INSTRUCTORSCarleen EatonGrant Fraser

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Dr. Carleen Eaton continues on to Algebra 2, and brings with her 10 years of experience in teaching math and science. This course meets or exceeds all state standards and is essential to those having trouble with Algebra in high school or college. With her clear explanations and examples of commonly seen problems, Dr. Eaton will make sure you understand all the confusing concepts in Algebra 2, ranging from Quadratic Inequalities to Matrices and Conic Sections. Dr. Carleen Eaton has an M.D. from the UCLA School of Medicine and in her teaching career has won numerous "Teacher of the Year" awards. She is also continually ranked as one of the top instructors in California.

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I. Equations and Inequalities
  Expressions and Formulas 22:23
   Intro 0:00 
   Order of Operations 0:19 
    Variable 0:27 
    Algebraic Expression 0:46 
    Term 0:57 
    Example: Algebraic Expression 1:25 
    Evaluate Inside Grouping Symbols 1:55 
    Evaluate Powers 2:30 
    Multiply/Divide Left to Right 2:55 
    Add/Subtract Left to Right 3:35 
   Monomials 4:40 
    Examples of Monomials 4:52 
    Constant 5:27 
    Coefficient 5:46 
    Degree 6:25 
    Power 7:15 
   Polynomials 8:02 
    Examples of Polynomials 8:24 
    Binomials, Trinomials, Monomials 8:53 
    Term 9:21 
    Like Terms 10:02 
   Formulas 11:00 
    Example: Pythagorean Theorem 11:15 
   Example 1: Evaluate the Algebraic Expression 11:50 
   Example 2: Evaluate the Algebraic Expression 14:38 
   Example 3: Area of a Triangle 19:11 
   Example 4: Fahrenheit to Celsius 20:41 
  Properties of Real Numbers 20:15
   Intro 0:00 
   Real Numbers 0:07 
    Number Line 0:15 
    Rational Numbers 0:46 
    Irrational Numbers 2:24 
   Venn Diagram of Real Numbers 4:03 
    Irrational Numbers 5:00 
    Rational Numbers 5:19 
    Real Number System 5:27 
    Natural Numbers 5:32 
    Whole Numbers 5:53 
    Integers 6:19 
    Fractions 6:46 
   Properties of Real Numbers 7:15 
    Commutative Property 7:34 
    Associative Property 8:07 
    Identity Property 9:04 
    Inverse Property 9:53 
    Distributive Property 11:03 
   Example 1: What Set of Numbers? 12:21 
   Example 2: What Properties Are Used? 13:56 
   Example 3: Multiplicative Inverse 16:00 
   Example 4: Simplify Using Properties 17:18 
  Solving Equations 19:10
   Intro 0:00 
   Translations 0:06 
    Verbal Expressions and Algebraic Expressions 0:13 
    Example: Sum of Two Numbers 0:19 
    Example: Square of a Number 1:33 
   Properties of Equality 3:20 
    Reflexive Property 3:30 
    Symmetric Property 3:42 
    Transitive Property 4:01 
    Addition Property 5:01 
    Subtraction Property 5:37 
    Multiplication Property 6:02 
    Division Property 6:30 
   Solving Equations 6:58 
    Example: Using Properties 7:18 
   Solving for a Variable 8:25 
    Example: Solve for Z 8:34 
   Example 1: Write Algebraic Expression 10:15 
   Example 2: Write Verbal Expression 11:31 
   Example 3: Solve the Equation 14:05 
   Example 4: Simplify Using Properties 17:26 
  Solving Absolute Value Equations 17:31
   Intro 0:00 
   Absolute Value Expressions 0:09 
    Distance from Zero 0:18 
    Example: Absolute Value Expression 0:24 
   Absolute Value Equations 1:50 
    Example: Absolute Value Equation 2:00 
    Example: Isolate Expression 3:13 
   No Solution 3:46 
    Empty Set 3:58 
    Example: No Solution 4:12 
   Number of Solutions 4:46 
    Check Each Solution 4:57 
    Example: Two Solutions 5:05 
    Example: No Solution 6:18 
    Example: One Solution 6:28 
   Example 1: Evaluate for X 7:16 
   Example 2: Write Verbal Expression 9:08 
   Example 3: Solve the Equation 12:18 
   Example 4: Simplify Using Properties 13:36 
  Solving Inequalities 17:14
   Intro 0:00 
   Properties of Inequalities 0:08 
    Addition Property 0:17 
    Example: Using Numbers 0:30 
    Subtraction Property 1:03 
    Example: Using Numbers 1:19 
   Multiplication Properties 1:44 
    C>0 (Positive Number) 1:50 
    Example: Using Numbers 2:05 
    C<0 (Negative Number) 2:40 
    Example: Using Numbers 3:10 
   Division Properties 4:11 
    C>0 (Positive Number) 4:15 
    Example: Using Numbers 4:27 
    C<0 (Negative Number) 5:21 
    Example: Using Numbers 5:32 
   Describing the Solution Set 6:10 
    Example: Set Builder Notation 6:26 
    Example: Graph (Closed Circle) 7:08 
    Example: Graph (Open Circle) 7:30 
   Example 1: Solve the Inequality 7:58 
   Example 2: Solve the Inequality 9:06 
   Example 3: Solve the Inequality 10:10 
   Example 4: Solve the Inequality 13:12 
  Solving Compound and Absolute Value Inequalities 25:00
   Intro 0:00 
   Compound Inequalities 0:08 
    'And' and 'Or' 0:13 
    Example: And 0:22 
    Example: Or 1:12 
   'And' Inequality 1:41 
    Intersection 1:49 
    Example: Numbers 2:08 
    Example: Inequality 2:43 
   'Or' Inequality 4:35 
    Example: Union 4:45 
    Example: Inequality 5:53 
   Absolute Value Inequalities 7:19 
    Definition of Absolute Value 7:33 
    Examples: Compound Inequalities 8:30 
    Example: Complex Inequality 12:21 
   Example 1: Solve the Inequality 12:54 
   Example 2: Solve the Inequality 17:21 
   Example 3: Solve the Inequality 18:54 
   Example 4: Solve the Inequality 22:15 
II. Linear Relations and Functions
  Relations and Functions 32:05
   Intro 0:00 
   Coordinate Plane 0:20 
    X-Coordinate and Y-Coordinate 0:30 
    Example: Coordinate Pairs 0:37 
    Quadrants 1:20 
   Relations 2:14 
    Domain and Range 2:19 
    Set of Ordered Pairs 2:29 
    As a Table 2:51 
   Functions 4:21 
    One Element in Range 4:32 
    Example: Mapping 4:43 
    Example: Table and Map 6:26 
   One-to-One Functions 8:01 
    Example: One-to-One 8:22 
    Example: Not One-to-One 9:18 
   Graphs of Relations 11:01 
    Discrete and Continuous 11:12 
    Example: Discrete 11:22 
    Example: Continuous 12:30 
   Vertical Line Test 14:09 
    Example: S Curve 14:29 
    Example: Function 16:15 
   Equations, Relations, and Functions 17:03 
    Independent Variable and Dependent Variable 17:16 
   Function Notation 19:11 
    Example: Function Notation 19:23 
   Example 1: Domain and Range 20:51 
   Example 2: Discrete or Continuous 23:03 
   Example 3: Discrete or Continuous 25:53 
   Example 4: Function Notation 30:05 
  Linear Equations 14:46
   Intro 0:00 
   Linear Equations and Functions 0:07 
    Linear Equation 0:19 
    Example: Linear Equation 0:29 
    Example: Linear Function 1:07 
   Standard Form 2:02 
    Integer Constants with No Common Factor 2:08 
    Example: Standard Form 2:27 
   Graphing with Intercepts 4:05 
    X-Intercept and Y-Intercept 4:12 
    Example: Intercepts 4:26 
    Example: Graphing 5:14 
   Example 1: Linear Function 7:53 
   Example 2: Linear Function 9:10 
   Example 3: Standard Form 10:04 
   Example 4: Graph with Intercepts 12:25 
  Slope 23:07
   Intro 0:00 
   Definition of Slope 0:07 
    Change in Y / Change in X 0:26 
    Example: Slope of Graph 0:37 
   Interpretation of Slope 3:07 
    Horizontal Line (0 Slope) 3:13 
    Vertical Line (Undefined Slope) 4:52 
    Rises to Right (Positive Slope) 6:36 
    Falls to Right (Negative Slope) 6:53 
   Parallel Lines 7:18 
    Example: Not Vertical 7:30 
    Example: Vertical 7:58 
   Perpendicular Lines 8:31 
    Example: Perpendicular 8:42 
   Example 1: Slope of Line 10:32 
   Example 2: Graph Line 11:45 
   Example 3: Parallel to Graph 13:37 
   Example 4: Perpendicular to Graph 17:57 
  Writing Linear Functions 23:05
   Intro 0:00 
   Slope Intercept Form 0:11 
    m and b 0:28 
    Example: Graph Using Slope Intercept 0:43 
   Point Slope Form 2:41 
    Relation to Slope Formula 3:03 
    Example: Point Slope Form 4:36 
   Parallel and Perpendicular Lines 6:28 
    Review of Parallel and Perpendicular Lines 6:31 
    Example: Parallel 7:50 
    Example: Perpendicular 9:58 
   Example 1: Slope Intercept Form 11:07 
   Example 2: Slope Intercept Form 13:07 
   Example 3: Parallel 15:49 
   Example 4: Perpendicular 18:42 
  Special Functions 31:05
   Intro 0:00 
   Step Functions 0:07 
    Example: Apple Prices 0:30 
   Absolute Value Function 4:55 
    Example: Absolute Value 5:05 
   Piecewise Functions 9:08 
    Example: Piecewise 9:27 
   Example 1: Absolute Value Function 14:00 
   Example 2: Absolute Value Function 20:39 
   Example 3: Piecewise Function 22:26 
   Example 4: Step Function 25:25 
  Graphing Inequalities 21:42
   Intro 0:00 
   Graphing Linear Inequalities 0:07 
    Shaded Region 0:19 
    Using Test Points 0:32 
    Graph Corresponding Linear Function 0:46 
    Dashed or Solid Lines 0:59 
    Use Test Point 1:21 
    Example: Linear Inequality 1:58 
   Graphing Absolute Value Inequalities 4:50 
    Graph Corresponding Equations 4:59 
    Use Test Point 5:20 
    Example: Absolute Value Inequality 5:38 
   Example 1: Linear Inequality 9:17 
   Example 2: Linear Inequality 11:56 
   Example 3: Linear Inequality 14:29 
   Example 4: Absolute Value Inequality 17:06 
III. Systems of Equations and Inequalities
  Solving Systems of Equations by Graphing 17:13
   Intro 0:00 
   Systems of Equations 0:09 
    Example: Two Equations 0:24 
   Solving by Graphing 0:53 
    Point of Intersection 1:09 
   Types of Systems 2:29 
    Independent (Single Solution) 2:34 
    Dependent (Infinite Solutions) 3:05 
    Inconsistent (No Solution) 4:23 
   Example 1: Solve by Graphing 5:20 
   Example 2: Solve by Graphing 9:10 
   Example 3: Solve by Graphing 12:27 
   Example 4: Solve by Graphing 14:54 
  Solving Systems of Equations Algebraically 23:53
   Intro 0:00 
   Solving by Substitution 0:08 
    Example: System of Equations 0:36 
   Solving by Multiplication 7:22 
    Extra Step of Multiplying 7:38 
    Example: System of Equations 8:00 
   Inconsistent and Dependent Systems 11:14 
    Variables Drop Out 11:48 
    Inconsistent System (Never True) 12:01 
    Constant Equals Constant 12:53 
    Dependent System (Always True) 13:11 
   Example 1: Solve Algebraically 13:58 
   Example 2: Solve Algebraically 15:52 
   Example 3: Solve Algebraically 17:54 
   Example 4: Solve Algebraically 21:40 
  Solving Systems of Inequalities By Graphing 27:12
   Intro 0:00 
   Solving by Graphing 0:08 
    Graph Each Inequality 0:25 
    Overlap 0:35 
    Corresponding Linear Equations 1:03 
    Test Point 1:23 
    Example: System of Inequalities 1:51 
   No Solution 7:06 
    Empty Set 7:26 
    Example: No Solution 7:34 
   Example 1: Solve by Graphing 10:27 
   Example 2: Solve by Graphing 13:30 
   Example 3: Solve by Graphing 17:19 
   Example 4: Solve by Graphing 23:23 
  Solving Systems of Equations in Three Variables 28:53
   Intro 0:00 
   Solving Systems in Three Variables 0:17 
    Triple of Values 0:31 
    Example: Three Variables 0:56 
   Number of Solutions 5:55 
    One Solution 6:08 
    No Solution 6:24 
    Infinite Solutions 7:06 
   Example 1: Solve 3 Variables 7:59 
   Example 2: Solve 3 Variables 13:50 
   Example 3: Solve 3 Variables 19:54 
   Example 4: Solve 3 Variables 25:50 
IV. Matrices
  Basic Matrix Concepts 11:34
   Intro 0:00 
   What is a Matrix 0:26 
    Brackets 0:46 
    Designation 1:21 
    Element 1:47 
    Matrix Equations 1:59 
   Dimensions 2:27 
    Rows (m) and Columns (n) 2:37 
    Examples: Dimensions 2:43 
   Special Matrices 4:22 
    Row Matrix 4:32 
    Column Matrix 5:00 
    Zero Matrix 6:00 
   Equal Matrices 6:30 
    Example: Corresponding Elements 6:36 
   Example 1: Matrix Dimension 8:12 
   Example 2: Matrix Dimension 9:03 
   Example 3: Zero Matrix 9:38 
   Example 4: Row and Column Matrix 10:26 
  Matrix Operations 21:36
   Intro 0:00 
   Matrix Addition 0:18 
    Same Dimensions 0:25 
    Example: Adding Matrices 1:04 
   Matrix Subtraction 3:42 
    Same Dimensions 3:48 
    Example: Subtracting Matrices 4:04 
   Scalar Multiplication 6:08 
    Scalar Constant 6:24 
    Example: Multiplying Matrices 6:32 
   Properties of Matrix Operations 8:23 
    Commutative Property 8:41 
    Associative Property 9:08 
    Distributive Property 9:44 
   Example 1: Matrix Addition 10:24 
   Example 2: Matrix Subtraction 11:58 
   Example 3: Scalar Multiplication 14:23 
   Example 4: Matrix Properties 16:09 
  Matrix Multiplication 29:36
   Intro 0:00 
   Dimension Requirement 0:17 
    n = p 0:24 
    Resulting Product Matrix (m x q) 1:21 
    Example: Multiplication 1:54 
   Matrix Multiplication 3:38 
    Example: Matrix Multiplication 4:07 
   Properties of Matrix Multiplication 10:46 
    Associative Property 11:00 
    Associative Property (Scalar) 11:28 
    Distributive Property 12:06 
    Distributive Property (Scalar) 12:30 
   Example 1: Possible Matrices 13:31 
   Example 2: Multiplying Matrices 17:08 
   Example 3: Multiplying Matrices 20:41 
   Example 4: Matrix Properties 24:41 
  Determinants 33:13
   Intro 0:00 
   What is a Determinant 0:13 
    Square Matrices 0:23 
    Vertical Bars 0:41 
   Determinant of a 2x2 Matrix 1:21 
    Second Order Determinant 1:37 
    Formula 1:45 
    Example: 2x2 Determinant 1:58 
   Determinant of a 3x3 Matrix 2:50 
    Expansion by Minors 3:08 
    Third Order Determinant 3:19 
    Expanding Row One 4:06 
    Example: 3x3 Determinant 6:40 
   Diagonal Method for 3x3 Matrices 13:24 
    Example: Diagonal Method 13:36 
   Example 1: Determinant of 2x2 18:59 
   Example 2: Determinant of 3x3 20:03 
   Example 3: Determinant of 3x3 25:35 
   Example 4: Determinant of 3x3 29:22 
  Cramer's Rule 28:25
   Intro 0:00 
   System of Two Equations in Two Variables 0:16 
    One Variable 0:50 
    Determinant of Denominator 1:14 
    Determinants of Numerators 2:23 
    Example: System of Equations 3:34 
   System of Three Equations in Three Variables 7:06 
    Determinant of Denominator 7:17 
    Determinants of Numerators 7:52 
   Example 1: Two Equations 8:57 
   Example 2: Two Equations 13:21 
   Example 3: Three Equations 17:11 
   Example 4: Three Equations 23:43 
  Identity and Inverse Matrices 22:25
   Intro 0:00 
   Identity Matrix 0:13 
    Example: 2x2 Identity Matrix 0:30 
    Example: 4x4 Identity Matrix 0:50 
    Properties of Identity Matrices 1:24 
    Example: Multiplying Identity Matrix 2:52 
   Matrix Inverses 5:30 
    Writing Matrix Inverse 6:07 
   Inverse of a 2x2 Matrix 6:39 
    Example: 2x2 Matrix 7:31 
   Example 1: Inverse Matrix 10:18 
   Example 2: Find the Inverse Matrix 13:04 
   Example 3: Find the Inverse Matrix 17:53 
   Example 4: Find the Inverse Matrix 20:44 
  Solving Systems of Equations Using Matrices 22:32
   Intro 0:00 
   Matrix Equations 0:11 
    Example: System of Equations 0:21 
   Solving Systems of Equations 4:01 
    Isolate x 4:16 
    Example: Using Numbers 5:10 
    Multiplicative Inverse 5:54 
   Example 1: Write as Matrix Equation 7:18 
   Example 2: Use Matrix Equations 9:12 
   Example 3: Use Matrix Equations 15:06 
   Example 4: Use Matrix Equations 19:35 
V. Quadratic Functions and Inequalities
  Graphing Quadratic Functions 31:48
   Intro 0:00 
   Quadratic Functions 0:12 
    A is Zero 0:27 
    Example: Parabola 0:45 
   Properties of Parabolas 2:08 
    Axis of Symmetry 2:11 
    Vertex 2:32 
    Example: Parabola 2:48 
   Minimum and Maximum Values 9:02 
    Positive or Negative 9:28 
    Upward or Downward 9:58 
    Example: Minimum 10:31 
    Example: Maximum 11:16 
   Example 1: Axis of Symmetry, Vertex, Graph 12:41 
   Example 2: Axis of Symmetry, Vertex, Graph 17:25 
   Example 3: Minimum or Maximum 21:47 
   Example 4: Minimum or Maximum 27:09 
  Solving Quadratic Equations by Graphing 27:03
   Intro 0:00 
   Quadratic Equations 0:16 
    Standard Form 0:18 
    Example: Quadratic Equation 0:47 
   Solving by Graphing 1:41 
    Roots (x-Intercepts) 1:48 
    Example: Number of Solutions 2:12 
   Estimating Solutions 9:23 
    Example: Integer Solutions 9:30 
    Example: Estimating 9:53 
   Example 1: Solve by Graphing 10:52 
   Example 2: Solve by Graphing 15:10 
   Example 1: Solve by Graphing 17:50 
   Example 1: Solve by Graphing 20:54 
  Solving Quadratic Equations by Factoring 19:53
   Intro 0:00 
   Factoring Techniques 0:15 
    Greatest Common Factor (GCF) 0:37 
    Difference of Two Squares 1:48 
    Perfect Square Trinomials 2:30 
    General Trinomials 3:09 
   Zero Product Rule 5:22 
    Example: Zero Product 5:53 
   Example 1: Solve by Factoring 7:46 
   Example 1: Solve by Factoring 9:48 
   Example 1: Solve by Factoring 12:34 
   Example 1: Solve by Factoring 15:28 
  Imaginary and Complex Numbers 35:45
   Intro 0:00 
   Properties of Square Roots 0:10 
    Product Property 0:26 
    Example: Product Property 0:56 
    Quotient Property 2:17 
    Example: Quotient Property 2:35 
   Imaginary Numbers 3:12 
    Imaginary 'i' 3:51 
    Examples: Imaginary Number 4:22 
   Complex Numbers 7:23 
    Real Part and Imaginary Part 7:33 
    Examples: Complex Numbers 7:57 
   Equality 9:37 
    Example: Equal Complex Numbers 9:52 
   Addition and Subtraction 10:12 
    Examples: Adding Complex Numbers 10:25 
   Complex Plane 13:32 
    Horizontal Axis (Real) 13:49 
    Vertical Axis (Imaginary) 13:59 
    Example: Labeling 14:11 
   Multiplication 15:57 
    Example: FOIL Method 16:03 
   Division 18:37 
    Complex Conjugates 18:45 
    Conjugate Pairs 19:10 
    Example: Dividing Complex Numbers 20:00 
   Example 1: Simplify Complex Number 24:50 
   Example 2: Simplify Complex Number 27:56 
   Example 3: Multiply Complex Numbers 29:27 
   Example 3: Dividing Complex Numbers 31:48 
  Completing the Square 27:11
   Intro 0:00 
   Square Root Property 0:12 
    Example: Perfect Square 0:38 
    Example: Perfect Square Trinomial 3:00 
   Completing the Square 4:39 
    Constant Term 4:50 
    Example: Complete the Square 5:04 
   Solve Equations 6:42 
    Add to Both Sides 6:59 
    Example: Complete the Square 7:07 
   Equations Where 'a' Not Equal to 1 10:58 
    Divide by Coefficient 11:08 
    Example: Complete the Square 11:24 
   Complex Solutions 14:05 
    Real and Imaginary 14:14 
    Example: Complex Solution 14:35 
   Example 1: Square Root Property 18:31 
   Example 2: Complete the Square 19:15 
   Example 3: Complete the Square 20:40 
   Example 4: Complete the Square 23:56 
  Quadratic Formula and the Discriminant 22:48
   Intro 0:00 
   Quadratic Formula 0:21 
    Standard Form 0:29 
    Example: Quadratic Formula 0:57 
   One Rational Root 3:00 
    Example: One Root 3:31 
   Complex Solutions 6:16 
    Complex Conjugate 6:28 
    Example: Complex Solution 7:15 
   Discriminant 9:42 
    Positive Discriminant 10:03 
    Perfect Square (Rational) 10:51 
    Not Perfect Square (2 Irrational) 11:27 
    Negative Discriminant 12:28 
    Zero Discriminant 12:57 
   Example 1: Quadratic Formula 13:50 
   Example 2: Quadratic Formula 16:03 
   Example 3: Quadratic Formula 19:00 
   Example 4: Discriminant 21:33 
  Analyzing the Graphs of Quadratic Functions 30:07
   Intro 0:00 
   Vertex Form 0:12 
    H and K 0:32 
    Axis of Symmetry 0:36 
    Vertex 0:42 
    Example: Origin 1:00 
    Example: k = 2 2:12 
    Example: h = 1 4:27 
   Significance of Coefficient 'a' 7:13 
    Example: |a| > 1 7:25 
    Example: |a| < 1 8:18 
    Example: |a| > 0 8:51 
    Example: |a| < 0 9:05 
   Writing Quadratic Equations in Vertex Form 10:22 
    Standard Form to Vertex Form 10:35 
    Example: Standard Form 11:02 
    Example: 'a' Term Not 1 14:42 
   Example 1: Vertex Form 19:47 
   Example 2: Vertex Form 22:09 
   Example 3: Vertex Form 24:32 
   Example 4: Vertex Form 28:23 
  Graphing and Solving Quadratic Inequalities 27:05
   Intro 0:00 
   Graphing Quadratic Inequalities 0:11 
    Test Point 0:18 
    Example: Quadratic Inequality 0:29 
   Solving Quadratic Inequalities 3:57 
    Example: Parameter 4:24 
   Example 1: Graph Inequality 11:16 
   Example 2: Solve Inequality 14:27 
   Example 3: Graph Inequality 19:14 
   Example 4: Solve Inequality 23:48 
VI. Polynomial Functions
  Properties of Exponents 19:29
   Intro 0:00 
   Simplifying Exponential Expressions 0:09 
    Monomial Simplest Form 0:19 
   Negative Exponents 1:07 
    Examples: Simple 1:34 
   Properties of Exponents 3:06 
    Negative Exponents 3:13 
    Multiplying Same Base 3:24 
    Dividing Same Base 3:45 
    Raising Power to a Power 4:33 
    Parentheses (Multiplying) 5:11 
    Parentheses (Dividing) 5:47 
    Raising to 0th Power 6:15 
   Example 1: Simplify Exponents 7:59 
   Example 2: Simplify Exponents 10:41 
   Example 3: Simplify Exponents 14:11 
   Example 4: Simplify Exponents 18:04 
  Operations on Polynomials 13:27
   Intro 0:00 
   Adding and Subtracting Polynomials 0:13 
    Like Terms and Like Monomials 0:23 
    Examples: Adding Monomials 1:14 
   Multiplying Polynomials 3:40 
    Distributive Property 3:44 
    Example: Monomial by Polynomial 4:06 
   Example 1: Simplify Polynomials 5:47 
   Example 2: Simplify Polynomials 6:28 
   Example 3: Simplify Polynomials 8:38 
   Example 4: Simplify Polynomials 10:47 
  Dividing Polynomials 31:11
   Intro 0:00 
   Dividing by a Monomial 0:13 
    Example: Numbers 0:26 
    Example: Polynomial by a Monomial 1:18 
   Long Division 2:28 
    Remainder Term 2:41 
    Example: Dividing with Numbers 3:04 
    Example: With Polynomials 5:01 
    Example: Missing Terms 7:58 
   Synthetic Division 11:44 
    Restriction 12:04 
    Example: Divisor in Form 12:20 
   Divisor in Synthetic Division 15:54 
    Example: Coefficient to 1 16:07 
   Example 1: Divide Polynomials 17:10 
   Example 2: Divide Polynomials 19:08 
   Example 3: Synthetic Division 21:42 
   Example 4: Synthetic Division 25:09 
  Polynomial Functions 22:30
   Intro 0:00 
   Polynomial in One Variable 0:13 
    Leading Coefficient 0:27 
    Example: Polynomial 1:18 
    Degree 1:31 
   Polynomial Functions 2:57 
    Example: Function 3:13 
   Function Values 3:33 
    Example: Numerical Values 3:53 
    Example: Algebraic Expressions 5:11 
   Zeros of Polynomial Functions 5:50 
    Odd Degree 6:04 
    Even Degree 7:29 
   End Behavior 8:28 
    Even Degrees 9:09 
    Example: Leading Coefficient +/- 9:23 
    Odd Degrees 12:51 
    Example: Leading Coefficient +/- 13:00 
   Example 1: Degree and Leading Coefficient 15:03 
   Example 2: Polynomial Function 15:56 
   Example 3: Polynomial Function 17:34 
   Example 4: End Behavior 19:53 
  Analyzing Graphs of Polynomial Functions 33:29
   Intro 0:00 
   Graphing Polynomial Functions 0:11 
    Example: Table and End Behavior 0:39 
   Location Principle 4:43 
    Zero Between Two Points 5:03 
    Example: Location Principle 5:21 
   Maximum and Minimum Points 8:40 
    Relative Maximum and Relative Minimum 9:16 
    Example: Number of Relative Max/Min 11:11 
   Example 1: Graph Polynomial Function 11:57 
