we can definitely see a lot of points even sometimes we do not need to graph all the points,1094

we should graph enough of them that you have a good sense of what it looks like.1100

Also if your graphing a line and you are going through this table and you got one that is a little off from all the rest, it is okay.1104

But if it is off from all the rest then we will check your work on now to make sure that you have done it correctly.1110

In this next one, we want to graph the equation again using a table of values.1121

Like last time, I will go ahead and solve this for y so we have a little bit easier of the time working with it.1126

I will move that 5x together just by adding 5x to both sides then I will go ahead and divide by 2.1133

This is the equation that I will be using it is the same as the original.1149

I have just manipulated it a little bit so I will have an easier time working with it.1153

Let us pick some values to put in there for x so we can see what y needs to be.1158

We will pick some nice ones like -4, -2, 0, 2 and 4.1165

This one I’m picking values that are multiples of 2 since all of them we have to multiply by 5/2 first.1172

That will make our lives a little bit easier.1179

y =5/2, let us plug back in -4 then we will add 10.1184

-4 × 5/2 =-10 when we add 10 to that we get 0 so our first point is -4, 0.1194

Let us try another one, this one is -2, I will plug that one in there -2 × 5/2 =-5.1212

I will add 10 to that and get a total of 5 so this point -2, 5.1229

What happens if we go ahead and put in a 0, 0 × 5/2 = 0 + 10 = 10.1250

I have another point for my chart here and this one is off the chart 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 way up here.1263

We can already see that after doing a bunch of these, the rest of my graph is going to go completely off of my graph paper.1275

We will go ahead and do the other 2 just so you can see what their value will be but we already have plenty of points so we go ahead and graph these line for 1, 2.1286

If I put in 2 I will have 5 +10 = 15.1293

If we put in 4, 4 × 5/2 = 10 + 10 =20.1305

Now we have plenty of points, let us go ahead and graph this out.1325

I will go ahead and connect the dots.1328

The graph of this equation looks like this and with this one we have taken it definitely farther and mark out its x intercept.1332

This one is at -4, 0 and its y intercept way up here will be at 0, 10 and see they show up in the chart here and here when one of the variables is 0.1342

We will look at more next time with being able to do some shortcuts for graphing these lines 1362

But with many types of equations you can often use a table to generate some values and put on some points.1367

Thank you for watching www.educator.com1373

Welcome back to www.educator.com.0000

In this lesson we will be taking a look at more about the slopes of lines and how we can use that to better graph.0002

Specifically some of the information that we went up looking at is how we can first determine the slope of line0012

whether we are just given a couple of points or whether we have the entire graph of that line.0017

Once we know more about slope we will be able to learn how to graph the entire thing using its slope and its y-intercept.0026

This will bring about many different forms that you can represent a line.0032

We will learn about switching back and forth between these many forms.0035

We will learn about some very special lines, the ones that are vertical and horizontal.0040

Look for those equations so you can easily recognize them in the future.0043

When it comes to a line, there is actually many ways that you can go about graphing or representing that line.0051

You can write it in standard form, slope intercept form and point slope form.0056

For now I will be mainly concerned with these first 2 forms, the standard form and the slope intercept form.0062

We will get more into the point slope form in another lesson.0068

When it comes to standard form, it looks a lot like this ax + by = c.0074

The way you want to recognize a standard form if you ever come across it, is that both of your x’s and y’s will be on the same side of the equation.0082

We like to usually put them on the left side.0089

There are no fractions or decimals present in the equation so the a, b, and c, those are nice whole numbers in there.0093

The x term here will be positive that is the a value.0102

We do not want to be a -6 or -7, we usually like to be 3, a nice positive value. 0106

The reason why this form of the line will be so important is that if you have a line in standard form, it usually easy to graph.0114

The way we go about graphing something in standard form is we use its intercepts.0122

That is where it crosses the x and y axis.0127

The reason why the intercepts are so nice for our standard form is because when it does cross one of those axis, one of the values either x or y will be 0.0130

It will be making a table, but it would not be that big.0140

Let us do a quick example of something in standard form to see how easy it is to graph.0145

In this one I have 7x + 2y = 14.0151

You can see that it is in a standard form because I have both of my x’s and y’s on the same side.0154

I do not see any fractions, no decimals and the coefficient of x here is positive, a nice 7.0160

In order to graph this, I will find its x and y intercepts.0169

I will make my chart here rather than picking a lot of different points, I will look at where x = 0 and where y = 0.0174

Watch what this does to the equation.0184

If I use 0 for x, then it is going to get rid of the entire x term.0187

Since 7 × 0 right here, all of that is just going to go away.0196

I just have to solve the nice simple equation 2y = 14 which I can do by dividing both sides by 2.0201

I know that one of my intercepts, my y-intercept is at 0, 7, nice and simple.0212

Let us do the same thing by putting in a 0 for y.0219

7x + 2 × 0 = 14.0222

You will see that 0 is in there and again it is going to get rid of this term entirely since 2 × 0 is 0.0229

Then I have 7x = 14, divide both sides by 7 and we will get x = 2.0237

Now that I have both of these points, we will put them on our graph.0249

0-7,0, 1, 2, 3, 4, 5, 6, 7 way up here and the other 1, 2, 0 over here.0253

Connect those intercepts and I will get the graph of the entire line.0273

One disadvantage with using just the intercepts to graph a line is if you make a mistake on one of them, it is often hard to catch.0277

If you want to get around that problem, it might not be a bad idea to actually put in an additional point to see what happens with the graph.0284

That way if you do make a mistake with one of them you be able to quickly see that they are not all in a straight line.0292

To understand some of the other forms like slope intercept form, you have to understand a lot more about slope.0301

What exactly is this slope?0307

If I had to describe it, it is a measure of the steepness of a line, how steep is a line, if it is shallows or more steep?0310

Can you attach a number to that steepness?0319

What we do and we call it the slope.0322

In many equations we will use the letter m to represent that slope.0324

How do we attach a number to the steepness?0329

What we will do is we take 2 points from the line and we end up looking at the difference in the y values over the difference in the x values.0331

This gives us a nice equation for figuring out the slope of a line.0340

You will see these whole numbers in here and you can interpret that as each of these points.0345

These x’s and y’s come from point number 1 and this other one, these values come from point number 2.0351

In our work later, it is often a good idea to label one of your points as .2, and one of this is .1, just we do not mix up things in the slope formula.0360

You may have also heard of other ways to describe slope.0373

One of the most common is to call slope the rise of the line divided by the run of the line or simply rise over run.0376

That is actually a good way to remember how it is related to its steepness.0384

Its change in the rise would be the y direction and this change in the run would be the x direction.0388

One thing that may throw you off is that sometimes there is a negative sign in the slope.0401

You can interpret that negative line sign as whether you are going up in your rise or down, or do you need to go left or right in your run.0405

Let me go ahead and pick this is apart.0414

If you see a positive sign in the top then think of going up on your rise.0417

If you see a negative sign then think of going down.0424

With the run if it is positive, you will end up going right and if it is negative, then go left.0431

To make it a little bit more sense once we start seeing some more lines.0439

Maybe I can give me a quick example right here.0445

A line like this, I can say that maybe the rise is 5 and run over here is 4 so the slope will be equal 5/4.0450

If I had a different line like this, I can see them going down and then I have to go right.0466

Since I’m going down, I will mark this one as -3 maybe this one over here as 5, then I will have a slope of -3/5.0477

In practice 1, if I do have a negative sign in my slope, I usually just give it to the top part of the fraction.0488

That way I only have to remember about going up and down.0492

I do not even have to worry about left and right since the bottom is positive.0496

Another word of warning, be careful not to mix up all your points.0503

If you have your y values from point 2 being first then take the x values from point 2 being first as well.0507

I also use the sign of this to give you a little bit more intuition as to which direction that line should be facing.0516

There are only a few instances of what your line can look like.0523

It could be going from the lower left to the upper right and that is an indication that your slope is going to be positive,0527

since both your rise and your run are going to be positive numbers.0534

If your line is going from the upper left to the lower right and you are going to have a negative slope.0543

This will be because your rise is negative, but your run is positive.0551

The other 2 special cases that you have to come and watch out for is what happens when your slope is 0 0557

and what happens when it is undefined.0562

In these 2 instances, you either have the alignment as completely horizontal.0565

This is when you have a 0 slope and it is completely up and down if you have an undefined slope.0570

Let us get into our examples and see how we can start finding our slope just from a couple of points.0580

With these ones we will use the formula for the slope of the line to pick it apart.0587

I’m going to try and keep things a little bit together by marking these out as point number 2 and I will mark the other one out as point number 1.0593

In the formula, here is what we are looking at.0610

Our slope should take our y values and subtract them all over our x values and subtract those.0613

Notice how I’m keeping things all lined up, I have both of my x and y from point 2 over here and both my x and y from point 1over here.0622

Let us put in some values.0636

I need the y value from point number 2, that is a 4 then we will minus our y value from point number 1 that is -1.0638

All over x value from point number 2, -2 and x value from point 1, -1.0654

Be very careful on your signs with that.0672

Looking at the top, 4 - -1 is the same as 4 +1 = 5.0675

-2 - -1is the same as -2 + 1= -1.0683

It looks like this slope for the first one 5/-1 I will just get a slope of -5 between these 2 points.0690

It does not matter which one you will label as point number 2 or point number 1 just as long as you keep them straight.0702

I’m going to switch which ones I’m calling point number 2 and point number 1 just to highlight this, but you will get the same either way.0712

Let us start off, I want to subtract my y2 from my y1.0721

Y2 is 6 – y1 6, my x2 is -5 and my x1 is 3.0727

On the top of this, I have 6 - 6 giving us 0.0752

On the bottom I have -5 - 3 - 8 and 0 divide by anything is 0.0757

This indicates that our slope is 0.0764

This is one of our special cases where we have a horizontal line, it is completely horizontal.0767

Let us do one more, point number 2 and point number 1.0772

The slope for this I will take my y value from the second point I will subtract the y value from the first one.0789

Then to our x’s, x from our second point minus x from our first point.0804

Now we will work to simplify, 5 - -4 is the same as 5 + 4 =9, 3 – 3 =0.0814

Be very careful with this one, we can not divide by 0.0826

Whatever that mean, it will get us around the bottom.0839

This is indicating that our slope is undefined.0841

It is not that there is not a line there, there is a line.0845

The line is just completely straight up and- own, it is our vertical line.0847

Just to make this a little more clear, I will say slope is undefined or sometimes we might say that there is no slope.0851

In order to know a lot more about slope, we will get a little bit more into the slope intercept form.0868

Slope intercept form looks like this, y = mx + b.0875

The way you can recognize this form is that the y will be completely alone on one side of the equation.0880

Usually I like to put it on the left side.0886

Our slope will usually be represented by that m and we will put it right next to the x.0889

The b in this equation stands for our y-intercept of the line.0896

That is where it crosses the y axis on our Cartesian coordinate system.0900

The reason why that this form is usually everyone's favorite is because it makes graphing a nice and simple process.0906

The way it makes a graphing so nice for us, is we start at the y intercept.0913

Just rely from the graph by whatever that d value is.0919

What we do is we use the slope as directions on how to get to another point on our graph.0922

Let me give you a real quick example of how this would work in practice.0929

I have 1/7x + 3.0934

The very first thing that I would do is I would look over here at the y intercept and I would take its value.0945

I know that this particular line crosses the y axis 3 and I would put a point on the y axis right at 3.0952

Starting at that y-intercept, I would use the slope as directions to get to another point.0967

Keep in mind that it is the same as rise over run.0976

Starting at that y-intercept, I will go up one into the right 7, up 1 to the right 1, 2, 3, 4, 5, 6, 7.0980

Now that I have 2 points, I can go ahead and connect them and make the entire graph of this equation.0995

One you start with the y-intercept and two you use the slope as directions to get to a second point.1006

Let us see this in action by actually graphing out some linear equations.1016

They said before let us start here on the nth with our y-intercept and make that our first point that we put on the graph.1022

This one is -5, it crosses the y axis down here at -5, it looks good.1029

In terms of our slope, we want to think of this as rise over run so starting at that y intercept, we go up 3 and to the right 4.1038

1, 2, 3 to the right, 1, 2, 3, 4 and now we have a second point.1050

Now that we have 2 points, simply connect them to a nice solid line and there is your entire graph.1063

It makes the graphing process much easier.1073

You do not even have to worry about the table and doing all of those values.1075

Let us try this one.1082

Graph the line using the y intercept and the slope.1083

On the back in here, I see that my y-intercept is 3 and that will be the first point on my graph.1087

My slope is -2, how will that work with rise over run?1099

It does not look like a fraction like in some of my other examples.1107

It does not feel free to turn it into a fraction by simply putting it over 1.1111

This tells me that I need to go down 2 since that is negative and to the right 1, down -2 and right 1, now I have a second point right there.1116

I can draw the entire graph, very nice.1132

We have both of these forms under our standard form and our slope intercept form,1142

you may be curious which one should you be using most of the time.1152

Both of them are good for graphing.1159

And what I often recommend is if you have to graph something and it is already in standard form, just go ahead and use the intercept.1163

It is usually one of the quickest ways to do it.1169

If it is already in slope intercept form then use its y intercept and slope as a direction to the second point and graph it that way.1171

That is usually the quickest.1178

If you do have to switch back and forth between these 2 maybe you are more comfortable with slope intercept form, then feel free to do so.1179

If you have something that is in some other form and you want to get in the slope intercept form then the process is pretty quick.1188

What you should do is simply solve for y and get it all by itself on one side of the equation.1195

In doing so, you will be able to better see what it slope is and the y-intercept.1200

Not very many people go the other direction, but potentially you could end up rewriting something into standard form.1206

There is a lot more criteria that go in there.1212

One of the first things is you should get both of your x’s and y’s on the same side.1215

Then you want to make sure that your constant is on the other side.1221

Only x’s and y’s on one, constant on the other.1226

Try and clear out your fractions by multiplying by a common denominator.1231

a, b, and c should not be fractions.1233

Then look at the coefficient in front of x and it should be positive.1237

If that is not positive, then multiply it by -1 and make it positive.1243

More practice switching back and forth so we can see how this process works.1250

Even though we have our standard form and our slope intercept form.1259

You want to be aware that there are 2 special cases and we seen them come up but once before.1262

We have some lines that are completely vertical and some that are horizontal.1268

The vertical ones are straight up and down, and the horizontal ones are left and right.1275

The way you can recognize their equations are they are simply x equals a number or y equals a number.1281

You may see something like x = 2, maybe this is like y = 15.1289

For the one it says x equals these are your vertical lines.1296

For the one that says y equals these are your horizontal lines.1304

The way that they can keep in track of which one should be horizontal and which one should be vertical is the way you interpret them. 1312

If you have an equation like x = 2, what that is trying to tell you is that the x value no matter where you are in that line is always 2.1318

If my line looks something like that and I decide to pick up some individual points,1330

no matter what point I pick out I can be sure that the x value will be 2, no matter where I am on that line.1335

In a similar fashion, if I’m looking at y = 15, no matter where I am on that line I should end up with the y value being 15.1346

Watch for these 2 special cases to come up in my examples.1364

Let us first work on switching back and forth between these 2 different forms.1371

What I have here is a line in standard form and we want to put it into slope intercept form and we want to put it into slope intercept form.1375

I want to actually go through the process of graphing it and I’m more familiar with slope intercept.1383

When I already did take this and put it in to that other form, I need to solve for y.1388

Let us start by moving the 7x to the other side, 2y = -7x + 14.1395

And then we will divide everything by 2 and that should get our y completely by itself.1407

Notice on the right side there, I have to divide both of those terms by 2, -7/2 x + 7.1416

The most important part about writing it in this new form is now I can easily identify what my y intercept is.1427

It looks like it 7 and I can more easily identify what my slope is.1435

It has a slope of -7/2 and I know it is facing down from the upper left to the lower right.1441

Let us go to the other direction.1450

Let us take a line that is written in slope intercept form and put it into standard form.1452

It requires a little bit more work but we can do it.1458

The first thing I’m going to do is try and work to get my x’s and y’s on the same side.1461

I will subtract 1/2x on both sides, -1/2x + y = 3.1466

I want to make sure that my constant is on the other side, that is the 3 and it is already there.1481

I want to get rid of all fractions so I need to get rid of that ½.1487

I can do this if I multiply both sides of the equation by a common denominator and in this case that would be 2.1492

-x + 2y = 6, we are almost there.1500

You can see that it certainly look a lot more like that standard.1507

The last thing we need to make sure is that the coefficient in front of x is positive.1510

Right now looks like it is negative, I already fixed that.1515

I will multiply everything through by -1.1518

-1 × -x would be x and -1 × 2y = -2y, equals -6.1528

This form is in standard form.1544

They might be looking at it and say what good is that? Why would you want it in standard form?1547

Remember that you can graph it in standard form now by simply looking at its intercepts plugging in at 0 for x and 0 for y.1551

Let us get into some very special cases.1560

We want to graph the line y =5.1563

There is not much of the equation to look at.1568

What should be the slope? What should be the y intercept?1570

This is one of our special cases, y equals a number.1574

Since it is y over here, this is going to help me indicate that this is going to be a horizontal line.1578

1, 2, 3, 4, 5 would be one point.1584

I will just make a giant horizontal line with all points y = 5 and double checking this make sense.1588

If I was to pick a point on a line at random, in this case it is 1, 2, 3, 4, 5, 6 its y value is 5.1596

If I pick something over here, its y value is still 5 no matter where you go on this line its y value will always be 5.1606

One last special case, this is x = -2.1618

This will be a vertical line straight up and down because we are dealing with x over here.1623

I’m at x = -2 and we will make it straight up and down.1631

The reason why this makes sense is no matter what point you choose on line, as a way up here at 1, 2, 3, 4, 5, 6, 7 the x value will always be 2.1641

With the slope of this one, remember that its slope is undefined.1663

There you have it some have very nice techniques you can use for graphing lines and now 2 forms that you can use to represent lines.1668

Thanks for watching www.educator.com.1676

Welcome back to www.educator.com.0000

In this lesson we are going to continue on with our linear equations and look at how we can build these linear equations for ourselves.0002

Specifically, some of things that we will look at is we will look at the point slope form on a line.0014

It will help us when building those equations.0020

For some of our examples, we will also have to know a little bit more about parallel and perpendicular lines.0023

Watch out for my explanation on what these are and how you can tell if two lines are parallel or perpendicular.0029

Earlier, we looked at two forms of a line.0038

We looked at standard form and we looked at slope intercept form.0041

Remember both of these forms are important because they were good for graphing.0045

When we want to build our own line, sometimes we can do this as long as we have enough information.0057

To form that we use for building our line is known as point slope form.0064

The reason why it is called that is because those are the two bits of information that we need. 0068

We need to know at least one point on a line and we need to know what the slope of that line is.0072

Once we have both those bits of information and we can drop it into point slope form.0077

This form here requires a little bit of explanation.0086

You will see that there is an x and y that has some subscripts on it and that is where the point goes in terms of its x and y.0090

The slope is the n here so we can just put that in and then we also have a couple of other x’s and y’s which do not have subscripts.0098

With the ones that do not have any subscripts whatsoever, you will not be putting any values in for those.0108

Those simply stay in our equation. 0114

One last thing to note is even though we will use point slope to actually build the equation on a line.0116

Usually you take point slope and end up converting it into slope intercept form because you are looking to do some other things without a line,0122

other than just than build it or maybe you want to eventually graph it.0129

It is good to go ahead and move it in to one of those other forms.0132

To build lines using point slope form, we want to substitute in our point and we will go ahead and substitute in our slope.0143

I already have point slope form, I will put it over here and I have a point and I have a slope.0151

Let us go ahead and just substitute these in and see how it works.0157

I will have y - -3 then the m represent the slope, I will put in ¾, x – xy.0161

Notice how the x and y which did not have any subscripts are still in there.0176

Once we have everything substitute in here, we want to go ahead and start rewriting this0181

either into standard form or slope intercept form so we can do some other stuff with it.0186

I’m going to go ahead and put this into slope intercept form by just getting y all by itself and maybe cleaning up a few other terms.0190

I'm subtracting a negative on the left side and that will be the same as adding 3.0198

I will go ahead and distribute my ¾ here.0203

I have ¾ x and -3, looking much better.0206

I will go ahead and subtract 3 from both sides.0216

Y = 3/4x – 3 – 3 is 6.0219

Our point slope gets our foot in the door so we can actually build the equation and then put it into a different form, maybe a slope intercept down here.0227

We can go ahead and graph it and do some other things with it.0241

Sometimes we will be given information about another line in order to build the one we want.0247

And information about that other line, we may know that it is parallel or perpendicular to the one we want.0253

What exactly does that mean? 0258

We can say that two lines are parallel if they had exactly the same slope.0261

In my little picture here, you can see what that does.0266

You will end up with two lines and they usually have a little bit of a gap or space in between them, but they have exactly the same slope.0269

They are going in the same direction.0275

You might also have lines that are perpendicular.0279

With this one, they end up meeting at a right angle.0282

We do not talk much about angles in this course, another way that you can say two lines or perpendicular is 0286

if their slopes are negative reciprocals of one another.0291

Let me show you what that means.0295

I suppose this blue line had a slope of 4/5, if the red one was perpendicular then I wanted to be the negative reciprocal of the blue line.0296

I made it negative and I flipped it over.0312

With perpendicular lines, they will always be different in sign, one will be positive and one will be negative and they will be reciprocal when flipped over.0314

Another way that you can test if two lines are perpendicular or not is actually you take both of their slopes and just multiply them together.0325

Two things could happen if you multiply them together and their slope is negative or their combined value is -1 0333

then you will know for sure that they are perpendicular.0342

If you multiply them together and you get anything else other than -1 then you will know that they are not perpendicular.0346

It is a nice and easy task you can use to know how the two lines are related.0352

We will definitely know this as we get into more of those examples.0358

Let us start off with example 1, in this one we want to write the equation on a line using the given point and the slope.0363

Since the two bits of information that I need for point slope, I will just go ahead and nearly drop it into the formula.0372

y -y1 equals slope x - x1.0378

Let us put in the y value first, y - 3 equals my slope is -2/3x - -6.0386

You will know the formula has a minus sign in there since the x value is a negative, go ahead and put that negative sign in there as well.0399

Now all we got to do is clean it up a little bit and maybe turned it into a different form.0408

I will see what we can do.0412

y – 3 = -2/3, when I subtract the negative that is the same as addition.0413

x + 6 and I think I will go ahead and distribute this -2/3 in there.0422

-2/3x – 12 ÷ 3 and a -12 ÷ 3 = -4, it looks pretty good. 0430

Let us add 3 to both sides and now we have that same line written into slope intercept form.0449

Again we use point slope form to go ahead and create the equation of the line and then probably put it into some other form.0462

This one is a little different, in this one we simply want to determine whether the two lines are parallel0472

or maybe the perpendicular or maybe they do not fall into either of those two cases.0477

With these ones, notice how many of them are written not in one of the forms that we have covered. 0482

In other words they are not in slope intercept form but the second one is in standard form but we have to be able to figure out what their slope is.0489

A way to determine what the slope of the line is to go ahead and rewrite it into our slope intercept form and we can just read it out of the equation.0501

Let us do that first before we actually compare what these slopes are.0509

Starting with this first one here, I can move the 2 to the other side by subtracting 2 so y =5x – 2.0512

Let us see what the second one, let us go ahead and move the x to the other side, -x – 15 and then we will divide it by 5.0521

Here is our first line, here is our second one.0539

Now they are both in slope intercept form we can say that the first one has a slope of 5 0545

since it is right next to x and the other one has a slope of -1/5.0550

Looking at the two slopes, they are definitely not the same, they are not parallel.0555

It looks like one is positive and one is negative and they are reciprocals.0562

I think these two are perpendicular.0567

If you want to test out feel free to just take the two slopes and multiply them together and see that when you do this you get a -1.0570

You will know that these two lines are perpendicular.0579

Let us try this process with the next pair of lines.0590

I will begin by just putting this into slope intercept form.0595

This one is almost in slope intercept form.0599

I just have to reverse the y and putting it first.0602

This one I will go ahead and move the 3x to the other side, so now I have my two lines right here.0606

y = 2x + 1 and y = -3x + 4, in this form, I can read off what both of their slopes are.0616

Looking at their slopes I can see that they are not the same, they are definitely not parallel.0627

One is positive and one is negative, but they are not reciprocal so they are not perpendicular either.0633

They are not parallel or not perpendicular, I will put this in the neither categories.0638

They are just two lines hanging around on the graph.0643

Let us try that same process with another pair of lines.0650

Let us see if we can find a couple of that which actually are parallel.0652

We will begin by putting these into a better form so we can read off that slope.0656

I'm moving the 4x to the other side, this one the y is still not completely all by itself.0662

Let us go ahead and divide everything through by 2, now we have one of our lines.0668

The second one let us go ahead and rearrange things.0678

The y is on the left and let us get it completely isolated by multiplying everything through by -1.0683

Now we have both of our lines.0693

Now that they are in this form, the slope of the first one is -2 and the slope of the second one is -2.0702

I can see that they are exactly the same and we will call these pair of lines parallel.0709

Onto one more pair of lines, let us see what they are, parallel, perpendicular, or neither, we will find out.0719

I need to get my y isolated I will start off by moving the 4x to the other side.0726

I will divide both sides by 3, y =4/3x +2.0734

The second one, let us go ahead and move the 2 over.0745

I have 3x + 2 and I will divide everything by the 4, ¾ x + ½.0749

Here is equation 1 and here is equation 2.0760

If you look at them in this form, we can pick out what each of their slopes are, I have 4/3 and 3/4.0767

These ones are pretty close, they are definitely reciprocals of one another since you would take 4/3 and flip it over and get ¾.0775

When those have been both positive so we can not say that these are perpendicular.0783

Even though they are reciprocals they are not negative reciprocals of one another.0788

They are definitely so they are not parallel.0792

Even though they are close, they do end up in the neither category.0795

They are not parallel or not perpendicular, they are simply neither.0799

They are just two lines.0802

With this one we want to write the equation of a line that happens to be parallel to the given line and also goes through the given point.0809

This one is a little different, notice how we need to know what the slope of the line is0818

and we need to know what the point is but they have not quite given us what the slope is.0822

We are going to have to extract that out of this other line that they have given us by knowing that it is parallel to the one we want to build.0827

Let us hunt down what the slope of this other line is.0834

We will do that by dividing everything through by 4.0837

It looks good so y = ¼ x + 5.0845

Let us see, I know that the slope is ¼.0850

Now that I have a point, I have a slope, I can use point slope form and build our line.0855

Y - -3 = ¼ x – xy and now let us clean it up and see what line we have.0861

y + 3 = ¼ x, ¼ of 2 would be -1/2.0872

Now we will subtract 3 from both sides.0890

y = ¼ x – ½ - 3, get a common denominator over there.0893

Our line is y = ¼ x – 7/2.0908

The way we built this line, I know for sure that it goes through the given point 2, -3.0917

I already explored what the slope of the other line was so I know it has a slope of 1/4.0923

Let us try this one, write the equation of a line which is perpendicular to the given line and it goes through the given point.0932

Similar to the other one, but it have not given us the slope directly.0939

It just told us it is perpendicular to this other line over here.0943

Let us figure out what slope is, y equals, I will move the 8 to the other side, -8x + 3.0948

The slope of this line is -8 which is good but that is not exactly the slope we want to use.0958

Our line is perpendicular to this line, so we need to use the slope -1, 1/8. 0966

That way it is the negative reciprocal of the other.0979

Now we will drop it into our formula.0984

Let us go ahead and put in our y value. 0994

We have our slope and our x value.0997

We will clean it up and see what line we have.1002

y - 8 = 1/8 - 4/8, my x in there.1005

y – 8, 1/8x – ½ and we will go ahead and 8 to both sides, find a common denominator and combine the last of our like terms.1017

1/8x + 15/2.1044

We can see that its slope is the negative reciprocal of the other line.1053

I know it is perpendicular and the way we build that it, for sure it goes through the point 4/8.1057

This last one is a little bit more of the word problem but you can see how we can still use these techniques in order to build the equation of the line.1066

We want to build a line that represents the following problem.1074

It cost a \$20 flat fee to rent a drill + \$2 every day starting with the first day.1077

Let x represent the number of days that we rent this drill and y represent the charge to the users.1085

How much will we end up willing to get from them?1091

When we are all done, let us see if we can write this line in a slope intercept form.1095

Using the techniques that we know about so far, we need to figure out what our slope is and what point this will go through.1100

Let us see, one thing that we can think of is the slope being what changes or your variable costs.1108

The \$20 that you have to pay up front is not a variable.1120

It is going to be there no matter what, what does change is a \$2 every single day.1123

The variable cost, we can think of that as our slope.1134

Let us see what we can do with that.1144

What happens with that flat cost?1146

We will be charged of that no matter what, that is like our y intercept.1151

If this one, we can just drop in our variable cost and we can drop in our flat fee.1158

What we are left here is the equation of the line but it represents how much we end up charging the person1177

and we can see this by substituting some values.1183

Suppose they rent this drill for one day, it cost \$2 for that day + \$20 flat fee, it cost them \$22.1186

If they rent it for two days, we can put that in for x, \$4 for the variable + \$20 flat fee, \$24.1196

You can go on and on figuring out how much it cost for each of the different days.1206

But in the end, this formula right here would represent how much we need to charge them.1211

Just simply plug in the number of days for x and y would give you the cost.1215

You can see that building a line is not so bad and using point slope form is handy in that entire process.1221

If you do use point slope form, you need to know the slope of the line and you need to know a point on that line.1228

Thanks for watching www.educator.com.1234

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at an introduction to functions. 0002

There is quite a bit to cover when it comes to functions, we will go over this bit by bit.0008

Some of the first things I will do with functions is look at some of the terms. 0014

In other words, what exactly makes a function and why is it a special type of a relationship. 0019

What is it that makes a function a function?0024

We will get into a term such as the independent and dependent variable so they can find those parts in a function.0028

We will talk about how functions can be like little machines and how they have input and output.0035

We will call this out its domain and its range.0040

We will get into a little bit on how you can represent function using diagrams that we can better keep track of the inputs and outputs0044

and also how you can work with functions by evaluating them.0051

Let us take a look.0056

A function is a very special type of relation.0061

What a relation does is it connects 2 variables and usually we use a pair of ordered pairs to describe that connection.0065

Maybe I have an x variable and a y variable and one other connected.0074

I can show you that anytime I use a 2 for x, I get a 3 for y.0078

There are many other types of connections that you can consider as part of a relation.0084

Some of our equations can be considered a type of relation.0091

A function is just a special kind of one of these relations.0097

It is special because for every x value it has, there is one and only one y value associated with it. 0101

What that means is that if you use a particular x value, you are never going to get two different values using that one x value. 0111

When we represent functions, we usually identify some sort of independent variable.0120

This is what we get to choose for in terms of the x.0126

The dependent variable is usually represented by y and it depends on whatever we choose for x.0131

Some of the notation you may see for functions looks like this.0142

It is important that we can pick out each of the different parts here.0147

The first thing is what is this f out in front? f represents the name of the function.0152

This function is called f.0159

The x inside the parentheses right next to it is not multiplication, what that is indicating is what the variable is.0168

We have our function called f and it takes a variable of x.0180

Everything after that on the other side, that represents the relation or how things are connected. 0186

You can also consider this as the rule of the function or what it does when you use a particular x.0196

The way you read this notation is f(x) = 3x + 9.0204

Our f part helps us identify what variables are being used inside the function.0215

To better understand what type of relations our function is, I like to tell my students that they are a lot like little machines.0226

They take some input and you put into and they produce some output.0233

A nice little diagram would look something like this down here.0240

Maybe my function f here takes an input of 3 and produces an output of 5.0246

If we start to gather up together more of its inputs and outputs, we can represent those as a whole bunch of ordered pairs.0254

What this list of ordered pairs would say is that my function takes an input of 1 and produces an output of 2.0262

If I give it an input of 3 then I get an output of 5.0272

Every ordered pair is telling me what the input and output was for that particular value.0278

Now, if we collected together all of the inputs, this would be known as the domain of the function.0288

If we collected together all of the outputs of the function, this be known as the range.0294

Using the example from a whole list of ordered pairs, we can actually identify all the things used as an input and all the things used as an output.0301

Remember this first value here, those are our x’s, those are inputs and the second values, those are our outputs.0311

Let us simply just list out all of the inputs and call that our domain and list out all the outputs and call that our range.0329

We used 1, 3, 4, 7 for input and what we got as output was 3, -2, 5, 10.0336

You can also represent a relation in the same way, so it is important to point out that this particular relation here is actually a function.0351