   Example 2: Graph Polynomial Function 16:19 
   Example 3: Graph Polynomial Function 23:27 
   Example 4: Graph Polynomial Function 28:35 
  Solving Polynomial Functions 21:10
   Intro 0:00 
   Factoring Polynomials 0:06 
    Greatest Common Factor (GCF) 0:25 
    Difference of Two Squares 1:14 
    Perfect Square Trinomials 2:07 
    General Trinomials 2:57 
    Grouping 4:32 
   Sum and Difference of Two Cubes 6:03 
    Examples: Two Cubes 6:14 
   Quadratic Form 8:22 
    Example: Quadratic Form 8:44 
   Example 1: Factor Polynomial 12:03 
   Example 2: Factor Polynomial 13:54 
   Example 3: Quadratic Form 15:33 
   Example 4: Solve Polynomial Function 17:24 
  Remainder and Factor Theorems 31:21
   Intro 0:00 
   Remainder Theorem 0:07 
    Checking Work 0:22 
    Dividend and Divisor in Theorem 1:12 
    Example: f(a) 2:05 
   Synthetic Substitution 5:43 
    Example: Polynomial Function 6:15 
   Factor Theorem 9:54 
    Example: Numbers 10:16 
    Example: Confirm Factor 11:27 
   Factoring Polynomials 14:48 
    Example: 3rd Degree Polynomial 15:07 
   Example 1: Remainder Theorem 19:17 
   Example 2: Other Factors 21:57 
   Example 3: Remainder Theorem 25:52 
   Example 4: Other Factors 28:21 
  Roots and Zeros 31:27
   Intro 0:00 
   Number of Roots 0:08 
    Not Nature of Roots 0:18 
    Example: Real and Complex Roots 0:25 
   Descartes' Rule of Signs 2:05 
    Positive Real Roots 2:21 
    Example: Positive 2:39 
    Negative Real Roots 5:44 
    Example: Negative 6:06 
   Finding the Roots 9:59 
    Example: Combination of Real and Complex 10:07 
   Conjugate Roots 13:18 
    Example: Conjugate Roots 13:50 
   Example 1: Solve Polynomial 16:03 
   Example 2: Solve Polynomial 18:36 
   Example 3: Possible Combinations 23:13 
   Example 4: Possible Combinations 27:11 
  Rational Zero Theorem 31:16
   Intro 0:00 
   Equation 0:08 
    List of Possibilities 0:16 
    Equation with Constant and Leading Coefficient 1:04 
    Example: Rational Zero 2:46 
   Leading Coefficient Equal to One 7:19 
    Equation with Leading Coefficient of One 7:34 
    Example: Coefficient Equal to 1 8:45 
   Finding Rational Zeros 12:58 
    Division with Remainder Zero 13:32 
   Example 1: Possible Rational Zeros 14:20 
   Example 2: Possible Rational Zeros 16:02 
   Example 3: Possible Rational Zeros 19:58 
   Example 4: Find All Zeros 22:06 
VII. Radical Expressions and Inequalities
  Operations on Functions 34:30
   Intro 0:00 
   Arithmetic Operations 0:07 
    Domain 0:16 
    Intersection 0:24 
    Denominator is Zero 0:49 
    Example: Operations 1:02 
   Composition of Functions 7:18 
    Notation 7:48 
    Right to Left 8:18 
    Example: Composition 8:48 
   Composition is Not Commutative 17:23 
    Example: Not Commutative 17:51 
   Example 1: Function Operations 20:55 
   Example 2: Function Operations 24:34 
   Example 3: Compositions 27:51 
   Example 4: Function Operations 31:09 
  Inverse Functions and Relations 22:42
   Intro 0:00 
   Inverse of a Relation 0:14 
    Example: Ordered Pairs 0:56 
   Inverse of a Function 3:24 
    Domain and Range Switched 3:52 
    Example: Inverse 4:28 
   Procedure to Construct an Inverse Function 6:42 
    f(x) to y 6:42 
    Interchange x and y 6:59 
    Solve for y 7:06 
    Write Inverse f(x) for y 7:14 
    Example: Inverse Function 7:25 
    Example: Inverse Function 2 8:48 
   Inverses and Compositions 10:44 
    Example: Inverse Composition 11:46 
   Example 1: Inverse Relation 14:49 
   Example 2: Inverse of Function 15:40 
   Example 3: Inverse of Function 17:06 
   Example 4: Inverse Functions 18:55 
  Square Root Functions and Inequalities 30:04
   Intro 0:00 
   Square Root Functions 0:07 
    Examples: Square Root Function 0:16 
    Example: Not Square Root Function 0:46 
    Radicand 1:12 
    Example: Restriction 1:31 
   Graphing Square Root Functions 3:42 
    Example: Graphing 3:49 
   Square Root Inequalities 8:47 
    Same Technique 9:00 
    Example: Square Root Inequality 9:20 
   Example 1: Graph Square Root Function 15:19 
   Example 2: Graph Square Root Function 18:03 
   Example 3: Graph Square Root Function 22:41 
   Example 4: Square Root Inequalities 25:37 
  nth Roots 20:46
   Intro 0:00 
   Definition of the nth Root 0:07 
    Example: 5th Root 0:20 
    Example: 6th Root 0:51 
   Principal nth Root 1:39 
    Example: Principal Roots 2:06 
   Using Absolute Values 5:58 
    Example: Square Root 6:18 
    Example: 6th Root 8:40 
    Example: Negative 10:15 
   Example 1: Simplify Radicals 12:23 
   Example 2: Simplify Radicals 13:29 
   Example 3: Simplify Radicals 16:07 
   Example 4: Simplify Radicals 18:18 
  Operations with Radical Expressions 41:11
   Intro 0:00 
   Properties of Radicals 0:16 
    Quotient Property 0:29 
    Example: Quotient 1:00 
    Example: Product Property 1:47 
   Simplifying Radical Expressions 3:24 
    Radicand No nth Powers 3:47 
    Radicand No Fractions 6:33 
    No Radicals in Denominator 7:16 
   Rationalizing Denominators 8:27 
    Example: Radicand nth Power 9:05 
   Conjugate Radical Expressions 11:47 
    Conjugates 12:07 
    Example: Conjugate Radical Expression 13:11 
   Adding and Subtracting Radicals 16:12 
    Same Index, Same Radicand 16:20 
    Example: Like Radicals 16:28 
   Multiplying Radicals 19:04 
    Distributive Property 19:10 
    Example: Multiplying Radicals 19:20 
   Example 1: Simplify Radical 24:11 
   Example 2: Simplify Radicals 28:43 
   Example 3: Simplify Radicals 32:00 
   Example 4: Simplify Radical 36:34 
  Rational Exponents 30:45
   Intro 0:00 
   Definition 1 0:20 
    Example: Using Numbers 0:39 
    Example: Non-Negative 2:46 
    Example: Odd 3:34 
   Definition 2 4:32 
    Restriction 4:52 
    Example: Relate to Definition 1 5:04 
    Example: m Not 1 5:31 
   Simplifying Expressions 7:53 
    Multiplication 8:31 
    Division 9:29 
    Multiply Exponents 10:08 
    Raised Power 11:05 
    Zero Power 11:29 
    Negative Power 11:49 
   Simplified Form 13:52 
    Complex Fraction 14:16 
    Negative Exponents 14:40 
    Example: More Complicated 15:14 
   Example 1: Write as Radical 19:03 
   Example 2: Write with Rational Exponents 20:40 
   Example 3: Complex Fraction 22:09 
   Example 4: Complex Fraction 26:22 
  Solving Radical Equations and Inequalities 31:27
   Intro 0:00 
   Radical Equations 0:11 
    Variables in Radicands 0:22 
    Example: Radical Equation 1:06 
    Example: Complex Equation 2:42 
   Extraneous Roots 7:21 
    Squaring Technique 7:35 
    Double Check 7:44 
    Example: Extraneous 8:21 
   Eliminating nth Roots 10:04 
    Isolate and Raise Power 10:14 
    Example: nth Root 10:27 
   Radical Inequalities 11:27 
    Restriction: Index is Even 11:53 
    Example: Radical Inequality 12:29 
   Example 1: Solve Radical Equation 15:41 
   Example 2: Solve Radical Equation 17:44 
   Example 3: Solve Radical Inequality 20:24 
   Example 4: Solve Radical Equation 24:34 
VIII. Rational Equations and Inequalities
  Multiplying and Dividing Rational Expressions 40:54
   Intro 0:00 
   Simplifying Rational Expressions 0:22 
    Algebraic Fraction 0:29 
    Examples: Rational Expressions 0:49 
    Example: GCF 1:33 
    Example: Simplify Rational Expression 2:26 
   Factoring -1 4:04 
    Example: Simplify with -1 4:19 
   Multiplying and Dividing Rational Expressions 6:59 
    Multiplying and Dividing 7:28 
    Example: Multiplying Rational Expressions 8:36 
    Example: Dividing Rational Expressions 11:20 
   Factoring 14:01 
    Factoring Polynomials 14:19 
    Example: Factoring 14:35 
   Complex Fractions 18:22 
    Example: Numbers 18:37 
    Example: Algebraic Complex Fractions 19:25 
   Example 1: Simplify Rational Expression 25:56 
   Example 2: Simplify Rational Expression 29:34 
   Example 3: Simplify Rational Expression 31:39 
   Example 4: Simplify Rational Expression 37:50 
  Adding and Subtracting Rational Expressions 55:04
   Intro 0:00 
   Least Common Multiple (LCM) 0:27 
    Examples: LCM of Numbers 0:43 
    Example: LCM of Polynomials 4:02 
   Adding and Subtracting 7:55 
    Least Common Denominator (LCD) 8:07 
    Example: Numbers 8:17 
    Example: Rational Expressions 11:03 
    Equivalent Fractions 15:22 
   Simplifying Complex Fractions 21:19 
    Example: Previous Lessons 21:36 
    Example: More Complex 22:53 
   Example 1: Find LCM 28:30 
   Example 2: Add Rational Expressions 31:44 
   Example 3: Subtract Rational Expressions 39:18 
   Example 4: Simplify Rational Expression 38:26 
  Graphing Rational Functions 57:13
   Intro 0:00 
   Rational Functions 0:18 
    Restriction 0:34 
    Example: Rational Function 0:51 
   Breaks in Continuity 2:52 
    Example: Continuous Function 3:10 
    Discontinuities 3:30 
    Example: Excluded Values 4:37 
   Graphs and Discontinuities 5:02 
    Common Binomial Factor (Hole) 5:08 
    Example: Common Factor 5:31 
    Asymptote 10:06 
    Example: Vertical Asymptote 11:08 
   Horizontal Asymptotes 20:00 
    Example: Horizontal Asymptote 20:25 
   Example 1: Holes and Vertical Asymptotes 26:12 
   Example 2: Graph Rational Faction 28:35 
   Example 3: Graph Rational Faction 39:23 
   Example 4: Graph Rational Faction 47:28 
  Direct, Joint, and Inverse Variation 20:21
   Intro 0:00 
   Direct Variation 0:07 
    Constant of Variation 0:25 
   Graph of Constant Variation 1:26 
    Slope is Constant k 1:35 
    Example: Straight Lines 1:41 
   Joint Variation 2:48 
    Three Variables 2:52 
   Inverse Variation 3:38 
    Rewritten Form 3:52 
    Examples in Biology 4:22 
   Graph of Inverse Variation 4:51 
    Asymptotes are Axes 5:12 
    Example: Inverse Variation 5:40 
   Proportions 10:11 
    Direct Variation 10:25 
    Inverse Variation 11:32 
   Example 1: Type of Variation 12:42 
   Example 2: Direct Variation 14:13 
   Example 3: Joint Variation 16:24 
   Example 4: Graph Rational Faction 18:50 
  Solving Rational Equations and Inequalities 55:14
   Intro 0:00 
   Rational Equations 0:15 
    Example: Algebraic Fraction 0:26 
    Least Common Denominator 0:49 
    Example: Simple Rational Equation 1:22 
    Example: Solve Rational Equation 5:40 
   Extraneous Solutions 9:31 
    Double check 10:00 
    No Solution 10:38 
    Example: Extraneous 10:44 
   Rational Inequalities 14:01 
    Excluded Values 14:31 
    Solve Related Equation 14:49 
    Find Intervals 14:58 
    Use Test Values 15:25 
    Example: Rational Inequality 15:51 
    Example: Rational Inequality 2 17:07 
   Example 1: Rational Equation 28:50 
   Example 2: Rational Equation 33:51 
   Example 3: Rational Equation 38:19 
   Example 4: Rational Inequality 46:49 
IX. Exponential and Logarithmic Relations
  Exponential Functions 35:58
   Intro 0:00 
   What is an Exponential Function? 0:12 
    Restriction on b 0:31 
    Base 0:46 
    Example: Exponents as Bases 0:56 
    Variables as Exponents 1:12 
    Example: Exponential Function 1:50 
   Graphing Exponential Functions 2:33 
    Example: Using Table 2:49 
   Properties 11:52 
    Continuous and One to One 12:00 
    Domain is All Real Numbers 13:14 
    X-Axis Asymptote 13:55 
    Y-Intercept 14:02 
    Reflection Across Y-Axis 14:31 
   Growth and Decay 15:06 
    Exponential Growth 15:10 
    Real Life Examples 15:41 
    Example: Growth 15:52 
    Example: Decay 16:12 
    Real Life Examples 16:30 
   Equations 17:32 
    Bases are Same 18:05 
    Examples: Variables as Exponents 18:20 
   Inequalities 21:29 
    Property 21:51 
    Example: Inequality 22:37 
   Example 1: Graph Exponential Function 24:05 
   Example 2: Growth or Decay 27:50 
   Example 3: Exponential Equation 29:31 
   Example 4: Exponential Inequality 32:54 
  Logarithms and Logarithmic Functions 45:54
   Intro 0:00 
   What are Logarithms? 0:08 
    Restrictions 0:15 
    Written Form 0:26 
    Logarithms are Exponents 0:52 
    Example: Logarithms 1:49 
   Logarithmic Functions 5:14 
    Same Restrictions 5:30 
    Inverses 5:53 
    Example: Logarithmic Function 6:24 
   Graph of the Logarithmic Function 9:20 
    Example: Using Table 9:35 
   Properties 15:09 
    Continuous and One to One 15:14 
    Domain 15:36 
    Range 15:56 
    Y-Axis is Asymptote 16:02 
    X Intercept 16:12 
   Inverse Property 16:57 
    Compositions of Functions 17:10 
   Equations 18:30 
    Example: Logarithmic Equation 19:13 
   Inequalities 20:36 
    Properties 20:47 
    Example: Logarithmic Inequality 21:40 
   Equations with Logarithms on Both Sides 24:43 
    Property 24:51 
    Example: Both Sides 25:23 
   Inequalities with Logarithms on Both Sides 26:52 
    Property 27:02 
    Example: Both Sides 28:05 
   Example 1: Solve Log Equation 31:52 
   Example 2: Solve Log Equation 33:53 
   Example 3: Solve Log Equation 36:15 
   Example 4: Solve Log Inequality 39:19 
  Properties of Logarithms 28:43
   Intro 0:00 
   Product Property 0:08 
    Example: Product 0:46 
   Quotient Property 2:40 
    Example: Quotient 2:59 
   Power Property 3:51 
    Moved Exponent 4:07 
    Example: Power 4:37 
   Equations 5:15 
    Example: Use Properties 5:58 
   Example 1: Simplify Log 11:17 
   Example 2: Single Log 15:54 
   Example 3: Solve Log Equation 18:48 
   Example 4: Solve Log Equation 22:13 
  Common Logarithms 25:23
   Intro 0:00 
   What are Common Logarithms? 0:10 
    Real World Applications 0:16 
    Base Not Written 0:27 
    Example: Base 10 0:39 
   Equations 1:47 
    Example: Same Base 1:56 
    Example: Different Base 2:37 
   Inequalities 6:07 
    Multiplying/Dividing Inequality 6:21 
    Example: Log Inequality 6:54 
   Change of Base 12:45 
    Base 10 13:24 
    Example: Change of Base 14:05 
   Example 1: Log Equation 15:21 
   Example 2: Common Logs 17:13 
   Example 3: Log Equation 18:22 
   Example 4: Log Inequality 21:52 
  Base e and Natural Logarithms 21:14
   Intro 0:00 
   Number e 0:09 
    Natural Base 0:21 
    Growth/Decay 0:33 
    Example: Exponential Function 0:53 
   Natural Logarithms 1:11 
    ln x 1:19 
    Inverse and Identity Function 1:39 
    Example: Inverse Composition 1:55 
   Equations and Inequalities 4:39 
    Extraneous Solutions 5:30 
    Examples: Natural Log Equations 5:48 
   Example 1: Natural Log Equation 9:08 
   Example 2: Natural Log Equation 10:37 
   Example 3: Natural Log Inequality 16:54 
   Example 4: Natural Log Inequality 18:16 
  Exponential Growth and Decay 24:30
   Intro 0:00 
   Decay 0:17 
    Decreases by Fixed Percentage 0:23 
    Rate of Decay 0:56 
    Example: Finance 1:34 
   Scientific Model of Decay 3:37 
    Exponential Decay 3:45 
    Radioactive Decay 4:13 
    Example: Half Life 5:33 
   Growth 9:06 
    Increases by Fixed Percentage 9:18 
    Example: Finance 10:09 
   Scientific Model of Growth 11:35 
    Population Growth 12:04 
    Example: Growth 12:20 
   Example 1: Computer Price 14:00 
   Example 2: Stock Price 15:46 
   Example 3: Medicine Disintegration 19:10 
   Example 4: Population Growth 22:33 
X. Conic Sections
  Midpoint and Distance Formulas 32:42
   Intro 0:00 
   Midpoint Formula 0:15 
    Example: Midpoint 0:30 
   Distance Formula 2:30 
    Example: Distance 2:52 
   Example 1: Midpoint and Distance 4:58 
   Example 2: Midpoint and Distance 8:07 
   Example 3: Median Length 18:51 
   Example 4: Perimeter and Area 23:36 
  Parabolas 41:27
   Intro 0:00 
   What is a Parabola? 0:20 
    Definition of a Parabola 0:29 
    Focus 0:59 
    Directrix 1:15 
    Axis of Symmetry 3:08 
   Vertex 3:33 
    Minimum or Maximum 3:44 
   Standard Form 4:59 
    Horizontal Parabolas 5:08 
    Vertex Form 5:19 
    Upward or Downward 5:41 
    Example: Standard Form 6:06 
   Graphing Parabolas 8:31 
    Shifting 8:51 
    Example: Completing the Square 9:22 
    Symmetry and Translation 12:18 
    Example: Graph Parabola 12:40 
   Latus Rectum 17:13 
    Length 18:15 
    Example: Latus Rectum 18:35 
   Horizontal Parabolas 18:57 
    Not Functions 20:08 
    Example: Horizontal Parabola 21:21 
   Focus and Directrix 24:11 
    Horizontal 24:48 
   Example 1: Parabola Standard Form 25:12 
   Example 2: Graph Parabola 30:00 
   Example 3: Graph Parabola 33:13 
   Example 4: Parabola Equation 37:28 
  Circles 21:03
   Intro 0:00 
   What are Circles? 0:08 
    Example: Equidistant 0:17 
    Radius 0:32 
   Equation of a Circle 0:44 
    Example: Standard Form 1:11 
   Graphing Circles 1:47 
    Example: Circle 1:56 
   Center Not at Origin 3:07 
    Example: Completing the Square 3:51 
   Example 1: Equation of Circle 6:44 
   Example 2: Center and Radius 11:51 
   Example 3: Radius 15:08 
   Example 4: Equation of Circle 16:57 
  Ellipses 46:51
   Intro 0:00 
   What Are Ellipses? 0:11 
    Foci 0:23 
   Properties of Ellipses 1:43 
    Major Axis, Minor Axis 1:47 
    Center 1:54 
    Length of Major Axis and Minor Axis 3:21 
   Standard Form 5:33 
    Example: Standard Form of Ellipse 6:09 
   Vertical Major Axis 9:14 
    Example: Vertical Major Axis 9:46 
   Graphing Ellipses 12:51 
    Complete the Square and Symmetry 13:00 
    Example: Graphing Ellipse 13:16 
   Equation with Center at (h, k) 19:57 
    Horizontal and Vertical 20:14 
    Difference 20:27 
    Example: Center at (h, k) 20:55 
   Example 1: Equation of Ellipse 24:05 
   Example 2: Equation of Ellipse 27:57 
   Example 3: Equation of Ellipse 32:32 
   Example 4: Graph Ellipse 38:27 
  Hyperbolas 38:15
   Intro 0:00 
   What are Hyperbolas? 0:12 
    Two Branches 0:18 
    Foci 0:38 
   Properties 2:00 
    Transverse Axis and Conjugate Axis 2:06 
    Vertices 2:46 
    Length of Transverse Axis 3:14 
    Distance Between Foci 3:31 
    Length of Conjugate Axis 3:38 
   Standard Form 5:45 
    Vertex Location 6:36 
    Known Points 6:52 
   Vertical Transverse Axis 7:26 
    Vertex Location 7:50 
   Asymptotes 8:36 
    Vertex Location 8:56 
    Rectangle 9:28 
    Diagonals 10:29 
   Graphing Hyperbolas 12:58 
    Example: Hyperbola 13:16 
   Equation with Center at (h, k) 16:32 
    Example: Center at (h, k) 17:21 
   Example 1: Equation of Hyperbola 19:20 
   Example 2: Equation of Hyperbola 22:48 
   Example 3: Graph Hyperbola 26:05 
   Example 4: Equation of Hyperbola 36:29 
  Conic Sections 18:43
   Intro 0:00 
   Conic Sections 0:16 
    Double Cone Sections 0:24 
   Standard Form 1:27 
    General Form 1:37 
   Identify Conic Sections 2:16 
    B = 0 2:50 
    X and Y 3:22 
   Identify Conic Sections, Cont. 4:46 
    Parabola 5:17 
    Circle 5:51 
    Ellipse 6:31 
    Hyperbola 7:10 
   Example 1: Identify Conic Section 8:01 
   Example 2: Identify Conic Section 11:03 
   Example 3: Identify Conic Section 11:38 
   Example 4: Identify Conic Section 14:50 
  Solving Quadratic Systems 47:04
   Intro 0:00 
   Linear Quadratic Systems 0:22 
    Example: Linear Quadratic System 0:45 
   Solutions 2:49 
    Graphs of Possible Solutions 3:10 
   Quadratic Quadratic System 4:10 
    Example: Elimination 4:21 
   Solutions 11:39 
    Example: 0, 1, 2, 3, 4 Solutions 11:50 
   Systems of Quadratic Inequalities 12:48 
    Example: Quadratic Inequality 13:09 
   Example 1: Solve Quadratic System 21:42 
   Example 2: Solve Quadratic System 29:13 
   Example 3: Solve Quadratic System 35:02 
   Example 4: Solve Quadratic Inequality 40:29 
XI. Sequences and Series
  Arithmetic Sequences 21:16
   Intro 0:00 
   Sequences 0:10 
    General Form of Sequence 0:16 
    Example: Finite/Infinite Sequences 0:33 
   Arithmetic Sequences 0:28 
    Common Difference 2:41 
    Example: Arithmetic Sequence 2:50 
   Formula for the nth Term 3:51 
    Example: nth Term 4:32 
   Equation for the nth Term 6:37 
    Example: Using Formula 6:56 
   Arithmetic Means 9:47 
    Example: Arithmetic Means 10:16 
   Example 1: nth Term 12:38 
   Example 2: Arithmetic Means 13:49 
   Example 3: Arithmetic Means 16:12 
   Example 4: nth Term 18:26 
  Arithmetic Series 21:36
   Intro 0:00 
   What are Arithmetic Series? 0:11 
    Common Difference 0:28 
    Example: Arithmetic Sequence 0:43 
    Example: Arithmetic Series 1:09 
    Finite/Infinite Series 1:36 
   Sum of Arithmetic Series 2:27 
    Example: Sum 3:21 
   Sigma Notation 5:53 
    Index 6:14 
    Example: Sigma Notation 7:14 
   Example 1: First Term 9:00 
   Example 2: Three Terms 10:52 
   Example 3: Sum of Series 14:14 
   Example 4: Sum of Series 18:13 
  Geometric Sequences 23:03
   Intro 0:00 
   Geometric Sequences 0:11 
    Common Difference 0:38 
    Common Ratio 1:08 
    Example: Geometric Sequence 2:38 
   nth Term of a Geometric Sequence 4:41 
    Example: nth Term 4:56 
   Geometric Means 6:51 
    Example: Geometric Mean 7:09 
   Example 1: 9th Term 12:04 
   Example 2: Geometric Means 15:18 
   Example 3: nth Term 18:32 
   Example 4: Three Terms 20:59 
  Geometric Series 22:43
   Intro 0:00 
   What are Geometric Series? 0:11 
    List of Numbers 0:24 
    Example: Geometric Series 1:12 
   Sum of Geometric Series 2:16 
    Example: Sum of Geometric Series 2:41 
   Sigma Notation 4:21 
    Lower Index, Upper Index 4:38 
    Example: Sigma Notation 4:57 
   Another Sum Formula 6:08 
    Example: n Unknown 6:28 
   Specific Terms 7:41 
    Sum Formula 7:56 
    Example: Specific Term 8:11 
   Example 1: Sum of Geometric Series 10:02 
   Example 2: Sum of 8 Terms 14:15 
   Example 3: Sum of Geometric Series 18:23 
   Example 4: First Term 20:16 
  Infinite Geometric Series 18:32
   Intro 0:00 
   What are Infinite Geometric Series 0:10 
    Example: Finite 0:29 
    Example: Infinite 0:51 
    Partial Sums 1:09 
    Formula 1:37 
   Sum of an Infinite Geometric Series 2:39 
    Convergent Series 2:58 
    Example: Sum of Convergent Series 3:28 
   Sigma Notation 7:31 
    Example: Sigma 8:17 
   Repeating Decimals 8:42 
    Example: Repeating Decimal 8:53 
   Example 1: Sum of Infinite Geometric Series 12:15 
   Example 2: Repeating Decimal 13:24 
   Example 3: Sum of Infinite Geometric Series 15:14 
   Example 4: Repeating Decimal 16:48 
  Recursion and Special Sequences 14:34
   Intro 0:00 
   Fibonacci Sequence 0:05 
    Background of Fibonacci 0:23 
    Recursive Formula 0:37 
    Fibonacci Sequence 0:52 
    Example: Recursive Formula 2:18 
   Iteration 3:49 
    Example: Iteration 4:30 
   Example 1: Five Terms 7:08 
   Example 2: Three Terms 9:00 
   Example 3: Five Terms 10:38 
   Example 4: Three Iterates 12:41 
  Binomial Theorem 48:30
   Intro 0:00 
   Pascal's Triangle 0:06 
    Expand Binomial 0:13 
    Pascal's Triangle 4:26 
   Properties 6:52 
    Example: Properties of Binomials 6:58 
   Factorials 9:11 
    Product 9:28 
    Example: Factorial 9:45 
   Binomial Theorem 11:08 
    Example: Binomial Theorem 13:48 
   Finding a Specific Term 18:36 
    Example: Specific Term 19:26 
   Example 1: Expand 24:39 
   Example 2: Fourth Term 30:26 
   Example 3: Five Terms 36:13 
   Example 4: Three Iterates 45:07 

Welcome to Educator.com.0000

Today is our first lesson for the Algebra II series, and we are going to start out with some review of concepts from Algebra I.0002

If you need more detail about any of these concepts, please check out the Algebra I series here at Educator.0008

The first session is on expressions and formulas.0014

Recall the earlier concepts of variables and algebraic expressions:0020

starting out with some definitions, a variable is a letter or symbol that is used to represent an unknown number.0025

It could be any letter; frequently, x, y, and z are used, but you could choose n or s or w.0034

Algebraic expressions means that terms using both variables and numbers are combined using arithmetic operations.0045

Remember that a term is a number, or a variable, or both.0057

So, a term could be 4--that is a constant, and it is a term; it could be 2x; it could be y2.0065

And when these are combined using arithmetic operations, then they are known as expressions.0076

And when variables are involved, then they are algebraic expressions.0082

For example, 4x3+2xy-1 would be an example of an algebraic expression.0085

The rules specifying order of operations are very important; and they are used in order to evaluate algebraic expressions.0097

Recall the procedure to evaluate an expression using the order of operations.0107

First, evaluate expressions that are inside grouping symbols: examples would be parentheses, braces, and brackets.0116

The next thing, when you are evaluating an algebraic expression, is to evaluate powers.0151

So, if a term is raised to a power (such as 42 or 34), you need to evaluate that next; that is the second step.0160

Next is to multiply and divide, going from left to right.0175

You start out at the left side of an expression; and if you hit something that needs to be divided, you do that.0193

And you proceed towards the right; if you see something that needs to be multiplied, you do that.0198

It is not "multiply all the way, and then go back and divide"; it is "start at the left; any multiplication or division--do it."0203

Move to the next step; move towards the right; multiply or divide...and so on, until all of that has been taken care of.0208

Finally, you do the same thing with addition and subtraction: you add and subtract from left to right.0216

And we will be illustrating these concepts in the examples.0226

One thing to recall is that a fraction bar can function as a grouping symbol.0231

For example, if I have something like 3x-2x+2, all over 4(x+3)+3, I would treat this as a grouping symbol.0236

And I would simplify this as far as I could, going through my four steps; and then I would simplify this;0248

and then I would divide this simplified expression on the top by the simplified expression on the bottom.0253

And remember, the reason that we use order of operations is that, if we didn't, and everybody was just doing things their own way,0259

we couldn't really communicate using math, because people would write something down,0265

and somebody might do it in a different order and come up with a different answer.0269

So, this way, it is an agreed-upon set of rules that everyone follows.0273

Monomials: a monomial is a product of a number and 0 or more variables.0281

Again, refreshing your memory from Algebra I: examples of a monomial would be 5y, 6xy2, z, 5.0288

So, it says it is a product of a number and zero or more variables.0304

Here, there aren't any variables, so that actually is simply a constant; but it is still called a monomial, also.0307

Here, I have 5 times one variable; here I have multiple variables.0314

This is examples of...these are all monomials.0321

A constant is simply a number; so, it could be -3 or 6 or 14; those are constants.0328

Coefficients: a coefficient is the number in front of the variable.0346

Up here, I said I had 5y and 6xy2: this is a coefficient: 5 is a coefficient, and 6 is a coefficient.0363

When you see something like z, it does have a coefficient: it actually has a coefficient of 1.0372

However, by convention, we usually don't write the 1--we just write it as z, but it actually does have a coefficient of 1.0378

Next, degree--the degree of a monomial: the degree of a monomial is the sum of the degrees of all of the variables.0386

So, it is the sum of the degree of all of the variables.0394

For example, 3xy2z4: if I want to find the degree, I am going to add the degree of each variable.0408

This is x (but that really means x to the 1--the 1 is unstated) plus y2 (the degree is 2), plus z4 (the degree is 4).0418

Adding these up, the degree for this monomial is 7.0429

When we talk about powers: powers refer to a number or variable being multiplied by itself n times, where n is the power.0435

For example, if I say that I have 52, what I am really saying is 5 times 5.0447

So, 5 is being multiplied by itself twice, where n equals 2.0455

I could say I have y4: that equals y times y times y times y; and here, the power is 4.0461

OK, continuing on with more concepts: a polynomial is a monomial or a sum of monomials.0477

Recall the concepts of term, like terms, binomial, and trinomial.0485

A polynomial is simply an expression in which the terms are monomials.0492

And we say "sum," but this applies to subtraction, as well--a polynomial can certainly involve subtraction.0496

For example, 4x2+x or 2y2+3y+4: these are both polynomials.0504

We also could say that 5z is a polynomial, but it is also a monomial; there is only one term, so it is a polynomial, but we usually just say it is a monomial.0519

OK, so looking at these other words: a binomial is a polynomial that contains two terms.0532

So, it is the sum of two monomials, whereas the trinomial is the sum of three monomials.0544

A monomial is simply a single monomial.0552

Recall that, as discussed, a term is a number or a letter (which is a variable) or both, separated by a sign.0561

Terms could be a number; they could be a variable; or they could be both.0572

3x-7+z: here, I have a number and a variable; here, I just have a number (I have a constant); here, I just have a variable.0582

And they are separated by signs--by a negative sign and a positive sign--so each one of these is a term.0594

The concept of like terms is very important, because like terms can be added or subtracted.0602

Like terms contain the same variables to the same powers.0609

For example, 1 and 6 are like terms; they don't contain any variables, so they are like terms.0626

3xy and 4xy are like terms; they both contain an x to the first power and a y to the first power.0636

2y2 and 8y2 are also like terms: they both contain a y raised to the second power.0645

And so, these can be combined: they can be added and subtracted.0653

A formula is an equation involving several variables (2 or more), and it describes a relationship among the quantities represented by the variables.0661

And we have worked with formulas previously: and just to review, one formula that we talked about is the Pythagorean Theorem.0670

That is a2+b2=c2, where c is the length0678

of the hypotenuse of a right triangle, and a and b are the lengths of the two sides.0683

And this tells us the relationship among the three sides of the triangle.0689

And that is really what formulas are all about, and really what algebra is all about: describing relationships between various things.0697

And of course, during this course, we are going to be working with various formulas.0706

OK, in this example, we are asked to simplify or evaluate an algebraic expression.0711

5x2...and they are telling me x=3, y=-3; so I have some x terms and some y terms.0718

My first step is to substitute: so, everywhere I have an x, I am putting in a 3; everywhere I have a y, I am putting in a -3.0724

So, here I have 3xy, so here it is going to be 3 times 3 times -3.0736

Recall the order of operations: the first thing I am going to do is to get rid of the grouping symbols.0740

Take care of the parentheses; and looking, I do have parentheses.0746

In here, I have a negative and a negative; so I am simplifying that just to positive 3.0755

OK, continuing to simplify inside the parentheses: 3 plus 3 is 6.0766

I completed my first step in the order of operations.0775

The next thing to do is evaluate powers; and I do have some terms that are raised to various powers.0778

32 is 9, minus 2 times 63; so, 6 times 6 is 36, times 6 is 216.0785

That took care of my powers; and the next thing is going to be to multiply and divide.0801

And when we do that, we always proceed from left to right.0807

2 times 216 is 432; OK, now I have: 3 times 3 is 9; 9 times -3 is -27.0812

I am going to rewrite this as 9 minus 432 minus 27.0825

Finally, add and subtract; and this is going from left to right.0830

9 minus 432 gives me -423, minus 27 (so now I have another bit of subtraction to do--that is -423-27) gives me -450.0839

So again, the first step was substituting in 3 and -3 for x and y.0854

The next step was to get rid of my grouping symbols; evaluate the powers;0860

multiply and divide, going from left to right (and I just had multiplication here);0867

and then, add and subtract, going from left to right, to get -450.0873

In this second example, again, we are asked to evaluate an algebraic expression; and here, we have three variables: a, b, and c.0879

So, carefully substituting in each of these, a=-1...so -12, minus 2, times b (b is 2), times c (c is 3), plus 33.0886

Here, I have c2 in the denominator; so that gives me 32, minus 2, times a (which is -1), times b (which is 2).0907

Since there was a lot of substituting, it is a good idea to check your work.0918

a is -1 (that is -12), minus 2, times b, times c, plus c3;0923

all of that is divided by 32 (so that is c2) minus 2, times a, times b.0934

Everything looks good; now, the first thing I want to do is eliminate grouping symbols.0941

Recall that, in this type of a case, the fraction bar is functioning as a grouping symbol.0945

So, the whole numerator should be simplified, and the denominator should be simplified; and then I should divide one by the other.0951

Starting with the numerator: within the numerator, there are not any grouping symbols,0958

so I am going to go ahead and go to the next step, which is to take care of powers.0964

And -1 times -1 is 1; and then, I have 33; that is 3 times 3 (is 9), times 3 (is 27).0971

OK, I can do the same thing in the denominator; I can just do these both in parallel.0991

And so, I am going to evaluate the powers in the denominator.0997

3 times 3 is 9; and then, I don't have any more powers--OK.1000

So, I took care of that; my next step is going to be to multiply and divide.1005

OK, so I have, in the numerator, 1 minus 2 times 2 (is 4), and then 4 times 3 (is 12), plus 27.1016

So, that took care of that step; now, in the denominator, I have 9, and then I have minus 2, times -1.1035

So, 2 times -1 is going to give me -2; -2 times 2 is going to give me -4.1046

OK, so now, I have taken care of all of the multiplication and the division.1063

The next step is to add and subtract--once again, going from left to right.1068

So, starting up here, the next step is going to be 1 minus 12; 1 minus 12 is going to give me -11.1073

So, it is -11 plus 27; that is going to leave me with 16 in the numerator.1086

In the denominator, I have 9, minus -4; well, a negative and a negative gives me a positive,1092

so in the denominator, I actually have 9 plus 4, which gives me 13.1099

The result is 16 over 13.1109

Again, starting out by substituting values for a, b, and c...I have done that in this first step.1112

And then, I treat this fraction bar as a big grouping symbol, and then I take care of the numerator and the denominator separately.1120

You could have done them one at a time, or you can do steps at the same time.1128

So, first, evaluate the powers; I did that in the numerator; I did that in the denominator (I am treating them separately).1132

Multiplying and dividing: I did my multiplication here and in the denominator.1138

And finally, adding and subtracting to get 16/13.1146

Example 3: The formula for the area of a triangle is Area equals 1/2 bh.1152

So, this is actually that the area equals one-half the base times the height of the triangle.1158

Find the height if a is 32, and the base (b) is 8.1164

OK, here we are being asked to find the height, and we are given the other two variables.1173

So, let me rewrite the formula: area equals 1/2 base times height.1179

Now, I am going to substitute in what I was given.1183

I am given the area; I am given the base; and I need to find the height.1185

What I need to do is isolate h; so, first simplifying this: 32=...well, 1/2 of 8 is 4, so that gives me...4h.1195

Next, divide both sides of the equation by 4 (32/4 and 4h/4) in order to isolate that.1206

Well, 32 divided by 4 is 8; the 4's cancel out on the right; and then just rewriting this in a more standard form, with the variable on left, the height is 8.1216

So again, first just write out the formula; substitute in a and b (which I was given).1228

And then, solve for the height.1237

The temperature in Fahrenheit is F=9/5C+32, where C is the temperature in Celsius.1242

If the temperature is 78 degrees Fahrenheit, what is it in Celsius?1250

Rewrite the formula and substitute in what is given.1255

What is given is that the temperature in Fahrenheit is 78.1263

And I am looking for Celsius (I always keep in mind what I am looking for--what is my goal?).1267

And that is +32; my goal here is going to be to solve for C--to isolate that.1272

Subtracting 32 from both sides gives me 46=9/5C.1281

Now, in order to isolate the Celsius, I am going to multiply both sides by 5/9.1290

When I do that, I am going to get 5 times 46 (is 230), and that is divided by 9.1302

Here, that all cancels out; so, rewriting this, Celsius equals 230/9.1310

That is not usually how we talk about temperature; so simplifying that, if I took 230 and divided it by 9, I would get approximately 25.5 degrees Celsius.1317

So again, the formula for converting Fahrenheit into Celsius (or vice versa) is given.1327

I substituted in 78 degrees and figured this out: so, 78 degrees would be equal to approximately 25.5 degrees Celsius.1335

That concludes today's lesson on Educator.com; and I will see you again for the next lesson.1346

Welcome to Educator.com.0000

In today's lesson, we will be discussing properties of real numbers.0002

A real number corresponds to a point on the number line, and real numbers may be either rational or irrational.0007

So first, just talking about the number line: some examples of real numbers would be 0, 1, 1/2,0016

or (expressing a number as a decimal) it can be 2.8; negative numbers--maybe -2.18 or -4.0032

OK, these are all real numbers, and they are rational; so let's talk about the difference between rational and irrational numbers.0042

Rational numbers can be expressed as a fraction.0051

So, rational numbers are expressed in the form a/b, where a and b are integers, and b does not equal 0,0062

because as you will recall, we cannot have 0 in the denominator, because that would result in an undefined expression.0075

When a rational number is expressed as a decimal, it will either be terminating or repeating.0084

Let me explain what I mean by this: if I have an example, such as 1/2, and I convert that to decimal form, it is going to terminate; it is .5.0097

I might have another number, 1/3, which is also rational; and it is going to end up being .3333, and on indefinitely.0108

This could also be written as .3 with a bar over it, indicating it is repeating.0122

So, this is also rational, because it repeats; the repeating pattern could be longer--it could be 2.387387387, so this is repeating.0126

The point is that they either terminate or repeat when expressed in decimal form.0139

Irrational numbers cannot be expressed in the form a/b, so they are expressed as decimal form.0145

But they neither terminate nor repeat in decimal form; they just go on indefinitely.0156

For example, I could have 4.871469837246, and on and on and on, with no pattern--no repeating and no ending.0174

Another example would be π; we often express π as 3.14, but it actually goes on and on indefinitely; it is just approximately equal to 3.14.0189

Therefore, we can add π up here (that is an irrational number) to represent some irrational numbers up here.0201

In addition, the square root of numbers, other than the square root of numbers that are perfect squares, are irrational numbers.0209

So, the square root of 3 is, or the square root of 2.0217

If it is a perfect square, that means it is the result of multiplying an integer by itself.0221

For example, 2 times 2 is 4, so that is a perfect square; and the square root of 4 is rational--in fact, it just equals 2.0227

All other square roots are irrational numbers; so again, both rational and irrational numbers are real numbers, but they have different properties.0234

Sometimes, we express the relationship between the various types of real numbers using a Venn diagram.0244

So, I am going to go ahead and break down the real number system into its subsets, and then show you how this works as a Venn diagram.0252

We have the real number system, and we have irrational numbers, and we also have rational numbers.0261

And a Venn diagram is a visual way of understanding the relationship between these and their various subsets.0282

You could use circles; I am using rectangles and squares...whichever.0295

Irrational numbers: this is sometimes just known as Q with a line over it, and some examples that I just discussed--0300

the square root of 2, π, the square root of 5--these are irrational numbers.0313

And then, I have rational numbers, sometimes expressed as Q; OK, this is the real number system.0319

Within rational numbers are some subsets; the first one is the natural numbers.0333

And the natural numbers are the numbers that we use to count things (1, 2, 3, and on); they are the numbers we use in counting.0344

There is a slightly bigger set of numbers known as the whole numbers.0354

The whole numbers include the natural numbers (this square is around the natural numbers, because it includes all of those).0362

And it also includes 0; so, we add 0 to this set; and it does not include negative numbers.0368

Next are the integers: the integers include natural numbers, whole numbers, and also negative numbers.0379

So, 0 is included, and on.0399

OK, so then, you can come out to just rational numbers (that are not whole numbers, natural numbers, or integers).0403

And you can get fractions included, like -1/2, 0, 1, 2, 3/2, and on.0411

The real number system is broken down into rational and irrational; and rational numbers0422

are further broken down into the natural numbers, the whole numbers, and the integers.0426

In algebra, it is important to understand the properties of real numbers--what it is0436

that you are allowed to do, and not allowed to do, when working with real numbers.0441

So, we call the various properties commutative, associative, identity, inverse, and distributive.0446

Reviewing those: the commutative property applies both to addition and to multiplication.0452

And what the commutative property tells us is that two terms can be added in either order, or multiplied in either order.0463

So, if I have two numbers, a+b, I can change that order, b+a, and it is not going to change my result.0471

For multiplication, the commutative property under multiplication, I could say a times b equals b times a.0478

The associative property also applies to both addition and multiplication.0487

The associative property tells you that, when you are adding or multiplying, the terms can be grouped in any way, and the result will be the same.0495

Remember that we sometimes use grouping symbols with expressions and equations.0504

So, looking at this, I could group it as (a+b)+c, or I could group it as a+(b+c)--group those together.0510

Either way, they are equivalent; it is not going to change my result.0522

The same holds true for multiplication: if I have (ab)c, that equals a(bc), and it doesn't matter0526

if I decide to group it like this or like this--if I multiply these first or these first.0536

The identity property: when I think of this, I just remember that the identity property tells me that the number maintains its identity.0544

It doesn't change; so, for addition, what this says is that the sum of any number and 0 is the original number.0553

So, a+0 is still a; so, the identity of the number does not change just because you add 0 to it.0565

For multiplication, the product of any number and 1 is the original number; that is the identity property under multiplication.0575

So, a times 1 is still the original number, a.0586

The inverse property, as applied to addition, says that, when you add the same number,0593

but with the opposite sign (a + -a, or say, 3 + -3), the result is 0.0607

A number plus its additive inverse is equal to 0.0617

This property can also be applied to multiplication; and this only applies to real numbers other than 0, when applied to multiplication.0625

And you will see why.0640

What this says is that, if you multiply a number by its reciprocal with the same sign, the result will be 1.0641

We can't apply 0 here, because that would give us a 0 in the denominator, which is an undefined expression.0650

So, a number times its reciprocal gives you 1.0657

Finally, the distributive property: you can review more about this in the Algebra I lectures.0663

This is a very important property when working with equations, solving equations, and working with algebraic expressions.0669

Recall that a(b+c) equals ab + ac.0677

So, we go forward to multiply; and when we go in the reverse direction, recall that that is factoring.0685

This property also applies to multiplying a number by terms that are subtracting (ab-ac).0691

If you put the numbers you are adding first, it doesn't change the property--it still applies; you get ab + ac.0704

The same for subtraction.0716

And finally, recall that this property can be applied to several numbers.0722

You can have more than two numbers in the parentheses: this could be a(b + c + d).0725

And then, you just multiply each one out: ab + ac + ad.0732

And again, this is something we are going to be using a lot throughout the course.0737

All right, applying some of these concepts to the examples: Example 1: What sets of numbers do these belong to (starting with 6)?0741

Well, 6 is a real number; it is also a rational number (I can easily express this as a fraction: 6/1).0750

And so, it is a rational number; then, I think about the subsets.0760

Is it a natural number? Yes, it is a number that can be used in counting (1, 2, 3, 4, etc.), so it is a natural number.0767

And it is therefore also a whole number; the whole numbers encompass the natural numbers; and it is an integer.0776

6 belongs to all of these categories.0786

The square root of 20 is a real number; however, recall that, unless you are talking about the square root of a perfect square, it is irrational.0789

Square roots of perfect squares are rational; other square roots are irrational.0799

-4/5: this is expressed as a fraction, so...well, it is a real number; and it is also rational, because it can be expressed as a fraction.0808

It is not a natural number, because it is negative, and it is a fraction.0821

The same thing: it is not a whole number, and it is also not an integer.0826

So, this belongs to the two groups real and rational.0830

OK, Example 2 asks what properties are used: so, there is an expression here--a mathematical expression--and various steps are taken.0838

We need to determine which properties were used that allowed those steps to be taken.0849

Well, looking at what happened between here and here, we started out with 2, times 4 plus 3 plus 7.0855

The order of these was switched: 3+7 is still grouped together in parentheses, but it was put before the 4.0863

Remember that the commutative property is the property that says that, when adding,0872

you can change the order that you are adding terms in, and still get the same result; so this is the commutative property.0877

OK, the next step: I look at what happened, and this big set of parentheses is gone.0884

And the way it was removed is by use of the distributive property.0891

Recall that the distributive property says that a, times (b + c), equals ab + ac; and that is what was done here.0894

2 times the whole expression (3+7), and then 2 times 4; this is the distributive property.0904

In the next step, the order of these two numbers was changed; so that is commutative.0919

And a 0 was also added to 4; and remember that, according to the identity property,0926

you can add 0 to a number, and the original number is unchanged (4 + 0 is 4).0931

So, this is commutative and identity.0937

Finally, it is getting rid of the rest of these parentheses by using the distributive property: 2 times 7, plus 2 times 3, plus 2 times 4, plus 2 times 0.0940

So again, we are using the distributive property.0951

Example 3 asks what is the multiplicative inverse of -6 and 7/8.0961

Recall: multiplicative inverse, the definition, is a number times the reciprocal of that number; and recall that that equals 1.0967

OK, so the multiplicative inverse--I have -6 and 7/8, so I am going to change this from a mixed number to a fraction.0979

6 times 8 is 48, plus 7 is 55; and this is negative, so that is -55/8.0987

So, looking at it as a fraction makes it much simpler; and I just need to take the multiplicative inverse of that.0996

And so, I would change that to -8/55.1007

And it does satisfy this formula right here, because if I took -55/8 (which is my original number),1012

and I multiplied it by -8/55, two negatives (a negative times a negative) gives me a positive;1020

the 8's cancel; then, the 55's cancel to give me 1.1028

So, I was able to check my work by seeing that it satisfies this equation.1033

OK, in Example 4, we are asked to simplify and state the property used for each step of simplification.1038

First, I want to get rid of the parentheses; so I am going to use the distributive property.1044

Multiplying out, recall that the distributive property is: a(b+c)=ab+ac.1051

OK, this is 2(6x) + 2(3y + 4z).1059

So, right now, I am just removing these outer parentheses; I am keeping these intact--that will take a second round.1068

So, this is plus -3, times the entire expression in the parentheses, plus -3, times z.1074

OK, I am going to apply the distributive property again, in order to remove the remaining parentheses.1084

This gives me: 2(3y), plus 2(4z), plus -3(3x), plus -3(-y), plus -3(z).1095

Now, I am going to multiply these out: this is 12x + 6y + 8z - 9x + (a negative and a negative is a positive, so that gives me) 3y - (this is -3z).1119

OK, now I am going to group together like terms: and that is using the commutative property--I can change the order of these terms.1145

I have my x's, 12x-9x; I have my y's, and that is 6y and 3y; and then finally, z's: 8z and -3z.1157