The reason why we know it is a function is that every time we are looking at any particular output, it only has one output associated with it. 0360

Input of 1 output of 3, input of 3 output of -2.0370

You do not see anything on this list like 1, 7 because that would be a problem 0377

because the 1 would go to 3 and 1 would go to 7, you will get 2 outputs.0384

I just want to make a little note here that this is a function because it satisfies our definition.0390

A great way to visualize what is happening with the inputs and outputs is to use a diagram.0403

One way to do that is to list all of your domains and group it together in a circle or a box.0409

And you group together all of your range in the same way.0415

You can show the relation by using arrows to connect the two. 0419

That is exactly what I have done down below.0423

Here I have two different relations, you can see my inputs and you can see my outputs.0426

We will consider these our domain and our range. 0436

The big question is does it satisfy the definition of a function or is it just simply some relation.0443

Let us take a look at it and see what happens to those inputs and outputs.0449

Remember, it is a function if every input you give it goes to exactly 1 output.0453

Let us see what we got.0459

Looking at this input here of 1, it looks like it has an output of 4 and that is the only thing it has for now, I will put just 4.0461

Let us check out the 3, if I use an input of 3 it looks like I will get an output of 4 and I will get an output of 0. 0470

Let me highlight that.0479

Notice how this one has an arrow that goes right over here to 4 and has another one that goes out to 0. 0483

When I use an input of 3, it goes to two different outputs then I can say that this is not a function.0493

It is close but it does not work.0509

Since we know that one was not a function, let us take a look at the other relation.0512

When I use an input of 1 it goes to 8, I will use an input of 3 it goes to 5, input of 6 goes to 1, input of 7 goes to 0. 0519

Every time I use an input here, it only goes to one output.0530

This one was okay, this one is a function.0534

You can see these diagrams help out with looking at what is in the domain and range, it is all grouped together nice and simple.0542

In some functions, they may not actually specify what domain is being used and do not actually says what the domain is.0555

We just assume that is known as the natural domain.0563

What this natural domain represents is all of the possible real numbers that you could use as input 0567

as long as it does not make the function itself undefined.0574

You might start with thinking about all possible numbers and eliminating ones that you can not use.0578

To help better identify what you can not use in the domain, look for stuff like fractions or even roots. 0586

The reason for this is, you cannot divide by 0.0595

If any value would make the bottom of a fraction 0 then you toss them out, they are not in the domain. 0600

Also, we do not want to have to deal with imaginary numbers, we do not want negative numbers underneath an even root. 0607

If that happens, we will toss those out of the domain as well.0615

When it comes to writing down a lot of different numbers in the domain, there is a special way that we go about doing that.0620

We often use what is known as interval notation. 0628

What interval notation is, it is a way of packaging up all of the range of numbers on a number line.0631

Let me give you a nice quick example.0640

I wanted to describe all the numbers between 2 and 3. 0641

I could do that by shading in those numbers on a number line so you can see that I have shaded in 2, 3, and everything in between. 0647

To use interval notation to do this, I would list out the starting point and the endpoint of everything that I have shaded.0658

Then I would use a bracket to say that the endpoints are included.0670

Everything between 2 and 3, I’m using that square bracket because 2 and 3 shall also be included. 0676

If we can go all the way up to a point but maybe not included, we will still use interval notation for this.0685

But usually we use a parenthesis when it is not included.0698

I have re adjusted the number line here, now I shaded everything from to 2 to 3 but not including 3.0704

I have made the corresponding interval notation to reflect that. 0711

I started at 2 stop at 3, includes 2, but does not include 3.0715

How we work with these functions?0724

One of the more common things we do with functions is to simply evaluate them.0726

To evaluate a function, what we simply do is substitute in the value given in place of x.0734

If I'm dealing with this function f(x) = 3x + 9 and maybe I want to evaluate it at the number 7. 0742

The way I do this is I just replace all x’s with 7.0750

You will see that the process itself is not that difficult.0759

It is sometimes just the notation that throws people off because when you look at the left side over here, 0763

your brain sometimes looks at that and you want to think of multiplication, but it is not multiplication.0769

When you have that f(7), what you are saying is that you used an input of 7 and then over here is what you got as your output.0776

Keep that in mind when you are evaluating functions.0789

Let us get into some examples and see if we can figure out whether certain relations are functions0795

and maybe identify a domain, range, and all that fun stuff.0803

In this relation, I can see that my inputs would be the first values and my outputs those are the second values.0808

Let us represent this using a diagram just to help out. 0820

We will group together all of our inputs into a giant circle over here.0824

We will group together all of our outputs into a giant circle over here.0828

First the input, I have -3, 6, I have another -3 so I do not want to list it twice and I have 5.0834

Our outputs are 7, -2, 4, and 9.0845

When I use an input of -3 it goes over here to 7 and when I use an input of 6, this heads over to -2.0853

I have another -3, -3 also goes to 4 and 5 goes to 9.0866

That is a bit of a problem isn’t it?0877

Notice how we have a single input and it goes to two different spots. 0879

If we have a single input going to 2 outputs then we can say that this is not a function.0887

Here I see another diagram.0905

Let us see if we can figure out if this one is a function, 13 goes to 7, 81 goes to 7, 32 goes to 60, 27 goes to 19 and 45 goes to 55.0906

This one actually looks pretty good. 0920

This is a function.0926

There is an interesting feature here that I want to point out, 0932

you noticed in the last one we had one input that went to 2 different outputs and that made it not a function.0937

You may be curious if this would also make it not a function? After all, 13 and 81 go to the same spot. 0943

That is okay, you can have two inputs go to the same spot 0954

but what you do not want to happen is to have one input split off into two different places. 0958

This right here, this is okay.0965

Let us see if we can identify the domain of a function just from looking at its equation.0980

Since no domain is specified, we are going to look for the natural domain.0986

That means we will assume that all real numbers can be used unless it makes the function undefined.0990

In these examples, I put one with a fraction because we want be concerned about dividing by 0.0997

I put another one with a square root because we do not want negatives underneath an even root, like the square root.1002

In this first one, the way that it would not work out is if I had a 0 in the bottom.1010

Are there any restricted values for x?1019

Anything can x can not be would that give us a 0. 1022

Well, if I had to solve just the bottom part, I can see that if x was equal to -1 I would definitely have a problem. 1026

I think that is the only issue that we would end up with.1036

Let us just write them out, x can not be -1.1039

Anything else is perfectly acceptable.1047

If I was looking at a number line and trying to shade in all the different things x could or could not be, 1052

the number line would look something like this.1059

It could definitely be any of these negative values over here. 1064

I’m digging a big open hole at -1 since it can not be that and it could be anything greater than that.1069

It can be any value but not just -1.1076

I will represent that using some intervals.1080

It could be anything from negative infinity all the way up to -1, I would not include that.1085

It can be anything from -1 up to infinity, I will use little u’s to connect those.1089

It can be anything just not -1.1096

Let us try another one, f(x) = √x+ + 5.1100

I want to make sure that whatever x + 5 is, we do not want it to be negative.1108

It is okay if it is 0 but definitely we do not want it to be negative. 1115

The only way it would end up being negative is if x was less than -5.1126

Let us say x can not be less than -5.1134

We will represent this on a number line as well so we can end up making an interval for it.1147

Let us see, here I have -5, -4, -3, -2, -1, 0, I have some things that are less than -5.1153

x can be anything, it can even be -5 because you can take the square root of 0, it can be anything greater.1169

The domain of this one, the natural domain would be from -5 up to infinity and it is okay to include that -5.1182

For this one, we want to work on evaluating the function for some given values of x.1195

Be careful on this notation over here, we are not multiplying or simply substituting 3 into our function and seeing what the result is.1201

I'm going to replace everywhere I see an x with this number 3.1215

Once I substituted it in there, we go through and clean it up a bit, I have -12 + 9 or -3.1224

Our input was 3 and our output is -3.1237

Let us do the same thing for f(-5).1243

Same function, we will just put in a different value, -5.1248

We will work to simplify it.1257

-4 × -5, negative × negative will be positive, 4 × 5 is 20, 20 + 9 and we would get 29.1260

In evaluating functions, just simply substitute a value in there for x and end up simplifying it. 1274

That is all I have for now, thank you for watching www.educator.com.1281

Welcome to www.educator.com.0000

In this lesson, we are going to work more with graphing functions now that we know a little bit more about them.0002

What you will see is that when it comes to graphing functions, we use a lot of the tools that we use with graphing just a normal equation.0012

We will see first how you can use a chart to plot a whole bunch of different points and graph out an entire function. 0018

The good news is if you have a linear function, we can often use our tools for lines to shortcut that process. 0027

Once we can look at the graph of the function, we will be able to determine whether it is truly a function using something like the vertical line test.0035

More importantly, once we have the graph of the function we can test out what its domain and range is just from looking at its graph.0044

Let us go ahead and take a look.0052

Graphing a function is the same process as graphing an equation.0058

We want to look at the relationship between the variables.0063

In order to do that, we can simply develop a table of values. 0066

Remember doing this in our graphing of linear equation section, we may pick a lot of different values for x and see what the corresponding y value is. 0071

The only difference that we might make with function is simply use different notation.0081

I would still pick a lot of different values for x but then I would simply see what the corresponding output is for my y values. 0086

Always keep in mind that when we are dealing with our inputs, those are going to be our x values.0096

When we are dealing with our outputs, those are going to be our y values.0102

We will still be able to plot them on a coordinate axis.0107

Our x and y will correspond to the inputs and outputs of that function. 0109

Now if a function represents a line then I have some good news for you.0119

You can use a lot of your techniques, especially about slope intercept form. 0122

Here is a function written in slope intercept form, you can see that it still has the n being slope and it still has b being the y intercept.0127

The only change that I have done here is instead of writing out the y, I'm using my function notation. 0141

This just gives us the name of the function and tells us that our independent variable here is x.0153

Since it is in slope intercept form, I could simply graph something like this by first using the y intercept as a starting point0161

and then using the slope to identify another point on that graph.0167

I have 2 points that I could connect them and then I'm good to go.0171

To determine if the graph is a function or not we can use what is known as the vertical line test.0182

The way the vertical line test works is you imagine a vertical line on the graph.0189

As long as it only crosses the graph in one spot, then you can consider it a function.0195

If the vertical line crosses the graph in only one spot and then we can consider it a function.0201

If it crosses in more than one spot that is where we will get into trouble.0208

Here I have two little diagrams, this one crosses here and here, we would say that this is not a function.0212

This one, it only crosses in one spot and if I was to move that dotted line into a different spot over here, it still crosses only one spot.0226

In that situation, I would say that this is a function.0247

To determine the domain and range of a function when you are looking at its graph,0262

think of tracing back all of the values that we used back to the x values on the x axis and back to the y values on the y axis. 0266

This looks a little difficult to do at first but it is not that bad.0276

I imagine picking out some point out on the graph but if I am looking for the domain, I will trace it back to figure out what value on the x axis it came from.0281

If I pick another point, trace that back where did it come from on the x axis.0293

I do this for many, many different points I’m always tracing it back 0299

What I'm looking to do is trace back essentially every single point on that graph.0303

Now what this would end up doing is I will end up plugging back many different points 0310

and they would end up shading in the domain of all the x values that we have used in the function.0315

This one, if I trace this back trace it back, you can see that it creates that entire line. 0327

In this part of the line out here comes from tracing back values on this side.0332

Even the ones that it can not see it, sure enough they traced all the way back.0338

In a similar fashion, you can figure out the range by taking these points and going to the y axis.0343

It is because the y’s represents the outputs, shading the axis, so you knew what was in your range.0349

That way we get a little bit better sense of how to graph functions and things about them.0365

Let us practice.0371

Let us use the vertical line test to see whether these following relations are functions or not.0373

The way a vertical line test works is we imagine a vertical line or test it to see if it crosses in only one spot.0379

On this first one over here, let us go ahead and put down a vertical line.0387

No matter where we move that vertical line, it looks like it will only end up crossing in one spot.0394

Since it only crosses once, we will say that it is a function.0402

With this one over here, when I put down a vertical line it is easy to see that it crosses in 2 spots.0412

It crosses in two spots, not a function.0422

The vertical line test says it must only cross in one spot at the most.0430

Let us go ahead and see if we can graph one of our linear functions.0440

In this one, it is a special type since it is linear.0443

We want to look at the form to see if that will help us out.0448

This one is written in slope intercept, I know that the y intercept is the 3 and I have a slope of -1/4.0451

Let us start with our first being right at 3 and from that point I will go down 1 to the right, 1, 2, 3, 4.0464

I have a second point, so I will connect the two.0473

There is my line.0483

I can also graph out this line by simply choosing a whole bunch of different values for x and evaluating them one at a time. 0485

I simply use the slope intercept form because it will be a lot quicker.0495

I do want to point out that either way would be fine, just pick out some different things for x, plug them in and see what you will get for y.0500

You use the graph to determine the domain and range of the function.0513

This is unusual in that with last time we actually looked at the equation and tried to pick out what the natural domain was.0518

In here, I just have the graph and I have no idea what is being used in here but I can see the inputs and outputs.0525

Remember that is every single point on this graph here.0531

To first figure out the domain, I will imagine all the points and trace them back to this x axis.0536

What I'm doing is I’m figuring out what points we get shaded in on that x axis when I start tracing them all back.0545

What it looks like it is doing is it is tracing out a lot of different values here.0555

In fact, even my little point way out here we get trace back.0561

I will end up shading quite a bit of the x axis. 0567

One thing to note is nothing is over on the side.0571

The reason why I have nothing over there, is there is no graph to trace back to the x axis.0576

Our domain looks like it would start here at -3 and it will go on and on forever from there.0581

We can use the same process to figure out what the range is.0596

We will simply take all our points now and trace them back to the y axis.0600

We will see what this shades in as we do this, bring that one back and you can see I’m shading a whole bunch of different values here.0605

In some places you might have more than one spot it traces back, but that is okay. 0614

Let us see, keep shading going to the y axis, this guy will go back to -2.0620

Notice how below that I'm not going to shade anything on that part because there is no graph to trace back to the y-axis.0630

What do we have for our range? Well, the lowest value I have here is that -2.0642

We will start there and then it keeps going on from there since the rest of it is shaded.0648

Let us do one more domain and range.0660

Graph the relation and determine if it is a function then state its domain and range.0665

We got a little bit to do with this one, let us first just develop a graph in it.0670

This one is not a linear equation so I do not have too many shortcuts at my disposal.0675

I’m just going to end up creating a table of values to help me out.0680

Let us choose some different values like 4, 5, 8, and 13.0687

These values will make it a little bit easier to evaluate this.0698

If I was to use 4, I would end up with 2 × √4 - 4 or 2 × √0 which is 0.0702

That is one point I know is on my graph, at 4, 0.0721

Let us go ahead and put in our next value and that would be 5.0731

2 × √5-4 that would be 1, √1 = 1 so I have 2 as this value.0736

I will plug in 5 and I have 2 as my output.0748

Looking good, let us try some more.0752

Let us go ahead and put in the 8.0756

8 - 4 would be 4 and the √4 is 2, 2 × 2 =4.0762

This shows that when I put in 8 my output is 4.0773

Let us put in 1, 2, 3, 4, 5, 6, 7, 8, 4, there you go.0792

It looks like 13 is going to be off my-0803

13 - 4 would be 9, 2 × √9, 2 × 3 or 6.0815

That is another point on our graph.0823

Looking at our points, I might as well start putting them together so we can have a nice little curve right here.0827

Onto our first question, I was able to graph the relation, but is it a function or not?0835

Does it pass the vertical line test? 0844

That is our question that we should be asking. 0846

If we imagine a vertical line on here, does it cross once, more than once? What is happening?0849

What was this vertical line? I can see that no matter where I put it, it is only going to cross this graph in one spot.0856

I will say yes it is a function since it passes the vertical line test.0862

Now we have to figure out what is its domain and range.0877

What are all the inputs we could use and what are all the outputs?0880

In making this chart, I can already see some of the inputs that I used that you could potentially use even more than that.0884

If we trace back all these values, it also includes all the numbers between the ones we used.0890

I’m tracing things back to the x axis and it looks like I would shade in all of this.0896

This starts at 4 and continues on from there.0903

The domain would start at 4 and just go on from there, 4 to infinity.0907

If I take all of the same values then I start to trace them back to the y axis it will shade in a lot of other values but it looks like nothing less than 0.0917

We would shade in all that and now we have our range from 0 up to infinity.0933

You can see that graphing functions are the same process as just graphing any type of relation. 0944

Keep track of your inputs and outputs. 0949

If it is a special type of function like a linear function, then use your tools for graphing lines.0953

When it comes to the domain and range, look at your inputs and outputs by tracing all of the values back0959

and then show the intervals of all the numbers that should be included. 0967

Thanks for watching www.educator.com.0970

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at systems of linear equations, pairs of linear equations 0002

and get into their solutions and how we can graph them.0008

More specific things that we want to know is, what exactly is a system of equation and how can we find their solutions?0015

One of the greatest ways that we can do to find their solutions is just looking at their graphs.0023

Some special things that we want to know in terms of their types of solutions is how do we know when there is going to be no solution0030

and how we will know when they will actually be an infinite amount of solutions.0036

Watch out for those two as we get into the nuts and bolts of all this.0041

What is a system of equations?0049

A system is a pair or more of an equation coupled together.0052

Sometimes you will see a { } put onto equations just to show that they are all being combined and coupled together.0059

Here I have a pair, but I could have 3, 4, 5 and many more than just that. 0070

Now if all of the equations that are being coupled together are lines, then we will call these linear equations.0076

You will notice in the examples that I have up here, both of these are lines in standard form.0083

I have a system of linear equations.0089

In order to be a solution of a system, it must satisfy all of the equations in that system all simultaneously, all at once.0094

If it only satisfies the first equation then that is not good enough to be a solution of the system, it must satisfy all of them.0101

Let us quickly look at how we can determine if a point is a part of the solution or not.0116

We are going to do this by substituting it into an actual system.0123

My system over here is 2x + 3y = 17 and x + 4y = 21.0126

Is the point 7, 1 a solution or not?0133

Let us go ahead and write down our system.0138

Let us first put it into equation 1, we have (2 × 7) + (3 × 1) is it equal to 17, I do not know.0144

Let us find out.0157

2 × 7 would be 14 + 3 and sure enough I get 17.0158

I know it satisfies the first equation just fine.0168

Let us take it and substitute it into this second equation.0174

I have 7x + (4 × y value × 1).0182

4 × 1 = 4, 7 + 4, that one does not work out because 11 is not equal to 21.0193

Even though it only satisfies one of the equations, I can say that it is not a solution.0206

It only satisfies one equation.0223

Let us try this with the point 1, 5 and let us see if that is a solution of the system.0234

We will put in 2 or 3 if we want to know if this is equal to 17.0241

I will put in 1 for x and we will put in 5 for y and let us simplify it.0248

2 × 1 is 2 ,3 × 5 is 15 and 2 + 15 =17 and this one checks out.0258

Now let us substitute it into the second equation.0273

I will take 1 and put in for x, we will take 5 and put it in for y.0278

I have 1 + 20 this that equal 21? It does 21 equals 21.0287

I can say it is a solution and it satisfies both of the equations.0296

There are few ways that you can go about solving a system of equations and we will see them in later lessons0316

or the first ways that you can sense what the solution should be is to the graph the equations that are present.0321

If these are linear equations that you are dealing with then you can use the techniques of graphing your lines to make this a much easier process.0328

Let us work on finding our solutions graphically.0335

To do this, all you have to do is graph the equations and then find out where those equations cross, when they do that is going to be our solution.0339

There is one downside to this so be very careful.0349

In order for this process to work out, you must make accurate graphs. 0353

If your lines are a little wavy or you do not make them very accurate then you might think you know where it cross, but actually identify a different point.0358

Be careful and make accurate graphs. 0369

Since finding solutions using a graph is as nice and visual, we definitely will start with it0381

but we will use some of those later techniques for solving because the accuracy does tend to be an issue with the graphs.0388

Let us get into the graphing and finding a solution that way, I will try.0400

Here I have a system of linear equations and it looks like both of them are almost written in standard form. 0405

The second one is not quite as in that form because it has a negative sign out front.0412

I think we will be able to graph it just fine.0417

I'm going to graph it by using its intercepts.0419

We will make a little chart for our first equation and I'm going to identify where it crosses the y axis and the x axis by plugging in an appropriate value of 0.0424

If I use 0 in my equation for the first one, I just need to solve and figure out what y is.0436

Putting in a 0, we will limit that term and solving the rest out, divide it by 3 and I will get that y =2.0445

I know that is going to be one point on my graph, 0, 2.0457

We will go ahead and plug in 0 for y and we can see that it will eliminate our y terms and now we just have to solve 2x = 6.0467

If we divide both sides of that by 2, we will get x =3.0484

We have a second point that we can go ahead and put on our graph, 3, 0.0492

Now that we have two points, let us be accurate on graphing this out.0501

If you want to make it even more accurate, one thing you can do is actually use more points to help you graph this out.0517

We have one line on here, it looks pretty good.0526

Let use another chart here and see if we can graph out the second one. 0528

Once we have both lines on here, we will go ahead and take a look at where they cross.0540

This will be for line number two, we will get x and y. What happens when we put in some 0?0545

-0 + y = 7 that is a nice one to solve.0554

-0 is the same as 0, the only thing left is y = 7.0560

Let us put on the point 0, 7 right up there.0567

If we put in 0 for y, -x + 0 = 7, I will get that –x =7 or x = -7.0578

There is another point I can put on there.0591

Now that I have two points here, let us go ahead and connect the dots.0599

We can see where our two lines cross.0614

This is from our first line and the blue line is from our second equation.0618

Right here, it looks like they definitely cross.0626

That is the point -3, 1, 2, 3, 4. 0630

The reason why we went over a little bit of work on testing solutions is, if our graph was inaccurate and this was incorrect, 0638

we can take it and put it into our system just to double check that it actually works out. 0645

If you want you can take -3 and 4, plug it back in and let us see what happens. 0650

Starting with the first equation 2, 3y = 6 and let us plug in our x and y.0664

Checking to see if this is the solution, to see if it satisfies our first equation.0678

-6 + 12 does that equal 6? -6 + 12 does equal 6.0683

It checks out for our first equation.0691

Let us put it into the second one, our x value is -3 our y value is 4.0696

Negative times negative would be 3.0710

I’m sure enough 3 + 4 does equal 7 so it satisfies the second equation as well.0716

I know that -3, 4 is definitely my solution.0721

Let us try and find a solution to another system. 0730

Here I have -2x + y = -8 and y = -3x + 2.0735

The first one I can go and ahead and make a chart for and maybe figure out to where its intercepts are.0744

That seems like a good way to find out that one.0752

What happens when x = 2, what value do we get for y?0755

This term would drop away because of the 0 and I have only be left with y = -8.0765

I know that is one point I will need.0772

Let us go ahead and put a dot on there.0775

x =0, y = -1, 2, 3, 4, 5, 6, 7, 8 way down here.0777

Now we put in a 0 for y, -2x + 0 = -8, let us see how this one turns out.0786

My 0 term will drop away and we continue solving for x by dividing both sides by -2, -8 ÷ -2 is 4.0799

There is another point, 4, 0, 1, 2, 3, 4, 0.0818

Now that we have two points, let us go ahead and connect them together and we will see our entire line.0824

We got one of them down and now let us graph our second one.0843

Notice how the second equation in our system is written in slope intercept form0847

which means we will take a bit of a short cut by using just the y intercept and also using its slope.0852

I will make things much easier. 0862

Our y intercept is 2, I know it goes through this point right here.0864

Our slope is -3 which I can view as -3/1.0868

Starting at our y intercept we will go down 3 and to the right 1.0873

1, 2, 3 into the right 1, there is a point.0878

I’m bringing out my ruler and we will connect these.0885

Now we can check if each of them cross.0897

Let us highlight which equations go with which line, there we go. 0902

Also, it looks like they do cross and I would say it crossed down here at 2, -4. 0906

Even though I made some pretty accurate graphs, it does look like it is a hair larger than 2, and sometimes this happens.0916

Even when we do make some pretty accurate graphs, sometimes where they cross it looks like it is just a hair off.0923

The frustrating part is the actual solution may be different from 2, -4.0932

About the only way I'm sure when using this graphing method is to check it by putting it in to the system.0937

Later on we will look at more accurate systems where we would not have to check quite as much.0944

Let us go ahead and put in our values for x and y to see if it satisfies both equations.0950

x is 2 and y is -4, -2 × 2 = -4 + -4 is a -8.0958

It looks like the first one checks out, everything is nice and balanced so it satisfies the first equation.0978

Let us go ahead and put it into the second one and see if that one also works.0986

X is 2 and y is -4, let us see if this balances out.0994

-4 is not equal to -6 + 2 I think it is, because when you add -6 and 2 you will get -4.1003

It satisfies both equations.1014

2, -4 is a solution. 1017

When you are going through finding different solutions for a system, several different things could happen. 1023

The first situation that could happen is a lot like the examples we just covered.1030

You are going to go through the graphing process and you are going to find one spot where the two actually cross.1035

You will notice this kind of case that you like to be in. 1040

However, when graphing these lines, you might also find that the lines are completely parallel.1043

It does pose a little bit of a problem because it means that the lines would not cross whatsoever.1049

With parallel lines, we will not have a solution to our system. 1065

Another thing that could possibly happen is you go to graph the lines and they turn out to be exactly the same line.1071

In that situation, we will not only get solutions but we will get lots of solutions because if they are the same line, they will cross an infinite amount of spots.1078

Basically everywhere on a line will be a solution.1087

Watch for these to show up when you are making these graphs, you will either find one spot where they cross then you have a solution. 1090

You will see that the lines are parallel and they do not cross so you have no solution.1098

You will find that they are exactly the same line and then you will have an infinite number of points as your solution.1103

In this next example, I want to highlight that the two special cases that could happen. 1114

One thing I want to point out is that when you are looking at the system, 1119

it is sometimes not obvious whether it has no solution or an infinite amount of solutions. 1122

You got to get down into the graphing process or the solving process before you realize that.1127

My first equation is x +2, y= 4 and the other one is 3x + 6, y =18.1133

I’m going to graph these out using my x and y intercepts.1140

What happens when x =0, what happens when y= 0?1146

For the first one, I put in 0 for x and we will go ahead and we start solving for y.1149

You can see my 0 term is going to drop away, which is good.1157

I will divide both sides by 2 and I will get that y = 2.1161

There is one point that I will mark on my graph 0, 2.1169

I will put in 0 for y and let us see what happens with that one. 1179

The 0 term is going to drop out and the only thing I'm left with is x = 4.1189

I will put that point in there, 4, 1, 2, 3, 4, 0.1199

I will go ahead and graph it out.1204

Now that we have one line, let us go ahead and graph the other.1219

We will get its intercepts by putting in these 0 for x and 0 for y.1228

(3 × 0) + 6 time to solve for y.1243

Our 0 term will go away and I will just be left with 6y =18.1251

6 goes into 18 three times so I have 0, 3 as one of my end points.1259

Putting in 0 for y, let us see what happens there.1273

3x + 6 × 0 = 18.1275

It looks like that term will drop away and I will just have 3x =18.1282

3 goes into 18 six times, x = 6.1290

1, 2, 3, 4, 5, 6 I have that one.1300

Let us go ahead and graph it out now.1306

What we can see from this one is that it looks like the two lines are parallel and they are not crossing whatsoever.1320

If it is a little off and it looks like one of your lines could potentially cross but maybe off the graph, 1330

one thing you can do is you can rewrite the system into slope intercept form so you can better check the slopes.1335

I'm pretty sure that these are parallel, I’m going to say that there is no solution to the system. 1344

They do not cross whatsoever.1353

Let us try one more and see if this one has a solution. 1359

This is y = 4x - 4 and 8x -2y =8.1362

The first one is written in slope intercept form so I will graph it by identifying the y intercept and its slope.1370

This starts at -4 and starting at the - 4 I will go up 4/1and just like that we can go ahead and graph it out.1382

There is our first line.1406

The second one is written more in standard form so I will go ahead and use its intercepts to track that one down.1408

What happens when x =0 and what happens when y = 0.1415

When x =0, that will drop away that term right there and I'm looking at -2y = 8.1422

We will be dividing both sides by -2 and I will end up with -4.1434

There is one point I can put on there, notice it is on the same spot.1443

Putting in 0 for y, I can see the term that will go away.1451

I am left with 8x = 8.1468

Dividing both sides by 8, I am simply be left with x = 1.1471

We will put that point on there and both points ended up on the other line.1482

When I go to draw this out, you will see that one line actually ends up right on top of the other one. 1487

They are essentially the same line. 1492

In this case, we end up with an infinite amount of solutions because they still cross, but they cross in lots of different spots. 1501

I could say they cross everywhere, there are infinite number of solutions.1508

One of the best ways that you can figure out a solution to a system of equations 1530

is simply graph both of the equations that are present in the system and see where they cross.1533

We will look at some more accurate methods in the next lesson. 1539

Thank you for watching www.educator.com.1542

Welcome to www.educator.com.0000

In this lesson, we are going to work on solving a system of equations using the substitution method.0002

What we want to focus on is how we use the substitution method in order to find solutions and I will walk into that process.0013

We will also keep in mind that since sometimes there is no solution or there is an infinite amount of solutions, 0021

how we can recognize those cases when using the substitution method.0026

We are called the reason why we need more methods for solving a system is that when we use the graphing method, 0035

you have to be very accurate in order for that method to work and if your lines are a little bit wavy 0042

or you do not make them just right then you may not find a solution of that system.0048

Since, accuracy is such a problem that is why we are going to focus more on those algebraic methods.0054

How can we manipulate the system in such a way so that it gives us a solution.0060

Two of these methods of our algebraic methods are substitution and elimination.0067

In this lesson, we focus on substitution and then look at elimination in the next one. 0073

Let us see how the substitution method works.0080

In the substitution method, what we first end up doing is taking one of our equations and solving it for one of the variables.0086

Once we have taken one equation and we have solved it for variable, we will take that and substitute it into the other equation.0095

What this will do is it will give us an entirely new equation with only one type of variable in it 0105

and will end up solving for that one remaining variable that is in there.0110

We will have half of our solution but since we are looking for a system and we need an x and y coordinate, 0117

then we will end up using back substitution to find the rest of our solution.0123

We will take what we have found and end up substituting back into one of the original equations. 0129

This will help us find the other variable. 0135

As long as everything works out good, that it should be our solution, but it is not a bad idea to just take that and check it in the original system. 0138

If you do make a mistake, sometimes you would not catch it, so playing it back into the original system is always a good idea.0147

Let us walk through this in substitution method using the following example.0157

I have - x + 3y = 10 and 2x + 8y = -6.0162

I want to start solving for at least one of the variables in one of the equations.0170

I have lots of different choices that I can do. 0175

I can solve for x over here. 0178

I can solve for y in the first one or you can even pick on the second equation a little bit and solve for its x or y. 0180

It does not matter which one you solve for, just pick a variable and go ahead and solve for it.0187

I only choose x in the first equation, it looks like it will be one of the easier ones to get all by itself. 0194

What I will do is solve for x and I'm moving everything to the other side.0200

I'm subtracting 3y on both sides.0209

It is almost all by itself but it looks like I have to multiply it by -1 to continue on from there. 0213

I’m multiplying through by -1 and I will end up with x = 3 and y -10.0220

This entire thing right here is equal to x.0228

For the substitution method happens is I will take in the x in the other equation and replace it with what it is equal to, 3y – 10.0235

It will leave lots of room for this so you can see what it is going to turn that second equation into.0250

Instead of writing x, I’m going to write 3y -10.0258

I have taken that second equation, substituted in those values and notice how this new equation we created it only has y.0269

Since this new equation only has y’s in it, we will be able to solve for y and be able to figure out what that is.0279

We still have a little bit of work to do, of course we will have to simplify and combine terms but we will definitely be able to solve for that y.0287

Let us give it a shot.0293

I’m going to distribute through by my 2 and put 6y – 20 + 8y = -6.0295

Let us go ahead and combine our y terms, so 6 + 8 is a 14y and then let us go ahead and add 20 to both sides.0307

14y – 6 + 20 is 14 divide both sides by 14 and get y = 1.0319

It looks like I have a solution but remember that this is only half of our solution.0330

We still need to figure out what x is.0333

What this point, now that we know what y is I can substitute it back into one of my equations and figure that out. 0336

Let us do that, -x + 3 and I will put in what y is equal to 1.0346

I will end up solving this equation for x, -x + 3 = 10, so I'm going to subtract 3 from both sides.0355