All that is left to do is add like terms; so, 12x-9x is 3x; 6y and 3y gives me 9y; and 8z-3z is 5z.1179

So, we are simplifying this, using first the distributive property (to remove the outer parentheses),1193

then the distributive property to remove these other sets of parentheses,1198

and the commutative property to re-order this to group like terms, and then simply adding or subtracting.1203

That concludes this lesson from Educator.com; see you next lesson.1211

Welcome to Educator.com.0000

In today's lesson, we are going to be talking about solving equations.0002

Recall that verbal expressions can be translated into algebraic expressions, and vice versa.0007

So, you can go from verbal expressions to algebraic, and from algebraic back to verbal.0014

For example, if I were given a sentence (a verbal expression) such as, "The sum of two numbers is equal to 7,"0019

I could translate that into an algebraic expression.0041

And the technique is to first assign variables; so I am looking, and I see that I have two numbers.0043

And they don't tell me what the numbers are, so I need to assign variables; and I am going to assign x and y.0049

This is actually an equation, not just an expression; so if I see that I have an equal sign (like this is equal),0057

I note where that is, because that is going to divide it into the left and right sides of the equation.0064

So, I have an equal sign; I am going to first deal with one side of the equation, and it says "the sum of two numbers."0068

So, sum means adding; and my two numbers I am going to call x and y; "is equal to 7."0075

And then, double-check: the sum of two numbers (x + y--that works out) is equal to (I have that taken care of) 7.0084

Or another example: A number is 5 less than the square of another number.0094

And you need to be careful with this "is less than," because it is easy to get that backwards.0113

So, first I notice that I have two numbers; I have "a number" and "another number."0118

Once again, I am going to use x for one of the numbers; and for the other number, I am going to use y; so x and y are my variables.0123

Now, it says "a number is," so I have "is"--that tells me where the equal sign is.0132

So, the left side of the equation is just a number.0141

With "5 less than" or "something less than," the temptation is sometimes to say "5 minus."0146

But that is not correct; what this is saying is to look at what comes after (the square of another number) and subtract 5 from that.0152

So, a number is, and then 5 less than...what is it 5 less than? the square of another number.0161

I am calling my other number y, so y2 - 5, not 5 - y2.0168

So, checking it: A number (x) is (=) 5 less than the square of another number (y2 - 5).0176

And later on in the examples, we will also see a situation where we are given an algebraic expression and asked to translate it into a verbal expression.0188

As you are solving equations, you need to keep in mind the properties of equality that you have used previously (but it is good to review).0201

The first property is the reflexive property: the reflexive property states that, for every real number a, a is equal to a.0209

So, a number is equal to itself.0219

The symmetric property states that, for all real numbers, if a equals b, then b equals a.0222

So, I can switch the left and right sides of the equation, and I am not going to change anything fundamental.0233

So, those are the reflexive and symmetric properties.0239

Another property is the transitive property: the transitive property states that, if a equals b, and b equals c, then a equals c.0241

Think of it this way: if a actually was 5 (let a equal 5), if I told you that a equals b, then the only way it is going to equal b is if b is also 5.0257

So, b must be 5; and then, if I went on and said, "OK, well, b equals c," if that is 5, then c must also be 5.0270

So, I go back; then 5 equals 5, or a equals c--it just follows through.0277

OK, that is the first three properties discussed up here.0284

But equality also satisfies the addition, subtraction, multiplication, and division properties.0288

And these are properties that you have used before in Algebra I; and again, you can review these in detail in the Algebra I lessons.0293

To refresh them here: the addition principle says that, if the same number is added to both sides of the equation, the resulting equation is true.0299

For example, if I have x - 5 = 10, and I want to solve for x, the way I am going to do that is: I am going to add 5 to both sides.0309

And these 5's cancel out; that will give me x = 15.0325

So, the resulting equation after I add 5, if I add 5 to both sides--this equation is also true.0329

The subtraction principle is the same idea, that if I have an equation (x + 3 = 5), I can subtract 3 from both sides;0338

I can subtract the same number from both sides, and the resulting equation is also true.0357

Multiplication principle: again, this is something you have used before to solve equations.0363

And it states that, if the same number is multiplied by both sides of the equation, the resulting equation is true.0367

So, I might have x/2 = 12; to solve that, I need to multiply both sides by 2; and that is allowable.0376

Division: the same idea--in this case, I may have 4x = 6; to solve that, I am going to divide both sides by 4.0391

So, the important thing is: if you do something to one side, you need to do the same thing to the other side of the equation.0406

OK, knowing those properties and the properties of real numbers (the properties of equality and the properties of real numbers), you can solve equations.0417

And we are going to use these properties frequently, even if you don't always know them by name.0427

You know how to do them, and you are going to be applying them throughout the course.0432

For example, if I have 19 = 3x + 4, since usually we have the variable on the left,0437

I can apply the symmetric property and just change this to 3x + 4 = 19.0447

That is the symmetric property that allows me to flip around the two sides of the equation.0451

Next, I need to isolate x, so I am going to subtract 4 from both sides.0459

And that utilizes the subtraction property, which is going to give me...subtracting 4 from both sides, I get 3x; the 4's cancel out, and I get 3x = 15.0465

Well, now I need to isolate x with one last step; and I am going to do that by using the division property.0480

If I divide both sides by 3, I am using the division property; and that is going to give me...the 3's will cancel out, and I will get x = 5.0487

So, by applying those principles, I was able to solve this simple equation.0499

Sometimes, in algebra, you will be asked to solve for a variable.0505

These properties can be used to solve a formula for a variable.0509

I may be given a formula such as 4x - 2y + z = 8.0514

And I might be asked to solve for z; so what I am going to do is solve for z in terms of x and y.0521

While I may not figure out the actual numerical value of z, I can solve for it; I can isolate it and solve for it in terms of the other variables.0532

So again, I am going to apply these same principles.0541

First, I am going to use the subtraction principle; and I can start out by subtracting 4x from both sides.0545

That is going to give me...these 4x's will cancel out; -2y + z = 8 - 4x.0561

Now, I need to add 2y to both sides, because I had -2y; so then, I am going to add 2y to both sides,0570

so that my 2y's cancel out to give me z = 8 - 4x + 2y.0590

And we usually write this so that we have the x's, then the y (in alphabetical order), and then the constant.0600

So, z equals -4x + 2y + 8; this is solving for z in terms of x and y.0607

First example: write an algebraic expression for the sum of three times the cube of one number and twice the square of another number.0617

OK, starting out, I am figuring out how many variables I have.0627

The sum of three times the cube of one number and twice the square of another number...x an y.0632

I am going to assign those as my variables, because I have two variables.0640

And there is no equal sign; this is an expression, not an equation, so I don't have an equal sign in it.0643

And it says "the sum," so I am doing addition.0651

"The sum of three times the cube of one number"--that is 3x3,0656

"and"--so that tells me that that is where the addition goes--"twice the square of another number";0663

so here is the other number: "twice the square of the other number."0670

Checking that: the sum of three times the cube of one number (3x3) and (+) twice--two times the square of another number (2y2).0675

Next, we are asked to do the opposite, which is to write a verbal expression for this algebraic expression.0692

OK, so looking at what I have: here, first, I have the difference of two squares;0699

and I am also looking and seeing that I have two variables.0706

So, the difference of the square of a number (x--I am just calling it "a number") and the square of another number (that is my y2)...0709

Next, I have an equal sign: so "is equal to"...0737

the sum of the square (we have sum--we are adding), and I want to make it clear0743

that we are talking about the same number here and here, so I am going to say "the square of the first number,"0759

and (I am adding again) 3 times (multiplying three times x times y) the product of0770

(my first number and my second number) the first number and the second number, plus the square of the second number.0787

OK, so this was quite long, but you can check it.0820

"The difference of the square of a number and the square of another number" (I have that right here)0823

"is equal to" (there is my equal sign) "the sum" (it's a sum) "of the square of the first number,0827

and three times the product of the first number and the second number, plus the square of the second number."0833

So, it is pretty long, but we got everything covered clearly.0839

The third example is to solve: and we can use the properties that we discussed today in order to solve this.0845

First, I am going to need to get rid of these parentheses; and I can do that by using the distributive property.0852

So, -2 times x gives me -2x; plus -2, times -8, minus...actually, let's do plus -3 times 9, plus -3 times -2x, equals 4 times 2x plus 4 times -9.0858

So, just checking: I have -2x + -2(-8) + -3(9) + -3(-2x) = 4(2x) + 4(-9).0892

Multiplying everything out to simplify: -2x; and then -2 times -8 gives me +16.0909

And then, I have -3 times 9; that is actually a negative, -27.0919

-3 times -2x; that is positive, +6x.0928

That equals...4 times 2x is 8x; 4 times -9 is -36.0934

The next step is to combine like terms; and I have -2x and 6x--those can be combined; that is going to give me 4x.0941

I have constants: I have 16 - 27, which is going to give me -11.0951

I can't really do anything further with that right side.0957

Now that I am this far, what I want to do is isolate the x (isolate the variable); I can do that by using my addition property.0961

I am going to add 11 to both sides; the 11's cancel out--that is going to leave me with 4x = 8x, and -36 + 11 is -25.0969

Next, I am going to subtract 8x from both sides--again, trying to get like terms together and isolate the x.0984

4x - 8x is going to give me -4x; equals...these cancel; that is -25.0994

Now, I am going to divide both sides by -4 to get x = -25/-4.1003

And the negative and the negative is a positive, so x = 25/4; I can either leave it like that, or I can go on and say this equals 6 and 1/4.1011

So again, starting out, getting rid of the parentheses by using the distributive property and carefully multiplying out each section got me to here.1020

We're adding like terms to get to this step, then isolating the x through using the addition property,1031

the subtraction property, and the division property to get x = 6 1/4.1039

OK, in this example, we are asked to solve for h.1047

And we are given...this is actually the surface area of a cylinder; the surface area of a cylinder1050

equals 2π, times the radius, times the height, plus 2πr2.1055

So, I need to solve for h in terms of these other variables.1061

And I am going to start out by using the symmetric property to rewrite this with the h on the left.1069

Let me rewrite this as 2πrh + 2πr2 = s.1077

To get the h by itself, I can start out by subtracting 2πr2 from both sides.1083

OK, once I have done that, then this is going to be gone from this side; and I will have 2πrh.1095

2πrh = s - 2πr2.1106

In order to get the h by itself, I need to divide both sides by 2πr.1111

2πr will cancel out, leaving h isolated on the left, and s - 2πr2 all divided by 2πr.1121

And you really can't go any farther with this.1130

I started out by rewriting this with the value, the h, which we are trying to isolate on the left,1132

using the subtraction property and the division property to isolate the h.1140

That concludes this session of Educator.com, and I will see you next lesson.1147

Welcome to Educator.com.0000

In today's lesson, we are going to be covering solving absolute value equations.0002

Recall that the absolute value x, written as |x| (this is the symbol for absolute value), is the distance from x to 0 on a number line.0009

For example, if you have the absolute value of x equals three, what you are saying is0022

that the absolute value of this number is 3 away from 0 on the number line (1, 2, 3).0036

OK, so looking at this: if I go from 0 to 3, this is 3 units away from 0 on the number line; therefore, x could equal 3,0052

because 3 is 3 away from 0 on the number line.0070

However, consider this: -3 is also 3 units away from 0 on the number line.0074

So, x could also equal -3; so, if the absolute value of x equals 3, x could be 3 (since the absolute value of 3 is 3),0081

and x could be -3 (since the absolute value of -3 is also 3).0093

Knowing this, and really understanding this definition, will allow you to solve equations involving absolute value.0102

Again, absolute value equations: you can solve equations containing absolute values, using the definition of absolute value.0110

For example, |9x + 2| = 29; well, I already said that, if the absolute value of x is 3, that means that x equals 3, or x equals -3.0121

So, you can apply that same concept right here: 9x + 2 = 29, or 9x + 2 = -29.0134

So, once you remove the absolute value bars (and you can do that by turning it into two related equations,0144

one where it equals the positive, and one where it equals the negative value)--once you have done that,0151

all you need to do is solve each equation.0155

So, I am going to solve, just using my usual techniques: subtract two from both sides, and that gives me 9x = 27.0159

Over here, if I subtract 2 from both sides, I am going to get 9x = -31.0169

Then, I am going to divide both sides by 9.0176

OK, so you handle the absolute value equations the same way, even if it is more complex.0186

The other thing to keep in mind is that sometimes, you have to first isolate the absolute value expression on the left side of the equation.0193

For example, if you were given 3 times |5x + 4| equals 6, the first step would be to divide both sides by 3,0199

because once you have done that, then you have the absolute value isolated,0210

and you can proceed by saying 5x + 4 = 2 or 5x + 4 = -2, and then solving both of those.0214

In some situations, some absolute value equations, where the absolute value equals c, if c is less than 0, these have no solution.0227

So, if it says that the absolute value of x is a negative number, then there is no solution; and we just say that the solution is the empty set.0236

And to make this more concrete: if I said that the absolute value of x equals -4...well, there is no situation0249

where an absolute value is going to be a negative number, because remember: the absolute value is defined0258

as the distance between that absolute value and 0 on the number line.0266

And you can't have negative distance; so there is no solution here.0271

And instead, we either just write the empty set as such, or like this, indicating that there is no solution.0275

We just talked about a situation where there is no solution.0287

An absolute value equation can have 0 solutions (which we just discussed--it is the empty set), 1 solution, or 2 solutions.0290

And it is important to check each answer to make sure that it is a valid solution.0298

For example, if I have |x + 4| = 9, I am going to go ahead and solve that,0304

using the usual technique of turning it into two related equations, x + 4 = 9 and x + 4 = -9.0311

I am going to solve that: x = 5; here, I am going to get x = -13.0321

So, let's check this by going back to the original, |x + 4| = 9.0329

If x = 5, then I am going to get |5 + 4| = 9; so, the absolute value of 9 equals 9--and that is true.0335

Since this is true, this is a valid solution; x = 5 is a valid solution.0345

OK, doing the same thing for my other solution: |x + 4| = 9...0350

trying this |-13 + 4| = 9, well, -13 + 4 is -9, so the absolute value of -9 is 9.0358

This is also a valid solution, since this is a true statement.0368

So here, I have two solutions: previously, we discussed that, if you end up with something like |x| = -6, there is no solution; it is the empty set.0372

The situation where you can get one solution is if you end up with |x| = 0.0390

So, if I went to solve this, I would say, "x equals +0, and x equals -0"; but that doesn't really just make sense; they are just 0.0400

Therefore, there is only one solution: x = 0; so here, I have one solution.0408

Three possibilities: no solution, one solution, or two solutions.0414

Or you may think you have two solutions, and then you go back and plug them in, and find out one is not valid.0419

And in that case, you could have something like this, that appears that it would end up with two solutions,0425

and it only ends up with one, or possibly even none.0430

OK, Example 1: Evaluate for x = -3; and this is multiple absolute value terms here--0437

substituting in -3 for x, solving for the absolute value of this would be 3 times -3...that is going to give me |-9 - 4|;0447

here, I have 2 times -3; that is |-6 + 3|; minus 3 times...the absolute value of a negative, times a negative, is a positive.0469

OK, this is |-9 - 4|; that is the absolute value of -13, plus |-6 + 3|; that is -3; minus 3...absolute value of 3.0480

Now, I just need to find the absolute value for each of these.0493

Well, the absolute value of -13 is 13; the absolute value of -3 is 3; the absolute value of 3 is 3;0497

but this time, we are multiplying it times a -3; this is -3 times |3| (which is 3).0512

So here, it's 13 + 3...-3 times 3 is -9, so 13 + 3 is 16, minus 9 is 7.0523

OK, so we are solving this by substituting in -3 for x, finding the absolute values for these three, and then this one is multiplied by -3, and then simply adding.0536

Example 2: I have an absolute value expression on the left, but it is not isolated.0550

So, the first step is to isolate the absolute value, and then find the two related equations and solve them.0555

Divide both sides by 4 to get that isolated.0562

Now, recall that the absolute value of x equals 2, for example: this means that x could equal 2,0570

or x could equal -2--to just illustrate the definition of absolute value.0578

In order to get rid of these absolute value symbols, I am going to create two related equations,0585

3x + 4 = 12, or 3x + 4 = -12: then I am going to solve both, and check to make sure that they are valid solutions.0590

3x = 8; so divide both sides...I had 3x + 4, so I subtracted 4 from both sides.0602

And then, over here, I have 3x = -16, subtracting 4 from both sides--I am just doing these together.0619

Now, dividing both sides by 3 is going to give me x = 8/3; dividing here, I get -16/3.0631

Now, check each of these in the original: that is this 4 times the absolute value of 3x + 4 equals 48.0641

First checking this one: 4, and this is 3 times 8/3 + 4, equals 48...so the 3's cancel out, and that gives me 8 + 4...equals 48,0652

so 4 times the absolute value of 12 equals 48; the absolute value of 12 is 12, so 4 times 12 equals 48; and it does, so this is valid.0672

This first solution is valid.0685

Now, substituting in -16/3, again, in this original: that is 4 times |3(-16/3) +4| = 48.0688

The 3's cancel out; that gives me 4|16 + 4| = 48.0704

-16 +4 is -12...equals 48...the absolute value of -12 is 12; so again, I come up with 4 times 12 equals 48, and that is valid.0710

That is a true statement; so both of these solutions are valid.0724

So this time, I had two solutions, 8/3 and -16/3, that satisfy this absolute value equation.0729

Example 3: again, my first step is to isolate the absolute value expression on the left side of the equation.0740

And I am going so start out by subtracting 15 from both sides, and that is going to give me -3 on the right.0748

Now, I want to divide both sides by 3, and that is going to give me |2x - 4| = -1.0756

And I don't even need to go any farther, because what this is saying is that0766

the absolute value of whatever this expression ends up being (once I solve for x)--the absolute value of this equals -1.0769

Well, that is not valid; you cannot have an absolute value equal a negative number, because it violates the definition of absolute value.0778

Therefore, I don't even need to go any farther; I can just say that there is no solution, or that it is the empty set.0789

So, there is no solution to this absolute value equation.0795

The important thing is to just look carefully at your work and make sure you didn't make any mistakes.0802

And if you did all the math correctly and handled this correctly, and you come up with something like this, then you didn't do anything wrong.0806

It is just that there is no solution.0813

OK, another absolute value expression: this time, the absolute value is already isolated on the left.0817

So, looking at this, it looks more complex; but we just use the same logic that we did with the simple case.0824

If the absolute value of x is 2, x equals 2, or x equals -2.0834

So, I do the same thing here: I get rid of the absolute value bars, and this is my positive permutation.0838

And then, I also have 2x - 7 = -(3x + 8).0846

OK, solving each of these: I am going to add 7 to both sides; that is going to give me 2x = 3x +15.0854

Then, I am going to subtract 3x from both sides, which is going to give me -x = 15.0867

I am going to multiply both sides by -1 to get x = -15.0876

So, that is my first solution: solving this...this is 2x - 7 = -3x - 8.0884

Adding 7 to both sides is 2x = -3x -1.0893

Adding 3x to both sides: 5x = -1; divide both sides by 5: x = -1/5.0900

OK, check: with absolute value equations, you always have to check your solutions.0909

So, checking this back in the original equation: 2 times -15, minus 7--the absolute value of that--equals 3 times -15, plus 8.0917

OK, so I have 2(-15), which is -30, minus 7, equals 3(-15), which is -45, plus 8.0931

This gives me |-37| equals...well, -45 + 8 is -37; the absolute value of -37 is 37.0945

Well, 37 does not equal -37, so this is not a valid solution; this is not true--this did not satisfy this equation.0959

So, x = -15 is not a valid solution; let's try this one, x = -1/5, substituting it in here.0970

|2(-1/5) - 7| = 3(-1/5) + 8: that is going to give me |(-2/5)-7| = (-3/5) + 8.0977

So then, adding these two together, I am going to get |-7 2/5| = 8 - 3/5...is 7 2/5.0999

OK, the absolute value of -7 2/5 is 7 2/5, equals 7 2/5; and that is true; this is a valid solution.1010

So, I actually have only one solution to this equation, and it is x = -1/5.1022

We are starting out by breaking this into two related equations and removing the absolute bar,1030

solving each, getting two solutions, and then checking and finding out that the first one is not valid, and that the second one is valid.1036

That concludes this session for Educator.com, and I will see you for the next Algebra II lesson.1046

Welcome to Educator.com.0000

In today's lesson, we are going to be discussing solving inequalities.0002

Recall the properties of inequalities: these are needed to solve inequalities.0008

The first one is that, if a is greater than b, and you add the same number to both sides, this inequality has the same solution set as the original inequality.0015

And you can think of this using numbers to make it clearer.0030

If I said that 6 is greater than 5, and then I decided to add 2 to both sides, 8 is greater than 7; so this still holds up.0035

You are allowed to add the same number to both sides of an inequality; and that is true for both greater than and less than.0048

And it is also true for greater than or equal to and less than or equal to.0056

The subtraction property states that, if a is greater than b, and I subtract the same number from both sides,0064

then the resulting inequality has the same solution set as the original inequality.0072

The same idea as up here: if 6 is greater than 5, and I subtract 3 from both sides, I am going to end up with 3 > 2; that still holds up.0079

And that also is valid for less than, and for greater than or equal to and less than or equal to.0093

With multiplication properties, if a is greater than b, and a positive number (c > 0), the same positive number,0104

is multiplied by both sides of the inequality, then the solution set of the resulting inequality is the same as that of the original inequality.0113

If I have 3 > 2, and I multiply both sides by 4, I am going to end up with 12 > 8, which is still valid.0125

The same for less than: a < b--you are allowed to multiply both sides of the inequality by the same positive number, without changing the solution set.0142

Again, this applies to greater than or equal to and less than or equal to, as well as to strict inequalities.0152

The case is different for negative numbers: if c is less than 0 (if you were multiplying both sides of an inequality0161

by a negative number), you must reverse the inequality sign.0167

If both sides of an inequality are multiplied by the same negative number, the direction of the inequality must be reversed.0172

And if you do that, then the resulting inequality has the same solution set as the original inequality.0179

If you don't reverse them, the solution set will not necessarily be the same.0184

For example, if I have 8 > 4, and I multiply both sides by -2, let's say I didn't reverse the sign:0190

then, I am going to end up with -16 > -8; and this is clearly not true.0205

So, as soon as I multiply by a negative number, this is incorrect; I immediately have to reverse the direction of the inequality.0212

And that will give me -16 < -8, which is valid; and again, the same for greater than or less than or equal to.0226

So, multiplying by a positive number, the solution set of the resulting inequality is the same as the original.0235

Multiplying by a negative number, you need to reverse the inequality sign in order for the solution set for this inequality to be the same as for the original.0241

A similar idea with division: if you are dividing both sides of an inequality by a positive number,0251

the resulting inequality has the same solution set as the original.0259

That is only if it is for a positive number; so 15 > 20--if I wanted to divide both sides by 5, I could do that.0264

And I don't need to do anything to the inequality; I just keep it the same.0275

And that is going to give me 3 >...well, I need to start out with an inequality that is valid!0279

So, if 20 > 15, and then I divide the same number by both sides...so I am starting out with 20 > 15;0288

20 divided by 5 is greater than 15 divided by 5, which is going to give me 4 > 3, which is true.0305

The same holds up for less than, and greater than or equal to/less than or equal to.0314

Now, if we are dealing with dividing by a negative number, you have to reverse the inequality sign.0320

And then, that resulting inequality is the same solution set as the original inequality.0325

So, if I have 4 < 6, and I divide both sides by -2, I need to immediately reverse this inequality symbol.0330

And then, I am going to get 6 divided by -2, and I am going to end up with -2 > -3; and that is valid.0344

If I hadn't reversed this, I would have gotten something that is not valid, or does not have the same solution set as the original.0354

So, you need to be very careful, as soon as you multiply or divide by a negative number,0360

when you are working with inequalities, that you reverse the inequality symbol.0366

There are multiple ways of describing the solution set of an inequality.0371

You can either express it as a graph, a set, or simply as an inequality.0379

For example, if I came up with the solution set x ≥ 2, I could just leave it as an inequality like that; that is a little less formal.0386

I could describe it as a set, using set builder notation.0396

And I would put it as follows: what this is saying is "the set of all x, such that x is greater than or equal to 2."0400

So, this is set builder notation: the set of all x, such that x is greater than or equal to 2.0414

You will see the same things, sometimes, with two dots here; this is more common, but you will see this sometimes; and it means the same thing.0420

You can also use a graph on the number line; you can express it as a graph.0429

And remember that, if you are saying "greater than or equal to," you are including this number 2 in the solution set.0436

So, in that case, you would want to use a closed circle, and then continue on to the right.0443

Now, let's say I was going to say x is less than 3.0450

In this case, it is a strict inequality; and 3 is not part of the solution set, so I am going to use an open circle.0458

So, you can express an inequality's solution set either using set builder notation, using a graph, or less formally, just as a simple inequality.0465

Looking at Example 1, we have -3x < -27, so I need to solve that.0478

I need to isolate this x; and in order to do that, I am going to need to divide by -3.0489

But as soon as I start thinking about dividing by a negative number, I have to reverse the inequality symbol.0494

So, instead of less than, this becomes greater than.0503

This is going to give me x > 9.0507

I can leave it just as an inequality; I can use set notation to show this; I can also graph it on the number line.0511

OK, and graphing it on the number line, I am going to use an open circle (since it is a strict inequality)0532

and show that x is greater than 9, but that 9 is not part of the solution set.0538

Solve -4x - 7 ≥ 9: the first step is to add 7 to both sides, to get -4x ≥ 16.0547

Now, to isolate x, I need to divide both sides by -4; and as soon as I do that, I am going to change this direction to less than or equal to.0561

16 divided by -4 is -4, so the solution is x ≤ -4.0570

Using set notation, and graphing on the number line--multiple ways...-1, -2, -3, -4.0578

And this is saying "less than or equal to"; therefore, -4 is going to be included in the solution set, as indicated by a closed circle.0596

OK, in Example 3, it is a little bit more complicated; but you just handle it using the same principles.0611

When you are dealing with fractions, the best thing to do is to get rid of them first, because they are difficult to deal with.0621

So, I am going to multiply both sides of the equation by -9, and (since that is a negative number)0628

immediately reverse the inequality symbol, so that you don't forget to do that.0633

This is going to be -9 times -2 times 3x - 6; I'll move this over a bit...and that is +(-9), times 4x, divided by -9.0637

It is greater than -3 times -9.0657

OK, so -9 times -2 is 18, times 3x - 6...these -9's cancel out, and that is going to give me...+4x, is greater than (-3 times -9 is) 27.0660

Using the distributive property, I am going to multiply this out; and 18 times 3x is 54x; 18 times -6 is -108; plus 4x, is greater than 27.0677

Now, I need to isolate this x; so first, I am going to combine like terms.0696

I have 54x + 4x, is 58x, -108 is greater than 27.0702

Adding 108 to both sides will give me 58x > 27 + 108, so that is 58x > 135.0712

So, x is greater than 135/58; I can leave it like that, or I can use set notation.0729

To graph this, you need to figure out...if you divide 135/58, it is a little bit larger than 2; it is approximately 2.3.0746

So, I could go ahead and graph that out, as well.0754

2 is here; 2.5 is about there; 2.3 is about right here; open circle at 2.3, or actually right over here;0762

1 is here; 2 is here; 2.5 is about here; so, 2.3 is going to be right about here; and that is open circle, like that.0771

There are three ways to express the solution.0788

OK, in Example 4, again, it is a little bit more complex, and there is a fraction, so we are going to get rid of that first.0794

Start out by multiplying both sides by -7, which tells me that I immediately need to reverse the sign.0803

-7 times -3 times (x - 4), over -7, plus -7 times 3 times (9 - 2x); reverse the inequality symbol--greater than or equal to -7 times (4 - x).0811

OK, this cancels; that gives me -3 times (x - 4), and this is -7 times 3, so that is -21, times (9 - 2x), is greater than or equal to...0843

-7 times 4; that is -28; -7 times -x is plus 7x.0860

So now, I just need to do some more simplification.0868

-3 times x is -3x; -3 times -4 is +12; -21 times 9 is -189; -21 times -2x is +42x.0872

And this is -28 + 7x; I can't really do anything with that.0888

Now, I am going to combine like terms (and I do have some like terms).0905

I have a -3x and 42x; combining those, I am going to get 39x; -189 +12 is -177.0908

The next step is to isolate the x, so I am going to add 177 to both sides; that is going to give me 177 - 28 + 7x.0921

Subtract 7x from both sides: 39x - 7x ≥ 177 - 28.0934

Combine like terms to get (39x - 7x is) 32x ≥ 149.0945

Now, I am going to divide both sides by 32; and it is a positive number, so it just becomes x ≥ 149/32.0956

Using set notation, x is greater than or equal to 149/32.0970

You could also graph this; this is approximately equal to 4.7, so 0, 1, 2, 3, 4, and 5;0977

it is going to be a little over halfway between 4 and 5; closed circle; and graph it.0989

OK, so the first step was to eliminate the fraction by multiplying both sides of the equation by -7 and reversing the inequality symbol.0996

Once that was done, we are using the distributive property to multiply everything out and get rid of the parentheses.1005

Then, we are using the addition and subtraction principles and combining like terms to simplify this inequality.1012

And the result was x ≥ 149/32.1021

That concludes this lesson on solving inequalities at Educator.com; see you again!1030

Welcome to Educator.com.0000

In today's lesson, we are going to be solving compound and absolute value inequalities.0002

A compound inequality consists of two or more inequalities combined by either "and" or "or."0009

To solve a compound inequality, solve each part.0016

For example, if you are given 4x - 2 < 10, and x - 1 ≥ 4, it is connected by the word and, so it is a compound inequality.0022

So, I am going to solve each part of that.0040

4x...add 2 to both sides; that would be less than 12; divide both sides by 4, so x is less than 3.0044

And if I add 1 to both sides, I am going to get that x is greater than or equal to 4.0054

We will talk, in a second, about how the solution set works for this; but the first step is to just solve both parts of the inequality.0063

The same idea applies if an inequality is a compound inequality connected by the word or.0074

For example, you might be given 3x > 6, or 4x < 4.0080

Again, solve each part: x > 2, or x < 1.0090

Breaking this down and talking about the solution sets for each: if "and" is used, the solution set is the intersection of the solution sets of the individual inequalities.0103

So, I am going to review this here--the idea of intersection--and this is also covered in detail in the Algebra I series.0113

Intersection means the common members of both solution sets.0121

Just looking at numbers: if I have 2, 4, 6, 8, and 10, and that is my first set; and then I have a second set0127

that is 6, 8, 10, 12, and 14; the intersection would be the elements common to both.0136

So, I see that I have 6 in both; 8; and 10; so the intersection would be 6, 8, and 10.0149

Apply that to inequalities connected by the word and: for example, x - 4 < 10, and 2x ≥ 4.0158

OK, solve each part as discussed; so add 4 to both sides...this means that x is less than 14;0173

and (both conditions must hold) if I divide both sides by 2, I get x ≥ 2.0179

To visualize this, look at the number line: if I have a 0 right here, and then 2, 4, 6, 8, 10, 12, 14,0187

this is telling me that x is less than 14; and that would just go on and on and on.0200

This is telling me x is greater than or equal to 2, which is going to start here and go on and on and on and on.0207

But I am looking for the intersection, or the common elements; and the common elements would be greater than or equal to 2, and less than 14.0215

So, the intersection would come out as 0, 2, 4, 6, 8, 10, 12, 14--greater than or equal to 2, and less than 14.0225

And you could write that out as an inequality, or using set notation, that x is greater than or equal to 2, and less than 14.0240

So, this is actually just a more efficient way of writing, instead of two separate sections here with the word and;0255

you can just write it out more efficiently like this; x is greater than or equal to 2 and less than 14; or showing it on the number line like this.0262

So, that is for inequalities joined by the word and.0271

If or is used in a compound inequality, the solution set is the union of the solution sets of the individual inequalities.0276

Now, reviewing what a union is, just using numbers: if you have a set, such as 4, 5, 6, 7, 8; and then you have another set,0285

-2, -1, 2, 3, 4, 5; and you are asked to find the union; well, the union is any elements that are in either one of these, or both.0298

So, if it is in one; if it is in the other; or if it is in both; that would be the union.0315

Now, I am looking, and I have 4, and that is in both; I don't need to write it twice; 5; 6 is just in this one, but it is included;0321

7, just in that one; 8; and then, I already covered 4 and 5, but I also have to include -2, -1, 2, and 3.0332

And I could rewrite this in ascending order, but it is all included here.0344

The union means that it includes the elements that are in either one of these, or both.0348

Applying this concept to inequalities: 3x + 2 > 8, or 4x - 3 < 1 is a compound inequality joined by the word or.0353

First, solve each inequality: 3x >...subtract 2 from both sides...6; divide both sides by 3 to get x > 2.0366

Or: solve this one also--if I add 3 to both sides, I am going to get 4x < 4; I am going to divide both sides by 4 to get x < 1.0380

Now, looking at this on the number line, what this is saying is that...0, 1, 2, 3, 4...x is greater than 2.0391

This is saying x is less than 1; and there is no overlap here, but that is OK, because the union0403

means that, if something is in either one of these or both, it is included.0409

This is saying that the solution set is that x is greater than 2, or x is less than 1.0414

And this can be written in set notation; so, I would write it as x is greater than 2, or x is less than 1.0420

So, when you use "and," it is the intersection of the two solution sets.0429

When "or" joins a compound inequality, the solution set is the union of the two solution sets.0432

Now, we have worked with absolute value equations; and this time, today, we are working with absolute value inequalities.0440

To solve an absolute value inequality, you need to use the definition of absolute value.0448

And recall that the definition of absolute value is the distance that a value is from 0 on the number line.0453

So, the absolute value of 3 is 3, because the distance between this and 0 on the number line is 3.0461

The absolute value of -3 is also 3, because the distance between -3 and 0 on the number line is 3.0481

So, the absolute value of 3 is 3, because it is 1, 2, 3 away from 0 on the number line.0489

The absolute value of -3 is 3, because -3 is also 3 units from 0 on the number line.0494

For more complicated problems, you may need to rewrite the inequality as a compound inequality.0505

Talking about this in a little bit more detail: you may see two different forms.0511

You may see inequalities in the form |x| < n.0516

If you see that, rewrite the inequality as a compound inequality, as follows.0523

Looking at this, I have an example: the absolute value of x is less than 4.0532

Think about what this is saying; it is saying that the absolute value of x is less than 4 units from 0.0541

OK, so anything that is less than 4...it could be 3, 2, 1, but it is less than 4 away from 0.0552

That could be satisfied by anything between 0 and 4; but it can also be satisfied by anything between 0 and -4,0562

because the distance between -4 and 0 on the number line is less than 4.0580

-3: the distance between that and 0 is less than 4.0585

So, anything in this range is going to satisfy it; so what this is really saying is that x is less than 4, and x is greater than -4.0589

So, in order for the answers to fall in this range, they can't just be less than 4 and go all the way down;0605

because then you will get numbers way over here, that have an absolute value that is much bigger than 4.0610

So, it is in this range, where x is greater than -4, but less than 4.0614

So, in general, when you see an absolute value in this form, where it is less than something, you rewrite it0621

as two related inequalities: x < 4, and x > -4.0630

We are just generalizing out: |x| < n can be rewritten as x < n, and x > -n.0637

OK, the other possibility is that you have inequalities that are in the form |x| > n.0647

For example, |x| > 3; well, what that is saying is that the absolute value of the number is more than 3 away from 0 on the number line.0660

So, anything bigger than 3 is going to have an absolute value of greater than 3.0673

4 has an absolute value greater than 3; and on up.0680

In addition, however, if I look over here on the left, any number smaller than -3 is also going to have an absolute value that is greater than 3.0685

If I took -4, the absolute value of that is 4, which is greater than 3.0695

So, this would translate to x > 3, or x < -3.0700

You need to memorize these two forms and be familiar with them.0709

And generalizing this out, this would say that |x| > n could be rewritten as x > n, or x < -n.0712

So, you need to keep these in mind: when you see |x| < n, you rewrite it as x < n, and x > -n.0722

When you see |x| > n, then x > n, or x < -n.0731

And you can apply these to more complex inequalities, such as |4x + 1| ≥ 12.0741

I would recognize that it is in this form, and I would rewrite it as 4x + 1 ≥ 12, or 4x + 1 ≤ -12.0752

So, just follow this pattern; and we will work on this in the examples.0770

OK, starting with Example 1: this is a shorthand way of a compound inequality that is joined by the word "and."0775

Instead of writing out 7 < 14x - 42, and 14x - 42 ≤ 35, they just combine the same term, put it in the middle, and left these on the outside.0785

But this is really a compound inequality that is joined by the word "and."0805

And remember that, in order to solve these, you solve each inequality; and then you find the intersection of their solution sets.0810

So, for absolute inequalities joined by the word "and," we are going to need to find the intersection of the solution sets.0818

All right, let's solve each of these; that is the first step.0824

I am going to rewrite this with the x on the left; 14x - 42 ≥ 7--more standard.0828

It is saying the same thing: 14x - 42 ≥ 7; I just reversed the sides of the inequality.0845

Next, add 42 to both sides; and that is going to give me 49; 14x ≥ 49.0854

Now, divide both sides by 14; and since that is a positive number, I don't need to reverse the inequality symbol.0869

I can simplify this, because they have a common factor of 7; so to simplify this, remove the common factor; that is going to give me 7/2.0878

I just pulled out the 7 from both the numerator and the denominator--factored it out.0890

OK, solving this inequality--my second inequality: adding 42 to both sides is going to give me 14x ≤ 77.0895

Again, dividing both sides by 14 is going to give me 77/14; again, I have a common factor of 7, so this is going to give me 11/2.0909

Now, the intersection of these, graphing this out: well, this is (2, 4, 6)...about 3.5, 3 and 1/2; and this is equal to 5 and 1/2.0922

OK, to help me graph it out, I am going to write it in decimal form: 1, 2, 3, 4, 5...a little more room....6.0945

OK, so what this is saying is x ≥ 3.5, which is right here; and it includes that point 3.5, so I am going to put a closed circle.0961

Here, the other restriction is that x ≤ 5.5; so this is greater than--it is going up this way;0975

but I have to stop when I get to here, because the intersection of the solution set has to meet both conditions.0983

It has to be greater than or equal to 3.5, and less than or equal to 5.5.0993

So, I can write it like this as an inequality; I can graph it here, or do set notation: where x is greater than or equal to 7/2 and less than or equal to 11/2.0998

OK, again, recognize that this is really a compound inequality; and it is joined by the word and, but it is just a shorthand way of writing it.1025

I wrote this out as two related inequalities, solved each, and found the intersection of their solution sets.1033

The second example is a compound inequality joined by the word or.1042

I am going to go ahead and solve both of these and find the union of the solution sets.1045

Adding 7 to both sides gives me 3y ≤ 15; divide both sides by 3: y ≤ 5.1058

On this side, I am subtracting 10 from both sides to get 2y > 16; dividing both sides by 2 gives me y > 8.1067

OK, since this is or, it is the union of the solution sets; so any element of either set,1082

or that is in both sets, would be included in the solution set for this compound inequality.1089

So here, I have y ≤ 5, and here I have y > 8.1104

So, all of this is included, and all of this is part of the solution set.1115

It could also be written as y ≤ 5, or y > 8.1122

OK, the third example involves an absolute value inequality.1132

When I look at this, I just want to think about which form this is in; and this is in the general form |x| < n.1138

And recall, for those, that you could rewrite this and get rid of the absolute value bars by saying x < n, and x > -n.1145

That is the general form; so I am going to rewrite this as 2z - 6 ≤ 8 (that is this form), and 2z - 6 ≥ -8 (which is this form).1158

It is really important to memorize these or understand them well enough that you can apply them to more complex situations.1174

Now, I am going to solve each.1181

2z - 6 ≤ 8: I am adding 6 to both sides to get 14, then dividing both sides by 2 to get z ≤ 7.1184

And I need to solve this other one.1200

I am going to add 6 to both sides to get 2z ≥ -6, z ≥ -3.1205

And since I am dividing by a positive number, I don't have to worry about reversing the inequality symbol.1218

OK, I end up with...again, I had an absolute value inequality in this form; I solved both inequalities.1226

Actually, I made a little mistake here; let me go ahead and correct that.1246

2z - 6 ≥ -8, so 2z ≥...this is actually going to be -2; so adding 6 to both sides is going to give me -2.1250

Now, if I divide both sides by 2, that is going to give me z ≥ -1.1263

OK, you can either leave this as it is, or you can go ahead and graph it out.1271

And since this is "and," the solution set needs to meet both of these conditions.1277

So, -1, 0, 1, 2, 3, 4, 5, 6, 7...a little more...let's go to 8; OK, z is less than or equal to 7, so this is going to go on and continue on;1284

but it also has to meet the condition that z is greater than or equal to -1.1301

So, anything in this range is going to be the intersection of the solution set for these inequalities.1305

Also, writing it out using set notation: what we have is that z is greater than or equal to -1, and less than or equal to 7.1313

These are three different ways of writing out the solution set.1331

OK, the next example is also an absolute value inequality.1335

And this one is in the form |x| > n, in this general form, which can be rewritten as x > n or x < -n.1340

So, rewriting this and removing the absolute value bars as two different inequalities: 3w - 8, divided by 5, is greater than 4;1354

so that is this first form; or 3w - 8, divided by 5, is less than -4.1364

OK, the first thing to do is get rid of the fraction: multiply both sides of this inequality by 5.1374

Next, add 8 to both sides; and finally, divide both sides by 3.1385

For this inequality, again, get rid of the fraction; multiply both sides by 5; add 8 to both sides; and divide by 3.1398

This is joined by the word "or"; and 28/3...I am going to rewrite that as 9 and 1/3.1415

So, w is greater than 9 and 1/3, or w is less than -4.1425

To graph this, -5, -4, -3, -2, -1, 0, 1...all the way to 9 and 1/3.1430

OK, so this is saying that w is greater than 9.3, or it is less than -4.1445

Open circle, because it is a strict inequality...1456

Set notation would be w is greater than 28/3, or w is less than -4.1460

Recognizing that this is in the general form |x| > n, I rewrote this as 3w - 8, divided by 5, is greater than 4, or the same expression is less than -4.1472

Solve each one: and the solution set is the intersection of the solution sets of those two inequalities.1487

That concludes this lesson of Educator.com; and I will see you soon!1496

Welcome to Educator.com.0000

Today is the first in a series of lectures on conic sections.0002

And we are going to start out with a review of two formulas that you will be applying later on, when we work with circles and hyperbolas and other conic sections.0005

Starting out with the midpoint formula: this formula gives you the midpoint of line segments with endpoints (x1,y1) and (x2,y2).0015

So, let's take an example, just to illustrate this: let's say you were asked to find the midpoint of a line segment with endpoints at (2,3) and (5,1).0028

If these were given as the endpoints, and you were asked to find the midpoint, you could go ahead and apply this formula.0044

So, to visualize this, let's draw out this line segment: this is at (2,3); that is one endpoint, and (5,1) is the other endpoint.0053

I draw a line between these two; and I am looking for the midpoint, which is somewhere around here.0065

Looking at this formula, it is actually somewhat intuitive, because what you are doing is finding0072

the average of the x-values and the average of the y-values, which will give you the middle of each (the midpoint).0078

So, x1 + x2 would give me 2 + 5 (the two x-values), divided by 2.0085

That is going to give me the x-coordinate of the midpoint of this line segment.0095

For the y-coordinate, I am going to add the two y's and divide by 2; so I will average those two y-values, which are 3 and 1.0099

2 + 5 gives me 7/2; 3 + 1 is 4/2; I can simplify to (7/2,2); and I could rewrite this, even, as (3 1/2,2) to make it a little easier to visualize on the graph.0110

This is the midpoint; it is going to be at 3 1/2 (that is going to be my x-coordinate); and 2 is going to be my y-coordinate.0128

And that is the midpoint; this is a fairly straightforward formula; you will need to apply it in a little while, when we start working with circles.0138

Also, to review the distance formula: if you recall, the distance formula is based on the Pythagorean theorem.0151

And the distance formula tells us that we can find the distance between two sets of points0159

if we take the square root of x2 - x1, squared, plus y2 - y1, squared.0163

Now recall: if I am given two points; if I am asked to find the distance between a set of two points, (2,1) and (5,6),0172

I can assign either one to be...I could assign this as (x1,y1), (x2,y2);0192

or I could do it the other way around: I could say this is (x2,y2); this is (x1,y1).0198

It doesn't matter, as long as you assign it and stick with that; you don't mix and match and say this is (x1,y2).0202

Just assign either set; it doesn't matter which set you call which; just stay consistent with the order that you use the points in.0209

OK, so let's look at what this would graph out to.0216

(2,1) and (5,6) would be right about up there; so I have a line segment, and I am asked to find the distance.0221

So, the distance is going to be equal to the square root of...I am going to call this (x1,y1) and this (x2,y2).0231

So, x2 is 5, minus x1, which is 2; take that squared,0239

and add it to y2, which is 6, minus y1, which is 1, squared.0247

Therefore, distance equals 32, plus 6 minus 1 is 52.0253

The distance equals the square root of 32, which is 9, plus 25.0259

Therefore, distance equals √36; distance equals 6.0269

The distance from this point to this point is equal to 6.0277

And again, this is review from an earlier lecture, when we discussed the Pythagorean theorem in Algebra I0282

and talked about how the distance formula comes from that.0288

So, you can always go back and review that information; but this is the application of the distance formula.0291

In the first example, we are asked to find the midpoint and distance of the segments with these endpoints (-9,-7) and (-3,-1).0299

So, recall that the midpoint formula just involves finding the average of the x-coordinates0310

of the two points, and the average of the y-coordinates of the two points.0315

Applying these values to this set of equations gives me -9 + -3, divided by 2, and -7 + -1, divided by 2.0324

Adding -9 and -3 gives me -12, divided by 2; and that is -8 divided by 2.0349

This becomes -6 (-12/2 is -6), and then -8/2 is -4.0359

So, that is the midpoint; that was the midpoint formula.0367

Recall that the distance formula is (x2 - x1)2 + (y2 - y1)2.0377

I am going to assign this as (x1,y1) and this as (x2,y2).0392

Again, it doesn't matter; you can do it the other way around.0404

Distance equals...I have x2 as -3, minus -9, all of this squared; plus y2, which is -1, minus -7, and that whole thing squared.0406

This gives me -3; a negative and a negative is a positive, so plus 9, squared, plus -1...and a negative and a negative gives me + 7...squared.0433

Therefore, distance equals the square root of...9 - 3 is 6, squared, and then 7 - 1 is also 6, squared.0444

Distance equals √(36 + 36), or radic;72.0453

But I can take this a step farther, and say, "OK, this is the square root of 36 times 2," and then simplify that,0461

because this is a perfect square, to 6√2.0467

So, I found the midpoint; it is (-6,-4); that is the midpoint of this segment with these endpoints.0475

And the distance between these endpoints is 6√2.0481

In Example 2, we are asked to find the midpoint and distance of the segment with these endpoints.0488

We go about it as we usually do; but this time we are working with radicals,0493

so we have to be really careful that we keep everything straight.0496

Recall that the midpoint formula is the average of the x-coordinates, and then the average of the y-coordinates.0499

Therefore, the midpoint equals x1...the square root of 2 + 3, plus x2, 2√2 - 4√3, divided by 2;0512

for the y-coordinate of the midpoint, we are going to get √3 - 5, plus 4√3, plus 2√5.0526

And then, we simplify this as much as we can.0536

Here, I have two like radicals: they have the same radicand, so I can combine those to make this 3√2.0540

I have a constant, and then this -4√3.0549

So, this is the x-coordinate of the midpoint; the y-coordinate is √3 here and 4√3; those can be combined into 5√3;0555

and then, let's see: this actually should have a radical over it, because that is a radical up there;0567

I have -√5 + 2√5; that is going to leave me with just + √5, divided by 2.0577

And the midpoint is given by this; it is a little bit messy-looking, but it is correct--you can't simplify, really, any farther.0585

So, we are just going to leave it as it is.0592

Now, we are also asked to find the distance; so I am going to go ahead and work that out here.0594

We found the midpoint; we are working with these same endpoints, but we are finding the distance.0599

Recall that the distance formula is the square root of (x2 - x1)2 + (x2 - y1)2.0603

So, I am going to let this be (x1,y1), (x2,y2).0617

I could have done it the other way around; it doesn't matter, as long as you are consistent.0625

Therefore, the distance is going to be equal to y2, which is 2√2, minus 4√3;0630

and I am going to take that, and from that I am going to subtract y1, which is √2 + 3.0645

So, this covers my (x2 - x1)2.0657

I am going to add that to (y2 - y1): I go over here, and I have y2; that is 4√3 + 2√5.0664

And I am going to subtract y1 from that, which is over here; and that is √3 - √5; and this whole thing is also going to be squared.0678

Let's apply these negative signs to everything inside the parentheses, so we can start doing some combining.0692

This is going to give me 2√2 - 4√3; negative...that is going to give me -√2;0700

apply the negative to that 3--it is going to give me -3; all of this is going to be squared.0713

Plus 4√3, plus 2√5...apply the negative to each term inside the parentheses to get -√3;0718

and this negative, and then the negative √5, gives me a positive √5; and this whole thing is squared.0728

All right, so let's see if I can do some combining to simplify a little bit before I start squaring everything,0736

because that is going to be the most difficult step.0742

All right, this gives me 2√2, and this is minus √2; so I can combine these two to get √2.0747

-4√3, minus 3; this whole thing squared, plus 4√3 - √3--that simplifies to 3√3.0757

2√5 + √5 gives me 3√5; squared.0773

Now, I need to just square everything, and then combine it.0780

There is no easy way around this first one; what I am going to do is multiply √2 times itself, times this, times this,0784

and then go on with the second, and finally the third, term.0792

So, this √2 times √2 is simply going to give me 2; √2 times -4√3 is going to give me -4... 2 times 3 is √6.0795

Then, √2 times -3 is going to give me -3√2.0817

OK, now I take -4√3 and multiply it by this first term, √2, to get -4; 3 times 2 is 6.0828

When I multiply this times itself, I am going to get -4 times -4; that is going to give me 16,0840

times √3 times √3 is going to give me 3.0846

Then, I multiply this times -3 to get -4 times -3 is 12√3.0851

Finally, multiplying -3 times √2 gives me -3√2; -3 times -4 is 12√3; -3 times -3 is 12√3.0861

And then, -3 times -3 is 9; all of this is the trinomial squared; now let's square the binomial.0878

3√3 times 3√3 is going to be...3 times 3 is 9; and then √3 times √3 is just going to be 3.0887

So, looking up here, just to make this a little clearer: 3√3 + 3√5, times itself (squared)...0902

I am going to do my first terms, and I get this; then I do the outer terms, plus the inner terms,0919

which would just be this times this times 2; so that is going to give me 2 times 3 times 3, which is 9,0924

times 3 times 5, with a radical over it; so that is this.0941

So, what I did is took...really, it is just outer plus inner, which is going to give me the outer, which is 9√15, plus the inner, which also 9√15.0951