I will go ahead and multiply both sides by -1 and get x = -7.0372

I have both halves of my solution.0383

The x is written first, so -7 and y is written second.0387

My solution is -7, 1.0393

If I'm looking at that and I'm curious if it is the solution or not, it is not a bad idea to just take it and plug it back into the original.0396

Let us see how that works out.0404

My value for x is -7 + 3 and my value for y is 1.0408

Does that really equal 10? Let us find out.0413

Also a- -7 would be 7 + 3 = 10, sure enough, 7 + 3 is 10.0416

That works out in the first equation.0427

Right now, it is still the same way with the second one.0430

(2 × -7) + (8 × 1) is that =equal to -6, let us find out.0432

-14 + 8 sure enough -6 is equal to -6, it checks out in both the equations. 0440

I know that -7, 1 is my solution.0449

There is a lot of substitution, of course, in the substitution method, but is not as bad as you think.0454

You do have some choices on what you will solve for and where it goes.0460

One thing that we saw when we are finding solutions graphically is that several things could happen.0472

We might have one solution, we might have no solution or we actually might have an infinite amount of solutions. 0477

Now when we are looking at all of these graphically, it was pretty simple on figuring out which case we were at. 0483

We either looked at both our lines, actually cross and then we said they have a solution. 0489

We could see if there was no solution if the lines were parallel and we can see that it was an infinite amount solution if they were on the same line.0495

Now that we are doing things more algebraically, we would not be able to visually see what case we are in.0503

How is it we figure out if it has no solution or an infinite amount of solutions?0509

It all depends on what values you will get while going through that solving process.0514

If you go through the substitution method and you can find an x and y that work, then you will know that it has a solution.0520

It has one solution, nothing strange happens.0526

If however you go through that substitution method and end up creating a false statement, it means that there is actually no solution to the system.0530

Now if you do come across a false statement, what I always tell everyone is check your work first to make sure that you have made no mistakes.0539

If all your work looks great and you still create a false statement then you can be sure that it has no solution.0548

If you go through and you create a true statement then you can not see what your solution is.0555

That is a good indication that the two lines are exactly the same and you have an infinite amount of solutions.0560

Again, check your work with this one, but if all your work looks good, then you know that you have an infinite amount. 0566

Let us look at some examples like these ones and see how we can recognize what I mean by a false statement and what I mean by a true statement. 0576

We want to solve the following system 5x - 4y = 9 and 3 - 2y = -x.0589

The first thing I need to do is solve for either x or y.0597

It does not matter which one we choose, it looks like it might be a good idea to solve for x down here since it is almost all by itself anyway.0601

We will just go ahead and multiply everything through by -1, -3 + 2y = x.0609

Now that we have what x is equal to, we will substitute it into the first equation for that x.0619

It will leave lots of space and we will just go ahead and put it in there, -3 + 2y.0627

We can see in this new equation it only has y, so I need to work on getting them together and isolating them.0636

-15 it will be from distributing through +10y -4y = 9.0645

It have some like terms I can put together 10y and -4 that will be 6y =9 and adding 15 to both sides, + 15 = 24.0656

Now dividing both sides by 6, I will get that y = 4.0680

We have half of our solution and we always figure out what the other half is by putting it back in one of the original equations. 0685

I often have students ask me should you put it back in the original, after all could not I just stick that y back and up here.0692

What I tell them is yes you could put it back into this equation for y, but be very careful because that equation, we already have manipulated in some sort of way.0700

If you want to be sure that you are not making mistakes on top of mistakes, take that and plug it into of the original ones, 0711

something that you have not mess around with too much.0718

Let us see 3 – 2 and I will put the 4 in there.0721

We take this and solve for x, 3 -8 = -x.0729

3 - 8 would be -5 and dividing both sides by negative, I have that x = 5.0737

I have both halves of my solution x = 5, y = 4.0747

My solution is 5, 4.0755

Again, if you want to make sure that is the solution, feel free to plug it back into both equations and check it to make sure it works out.0759

In this next example, we will see if we can solve this one using the substitution method.0769

It says 4x - 5y = -11 and x + 2y = 7.0773

Let us see what variable shall we solve for.0780

This x looks like it would be nice and easy to get all alone, let us do that one.0784

We will go ahead and say x = -2y + 7.0790

I took the 2y and subtracted it from both sides.0797

I will take all of this and substitute it into the first one.0801

It will give me 4 - 5y = -11.0805

I left a big open spot so I can drop what x is equal to in there, 2y + 7, it looks good.0812

Now I’m going to distribute through with my 4, -8y + 28.0820

And I'm going to continue trying to solve for y.0832

I will also see what I got here, -8 - 5 putting those together I will end up with -13, subtracting 28 from both sides, I will end up with -39.0835

It looks like I can finally do dividing both by it sides by -13, in doing that y = 3.0853

Let us go ahead and take that and substitute it back into one of our original equations.0863

Let us go ahead and put in here, x + (2 × y).0867

Multiplying together the 2 and the 3 would be 6 and I'm left with x + 6 = 7.0878

Subtracting 6 from both sides x = 1.0886

My solution for this one is x = 1 and y = 3.0892

In some systems, remember we could have that situation of having no solution or an infinite amount of solutions.0903

Sometimes when you are looking at an equation, it is tough to tell whether you have that situation or not.0909

Watch very carefully what happens with this one.0914

I’m going to start off like I what did with the other examples.0917

We just want to take these and solve it for one of our variables.0920

I’m going to solve the first one for x since that one looks like it can be nice and easy to get isolated.0924

That would give me x - 2y from both sides, + 4.0932

Now that I have x isolated, let us substitute it into the second equation.0940

3 + 6y = 13 and we will drop in -2y + 4.0947

Now I only have y in this new equation so we will continue solving it by getting y all by itself. 0959

Let us go ahead and distribute -6y + 12, + 6y = 13.0965

There is something interesting happening here.0975

Over here I have -6y and over here I have 6y.0977

Since they are like terms, I can put them together by -6 and 6 would give me 0y, it cancel each other out.0982

That means the only thing left is 12 on the left and 13 on the right but that does not make any sense.0993

12 is not equal to 13 so I’m going to write not true.0999

If you check through back all of my steps, I did not make any errors.1008

I moved the -2y to the other side that turnout okay and I did my distribution fine.1012

I know my steps worked out perfectly good.1019

What this statement is telling me at the very end, since it is not true is that these two lines are actually parallel. 1023

There is no solution and they do not cross.1030

Watch out for these false statements that might come along the way and they will help you recognize when a system does not have a solution.1039

One more system, this one is 2y = 4x and the other equation is 4x - 2y = 0.1051

Let us take the first one and solve it for y.1061

We can easily do it just by dividing both sides by 2, y = 2x1065

Now that it is solved for y, let us go ahead and substitute that into the second equation.1075

You will see that this is turning out a lot like the previous example.1091

If I combine my x’s, the other since they are like terms, they will cancel each other out.1097

This is slightly different though because after they do cancel out, all that I get is 0 = 0, which happens to be a true statement 0 does equals 0.1103

Let us check our work and make sure you did everything okay.1118

I divide both sides by 2 and that worked out good and I combined my 2’s together and cancelled.1122

All my work looks great.1127

The fact that I have a true statement and all my steps were good means that what is going on here is that the lines themselves are exactly the same line.1129

We might even see that at the very beginning if we end up rewriting them.1139

When I moved the 2y to the other side on the second equation and then end up rearranging them, they are exactly the same equation.1146

Visually these would be right on top of one another.1160

This tells us we have an infinite number of solutions and everywhere on that line is a solution.1164

These more algebraic methods are definitely more accurate when it comes to finding solutions.1179

You want to be very familiar and being able to go through them.1184

Be on the watch out for these two special cases, if you get a false statement then you know you have no solution.1188

If you get a true statement then know you have an infinite amount of solutions.1195

Thank you for watching www.educator.com.1199

Welcome back to www.educator.com.0000

In this lesson we are going to work on solving a system of linear equations using the method of elimination.0003

We focus on how to use this elimination method or how it differs from substitution 0011

and how we can recognize the many different types of solutions that could happen.0017

Do we get one solution, and infinite amount of solutions, or possibly no solution or whatsoever. 0021

The goal with the elimination method is to combine the equations in such a way that we eliminate one of our variables.0030

In that way, we will only have one that we will need to solve for.0037

In order to do this, sometimes we will have to multiply one of the equations or even both of the equations by some sort of a constant. 0042

Now what we are looking to do when we multiply is we want a pair of coefficients that will end up canceling each other out.0050

Do I have - 3x and 3x?0059

When we combine those they would end up canceling out and that is what I want.0062

Once I get a pair of coefficients that I know will cancel, we will go ahead and actually add the 2 equations together.0067

In fact, some people call the elimination method the addition method because we end up adding the equations.0073

We will get a new equation after doing that and we will only have one type of variable in it.0081

We will end up solving that new equation.0086

We will have half of our solution at this point so we will be able to take that and substitute it back into one of the originals.0090

In fact, if you look at all the rest of stuffs from here on out, these are the same as the substitution method.0096

We simply have half of our solution and we are back substituting so we can find the other value. 0104

Once we are all the way done through this entire process, it is not a bad idea to check the solution to make sure that does satisfy both of the equations.0110

Let us walk through this method, so you get a better sense of how it works.0123

Here I have 2x – 7y = 2 and 3x + y = -20.0129

What my goal is to either eliminate these x values here or I need to eliminate my y values. 0136

If I was just going through and I decide that I was going to add them together as they are, 0145

you will see that this would not be a good idea because nothing happens.0150

2x + 3x would give me 5x and – 7y + y would give me -6y.0154

2 and -20 is -18, so I definitely created a new equation, but it still has x’s and it still has y’s.0163

There is not much that I can do with it.0171

What I want to happen is for some things to cancel out, so I only have one type of variable to solve for. 0174

Let us go ahead and a do a little work on this so that we can get something to cancel out.0181

The first thing is we get to choose what we want to cancel out.0188

Is it going to be the x's or y’s?0191

I want to choose my y since one of them is already negative and the other one is positive.0194

In order to make these guys came flat, let us take everything in the second equation and multiply it by 7.0200

The first equation exactly the way it is, no changes, and everything in the second equation will get multiplied by 7.0210

3 × 7 = 21x, I have 7y, -20 × 7 would be -140.0218

I still have a system and I still have a pair but now I noticed what is going on here with the y.0231

Since 1 is -7 and one is 7, when I add those together, we will end up canceling each other out.0236

Let us do that, let us go ahead and get to the addition part of the elimination method.0244

Adding up our x's, we will get 23x.0248

When we add up the y, we will get 0y so that term is gone.0254

Over on the other side - 138 and in this new equation the only thing I have in here is simply an x.0259

I can solve this new equation for x and figure out what it needs to be which I can do by dividing both sides by 23.0270

How many times 23 is going to 138?0292

6 × 3 = 18, looks like 6 times so x = -6.0303

That is half of our solution and now we need to work on back substitution.0313

We will take this value here for our x and plug it back into one of the original equations.0318

It does not matter which one you plug this into, just as long as you put it into one of them it should turn out okay.0323

I end up solving this one for y, 2 × -6 = -12 and we are adding 12 to both sides would give me 14 and y divided by -7, I have -2.0335

I have both halves of the solution.0360

I have x = -6 and that y = -2.0362

In the elimination method, we are working to eliminate one of our variables.0370

You might be curious what would have happened if we would have chosen to get rid of those x's instead.0377

We could have done it, but you would have to multiply your equations by something different.0383

I’m not going all the way through this, but just to show you how you could have started.0390

For example, if you multiply the first one by 3, it would have given you 6x - 21y = 6.0394

If you multiply the second equation by -2, - 6x - 2y = 40.0405

You would see that using those multiplications, the x's would have cancel out when you add them.0414

In the elimination method, eliminate one of your variables.0421

Onto the next important thing with the elimination method.0428

We have many different things that could happen. 0432

We could have one solution, no solution or an infinite amount of solutions.0433

Visually, we can see that the lines either cross, do not cross, or perhaps they are the same line.0439

The way you recognize this when using the elimination method is that you might go through that method and everything will work out just fine.0457

You actually be able to find your solution.0464

That is when you know it has one solution, everything is good.0466

If you go through the system and all your work looks good, but you create a false statement 0471

then you will know that there is actually no solution to the system.0476

In fact, they are 2 parallel lines and they never cross. 0479

I think I mentioned this earlier, but go ahead and check your work if it looks like you are getting a false statement0483

just to make sure that the false statement is created from them being parallel and not from you making a mistake.0488

If you go through and you create a true statement, then this is an indication they have an infinite amount of solution.0496

Again, check your work, but make sure that you know whether it has an infinite amount or not.0503

These are exactly the same criteria that using the substitution method, so they are nice and easy to keep track of.0512

False statement, no solution, true statement, infinite amount of solutions and everything works out normally, then you just have one solution.0518

Let us get into some examples and see this in action.0526

In this first example, we will look at solving 2x - 5y = 11 and 3x + y = 8.0534

Think about our goal with the elimination method.0543

We want to get rid of these x's or we want to get rid of the y.0546

You get to choose which one you want rid of just a matter how you will end up manipulating these to ensure that they do cancel each other out.0550

I think the better ones to go after are probably these y, since one is already positive and one is already negative.0561

The way we are going to do this is when I take the second equation and we are going to multiply everything there by 5.0570

Let us see the result this will have.0579

We have not touched the first equation and everything in the second equation by 5 would be 15x + 5y = 40.0582

I can see that when we add together our y values, they will cancel each other out.0597

2x + 15x that is 17x, -5y + 5y they would cancel each other and get 0y.0605

11 + 40 =51, so in this new equation I can see that we only have x to worry about and we can continue solving for x.0617

We will do that by dividing both sides by 17, x =3.0625

Now that I have half of my solution and I know what x is, let us take it and end up substituting it back into one of our original equations.0636

(2 × 3) - 5y = 11.0647

Our goal here is to get that y all by itself.0656

We will multiply the 2 and 3 together and get 6, then we will subtract 6 from both sides, - 5y is equal to 5.0660

Now, dividing both sides by -5, we will have y = 8, -1.0680

We have both halves of our solutions and I can say that our solution is 3, -1.0693

Let us try another one.0703

For this next one we will try solving 3x + 3y = 0 and 4x + 2y = 3.0709

In this one, it looks like all of my terms on x and y are all positive.0716

It is not clear which one will be the easier one to get rid off, we just have to pick one and go with it.0721

Let us go ahead and give these x's a try over here and see if we can eliminate them.0726

Some things that you can do to help you eliminate some of these variables, is first just try and get them to be the exact same number.0732

What some of my students realizes is that if you just take this coefficient and multiply it up here, 0740

then take the other coefficient and multiply it into your second equation.0746

That is usually enough to make them exactly the same .0751

I’m going to do that here. 0753

I’m going to take the entire second equation and we will multiply it by 3.0755

We will take everything in the first equation and we will multiply that by 4.0762

Let us see, 4 × 3 = 12x + 12y, 0 × 4 = 0, so that will be my new first equation.0768

Everything in the second one by 3, 12x + 6y = 9 will be the second one.0783

We are off to a good start and I can see at least that the x's are now exactly the same.0792

We want them to cancel out, let us go ahead and take one of our equations and multiply it entirely through by a negative number.0798

Let us do it to the second one, everything through in the second one by -1.0806

It will give us -12x - 6y = -9, the first equation is exactly the same so we would not touch that one.0810

By multiplying it just that way, we are doing a little bit of prep work, now I can be assured that my x's will definitely cancel out.0823

Let us add these 2 equations together and end up solving for y.0831

12x - 12y that will give us 0x, that is gone.0836

12y - 6y will be 6y and 0 + -9 = -9.0841

y = -9/6 which we might as well just reduce that call it to say – 3/2 and there is half of our solution.0852

Let us go ahead and take this, plug it back into the original and see if we can find our other half.0866

I’m just going to choose this one right here, 4x + 2 =- 3/2 equals 3.0873

In this equation, the only thing I need to solve for is that x, let us go ahead and do that.0886

2 × -3/2 = - 3, adding 3 to both sides here I will be left with 4x = 6 and dividing both sides by 4, I will have that x = 3/2.0895

My final solution here, x = 3/2 and y = -3/2.0917

As long as you do proper prep work, you should be able to get your final solution just fine.0926

Remember that it has 2 parts, the x value and y value.0931

Let us see if we can solve this one using elimination.0939

I have my x's that are both positive and my y’s are both negative, but they are almost the same.0947

Let us start off by multiplying the second one by 5 and see if it gets a little bit closer to canceling.0955

5x - 5y, 5 × 12 = 60 we multiply the entire second equation by 5.0965

It still looks like nothing will cancel out, so I will also multiply the second equation by -1.0977

-5x + 5y = - 60 looking pretty good here.0983

I will add these together and I can eliminate my x's but you will notice that it looks like the y will also eliminate.0998

If the x's and y’s are gone, the only thing on the left side is just 0.1007

What is on the other side? 3 - 60 would be a -57 and that is a bit of a problem because the 0 does not equal -57.1013

Let us say that this is a false statement.1026

This is one of these situations and it is a good idea to go back through your work, check it and make sure it is all correct.1031

All of our steps do look good here, we multiplied it by 5, we multiplied by - 1.1038

We do not actually have a whole lot of spots where we could make mistakes.1042

What this false statement is telling us here is that there is no solution to the system.1046

These 2 lines are actually parallel and they are not crossing whatsoever. 1057

Watch out for those special cases.1061

Let us try one last example, x – 4y =2 and 4x – 16y = 8.1068

I want to try and maybe eliminate, let us do the x's.1077

We will do this by multiplying the first equation by -4.1081

-4x, -4 × -4 =16y equals -8.1094

In the second equation I do not need to manipulate that one, I will leave it exactly the same.1105

This is a lot like our previous example.1113

The x's are going to cancel out when they are added together, but then again, so are the y values. 1117

Both of them are going to cancel out.1122

I have 0 on one side of my equal sign.1124

This one actually is not quite so bad because if I look at my -8 and 8, they will cancel each other out as well.1128

I will get 0 = 0, which happens to be a true statement. 1137

Now, one statement is true but what is your x and y?1141

In indicates that the lines are exactly the same and there are on top of one another.1145

We have an infinite number of solutions.1151

Using one of these algebraic methods is a great way to figure out what the solution is.1164

Keep in the back of your mind what the graphs of these look like1169

so we can interpret what it means to have one solution, no solution, or an infinite amount of solutions.1172

Thank you for watching www.educator.com.1178

Welcome back to www.educator.com.0000

In this lesson we are going to look at applications of systems of equations. 0003

Think about word problems involving the systems of equations that we saw earlier. 0007

Our main goal will be looking at these word problems and figuring out how we can interpret them and build the system from there.0015

Here is a part of this that we will be actually looking at our solutions and be able to figure out how it fits in the context of that word problem.0022

A good way to think of how to start picking these apart is to take every little bit of information in there.0033

Say 2 different situations and create an equation for each of those situations.0039

You will notice some of these problems look similar to ones we did in the earlier applications of linear equations.0045

In that lesson we tried to write everything in terms of one variable. 0051

Now that we know more about systems, we can use 2 variables to help us out and create an equation for each of the situations present.0055

It will actually make things a little bit neater.0062

Once we have built our system of equations, we are getting down to being able to solve that system.0067

We will use our techniques for solving a system such as elimination method or the substitution method.0072

We can also use tables like we did earlier to help you organize this information 0079

but I will mainly focus on just the methods of substitution and elimination to tackle the systems of equations.0083

One very important part to remember is that all of these are going to be word problems.0094

Even though we get a solution like x = 3 and y = 5, we have to take those and interpret them and say what is x and y.0099

Always interpret these in the context of the problem. 0108

One thing that will make this very easy and not so bad, is to go ahead and write down what your variables are at the very beginning.0112

In that way, once you are all done with the problem you can reference that and say okay my x represents this and my y represents this.0122

That way you can easily say here is how it fits in the context of the problem.0130

Let us jump into our examples and see how we can actually start creating a system of equations.0140

In this first one, we have a restaurant that needs to have 2 seat tables and 3 seat tables.0146

According to local fire code, it looks like the restaurants maximum occupancy is 46 costumers.0151

If the owners have hardly enough servers to handle 17 tables of customers, how many of each kind of table should they purchase?0158

Notice that we are under a few different constraints.0165

One has to do with simply the number of people that you can fit in this restaurant.0168

We want to make sure that we have no more than 46 customers.0173

Another restriction that we are under is the number of tables that our servers can handle and provide good service and we can only do 17 tables.0177

From the owner's perspective, we do not want to go less than these numbers. 0187

After all, if we have less than 46 customers, we are not making as much money.0190

If you have less than 17 tables then we have servers that are not doing a lot of work.0195

We are going to try and aim to get these exact when we set up our equations.0200

Let us hunt down some unknowns and first write those down.0205

I'm going to say let x that will be the number of 2 seat tables.0210

Let y be the number of 3 seat tables.0228

We can use these unknowns to start connecting our information gathered.0243

In the first situation, we are going to focus on the number of people that we want to fit in this restaurant.0248

We are looking to put 46 people in this restaurant and it will come from seating them either at these 2 seat table or at the 3 seat table.0254

There will be 2 people for every 2 seat table so we can represent that by saying 2 × X.0264

There will be 3 people at every 3 seat table or 3y.0272

That expression right there just represents a number of people and we want it equal to 46 customers.0277

That is dealing with our people, maybe I will even label that people.0285

The other restriction is what we can do with our servers in handling all of these tables.0291

We want the total number of tables and that is our 2 seat tables and our 3 seat tables to equals 17.0298

You can say we have encapsulated all the information to one equation or the other.0311

We have actually set up the system and it is just a matter of going through and solving it.0316

You could use the elimination method or you could use the substitution method, both of them should get you to the same answer.0321

I’m going to go through the elimination method.0327

I will do this by taking the second equation here and we will multiply that one by -2.0331

Let us see what the result will be.0340

The first equation will remain unchanged and everything in the second one will be multiplied by -2.0341

We have done that, we can go ahead and combine both of those equations together.0365

You will notice that the x’s are canceling out since one is 2 and one is -2, that is exactly what we want with the elimination method.0371

3y + - 2y will give us 1y, and now we have 46 - 834 and that will give us 12.0379

One of the great thing is about writing down our variables at the very beginning is not only do I know that y = 12, 0397

but I can interpret this as the number of 3 seat tables since I have it right here that y is the number of 3 seat tables.0402

We are going to continue on, and I will figure out what x is so we can figure out the number for both of the tables.0410

It is not so bad, just always remember that the total number of tables must be equal to 17.0415

I will borrow one of our original equations down here and just substitute in the 12 for y.0421

To solve that one we will subtract 12 from both sides and get that x = 5.0433

Now we can fully answer this problem.0441

The number of 3 seat tables is 12 and the number of 2 seat tables that will be 5.0446

We will keep that as our final answer.0462

Set down your unknowns, set up your system, solve it and interpret it in the context of the problem.0468

Let us try another problem, it has a lot of the same feel to it.0478

This one says that at a concession stand, if you buy 5 hotdogs and 2 hamburgers the cost is \$9.50, 2 hotdogs and 5 hamburgers cost \$13.25.0483

We are interested in finding the total cost of just one hamburger.0495

First, we need to go ahead and label our unknown. 0499

We have the number of hotdogs and we have our hamburgers.0504

Let us work on that. 0509

I will say that we will use and see d for hotdogs and let us use h for our hamburgers.0512

Both of these situations here are involving costs.0529

In the first one, we are looking at so many hotdogs, 5 of them and 2 hamburgers all equal to \$9.50.0534

In the next situation, we have a different combination of hotdogs and hamburgers to equal the \$13.25.0543

We will take each of these and package them into their own equation. 0549

5 hotdogs 2 hamburgers equals to \$9.50.0554

2 hotdogs 5 hamburgers equals to \$13.25.0569

There is our system of equations.0581

When we are done solving we should be able to get the price for just one hotdog and a price for just one hamburger. 0583

In this system, you could solve it using elimination or substitution. 0591

I think I'm going to attack this using our elimination method.0595

I will have to multiply both equations by something to get something to cancel out.0599

Let us multiply the first one here or multiply everything in there by -2.0604

We will take our second equation and I will multiply everything there by 5.0614

Let us see what the result is, -10 - 4h -19.0621

Then we will take our second equation 10 d + 25h (5 × 13.25).0639

Adding these 2 equations together, let us see what our result is.0661

The 10 d and -10 d both will cancel each other out and we will have 25h - 4h = 21h.0665

We can add together the 66 + 25 -19.0679

In this new equation, we only have h to worry about and we can get that all by itself just by dividing both sides by 21.0689

This will give me that h = \$2.25.0702

By looking at what we have identified earlier, I know that this is the cost of one hamburger \$2.25.0708

Even though the problem is not asking for it, let us go ahead and figure out how much a hotdog is0715

by taking this amount and substituting it back into one of our original equations.0719

You can see that I have put this into the first equation, now I need to multiply 2 by \$2.25 that will \$4.50.0737

Subtract \$4.50 from both sides and divide both sides by 5.0750

Now we have our entire solution. 0768

I know for sure that the hamburgers will cost \$2.25.0769

I also know that one hotdog will cost exactly \$1. 0774

Try and pick out each situation there and interpret it into its own equation.0780

Let us try this one out.0789

In the U.S. Senate, there are 100 representatives, if there are exactly 10 more democrats than republicans, how many of each are in the senate?0791

One assumption that we are going to use here is that all representatives are either democrats or republicans.0799

We are not even going to consider any third parties here.0804

Let us see if I can break this down but first let us identify some variables.0808

Let us say r is the number of republicans and we will let d be the number of democrats.0813

We need to set up an equation for each of these and one of the first big bits of information I get is that the total number of representatives should be 100.0846

We can interpret that as the number of republicans + the number of democrats that should be equal to 100.0856

We need to interpret that there are exactly 10 more democrats than republicans.0870

To look at that, let us compare the 2.0879

If I was to take republicans and democrats and put them on each side of the equal sign,0881

they would not exactly be equal because I have 10 more democrats than republicans.0887

The way you want to interpret is that right now this side got a lot more, it has 10 more than the republican side.0893

If I want to balance out this equation, I need to take something away from the heavier side.0901

Since it is exactly 10 more, I will take away 10 and that should balance it just fine.0906

My second equation is r = d -10.0914

I will get that relationship that there are exactly 10 more democrats than republicans.0917

When we are done setting up the system, notice how it set up a good for substitution.0924

The reason is if you notice here our r is already solved.0929

Let us use that substitution method to help us out.0940

We will take the d – 10 and substitute it into the first equation.0943

In this new equation, we only have these to worry about.0958

Let us combine the like terms, 1d and another d will give us a total of 2d -10 =100 and we can add 10 to both sides.0961

One last step, let us go ahead and divide both sides by 2.0979

This will give me that the number of democrats should be 55.0987

Since I also know that there are exactly 10 more democrats than republicans, 0992

I can take this number then subtract 10 and that will give me my republicans.0997

Let us say we have 55 democrats, 45 republicans.1011

I have one more example with these systems of equations and this involves a little bit more difficult numbers.1038

We are going to end up rounding these nice whole numbers, so we do not have to deal with too much decimals.1045

Be careful along the way because some of the numbers do get a little messy.1050

In this problem, I have a Janet that blends coffee for a coffee house1054

and what she wants to do is she wants to prepare 280 pounds of blended coffee beans and sell it for \$5.32 per pound.1058

The way she plans on making this mixture is she is going to blend together 2 different types of coffees.1066

She is going to blend together a high-quality coffee that cost \$6.25/pound.1072

And she is also going to blend together with that a cheaper coffee that only costs \$3/ pound.1080

The question is to the nearest pound, how much high quality coffee bean and how much cheaper quality coffee bean should she mix in order to get this plan.1088

Let us set down some unknowns and see if we can figure this out.1098

I will call (q) the number of pounds and this will be for our high quality coffee.1105

Let (c), even number of pounds for our cheap quality coffee.1126

We are going to try and set up a how many pounds of each of these we need.1149

One of the first bits of information that will help out with that is we know that we have a total of 280 pounds.1153

The number of pounds for my high quality coffee + the number of pounds for my cheaper coffee better equal to 280 pounds. 1160

This entire equation just deals with pounds.1170

The second part will have to deal with the cost.1176

We want the final mixture to be \$5.32, so we will use the cost of the high quality coffee bean and the cost of the low quality coffee bean1178

and blend the 2 together and get that final cost.1186

Let us see what we have here.1191

Normally, high quality coffee costs \$6.25 every pound of high quality coffee.1191

The cheaper coffee usually costs \$3/pound.1200

We are hoping to make is \$5.32 for that final 280 pounds of coffee.1206

Let us mark this other one as just costs.1215

In our system here, you can see the numbers are not quite as nice but we can definitely still work with them.1221

Let us do a little bit of cleaning up then we will go ahead and try the substitution method on the system.1227

I’m going to multiply these together first, so \$5.32 × 280,1234

I'm doing the new system 625q + \$3, c =1489.6.1248

To use the substitution method, I’m going to take the first equation and let us go ahead and solve it for q.1265

Q = 280 – c.1274

Once we have that, we can go ahead and substitute it into our second equation.1280

\$6.25 q is 280 – (c + \$3.04c) = 1489.6.1289

Our new equation only has c and we can go ahead and solve for that.1307

Let us distribute through by the \$6.25 and see what we will get.1312

17.50 – 6.25c + 3c = 1489.6.1323

Let us combine our c terms, -3.25c = 1489.6.1338

We almost have c all by itself, let us get there by subtracting the 17.50 from both sides, -260.4.1354

One last step, let us go ahead and divide both sides by the -3.25.1378

I’m getting 80 and then something after the decimal 12307692, it just keeps going on and on.1396

I can see that I’m going to have just a little bit more than 80 pounds, but we want to round this to the nearest whole pound.1407

What I will say is that c is about equal to 80 pounds.1415

Now, once I know how many pound the cheaper coffee I will need, we can quickly go back to one of our originals1421

and figure out how much of the high quality coffee we need.1427

I need the total poundage to be 280 and 80 of that will be in the cheaper coffee, then I know that the high quality coffee I will need 200 pound.1433

Our rounded solution, I have that 80 pounds of cheap coffee and 200 pounds of high quality coffee.1446

If you pick apart your word problems and try and set down an equation for some of the situations in there,1454

you should be able to develop your system of equations just fine.1459

Then use one of the methods that we have learned especially those algebraic methods to solve the system 1463

and do not forget to interpret your solution in the context of the problem.1467

Thanks for watching www.educator.com.1472

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at solving linear inequalities in one variable.0003

To solve linear qualities, you will see that the process is not so bad. 0010

But we are willing to take a little bit of time just to highlight the difference between inequalities and equations, 0013

after looking at the difference between the two and that will cause a little bit of a problem.0020

You have to be careful on how we describe our solutions for inequalities.0025

Then we will get into a bunch of different examples on how we can finally solve these linear inequalities. 0030

Let us get started.0035

When you are looking at an inequality versus an equation and you try to figure out what is the difference.0040

One of the main differences, of course you will see a different symbol in there, one of our inequality symbols.0045

It is these guys that we saw earlier.0051

We have our greater than symbol, greater than or equal to symbol, our less than symbol or less than or equal to symbol.0054

It is a little bit more than just having an additional symbol in there that makes this different. 0063

What makes an inequality different from an equation has to do with the solutions that you get out of them.0068

To highlight this I will go over some quick examples of equations and get to one of those inequalities.0075

Consider this first one here, 3x = 15.0081

If you are going to solve that, it would not take probably that long.0087

Just simply divide both sides by 3 and you get that x = 5. 0091

Now what that says the only number, the only one out of all possible numbers that will make that equation true is just 5.0099

It is like this little isolated point out on the number line.0106

Some equations might actually have more than one solution, for example, this next one x2 = 9. 0110

If you are going to look for numbers that will make that equation true, you will actually find two of them.0117

When x = 3 and you square it you will get 9, but also when x = -3 and you square it you still get 9.0123

You know it is a little different that you have two solutions, but both of the solutions those are the only two things that will make it work.0132

It is like two little isolated points on your number line.0138

To contrast that to this first inequality over here, 4x is greater than or equal to 8.0142

If you try to figure out what numbers make that true, you would end up with a huge, huge list of numbers that work.0149

Let us start picking out some examples. 0157

Also, what if I said that x was equal to 2.0159

If you substitute that in, you would have 4 × 2 is greater than or equal to 8, 0163

which would simplify to 8 is greater than or equal to 8, which is, of course true.0167

But it is not the only thing that works out. 0174

You could also put in 3, 4, you can get a little creative and put in some other ones like 4 1/2, and all of these would be solutions.0176