And this is going to end up giving me 18√15, which is the same as what I have here.0969

And then, finally, the last terms are going to give me 3 times 3 is 9, and then √5 times √5 is just going to give me 5.0976

OK, now combining what we can in order to just make this a bit simpler: I have a constant here;0989

I am just going to cluster all of my constants together in the beginning; that is 2; 16 times 3 is 48, so that is 48;1007

what other constants do I have?--I have a 9, and then I have 27, and I have 45.1016

All right, so those are my constants.1035

Now, for terms with a √6 in them, I have 2 of those: -4√6 and -4√6 is going to give me -8√6.1037

So, I took care of the constants and the terms with a radicand of 6.1048

Next, let's look at terms with a radicand of 2: -3√2, and I have -3√2 here; and that is it.1052

I add those two together, and I am going to get -6√2.1062

So, these are all taken care of: √2, √6, constant: now I have √3.1066

12√3--do I have any other terms like that?--yes: 12√3, so I have 2 of those; and that is going to be...12 and 12 is 24√3.1072

Let's see what else I have left: I took care of those, those...that just leaves me with...2 times 9 is 18 times √15.1083

Therefore, the distance equals...putting this all together, if you added these up, you will get 131 - 8√6 - 6√2 + 24 √3 + 18 √15.1094

So, this is the distance; and this was really a lot of practice just working with multiplying and adding and subtracting radicals.1115

We found the midpoint in the previous slide; and here is the distance for the segment with these endpoints, applying the distance formula.1123

Example 3: Triangle XYZ has vertices X (4,9), Y (8,-9), and Z (-5,2).1132

Find the length of the median from X to line YZ.1143

Definitely, a sketch would help us to solve this; so let's sketch this triangle.1146

And it has...we will call this X at (4,9); at (8,-9), we are going to have Y; and then, Z is going to be over here at (-5,2).1152

The median is going to be a line going here from here right to this midpoint.1173

So, what we are asked to find is the length: we can use the distance formula to find the length.1179

But in order to use the distance formula, I need this endpoint and this endpoint.1185

But I don't have this endpoint; however, I can find it, because if you look at this, this is going to draw out right to here, right in the middle.1190

So, this is the midpoint; if I find the midpoint of YZ, then I have the endpoint of this median.1197

Using the midpoint formula: (x1 + x2)/2, and then for the y-coordinate, (y1 + y2)/2.1208

So, I am looking for the midpoint of YZ: that is going to give me...1227

for Y, the x-value is 8; for Z, it is -5; divided by 2; for Y, the y-value is -9; for Z, the y-value is 2; divided by 2.1235

This is going to give me a midpoint of 8 - 5; that is 3/2; -9 + 2 is -7/2.1252

This is the midpoint, which is (3/2, -7/2); now I can use my distance formula,1262

distance equals √((x2 - x 1)2 + (y2 - y1)2).1271

I can use that distance formula to find this.1282

And I am going to call (4,9)...I need the distance from x to this point m...I am going to call this1283

(x1,y1), and this (x2,y2), applying my distance formula.1292

So first, x2, which is 3/2, minus x1, which is 4--I am going to rewrite that as 8/2 to make my subtraction easier.1299

But this is just 4; squared; plus y2, which is -7/2, minus 9, which I am going to rewrite as 18/2.1311

Again, I am just moving on to the next step of finding the common denominator.1324

But this is my (x1,y1), (x2,y2); that is where these came from.1328

Therefore, the distance equals...3/2 - 8/2 is -5/2; and we are going to square that;1339

plus -18 and -7 combines to -25/2, squared.1347

Therefore, the distance equals the square root of...this is 25 (5 squared), over 4, plus...25 squared is actually 625, divided by 4,1357

which equals the square root of 625 and 25 is 650, and they have the common denominator of 4.1370

And you could leave it like this, or take it a step farther and write this as 650 times 1/4, which equals 1/2.1379

And you could even look and see if there are any other perfect squares that you could factor out.1389

But this does give us the length of the median; in order to find the length of the median, we had to find this other endpoint,1393

which conveniently was the midpoint of YZ, so we found the midpoint of YZ and used that as the other endpoint.1401

Then I had X as an endpoint and the midpoint as the endpoint.1408

Plug that into the distance formula to get the length.1412

Working with triangles again: find the perimeter and area of the triangle with vertices at (1,-4), (-1,-2), and (6,1).1418

So, to help visualize this, we are going to draw it out on the coordinate plane.1427

(1,-4) is my first point; (-1,-2) is my second point; and then 6 is going to be about here; this is 6; 1 will be right here.1433

Let's see what this triangle looks like, and if the graph can help me.1449

Now, I am asked to find the perimeter and area: if I want to find the area of a right triangle,1459

I know that it is going to be 1/2 the base times height.1462

The problem is that I don't know if I am working with a right triangle; I may be; but this definitely does not look like a right angle.1465

This may be a right angle, but I am not sure.1472

In order to determine if it is a right angle (let's call this A, B, and C)--is this a right angle?--1475

well, if it is a right angle, that means that this AC and BC are going to be perpendicular.1485

And recall that perpendicular lines...the product of their slope equals -1.1492

So, what I am going to do is go ahead and find the slope of these two.1510

And I am going to determine what this slope is of AC; what BC is; find their product; and if they are a right angle, then I can proceed.1515

So, this point right here is (-1,-2); this point is (1,-4); and this point is (6,1).1523

Slope is just change in y, divided by change in x.1534

So, let's find the slope of AC: that is going to be the change in y (-2 - -4), divided by change in x (which is -1 - 1).1538

So, this is going to be -2 plus 4, divided by -2; this is going to give me 4 divided by -2, which is -2.1561

I just double-check this: -2 minus -4...actually, correction: 4 minus 2 is going to give me, of course, 2; and the slope, therefore, will be -1.1579

OK, now I also need to find, over here (to keep this separate), the slope of BC.1596

So, change in y: y is 1 - -4, over change in x: x is 6 - 1; this is going to be equal to 1 + 4, divided by 6 - 1 is 5, which equals 5/5.1606

So, the slope of BC equals 1.1630

Now, this means that, if I have the slope of BC, times the slope of AC, equals 1 times -1, or -1.1634

This is a right angle; I have a right triangle; I can use my formula, 1/2 the base times the height.1649

What I need to do next is find the length of these sides, using the distance formula.1656

So, let's start out by finding the length of side AC.1661

Recall your distance formula: for the distance formula, I am going to take x2, side AC,1668

and I am going to call...it doesn't matter which one...I am going to call this, for this first one, (x1,y1), (x2,y2).1678

x2 is -1, minus x1, which is 1, squared, plus y2, which is going to be -2, minus y1, which is -4.1689

So, the length of AC equals...-1 and -1 is -2, squared; plus -2...and a negative and a negative is a positive, so -2 + 4 is going to give me 2, squared.1705

Therefore, the length of AC is going to equal the square root of 4 plus 4, which equals the square root of 8.1723

So, that is the length of AC.1733

Now, I need to find the length of another side to find the area.1735

I found AC; let's go for BC next--the length of BC.1739

I am going to make this (x1,y1),(x2,y2).1748

So, I am starting out with my x2, which is 6, minus x1, which is 1, squared,1753

plus x2, which is 1, minus -4 (that is y2 - y1), squared,1761

equals the square root of...6 minus 1 is 5, squared; 1 minus -4...this becomes 1 + 4, so that is 5, squared;1773

So, this equals the square root of 25 + 25, or the square root of 50.1783

All right, so now I have two sides: I know that this side, AC, has a length of √8; and I know that BC is √50.1789

So, the area equals 1/2 the base times the height; so the area equals 1/2 (√8)(√50), which equals 1/2√400.1804

All right, and you could go on and then simplify this, because this would give you the perfect square of 20 times 20,1828

because 202 would give you 400; so then I could make this 1/2(20) is 10.1845

All right, now we still need to find the perimeter.1856

In order to find the perimeter, I need this third side; and I can use the Pythagorean theorem, because this is the hypotenuse.1859

And I know that a2 + b2 = c2.1864

So, a2 is...one side is the square root of 8, squared, plus b2, the square root of 50 squared, equals c2.1869

Well, the square root of 8 squared is 8, plus 50 (because the square root of 50 squared is going to be 50), equals c2.1879

So, 58 = c2; therefore, c equals √58.1890

All right, so I found the area right here; now, the perimeter.1895

The perimeter equals the sum of the three sides: so that is √8 + √50 + √58.1907

And you could go on and do a little simplification.1920

You could pull the perfect squares out of here.1922

So, I could go a little farther with this and say, "OK, 8 is equal to 4 minus 2; 4 is a perfect square, so this is 2√2."1927

50 is the square root of 25 times 2, so that is going to give me 5√2, plus √58.1935

But you can't combine any of these, because they are not like radicals.1942

So, in this example, we applied the distance formula, as well as the Pythagorean theorem and some geometry,1947

to find the area and the perimeter of this triangle.1953

Thanks for visiting Educator.com; and that concludes this lesson on the midpoint and distance formulas.1957

Welcome to Educator.com.0000

Today, we are going to talk about parabolas.0002

And in some earlier lectures in this series on quadratic equations, we talked about parabolas and did some graphing.0004

But now, we are going to go on and give a specific definition to parabolas, and learn about some other features of parabolas.0011

Although you have seen parabolas previously, when we graphed, we didn't form a specific definition of them.0021

So, the definition of a parabola is that it is the set of points in the plane whose distance from a given point,0027

called the focus, is equal to its distance from a given line, called the directrix.0033

Let's talk about that before we go on to talk about the axis of symmetry.0041

So, if you had a parabola (let's say right here; and we will do an upward-facing parabola), you would have some point,0045

which is known as the focus, and a line (I'm going to put that right about here) called the directrix.0059

By definition, every point on this parabola is equidistant from the focus and the directrix.0080

So, if I took a point right here, and I measured the distance from the focus, it would be equal to the distance from the directrix.0085

And this is just a very rough sketch; but these distances actually would be equal; they are theoretically equal.0095

Looking right here at the vertex, these distances would be equal; so that would be, say, y.0108

If I took some other point, say here, and I measured here to here, these two distances would be equal.0117

So, a couple things to note: the focus is inside the parabola; the directrix is outside.0128

And this is because the focus and the directrix are on the opposite sides of the vertex.0150

So, you could have a parabola facing downward, and then it would have a focus here and a directrix up here.0155

We are also going to talk, today, about parabolas that face to the left and right--horizontal parabolas.0165

But right now, we are going to stick with just (for this discussion) focusing on vertical ones,0173

the definition being that every point in the parabola equidistant between the focus and the directrix.0180

The axis of symmetry of the parabola passes through the focus; and it is perpendicular to the directrix.0189

In this case, the y-axis is the axis of symmetry; it is right here.0195

And you see that it passes through the focus, and it forms a right angle; it is perpendicular to the directrix.0204

Again, we talked about some of these concepts in earlier lectures.0214

But to review, vertex: the vertex of a parabola is the point at which the axis of symmetry intersects the parabola.0216

And it is a maximum or minimum point on the parabola, if the axis of symmetry is vertical.0224

If the axis of symmetry is horizontal (say we have a parabola like this, then the axis of symmetry would be horizontal),0230

we still have a vertex, but it is not a maximum or minimum.0242

And again, we are going to focus a little more on vertical parabolas right now, and then we will talk about horizontal parabolas.0247

So, if I have a downward-facing parabola, the vertex is here; the axis of symmetry is right here.0254

And this vertex is the maximum; this is as large as y gets--it is the largest value that the function attains.0265

If I am looking at a vertex that is upward-facing, then the axis of symmetry...we will put it right here; and the vertex is here.0274

In this case, the vertex is a minimum; this is the smallest value that the function will attain.0286

The standard form of a parabola with vertex at (h,k) is y = a(x - h)2 + k.0299

And this is for vertical parabolas; there is a slightly different form when we are talking about horizontal parabolas.0308

And you might recall this form of the equation that we covered earlier on, under the lecture on quadratic equations.0314

And we called this the vertex form of the equation; now we are going to refer to it as standard form.0319

And it is a very useful form, because it tells you a lot about the parabola.0324

The axis of symmetry is x = h: so I know a few things just from looking at this.0330

I know the vertex, because it is (h,k); I know the axis of symmetry--it is at x = h;0335

and if I look at a, I will know if the parabola is upward- or downward-facing.0341

If a is greater than 0, the parabola will open upward; and k gives you the minimum.0348

If a is a negative value--if it is less than 0--the parabola opens downward, and k is the maximum value of the function.0357

Let's look at an example: y = 2(x - 1)2 + 4.0367

So, this is in standard form: this means that I have h = 1, k = 4, and a = 2.0374

So, I know that my vertex is going to be at (1,4); the axis of symmetry is going to be at x = h, so at x = 1.0384

And since a is greater than 0, this opens upward.0404

So, I can sketch this out: I have a vertex at (1,4), right here, and it opens upward.0411

And the axis of symmetry is going to be right here at x = 1.0421

Here is my vertex at (1,4); and this vertex is a minimum, because this opens upward.0428

The minimum value is k, which is 4.0433

If I were to take a similar situation, but say y = -2(x - 1)2 + 4,0441

I would have, again, an h equal to 1 and a k equal to 4, but this time a would be -2, so this would open downward.0453

What I would end up with would be a parabola here, again, with the vertex at (1,4).0465

But it would open downward, and therefore, this would be a maximum.0473

Also, if the absolute value of a is greater than 1, you end up with a relatively narrow parabola.0481

If the absolute value of a is less than 1, you end up with a relatively wide parabola.0490

So, this form is very useful, because just by having the equation in this form, we can at least sketch the graph.0500

Let's talk a little bit more about graphing parabolas.0508

You can use symmetry and translations to graph a parabola: and by translations, we mean a shift.0511

Looking at the standard form: what this really is: if you took a graph of y = ax2, this is letting h equal 0 and k equal 0.0519

And then, if you altered what h is, it is going to shift the graph horizontally by that number of units.0530

If you alter what k is, it is going to translate or shift that graph upward and downward by a certain number of units.0538

In order to graph a parabola, you often need to put it in standard form.0547

Let's start out by just talking about putting an equation or a parabola in standard form.0552

And then we will go on and look at some graphs, and how different values of h and k can affect the graph.0556

So, in order to put the equation into standard form...let's say you are given an equation such as this, y = x2 + 6x - 8,0562

and I want it in this standard form, y = a(x - h)2 + k.0572

The first thing to do (and this is, again, review from an earlier lesson--you can go back and look at the lesson0580

on completing the square as part of this lecture series, but we will review it again now): first, I am going0586

to isolate the x variable terms on the right side of the equation.0592

I am going to add 8 to both sides: now I am going to complete the square.0596

I am going to focus on this, and I need to add a term to it to make this a perfect square trinomial.0602

The term I am going to add is going to be b2/4.0608

In this case, b is 6, so this is going to give me 62/4, which is 36/4, which is equal to 9.0613

So, that is what I need to add in here: y + 8, plus I need to add 9 to both sides.0627

It is easy to forget to add it to the other side, because you get so focused on completing the square.0640

But if you don't, the equation will no longer be balanced.0645

So, I am going to add 9 to both sides.0648

And I want this to end up in this form; so I am going to rewrite this.0653

First I will add these two together to simplify to get y + 17 =...well, this is a perfect square trinomial, so I just take (x + 3)2.0657

And I look at what I have, and it is almost in this form, but not quite.0668

I want to isolate y on the left, so I am going to subtract 17 from both sides to get y = (x + 3)2 - 17.0671

And this is in this form: a happens to be equal to 1 in this case.0679

And so, if you are given an equation that is not in standard form, and you want to get it in standard form,0683

isolate the x variable values on the right (although if we are working with horizontal parabolas,0690

it is going to be the other way around, as we will see in a minute--we are actually going to end up0697

getting the y variable terms on the right; but for now, the x variable terms on the right); complete the square0701

by adding the b2/4 term to both sides of the equation; and then simplify;0707

shift things around as needed to get it in this form.0716

Remember, also, that if you have a leading coefficient that is something other than 1,0718

when you get to this step after isolating the x variable terms, you are going to need to factor out that term before completing the square.0724

All right, assuming that you have gotten your equation in standard form, and you are ready to graph the parabola, you are going to use symmetry.0734

The two halves of the parabola are symmetrical; if you graph half the points, you can use reflection across the axis of symmetry to graph the other points.0741

And translation is knowing how h and k, and changes in h and k, affect the graph, in order to graph.0749

All right, so let's just start out with something in this form--a very basic equation for a parabola.0760

Let's let f(x) equal x2, so it is in this form: y = ax2.0767

And so, here, what is happening is: if you think about what we have, we have a = 1, and then h is 0 and k is 0.0775

What this tells me is that the vertex is going to be at (0,0), and the axis of symmetry is going to be at x = 0.0784

And you can also very easily find some points to graph this.0796

right now, I am just going to sketch it out, and not worry about exact points, just so you get the idea.0800

So, since a = 1, this is going to open upward; this is going to be upward-opening, so the vertex is here at (0,0);0805

it is upward-opening; and it is going to look something like this.0819

So, this is my graph here of y, or f(x), = x2.0833

Now, let's say I change this slightly: let's say I have another function, g(x) = x2 + 2.0838

So, looking at this form, h is still 0; but now I have k = 2.0847

And according to this, this is going to shift the graph up 2 units; so k is going to translate this graph up 2 units.0855

I have a similar graph, but it is going to be with the vertex right here at (0,2).0867

And remember that the axis of symmetry is at x = h, so the axis of symmetry is going to still be at x = 0; right here--this is the axis of symmetry.0880

This is shifted upward; it still opens upward, because a is positive.0888

So, now I am just going to have a similar idea, but shifted upward by 2.0892

So here, I have y = x2 + 2.0901

If this had been a -2, then it would have been shifted down by 2, and I would have had a graph right here.0907

So, let's see what happens when I change h.0912

Let's get a third function: we will call it h(x) = (x - 1)2.0917

OK, now what I have here is h = 1; k, if I look here, is 0.0935

Therefore, the vertex (this is the vertex right here) equals (1,0), and the axis of symmetry is going to be at x = 1.0947

So, this is going to be shifted to the right; so I am going to have a graph something...let me move this out of the way...like this.0962

So, this one is y = x2, and this is y = (x - 1)2.0979

Important take-home points: a change in h will shift the graph horizontally, to the right or left.0991

A change in k will shift the basic graph either up or down, by k number of units.1001

Using symmetry: if I were to graph these out exactly, I would need to find points.1009

And I don't need to find all of the points: for example, if I had a parabola that was a downward-facing parabola1013

somewhere, then I could use the axis of symmetry, and I could just find the points over here and reflect across that axis in order to graph.1021

All right, this concept is another one adding on to our knowledge of parabolas from prior lessons.1034

And it is defining a segment called the latus rectum.1040

The latus rectum is the segment passing through the focus and perpendicular to the axis of symmetry.1044

Let's see what that means--let's visualize that.1051

Let's say I have a parabola like this, and let's say the focus is here.1054

So, this passes through the focus, and is perpendicular to the axis of symmetry.1065

This is the focus, and here we have the axis of symmetry.1072

That means the latus rectum is going to pass through here, and it is going to be perpendicular to the axis of symmetry.1083

So, that is this line; this is the latus rectum.1090

The equation for its length is the absolute value of 1/a; and if you have the equation of the parabola in standard form,1095

then this a is the same a as you will see in that formula.1108

So, this is something you might occasionally need to use.1112

For example, if I were given an equation of a parabola y = 2(x - 3)2 + 5,1115

and I was asked to find the length of the latus rectum of this parabola, then I would just say,1122

"OK, a equals 2; therefore, the length equals the absolute value of 1/2."1127

Horizontal parabolas: I mentioned that you can also have parabolas that open to the right or left, not just up and down,1138

although up to this point in the course, we have just talked about vertical parabolas, or parabolas that open upward or downward.1143

For parabolas whose axis of symmetry is horizontal, we end up with equations in this form: y = a(x - k)2 + h.1150

So, one thing to note: the positions of the x's and y's are reversed, but so are the h's and k's.1160

In the vertical formula, the h was in here, and the k was out here.1168

So, be careful when you are working with this formula to notice that the positions of h and k are reversed.1171

And there are translations of x = ay2, and then again, h and k shift this graph around horizontally and vertically.1177

So, it would look something like this, for example: the axis of symmetry would be right here;1189

and it would be a horizontal axis of symmetry; or maybe I have one that opens to the left, and it has an axis of symmetry right here.1197

These do not represent functions; and you can see that they don't represent functions1208

by trying to pass a vertical line through them: they fail the vertical line test.1212

Remember: with a function, the vertical line test tells us that a vertical line drawn...you could try1216

any possible area of the curve, and the vertical line will only cross the curve once.1224

If the vertical line crosses the curve more than once, it is not a function.1230

So, this fails the vertical line test.1233

It is not a function; it is still an equation--you can still make a graph of it; but horizontal parabolas do not represent functions.1241

I am working on graphing some horizontal parabolas.1250

When you look at the equation in standard form, y = a...and remember, the k and h are in opposite positions;1253

they are reversed...looking at a, if a is positive (if a is greater than 0), then the parabola is going to open to the right.1263

If a is negative, then the parabola is going to open to the left.1273

So, let's look at a very simple horizontal parabola, x = y2.1278

OK, the vertex is at (h,k); and I can see that h and k are both 0, so the vertex equals (0,0).1284

The axis of symmetry is at y = k, so that is going to be at y = 0.1294

And the a here is 1: a = 1, so this opens to the right.1298

So, you are going to have a parabola that looks something like this.1308

You could have another parabola, x = -y2.1321

Here we would have the same vertex and the same axis of symmetry; here the x-axis is actually the axis of symmetry.1325

And I look at a now, and a equals -1, so this parabola is going to open to the left.1333

So, I am going to end up with a parabola like this.1343

Now again, change in h or change in k is going to shift this parabola a bit.1350

Let's change h and see what happens: let's let x equal y2 + 2.1358

Here I have h = 2, k = 0; so (2,0) is the vertex; a = 1, so it is positive, so this still opens to the right.1366

If I look at this, x = y2...here is my graph of x = y2; over here is x = -y2.1379

Now, I am going to have h = 2, so that is going to shift horizontally by 2.1386

(2,0) will be the vertex; and it is going to open to the right.1397

So, this is x = y2 over here; right here, this is actually x = y2 + 2 now.1402

And k, as discussed before, shifts the graph of a parabola vertically.1416

The same idea here: if I were to change k, then I would shift this graph up or down by k units.1423

So, with horizontal parabolas, you need to be familiar with this equation.1430

You need to know that they open to the right if a is greater than 0; they open to the left if a is less than 0.1435

The vertex is at (h,k), and the axis of symmetry is y = k.1441

And you also need to keep in mind that these do not represent functions.1446

In the beginning of today's lesson, we talked about the focus and directrix.1452

And here are formulas to allow you to find those if you need to.1455

If you have a vertical parabola, the coordinates of the focus are h for the x-coordinate, and k plus 1/4a.1460

And the equation for the directrix is y = k - 1/4a; remember that the directrix is a line, so this is giving you the equation for that line.1473

And this would be for a vertical parabola; for a horizontal parabola, the focus is found at the coordinates h + 1/4a;1485

and then the y-coordinate is k, so the focus is a point, and this gives the coordinates of that point.1493

The directrix is a line, and the equation for this line for a horizontal parabola is x = h - 1/4a.1498

And you might need to occasionally use these when we are working problems.1505

And we will see that in one of the examples, actually, shortly.1508

Starting out with Example 1: Write in standard form and identify the key features: x = 3y2 - 12y + 10.1513

We have x equal to all of this; so this tells me, since I have x set equal to this y2 term, that I am looking at a horizontal parabola.1524

So, the standard form of this equation is going to be x = a(y - k)2 + h.1536

Remember, h and k are going to be in opposite positions.1547

In order to get this equation in standard form, we need to complete the square.1550

This time, since I am working with a horizontal parabola, I am going to isolate all of the y variable terms on the right.1554

And I am going to do that by subtracting 10 from both sides to get x - 10 = 3y2 - 12y.1561

This leading coefficient is not 1, so I have to factor it out.1569

And then, I have to be really careful when I am adding to both sides of the equation, because this is factored out.1573

So, factor out a 3 to get y2 - 4y.1580

I need to complete the square: that means I need to add something over here.1585

And the term that I need to add is going to be b2/4.1589

b is actually 4; so this is going to be 42/4, equals 16/4, equals 4.1594

Here is where I need to be careful: on the right, I am adding 4 inside these parentheses, which is pretty straightforward.1604

But what I need to do on the left is realize that I am actually going to be adding 3 times 4, which is 12.1613

So, if I were just to add 4, this equation would not be balanced,1628

because in reality, what I am doing over here is adding 3 times 4.1631

So, on the right, I am going to add 12; and I got that from 3 times 4.1635

Simplifying the left: 12 - 10 is 2; on the right, inside here, I now have a perfect square.1640

And I want this to end up in this form, so I am going to write this as (y - 2) (and it is negative, because I end up with a negative sign in here) squared.1649

I am almost done; I just need to move this constant over to the right to have it in this form.1661

x = 3 times (y - 2)2, minus 2.1666

So, now that I have this in standard form, I can identify key features.1671

Key features: 1: this is a horizontal parabola, as you can see from looking at this equation.1677

2: The vertex is at (h,k); h is 2, and k is also 2.1686

Actually, being careful with the signs, h is actually -2, because remember, standard form has a plus here.1703

I don't have a plus here; I could rewrite this so that I do, and that would give me + -2.1710

And it is good practice, actually, to write it exactly in this form, although this is correct--you could leave it like this.1718