Trying to list out solutions for an inequality is simply not the way to go.0191

The reason for that is if you look at the numbers that would make an inequality true, you will get an entire range of values.0196

So, not just in a little isolated point, they are in whole ranges of values on our number line.0204

Since we are after a whole range of values other than these isolated points, we are going to have to be a little bit careful on how we describe those.0211

In fact, there are many ways that you can describe the solution to an inequality.0223

Some you seen before and some might be a little new, let us go over them.0227

The first way that you can describe the solution is to simply use our inequality symbols that will be like this guy down here.0231

Maybe I’m trying to describe all the numbers that are less than 4, I can simply write x is less than 4.0239

We can also use our interval notation.0246

This is a good way to mimic the number line and that it gives us the lowest number or the starting point and the endpoint. 0250

It also gives us information on whether we should include the number or not.0262

Since this one has parentheses, I know that the 4 is not included.0266

It is simply a different way of writing the same information.0273

This interval says I'm looking at all the values that are less than 4. 0276

We could also describe our solutions a little bit more visually using a number line.0283

The way we do this is we have a number line and we shade in the solution.0288

To indicate whether the endpoint is included or not, you can use an open circle or close circle.0300

The open circle here means it is not included, so I would have used a close circle if I had x is less than or equal to 4.0310

And one last way that you might see, I do see this in a few Math books is known as set builder notation. 0323

It looks a little clunky and that we have a couple of curly brackets thrown in there 0329

but let me show you how you can interpret the set builder notation.0335

First of all, we identify some sort of variable at the very beginning. 0341

We are looking for all x values.0346

And after that you will notice that we borrow our inequality. 0352

This is the rule of all of the numbers that we will include.0356

That little line you see in between, we say that is L and it is just the divider or it separates both those sections. 0370

The way you would read the set builder notation is the set of all x’s such that x is less than 4.0384

We know that we are building for x values and we know which x values would get in there.0393

I’m not going to focus a whole lot on that set builder notation but you will definitely see me use some inequalities, some number lines 0397

and some interval notation just to describe our solutions.0404

Now that we know little bit about what makes an inequality different, we know little bit more on how to describe the solutions, 0410

let us get into how we can actually start solving these things.0416

The good news is when it comes to solving linear inequality, we use the same techniques as we do with solving a normal linear equation.0421

There is one additional rule that you want to be aware of.0432

When you multiply or divide by a negative number then you want to flip the inequality sign.0435

Now that rule applies to multiplying and dividing, so be careful not to try and use it if you are only adding and subtracting.0444

Use it only for multiplying and dividing.0453

I often get lots of questions like why is it that you have to flip the sign when your doing the multiplication and division by negative numbers.0458

I think a quick example will help you understand why that is.0465

Let us take a nice inequality like 2 < 3, so seems pretty simple. 0469

We know that 2 < 3 but watch what happens if I multiply both sides by a negative number.0476

-2 and -3, now which one is less than the other one?0483

If we take a peek at our number line, you will see that it is the -3 that is now the smaller number since it is on the left side of -2.0491

This means that if we want to preserve the trueness of our statement, we also must flip that sign to compensate for that negative that we threw in there.0510

It is easy to see with numbers like this but in a moment when we start dealing with x’s and unknowns, 0520

we still have to remember to flip the sign when we divide or multiply by that negative number.0525

I think we have all the information we need.0534

Let us go ahead and get into the solving process and see if we can tackle some of these inequalities.0537

We want to solve the following inequality then write our answer using a number line and using interval notation.0543

Okay, when I solve an equation usually I try and isolate the x's and I'm going to do the same thing here. 0549

I will do that by first moving the 5 to the other side, 3x < 11 - 2x and then we will go ahead and we will add 2x to both sides. 0557

5x < 11, let us go ahead and divide both sides by 5.0578

I will get that x < 11/5.0593

This could represent my solution using just the inequality symbol. 0597

Now I want to go and have a look at the number line and its interval notation.0601

Here is a nice quick sketch of a number line. 0615

I'm looking to shade in all values that are less than 11/5.0620

If I want to rewrite that, 5 goes into 11 twice with the remainder of one, so it is the same as 2 1/5.0625

It is a little bit greater than 2, but not much.0638

Notice how I’m using an open circle there because our inequality is strict. 0643

We do not want to include that value.0650

I will shade in everything that is less than my 2 1/5.0653

This number line represents all of the solutions to my inequality, I could use anything on that side and it worked. 0662

As long as we have our number line, let us go ahead and also represent our solution using interval notation.0670

We want to think of where we start on the left side, it looks like we are starting way down at negative infinity.0680

We go all the way up to 11/5 but I will use a parenthesis since we do not include that value.0686

I have represented my solution now in 3 different ways. Not bad.0694

Let us look at another, for this one we are looking to solve 13 - 7x = – 4.0697

Let us start off by trying to get those x's isolated and we will do that by first getting them together. 0710

I'm going to move this 10x to the other side, so I'm subtracting here. 0716

I do not have to worry about flipping any signs nothing like that yet, 13 - 17x = -4.0722

Let us go ahead and subtract 13 from both sides, we have -17x = -17.0735

Be very careful in this next step.0750

I need to get the x all by itself, in order to do that I will divide by -17.0751

I will do all of the algebra normally -17 ÷ -17 = 1. 0763

I'm also going to remember what to do with that sign, we are going to flip it around.0770

I have it x = 1.0777

Let us get to our number line and go ahead and represent the solution.0784

I have the number 1,we will make a nice solid circle since this is or equal to 0793

and we will go ahead and shade in everything less than that since it says x < 1.0804

Now that I have my number line, let us represent this using our interval notation.0811

On the left side we are starting at way down to negative infinity and going all the way up to 1.0815

We want to include the one so we will use a bracket.0821

Let us try one more example.0838

In this last one I picked a little bit more of a word problem so you can see that inequalities are important even in applications.0843

This one says that Joanna is hiring a painting company in order to get her house painted0850

and the company has two plans to choose from, you can choose plan A.0855

In plan A they charge you a flat fee of \$250 and \$10 for every hour that it takes to paint the house.0861

For plan B, they do not charge a flat fee but they will charge \$20 per hour.0869

The question is when will plan B be more expensive than plan A?0874

Pause for a moment and think why would we be using an inequality here?0878

Why do not we just set up an equation?0882

What I'm looking for is not one specific time, but all times when plan B will be more expensive.0885

It makes a little bit more sense to go ahead and set up an inequality for this word problem.0891

Let us go ahead and hunt down both of our situation.0896

Let us have plan A and let us go ahead and have plan B.0899

We need an unknown here, let us say x is the time to paint the house.0907

Plan A cost a flat fee of \$250 + \$10/hour.0924

\$10/ hour is like our variable cost, that would be 10x, flat fee does not vary so +250.0929

That expression will just keep track of the cost for plan A depending on how many hours it takes.0938

Plan B has only a variable cost of \$20/hour, I will just say 20x.0944

Now comes the big question, when is plan B more expensive than plan A.0951

B would be more expensive, I would use an inequality symbol of plan A < plan B 0958

and that is exactly how we will connect our expressions as well. 0964

10x + 250 < 20x and to completely answer this problem we just have to solve the inequality.0967

Let us work on getting our x's together by subtracting 10x from both sides.0976

250 < 10x, now we will divide both sides by 10.0986

25 < x, now it is time to interpret exactly what we have here.0999

x represents the time that it will take to paint the house.1006

What I can see here is that anytime my time is more than 25 hours then I can be sure that plan B will be more expensive, there you have it. 1011

Also, be careful when setting up and solving these inequalities.1021

Remember to go ahead and flip your inequality anytime you multiply or divide by negative number.1024

Thank you for watching www.educator.com.1031

Welcome back to www.educator.com.0000

In this lesson, we are going to take care of compound inequalities.0002

In compound inequalities we will look at connecting inequalities using some very special words and, and or.0010

We will have to first recognize some subtleties between using both of these words.0017

Blabbering about some nice new vocabulary, we will learn about union and intersection.0023

Once we know more about how to connect them then we can finally get to our compound inequalities and how we can solve these.0029

In some situations, you have more than one condition and they could be connected using the word and or they could be connected using or.0040

To see the subtle difference between using one of these things, let us just try them out on a list of numbers.0050

I have the numbers 1 all the way up to 10 and we will see what number should be included in the various different situations below.0056

Let us see if we can first take care of all the numbers greater than 3 and less than 7.0066

Notice in this one I’m using that word, and.0071

Numbers that are greater than 3, we will have 4 or 5 and 6 and all of these are also less than 7.0075

The things I will include in my list 4, 5, and 6.0083

Let us think on how we will approach this problem a little bit different for the next one.0090

Maybe list out all numbers that are greater than 4 or they are less than 2.0095

The numbers greater than 4, that would we 5, 6, 7, 8, 9, and 10 and the numbers that are less than 2, 1.0101

Notice how in that situation I was looking for 2 things.0112

I was looking for all numbers that were greater than 4 and I have highlighted those first.0115

And then I went back and I looked for ones that were less than 2.0119

Let us give it a few little marks here and try the third situation. 0127

All numbers greater than 6 and they are less than 3.0132

That is a tough one to do because I can go ahead and highlight the numbers greater than 6, that will be 7, 8, 9, 10 but they are not less than 3.0137

I can not include them.0148

Notice how in that situation since I'm dealing with and, it is like I have to satisfy both of these conditions must be greater than 6 must be less than 3.0150

If it does not satisfy both of them then I cannot include them.0164

Let us try another one and see how it works out.0167

All numbers greater than 4 so I have 5, 6, 7, 8, 9, 10 or they are less than 7. 0170

What numbers less than 7 would be 6, 5, 4, 3, 2,1.0178

All the numbers fall into one category or they have fallen to the other one. 0183

In fact, using or seems like it is a little bit more relaxed as long as it satisfies one of my conditions, then I will go ahead and include it in my list. 0188

In fact, that highlights the difference between using and using the word or to connect these 2 conditions.0197

Let us make it even more clear.0204

When you use and to connect two conditions, then you must have both conditions met in order to include it.0207

However, when you are using or, then as long as you have one of the conditions met then you can go ahead and include that.0214

The way you will see and and or used is when we combine our intervals in our solutions.0222

If an object meets both conditions using our and connection, then it is said to be in the intersection of the conditions 0230

and we can use the symbol to connect those.0239

Some people might think of this as and but it stands for the intersection.0242

On the flip side, if an object meets at least one of the conditions using the word or, 0253

then it is said to be in the union of the conditions.0257

We will use the symbols, think of or and union.0261

You will see these symbols for sure and watch how we connect our solutions using either and or or.0267

We know a little bit more about the connections, let us get into how we can actually solve these things.0277

We will connect these things using and or or and the way you go about solving is you can actually solved each of the inequalities separately.0285

Just take care of one at a time and the part where these and or or come into play is when we want to connect together our solutions. 0294

There are some situations where we can actually connect the inequalities at the very beginning.0305

One of those situations is when you are dealing with and and 2 of the parts are exactly the same.0310

Just like this example that I have highlighted below.0316

I have 5x < or = 3 + 11x and I also have a -3 + 11x on the other side over here.0321

We are connected using and. 0329

I'm going to put these together into what is known as a compound inequality.0332

You will notice that everything on this side of the inequality comes from the left side.0336

Everything on the other side of inequality over here comes from this inequality.0344

It encapsulates both of them at the same time.0351

If you do connect one like this, the way you end up solving it is just remember that whatever you do to one part, do it to all 3 parts.0354

If you subtract 3, subtract 3 from all parts.0362

This also includes if you have to flip a sign.0366

If you multiply by a negative number then flip both of those inequality signs that are present then you should be okay.0368

Just remember if you want to solve each one separately, that works to.0377

Let us see some of our examples, see the solving process in action.0383

We want to solve the following inequality and of course write our answer using a number line and using interval notation.0388

I can see I have 2x -5 < or = - 7 or 2x – 5 > 1.0395

I’m going to solve these just separately, just take care of one at a time.0403

Looking at the left, I will add 5 to both sides, 2x > or = -2.0406

Now I can divide both sides by 2 and get that x < or = -1.0418

There is one of my solutions, let us focus on the other one.0429

You will add 5 to both sides of this inequality, 2x > 6, now divide by 2 and get that x > 3.0434

I have 2 intervals and I will be connecting things, let us drop this down using or.0450

I want to think of all the numbers that satisfy one of these, or satisfy the other one.0458

As long as they satisfy one of these conditions, I will go ahead and include it in my overall solution.0463

Let us take this a bit at a time.0469

I will do a little number line here, a number line here and I work on combining them into one number line.0472

First I can look at all the numbers that are less than or equal to 1.0481

Here is -1, I'm using a solid circle because it is or equals to and we are less than that so we will shade in that direction.0489

For the other inequality, I'm looking at numbers like 3 but not included or greater so I’m using an open circle there.0501

What will I put on my final number line is all places where I shaded at least once. 0509

Let us see how that looks, I have my -1 and everything less and I have 3 or everything greater.0523

This number line which has both of them shows one condition or the other, and I can include both.0535

Let us go ahead and represent this using our interval notation.0543

We have everything less than - 1 and we include that -1 and we have everything from 3 up to infinity. 0546

Since we are dealing with or, let us go ahead and use our union symbol to connect the two.0558

I have the many different ways that you can represent the solution for this inequality.0563

Let us take a look at another one, solve the following inequality and write your answer using a number line and interval notation.0571

This is a special inequality, this is one of our compound inequalities because it have 1, 2, 3 different parts to it.0578

As long as you remember that whatever we do to one part, we should do it to all three,I think it will turn out okay.0585

Let us work on getting that x all by itself in its particular part.0591

We will go ahead and subtract 5 from all three parts and get – 8 < 2x is < or = 2.0597

Let us divide everything by 2.0612

-8 ÷ 2 = -4, x < or = 1.0620

In this one I am looking for all values between -4 and 1.0628

I think we can make a number line for that.0633

Okay, I need to shade in everything in between -4 and all the way up to 1. 0649

It looks like the -4 is not included, I will use a nice open circle, but the 1 is included, so we will shade that in.0657

Now that we have our number line, we can describe this using our interval notation.0667

We are starting way down at -4 not included and going all the way up to , that is included. 0671

In some of these inequalities, you have to be careful on which conditions it satisfies.0689

If you end up with no numbers that simply do not work, watch how that turns out.0695

We want to solve the following inequality and write our answer using a number line and interval notation.0701

I have 4x + 1 > or = -7 or -2x + 3 > or = 5.0706

Let us begin by solving each of these separately.0716

With 1 on the left, I will start by adding 1 to both sides, 4x < or = -6 and now divide both sides by 4.0721

We are reducing that, I get that x > or = -3/2.0737

We will set that off to this side and solve the other one.0744

Let us subtract 3 from both sides and we will go ahead and divide both sides by -2.0749

Since we are dividing by negative, I’m going to flip my inequalities symbol.0758

This one is very interesting.0766

Notice how I'm looking for numbers that are less than or equal to 1, but I’m also looking for numbers that are greater than or equal to -3/2.0768

If I look at where those two are located, here is -3/2 and here is 1.0779

You will see that we get actually all numbers because I'm looking for things that are greater than or equal to -3/2, 0785

that will shade everything on the right side of that - 3/2.0792

Everything less than or equal to 1 would shade everything in the other direction.0796

It will end up shading the entire number line.0801

We will say we would not include all numbers from negative infinity up to infinity. 0805

I might make a note, all real numbers.0813

Watch for a very similar situation to happen when we deal with and.0824

Solve the following inequality and write our answer using a number line and interval notation.0830

I have two inequalities and a bunch of different x's, let us work on getting them together first. 0837

I'm going to subtract an x from both sides on the left side here, 1x < -5.0842

I can add up 8 on both sides and we will get that x < 3.0851

Let us go over to the right and see what we can do there.0861

I will subtract 15 from both sides, that will give me –x < -10.0864

Now divide both sides by -1, since we are dividing by a negative number, let us flip out the inequality sign.0871

I have x < 3 and x > 10.0882

This one actually proves that when we are dealing with and, it must satisfy both of these conditions in order to be included.0889

If you start thinking what numbers are less than 3 and also greater than 10, you will that you do not get any number.0896

You can not be both things at the same time.0902

They are both in completely different spots on a number line.0906

Let us make another one.0913

We can not shade in numbers that are both less than 3 and greater than 10.0917

What is that mean to our solution then?0942

I mean what can we put down?0943

This is where we say there is no solution.0947

Be careful on using those connectors and how it affects your solution.0953

Remember that when using the word or, it must satisfy one of the conditions.0957

However, when using the word and, it must satisfy both of those conditions before you can include it in your solution.0964

Thanks for watching www.educator.com.0971

Welcome back to www.educator.com.0000

In this lesson we are going to go ahead and look at solving equations that have absolute values in them.0002

We only have one goal, how do you get down to the nuts and bolts of solving it when you see an absolute value.0010

When you have a number that is equal to an absolute value or any algebraic expression, then you can break it down into the following.0021

That expression is equal to the number or that expression is equal to the negative of that number.0030

What this is trying to say is that when you are dealing with this situation, absolute value equals number, is that you can break it down into 2 situations.0035

It takes care of 2 possibilities that whatever was inside the absolute value could have been positive 0045

or maybe what is on the inside could have been negative.0050

The reason why we have 2 possibilities is the absolute values going to take away information about what the original sign was.0053

Maybe that original sign was positive, maybe it was negative, but we are not sure so we have to take them both into account.0060

To highlight how this works, let us look at some actual numbers.0067

Let us suppose that our algebraic expression was just x and we are going to look at the absolute value of x is equal to 2.0073

What would have to be inside that absolute value in order for this thing to work? 0081

One possibility is that maybe x = 2, so we get this case over here.0085

What if the x was negative and we lost all information about that negative sign.0091

That looks like more of like the situation on the right, what we see is that x could be 2 or x could actually be -2.0097

The only difference between these two is I multiply both sides by -1.0105

What I like to do when handling these absolute values is immediately split them into 2 different problems.0111

I will handle what I called my positive case where I just keep everything the same and drop the absolute values.0118

Then I will take care of what I call the negative case, 0125

I will put a negative sign on whatever was inside the absolute value and then continue solving from there.0127

When I’m all done, I will go ahead and take both of my solutions and check them in the original problem, just to make sure that they are correct.0133

Some other things to consider when solving an equation that have an absolute value is to make it fit the form.0143

You want to make sure that you isolate the entire absolute value first.0150

Do not worry about splitting it into 2 problems, until you have absolute value on one side and the one on all the numbers and other stuff on the other. 0155

Once you do have that absolute value, go ahead and split it into 2 problems.0165

And what you are looking to do with each of those problems is you want one of them to handle what I said positive possibilities,0172

it will be exactly the same, just no absolute value.0180

And one to handle the negative possibilities, what would happen if there was a negative sign on the inside of those absolute values.0183

When you have two new problems then simply solve each of them separately as normal. 0191

Follow all of your normal rules for algebra and you should be okay. 0196

Let us give this a try. 0201

The first one I'm looking at the absolute value x -3 -1 = 4. 0204

The very first thing I want to do is to find those absolute values here and get them all alone onto one side of my equation. 0211

I have the absolute value of x - 3 and isolate it then I will add one to both sides.0220

Once I get the absolute value completely isolated, this is the spot we are going to split this into 2 problems. 0230

Get some space and get our two problems.0237

On the left side here, I'm going to consider this my positive possibility that whatever the absolute value was encapsulating it was positive.0241

It is one of the changes that I need to make.0250

The other possibility is that maybe the x -3 maybe that entire thing was negative.0254

Now that we have two problems, we just go through each one separately and see how they turn out.0262

x - 3 = 5, add 3 to both sides and get that x = 8.0268

With the other one, it is okay if you want to distribute that negative sign first.0276

-x + 3 = 5, then we can subtract 3 and move that negative sign, x = -2.0288

I have two solutions to my equation that I started with.0300

It is a good idea to check these just to make sure they work out.0306

Let us take them individually and see what happens. 0309

8 - 3 would give us 5.0319

Our absolute values make everything positive so the absolute value of 5 would be 5.0326

I can see that sure enough, 5 – 1 =4 and that solution checks out.0333

Onto the other side, we will put a -2 – 3 working with the inside of the absolute values,-2 - 3 would give us -5.0340

Notice, as we start to simplify that we can see what I was talking about how the sign got me positive so we do have to handle both possibilities.0358

Cancel the absolute value of -5 is 5, 5 - 1 = 4 and sure enough, we know that is true so our solution checks out, that x = 8 or x = -2.0373

In this next example, we will have to do a little bit more on getting the absolute value isolated, but again you will see this is not too bad.0397

To get rid of that fraction out front, I'm going to multiply both sides by 2, 2 over there and 2 over there.0404

It will give us the absolute value of 5x + 2 = 12.0415

We have the absolute value on one side of the equal sign.0422

Let us go ahead and split this into two different problems. 0427

Here I will handle my positive possibility that whatever was inside absolute value it is positive.0435

Over here, maybe what was inside the absolute values was negative, so we will put that negative sign.0443

Let us solve both of these separately.0453

On the left, I will subtract 2 from both sides, and then we will go ahead and divide both sides by 5, I have x = 2.0457

On the other side, I’m going to do something a little bit different.0471

To move that negative sign at the very beginning, I will multiply both sides by -1. 0474

It is important to know that you can also distribute through being your first step and then go through the solving process.0481

I’m just doing this so I can take care of the negative sign at the very beginning.0487

Negative × negative on the left side would give me 5x + 2 =-12.0491

Let us subtract 2 from both sides, and finally divide by 5.0500

Now I have two solutions to my absolute value problem.0515

Let us take a little bit of time just to check them and make sure they work out.0521

½ of 5 × 2 + 2 does that equals 6?0527

Working on the inside of the absolute value, I have 5 × 2 = 10 + 2 =12.0536

The absolute value of 12 is 12, 1/2 of 12 = 6, that one checks out.0546

On to the other side, this one is -14/5 + 2.0556

Let us see if this one equal 6 or not.0570

14/5 × 5 = -14, -14 + 2 =-12, it is very similar to what we got last time.0573

We are just dealing with the negative on the inside.0593

The absolute value will make that -12 into a positive 12 and sure ½ of 12 = 6.0599

Both of our solutions work out just fine. 0607

There is still one more example to show you some of the things that might happen when going through the solving process.0615

This one is 3 times the absolute value of x + 2 + 3 = 0.0620

We will start off by trying to isolate that absolute value. 0626

I'm going to subtract 3 from both sides.0635

To get rid of that 3 out in front of the absolute value, let us divide both sides by 3.0639

I have the absolute value of x + 2 .0648

Now following our steps, let us split this into 2 problems.0651

Here we have the problem exactly the same, just no absolute value and over here, -x + 2 = -1.0660

Let us solve each of them separately.0670

I will subtract 2 from both sides given the x = -3. 0674

On the other side here, I will go ahead and distribute through with my negative sign and I will add 2 to both sides.0679

Lastly let us go ahead and multiply both sides by -1.0693

I have two possible solutions, x = -3 and x = -1.0699

Let us check them to see how they turn out.0705

Let us see, -3 + 2 = -1 then we will take the absolute value of -1, we get 1.0716

What I'm getting here is that 3 × 1 is 3 and when we add another 3, I get 6.0730

This shows that this one does not work out.0738

I'm going to get rid of that, it is not a solution.0742

Let us try it on the other side.0747

I will put in a -1.0750

I have 3, -1 + 2 = 1 and the absolute value of 1 is 1.0758

I have 3 × 1 + 3 I think that is the same situation as before or 6 = 0.0772

I think that does not work out as well so I have to throw away that solution. 0778

What this shows is that if none of the work then we actually have no solution.0784

Let us highlight that it is a good idea to always check your answers.0792

I will also highlight that some funny things could happen when working with absolute values.0796

Remember that an absolute value takes whatever you put into it and makes it positive.0800

In fact, we can see a problem in this step when we are working. 0806

I have that the absolute value of something is equal to a negative and that can not happen because our absolute value on one side will be positive.0813

On the other side we are setting equal to a negative and we can not get a positive and a negative, to agree.0826

That is why this one also has no solution.0836

If you catch it early on, it usually can save yourself a lot of work.0839

But if you do not, make sure your check your solutions also that you can definitely find these no solution problems.0842

Thank you for watching www.educator.com.0850

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at inequalities with absolute values.0002

In our previous lessons, we did take care of inequality separately and we did take care of absolute value separately.0009

Our goal for this one is to mix the two together and see if we can solve some new types of problems.0015

When you have some number A greater than 0 and you have some algebraic expression x as packaged together in the following way.0026

The absolute value of x < A, or the absolute value of x > A then you can actually split these into two different problems. 0035

Let us look at the first one, suppose that you have the absolute value of x < A, the way that we can put this is that the algebraic expression x > -A.0043

The algebraic expression of x could be less than A.0054

It is interesting to note that not only do we split it into two problems 0059

and we will see that we have picked up our connector and it actually connected the two.0063

On the other side, when we are dealing with the absolute value of x > A.0068

This will still split into two problems but now we have the x < -A or x > A.0073

This is a little bit different and that now we have a different connector between the problems.0084

We will definitely use our tools on how to deal with these connectors and solving both problems separately0091

and figure out how their solutions should be connected in the end.0096

One thing that you may be looking at and be a little bit worried is keeping track of all of your signs.0100

It might start with less than and then these guys one is greater than and one is less than.0107

These guys over here, I have greater than and then I have less than or greater than.0113

Watch for my tips on how to deal with dealing with the absolute value and the absolute value separately so we can keep them both straight.0118

To understand why it splits into all of these different cases and why the signs look so funny, you have to recall two things that we covered earlier.0131

One has to do with those absolute values.0140

When we are dealing with absolute values this will end up being split into two different problems.0145

One that handles the positive possibilities, and one that handles the negative possibilities.0157

That is why we have two different things.0162

Also, you have to remember the rule that when you multiply or divide by negative that you will end up flipping your inequality sign.0165

That is why the direction of the inequality ends up changing in both of our cases.0172

Since one is that negative possibility, the sign ends up getting flipped.0177

To help you keep these both straight this is what I recommend.0183

When you are looking at your connectors, if you have the absolute value being less than a number then connect using and.0187

If you have the absolute value greater than a number, then use or to connect the two.0198

You will see that if you use each of these rules bit by bit, rather than trying to jumble them altogether, 0205

it is not too bad being able to pick apart one of these absolute values and inequality problems. 0211

Let us start out with some very small examples and work our way up.0221

That way we can keep track of everything we need to do.0225

This one simply says that the absolute value of x < 3.0228

Here is what I'm going to do first.0232

I’m going to use my techniques for our absolute values to simply just split this into two problems.0234

Remember before that we have our positive possibility that x < 3 or we have our negative possibility -x < 3.0246

I have not done anything with that inequality symbol, I have not even touched it yet I simply split it into two problems.0258

Because we have the absolute value less than a number, I will connect these using the word and.0266

It will become especially important when we get tour final solution.0274

Now I just have to solve these separately.0278

x < 3 is already done, -x < 3 I have to multiply both sides by -1.0281

Since I’m multiplying by a negative, we need to flip our sign.0289

Here I have two things that x must be a number that is less than 3 and x must be a number greater than 3.0294

Since we are dealing with and, we are looking for all numbers that satisfied both of the possibilities.0302

Let us look at our number line first to see if we can hunt down everything that does that.0309

All numbers less than 3 would be on this side of 3 and we will put in a big open circle.0327

They must be greater than -3 and I will put in another big open circle and -3.0336

We will see that it shades in everything greater than that value.0341

Our overall solution is all numbers between -3 and up to 3.0346

Use your techniques for splitting up into two problems.0354

Be careful on how you connect them and remember that you should flip the sign when you multiply or divide by negative.0357

Let us try another small one, and again watch for this process to happen.0363

This one says the absolute value of z = 5.0369

The first thing I'm going to do is just split this into two problems. 0373

I’m doing this because I'm taking care of that absolute value.0380

z > or = 5, -z > or = 5 I have not touched my inequality symbol I just took care of my two cases.0384

Since I'm dealing with the absolute value being greater than or equal to a number over here and I will connect these using or.0397

This will become especially important when we get to our solution and we just solve each of them separately.0408

Z > or = 5 is already done and for the other one, I will multiply both sides by -1.0413

Now if I multiply it by a negative, this will flip its sign.0423

I have z > or = 5 or z < or = 5.0428

I think we are ready to start packaging things up here.0434

Since I’m dealing with the connection or, as long as it satisfies one of these possibilities I will include it in my final solution.0442

All the numbers greater than or equal to 5 that would be all of these numbers here on the number line.0450

All the ones less than or equal to 5, that would be these ones down here.0460

Now that I know more about the number line, let us go ahead and write this using our interval notation.0465

Negative infinity up to 5 brackets, because we are including from 5 to infinity, our little union symbol since it could be in either one of those intervals.0470

Now that we have the process down a little bit better for being able to split this up, let us look at ones that are just a little bit more complicated.0487

We want to solve the inequality involving our absolute value.0498

I want make sure that the absolute value is isolated on one side of my inequality sign and fortunately this one is.0503

Let us go ahead and split it into our two problems.0510

On this side we will have 3x + 2 < 5.0519

On the other side I have -3x + 2 < 5.0525

I will take note as to the direction of my inequality symbol, our inequality less than number.0533

Because of its direction I will connect these using and.0541

A little bit more work, let us go ahead and solve each of these.0547

For this one, I will subtract 2 from both sides and then divide both sides by 3 so x < 1.0551

On the other side, let us go ahead and distribute our negative sign and let us add 2 to both sides.0563

We can see that on this case I only did divide by -3 so because of that we will end up flipping our sign -7/3.0576

I need all numbers that are greater than -7/3 and they are also less than a 1.0589

On a number line I could see where -7/3 is, there is 0 and I can see where 1 is.0600

I’m just looking for all numbers in between. 0609

Since those would be the only ones that are greater than -7/3 and also less than 1.0614

We have our interval -7/3 up to 1 and that would be our solution.0620

Let us get into another one. Solve the inequality using an absolute value.0631

In this one, we do a little bit of work to isolate that absolute value first.0637

Let us go ahead and add 1 to both sides, 5 - 2x > or = 1.0642

Now that my absolute value is isolated, we will split it into two problems.0653

This will handle our positive possibility, so no changes, 0660

And the other one will take care of what happens when the inside part could have been negative.0666

Since I'm dealing with these absolute values and inequalities, let us look at the direction of everything.0674

I have the absolute value less than or equal to number.0680

This is that situation where we connect those using or.0684

I have two problems that are connected using or, let us continue solving.0689

On the left side, I will continue by subtracting 5 from both sides, then I can divide both sides by -2.0696

Since I'm dividing by negative, let us flip our sign, x < or = 2.0710

Onto the other side I'm going to distribute my negative sign in there first -5 + 2x > or = 1.0719

Let us get that x all alone and we are just a little bit closer being all alone by adding 5 to both sides.0733

And I will finally divide both sides by 2, x > or = 3.0743

I have what I’m looking for all numbers that are less or equal to 2 and all numbers that are greater than 3.0753

Let us see what that looks like on a number line.0760

All numbers less than or equal to 2 would include the 2 and everything less than that, so we will shade that in.0767

All numbers equal to 3 or greater than would be over here and so we have a big gap between 2 and 3 but that is okay. 0775

Let us graph this using our interval notation.0784

From negative infinity up to 2 included and from 3 up to infinity and we are looking at the union of both of those two.0787

That is not too bad as long as you isolate that absolute value first before splitting it into two problems, you should be fine.0799

Let us look at one last one, and this highlights why you have to be careful on how you package up your solution using those connections and and or.0807

In this one, we already have our absolute value isolated to one side 3 - 12x all in absolute value.0818

I'm going to split this immediately into two problems.0828

3 - 12x < or = -3 and 3 - 12x < or = -3 because of the direction of our inequality symbol.0838

Let us go ahead and connect these using and.0858

Solving the one on the left 3 - 12x < or = 3.0864

Let us subtract 3 from both sides then we will divide both sides by -12.0870

I’m flipping my signs because I'm dividing by negative.0883

We can reduce this fraction as normal, just ½ so x > or = ½.0889

Working with the other one, I will distribute through with my negative sign and I will add 3 to both sides. 0897

It looks like one last step and I can go ahead and divide both sides by 12, 0 ÷ 12 is simply 0.0918

Let us see exactly what this would include.0933

On a number line, I want to shade in all numbers that are greater than 1/2 and they must be less than 0.0936

Of course, that poses a huge problem.0945

There are lots of numbers greater than ½ and there is lots of numbers less than 0 but unfortunately I can not find numbers that satisfied both of them.0951

There are no numbers that satisfied both conditions.0962

In fact this one, I can say has no solution. 0966

Now, you might have caught that earlier and if you did that is okay, you can definitely save yourself a lot of work.0972

If we look all the way back here at the original problem, we have a bit of an issue.0980

Remember that the absolute value makes everything positive and sitting on the other side of our inequality is a negative number.0985

We simply can not have something positive being less than something negative, that is not going to work. 0999

That is another good reason why this one has no solution.1005

Be careful when going through these problems and make sure you split them up because of your absolute value.1009

Watch the direction of that absolute value so you know how to connect them using, say, either, and or or .1013

Remember to solve each of them separately and connect very carefully when you get to your solutions.1020

Thanks for watching www.educator.com.1025

Welcome back to www.educator.com.0000

In this lesson we are going to look at graphing inequalities using two variables.0002

In order to make this process work out , we will first briefly review how to just simply graph lines.0009

We will see how this works with our inequalities which involved entire regions of our graph.0016

The important parts for highlighting these regions will be shading in the proper part of the graph.0022

Watch for these things to come into play.0028

Recall it when you are dealing with any inequality it is not just usually an isolated point, it is usually a whole range of solutions.0035

When we are working with graphing, and we have more than one variable to worry about we are dealing with points like x and y.0043

If we are looking for the x and y that will make an inequality true, we are not talking about just one point on a graph,0052

we are actually talking about an entire region of the graph, and all the little points in that region.0058

The way we do this is when you start off by graphing the equation, corresponding with the inequality.0065

Let me show you what I mean by that.0071

I’m looking at the inequality y > or = 6/5x + 7 rather than trying to tackle on the entire inequality first, 0074

I looked at the corresponding equation like y = 6/5x +7.0086

In doing so, I can look at my techniques for graphing just that equation. 0093

Maybe find the slope and find the y intercept.0098

We will see how that equation connects back to our inequality.0106

What we are going to find is that equation is going to split our entire graph into two regions.0115

In one of the regions I will involve our solution and the other region will not.0121

I have cooked up another example, so you can see how this works.0126

We are going to start in graphing y = 2x -4.0131

The first thing we wanted to do is look at the corresponding equation y = 2x – 4 and focus on that first.0141

We can see that it has a y intercept and it has a slope.0151

I'm going to end up graphing this line, and I'm going to graph it using a nice solid line.0158

I will get into that y in just a bit.0163

y intercepts way down here at -4 and a slope of 2.0166

From there up 1, up to over 1 and now we have our line.0172

This line is going to split the entire graph in to two regions.0190

We had this region on the left side of it or the region on the right side of it.0195

One of those regions will be our solution for our inequality.0201

The question is, in which one should it be?0205

Since this is the entire region, we can pick out an entire or just one little point from that and see if that region should be included or not.0208

To get all points inside of the region is either solution or all points inside the region are not a solution.0217

We need one to the find out.0223

I'm going to borrow a point, let us take this guy right here.0226

I’m going to take a point and test it into my inequality, y =4 and x = -2.0236

I will put that little question mark in there because I want to see if this is true or not.0249

Is 4 > or = 2, 2 × -2 = -4 + 4?0253

That is 4 > or = -8, yes it is, 4 > or = -8.0260

I know that, that particular point should be in my solution region.0269

To highlight that it is in my solution I will go ahead and shade everything on that side of the line.0275

Everything on that side, shade it.0297

Everything that is shaded in red and including that line on my graph represents a solution to the original inequality.0301

When you have an inequality that has the or equals to it, we usually use a solid line to show that everything on that line is actually one of our solutions.0311

Even if I use to pick like a point right on that line it should satisfy the inequality.0322

If it was a strict then we will use a nice dotted line to show that the line should not be included.0328

Now that we have the basics to graphing out one of these regions, let us give it a few more try.0338

We know that our equation will split the region up in the two regions and that the equation will either have to be solid or dash.0348

It is determined by the type of inequality used.0358

If we are dealing with a strict inequality like strictly less than or strictly greater than, 0363

feel free to use a dotted line because you should not include the points on that line.0367

If you have the or equals to in there and use a nice solid line to show that the line is included. 0373

To determine the region, shade on one side or the other, use a test point to help you out,0381

that determines is that point and the region it is in included or not.0388

Pick something inside the region, you do not want to pick something on the line.0395

If the point satisfies your inequality, go ahead and shade that entire region that it is in.0399

If it does not satisfies then feel free to shade the other side, which it is not in.0405

One last thing if I could add to this, go ahead and pick something easy.0410

Potentially you could pick any point in one of these regions and you have some freedom, pick something that is nice and easy to evaluate.0420

I think we have our tools, let us go ahead and go to one of our examples.0428

We want to graph the inequality 3x -2y > 6.0435

First, I'm going to look at this as if it was a line.0440

Let us use our little equal sign in here and that is the corresponding equation.0446

The tools that I will use on this one is, maybe I will look at the intercepts.0450

What happens when x = 0 and y = 0.0456

Plugging in 0 for x, I would have 0 × 3 =0 and that term would drop away.0459

Then I divide both sides by -2 giving me -3.0470

I have one of my points, let us go ahead and put that on there 0, -3.0479

Let us put in a 0 for y.0487

It looks good -2 × 0 = 0 so that is gone and dividing both sides by 3 we will get 2.0495

I have another point I can go ahead and put on this graph 2, 0.0508

I’m going to use these points to graph the line but I’m going to be very careful to make it a dotted line since our inequality is a strict inequality.0514

I do not actually want to include this.0522

This looks good, nice big dotted line. 0541

We have to figure out which region we should shade, either on the left side of this or on the right of this.0545

To help out, I'm going to borrow a test point just to see what is going on.0551

We could choose anything on here but I’m going to choose something nice and simple.0556

We are going to choose the origin since they are 0 for x and 0 for y.0561

Let us see if this works.0566

0 for x and 0 for y is that greater than 6?0568

0 × 3 and 0 × 2 both of those would cancel.0576

Is 0 > 6? I am afraid not, it is not greater than 6.0580

That tells me that origin is not in my solution region and I should shade the other side of the line.0586

Let us do that then.0593

This entire region would be considered as our solution region and we have graphed the inequality.0604

Let us try another one of these.0619

This one is to 2y > or = 4x -2.0620

Let us start off like before, looking at the corresponding equation and see if we can graph that .0626

2y = 4x -2 this one is almost in slope intercept form, I’m going to divide everything by 2 and put it into that form y = 2x – 1.0635

Now I can see that it has a y intercept at -1 and a slope of 2.0649

Let us put those points on them.0656

-1 and from there I will go up to over 1 and get a few more points.0661

I can definitely graph out my line and I will make it a solid line, since my inequality has the or equals to. 0669

There we are, almost done.0687

We just have to figure out what side we should shade on.0690

Let us pick a nice easy point that we can evaluate and see where we go from there. 0695

I already used the origin last time, so I'm going to pick something a little bit farther away.0699

Let us try this point right here, -2, 2.0706

When y = 2 is this greater than or equal to 4 × -2 -2.0715

I’m going back to the original.0730

I have put in the point into x and y.0731

2 × 2 = 4, 4 × -2 = -8 -2, is 4 > or = -10.0735

4 is positive and -10 is negative, positive number should be greater than negative numbers.0747

I know that point is in my solution region, I should shade everything on that side of the line.0752

This one we are not only including the entire region that I'm shading in right now but we are also including everything on that line since it is nice and solid.0763

There is our inequality all graphed out.0782

In this last example, let us go ahead and work in the other direction.0786

You will notice that I have an inequality I already graphed.0790

We are going to try and create what that inequality was.0793

When we are going and building these from the ground up, we focus on the equation associated with it first.0798

We are going to start in the same spot.0805

When you look at this dotted line and I can see that it has a y intercept up here at 2.0807

I think I can also interpret it slope.0813

To get back on the line I have to go down 2 and to the right 3 to get back on the line.0816

I definitely know its slope as well. 0823

The equation of the line be y = -2/3x + 2, that worked out pretty good. 0826

I can see that it is dotted so whatever inequality I’m going to deal with next should be nice and strict, either less than or greater than.0838

The question is which one should I include?0848

I want to make sure that it has the proper solution region.0850

In order to help me figure this out, I’m going to borrow a point from inside that solution region.0854

Let us borrow the point 2, 2.0860

I’m going to plug it into both sides of the equation and figure out what inequality symbol should go in between and make the rest of the statement true.0863

I have 2 for y and I will put in 2 for x.0872

Let us just go ahead and simplify this and see how this turns out.0879

2 is already simplified, over in this side I have -4/3 + 2, I have 2.0883

-4/3 + 2 I will have 2/3 left.0892

What is greater 2 or 2/3?0899

That does not take too long to figure out, 2 is definitely greater.0903

We will go back and put it into all of the spots all the way back up here and that way we can actually build our inequality y > -2/3x + 2.0908

That is the one represented by this graph.0924

You can notice how I’m using a strict inequality because it is a dotted line.0927

Thank you for watching www.educator.com.0931

Welcome back to www.educator.com. 0000

In this lesson we are going to wrap things up with inequalities by looking at systems of inequalities.0002

We do not have a whole lot to cover but we will look at how you can satisfy a system of inequalities rather just individual inequalities.0011

That will bring us to looking at overlapping solution regions.0019

Recall it when we are usually dealing with the system, such as a system of equations,0027

we want to make sure that all of the equations in the system are satisfied.0031

When we are dealing with the system of inequalities, rather than equations, but it is the same goal. 0037

We want to make sure that every single inequality in here gets satisfied.0042

This makes it a little unusual because rather than just looking for individual points that would satisfy inequality0048

remember that each of these is a region on our graph.0054

What we are looking for then is where these regions for each of them would end up overlapping.0059

That overlapping region would be where the entire system is satisfied.0065

Like before, after we are done identifying an overlapping region, we can use a test point to go ahead 0070

and figure out whether that is the correct overlapping region.0077

It is not about process, we just have to get into figuring out what those regions are.0081

What we are going to do to basically figure out the overlapping region is, first we will graph each individual inequality in that system.0090

That breaks it down into figuring out the equation and figuring what side is shaded on that line.0099

Be very carefully that you are using either a dotted or dash line depending on the type of inequality in there.0107

We are using the dotted for our strict inequalities and we are using the nice solid line for our or equals to.0113

Since the line could be included as a solution.0122

Individually, we will make sure we shade on one side or the other and we can check that out using a test point.0126

Here is the big one that we have for our system, look for where the two regions overlap so we know where the solution to the system is.0132

That is a little bit of background, let us see these examples and see how this works in practice. 0145

In my system of inequalities, I have 2x + y < or = 1 and I have x > 2y so individually let us take a look at these.0150

The corresponding equation to the first inequality would be 2x + y = 1.0161

If you want to graph that one out, you could get that one into slope intercept form by moving the 2x to the other side by subtracting a 2x.0169

From this I can see it has a y intercept of 1 and the slope of -2.0181

From that point I go down 2/1, down 2/1.0186

Looking at the original I can see it involves an or equals to so I'm going to use a nice solid line. 0192

Now we have our equation, we need to figure out what side to shade it on.0212

We usually grab a test point to figure out what it should be and it is good to grab a nice test point that is to easy to evaluate.0216

I’m going to borrow the point 0, 0 here and see if it works in the original.0224

2x is 0, y is 0 is that less than or equal to 1, let us find out.0231

2 × 0 + = 0 and 0 < or = 1.0240

I know I should shade on that side of the line.0248

This entire region that we can see right now is only for the first inequality in our system.0255

Simply just going to go through the same process for the second inequality and figure out where both of their regions overlap.0265

Let us get some space and let us see this process for the second equation.0278

The second equation is x > 2y.0289

Maybe I will find this one in its slope intercept form so I will move the two together side by dividing everything by 2.0293

I could look at this as the equation y = ½ x, it has a y intercept of 0 and a slope of 1/2.0305

It goes through the origin 0 up 1/2, up 1/2, up 1/2, up ½.0317

Let us use a dotted line since the inequality is strict and see what we have.0325

We just need to shade on one side or the other.0348

We could test point for this particular line, let us just choose one way out here.0353

This one is at 3, 3.0358

Plug 3 in for y and plug 3 in for x and we will see if it is true not.0363

Is 3 < 3/2? That is not looking so good, that is not true.0376

What that means is we should actually shade on the other side of that line.0386

Let us shade everything below it.0390

Now comes the very important part of this.0402

We can see where each individual inequality is being satisfied but since we are interested in the region where the entire system is being satisfied,0406

we are looking at the overlapping regions.0418

I’m going to highlight that.0421

This region right down here is where the two overlap and I can see along one of the borders, it is dotted and along the other border it is solid.0424

This yellow area and its borders would be considered the solution to the system.0440

Let us try another of these and be careful as we walk through the process so we can see where that overlap is.0450

In this one we are dealing with the system 3x - 2y > 12 and 5x - y < 6. 0455

Let us grab the first one and look at its corresponding equation.0463

The form of this ones in would be nice if we just use our intercepts to go ahead and see where it crosses the x and y axis.0477

I will put in a 0 for x and 0 for y.0486

If x = 0 what is y?0491

This term would go away for sure and I can divide both sides by -2 so y = -6.0496

Let us go ahead and put that on our graph, 0, -1, 2, 3, 4, 5, 6 right there.0507

We will go ahead and put in a 0 for y.0515

Let us see how this one turns out, 2 × 0 = 0, 3x = 12 divide both sides by 3 now and we will get x = 4.0524

4, 0 is our other point.0536

Now that we have 2 points, let us go ahead and connect it.0541

We will connect it using a dotted line and then we will figure out what side we should shade this on.0544

To determine what side to shade it on, let us borrow a nice, good test point.0564

Let us borrow the origin add 0, 0 and see if it is true or not.0569

(3 × 0) – (2 × 0) > 12? That would simplify to 0.0574

Is 0 > 12? No.0583

Let us shade the other side.0587

That entire side would be our solution region or least for the first one.0602

Let us get some space and do the same exact process for our second equation.0608

The corresponding equation would be 5x - y = 6. 0622

This looks like maybe be easier to graph it if it was in slope intercept form.0629

I’m going to move the 5x to the other side and then multiply everything by -1. 0634

Okay, my y intercept is way down there at -6 and it has a slope of 5 so up 1, 2, 3, 4, 5/1, up 1, 2, 3, 4, 5/1.0649

I will graph this one using a dotted line.0669

I have a few regions that it might be if we want to figure out until we shade one side of the blue line or the other. 0692

Let us grab a test point for it.0701

I think origin 0, 0 is going to work out very nicely for this one as well.0702

(5 × 0) – 0 is that less than 6?0708

The entire left side simplifies to 0 and 0 < 6.0713

Let us go ahead and shade everything on that side of the line. 0720

There is a lot of region over here to shade.0734

Now comes the important part where exactly do both of these overlap?0744

This is a little tough to tell, but if you look near the bottom you can see that there is the small triangle here where both of them overlap.0751

We would have that as our solution region.0761

As long as it is in the overlapping portion we are good to go.0765

For this last example I wanted something that was a little bit more like a word problem.0773

You could see how inequalities have a lot of good applications in them 0777

and how some more problems lead to the rise of a system of inequalities rather than a system of equations. 0781

What we want to do for this one is graph the solution region for the system and interpret what that means in the context of the problem. 0789

What we are looking to do is recreate what it means when a person's heart rate, 0797

or a person's maximum heart rate 220 – x, where x is the person’s age.0802

This one is good as long as it is for people between 20 and 70 years old.0808

When a person exercises and are trying to lose some weight, it is recommended that the person strive for heart rate 0815

that is at least 60% of the maximum and at most 70% of the maximum.0820

There is a little bit of a range for that maximum heart rate that we are aiming for.0826

Let us see if we can interpret some of the things in here and create an inequality for what we have. 0832

First of all, let us talk about a person's age.0839

Earlier it says that x is our person's age and that we are only interested when x is in between 20 and 70.0844

Let us say x > 20 and x < 70.0856

There are two inequalities that we can put in our system.0864

We also want to look at a person losing weight. 0868

Also, it recommends that they strive for 60% of their maximum and no more than 70% of their maximum.0873

What exactly is their maximum?0880

It is way up here at this expression, 220 - x. 0883

60% of their maximum 3.60 (220 – x) and we want to make sure their heart rate is greater than that.0890

You do not want to stress them too much, so we will make sure that it is less than .70 of the maximum heart rate. 0915

What we now have here are 1, 2, 3, 4 different inequalities in our system that we will be looking to satisfy.0924

And we will graph this out so we can see the region it creates.0934

Let us end up rewriting this just so we have them handy.0943

x must be greater than 20, x is less than 70, 0947

and we also have two more, H is greater than (.6 × 220) - x and H is less than (.7 × 220) – x.0955

If I'm going to graph these other two here I want to get them in a better form. 0974

It is tough to see maybe what their slope is and what their y intercepts is.0980

Let us do a little bit of work with distributing that .6 and .7 in so we have a better idea.0985

This one will be H > .6 × 220 = 132 - .6x and for this other one we have 154 -.7x, that gives us a much better idea. 0995

I think I’m going to write them again so I can put that x term first and see that is my slope.1020

We will end up graphing these two inequalities in our system as well as these two on our system.1039

Let us see how we can do that.1046

With the first two that have to deal with the age of the person, we will use that along our x axis.1048

We want the age of the person to be between 20 and 70.1059

Let us use a nice vertical line and I’m going to make this dotted because my inequality is strict.1068

Let us do the same thing for our 70 here.1080

If I was looking to just satisfy those first two then I would have to be between of these two lines. 1092

Let us work in what we want their heart rate to be and see how that fits into this. 1104

In the first one I can see it has a y intercept of 132 and the other one has a y intercept of 154. 1113

My slope on the first one is a -.6x so since it is negative it should be going down, maybe something like this.1127

In the other one is very similar its slope is negative as well, but it is a little bit steeper since its .7 and we will put that on.1153

Now where should the shading for this one be?1174

We want our heart rate to be more than the .6x + 132.1177

We want to think of that as our minimum heart rate that we are aiming for in terms of exercise.1183

The other one would be a .7x + 157 we want them to be no greater than that value.1189

We are looking in between these 2 lines again. 1195

From looking at these two solution regions, it is this little box looking thing where they end up overlapping.1202

We can interpret what that solution region means in terms of the problem. 1216

If you we are to just pick a random point inside that box, what we are getting is maybe an age of a person, 1221

let us say 65 and we are also getting what the heart rate should be when they exercise to try and lose weight. 1231

As long as they stay inside that box they should do some good things to their heart.1239

If we try and fall outside of the solution region that means either the formula is for our maximum heart rate no longer apply1245

or we would end up overtaxing the heart and either one we definitely do not want.1254

When setting up a system of inequalities look at them individually and graph out each of the regions.1261

Then look at the overlap and see what would be included in both.1267

Thank you for watching www.educator.com.1271

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at integer exponents.0002

What does it mean? You take a quantity and raise it to a power.0005

A lot of other things we will also be picking up along the way.0010

Let us get a good overview.0013

We will start off by interpreting what it means to raise something to a power when that power is a nice whole number.0015

We will see that we can often combine things using the product and the power rule because all these things have the same base.0021

We will see what it means to raise things to a 0 power and get into the quotient rule.0028

And get into understanding what happens when you raise something to a negative power.0034

Watch for all of these things to play a part.0038

One way that we can look at multiplication is that it is a packaging up of addition.0046

I have the value of x and we know that I have added together 5 times.0053

Rather than trying to string out to x + x + x over and over again I can write that same thing by just saying 5 multiplied by x.0059

It is a great way to take a long addition problem and package it all down into a nice single term. 0074

Exponents are the things that we are interested in as long as we are dealing with whole numbers you can look at that as packaging up some multiplication. 0081

Here I have those x’s once again, I have 5 of them all being multiplied together but rather writing all that, I'm going to use my exponents.0090

Specifically the bottom here will be our base. 0105

Remember, that is what we are multiplying over and over again.0108

The exponent itself tells us how many times.0111

Let us get some practice with just understanding how exponents work before moving on. 0121

With this, I simply want to write them into their exponential potential form.0128

If I have 5 × 5 × 5, I can see that it is the 5 being repeatedly multiplied and I did it 3 times.0132

I could consider this one like 53.0139

If I wanted to take this one little bit farther, this is equal to 125.0143

You have to focus on what is being multiplied over and over and over again to properly identify your base.0150

In this one, we have -2 × -2 × -2, we can see that it is being multiplied by itself 4 times.0157

We want the entire value of -2 multiplied by it self 4 times so we will use an exponent of 4.0167

If we had to simplify this one, I have -2 × -2 × -2 × 2 and that would be 16.0177

Keep an eye on the base, so you know what is being multiplied and the exponent will tell you how many times to do that.0186

In this next two, we just want to evaluate the exponential expression.0196

Let us see if we can identify the base here.0200

This is a 34 that happens to be negative sign out front and my next example, it is almost the same thing but it is -34.0203

Watch how I treat both of these a little bit differently. 0212

In this first one, the only thing that is considered the base is actually that number 3, 3 × 3 × 3 × 3 multiplied by itself 4 different times.0215

What should I do with that negative sign out front, it is still out front, but since it is not part of the base and it only shows up once.0228

I can take care of it from here, 3 × 3 × 3 × 3 a lot of 3’s in there, but that will equal 81.0235

Of course that negative sign is out front along for the right.0244

In the next one, these parentheses are highlighting that the entire -3 is what is in the base so I'll repeat the entire -3. 0248

Look at this one, I'm dealing with a -3 multiplied by itself 4 times.0267

I know that a negative × negative would be positive and another negative × negative would be a positive, 0272

the answer to this one action turns out to be 81.0278

Use those parentheses to help you know what is in the base and it is good for all of those negatives.0283

One of our first rules that we can throw into the mix is the product rule for exponents.0293

This will help us when we have two things and both of them are raised to an exponent.0299

It is a great way to actually put them together so they only have one thing raised to an exponent.0306

The way this rule works is, we want to make sure that both the bases are exactly the same.0311

I have a and a as my base, if they are the same then it says we simply need to take each of their exponents and add them together. 0318

It seems like a little bit of an odd rule how you can always start with multiplication and end up with addition 0327

but let me show you an example of why this works.0333

I’m going to base that off just what you know about exponents that it is repeated multiplication. 0336

Here I have 24 power be multiplied by 23, let us pretend I knew nothing about the product rule.0342

I can start interpreting this by using repeated multiplication, so 24 will be 2 × 2 × 2 × 2 until I done that 4 times in a row.0350

I can do the same for 23 that would be 2 × 2 × 2.0364

You can see what we have here, the whole bunch of 2s are all being multiplied together and there is even no real need to put on those parentheses.0373

I will just write 2 × 2 × 2 until I got all of them written down. 0381

As long as I have all of these 2s written out, it was said earlier that exponents are just repeated multiplication, 0388

I can actually package these all backup.0394

If you count how many are here, it looks like we have 7 total.0398

I can say that this is 27.0404

Now comes the important part, if you look at our original exponents that we have used the 4 and 3 and you add them together 0409

you can see that sure enough, it does add to 7.0418

The reason why this is happening is because you are simply throwing in a few more numbers to multiply by and incrementing up that exponent.0421

That is why the product rule works.0431

Let us go ahead and practice with it now.0434

In our first one I have -75 × -73.0439

The first thing you want to recognize is both of those bases are the same, so we will just focus on their exponents the 5 and 3.0445

This one as -7^(5+3) and then I can go through and add those 2 things together, -78.0456

You could simplify it further from there if you want to but I’m just going to leave it like that so you can see the product rule in action.0472

The next one will involve a little bit more work and that is why I put in the -4 and 3.0479

Before I can get too far, I’m going to start rearranging the location of the 4 and the p's raised to the exponents.0488

What allows me to do this is the commutative property for multiplication that the order of multiplication does not matter.0495

This is just helping me organize things a little better so that I have my numbers all in one spot and my exponents on one spot as well.0504

That looks good.0513

Put the -4 and 3, I will just straight multiply those two together and get -12.0515

With the other two, I can see that they are both the same base of p.0521

I will look at adding their exponents together so p^(5+8).0526

Here is -12 p13 not bad.0535

One last, let us see if we can use the product rule in that one.0542

Looking at this one you will know that they both have a base of 6.0546

One is raised to the 4th power and one is raised to the power of 2. 0549

But there is a slight problem with this one.0553

This one is dealing with addition and the product rule, the one that we are interested in is for multiplication.0557

Since the product rule only applies to multiplication, it means we can not use the product rule here.0570

Even though that my bases are the same it simply not going to work here. 0584

If I did want to continue this one out, I would not use the product rule. 0589

I would just simply take 64 and see what value that is.0593

I will get 1296 then I will get 62 = 36 and I will add the two together 1332.0597

You can notice how we do not use the product rule since it has addition there on the bottom.0608

Let us try a few more examples and see if we can recognize our product rule in action.0619

I have thrown in a few more variables this time.0623

The first one is n × n4.0625

It is tempting to say hey maybe this one is just the n4, but if you see missing powers like on that first one, assume that they are 1. 0629

That way you actually have something to add.0638

This would be n1+ 4 and now that they actually have your exponents added together, we will get n5.0640

Now you know if you strictly look at the product rule it only talks about putting two things together, you can even apply it for more than two things.0655

In this one, I have z2 × z5 × z6 and we can apply the product rule as long as we do it two at a time.0664

Let us just take the first two z’s here. 0673

This would be z^(2+5), we will get to that other z in a bit.0677

I can see that I have a single z^(2+5) and z6, that simplifies into z7.0686

I can put these two together z^(7+6) and that would be z13.0696

We could have even taken a much bigger shortcut if we simply just added all 3 of them together.0707

You can definitely do this as long as all of their bases are the same, and you are dealing with multiplication.0716

One more, this one is 42 multiplied by 35.0723

We are definitely dealing with multiplication but again, this one has a little bit of a problem.0730

Notice how this one, the bases are not the same. 0737

The condition for using that product rule is that we need the base is the same 0748

and we need multiplication and so we simply can not use the product rule here.0752

If we are going to go any farther with this one, we have to evaluate these exponents.0756

42 would be 16 and then I will go ahead and multiply that by 35 = 243 and then multiply the two together, so 3888.0762

Be very careful that the conditions of the product rule are met before you try and use it.0787

Another good rule that we can use for combining things together is known as the power rule.0797

The way the power rule works is, we have something raised to an exponent and then we go ahead and re-raise all of that to another exponent.0804

It is like we have exponents of exponents.0812

The way the power rule handle these is we will take both of the exponents and we will actually multiply them together.0816

To convince you that this is the way it should work, we are going to deal with another example that has just numbers in it, 0826

and show you that this is the way you should combine your exponents in this case.0833

We are going to take a look at 43 and all of that is being squared. 0838

Say you knew nothing about the power rule, how could you end up interpreting this?0845

One way is you could use again repeated multiplication to interpret your exponents. 0849

In the first exponent I'm going to interpret is this two right here.0856

I’m going to take my base, 43 and it will be multiplied by itself twice, you can see I have two of them.0860

I want to go further with this, now I will interpret both of these 3s as 4 being multiplied by itself 3 times.0873

Now that I have expanded it out entirely, I’m going to work to go ahead and package it all up.0891

What I can see here is that I have a whole bunch of 2s and they are all going to be multiplied together.0897

In fact I have a total of 6, this is 46.0905

Go ahead and observe exactly what happened with the exponents.0913

Originally we started with 3 and 2 and if you multiply those together like the power rule says we should sure enough, you get 6.0917

If you fall along the process you can see why that happens. 0925

The power rule which we learned earlier can be applied to multiplication and division.0937

The way it does it is we can take two things that are being raised to a power and then that giving each of them that exponent.0942

When dealing with division, if we have things being divided, say, a ÷ b we can take that exponent and give it to each of them.0953

What I’m trying to say here is that if you have an exponential expression and it is raised to a power,0964

you are going to apply it to all the parts on the inside as long as those parts are being multiplied or divided.0968

This one helps us when we get to those larger expressions.0973

Let us try a few of these up, we simply want to use the power rule for each expression rated out.0981

The first one is 62 and all of that is being raised the fifth power.0988

I'm not going to expand that out, I’m just going to go directly to the rule.0993

This would be 6 then we have 2 × 5.0998

Taking care of just the exponents this would be 610.1004

I could continue to try and figure out what number that is but I’m only interested in considering what the power rule is.1010

Let us try another one.1017

This is z4and all of that is being raised to the 5th.1019

I can take my exponents and I will multiply them together. 1026

This will be z20.1031

Let us get into a few that are a little bit more complicated and see how we can tackle this.1036

In this next one I have 3 × a2 × b4.1042

One of the first things I need use is my power rule and give that 5 to all of the parts.1047

Let us write that to each of these.1057

I’m going to do the 53, I'm going to give it to the a2 and I'm going to give it to the b4.1061

It must go on to all of those parts and it is okay since all these parts were multiplied in the original.1070

Now that I have this, I will go ahead and start to simplify each of these.1077

My 35 is 243, then I will multiply these exponents together and get a10.1086

We will multiply the 4 and 5 together and would give us b20.1097

This one is a completely condensed down far as I can go.1104

Let us see one that involves division.1109

I'm going to give the 3 to the top and bottom.1116

The top will be 33 and the bottom will be x3.1121

This will give us 27 all over x3.1133

Let us try one more 4 + x2.1140

What does the power rule say we should do with this one?1145

I threw this one in there because you should not use the power rule on this one. 1149

Why not? Why can not we use the power rule here?1156

Notice the parts on the inside, they are being added 4 + x instead of being multiplied.1159

Since it is being added instead of multiplied, I can not simply give the 2 to all the parts.1170

This situation down here is completely different from this one because of that addition.1175

How could I take care of this?1181

What we will see in one of our future lessons is that we will handle this one using repeated multiplication1185

and we simply have to multiply those two terms together.1193

For now I want to point out is you should not use the power rule on that situation simply does not work.1196

All of these rules are extremely important and you should be very comfortable with them.1207

I recommend memorizing all of them.1212

Usually a lot of practice with them will get you comfortable with recognizing when you should and should not use them. 1214

In addition, you should be comfortable with these that you can use more than one of them in a single problem.1223

Maybe you will end up using the power rule and then end up using the product rule all in the same one to condense and simplify an expression.1229

That is how comfortable with these rules you should be. 1237

We can get a better handle and the least a little bit more comfortable with these we will give it a try.1243

We will try and use many of these rules together to simplify the following expressions.1248

Okay, this first one is (5k3 / 3)2.1253

One of the very first rules that I'm going to use, is I'm going give that 2 to the numerator and to the denominator of that fraction.1262

You got to be careful in doing this, the top is 5k3 and the bottom is just 3.1272

We will give 2 to each of those.1280

Now that I have taken care of that rule I can see I have a little bit more of work to do on top. 1285

I have multiple parts in there all being multiplied and I’m going to give the 2 to each of those parts. 1289

Here is my 5k3 so we will give the 2 to the 5 and we will give the 2 to the k3.1296

Now that we spread out our variable amongst all of the parts that are being multiplied and divided, let us see if we can continue.1308

52 that would be 25 and now I have k3 and all of that is being raised to the power of 2.1316

What rules should we use for that?1326

As we only have a single base in there, this is that situation where we multiply the two together so this will be k6.1330

Now I simply have 9 on the bottom from 32.1341

I do not see anything else that I can combine or any other rules that I can use, this is good to go.1346

Be careful in using multiple rules.1353

Let us take a peek at the second one here. 1357

This one is (-3 × x × y2)3.1359

In addition that is being multiplied by (x2 × y)4, lots and lots of different exponents flying around.1366

Let us recognize that we do have some multiplication on the insides of those parentheses, 1376

I will be able to spread out that exponent amongst all of its parts.1381

What parts do I have in there?1386

I have -3, x, y2 so we will put 3 on each of those.1388

Let us do the same thing for the other one.1403

I have x2 and y4, both of these looks like they need 4.1406

Now that I have applied that rule, let us go through and start cleaning other things up.1421

Starting at the very beginning I have -33.1427

That is -3 × -3 × -3 = -27.1431

Right now I have x3 just as it is.1440

It looks like this next part that is a good place where I can multiply the exponents together, y6.1445

Here is a couple of more situations where again I can just multiply those exponents together x8 y16.1456

It is important not to stop there because all of these pieces are still being multiplied together.1470

I can see that I have a couple of x and a couple of y.1476

Let us use our commutative property for multiplication to change the order of these parts.1481

x3 x8 y6 y16.1488

The reason why I’m doing this is so we can better see that I need to apply one more of our rules that will be the product rule.1494