By writing it in this form...and the same thing if I had ended up with a plus here--then I would need to rewrite that,1725

because here I need a negative to be in standard form; if I ended up with a plus here,1736

then I would have needed to rewrite that, as well, which would have been equal to minus -2.1741

Standard form, just like this, looking here, gives me a vertex at (-2,2).1748

And because a equals 3, that means that a is greater than 0; a is positive, so the parabola opens to the right.1755

OK, so key features: horizontal parabola; it has a vertex at (-2,2); a = 3, so this tells me that the parabola opens to the right.1773

We can also say that the axis of symmetry is at y = k, and therefore the axis of symmetry is at y = 2.1784

OK, in Example 2, we are asked to graph.1797

And you will notice that this is the same equation that we worked with in Example 1.1802

We already figured out standard form: and standard form is x = 3(y - 2)2 - 2.1806

And for clarity, we can actually write this as I did at the end, which is 3(y - 2)2 + -2,1816

so that we truly have it in standard form, with the plus here to make it easy to see what is going on.1826

To graph this, I want to know the vertex: the vertex is (h,k): h here is -2; k is 2.1831

The axis of symmetry is going to be at y = k; k is 2, so it is going to be at y = 2.1840

I know that this opens to the right, so I have a general sense of this graph.1854

But I can also just find a few points.1859

And we are used to working with a situation where x is the input and y is the output.1868

It is the opposite here, so we need to be really careful.1873

I also want to note that, since the vertex is here at (-2,2), and this opens to the right,1876

for this graph, we are not going to have values of x that are smaller than -2.1881

So, if I end up with something where an x is smaller than -2, then it is going to be off the graph.1885

Let's let y equal 1: if y is 1, 1 - 2 is -1, squared gives me 1; 1 times 3 is 3, minus 2 is 1; so, when y is 1, x is 1.1891

Let's let y equal 3: when y is 3, 3 minus 2 is 1, squared is 1; 1 times 3 is 3, minus 2 is 1.1906

And you can see, as I mentioned, that this is not a function; it failed the vertical line test (as horizontal parabolas do).1915

And you can see that there is an x-value, 1, that is assigned 2 values of y; so it does not meet the definition of a function.1921

So, just a couple of points...let's do one more: 0...0 minus 2 is -2; squared is 4; 4 times 3 is 12; 12 minus 2 is 10.1929

So, that is off this graph; but it gives us an idea of the shape.1941

So, I know that my axis of symmetry is going to be here; and I have a point at (1,1);1945

I have another point at (1,3); and then I have a point way out here at (10,0).1954

I know that this is going to be a fairly narrow graph, because a equals 3.1960

This is the graph of the horizontal parabola described by this equation; and here it is, written in standard form.1971

So, it opens to the right; it is fairly narrow, because a equals 3.1979

It has a vertex at (-2,2), and it has an axis of symmetry at y = 2.1984

Example 3: we are asked to graph; this is also going to be a horizontal parabola.1994

We are going to start out by putting it in the standard form, x = (y - k)2 + h.1999

We need to complete the square; start out by isolating the y variable terms on the right.2009

So, I am going to add 6 to both sides to get -2y2 + 8y.2014

Since the leading coefficient is not 1, I need to factor it out; so I am going to factor this -2 to get y2.2021

Factoring a -2 from here would give me a -4.2032

And I need to add something to this to complete the square.2035

What I need to add is b2/4.2040

b is 4, so I am going to be adding 42/4; that is 16, divided by 4; that is 4.2044

So, I am going to be adding 4 to the right; but to the left, I am actually adding -2 times 4, which is -8.2061

So, we subtract 8 from that side; to this side, since I am adding inside the parentheses, I am just adding 4.2073

But then, 4 times -2--that is how I got the -8 on the left.2082

This gives me x - 2 = -2; and I want it in this form, so I am going to rewrite this as (y - 2)2.2085

The last thing I need to do is add 2 to both sides; and I have it in standard form.2097

Now that I have this in standard form, it is much easier to graph.2106

The vertex is going to be at (h,k); so h is here; k is here; the vertex is at (2,2).2110

There is going to be an axis of symmetry at y = k, and so that is going to be at y = 2; my axis of symmetry is going to be at y = 2.2119

Now, to finish out graphing this, I am going to find a few points.2142

I have the vertex at (2,2); I also know that a is less than 0 (a is negative), so I know this is going to open to the left.2146

So, I know it is going to look something like this; but I will find a couple of points.2155

And I know that x is (actually, (2,2) is right here)...I know that this opens to the left, and that x is not going to get any larger than that.2158

The graph is just going to go this way.2172

So, I can't use values that end up giving me an x that is greater than 2.2174

Let's try some simple values: I am going to try 1 for y, and looking in standard form, 1 - 2 gives me -1, squared is 1, times -2 is -2, plus 2 is 0.2181

And 3: 3 minus 2 is 1, squared is 1; 1 times -2 is -2, plus 2 is 0.2195

So, I have a couple of points here: this is at 0...when x is 0, y is 1; when x is 0, y is 3.2204

And this is going to give me a parabola shaped like this, opening to the left with a vertex at (2,2).2214

The axis of symmetry would be right through here; and I have a couple of points, just to make it a bit more precise.2226

So, the first step in graphing a parabola is always to get it into this form by completing the square.2235

And then, using the features you can see from here, sketch it out, and finding a few points, make the graph more accurate.2240

Find the equation of the parabola with a vertex of (2,3) and focus at (2,7); draw the graph.2250

This is a very challenging problem: we are not given an equation--we actually have to find the equation based on some key points that we are given.2256

Well, I am given that the vertex is at (2,3); so I know that the vertex is right here; that is the vertex.2266

This time, I am also given the focus; the focus is at (2,7), which is going to be up here somewhere...5, 6, 7...about here.2278

So, the vertex is (2,3); the focus is (2,7).2290

Remember, in the beginning of this lesson, I mentioned that the focus is always inside the parabola.2296

Since the focus is inside the parabola, I already know that this has to open upward.2301

So, I know something about the shape of the graph.2306

Let's find the equation: now, I know that this is a vertical parabola, because the focus is inside the parabola.2311

That told me that this has to open upward, so I know I am dealing with a vertical parabola.2316

And that helps me to find the equation, because the standard form is going to be y = a(x - h)2 + k.2320

I am given the vertex, so I am given h and k: I know that h = 2 and k = 3.2330

In order to write this equation, I need a, h, and k; all I am missing is a.2340

I am given the piece of information, though, that the focus is (2,7).2347

And that is going to allow me to find a.2350

You will recall that I mentioned the formulas for focus and directrix.2353

And for a vertical parabola, the focus is at h...the x-coordinate is h, which we see here; and the y-coordinate is k + 1/4a.2359

And I know the focus is at (2,7): so 2 = h, and 7 = k + 1/4a, according to this definition.2377

Well, since I know that k is 3, then I can solve this.2392

So, I know k; so I can solve for a.2401

Subtract 3 from both sides to get 1/4a; 1/a equals 16; multiply both sides by a, and then divide both sides by 16,2408

or just take the reciprocal of each side (essentially, that is what you are doing) to get a = 1/16.2422

Now, I have h and k given; I was able to figure out a, based on the definition of focus.2427

So, I end up with the equation y = 1/16(x - 2)2 + 3.2433

So, this is the equation.2444

And as you know, once we have the equation, the graphing is pretty easy.2446

I know that this opens upward; and since I know what a is, I know that this is going to be a pretty wide parabola; the a is a small value.2451

I am going to have a parabola that opens upward, with a vertex of (2,3), and fairly wide in shape.2461

That was a pretty challenging problem, because you had to go back2470

and think about how you could use a formula to find the focus; and knowing the focus allowed you to find a.2473

That concludes this lesson on parabolas at Educator.com; thanks for visiting!2483

Welcome to Educator.com.0000

Today we are going to talk about circles, beginning with the definition of a circle.0002

A circle is defined as the set of points in the plane equidistant from a given point, called the center.0008

For example, if you had a center of a circle here, and you measured any point's distance from the center, these would all be equal.0018

And the radius is the segment with endpoints at the center and at a point on the circle.0031

The equation for the circle is given as follows: if the center is at (h,k) and the radius is r,0044

then the equation is (x - h)2 + (y - k)2 = r2.0052

And this is the standard form; and just as with parabolas, the standard form gives you a lot of useful information0060

and allows you to graph what you are trying to graph.0067

For example, if I were given (x - 4)2 + (y - 5)2 = 9, then I would have a lot of information.0071

I would know that my center is at (h,k), so it is at (4,5).0084

And the radius...r2 = 9; therefore, r = √9, which is 3.0091

So, based on this information, I could work on graphing out my circle.0099

Use symmetry to graph a circle, as well as what you discover from looking at the equation in standard form.0108

Looking at a different equation, (x - 1)2 + (y - 3)2 = 4: this equation describes the circle0114

with the center at (h,k), which is (1,3); r2 = 4; therefore, r = 2.0124

So, I have a circle with a radius of 2 and the center at (1,3).0134

So, if this is (1,3) up here, and I know that the radius is 2, I would have a point here; I would have a point up here.0139

Symmetry: I know that, if I divide a circle up, I could divide it into four symmetrical quarters, for example.0156

So, if I have this graphed, I could use symmetry to find the other three sections of this circle.0169

All right, if the center is not at the origin, then we need to use completing the square to get the equation in standard form.0187

Remember that standard form of a circle is (x - h)2 + (y - k)2 = r2.0198

With parabolas, we put those in standard form by completing the square.0210

But at that time, we were just having to complete the square of either the x variable terms or the y variable terms.0213

Now, we are going to be working with both; and as always, we need to remember to add the same thing to both sides to keep the equation balanced.0219

If I was looking at something such as x2 + y2 - 4x - 8y - 5 = 0,0226

what I am going to do is keep all the x variable and y variable terms together, and then just move the constant over to the right.0241

The other thing I am going to do is group the x variable terms together: x2 - 4x is grouped together, just like up here.0249

And then, I am going to have y2 - 8y grouped together, and add 5 to both sides.0258

Now, I have to complete the square for both of these.0266

This is going to give me x2 - 4x, and then here I need to have b2/4.0269

Since b is 4, that is going to give me 42/4 = 16/4 = 4; so, I am going to put a 4 in here.0277

For the y expression, I am going to have...let's do this up here...b2/4 = 82/4, which is going to come out to 16.0292

So, I am going to add 16 here; and I need to make sure I do the same thing on the right.0307

So, I need to add 4, and I need to add 16; if I don't, this won't end up being balanced.0312

Now, I want this in this form; so let's change it to (x - 2)2 +...here it is going to be (y - 4)2 =...0321

4 and 16 is 20, plus 5; so that is 25.0335

This gives me the equation in standard form; and the center is at (h,k), (2,4).0341

And the radius...well, r2 is 25; therefore, the radius equals 5.0350

And as always, you need to be careful: let's say I ended up with something in this form, (x + 3)2 + (y - 2) = 9.0355

The temptation for the center might be just to put (3,2); but standard form says that this should be a negative.0365

So, I may even want to rewrite this as (x - (-3))2 + (y - 2) = 9, just to make it clear that the center is actually at (-3,2).0372

And then, the radius is going to be the square root of 9, which is 3.0387

So, be careful that you look at the signs; and if the signs aren't exactly the same as standard form,0390

you need to compensate for that, or even just write it out--because -(-3) would give me +3, so these two are interchangeable.0396

All right, in this example, we are asked to find the equation of the circle which has a diameter with the endpoints (-3,-7) and (9,-1).0405

So, let's see what we are working with.0413

Just sketch this out at (-3,-7), right about there; over here is (9,-1); there we have the diameter of the circle.0418

We have a circle like this, and we want to find the equation.0433

Recall that the formula for the equation of a circle is (x - h)2 + (y - k)2 = r2.0439

Therefore, I need to find h; I need to find k; and I need to find the radius.0449

Recall that (h,k) gives you the center of the circle.0453

Since this is the diameter of the circle, the center of this line segment is going to be the center of the circle.0460

So, (-3,-7)...over here I have (9,-1); and here I have the center--the center is going to be equal to the midpoint of this segment.0467

Recall the midpoint formula equals (x1 + x2)/2, and then (y1 + y2)/2.0477

So, the center--the coordinates for that are going to be equal to (-3 + 9)/2, and then (-7 + -1)/2,0491

which is going to be equal to 6/2...-7 and -1 is -8/2; which is equal to (3,-4).0507

This means that h and k are 3 and -4; so I have h and k; I need to find the radius.0520

Well, half of the diameter...this is all the diameter; this is my midpoint; and I know that this is at (3,-4).0526

So, I just need to find this length--this is the radius.0535

And I now have endpoints; so I can use either one of these--I have a set of endpoints here and here, and here and here.0539

I am going to go ahead and use these two, and put them into the distance formula.0549

(-3,-7) and (3,-4)--I can use these in the distance formula: the distance here equals the radius,0553

which is the square root of...I am going to make this (x1,y1), and then this (x2,y2).0562

So, this is going to give me...x2 is 3, minus -3, squared, plus...y2 is -4, minus -7, squared.0574

So, the radius equals the square root of 3 + 3; a negative and a negative is a positive; all of this squared,0589

plus -4...and a negative and a negative is a positive, so -4 + 7, squared.0599

So, the radius equals the square root of...3 + 3 gives me 6, squared, plus...7 - 4 gives me 3, squared.0608

So, the radius equals √(36 + 9); 36 plus 9 is 45, so the radius equals √45.0619

But what I really want for this is r2, so r2 is going to equal (√45)2, which is going to equal 45.0632

Putting this all together, I can write my equation, because I now have h; I have k; and I have r.0651

So, writing the equation up here gives me (x - 3)2 + (y...I can either write this as - -4,0658

or I can rewrite this as (y + 4)2 = r2, which is 45.0672

So, this, or a little more neatly, like this: (y + 4)2 = 45--this is the equation for the circle.0679

And I found that information based on simply knowing the diameter.0689

Knowing the diameter, I could use the midpoint formula to find the center, which gave me h and k.0693

And then, I could use the distance formula to find the distance from the center to the end of the diameter, which gave me the radius.0698

And then, I squared the radius and applied it to that formula.0707

Example 2: Find the center and radius of the circle with this equation.0711

In order to achieve that, we need to put this equation in standard form.0716

And recall that standard form of a circle is (x - h)2 + (y - k)2 = r2.0719

So, we need to complete the square: group the x variable terms together on the left; also group the y variable terms on the left.0730

Add 8 to both sides to move the constant over.0742

I need x2 - 8x + something to complete the square, and y2 + 10y + something to complete the square, equals 8.0746

So, for the x variable terms, I want b2/4, and this is going to be 82/4, or 64/4, equals 16.0759

So, I am going to add 16 here.0773

For the y variable terms, b2/4 is going to equal 102/4, which is 100/4, which is 25.0774

I need to be careful that I add the same thing to both sides to keep this equation balanced, so I am also going to add 16 and 10 to the right side.0785

Correction: it is 16 and 25--there we go.0805

(x - 4)2 equals this perfect square trinomial, and I am trying to get it in this form; that is what I want it to look like.0809

Plus...(y + 5)2 comes out to this perfect square trinomial.0820

On the right, if I add 8 and 16 and 25, I am going to end up with 49.0830

8 and 16 is going to give me 24, plus 25 is going to give me 49.0840

Now, I have this in standard form: because the center of a circle is (h,k), I know I have h here,0848

and I know I have k here, this is going to give me...h is 4; k is -5.0861

Be careful with the sign here, because notice: this is (y + 5), but standard form is - 5, so this is equal to (y - -5)2--the same thing.0869

It is just simpler to write it like this; but make sure you are careful with that.0880

The radius here: well, I have r2; the radius, r2, I know, is equal to 49.0884

Therefore, r = √49, so the radius equals 7.0892

Therefore, the center of this circle is at (4,-5), and the radius is equal to 7.0898

Example 3: Find the radius of the circle with the center at (-3,-4) and tangent to the y-axis.0909

This one takes more drawing and just thinking, versus calculating.0919

The center is at (-3,-4), right about here.0924

The other thing I know about this circle is that it is tangent to the y-axis.0933

That means that, if I drew the circle, it is going to extend around, and it is going to touch this y-axis.0939

Well, the radius is going to have one endpoint on the circle, and the other endpoint at the center.0948

Therefore, the radius has to extend from (-3,-4) over here.0954

And at this point, we are able to then find the length, because, since x is -3, and it has to go all the way to x = 0,0965

then this distance must be 3; therefore, the radius equals 3.0976

And again, that is because I know the center is here at -3, and I know0984

that the other endpoint of the radius is going to be out here, forming the circle,0988

and that, because it is tangent to the y-axis, x is going to be equal to 0 right here.0996

So, I know that x is equal to 0 here; and I know that x is equal to -3 over here; so it is just 1, 2, 3 over--this distance here is going to be 3.1003

And that is going to be the same as the radius, so the radius is equal to 3.1013

Example 4: Find the equation of the circle that is tangent to the x-axis, to x = 7, and to x = -5.1018

We are given a bunch of information about this circle and told to put it in standard form.1026

The first thing we are told is that it is tangent to the x-axis; so this circle is touching the x-axis; let's just draw a line here to emphasize that.1040

It is also tangent to x = 7; x = 7 is going to be right here--it is tangent to this.1049

It is also tangent to x = -5, out here.1061

I am going to end up with a circle that is touching, that is tangent to, these three things.1068

Let's think about what that tells me.1074

I need to find h and k (I need to find the center).1077

I also need to find the radius, so I can find r2.1081

If this extends from -5 to 7, that gives me the diameter.1086

So, the diameter goes from 7 all the way to -5; so if I just add 7 and 5 (the distance from here to here,1097

plus the distance from here to here), I am going to get that the diameter equals 12.1104

The radius is 1/2 the diameter, so the radius equals 6.1110

I found that the radius equals 6.1117

The radius is going to extend from these endpoints to the center.1122

And I know that it is 6, so I know that the radius is going to go from 7 over here, 6 away from that.1127

7 - 6 is 1; it is going to go up to x = 1.1136

Again, that is because the length of the radius is 6, so the distance between the center and this endpoint has to be 6.1141

7 - 6 is 1; the radius is going to extend from there to there.1152

Therefore, the x-value of the center has to be 1.1156

Now, what is the y-value of the radius? The other thing I know is that this circle is tangent to this x-axis.1164

So, I know that it is going to have an endpoint on the x-axis.1171

And then, if it is going to extend from here to here, it is going to have to go up to 6; therefore, the center is at (1,6).1175

All right, so the radius equals 6, and the center is at (1,6); and that gives me an equation:1193

(x - 1)2 + (y - 6)2 = the radius squared.1200

If r = 6, then r2 = 36.1208

Again, that is based on knowing that this is tangent to x = -5, x = 7, and the x-axis.1212

So, I had the diameter, 12; I divided that by 2 to get the radius; I know that I have an endpoint here and an endpoint at the center.1221

So, that gives me the x-value of the center, which is 1.1231

I know I have an endpoint here, and also an endpoint at the center; so it has to be up at 6.1233

That gives me (1,6) for my value.1238

OK, and this is just drawn schematically, because the center would actually be higher up here.1243

This is just to give you...the center is actually going to be up here, now that I have my value: it is going to be at (1,6).1247

OK, that concludes this lesson of Educator.com on circles; thanks for visiting!1258

Welcome to Educator.com.0000

So far in conic sections, we have discussed parabolas and circles.0002

The next type of conic section we are going to cover is the ellipse.0006

First of all, what is an ellipse? Well, an ellipse is formally defined as the set of points in a plane0012

such that the sum of the distances from two fixed points is constant.0019

Well, what does that mean? First of all, this is the general shape of an ellipse.0024

And these two points here are the foci of the ellipse: we will call them f1 and f2.0029

And looking back at this definition, it is the set of points in the plane...0036

and if you pick any of these points (say right here) and measure the distance from this point to one focus0041

(we will call that d1), and then we measure the distance from that same point to f2,0051

to the other focus, we will call that distance d2.0059

This definition states that, if I add up these two distances, d1 + d2, they will equal a constant.0063

I could pick any point; I could then pick this point and say, "OK, here is another distance," calling that, say, d3, and this one d4.0074

And if I added up this distance and this distance, d3 + d4, I could get that same constant.0087

And I could do that with any point on the ellipse.0094

Continuing on with some properties of the ellipse: an ellipse actually has two axes of symmetry.0104

One is called the major axis, and the other is the minor axis, and these intersect at the center of the ellipse.0110

Here, we are going to look at an ellipse that is centered at the origin (it has a center at (0,0)).0117

And again, I have foci f1 and f2.0122

This is one of the vertices of the ellipse; here is a vertex; this is the second vertex of the ellipse.0127

And the major axis runs from one vertex to the other vertex.0135

And you can see that it passes through both of the foci; this is the major axis.0140

And what we are looking at here is an ellipse that has a horizontal major axis.0149

In a few minutes, we will look at ellipses that are oriented the other way--ellipses that are oriented with a vertical major axis.0154

So, that does exist; but right now we will just focus on this for the general discussion.0163

Here we have the major axis; and intersecting at the center is a second axis called the minor axis.0167

Looking more closely at the relationships between the major axis, the minor axis, the foci, and the distances relating them:0182

let's call the distance from one vertex to the center A: this distance is A.0193

Therefore, the length of the major axis is 2A.0200

And this is going to become important later on, when we are working with writing equations for ellipses,0209

or taking an equation and then trying to graph the ellipse.0214

That is my first distance: this is actually called the semimajor axis.0218

The length of the semimajor axis is A; the length of the major axis is 2A.0226

Now, looking at the minor axis: from this point to the center, this length is B.0230

Therefore, the length of the minor axis is equal to 2B.0237

Now, looking at the foci: the distance from one focus to the center is going to be C.0246

Therefore, the distance from one focus to another, or the distance between the foci, is equal to 2C.0255

There is also an equation relating these A, B, and C; and that is A2 = B2 + C2.0276

So, keep this equation in mind: again, it becomes important, because you might be given A and B, but not C;0287

or you may be given a drawing, a sketch of the ellipse, and then asked to write an equation based on it.0294

And sometimes you need to find this third component in order to write that equation.0300

And you can do that by knowing that A2 = B2 + C2.0304

Again, A is the length of the semimajor axis; B is the distance from here to the center of the minor axis0309

(2B is the length of the minor axis); and C is the distance between one focus and the center (2C gives the distance between the two foci).0319

Those are important relationships to understand when working with the ellipse.0328

Standard form: looking at what the standard form of the equation of an ellipse with a center at (0,0) and a horizontal major axis is:0333

it is x2/A2 + y2/B2 = 1.0343

This is the standard form; and again, we already discussed what A is and B is.0353

And if you figure out those from the graph, and are given those, then you can go ahead and write your equation.0359

Or, given the equation, you can graph the ellipse.0365

Let's take a look at an example: A2/9 + y2/4 = 1.0368

So again, this is just the equation for an ellipse with a horizontal major axis, so it is sketched out here that way.0379

We can make this more precise by saying, "OK, A2 = 9; therefore, A equals √9, or 3."0388

That means that the distance from here to here is going to be 3.0405

And since this is centered at the origin, this will actually be at the point (3,0).0409

So, let's write that as a coordinate pair, (0,3).0415

OK, that means that this length of A is 3; over here, the other vertex is going to be at (0,-3).0420

So, I have one vertex at (0,3) and the other at (0,-3); and I have f1 here and f2 here.0431

And the length of the major axis is actually 6: it is going to be equal to 2A.0437

2A equals 6, and that is the length of the major axis.0443

B2 = 4: since B2 = 4, B = √4, which equals 2.0452

Therefore...actually, this needs to be written the other way; this is actually (3,0), and (-3,0).0461

Now, up here, we have (0,2) and (0,-2).0474

And 2, then, is the length; that is B--that is the length of half of the minor axis.0485

2B = 4, and this is the length of the minor axis.0492

You can get a lot of information just by looking at the equation of the ellipse in standard form.0498

If I needed to, I could figure out the distance between the foci, and I could figure out what C is.0505

And remember, C is this length; because recall that A2 = B2 + C2.0512

And I know that A2 is 9; I know that B2 is 4; so I am looking for C2.0518

9 - 4 gives me 5, so C2 = 5; therefore, C = √5, and that is approximately equal to 2.2.0524

So, this distance here is roughly 2.2; and the distance between these foci would be about 4.4.0535

So, just by having this equation, I could graph the ellipse, and I could find this third component that was missing.0544

OK, so that was standard form for an ellipse that has a horizontal major axis.0555

For an ellipse that has a vertical major axis, you are going to see that the A2 is associated with the y2 term.0560

In the horizontal major axis, we had x2/A2.0567

Now, if you were given an equation, how would you know what you are dealing with--0572

if you were dealing with a vertical major axis or a horizontal major axis?0578

Well, A2 is the larger number; so let's say I was given something like y2/16 + x2/9 = 1.0581

When I look at this, I see that the larger number is associated with y2; so that tells me that I have a vertical major axis.0594

If it had been x2 associated with 16, then I would have said, "OK, that is a horizontal major axis."0602

Again, I can graph this ellipse by having this equation written in standard form.0609

I know that A2 equals 16; therefore, A equals the square root of 16; it equals 4.0614

This time, I am going to go along here for the major axis; and that makes sense,0622

because I have a focus here and another focus here, and the major axis passes through the two foci.0628

OK, so (0,4) is going to give me one vertex; (0,-4) will give me the other vertex.0638

And remember that 2A = 8, and that is going to be the length of the major axis.0647

And again, right now, we are limiting our discussion to ellipses with a center at (0,0).0658

Later on, we will expand the discussion to talk about the graphs of ellipses with centers in other areas of the coordinate plane.0664