Both of these have the same base and I can simply add their exponents together, the 3 and 8.1502

-27 x^(3+8) is 11 and we can do the same thing down here with our y since they have the same base as well and are being multiplied.1512

y^(6+16) would be 22.1531

We finally got down to our answer and apply that to all the rules that we could.1538

You can be very comfortable with mixing and matching and putting these rules together.1543

Just be very careful that you do them correctly.1548

A very important rule that we will need is the 0 exponent rule.1555

What this rule says is that when you raise anything to the 0 power like a over here, that what you get is simply 1.1561

This is probably one of the most curious rules that you will come across.1570

After all, usually when you are dealing with 0, you are familiar with multiplying by 0 and getting 0 or even adding 0 and do not change anything.1573

Why is it that when we take something to the 0 power, it becomes 1?1581

One reason that we do this is if we want to be consistent with the rest of our rules. 1587

After all, we have been building up a lot of other rules, combining things, our repeated multiplication, we wanted to the mesh well with those other rules.1593

Let us see how it actually does need to be defined as 1, so that it fits with everything else we are trying to do.1600

I’m going to look at 60 × 62 and we will do this twice. 1609

The first time that I go through this, I'm going to end up using my product rule.1617

The reason why I’m doing this is that I have the 6 and the 6 are both the same base and that rule says I need to add the exponents together. 1622

This would be 6^(0+2) now the addition is not so bad 0+2 is 2 so this would end up being 36.1631

For this rule to stay consistent with that, I need it to also equal 36.1648

You can see that if I look at just 62 by itself, this guy over here that does equal 36.1656

What should I call this thing to multiply by 36 so that my final result is 36.1666

You should not have to think on it to long, there is only one thing that I can call 60.1676

I must call it a 1 so that when it is multiplied by that 36, it still is 36 as the final result. 1683

It is the only way that we can define something to the 0 power so that it mixes well with all the rest of our properties.1691

Let us play around this rule a little bit.1701

This new rule is the quotient rule.1707

In this one we are dealing with two things that have the same base so a and a and they are being divided.1710

When we divide like this, it looks like we can simply take their exponents and subtract them.1720

It is a lot like our product rule only that one dealt with multiplication and addition, this one has division and subtraction. 1726

Some important things to know of course we must have the same base for this to work out.1735

We are dealing with division and subtraction.1740

This is a quick example to see why this might work.1743

Let us look at 54/ 52.1747

Suppose I knew nothing about the quotient rule, one way that I could just end up dealing with this, is looking at it as repeated multiplication 5 × 5 × 5.1754

Then I could look at the bottom and say okay what is 52?1768

That would be 5 × 5 and then I can go through canceling out my extra 5 in the top and in the bottom.1774

This would mean 5 × 5, which is the same as 52. 1784

If we look at the result of this, you know what happened to the original versus the answer in the end, 1792

you can see that we also have to is subtract those exponents.1798

4-2 does equal 2.1801

This is the way that we will handle things that have the same base and we are dealing with division.1804

We will simply subtract their exponents. 1810

Now, unfortunately this does lead to a very interesting situation and potentially it might give us some negative exponents.1816

To see why this might happen, let us use an example with numbers so 1825

we get a good sense of some of the strange things that we want to be prepared for.1829

This one I have 24 ÷ 25.1833

The first way I’m going to handle this, is I am going to use my rule for the quotient rule.1838

This will be 2^(4-5).1845

If you take 4 – 5, you will get 2-1.1852

Now I can see that is what the product rule tells me to do, but you know what exactly does that mean to have 2-1.1858

After all, when we are dealing with just whole numbers, then I would often look at this as repeated multiplication, but how can I multiply 2 by itself?1867

A -1 number of times and that just does not seem to make sense. 1875

To get a handle of what that means, I'm going to look at the original problem as repeated multiplication.1879

This would be 2 × 2 × 2 × 2 all over 2 × 2 × 2 × 2 × 2.1886

I have 4 of them on top and 5 of them on the bottom.1898

Let us go through and cancel out all of our extra 2s, 4 from the top and 4 from the bottom.1902

You can see from doing this, there is only one left and it is on the bottom so this would be ½ .1911

Now comes the important part that if we look at each sides of our work out, we have used valid rules and we have done things correctly.1918

What I have here are two different expressions which are the same thing.1927

This gives us a good clue on what we should do with those negative exponents. 1934

We want to interpret our negative exponents by putting the base in the denominator of our fraction.1938

It is how we will end up handling these things.1945

It will take a little bit more work to get comfortable with these, let us see what else we can do.1949

The rule that we have just developed is anytime we have a base raised to a negative exponent we will put it in the denominator of the fraction1960

and we will change that base, a rule change that exponent to a positive number. 1969

Another way to say that is if you have an expression raised to a negative power it can be rewritten in the denominator with a positive power. 1975

It leads to something very interesting. 1983

If you have a negative in the top of your fraction and a negative in the bottom, 1986

then what you can end up doing is changing the location of where those things end up.1991

I have a^-n and it was on the top but now it is in the bottom. 1997

I had b^-n to in the bottom, now it is in the top.2003

A good rule of thumb that I give my students to keep track of what to do with that negative in the exponent 2010

is think of it as changing the location of where something is.2016

If the negative exponent is in the numerator, think of the top, go ahead and move it to the bottom at your denominator and make it positive. 2020

The other situation, if it is already in the bottom at your denominator, then move it to the numerator the top and make it positive.2030

Now that we know a lot more about our exponents and especially those negative ones, let us get into using them a bit more.2047

Here I have ¼-3.2055

One of the first rules that I'm going to use is to spread out that -3 onto the top and bottom of my fraction.2062

I have 1-3 / 4-3.2070

Here is where I’m going to handle that negative exponent.2077

That 1-3 on the top, I’m going to move it to the bottom and make its exponent positive.2080

I’m going to do the same thing with a 4.2090

4-3 now it will go to the top and its exponent will now be positive as well.2091

From here, I just go through simplify it, 43 = 64, 13 = 1 so it is 64.2098

Let us try the same thing with the next one and notice how things are changing location.2116

2-3 will end up in the bottom as 23.2123

3-4 that one is going to go into the top as 34.2130

I will go ahead and simplify from here.2139

There may be a few situations where some things will have positive exponents and some things will have negative exponents. 2144

The only ones that will change locations will be the ones with negative exponents.2155

If I have an example like x2 and y-3, then the x2 will remain in the same spot but the y must go into the bottom since it had the negative exponent.2160

Watch out for those.2172

Let us see if we can definitely combine many more of our rules together and simplify each of these expressions.2177

We got lots of rules to keep track of so we will just do this carefully, bit by bit. 2184

In this first one I have the (x2 / 2y3)-3.2191

One of the biggest features I can see here is probably that fraction.2198

I’m going to give the -3 to the top and the bottom of my fraction.2204

I have x2 and will give it -3 and I will do the same thing with the bottom. 2209

Now one thing I can see on the top is I have x2 and that is being in turn raised to another power.2218

Our rule for that says we need to multiply the exponents x-6.2225

What to do with the bottom?2233

In the bottom we have some multiplication in there, so 2 × y3.2234

I need to give that -3 to each of the pieces down here.2239

Moving on, bit by bit, I will deal with that negative on the x and eventually that negative on the 2.2250

Let us see what we need to do with the y3 and y-3.2258

The rule says I need to multiply those things together and get -9.2262

I think I have used all of my product rules and power rules and quotient rule.2269

It is time to use that negative exponent rule.2273

Anything that has a negative exponent on it is on the top and I’m going to put in the bottom.2277

If it was on the bottom, it is going right to the top.2281

In the bottom, x6 and that guy is done.2287

In the top, 23 and y9.2292

One last thing to go ahead and clean this up, 23 is 8.2298

I will just ahead and put it in there.2303

Our final simplified expression for this one is 8y9 / x6.2307

It is quite a bit of work, but it is what happens.2314

Let us do another one. 2318

This one is for 4h-5 / m – (2 × k).2320

I do not see a whole lot of rules in terms of spreading things out over multiplication, but one thing I do want to do is take care of those negative exponents.2327

I have one of them up here on the h and another one here on the m.2336

Let me first write down the things that do not have negative exponents, they will be in exactly the same spot. 2341

The h-5 I need to put that in the bottom as h5.2348

The m-2 it now needs to go to the top as m2. 2354

Okay, I would continue simplifying from here if I could but I think I do not see anything else that has the same base.2360

I will leave it as it is and will call this one done.2368

It is time to put all of our rules in practice and see if we can do a much harder problem.2376

This one is (39 × x2 × y)-2 / 33 × x-4 × y.2381

One thing I want to point out with this one is that when you are applying the rules, you do have a little bit of freedom on which ones you do first. 2395

Experiment with trying the rules in a different order and see if you come up with the same answer.2403

As long as you apply the rules carefully and correctly, it should work out just fine.2408

What shall we do with this one?2414

I'm going to go ahead and take that -2 and spread it out on my x2 and y since both of those are being multiplied on the inside there.2417

39 now have an (x2)-2.2425

I also have a y-2 as well.2434

I will put that -2 in both of them.2438

Let us go ahead and multiply our exponents here, 39 then 2 × -2 would be x-4 y-2.2448

Let us see where can I go from here. 2471

I like dealing with those negative exponents and changing the location of things so I do not have to worry about negatives.2472

Let us do that first. 2479

Everywhere I see a negative exponent that part will change its location.2481

I’m going to leave the 39 for now and leave my 33 and let us take care of this one.2486

x-4 will be x4 in the bottom.2493

This one which is x-4, let us put that in the top.2499

y-2 is in the bottom and this y already has a positive exponent, so no need to change that one.2505

Continuing on, let us see what else we can do.2516

These 3s out here, they have exactly the same base so I can subtract their exponents, 9 – 3.2519

I can do the exact same thing with these x’s, subtract their exponents.2529

What should I do with those y’s on the bottom?2541

They have the same base, so I will add those exponents together.2543

Good way to start crunching things down.2550

This will be 36 x0 / y3.2554

36 = 729 x0, remember that is one of our special ones is 1, all over y3.2564

This one crunches down quite a bit, but in the end we are left with 729 / y3.2581

We do not have anything else to combine and so we will consider this one simplified.2589

Only one more to do this one, we want to make an expression that represents the area of the figure.2597

I want something that is a little bit more like a word problem, at least something that we have to drive out of.2604

We will simply use our rules along the way to crunch this down.2609

When I look at the area of a rectangle like this, it is formed by taking the width of that rectangle and multiplying it by the length.2613

I’m not sure what is the width and length here, since both of them are written as expressions.2625

I will write those in, area is the width 4x2 and the length is 8x4.2630

That means I can take this expression for the area and put it together using some of my rules.2640

Let us rearrange that.2646

I’m going to put my numbers 4 and 8 together, and my variables and exponents together. 2648

4 × 8 = 32 and I have x2 and x4.2657

I can do these by adding them together since they have the same base giving me x6.2665

This expression would be equal to the area of the rectangle.2671

It has quite a lot of rules to enjoy but again with a little bit of extra practice, they should become a little more familiar.2677

Watch for those special ones like raising something to the 0 power so that you know that is always 1.2683

Thank you for watching www.educator.com.2689

Welcome back to www.educator.com. 0000

In this lesson we are going to take a look at how you can add and subtract polynomials.0002

Before we get too far though or have to say what polynomials are and how we can classify them.0009

I think we will finally get into adding and subtracting them.0015

To watch for along the way, I will cover how you can evaluate polynomials for several different values.0018

Recall some vocabulary that we had earlier.0029

A term is a piece that is either connected using addition or subtraction.0033

In this little example down below, this expression I have four terms. 0040

A coefficient is the number in front of the variable.0051

The coefficient of this first term would be a 4 and I have a coefficient of 6, - 5 and 8 would be a coefficient.0055

We often like to organize things from the highest power to the smallest power.0068

The highest power here, it is a special name we will call this the leading coefficient.0073

This will be important terms you will often here me use later on when talking about polynomials.0083

Recall that things are like terms if they have the exact combination of variables with the same exponents. 0092

In this giant list here, I have lots and lots of examples of like terms. 0100

This first one is an example of like terms because they both have an m3 and notice it has no difference what the coefficient out front is.0106

I have 19, 14 but it does not matter that part. 0117

What does matter is that I they both have an m3.0121

The next two have both y9 and if you have more than one variable in there then both of those variables better matchup.0127

Both of them have a single x and they both have a y2.0136

If we understand polynomials and understand terms then you can start understanding polynomials.0144

A polynomial is a term or a finite sum of terms in which all the variables have whole number exponents and no variables up here in the denominator.0150

To make this a little bit more clear, I have many different examples of polynomials and many different examples of things that are not polynomials.0160

I will pick these over and make sure that they fit the definition.0168

This first one here, I know that it is a polynomial as I can see that has a finite number of terms that means that stops eventually.0174

If I look at all the exponents present like the two, then all of those are nice whole numbers and I do not see anything in the denominator.0182

In fact, there are no fractions there and we do not have any variables in the denominator.0191

That is why that one is a polynomial. 0196

I will put in this next example to highlight that you could have coefficient that are fractions0201

but the important part is that we do not have variables in the denominator, that would make it not a polynomial. 0206

It is okay if we have more than one variable like start mixing around m and p, just as long as the exponents on those state nice whole numbers.0216

Even the next one is a good example of this. 0227

I have y and x, but I have a finite number and eventually stops and it looks like I do not have any variables in the bottom. 0229

Now a lot of things get to be a polynomial, even very small things. 0239

For example, something like 4x is a type of polynomial, it is not a very big polynomial and has exactly one term, but is finite.0245

All of the exponents are nice whole powers and there are no variables in the denominator.0253

You can have even very, very short polynomials. 0259

This one is just the number 5 has no variables, or even consider it x0.0261

Compare these ones to things that are not polynomials and watch how they break the definition in some sort of way. 0267

In this first one is not very big, it is 1 ÷ x and it is not polynomial because we are dividing by x.0276

We do not want that x in the bottom. 0282

This next one is not a polynomial because of its exponents. 0286

It has an exponent of ½ and another exponent of 1/3 and because of those exponents, it is not a polynomial. 0290

The next one is a little tricky. 0299

It looks like it should be a polynomial, I mean I have 1, 2x to the first and 3x2.0301

The reason why this is not a polynomial has to do with this little…0307

That indicates that this keeps going on and on forever.0311

In order to be a polynomial, it should stop.0315

It should be finite somewhere.0318

That one is not a polynomial.0319

Other things that we want to watch out for is we do not have any of our variables and roots and none of those exponents should be negative.0323

That should give you a better idea of when something is a polynomial or not a polynomial.0334

We can start to classify the types of polynomials we have by looking at two aspects.0342

One is how many terms they have.0348

If it only has one term we would call that a monomial. 0351

In example 3x, it only has a single term, it is a monomial. 0357

If it has two terms then we can call that one a binomial, think of like a bicycle or something like that, two terms.0363

If I have 3 terms we will go ahead and call it a trinomial.0374

Usually if it has four or more terms you will get a little bit lazy we just usually call those as polynomials.0384

Technically, all of these are examples of polynomials but they are just a very specific type of polynomial. 0391

Let me write that one with a bunch of different terms, 5x4 - 3x3 + x2 - x +7 that would be a good example of a polynomial.0398

In addition to talking about the monomial and binomial, that fun stuff, you can also talk about the degree of a polynomial.0413

What the degree is, it is the highest power of any nonzero terms.0421

You are looking for that biggest exponent.0426

In this first one here, you can see that the largest exponent is 2, we would say that this is a 2nd degree polynomial.0429

In the next one, just off to the right, the largest power in there is a 3, so this would be a 3rd degree polynomial. 0440

Now you can combine these two schemes together and get specific on the types of polynomials you are talking about. 0452

Not only for that first one can I say it is a 2nd degree because it has the largest power of 2 in there, 0459

but I can say it is a 2nd degree binomial because it has two terms in it.0465

With the other one in addition to saying it is a 3rd degree polynomial, I can take a little further and say it is a 3rd degree trinomial.0473

That is because we have 1, 2, 3 terms present in a polynomial.0484

To evaluate a polynomial it is a lot like evaluating functions. 0494

We simply take the value that we are given and we substitute it for all copies of the variables.0498

Let us give that a try with 2y3 + 8y - 6 and we want to evaluate it for y = -1.0504

I will go through and everywhere I see a copy of y, we will put in that -1.0513

Let us go ahead and work on simplifying this.0527

-13 would be -1 × -1× -1,that would simply be -1.0532

8 × -1 would be -8 and then we have -6. 0539

Continuing on, I have -2 – 8 – 6 = -16.0546

If you are evaluating it, just take that value and substitute it in for all copies of that variable.0557

Onto what we wanted to, adding and subtracting polynomials.0568

When we get into addition and subtraction, what we are looking to do is add or subtract the like terms present in the polynomial.0573

There are two ways you should go about this and both of them are perfectly valid so you will use whichever method you are more comfortable with.0581

Here I want to add the following two polynomials.0590

The way I’m going to do this is I'm going to simply highlight which terms are like terms.0593

Here is 4x3 and 6x3 those are like terms - 3x2 2x2, those are like.0599

2x and - 3x and so one by one we will take these like terms and simply put them together.0610

4x3 and 6x3 = 10x3.0619

-3x2 and 2x2 would be -1x2.0626

2x - 3x =- 1x.0633

You can see I have all of the parts there and it looks like I am left with a 3rd degree trinomial.0640

The other way that you can do this, if you are a little bit more comfortable with it, is you can take one polynomial over the other polynomial.0645

If you do it this way, you want to make sure you line up what are your like terms.0660

It is essentially the same idea and that you go through adding all of the terms as you go along.0672

I will start over on the right side of this one, 2x and I'm adding - 3x, there is - x for that one.0678

Onto the next set of terms, - 3x2 + 2x2 = - x2 and 4x3 and 6x3 = 10x3.0688

You can see you get exactly the same answer but just use whatever method you are more familiar with.0701

Let us try some examples.0711

In this one we want to determine if these are polynomials or not.0712

Let us try that first one.0718

I see I have 2x + 3x2 – 8x3.0719

The things we are watching out for is, one does it stop.0723

I do not see any… out here, I know this is finite, that looks pretty good. 0727

I do not see any variables in the denominator, so that is good.0732

There are no fractions with the x's in the bottom. 0736

All of the exponents here, the 1, 2, and the 3 all of those are nice whole numbers.0740

This one is looking good. 0747

I will say that this one is a polynomial.0749

2/x + 5/4x2 – x3/6.0757

In this one I can think I can see a problem right away. 0762

Notice how we have variables in the bottom and because of that I will say that this one is not a polynomial.0765

Be careful and watch out for that criteria. 0776

This next example, we want to just go through and classify what types of polynomials these are.0782

We use two criteria for this, we look at its degree and we will see how many terms it has.0788

The first one, the largest power I can see in here is this 3.0794

I will say this is a 3rd degree.0798

It is a polynomial but let us be a little more specific. 0811

It has 1, 2, 3 terms, so I will say this is a 3rd degree trinomial.0814

Let us try another one, the largest power here is 4, 4th degree.0826

It only has 1, 2 terms, so it will be our binomial.0838

One more to classify, this one has a bunch of different exponents, but of course the largest one is the only one we are interested in.0847

This is a 3rd degree and now we count up all of its terms, 1, 2, 3, 4.0856

Since it has four terms, I will just keep calling this one a polynomial.0870

Let us get into adding the following polynomials together.0883

The way I’m going to do this is I’m going to line them up, one on top of the other.0886

Starting with the first one, I have 3x3, I do not have any x2, I have 4x and I have 1.0892

All of that would represent my first polynomial there.0905

Below that I want to write the second polynomial but I want to lineup the terms, -3x2 I will put it under the other x2.0910

I have a 6x I will put that under the other x and the 6 I will go ahead and put that with the 1.0922

You can now have things all nice and lined up.0931

1 + 6 = 7, -4 + 6x =2x, 0x2 + -3x2 = -3x2.0938

One more, this has nothing to add to it so just 3x2.0961

This would be our completed polynomial after adding the two together.0967

One last example is we are going to work on subtracting polynomials.0977

With these ones, what I suggest is being very careful with your signs. 0981

You will see that I’m going to start with lining one on top of the other one.0985

But I’m going to end up distributing my negative signs, I will turn this into an addition problem.0989

Let us give it a try.0995

The polynomial on the left is 7y2 -11y + 8 and right below that is -3y2 + 4y + 6.0996

Now comes the important part, we are subtracting these polynomials so I will put a giant minus sign up front.1015

Before getting too far, I could go through and try and subtract this term by term, 1023

but it is much easier to distribute my negative sign and just look at this like an addition problem. 1028

My top polynomial will stay unchanged and we will leave that as it is.1036

After distributing the bottom, here is what we get negative × - 3y = 3y2, negative × 4y = -4y and for the very last one -6.1043

We can take care of this as an addition problem and we know that the subtraction is taken care of because we put it into all of our terms.1060

8 + -6 = 2, -11y + -4y = -15y and then I have a 7y2 + 3y2 = 10y2.1071

This final polynomial represents the two being subtracted.1098

Now you know a little bit more about polynomials and have put them together.1105

Remember that you are just combining your like terms. 1109

Thank you for watching www.educator.com.1112

Welcome back to www.educator.com.0000

In this lesson we are going to take care of multiplying polynomials.0002

Specifically we will look at the multiplication process in general, so you can apply that to many different situations.0009

We will look at some more specific things like how you multiply a monomial by any type of polynomial 0015

and how can you multiply two binomials together.0021

I will also show you some special techniques like how you can organize all of this information into a nice handy table.0026

I will show you the nice way that you can multiply two binomials using the method of foil. 0033

Watch for all of these things to play a part. 0038

Now in order to multiply two polynomials together, what you are trying to make sure is that every term in the first polynomial gets multiplied0044

by every single term in the second polynomial.0052

That way you will know every single term gets multiplied by every other term. 0056

If you have one of your polynomials being a monomial, it only has one term and this looks just like the distributive property.0061

Let us do one real quick so you can see that it is just the distributive property.0071

We are going to take to 2x4 that monomial one term and multiplied by 3x2 + 2x – 5.0075

We will take it and multiply it by all of these terms right here.0083

We will have 2x4 × 3x2, then we will have 2x4 × 2x, and 2x4 × -5.0091

Each of these needs to be simplified, but it is not so bad.0115

You take the 2 and 3, multiply them together and get 6.0119

And then we will use our product rule to take care of the x4 and x2 by adding their exponents together x6.0124

We will simply run down to all of the terms doing this one by one.0134

2 × 2 =4, then we add the exponents on x4 and x1 power = x5.0137

At the very end, 2 × -5 = -10 and we will just keep the x4.0147

This represents our final polynomial after the two of them multiplied together.0156

Remember that we are looking so that every term in one is multiplied by every term in the other one. 0161

What makes this a little bit more difficult is, of course, when you have more terms in your polynomials.0171

As long as you make sure that every term in one gets multiplied by every term in the other one should work out just fine. 0177

Be very careful with this one and see how that turns out.0183

First, I’m going to take this first m3 term and make sure it gets multiplied by all three of my other terms. 0187

Let us put that out here.0194

I need to make sure m3 gets multiplied by 2m2.0198

Then I will have m3 × 4m and m3 × 3. 0206

Now if I stop to there I would not quite have the entire multiplication process down. 0221

We also want to take the -2m and multiply that by all 3.0226

Let us go ahead and put that in there as well.0232

We will take -2m × 2m2, then we will take another -2m × 4m and then -2m × 3.0234

That is quite a bit but almost there. 0264

Now you have to take the 1 and multiplied by all 3 as well. 0266

1 × 2m3, 1 × 4m,1 × 3.0272

That is a lot of work and we still have lots of simplifying to do, 0292

but we made sure that everything got multiplied so now it is just matter of simplifying.0295

Let us take it bit by bit.0301

Starting way up here at the beginning I have m3 × 2m2, adding exponents that would be 2m5.0302

Now I have m3 × 4m so add those exponents, and you will get 4m4.0313

Onto m3 × 3 = 3m3 and now we continue down the list over here.0324

-2m × 2m2, well 2 × -2 = -4 then add the exponents and I will get m3, that will take care of that.0333

We will go to this guy -2m × 4m = -8m2. 0345

Now this -2m × 3 = -6m and that takes care of those.0354

Onto the last where we multiplied one by everything.,0363

Unfortunately, 1 × anything as itself we will have 2m2, 4m and 3.0366

I have all of my terms and the resulting polynomial, but it still not done yet. 0376

Now we have to combine our like terms.0380

Let us go through and see if we can highlight all the terms that are like.0383

I will start over here with 2m5 and it looks like that is the only m5, it has no other like terms to combine.0387

We will go on to m4, let us see what do we got for that.0397

I think that is the only one, so m4.0403

3m3 looks like I have a couple of m3, I’m going to highlight those.0408

I have some squares, I will highlight those.0415

Let us see what else do we have in here, it looks like we have single m’s.0420

There is that one and there is that one and there is a single 3 in the m.0428

We can combine all these bit by bit.0433

2m5, since it is the only one.0436

4m4, since this the only one, let us check this after them.0440

Now I have 2m3, 3 – 4=-1m3 and that will take care of those ones.0446

-8m2 + 2m = -6m2.0459

That one is done and that one is done. 0467

-6m + 4m = -2m done and done.0471

And then we will just put our 3 in the end.0478

You can see it is quite a process when your polynomials get much bigger, 0483

but it is possible to take every term and multiply it by every other term. 0487

Now watch for later on how I will show you some special techniques to keep track of all of these terms that show up.0492

They will actually not be quite as bad as this one.0498

One way we can deal with much larger polynomials and keep track of all of those terms that multiplied together0503

is try and organize all of those terms in a useful way.0508

I’m going to show you two techniques that you can actually organize all that information. 0512

One of them we will be using a table and another one we will look like more standard multiplication where you stack one on top of the other.0517

What I'm trying to with each of these methods is ensure that every term in one polynomial gets multiplied by every term in the other polynomials.0524

I’m not are changing the rule while we are doing a shortcut.0531

We are just organizing information in a better way.0534

No matter which method you use, make sure you do not forget to combine your like terms at the end so you can see the resulting polynomial.0536

Let us give it a try.0543

I want to multiply x2 + 3x + 5 × x -4.0545

The way I’m going to do this is first I’m going to write the first polynomial right on top of the second polynomial.0550

From there I’m going to start multiplying them term by term and I'm starting with that -4 in the bottom,0566

now multiply it by all the terms in that top polynomial.0573

Let us give it a try.0580

First I will do -4 × 5 = -20 then I will take a -4 × 3x = -12x and I have -4 × x2 = -4x2.0582

That takes care of that -4 and make sure that it gets multiplied by all of the other terms.0603

We will do the same process with the x.0609

We will take it and we will multiply it by everything in that top polynomial.0612

x × 5 = 5x and I’m going to write that one right underneath the other x terms.0618

This will help me combine my like terms later.0626

x × 3x = 3x2.0629

And one more x × x2 = x3.0635

I have all of my terms it is a matter of adding them up and I will do it column by column. 0644

This will ensure I get all of my like terms -20 - 7x - 1x2 and at the very beginning x3.0650

That is my resulting polynomial. 0665

Now another favorite way that I like to combine the terms of my polynomial is to use a table structure.0669

Watch how I set this one up.0676

First, along the top part of my table I'm going to write the terms of the first polynomial.0679

My terms are x2, 3x and 5. 0687

Along the side of it I will write the terms of the other polynomials, so x, -4.0697

Now comes the fun part, we are going to fill in the boxes of this table by multiplying a row by a column.0706

In this first one we will take an x × x2.0713

It feels like you are a completing some sort of word puzzle or something, only guesses would be a math puzzle.0717

Also x × x2 = x3.0722

x × 3x = 3x2 and x × 5 = 5x.0728

It looks pretty good.0738

I will take the next row and do the same thing. 0739

-4 × x2 = -4x2, - 4 × 3x = -12x and -4 × 5 = -20.0743

You will get exactly the same terms that you do know using the other method in a different way of looking at them.0757

We need to go through and start combining our like terms.0764

Looking at my x3 that is my only x3 so I will just write it all by itself.0768

But I have a couple of x2's so I will write both of those and combine them together, -4x2 + 3x =-x2.0776

Here I have -12x + 5x =- 7x and of course the last one -20.0789

Oftentimes you will find your like terms are diagonals from each other, but it is not always the case that seems to be very common.0800

A good important thing to recognize in the very end is that you get the same answer either way.0807

Use whichever method works the best for you, and that you are more comfortable with.0813

Now that we have some good methods and above, let us try multiplying these polynomials again and see how it is a little bit easier. 0821

I will use my table method and we will take the terms of one polynomial write along the top. 0832

I will take the terms of the second polynomial and write them alongside.0843

You will see this will go much quicker m3 – 2m and 1, 2m2, 4m and 3.0848

Let us fill in the boxes.0864

2m2 + m3 = 2m5, 2 × -2 =-4m3, 1 × 2m2 = 2m2.0865

On to the next row, 4m4, 4 × -2 = -8m2, 4m × 1 = 4m.0880

Last row, 3 × m3 = 3m3, 3 × -2 =-6m and 3 × 1 =3. 0896

Let us go through and start combining everything.0909

I have a 2m5 I will write that as our first term, 2m5.0911

I’m onto my 4m4 and I think that is the only one I have floating around in there, 4m4.0918

We can call that one done.0930

3m3 – 4m3, two of those I need to combine, that will be -1m3.0934

I’m onto my squares, -8m2 + 2m2 = -6m2, -6m + 4m =-2m and the last number, 3.0949

The great part is that it goes through and combines all of your like terms and I know I got them off because they are all circle.0977

I’m going to fix this -1, so it is just a - m3 but other than that I will say that this is a good result right here.0984

Some other nice techniques you can use to multiply polynomials together is if both of those polynomials happen to be binomials.0995

Remember that they have exactly two terms, this method is known as the method of foil.1003

That stands for a nice little saying it tells you to multiply the first terms together, the outside terms, the inside terms and the last terms.1010

It is a great way of helping you memorize and get all of those terms combined like they should.1020

It also saves you from creating a large structure like a table when you do not have to.1027

Let us see how it works with this one. 1032

I have x -2 × x - 6 I’m going to take this bit by bit.1035

The first terms in each of these binomials would be the x and the other x.1042

Let us multiply those together and that would give us an x2.1047

Then we will move on to the outside terms.1056

By outside that would be the x and -6 we will multiply those together, - 6x.1060

Continuing on, we are on inside terms, -2 and the x, they need to multiply together -2x.1072

And then our last terms -2 × -6 = 12. 1084

We do get all of our terms by remembering first outside and inside last.1096

With this method, oftentimes your outside and inside terms will be like terms1101

and you will be able to combine them, and this one is no different. 1106

They combined to be 8x.1108

Once you have all of your terms feel free to write them out. 1112

This is x2 - 8x – 4 + 12 and the more you use this method, it will come in handy for a factoring a little bit later on.1115

Let us try out our foil method as we go through some of these examples.1129

Here I want to multiply the following binomials, 5x - 6 × 2y + 3.1134

First, I'm going to take the first terms together that will be the 5x and 2y, 5 × 2 = 10x × y.1142

That is as far as I can put those together since they are not like terms.1154

Outside terms that would be 5x and the 3 = 15x.1159

Onto inside terms, -6 × 2y = 12y and the last terms -6 × 3 = -18.1168

We got our first outside, inside, last and it looks like none of these are like terms.1186

I will just write them as they are 10xy + 15x -12y – 18 and we will call this one done. 1190

Let us try another one, and in this one you will see it has few more things that we can combine.1206

We are going to multiply - 4y + x and all of that will be multiple by 2y -3x.1211

Starting off with our first terms let me highlight them.1218

- 4y × 2y = -8 and y × y = y2.1223

That is the case here of our first terms.1232

Now we will do our outside terms, -4 × -3 = 12 and x × y.1235

Onto the inside terms, x × 2y = 2xy.1250

Of course our last terms, -3x2.1261

Now that we have all of our terms notice how our outside and inside terms, they happen to be like terms so we will put them together.1273

That will give us our final polynomial, 8y2 + 14xy - 3x2.1280

We can say that this one is done.1294

One more example and this one is a little bit larger one.1303

In fact, the second polynomial in here is a trinomial so we will not be able to use the method of foil.1307

That is okay, we will still be able to multiply it together, 1315

but I will definitely use something like a table to help me organize my information a little bit better. 1318

Okay, along the top of this table, let us go ahead and write our first polynomial, x – 5y.1330

Then along the rows we will put our second polynomial, I have an x2 – 2xy and 3y2.1340

Here comes the fun part, just fill in all of those blanks by multiplying a row and a column.1358

x2 × x =x3, x2 × -5y = -5x2y.1364

Onto the next row, -2xy × x, the x’s we can put those together as an x2 and the a y.1378

The last part here -2xy × -5y, let us put the y’s together, -2 × -5 =10xy2.1389

One more row, 3y2 × x =3xy2.1401

I have 3y2 × -5y -15y3.1412

We have all of our terms in there, now we need to combine the like terms.1422

Let us start here on the upper corner.1427

If we have any single x3 that we can put with this one.1429

It look like it is all by its lonesome, we will just say x3.1435

We are looking for x2y, they must have x2 and they must have y, I think I see two of them, here is one and here is that other one.1441

Let us put these together, -2 + -5 = -7 and they are x2y terms.1453

Continuing on, I have an xy2.1464

I have two of those so let us put them together, we will take this one and we will take that one, 10 + 3= 13xy2.1468

That takes care of those terms.1482

One more -15y3, I will put it in -15y3.1483

Now I have the entire polynomial.1491

Remember, at its core when you multiply polynomials you just have to make sure that every term gets multiplied by every other term.1493

Use these techniques such as foil or a table to help you organize all of those terms.1500

Thank you for watching www.educator.com.1506

Welcome back to www.educator.com.0000

In this lesson we are going to take care of dividing polynomials. 0003

I like to break this down into two different parts. 0009

First, we will look at the division process when you have a polynomial divided by a monomial.0013

We will look at what happens when you divide any two polynomials together.0017

In the very end, I will also show you a very special technique to make the division process nice and clean. 0022

To get into the basics of understanding the division of polynomials, I like to take it back to looking at the division process for real-world fractions. 0031

Suppose that when you were adding fractions together you know how that process will go.0042

One very important thing that you do when adding fractions is you would find a common denominator. 0047

Once you have a common denominator, then you can go ahead and just combine the tops of those fractions together.0053

Think of a quick example like 2/3 and are looking to add 5/3.0060

Since they have exactly the same bottoms then you will only add the 2 and 5 together and get 7/3 as your result.0067

Since this has a giant equal sign in between it, it means you can also follow this process in the other direction.0074

This may look a little unfamiliar, but it does work out if you go the other way. 0080

Suppose I had A + B and all of that was dividing by C.0085

The way I could look at this is that both the A and B are being divided by C separately.0090

This is the same equation I had earlier, I just turned it around. 0097

The reason why I show this is this will help us understand what happens when we have a binomial divided by a monomial.0102

In fact, that is the example that I have written out.0108

The top is a binomial and the bottom is an example of a monomial.0112

You can see that the way we handle it is we split up that monomial under each of the different parts of the polynomial in the top.0121

Let us see this process with numbers and see what polynomials and you will get an idea of how this works.0134

Working the other direction, if I see (2 + 5) ÷ 3, I want to visualize that as the 2 ÷ 3 and also the 5 ÷ 3.0140

For our polynomials if I have something like (x + 3z) ÷ 2y, then I will put that 2y under both of the parts.0152

In addition, you will notice splitting up over both of the parts in the top, you always want to make sure that you simplify further, if possible. 0162

This means if you use your quotient rule for exponents, go ahead and do reduce those powers as much as possible. 0172

If you have just any two general polynomials, then you want to think of how the process works with numbers.0183

In fact we are going to go over a long division process so that we can actually keep track of all the parts of what goes into what.0189

To make this process easier, remember to write your polynomial in descending power.0199

Start with the largest power and write it all the way down to the smallest power. 0203

An additional thing that will also help in the division process is to make sure you put placeholders for all the missing variables.0214

If I’m looking at a polynomial like 5x2 + 1, then I will end up writing it with a placeholder for the missing x.0221

It is not missing but it will help me keep track of where just my x terms go.0233

I'm going to show you this process a little bit later on0239

so you can see how it works with numbers and make some good parallels too doing this with polynomials.0242

Let us look at the division process for numbers.0251

Suppose I gave you 8494 and I told you to divide it by 3 and furthermore I said okay, let us see if you can do this by hand.0254

Some good news that you will probably tell me is that you do not have to take the 3 into that entire number all at once. 0263

No, you will just take the 3 into 8494 bit by bit.0269

In fact, the first thing that you will look at is how many times this 3 go into the number 8.0274

That will be the only thing you are worried about.0279

How many times does the 3 go into 8?0282

It goes in there twice and you will write that number on the top. 0285

Now after you have that number on the top, you do not leave it up there you go through a multiplication process, 0290

2 × 3 and you will write the result right underneath the 8.0296

With the new number on the bottom, you will go ahead and subtract it away, so 8-6 would give you a 2. 0305

That would be like one step of the whole division process.0315

You would continue on with the division process by bringing down more terms and doing the process again.0319

At this next stage, we will say okay how many times this 3 go into 24?0329

It goes in there 8 times.0335

Then we could multiply the 8 and 3 together and get a new number for that 3 × 8 = 24.0338

We can go ahead and subtract those away.0352

We will not stop there, keep bringing down our other terms and see how many times 3 goes into 9.0359

Write it onto the top and multiply it by your number out front.0369

Subtract it away and continue the process until you have exhausted the number you are trying to divide.0377

Let us see.0389

3 goes into 4, it looks like it goes in there once.0390

I will get 3, subtract them away and I have remainder of 1.0396

That is a lengthy process but notice the Q components in there.0401

You are only dividing the number bit by bit.0405

You do not have to take care of it all at once.0408

The way you take care of it is you are saying how many times 3 goes into that leading number.0410

You write it on the top, you go through the multiplication process and then you subtract it away from the number.0415

You will see all of those same components when we get into polynomials.0423

We have our answer and I could say it in many different ways but I’m going to write it out.0429

8494 if we divide this by 3 is equal to 2831 and it has a remainder down here of 1.0434

We could say +1 and still being divided by 3.0446

We have many different parts in here that are flying around, and you want to keep track of what these parts are.0451

The part underneath your division bar is the dividend. 0458

What you are throwing in there this is your divisor.0466

This guy down here is our remainder. 0477

Notice how those same parts actually show up in our answer.0483

Dividend, divisor, quotient, and remainder.0488

We put the remainder over the divisor because it is still being divided.0506

Now that we brushed up on the process with numbers, let us take a look at how we do this with polynomials. 0510

I want to divide 2x2 + 10x + 12 and divide that by x +3.0519

The good news is we do not have to take care of the entire polynomial all at once.0526

We are going to take it in bits and pieces.0530

We will first going to look at x and see how many times it will go into 2x2.0532

In fact one thing that I can do to help out the process is think to myself what would I have to multiply x by in order to get a 2x2.0537

1 × what would equal to 2? That would have to be 2.0548

What would I have to multiply x by to get an x2?0552

I have to multiply it by x.0555

I will put that on top, just like I did with numbers.0557

After I do that, we will run through a multiplication process.0563

We will take the 2x to multiply it by x, I will multiply it by 3.0568

We will record this new polynomial right underneath the other one.0573

2x × x = 2x2.0577

2x × 3 = 6x.0581

Now comes a very important step.0589

Now that we have this new one, we want to subtract it away from the original.0592

Notice how I put those parentheses on there, that will help me remember that I need to subtract away both parts and keep my signs straight.0599

Starting over here, I have 10x – 6x = 4x.0606

Then I have 2x2 – 2x2, that will cancel out and will give me 0x2.0616

If you do this process correctly, these should always cancel out.0622

If they do not cancel out, it means we need to choose a new number up here.0628

That is just one step of the division process, let us bring down our other terms and try this one more time.0635

I want to figure out how many does x goes into 4x?0648

What would I have to multiply x by in order to get 4x?0652

I think I have to multiply it by 4 that is the only way it is going to work out.0658

Now we have the 4, go ahead and multiply it by the numbers out front.0663

4 × x = 4x and 4 × 3 = 12.0669

Once you have them, put on a giant pair of parenthesis and we will go ahead and subtract it away.0678

12 – 12 =0 and 4x – 4x = 0.0687

What that shows is that there is no remainder and that it went evenly.0693

Let us write this out.0699

When I had 2x2 + 10x + 12 and I divided it by x + 3, the result was 2x + 4.0699

There was no remainder.0714

We can label these parts as well.0717

We have our dividend, divisor, quotient, and if I did have a remainder I will probably put it out here.0720

Here is our dividend.0739

Here is our divisor and our quotient.0744

Now that we know a lot more about dividing polynomials, let us look at a bunch of examples.0758

Some of them will take a polynomial divided by a monomial.0764

Some of them will take two polynomials and divide them.0767

We will approach both of those cases in two different ways.0770

Example 1, divide a polynomial by a monomial.0774

I will take (50m4 -30m3 + 20m) ÷ 10m3.0777

Since we are dividing by a monomial, I will take each of my terms and put them over what I’m dividing them by.0784

50m4 ÷ 10m3, 30m3 ÷10m3 and 20m ÷ 10m3.0792

Now that I have done that, I will go through and simplify these one at a time.0811

50 ÷ 10 = 5, m4 ÷ m3 = I can use my quotient rule and simply subtract the exponents and get m1.0816

Continuing on 30 ÷ 10 = 3, if I subtract my exponents for m3 and m3, I have m0.0830

Onto the last one, 20m ÷ 10m3, there is 2.0844

Let us see, if I subtract the exponents I will m-2.0852

Then I can go through and just clean this up a little bit.0857

5m – anything to the 0 power is 1, 1 × 3 + and I will write this using positive exponents, 2/m2.0860

This will be the final result of dividing my polynomial by a monomial.0872

Let us try this scheme with something a little bit more complicated. 0881

This one is (45x4 y3 + 30x3 y2 - 60x2 y) ÷ 50x2 y.0884

We have to put our thinking caps for this one.0894

We will start off by taking all of our terms in the top polynomial and putting them over our monomial, only one term.0897

Once we have this all written out we simply have to simplify them one at a time.0917

Let us start at the very beginning.0925

15 goes into 45 3 times, now I will simplify each of my variables using the quotient rule.0929

4 – 2 =2 and 3 – 1= y2.0939

There is my first term, moving on.0946

15 goes into 30 twice.0950

Using my quotient rule on the x, 3 – 2 = 1 and y2 – y1 = y1.0955

Both of these have an exponent of 1 and I do not need to write it.0964

One more, 15 goes into 60 four times.0969

I have x2/x2 which will be x0 and y/y will be y0.0975

Anything to the 0 power is 1.0983

I can end up rewriting that term 3x2 y2 + 2xy – 4.0987

There is my resulting polynomial.1000

In this next example, we are going to take one polynomial divided by another polynomial.1007

I would not be able to split it up quite like I did before, now we will have to go through that long division process.1012

Be careful you want to make sure that you line up your terms in descending order. 1018

If you look at the powers of the top polynomial you would not mix up.1024

I want to start with that 3rd power then go to the 2nd power, then go to the 1st power, just to make sure I have it all lined up.1027

Let us write it out.1034

2x3 + x2 + 5x + 13, now it is in a much better order to take care of.1035

We will take all of that and we will divide it by 2x + 3.1048

That looks good.1055

It is time to get into the division process.1057

Our first terms there, and what would I need to multiply 2x by in order to get 2x3?1060

The only thing that will work would be an x2.1068

Once we find our numbers up top, we will multiply them by the polynomial out front.1072

2x × x2 = 2x3, that is a good sign, it is the same as the number above it.1079

X2 × 3 =3x2, now comes the part that is tricky to remember.1088

Always subtract this away and do not be afraid to use this parenthesis to help you remember that.1097

x2 – 3x2, 1 – 3 = -2x2.1106

Then I have 2x3 – 2x3 and those will be gone, cancel out like they should.1116

The first iteration of this thing looks pretty good. 1122

Let us try another one.1126

I want to figure out how may times does 2x go into -2x2?1129

I have to multiply it by –x.1137

Let us bring down some more terms.1146

I will get onto our multiplication process.1149

-x × 2x =-2x2.1153

-x × 3 =-3x.1158

We have our terms in there, it is time to subtract it away and be careful with all of our signs.1164

5x - -3x when we subtract a negative that is the same as addition.1171

I’m looking at 5x + 3x = 8x.1179

-2x2 - -2x2 = that is a lot of minus signs.1188

That would be the same as -2x + 2x2 and they do cancel out like they should.1193

That was pretty tricky keeping track of all those signs but we did just fine.1201

We have 8x + 13 and I’m trying to figure out what would I have to multiply 2x by in order to get that 8x.1207

There is only one thing I can do I need to multiply by 4.1218

We will multiply everything through and see what we get.1224

4 × 2x = 8x, 4 × 3 =12.1228

I can subtract this away and let us see what we get.1238

13 – 12=1, 8x – 8x = 0.1245

Here I have a remainder of 1.1250

I could end up writing my quotient and I can put my remainder over the divisor.1257

In some polynomials you want to make sure you put in those placeholders.1275

That way everything lines up and works out good.1280

It is especially what we will have to do for some of those missing powers in this one. 1283

I’m missing an x2 and a single x.1287

Let us write this side and put those in.1291

x3, I have no x2, no x – 8.1294

All of that is being divided by x – 2.1305

I’m going to go through and let us see what I need to multiply x by in order to get x3.1310

I think I’m going to need x2.1318

Now that I found it, I will go ahead and multiply it through.1323

x2 × x =x3 and x2 × -2 = -2x2.1327

We have our terms, let us go ahead and subtract it away.1336

0x2 – 2x2, this is one of those situations where if we subtract a negative is the same as adding.1345

0x2 + 2x2 =2x2.1354

x3 – x3, that is completely gone.1360

Bring down our other term here and we will keep going.1366

What would I have to multiply x by in order to get 2x2?1374

I’m going to need 2x.1381

Let us multiply that through, 2x × x =2x2.1386

2x × -2 =-4x.1392

Now we found that, let us subtract that away.1399

Starting on the end, 0 – -4, that is the same as 0 + 4x = 4x.1405

Then 2x2 – 2x2, they are completely gone, you do not have to worry about it.1415

We will bring down our -8 and continue.1422

What would I have to multiply x by in order to get 4x?1428

That will have to be 4.1433

4 × x = 4x and 4 × -2 = -8.1437

It looks like it is exactly the same as the polynomial above it.1447

I know when I subtract, I would get 0.1452

There is no remainder for this one.1454

I will take (x3 – 8) ÷ x -2 and the result is x2 + 2x + 4.1457

Division process can take a bit but as long as you do the steps very carefully, you should turn out okay.1467

Let us try this giant one.1476

This is (2m5 + m4 + 6m3 – 3m2 – 18) ÷ m2 + 3.1478

There are a lot of things to consider in here.1489

One thing that I will be careful of is putting those placeholders for this guy down here.1493

Notice that it is missing an m, let us give it a try.1498

I will have m2 + 0m + 3 and all of that is going into our other polynomial 2m5 + m4 + 6m3 – 3m2 – 18.1501

Lots of things to keep track of but I think we will be okay.1523

What would I have to multiply the m2 by in order to get a 2m5?1527

That would be 2m3.1535

I can run through the multiplication and write it here.1539

2m3 × m2 = 2m5.1543

Now we can multiply it by our 0 placeholder but anything times 0 will give us 0 so 0m4.1549

By my last one, let us see 2m3× 3 + 6m3.1558

Let us subtract that away.1568

6m3 – 6m3 those are gone.1572

I have m4 – 0m4 I still have m4.1575

2m5 – 2m5 those are gone.1580

I dropped away quite a bit of terms.1584

Let me go ahead and write in my 0m3 as one of those placeholders so I can keep track of it.1587

Let us try this again.1598

m2 goes into m4, if I multiply it by another m2.1600

Multiplying through I have m2 × m2 = m4.1609

m2 × 0m = 0m3 and m2 × 3 = 3m2.1614

We will take that and subtract it away.1626

I need to go ahead and subtract these.1636

Be very careful on the signs of this one.1638

I have –m2 and I’m subtracting -3m2, the result here will be -6m2.1641

The reason why it is happening is because of that negative sign out there.1652

Now I have 0m3 – 0m3, 0 – 0 =0, m4 – m4 = 0.1659

It looks like I forgot an extra placeholder.1669

I need 0m and then I need my 18.1676

Let us bring down both of these.1684

I need to figure out what I have to what would I have to multiply m2 by in order to get -6m2.1691

-6 will do it, -6m2 6 × 0 = 0m and -6 × 3 =-18.1699

It is exactly the same as the polynomial above it.1718

Since they are exactly the same and I’m subtracting one from the other one, 0 is the answer.1723

There is no remainder, it went evenly.1729

The quotient for this one would be 2m3 + m2 – 6.1732

In this polynomial, I have (3x3 + 7x2 + 7x + 11) ÷ 3x + 6.1744

The reason why I put this one is because it can get a little bit difficult figuring out what you need to multiply to get into that second polynomial.1753

I’m going to warn you, this involves a few fractions.1762

Let us give it a shot.1767

(3x3 + 7x2 + 7x + 11) ÷ 3x + 6.1770

Let us start off at the very beginning.1788

What do I need to multiply my 3x by in order to get 3x3?1790

The only thing that will work will be an x2.1797

I will go through and I will multiply and get the result.1804

3x3 + 6x2 and let us subtract that away.1809

7x2 – 6x2 = 1x2.1821

Now comes the tricky part, I need to figure out what I need to multiply 3x by in order to get x2.1837

If I’m looking at just the variable part of this, I have to multiply and x by another x in order to get an x2.1845

We will go ahead and put that as part of our quotient.1851

What do I have to multiply 3 by in order to get 1 out front?1854

That is a little bit trickier.1859

3 × what = 1.1861

That is almost like an equation onto itself.1865

What we see is that x would have 1/3, a fraction.1869

It is okay, we can use fractions and end up multiplying by those.1874

3x × 1/3x = 1x2.1881

Let us multiply that through.1887

1/3x × 3x = 1x2 I will write it down and 1/3x × 6 = 2x.1888

Now we can take that and subtract it away.1900

7x – 2x = 5x.1904

Bringing down our extra terms and I think this one is almost done.1913

What would I have to multiply 3x to get 5x?1917

Let us see.1922

he only way I can get an x into another x is to multiply by 1, but I’m going to think of how do I get 3 and turn it into 5?1924

Let us do a little bit of scratch work on this one.1933

3 × what = 5?1939

If we divide both sides by 3 I think we can figure out it is 5/3.1945

That I can write on top 5/3.1951

We can go through multiplying.1956

5/3 × 3 = 5x.1958

5/3 × 6 =10.1966

We will go ahead and subtract this away.1977

11 – 10 = 1 and 5x – 5x = 0.1980

We have a remainder of 1.1985

Now that we have all of the quotient and the remainder, let us go ahead and write it out.1993

We have (x2 + 1/3x + 5/3 + 1) ÷ 3x + 6.1998

Definitely do not be afraid some of those fractions to make sure it goes into that second polynomial.2010

This process can get a little messy as you can definitely see from those examples.2018

I have a nice clean way that you can go through the division process known as synthetic division.2022

This is a much cleaner way for the division process so that you can keep track of all the variables.2027

It is much clean but be very careful in how you approach this. 2033

It works good when dividing by polynomials of the form x + or – number.2037

It will work especially with my little example right here (5x3 - 6x2 +8) ÷ x -4.2043

First watch how I will set this up. 2051

I'm going to create like a little upside down division bar and that is where I will end up putting the polynomial that I'm dividing.2054

But I will not put the entire thing I’m only going to put the coefficients of all of the terms.2062

The coefficient of the x3 is a 5. 2069

The coefficient of my x2 is -6.2073

I will put a 0 placeholder in for my missing x and then my last coefficient will be 8.2077

Once I have all of those I will put another little line. 2086

I want to put in the value of x that would make this entire polynomial 0, if x was 4 that would be 0.2090

I’m going to write 4 out here.2105

That turns to be a tricky issue and many students remember what to put out over here because it will be the opposite of this one.2109

If you see x – 4 put in a 4.2118

If you see something like x + 7 then put in -7.2121

We got that all set let us go through this synthetic division process.2127

It tends to be quick watch very carefully how this works.2131

The very first thing that you do in the synthetic division process is you take the first number here and you simply copy it down below.2137

This will be a 5.2146

Once you get that new number on the bottom, go ahead and multiply it by your number out front, 4×5 = 20.2149

That is one step of the synthetic division process.2161

To continue from there simply add the column -6 + 20 and get the result. 2165

This would be a 14.2173

Once you have that feel free to multiply it out front again.2177

14 × 4 = 56.2181

There are 2 steps now we will take the 56 and we will add 0, 56.2191

When we get our new number on the bottom, go ahead and multiply it right out front.2200

4 × 50 = 200.2205

4 × 6 = 24, 224.2212

One last part to this we got to do some addition.2222

8 + 224 = 232.2225

It does not look like we did much of any type of division. 2234

We did a lot of adding and we did a lot of multiplying but do you know what these new numbers stand for on the bottom.2236

That is the neat part, these new numbers I have here in green stand for the coefficients of our result.2243

You know what happens after the division.2249

The way you interpret these is the last number in this list will always be your remainder.2252

I know that my remainder is 232.2260

As for the rest of the values, the 5, 14, and 56, those are the coefficients on our variables. 2264

What should they be? Let me show you how you can figure that out.2270

Originally we had x3 as the polynomial that we are dividing and these new ones will be exactly one less in power.2274

That 5 goes with 5x2 and the 14 goes with the 14x and 56 has no x on it. 2282

The result for this one is 5x2 + 14x + 56 with a remainder of 232 which you can write over x – 4.2293

Since it is still being divided2312

It is a much cleaner and faster method for division.2315

Let us go ahead and practice it a few times just to make sure got it down.2318

We will go ahead and do (10x4 - 50x3 – 800) ÷ x – 6.2325

It is quite a large problem.2332

We will definitely remember to put in some of those placeholders to keep track of everything.2334

First I will write in all of the coefficients of my original polynomial.2339

I have a 10x4 - 50x3 I need to put in a placeholder for my x2 and another placeholder for my x.2344

We will go ahead and put in that -800.2356

What shall we put on the other side?2364

Since I'm dividing by x – 6, the value of 6 will be divided that would make that 0.2367

I will use 6 and now onto the synthetic division process.2376

The first part we will drop down to 10, just as it is.2381

Then we will multiply by the 6 out front 60.2387

Now that we have that, let us add the -50 and the 60 together, 10 again.2396

We will multiply this out front.2405

That result will be 60.2409

We will go ahead and add 0 + 60 = 60 and multiply 60 × 6= 360.2417

Now we will add 0 + 360 = 360.2430

I have to take 360 × 6.2435

think I have to do a little bit of scratch work for that one.2442

6 × 0 = 6 × 6 = 36 and then 3 × 6 = 18 + 3 = 21.2443

I have 2160, let us put it in.2455

Only one last thing to do is we need to add -800 to the 2160 and then let us do a little bit of scratch work to take care of that.2465

0 – 0= 0, 6 – 0= 6, 20 and we will breakdown and will say 11 – 8 =3.2478

I have 1, I have 1360.2489

Now comes the fun part, we have to interpret exactly what this means.2493

Keep in mind that this last one out here that is our remainder.2498

Originally our polynomial was x4 so we will start with x3 in our result.2505

10 x3 + 10x2 + 60x + 360 and then we have our remainder 1360.2513

It is all still being divided by an x – 6.2532

That was quite a bit of work, but it was a lot clean than going through the long division process. 2536

Let us see this one more time.2540

In this last one we will use (5 - 3x + 2x2 – x3) ÷ x + 1.2545

We will first go ahead and put the coefficients of our top polynomial in descending order.2555

Be very careful as you set this one up.2560

On that side you can write out the polynomial first in descending order and then go ahead and grab its coefficients.2564

-x3 would be the largest, then I have 2x2 and then I have 3x, and 5.2570

I need -1, 2, -3 and 5.2579

I’m dividing by x + 1 so the number I will use off to the left will be -1.2589

I think we have it all set up now let us run through that process.2597

The first I’m going to bring down is -1 then we will go ahead and multiply.2600

Negative × negative is positive.2608

2 + 1 =3, 3 × -1= -3.2615

-3 + -3 = - 6, -6 × -1 = 6 and 5 + 6 =11.2630

Now we have our remainder, we can finally write down the resulting polynomial.2647

We started with x3 so I know this would be x2.2657

-x2 + 3x – 6 and I still have my 11 being divided by x + 1.2662

Now you know pretty much everything that there is to know about dividing polynomials.2676

If you divide by a monomial, make sure you split it out among all the terms.2680

If you divide a monomial by a polynomial, you can go through the long division process 2685

or use this synthetic division process to make it nice and clean.2690

Thank you for watching www.educator.com.2694

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at the greatest common factor and the technique of factor by grouping.0002

First we will learn how to recognize a greatest common factor for a list of different terms0012

and how we can actually factor out that greatest common factor when you have something like a polynomial. 0017

This will lead directly into the technique of factor by grouping, a good way of breaking down a large polynomial.0023

You are going to hear me use that word factor quite a bit in this entire section.0032

You are probably a little curious about what exactly factoring means.0036

Technically it means to write a quantity as a product.0043

We will be breaking down into pieces that are multiplied.0047

In a more practical sense, the way I like to think of factoring is that you are ripping things apart.0051

It is almost like doing the opposite of multiplication. 0056

For example if I have 2 × 3 I could put them together and get 6. 0059

When I have the number 6 then I can factor it down into those individual pieces again 2 × 3.0066

It is like multiplication but we are going in the reverse direction. 0072

Now some expressions and numbers could have many different factors.0077

Let us take the number 12. 0082

If you break that one down, you could look at it as 2× 6 and you could continue in breaking down the 6 and 2 × 3.0085

You have factors like 2 would go in there, 4 would go into 12, 3 would be a factor, 6 would be factor.0095

All those things could go into 12.0103

The greatest common factor will be the largest number that will divide into all the numbers present.0110

I will show you this to you twice just to get a better feel for it.0116

Suppose I'm looking at 50 and 75, the greatest common factor would be the largest number that could divide into both of those evenly.0120

5 could divide it into both of those evenly and that would be a pretty good choice, but is not the largest thing that could go in there.0128

The largest thing that could go in would be the number 25.0136

You will see that if I do take them both and divide them by 25, true enough they both go in there evenly.0140

25 is my greatest common factor.0150

We may only do this for pairs but you could do it with an entire list of numbers.0155

In this one I have 12, 18, 26, 32, and again we are looking for the largest thing that divides into all of that. 0162

Sometimes a good way to find the largest thing that goes into all of them is just find something that works, 0169

maybe something like 2 and end up reducing them bit by bit.0177

If I divide everything in here by 2 I would get 6, 9, 13 and 16.0183

Nothing else will go into all of those since 13 is a prime number.0191

My greatest common factor is 2 in this case.0197

It is not a very big number but it happens to be the largest in there that will go into all for these numbers evenly.0201

In algebra we are not interested in just single numbers we also want to see this process work out when we have lots of variables.0209

Let us see how this works out. 0216

Maybe I’m looking at the numbers over the terms -16r9, -10r15, 8r12.0218

To find the greatest common factor, I’m first going to look at those numbers and see if there is a large number that could go into all of them. 0226

I could divide all of them by 2, that would work but actually I think 2 is the largest one.0234

I will say that 2 will divide it into all of these evenly.0242

Divided by 2, divided by 2, divided by 2 and I would get -8, -5, 4.0248

We can also do this thinking about our variables. 0258

What was the largest variable raise to the power that we can divide into all of them?0261

Think of using your quotient rule to help out.0266

With this one r9 is the greatest thing that I can divide all of them by.0272

I would not have to deal with negative exponents or anything like that.0279

Let us write that over here, r9.0284

r9 ÷ r9 would be a single r0 or 1.0288

r15 ÷ r9 = r6 and r12 ÷ r9 = r3.0293

Our greatest common factor would be 2r9.0303

Be careful when you are dealing with those variables and exponents.0307

Sometimes you will hear greatest common factor in your mind and you think you should go after the greatest exponent.0310

But as we can see in this example, it is not the case. 0315

We actually took the smallest exponent because it was simply the greatest thing that could go into all of them.0319

Put that in mind and let us try this again.0327

In this one I have nothing but variables.0330

h4 k6 h3 k6 and h9 k2.0332

What is the largest power of h that could go into all of them?0340

It is going to be h3.0345

I picked up on that because I can see that looking in the middle one it is the smallest power.0349

Looking at all the k’s, the largest exponent of k that could go into all them would be k2.0354

My greatest common factor will be h3 k2.0364

Let us see what this will end up reducing down to if I take that greatest common factor.0373

h1 k4 h3 and h3 will cancel each other out.0378

k4 I have h6 and I have no k that will end up cancelling each other out.0388

Being able to recognize the greatest common factor will help us in the next process.0396

Make sure it is pretty solid. 0400

If you can recognize the greatest common factor you can often factor it out of a polynomial. 0405

What I mean by factoring out is we are going to place it out front of a pair of parentheses 0413

and put the reduced terms inside of that parentheses .0418

3m + 12 this is like looking at 3m and looking at 12, what is the greatest common factor of those two terms?0423

What could potentially go into both of them evenly and what is the largest thing that will do that?0431

The number 3 will go into 3m and the number 3 will also go into 12.0438

3 will be my greatest common factor.0446

I’m going to write that out front on the outside of the parenthesis.0449

What I write on the inside, what happens when I take 3m ÷ 3?0453

I will get just single m and when I take 12 ÷ 3 I will get 4.0458

Here I have taken a polynomial and essentially factored it into a 3 and into another polynomial m + 4.0467

The neat part about this is since factoring is like the opposite of multiplication you can check this by running through the distribution process.0475

If you take 3 and you put it back in here do you get the original answer?0487

You will see that in fact you do get 3m + 12.0492

That is how you know this has been factored correctly.0496

In some instances, just like with numbers when you are looking for the greatest common factor0501

it turns out you would not be able to find greatest common factor to pull from all of the terms or the largest thing will actually be 1 or itself.0506

In cases like that, we say that the polynomial is prime.0513

In other words it does not break down into any other pieces.0517

Let me give you a quick example of a prime polynomial 5x + 7.0520

There is not a number that goes into 5 and into 7 so I would consider that one prime.0526

Individually those numbers are divisible, but there is not a number that goes in the 8 and 9 it would be a prime polynomial.0542

Let us work on factoring out the greatest common factor from various different polynomials, 0555

just so we get lots of good practice with it.0560

Starting with the first one I have 6x4 + 12x2.0562

Let us hunt down those numbers first. 0567

The largest number that would go into both 6 and 12 would be 6.0570

Let us look at those x.0578

What is the largest power of x that would divide into both of them?0581

That have to be x2.0586

We will write in the leftovers inside our parentheses 6x4 ÷ 6x2=x2 and 12x2 ÷ 6x2 = 2.0592

I factored out the greatest common factor for that polynomial.0611

Let us give this another shot with something has a bunch of terms to it.0615

30x6 25x5 10x4.0619

Looking at the numbers 30, 25, and 10, what is the largest thing that could go into all of them, 5 will do it.0625

Looking at our variables the largest variable that could go into all them would be x4.0634

It is time to write down what I left over 30 ÷ 5 = 6, x6 ÷ x4 = x2.0645

25x5 ÷ 5 = 5, x5 ÷ x4 =x and 10x4 ÷5x4 = 2.0659

That one has been factored out.0675

Continuing on, let us get into some of the trickier ones. 0678

This one has 3 terms and has multiple variables in there.0681

It has e and t.0685

Let us look at our numbers.0688

What could go into 8, 12 and 16?0689

2 would definitely go into all of them.0696

4 is larger and it would also go into all of them and I think that is largest thing, so we will go with 4.0698

Let us take care of these variables, one at a time.0707

Looking at the e’s, what is the largest power of e that could go into all of them?0710

This will be e4.0715

Onto the t’s, the largest exponent of t that would go into all of them is t2.0721

Now that we have properly identified our greatest common factor, let us write down what is left over.0732

8 ÷ 4 =2, e5 ÷ e4 =e and t2 ÷ t2 =1.0737

16 ÷ 4 = 4, e6 ÷ e4 = e2 and t3 ÷ t2 =t.0752

Onto the last one, -12 ÷ 4 = -3, e4 ÷ e4 = 1 and t7 ÷ t2 = t5.0766

We have our factor polynomial.0790

One last one, I threw this one in so we can deal some fractions.0794

We will first think of what fraction could divide into ¼ and into ¾ ?0800

¼ is the only thing that will do it.0808

What could go into y9 and y2?0812

Look for the small exponent that usually helps out.0817

That will be our greatest common factor, let us write what is left over.0823

¼ ÷ ¼ = 1 and y9 ÷ y2 = y7.0828

¾ ÷ ¼ = 3 and y2 ÷ y2 = 1.0839

At all of these instances you identify your greatest common factor first and simply write your leftovers inside the parentheses. 0849

One quick thing that can help out, you can double check your work by simply multiplying your greatest common factor0856

back into those parentheses using your distribution property.0861

If it turns out to be the original problem, you will know that you have done it correctly.0865

In some terms, there is a much larger piece in common that you can go ahead and pull out.0873

Even though it is larger, feel free to still factor it as normal.0880

That is the hard part.0884

To demonstrate this I have 6 × (p + q) –r (p + q).0887

It has that p + q piece again and both of its parts.0894

When I'm looking at the greatest common factor of this first piece and the second piece, it is that p + q.0899

I’m going to take out the p + q as an entire piece and that is my greatest common factor.0910

Inside the set of parentheses I will write what is left over.0919

From the first part there is a 6 that was multiplied by p + q so I will put that in there and there was a -r on the second piece.0924

This looks a little strange, I mean it seems weird that we can take out such a large piece, but unusually that it does work.0933

To convince you I will go ahead and multiply these things back together using foil.0941

What I have here is my first terms would be 6p, outside terms would be –pr.0947

Inside terms 6q and my last terms –qr.0955

Compare this after I use my distribution property on the original.0962

(6p + 6q) – (rp + rq).0970

What you see is that it does match up with the original. 0984

There is my 6p, I have my – pr and – rp, there are just in a slightly different order. 0988

I have 6q and 6q -qr and I forgot to distribute my negative sign.0995

This should be a -rq and so it is the same term.1007

What you find is that you can pull out that large chunk of p + q and that leads to what is known as factor by grouping.1013

The idea behind factor by grouping is you try and take out the greatest common factor from a few terms at a time.1024

Rather than looking at all of them, just take them in parts.1031

This grades usually a very large piece and you can collect it into two groups.1034

Then you factor within those groups, you take out one of those large pieces. 1040

This will allow you to factor the entire polynomial. 1045

Sometimes factor by grouping does not seem to work or usually you end up struggling with a little bit.1048

If that is the case, try rearranging the terms and try to factor by grouping one more time. 1053

It is a neat process and let us give it a try, when you factor by grouping on p × q + 5q + 2p +10.1060

Watch how this works.1071

Rather than looking at all the terms at once, I’m going to take them two at a time.1073

A part of my motivation for doing that is if I were to look at all the terms all at once, they do not have anything in common.1078

Let us take these first two.1086

What would be the greatest common factor for p × q + 5q.1088

q is in both of the terms so that is my greatest common factor for both of those.1094

What is left over? There is still a p in there and there is still 5.1101

Those are done. 1109

Let us look at the next two terms.1111

What is the greatest common factor of just those two. 1115

I can see a 2 goes into both of them and then let us see, the only thing left over would be a p + 5.1119

I will factor them into those little individual groups and now notice how I only have two things and they both have a p + 5 in common.1131

I'm going to take out the p + 5 as my greatest common factor of those two terms.1142

What will be left behind will be the q + 2.1151

This will represent my factor polynomials.1156

Let us try this again and see how it could work. 1162

I have 2xy + 3y + 2x + 3.1164

Looking at the first two.1170

These have a y in common, let us go ahead and take that out.1174

2x will still be left when I divide 2xy by y and I still have a 3 over here when I divide 3y by y.1187

Looking at the next two terms, it looks like these ones do not have anything in common.1198

I might consider that the only thing that you do have in common is just a 1.1205

I could divide them both by 1.1209

I still have a 2x and I still have a 3.1213

Notice how we have that common chunk in there 2x + 3.1218

We will take that out 2x + 3 and then we will write what is left over just the y +1.1223

We can consider this one factored.1237

In this next example, be on the watch out for some negative signs which could show up as we factor out that greatest common factor.1242

I’m going to take this two at a time.1251

Looking at the first two, they do not have a number in common, but they both looks like can be divided by x2.1255

Let us take that out.1262

After taking our x2 we would have left over a single x and a 3.1268

Let us look at the next two terms.1280

I can see that if I would go into -5 and 5 would divide into a -15 and they both have the negative sign. 1284

That is important because I’m thinking ahead and try to think how I also have this common x + 3 piece.1292

Especially if I’m trying to factor by grouping.1300

The only way that is going to work out is if I take out a greatest common factor that is a -5.1303

What would that does to our left over pieces now?1310

-5x ÷ -5 = x and -15 ÷ -5 = 3.1314

I get those left over pieces like I need to, that same x +3 on the other side. 1324

Now I have this I can take out my common piece of x + 3 and I can go ahead and write the leftovers x2 – 5.1332

Be on the watch out for certain situations where you need to rearrange things first.1352

In this one I have 6w2 - 20x + 15w - 8wx.1357

And it is tempting to just go ahead and jump in there and try and factor.1364

Watch what happens if you do so. 1367

First we take the first two and we would look for something common.1370

The only thing I can see that would be common is a 2 goes into 6 and into 20 and they do not have the same variables.1375

That is all I can do.1385

I have 3w2 and -10x.1386

Looking at the other terms the only thing that I see common over there is they have a w in common.1393

What would be left over? I still have a 15 and have 8x.1406

That is definitely a problem because these pieces are not the same. 1413

I need to do some rearrangement and retry this factor by grouping one more time.1419

Let us go ahead and rewrite this.1433

I will rewrite it as 6w2 + 15w - 8wx and we will put that -20x on the very end. 1435

Here is a little bit of my motivation for putting in that order. 1452

When looking at the w's, notice how this one is a w2 and this one is a single w.1456

I have the one is one more in power after the left.1460

And looking at the same w's, here is my w1, this is like w0.1464

It is like I have lined things up according to the w power.1470

Watch how I factor by grouping works out much better.1475

Taking the first two they both have a w in common and I can pull that out.1481

It also looks like I can pullout 3.1485

3w will be like our greatest common factor.1488

What would be left over on the inside?1494

6w2 ÷ 3w = 2w and 15w ÷ 3w = 5.1499

I have -8wx -20x, both of those are negative, I think I will go ahead and pull out a -4, that will go into both.1510

They both contain x so we will also take out an x.1523

-8wx ÷ -4x.1532

There are still 2w left in there.1536

-20x ÷ -4 = 5.1540

I can see there is that common piece that I wanted the first time.1546

We can go ahead and factor that out front.1552

2w + 5 and 3w - 4x and now this is completely factored.1556

Let us do one last one, and this is the one where we might have to do a little bit of rearranging just make sure it works out.1575

9xy – 4 + 12x – 3y.1581

What to do here?1587

Let us go ahead and put the things that have x together.1590

9xy + 12x - 3y - 4 and I have just rearranged it.1606

We will look at these first two terms here and take out their greatest common factor.1612

You can see that 3 goes into both of them and they both contain x.1618

It will take both of those out.1624

9x ÷ 3x, 9 ÷ 3 = 3, x ÷ x =1 and we still have a y sitting in there.1628

12 ÷ 3 = 4, x ÷ x = 1.1639

Those two would be gone.1643

That looks pretty nice and it is actually starting to match what I have over here. 1646

Notice that the only difference is a negative.1651

I'm going to take out a -1 from both of the terms that not should be able to flip my signs and make it just fine.1655

-3y ÷ -1 = 3y and -4 ÷ -1 = 4.1663

I have that nice common piece that I need. 1672

We will go ahead and factor that out .1677

3y + 4 and 3x - 1.1681

Now this is completely factored.1690

When using the technique of factor by grouping take it two at a time and factor those first.1692

Look for your common piece and factor that out and that should factor your entire polynomial completely.1698

Thank you for watching www.educator.com.1705

Welcome back to www.educator.com. 0000

In this lesson we are going to work on factoring trinomials.0003

We are not going to tackle up all types of trinomials just yet.0006

For the first part we are going to focus on ones where the squared term has a coefficient of 1.0010

We will also look at polynomials where we can factor out a greatest common factor.0015

In future lessons we will look at the more complicated trinomials.0020

One of our first techniques we have to dig back in our brains and recall how we used foil in order to multiply two binomials together.0028

For example, what did we do when we are looking at x -3 × x +1.0037

Using the method of foil we would multiply our first terms together and get something like x2.0042

We would multiply our outside terms together 1x then we would multiply our inside terms.0049

And finally we would multiply our last terms together. 0060

Sometimes we had to do a little bit of work to clean this up.0067

As long as we made sure that everything got multiplied by everything else. 0070

We where assured that we could multiply these two binomials out.0075

Since we are working with factoring and breaking things down into a product you want to think of this process, but do it in reverse. 0079

If you had a trinomial to begin with, how could you then break this down into two binomials?0087

Since this is the exact same one as before I will simply write down the two binomials that it will break up into.0096

This is the process that we are after of taking a trinomial and breaking it down into two binomials.0105

In doing so it is not quite a straightforward process.0114

The way we are going to attack this is to think of that foil process in our minds.0119

This will help us determine our first and last terms in those binomials. 0124

Now, after we have chosen something for those first and last terms 0129

we will have to check to make sure that the outside and inside terms combined to be our middle term.0133

Sometimes we will have to do some double checking just to make sure that it does combine and give us that middle term.0140

Sometimes there are lots of different options.0146

We may have to do this more than once until we find just the right values that make it work. 0149

Watch how that works with this trinomial.0154

I have x2 + 2x -8 and what we are looking to do is break this down into two binomials.0157

I’m going to go ahead and write down the parentheses just to get it started.0165

I first want to determine what should my first terms be in order to get that x2.0169

We are looking to multiply two things together and get x2.0177

The only two things that will work is x and x.0180

We have a good chance that those are our first terms.0184

Rather than worrying about the outside and inside just yet, we jump all the way to the last terms.0191

That will be here and here and I'm looking to multiply them and get -8.0197

Here is the thing there are lots of different options that we could have, it could be 1 and 8, 2 and 4.0203

We could also look at the other order of these maybe 4 and 2, 8 and 1.0211

We want to choose the proper pair that when combined together will actually give us that 2x in the middle.0217

Let us go ahead and try something.0224

Watch how this process works. 0226

Suppose I just tried the first thing on the list, this 1 and 8.0228

1, 8, with this combination I can be sure that my first terms work out and that my last terms work out.0233

I'm not completely confident until I check those outside and inside terms to make sure that they work out.0242

I will do some quick calculations.0250

Let us check our outside terms, 8 × x = 8x and inside terms 1 and x, and these would combine to give us a 9x.0252

If you compare that to the original it is not the same.0265

What that is indicating is that pair of 1 and 8, those are not the ones we want to use.0270

We can backup a little bit and try another pair of numbers.0277

Let me try something different.0287

I'm going to try 4 and -2.0290

Now when I do my outside and inside terms I will get -2x on the outside, 0298

4x on the inside and those combine to give me 2x, which is the same as my middle term.0305

I know that this is how it should be factored.0313

If you want you can go through the entire foil process, just to double check that all the rest of terms work out.0319

In fact it is not a bad idea when you are done with the factoring process, just to make sure it is okay. 0324

Now, there are a few tips to help you along the way when doing this reverse foil method.0333

It works good, as long as you are leading coefficient is 1 and you can take a look at the signs of the other two coefficients. 0339

In general, here is what you are trying to do.0351

You are looking for two integers whose product will give you c.0354

They multiply and give you c but whose sum is b.0358

It is what we did in that last example. 0363

Now you can get a little bit more information if you look at the signs of b and c.0367

If b and c are both positive then those two integers you are looking for must also be positive. 0371

One situation that might happen is you know both integers that you are looking for will be negative if c is positive, 0383

that is this one in the end and d is negative.0391

Watch for that to happen. 0396

Of course one last thing, you will know that the integers you are looking for are different in sign, one positive and one negative if c is negative.0400

That is the only way they could multiply together and give you a negative number here on the end.0410

Watch for me to use these shortcuts here in just a little bit.0416

We want to use this reverse foil method in order to factor the following polynomial y2 + 12y + 20.0426

I’m going to start off by writing set of parentheses this will break down into some binomials.0435

Let us start off for those first terms.0444

What times what would give us a y2?0447

One thing that will do it is just y and y.0451

Let us look for two values that would multiply and give us a 20.0457

It is okay for me to write down some different possibilities like 1 and 20, 2 and 10, 4 and 5.0462

You can imagine the same values just flipped around.0470

Let us see if we can use any information to help us out. 0474

Notice how this last term out here is positive and so is my middle term, both of them are positive. 0479

Now what is that telling me about my signs, I know that the two numbers I'm looking for will both be positive.0488

That actually is quite a bit of information because now when we look at our list I can pick two things that will add to be the middle term 12. 0496

I know they multiply between.0504

Let us drop those in there, 2 and 10.0507

I have factored the trinomial.0511

Let us quickly go through the foil process just to make sure that this is the one we are looking for. 0514

We are looking to make sure that this matches up with the original. 0520

First terms would be a y2, outside terms 10y, inside terms 2y and last term is 20.0524

These middle ones would combined giving us y2 + 12y + 20 and that shows that our factorization checks out.0534

We know that this is the proper way to factor it.0544

Let us try another one, this one is x2 - 9x – 22.0551

Let us start off in much the same way.0557

Let us write down a set of parentheses and see if we can fill in the blanks.0559

We need two numbers that when multiplied together will give us x2.0566

That must be an x and another x.0571

Now we need a pair of numbers that will multiply and give us a -22.0577

Let us write down some possibilities like 1 and 22, 2 and 11, I think that is it.0581

Let us get some information about the signs of these numbers. 0589

I’m looking at this last number here and notice how it is negative,0594

I know that these two numbers I’m looking for on my last terms, one of them must be positive and one of them must be negative. 0598

The question is which one is positive and which one is negative?0607

We will look to our middle term to help out. 0613

I need the larger term to be negative, so that I will get a negative in the middle, that -9.0617

We have plenty of information it should be pretty clear that it is actually the 2 and 11 off my list that will work.0624

2 and 11.0632

Let us just check it real quick by foiling things out to make sure that this is how it should factor.0637

First terms x2, outside terms -11x, inside terms 2x, and last terms -22.0643

Since I have a positive × negative, these middle terms combined I will get x2 - 9x -22.0656

That is the same as my original, so I know that I have factored it correctly.0666

Let us use the reverse foil method for this polynomial.0679

It is not very big, it is r2 + r + 2.0682

We are going to start off by setting down those two parentheses.0686

We are hunting for two first terms that will multiply to give us r2.0691

There is only one choice that will do that, just r and r.0697

We turn our attention to the last terms and we need them to multiply to be 2.0704

Unfortunately, there is only one possibility for that, 1 and 2.0711

Let us hunt down our signs.0717

The last term is positive, the middle term is positive that says both of the terms that we are looking for must both be positive.0720

Let us put them in and now we can go and check this using the foil process.0728

r × r =r2, the outside terms should be 2r, inside terms 1r, and the last term is 2.0735

Combining the two middle terms here, I get r2 + 3r + 2.0746

Something very interesting is happening with this one, let us take a closer look.0755

If we look at the resulting polynomial that we got after foiling out and we compare that with the original, they are not the same.0759

That tells us something. It tells us that this is not the correct factorization. 0769

Now if this is not how it should be factored then what other possibilities do we have?0783

If we look at all of our possibilities for those last terms, it must contain 1 and 2 if it is going to multiply and give us 2.0789

Since that did not work and I have no other possibilities, it tells us that this does not factor into two binomials using 1 and 2.0798

This is an indication that our original is actually prime.0807

Watch out for ones like this where when you try and factor it, it simply does not factor into those two binomials.0813

Let us try something with a few more variables in it.0824

This one is t2 – 6tu + 8u2.0827

Even though we have a few more variables in here, you will see that this process works out the same as before. 0833

Starting off with those first terms, something × something will give us t2.0841

That must be t and another t.0848

I need two things to multiply and give us 8u2.0853

Well I'm not quite sure about that 8 yet because it could be 1 and 8, 2 and 4, or could be those reversed.0860

I did know about the u, you better have a u and another u in order to get that u2.0867

Let us just focus on numbers for bit and see what information we can get from there.0874

Looking at the sign of my last term it is positive, but my middle term is negative.0881

The information I’m getting from there is that both of my numbers I’m looking for must both be negative.0887

I think we can pick it out from our list now and it looks like we must use the 2 and 4.0896

Finally let us check that to make sure this one works.0906

Let us be careful since we have both t’s and u’s.0909

First terms t2, outside terms -4tu, inside terms -2tu and my last terms – 2 × -4 = 8u2.0912

We can combine our middle terms giving us t2 – 6tu + 8u2.0931

And now we can see that yes, this has been factored correctly since the resulting polynomial is the same as my original.0940

This one is good.0947

One thing to watch out for is to make sure you pull out any common factors at the very beginning. 0953

That is what we will definitely need to do with this example before we even start the forming process.0959

Always check out for a good common factor. 0966

This is also good idea because it can potentially make your number smaller, so that you do not have to think of as many possibilities.0970

We are going to quickly factor 3x4 -15x3 + 18x2.0978

Look with this one, everything here is divisible by 3 and I can pull out an x2 from all of the variables.0985

Let us take it out of the very beginning, I have 3x2 and let us write down what is left.0996

3x4 ÷ 3x2 = x2, =15 ÷ 3 = -5x, 18 ÷ 3 + 6 and I think that is all my leftover parts.1004

With this one now, I will go and do the reverse foil process on that.1024

I need two binomials, let us break it down. 1030

What should my first terms be in order to get my x2?1035

That must be an x and an x.1042

I have to look at my last terms and then multiply together to give us a 6.1047

1 and 6, possibly 2 and 3, but you can use the signs to help you out. 1052

The last term is positive, the middle term is negative, so both of these will be negative.1058

It looks like I need to use that 2 and 3.1066

Some quick checking to make sure this is the correct factorization.1077

I have x2 - 3x - 2x + 6 and looks like my outside and inside terms do combine and give us that - 5x.1080

This is the correct factorization and let us go head and write out that very first 3x2 that we took out at the very beginning.1093

Now that we have all the pieces, we can say that this is the correct factorization.1103

Always look for a common factor that you could pull out from the very beginning before starting the foil method. 1109

Sometimes you can pull something out, sometimes you can not but it will make your life easier if you can find something.1116

Keep that in mind for this next example. 1125

This one is 2x3 - 18x2 - 44x.1128

Let us come at this over.1134

It looks like everything is divisible by 2 and they all have an x in common.1136

Let us take out a 2x at the very beginning. 1142

What do we have left?1148

2x3 ÷ 2x = x2, -18 ÷ 2 = -9x and -44x ÷ 2x = -22.1150

Now we want to factor that into some binomials.1168

Let us go ahead and copy over this 2x just we can keep track of it.1174

I need my first terms to multiply together and get an x2.1180

That will be an x and another x.1186

I need to look at my last terms so that they multiply together to give me a -22.1190

Some of our possibilities are 1 and 22, 2 and 11.1197

Since the last one is negative and my middle term is negative, I know that these will be different in sign.1204

Let us take the 2 and 11 off of our list, those are the ones we need, -11 and 2.1214

Let us quickly combine things together and make sure that it is the correct factorization.1224

x2 - 11x + 2x – 22.1229

Combining these middle guys x2 - 9x -22, so that definitely checks with this polynomial right here. 1236

Now if you want to go ahead and put in the 2x as well, this will take you back all the way to the original one.1249

Remember to use your distribution property so you can see how that will work out.1258

2x3 - 18x2 - 44x and sure enough that is the same as the original.1264

Everything checks out. I know that this is the correct factorization for our polynomial.1275

Just a few things, make sure that when you are using this method, always check for common factor to pull it out. 1286

Make sure you set down your first terms and then your last terms 1292

Thank you for watching www.educator.com.1303

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at factoring some trinomials using a method known as AC method. 0002

We have already seen factoring trinomials once before but these ones are going to be a little bit more complicated0010

and that our squared term will be something other than the number 1.0015

This is why we are going to pick up the AC method, we have a little bit more algebraic way to approach these types of problems.0019

Recall some of the earlier trinomials that we have been factoring so far.0029

The squared term in front has always been 1 and that made life pretty easy on us 0033

because when we went searching for those two binomials to break it down into.0039

We did not have a lot of options in order to get that first term.0044

It is probably something like y and y or x and x.0048

There is not a whole lot of other things it could be.0051

The reason why this made things a little bit easier is we only have to focus on the last term 0054

and making sure our outside and inside terms combined to give us the middle term. 0059

Now, we do not want to necessarily stick with those types of trinomials.0064

We want to go ahead and factor things where the initial term is something other than 1. 0068

Now this will end up making things a little bit more difficult.0073

The good news is with these ones you can still use something like the reverse foil method.0080

You have not only possibilities for your first term, but now you also have possibilities for your first and last terms, both of those.0086

This will make it a little bit more difficult when we are checking to make sure that the outside and inside terms combine to give us that middle term.0096

Let us do a reverse foil example, so you can see that we have to track down many more possibilities. 0108

This one is 2x2 + 7x + 6.0113

It is not that big but we will look at the two binomials that we are looking to break this down into.0117

Like before, I will be looking for two terms that multiplied to give us the 2x2.0124

Since that number is there I have to look at possible things that will give us 2.0129

This one is not too bad, it has to be 1 and 2 to get that x2, 1x and 2x.0136

We will look at the possibilities that will give us our second term.0149

6 could be 1 and 6, could be 2 and 3, or it could actually be those values flipped around.0154

The question is what should it be? What things are we looking for here?0163

I know that these two numbers whatever they are looks like they both better be positive since these are both positive.0170

It could be 1 and 6, could be 2 and 3.0177

Let us go ahead and put one of those in there just to see what happens. 0182

Let us suppose I'm trying out 1and I'm trying out 6. 0187

My first terms multiply out just fine. 0192

My last terms multiply out just fine.0194

Let us check out our outside and inside terms.0197

The outside would give us 6x and the inside would give us 2x.0202

When you combine those together, you get 8x which unfortunately is not the same as that middle term.0212

I need to come up with some other choice for those last terms.0219

Let us see, if it is not 1 and 6, I guess we have to try 2 and 3.0226

I will put those in there and we will double check our outside and inside terms.0232

Outside would be 3x, inside would be 4x and sure enough those do combine to give us that 7x.0240

The important part to recognize is that if your initial term is something other than 1, 0250

and play around with how they are ordered in order to get your proper factorization.0259

Let us try another one that has a few more possibilities for that first term just to make things a little bit more interesting.0270

Looking at the first term, we need something that will multiply and give us 6y2.0283

That could be 1 and 6, could be 2 and 3, but they both will definitely contain y because of that y2.0290

Looking at our last terms, many different things could multiply and give us the 10.0300

1 and 10, 2 and 5 or possibly those just reversed around.0307

I have lots of different options that I can end up packaging this together. 0313

Maybe this is 1 and 6 with the 1 and 10 or maybe I should use the 2 and 3 with the 1 and 10, 0319

or the 1 and 6 with the 2 and 5, or the 2 and 3 with the 2 and 5.0329

There are many different ways and I can also do these in different orders to make sure that they combine.0333

The key for figuring out which combination should you use is looking at those outside and inside terms.0340

In this one, we want them to combine to give us that 19.0349

We definitely have to do a little bit of work to figure out what that is.0352

Let us see for the current setup my outside would be 10y and my inside would be 6y,0355

which unfortunately does not combine enough to give us that 19y in the middle.0360

I know that my 1 and 6, and my 1 and 10 I need to change something around this one.0366

This one is just not going to work that way.0370

Let us play around with our first term.0374

Let us try something else for the beginning here.0378

Let us try 2 and 3.0381

What will that give us?0386

I can see the outside is 20y, the inside is 3y but that is a little too much, 23y not going to work out.0389

The good news is I did do this one earlier so I do have a combination that will actually factor.0406

What we are looking for this one is 2, 3, 5 and 2.0414

Let us check the outside and inside terms for this guy.0423

The outside would be 4y and the inside would be 15y and sure enough those combine to give us 19y.0427

I know that this is the proper factorization.0437

Notice what this highlight is that when your initial term is something other than 1, and you have lots and lots of possibilities to run through, 0441

it can be very difficult to find out just the right combination of numbers to use in order to make it all work out correctly.0449

If my numbers were even bigger I would have even more possibilities to run through.0456

This is a problem.0462

To fix this problem where our leading term is something other than 1, and our numbers could get fairly large,0467

we do not necessarily want to use the reverse foil method.0474

There are simply too many possibilities to consider for some problems and it gets too difficult. 0478

This is why I run to pick up something known as the AC method. 0483

This is a little bit more of an algebraic method that we can use and hunt down some of the possibilities we need for breaking it down. 0486

Let me quickly run you through how the AC method works, and then I will give you some quick tips on using it.0493

The very first thing that you want to do when using AC method is just to see if everything has a common factor or not.0499

And if it does have a common factor, go ahead and factor that out before beginning any other type of factoring process.0507

If they have anything in common, pull that out.0513

Then I will multiply the first and the last terms together, this is known as the A and C terms.0516

This is where the method gets its name. 0521

Once you get that new number, you will be looking for two numbers that multiply to give you AC and they actually add to get you B. 0524

This will have the feel of looking for those two integers, but it will be a little bit more straightforward than what you see in the past.0532

I do have a nice way to organize that step to keep track of the two numbers are looking for.0538

Here comes the interesting part, when you find those two numbers we will actually split up your middle term into two new numbers 0543

and then you will have four terms total and you will actually use factor by grouping to move from there.0550

The AC method is a way of splitting up your middle term and using a different approach that factor by grouping to handle it instead.0556

Let me give you some tips on using the AC method.0565

When you use the AC method you want to organize where your AC and your B terms go.0572

What I recommend is that you draw a small x.0578

In the top part of that x you put the values of A× C and in the bottom part of that actually put the value of B.0585

What you are looking to do is you want to fill out the rest of this x by putting in these two numbers.0595

The numbers that will go there, they must multiply to give you that top number and they must add to give you the bottom number.0604

You will feel like you are filling out a very small crossword. 0617

Be very careful in doing this and make sure that the signs matchup.0621

If you need to add to give you a negative number then make sure the two side numbers will add to give you that negative number.0625

Always be careful on your signs with this one, they should matchup.0633

You have heard a lot about the AC method and have not seen it yet, let us go ahead and do an example see can see it all in action. 0637

We want to factor 10q2 – 23q +12.0647

This is a good example of one that you want use the AC method on.0652

If you try to factor directly you have lots of possibilities for the first term, like 1 and 10, 2 and 5. 0656

You have lots of possibilities for the second term, 1 and 12, 2 and 6, 3 and 4.0662

Of course all of those reversed.0667

Let us tackle this using the AC method.0671

In step one, check all of your numbers to see if they have a common factor.0675

I get 10, 23 and 12 it looks like they do not have anything in common. 0679

Unfortunately, means I can not pull anything out and make the number smaller.0685

In to step two, I want to go ahead and multiply my a term and my c term together.0689

10 × 12 = 120.0697

I’m looking for two numbers that will multiply to give me 120 and add to be -23. 0702

This is where little box will come in handy.0709

Let me just put in our few little notes.0718

We want them to multiply that gives us our top number. 0720

We want them to add to give us that -23. 0725

To help us better find the numbers that will go ahead and do this, 0732

I’m going to start listing out all the pairs of numbers that will multiply to give us 120.0736

1 and 120, 2 and 60, I have 3 and 40, it just keep continuing making this list until you get as many numbers as possible.0742

I got 4 and 30, 5 and 24, 6 and 20, 8 and 15.0757

Now that we have a bunch of numbers on this list, let us see how we can use it. 0770

We want these two numbers to multiply to be 120 and when we built those list that all should multiply to give us 120.0775

But they must add to give us -23. 0784

The only way you want to add to get a negative number and multiply to get a positive number 0788

is if both of these new numbers here were negative.0792

Let us say if you are going to pick two things off this list that will give us -23 when added together.0797

I think it is going to be that last two, 8 and 15.0804

Notice how those will multiply negative × negative will give us that positive 120 and will definitely add to be -23.0812

Those are the two numbers we want.0820

Now comes the interesting thing.0824

What I'm going to do with those two numbers is end up splitting up my original middle term.0827

I'm writing down the numbers of my original polynomial but I'm not writing that -23 in there. 0833

This is where I split it into two terms.0841

This will be -8q – 15q so I have not changed my polynomial. 0845

I just take a look at it in a different way.0855

You will notice how this new polynomial has four terms 1, 2, 3, 4.0858

I'm going to now attack it using factor by grouping, which means I will take these terms two at a time.0863

Let us do the first two and see what they have in common.0871

Both are divisible by 2q.0875

Let us see what we got left over in here.0887

2q, I have 5q - 4 and let us see what is in common with the next two.0890

It looks like I can pull out -3 from both of them.0909

That would leave us 5q - 4 and notice how the signs do match up, -3 × 5 =-15.0923

-3 × -4 =12.0933

We can go ahead and wrapped this one up. 0939

They both have a 5q - 4 in common, I will write that for my first binomial with the leftover pieces being 2q and – 3.0940

It is quite a journey to get to those final two binomials, but notice how it is a little bit more methodically, 0956

not necessarily guessing or picking things out of here.0962

You have a better hunting way of going about it. 0964

Let us quickly check to make sure that this is the correct factor polynomial just by running through the foil process.0969

5q × 2q = 10q2, outside terms -15q, inside terms – 8q, last terms +12. 0974

It is already starting to look pretty good since it looks like that one.0987

Just combine my little terms here and I get 10q2 – 23q + 12, which is exactly the same as I had originally.0991

I know that this one is factored correctly using that AC method.1005

Let us see the AC method again just we can get more familiar with it.1012

In this one we want to factor 5t2 + 13t – 6.1016

I want to make sure that they have a greatest common factor of only one, 1025

which means do they have anything in common that I can factor out at the very beginning.1028

5, 13, and 6 do not have anything in common.1032

Let us move on to multiplying the A and C terms together.1035

5 × -6 = 30.1041

Two numbers that will multiply to give us 30, but add to give us a 13.1046

We will use our box to help us out.1051

It need to multiply to be 30 and add to be 13.1057

The two numbers that we put in here, they must multiply to give us 30 and they will add to give us 13.1064

To help out with the search, we will list down all the things that multiply to give us 30.1077

1 and 30, 2 and 15, 3 and 10, 4 and 15.1083

The only two things that are going to work from this list or the least that I can see I think will be our 3 and 10.1099

Let us go ahead and put those in.1109

Hold on, I think we forgot one of our signs here should be -30 on top.1122

We will try this again. 1131

We need two numbers from our list that will multiply to give us -30, but add to be 13.1133

I think the 2 and the 15 will have to be the one to do it because they have to be different in sign to multiply to give us that -30. 1141

We are ready to split up our middle term.1156

I have written the first term and the last term now we will write out that middle term split using these two new numbers.1159

– 2t + 15t looking pretty good.1169

Now that we have this, we want to factor by grouping.1177

We will take these two at a time. 1181

5t – 2t they only have one thing in common and that would be t.1186

Let us see what is left over.1199

5t - 2 looking at the next two terms 15t – 6 they have a 3 in common.1200

5t – 2.1219

We have the 5t -2 common piece, we will go ahead and take it out of both of them and write out our leftover pieces, t + 3.1224

This one did take quite a bit of work let us quickly check it again by our formula.1236

5t2 + 15t – 2t -6. 1242

These two little terms combine giving us 13t - 6, which is the same as the original.1251

I know that this is the correct factorization.1260

Let us go ahead and try to factor this one using the AC method. 1269

This one has a few more variables in it so notice I have x2 and y2.1273

Do not worry too much about those x and y, what you will see just focus mainly on those numbers and hunt down what those need to be in.1278

Is there anything I can take out at the very beginning?1288

Do they have anything in common?1291

It looks like the 11 is going to make it where they do not have anything in common.1295

I will multiply the A and C terms together.1301

6 × -10 = -60.1304

We need two numbers that will multiply to be -60, but add to be 11.1312

Let us try our box to help us out.1317

-60 and 11. 1322

They need to multiply and give us -60 and add to be the 11.1327

Starting off with writing down all the possibilities to give is that 60.1337

1 and 60 would do it, 2 and 30, 3 and 20, 4 and 15.1341

The two numbers that we use must multiply to be -60.1353

We want one of these numbers to be positive and the other one to be negative.1357

Since we are adding to be 11, the larger number must be positive.1362

I think I see a good option on this list, the 4 and the 15.1368

The 15 is the larger one, so it must be positive and the 4 is the smaller one, so we will make it negative.1372

We will write down our polynomial and split up the middle term into two new terms.1380

We will use 15xy and - 4xy.1391

Notice that we are using that xy here because those are what is on the middle t