OK, so now I have B2 = 9; therefore, √9 is going to equal B; this tells me that B equals 3.0672

So, since B equals 3, the length of half of the minor axis is going to be 3.0689

So, right here at (3,0) is going to be one point, and I am going to have the other point right here at (-3,0).0696

And the length of the minor axis...2B = 6, and that is the length of the minor axis.0704

Again, I can use the relationship A2 = B2 + C20719

to figure out what C is, and to figure out where the foci are located.0725

I know that A2 is 16; I know that B2 is 9; and I am trying to figure out C2.0730

So, I take 16 - 9; that gives me 7 = C2; therefore, C = √7.0738

And the square root of 7 is approximately equal to 2.6, so this is going to be up here at about (0,2.6).0746

And f2 is going to be down here at about (0,-2.6).0755

So again, using standard form, you can graph the ellipse, and you can find where the foci are, based on the values of A2 and B2.0758

We talked a little bit about graphing ellipses; but sometimes you are not given the equation in standard form.0771

If the equation is not in standard form, you have to put it that way.0776

And working with other conic sections, we have learned that you can put an equation into standard form for a conic section by completing the square.0780

You can also use symmetry, just as we did when graphing parabolas or circles.0788

So, let's say you are given an equation like this: 3x2 + 4y2 - 18x - 16y = -19; and you are asked to graph it.0794

You are going to put it in standard form; but it is nice, first of all, to know what kind of shape you are working with.0807

And you can tell that just by looking at this, even though it is not in standard form yet.0811

And the reason is: I see that I have an x2 term and a y2 term0815

on the same side of the equation, with the same sign, with different coefficients; that tells me that I am working with an ellipse.0820

And we are going to go into more detail in the lecture on conic sections.0827

We are going to review how to tell apart the equations for various conic sections.0831

But just briefly now: recall that a parabola would have either an x2 term or a y2 term, not both.0835

With a circle, the coefficients of x2 and y2 are the same; that is for a circle.0845

For an ellipse, the coefficients of the x2 and y2 terms are different.0861

Now, you see that these have the same sign; so with an ellipse,0876

it is important to note that the x2 and y2 terms have the same sign.0881

If they don't have the same sign, that is actually a different shape.0884

They have the same sign; but this coefficient (the x2 coefficient) is 3, and the y2 coefficient is 4.0887

That tells me I have an ellipse, not a circle.0894

My next step is to complete the square and write this in standard form,0898

so that, if I wanted to graph it, I would have all of the information that I need.0903

The first thing to do is to group the x variable terms and the y variable terms.0906

This gives me x2 - 18x + 4y2 - 16y = -19.0911

Now, when I am looking at this, I remember that, to complete the square, I want to end up with a leading coefficient of 1.0931

I am going to factor out this 3 to get 3(x2 - 6x); and I am going to need to add0937

something else over here to complete the square--a third term.0944

Here, factor out the 4 to give me (y2 - 4y) = -19.0949

Recall that, to complete the square, you are going to add b2/4 to each set of terms.0957

So, for this x group, we have b2/4 = -(6)2/4 = 36/4 = 9.0963

So, I am going to add a 9 here; and very importantly, to the right side, I am going to add 9 times 3, which is 27.0978

Add that to the right, because if I don't, the equation won't be balanced anymore.0995

Now, the y variable terms: I need to add something here to complete the square.1003

And I will work over here for this: b2/4 equals (-4)2/4 = 16/4 = 4.1013

So, I am going to add 4 here; but to the right, what I need to add is 4 times 4; so I need to add 16.1026

Now that I have this, my next step is to write this in standard form: 3...and this is x - 3, the quantity squared;1038

if I squared this, I would get this back; plus 4y - 2, the quantity squared, equals...1049

-19 plus 16 gives me, let's see, -3; 27 minus 3 gives me 24 on the right.1057

Well, recall that for standard form, I want a 1 over here on the right; I want this to equal 1.1069

So, this gives me...I need to divide both sides by 24; this is going to give me 3(x - 3)2/24 + 4(y - 2)2/24 = 24/24.1076

This finally leaves me with (x - 3)2...this 3 will cancel, and I am going to get 8 in the denominator;1101

this 4 will cancel, and I will get 6 in the denominator; equals 1.1109

Now, I have the bigger term associated with x; that tells me that I have a horizontal major axis.1115

So, the major axis is horizontal; and let me go ahead and write that up here.1126

x2/a2 + y2/b2 = 1.1144

OK, now you can see that this looks slightly different; and we have these extra terms here.1154

And what that tells us is something that we are going to discuss in a moment.1160

And what we are going to discuss is situations where the center of the ellipse is not at (0,0).1164

But we have the same information that we would have with the center being at (0,0).1174

And that is that I have A2, which equals 8; and I know I have a horizontal major axis; and I have B2 = 6.1180

Looking, now, at equations for ellipses with centers at (h,k), at somewhere other than 0:1198

if you look at the situation where we have a horizontal major axis, versus a vertical major axis,1206

you can again see that it is very similar to when the center is at (0,0).1212

The x is associated with the A2 term when the major axis is horizontal.1217

The y is associated with the A2 term when the major axis is vertical.1221

The only difference is that we now have these terms telling us where the center is.1227

And if the center was going to be 0, all that would happen is that you would have x - 0 and y - 0,1231

which then gives you back the equation we worked with before, x2/A2 + y2/B2,1237

or y2/A2 + x2/B2, all equal to 1.1244

So, let's talk about an example for this: (x - 4)2 + (y + 6)2 = 1.1252

This is all over 100, and this is all divided by 25.1264

What this tells me is that the center is at (h,k); so this is h (that is 4).1269

You need to be careful here, because what you have is a positive; but the standard form is a negative.1278

And it is perfectly fine to write this like this, but you need to keep in mind that this is really saying...1284

if you think about it, y + 6 is the same as y - -6.1291

So, when I look at it this way, I realize that, if it is in this form, k is actually going to be -6.1296

And if you are not careful about that, you can end up putting your center in the wrong place.1302

Let's let this be 2, 4, 6, -2, -4, -6; the center is at (4,-6), right here.1309

A2 = 100; my larger term is my A2term, and I have a horizontal major axis.1323

Therefore, A = √100, which is 10.1331

So, since A equals 10 and the length of the major axis is horizontal, then I am going to need to go 10 over from 4.1337

It is going to be all the way out here at 14.1351

So, this is where one vertex would be--this is going to be at 4 + 10 gives me (14,-6).1357

That is one vertex: -2, 4, 6, 8, 10, 12, 14...so -14 is going to be about here.1368

And I am going to have the other vertex here at (-14,-6).1376

I have a minor axis: B2 = 25; therefore B = 5.1384

So, what that tells me is that I come here at -6; I add 5 to that; and that is going to bring me right here at -1; that is going to be right about here.1389

And then, -8, -10, -12...-(6 + 5) is going to be down here at -11; so this ellipse is going to roughly look like this.1400

OK, and standard form allowed me to determine that the center is at (4,-6);1419

that I have a vertex; if I add 10, that is going to give me this vertex right over here1427

(let's draw this more at the vertex); that I have another vertex here; that the major axis is going to pass through here;1433

and then, I am going to have a minor axis passing through here.1441

All right, in the first example, we are asked to find the equation of the ellipse that is shown.1447

And I am going to go ahead and label some of these points.1453

This is going to be f1; and let's say we are given f1 as being at (0,3).1457

This is going to be 1, 2, 3: each mark is going to stand for one.1463

Down here, I have f2; that should actually be down here a little bit lower; so that is f2 at (0,-3).1467

Let's say we are also given a vertex at (0,5), and the other vertex at (0,-5).1480

All right, so I am asked to find the equation of the ellipse shown.1492

And the one thing I see is that there is a vertical major axis.1495

Since there is a vertical major axis, it is going to be in the general form (y - k)2/A2 + (x - h)2/B2 = 1.1503

Now, I notice that the center is at (0,0); so that tells me that h and k are (0,0).1520

So, this is actually going to become the simpler form, y/A2 + x/B2 = 1.1528

The next piece of information I have is the length of A.1537

So, this line is going to give me A; and I know that, since the center is here at (0,0), the length of this is 5; therefore, A = 5.1541

Since A equals 5, A2 is going to equal 52, or 25.1553

I don't know B; this is going to be B, but I don't know what it is.1561

However, I do know C; if I look here, this is C--from -3 to the center is 3, so C = 3.1567

Therefore, C2 = 32, or 9.1583

The other piece of information I have is the equation we have been working with,1587

that states that A2 = B2 + C2.1591

Therefore, I can find B2, which I need in order to write my equation.1602

So, I know that A2 is 25; I know that C2 is 9; and I am looking for B2.1606

25 - 9 is 16; therefore, B2 = 16, and B = 4.1620

Now, I can go ahead and write my equation; I have all the information I need, so let's put it all together right here.1628

y, divided by A2...I know that A2 is 25...(that is actually y2 right here--1633

let's not forget the squares)...plus x2 divided by B2...B2 is 16...equals 1.1647

So again, this had a center at (0,0); so remember that this is the general form of the equation.1655

When you have a center at (0,0), it becomes this form.1660

I had a vertical major axis, so I am using this form, where the A2 is associated with the y2 term.1663

And by having A2 and C2, I was able to determine what B2 is.1671

Example 2: let's find the equation of the ellipse satisfying a major axis 10 units long and parallel to the y-axis; minor axis 4 units long; center at (4,2).1678

Let's start with the center--the center is at (4,2)--the center is right here at (4,2).1690

And it says that the major axis is parallel to the y-axis.1699

What that tells me is that I have a vertical major axis.1703

Since this is a vertical major axis and I know the center (I know (h,k)), I am going to be working with the general form1711

(y - k)2/A2 + (x - h)2/B2 = 1.1719

I have h; I have k; I need to find A2 and B2.1728

The other piece of information I have is that the major axis is 10 units long; and I know that it is vertical.1732

Since it is 10 units long, that means that the major axis length is equal to 2A, and I know that that length is 10;1738

therefore, I just take 10 divided by 2, and I get that A = 5.1751

So, I am starting here; and if I take 5 + 2, I am going to get that I am going to have the vertex up here at 7.1757

5 + 2 is going to give me 7, right there; so I go to (4,7); that is where this is going to be.1775

And I got that by keeping the x where it was, and then adding 2 where the center was and adding 5 to that, which is the length of A.1793

Then, I am going to go down; again, it is going to be at 4.1803

But then, if I take 2 and subtract 5 from it, I am going to get -3.1810

So, right here at (4,-3) is the other vertex.1815

Now, I was not asked to graph this; but I am graphing it so that I have an understanding of what each of these means,1823

so that I can go ahead and write the equation.1829

What I have now is...I know what A is, which means I can figure out A2; and I know the center.1832

The last thing I know is that the minor axis length equals 2B: I am given that that is 4 units long; therefore, B = 2.1844

So, if I started here at 4, and I added 2 to that, I am going to end up with (6,2).1857

And I am going to end up with...if I start at 4 and I subtract 2, I am going to end up with (2,2).1867

And this is B; this is A.1881

I have a minor axis that has a length of 2B, which tells me that B = 2.1887

From this information, I can go ahead and write this equation.1891

I know A equals 5, so A2 equals 52, which is 25.1897

B = 2; B2, therefore, equals 22, or 4.1904

So, I have everything I need to write this: (y - k)...well, k is 2; the quantity squared, divided by A2;1911

A2 is 25; plus (x - h)...h is 4...the quantity squared, divided by B2, which is 4, equals 1.1920

So, this standard form describes the ellipse with the major and minor axis here.1936

And you could finish that out by just connecting these points and drawing the ellipse if you wanted to finish graphing it.1941

Find the equation of the ellipse satisfying endpoints of the major axis at (11,3) and (-7,3) and foci at (7,3) and (-3,3).1953

All right, endpoints of the major axis: let's do 2, 4, 6, 8, 10, 12, and -2, 4, -6...OK.1962

The endpoint is at (11,3): 11 is right here, and then we will have 3 be right here.1979

The other endpoint is at -7, which is going to be here, 3.1991

And that tells me the major axis: since the major axis is horizontal, we are going to be working with the general form1999

(x - h)2/A2 + (y - k)2/B2 = 1.2010

So, that is my first piece of information.2018

I can also figure out the length of the major axis.2020

So, since the major axis goes from -7 to 11, if I take -7 minus 11, I am going to get -18.2024

And a length is going to be an absolute value, so I am just going to take 18; the length is going to be the absolute value of this difference.2045

All right, I know that the major axis length is 18; the other thing I know is that the major axis length equals 2A, as we have discussed.2053

So, 2A = 18; therefore, A = 9; so I know that the distance from the center to this endpoint is 9.2061

Therefore, I could just say, "OK, 9 minus 11 gives me 2; and I know that the y-coordinate will be 3; so that is (2,3)."2076

Another way to solve this, without using all this graphing, would have simply been to find the midpoint.2087

I am going to go ahead and do that, as well, because the midpoint of the major axis is the center of the ellipse.2093

Let's try that, as well--the center, using the midpoint formula.2100

Recall the midpoint formula: we are going to take x1 + x2 (that is 11 + -7);2105

and we are going to take the average of that (we will divide it by 2); that is going to give me the x-coordinate.2112

For the y-coordinate, I am going to take 3 + 3, and I am going to divide that by 2.2117

And this will give you a center at...11 - 7 gives you 4; 4 divided by 2 is 2.2126

3 + 3 is 6, divided by 2 gives you 3; and that is exactly what I came up with using the graphing method.2138

So, just to show you: you could have figured this out algebraically (where the center is); or you could have figured it out using graphing.2144

This gives me (h,k), so I have (h,k), and I have A, which is 9, so I can get A2.2154

The next thing I need to do is figure out B, and they don't give me B; but what they do give me are the foci.2162

There are foci at (7,3) (7 is here--focus at (7,3)--we will call this one f1 at (7,3)); and this is the center right here.2167

There is another focus at (-3,3): f2 is going to be at (-3,3).2187

Recall that the distance from one focus to the other is 2C; the distance from one focus to the center is C.2201

Let's just work with this; this is C; 7 - 2 is 5; therefore, C = 5.2210

So, I have A = 9; I have C = 5 (again, that is the distance from the center to a focus);2221

or I could have figured out the distance from one focus to another--that is 2C--and then divided by 2;2229

So, I have A, and I have C, and I know that A2 = B2 + C2.2233

So, this gives me 92 = B2 + 52: 92 is 81, equals B2 + 25.2240

If I take 81 - 25, I am going to get 56 = B2.2254

And I don't even need to take the square root of that, because to put this into standard form, I actually need B2.2259

So, I get (x - h); remember, the center is (h,k), so that is 2; the quantity squared, divided by A2...2265

recall that A2 is 92, so it is 81; plus (y - k)2...k is 3, the quantity squared;2273

divided by B2...I determined that B2 is 56; all of this equal to 1.2284

Just by knowing the endpoints of the major axis and the location of the foci, I could figure out A22289

and B2, as well as the center, and then write this equation for the ellipse in standard form.2295

Finally, we are asked to graph an ellipse that is not given to us in standard form.2307

We have some extra work to do: we are going to actually have to complete the square in order to even graph this.2311

So, let's go ahead and start by grouping the x terms together and y terms together.2318

Also note that, since this has an x2 term and a y2 term on the same side of the equation,2329

with the same sign (they are both positive), but different coefficients, I know I have an ellipse.2335

It is not a circle, because for a circle, these coefficients would be the same.2341

Grouping the terms together gives me 14x2 - 56x + 6y2 - 24y = -38.2347

Looking at this, I see that I have a common factor of 2.2368

If I divide both sides by 2, I can simplify this equation; so the numbers I will work with will be smaller.2372

So, let's divide both sides by 2 to get 7x2 - 28x + 3y2 - 12y = -19.2378

The next thing to do is to factor out the leading coefficient, since it is not 1.2391

I am going to factor out a 7 to get x2 - 4x, plus...2397

over here, I am going to factor out a 3; that gives me y2 - 4y, all equals -19.2402

I need to complete the square, so I need to get b2/4.2411

In this case, that is going to give me 42/4 = 16/4 = 4.2415

So, over here, I am going to add a 4; very important--on the right, I have -19, and I am adding to the left 7 times 4; that is 28.2423

So, I need to add 28 to the right.2440

All right, over here, for the y terms: b2/4: again, we have a b term that is 4, so I am going to end up with the same thing, 4.2446

Now, here I am actually adding 4 times 3 (is 12), so I need to add that to the right, as well, to keep the equation balanced.2462

This is the easiest step to mess up on: you are focused here on completing the square,2470

and then you sometimes don't remember that you have to add the same thing to both sides.2474

Now, working on writing this in standard form: this is going to give me (x - 2)2 + 3(y - 2)2 = -19 + 12...2479

that is going to leave me with 28; -19 + 12 is going to be -7; 28 minus 7 equals 21.2493

To get this into standard form, I need to have a 1 on the right, so I am going to divide both sides by 21.2504

This is going to give me 7(x - 2)2/21 + 3(y - 2)2/21 = 21/21.2510

This cancels to (x - 2)2/3; this becomes (y - 2)2/7; and this just becomes 1.2528

Now, I have standard form; I can do some graphing.2547

I have a center at, let's see, (2,2); that is right here; the center is at (2,2), so that is h and k.2550

And I notice, actually, that the larger term is under the y; it is associated with the y.2577

That tells me that this has a vertical major axis; and therefore, I am going to keep that in mind--that it is going to be oriented this way.2587

The ellipse is going to end up like this, instead of like this.2600

I have my center at (2,2); therefore, A2 = 7; A = √7.2604

The square root of 7 is about 2.6, so it gets messy, as always, when you are working with radicals.2620

But it is about 2.6; so what I have to do is say, "OK, my vertex up here is going to be at x = 2, and then y is going to equal 2 + 2.6, which is 4.6."2626

So, (2,4.6): and again, I got that by saying the length of A (the length from the vertex to the center) is 2.6.2643

So, 2 + 2.6 gives me 4.6.2655

Over here, I am going to take 2, and I am going to subtract 2.6 from it; so x is still going to be 2; now y is going to be about here, which is (2, -.6).2658

Again, 2 is here; the length of A is 2.6, so it is going to be 2 - 2.6; this gives me my major axis.2674

I know, from vertex to vertex, where this ellipse is going to land.2685

Now, the minor axis, B: I know that B2 = 3; therefore, B = √3, which is approximately equal to 1.7.2690

This is not to do the same thing, but going along the x direction, the horizontal direction.2700

So again, this is A; OK, now to get B, I am going to have 2, and I am going to add 1.7 to it.2706

That is going to give me 3.7; it is going to land about here.2715

In this direction, I am going to subtract; I am going to say that I have 2 - 1.7, so that is going to land here, at about .3.2722

And in the y direction (in the vertical direction) it is still going to be up at 2.2739

So again, to get this, I said 2 + 1.7; that brought me to (3.7,2)--that is this point.2742

This point is at 2 minus 1.7, so that is (.3,2).2753

So, this gives me the general shape of this ellipse, like this; so I can get a good sketch, based on this equation.2761

I took this equation, and I recognized that it was an ellipse.2782

I completed the square to get it in standard form, and saw that it had a vertical major axis and that it had a center at (2,2).2787

I then found A to determine where the vertices would be, and then B to determine the width of the ellipse here; and then I could get a good sketch.2795

That concludes this session of Educator.com on ellipses; thanks for visiting!2806

Welcome to Educator.com.0000

Today, we are going to be discussing the last type of conic section, which is hyperbolas.0002

So far, we have covered parabolas, circles, and ellipses.0006

As you can see, hyperbolas are a bit different in shape than the other conic sections we have worked with.0012

And one thing that makes them unique is that there are two sections referred to as branches; there are two branches in this hyperbola.0018

The formal definition is that a hyperbola is a set of points in the plane,0025

such that the absolute value of the differences of the distances from two fixed points is constant.0029

What does that mean? First, let's look at the foci.0037

These two fixed points are the foci; and here is a focus, f1, and here is the other, f2.0041

If I take a point on the hyperbola, and I measure the distance to f1,0048

and then the distance to f2, that is going to give me d1 and d2.0056

Recall that, with ellipses, we said that the distance from a point on the ellipse--if you measured the distance to one focus,0067

and then the other focus, and then added those, that the sum would be a constant.0074

Here, we are talking about the difference: the absolute value of this distance, d1, minus (the difference) d2, equals a constant.0078

That is the formal definition of a hyperbola.0092

Again, I could take some other point: I could take a point up here on this other branch.0094

And I could find a distance, say, d3, and then the distance to f1 could be something--say d4.0100

Again, the absolute value of those differences would be equal to that same constant.0107

All right, properties of hyperbolas: A hyperbola, like an ellipse, has two axes of symmetry.0117

But these have different names: here you have a transverse axis and a conjugate axis, and they intersect at the center.0124

We are looking here at a hyperbola with the center at (0,0).0133

One thing to note is that you can also have a hyperbola that is oriented as such.0139

But right now, we are looking at this one, with a more horizontal orientation.0147

But just to note: this does exist, and we will be covering it.0153

All right, first discussing the transverse axis: the transverse axis is going to go right through here--it is going to pass through the center.0156

And this is the vertex, and this is the vertex of the other branch, and this is the transverse axis.0166

The distance from one vertex to the center along this transverse axis is going to be A.0179

Again, a lot of this is going to be similar from when we worked with ellipses; but there are some important differences, as well.0186

So, the length of the transverse axis equals 2A; from here to here would be 2A.0195

The foci: if you look at foci, say f1, f2...let's look at f2...0212

it would be the same over on f1: if I looked at the distance from one focus to the center, that is going to be C.0218

The distance between the foci is therefore 2C; if I measured from here to here, that length is going to be 2C.0229

There is a second axis called the conjugate axis; and the two axes intersect here at the center.0246

The length of half...if you take half the length of this conjugate axis, it is going to be equal to B.0260

The transverse axis lies along here; the conjugate axis, in this case, is actually along the y-axis (and the transverse axis is along the x-axis).0273

The length of the conjugate axis is 2B.0288

As with the ellipse, there is an equation that relates A, B, and C; but it is a slightly different equation.0297

Here, A, B, and C are related by C2 = A2 + B2.0303

This relationship will help us to look at the equation for a hyperbola and graph the hyperbola,0311

or look at the graph, and then go back and write the equation.0317

So again, there are two axes: transverse, which goes from vertex to vertex; and conjugate, which intersects the transverse axis0321

at the center of the hyperbola and has a length of 2B (the transverse axis has a length of 2A).0331

The distance from focus to focus (between the two foci) is 2C.0338

The standard form of the hyperbola is also going to look somewhat familiar, because it is similar to an ellipse, but with a very important difference.0346

Here we are talking about a difference instead of a sum.0352

So, if you have a hyperbola with a center at (0,0) and a horizontal transverse axis, the equation is x2/A2 - y2/B2.0355

And here, we again have that the center is at the origin, (0,0).0368

Although the center certainly does not have to be at the origin, right now we are going to start out0373

working with hyperbolas with a center at the origin, just to keep things simple.0377

And again, by being given an equation in standard form, you can look at it and get a lot of information about what the hyperbola looks like.0384

Therefore, the vertex is going to be at (A,0); the other vertex will be at (-A,0).0396

You are going to have a point up here that is going to be (B,0); and this length, B, gives the length of half of the conjugate axis.0408

And then, you are going to have another point...actually, that is (0,B), because it is along the y-axis...another point, (0,-B).0420

This distance is B, from this point to the center; this distance is A.0429

And then here, I have f1 and f2; and the distance from one of those to the center is C.0436

As I mentioned, you can have a hyperbola that is oriented vertically.0447

So, if the transverse axis is vertical, and the center is at (0,0), the standard form is such that y2 is associated with the A2 term.0451

And here, it is positive: so you are taking y2/A2 - x2/B2 = 1.0463

In this case, what you are going to have is a vertex right here at (0,A), the other vertex is here at (0,-A); here is the transverse axis.0470

And then, you are going to have the conjugate axis; the length of half of that is going to be B; the length of the entire thing is 2B.0482

So, this is going to be some point, (B,0); and B2 is given here, so you could easily find B by taking the square root.0495

And then, over here is (-B,0).0502

So again, there are two different standard forms, depending on if you are working with a hyperbola0505

that has a horizontal transverse axis or one that has a vertical transverse axis.0508

Something new that we didn't talk about with ellipses is asymptotes.0516

Recall that an asymptote is a line that a curve on a graph approaches, but it never actually reaches.0520

And asymptotes are very useful when you are trying to graph a hyperbola.0528

The equations are given here: let's go ahead and draw these first.0534

Now, recall that this vertex is at a point (A,0); this vertex is at (-A,0).0536

If I measure the length...this is the transverse axis, and it is horizontal...let's say that it turns out that B is right up here, (0,B).0545

And B is going to be the length from this point to the center; 2B will be the length of the conjugate axis.0555

Then, I am going to have another point down here, (0,-B).0564

What I can do is form a box, a rectangle; and the rectangle is going to have vertices...I am going to go straight up here and across here.0568

Therefore, this is going to be given by (-A,B); that is going to be one vertex.0578

I can go over here and do the same thing; I am going to go straight up from this vertex, and straight across from this point.0586

And that is going to give me the point (A,B).0592

I'll do the same thing here: I go down directly and draw a line across here; this point is going to be (-A,-B).0597

One final vertex is right here: and this is given by (A,-B).0608

OK, now you draw a box using these A and B points; and then you take that rectangle and draw the diagonals.0618

If you continue those diagonals out, you will have the two asymptotes for the hyperbola.0630

OK, so each of these lines is an asymptote.0650

And notice that the hyperbola is going to approach this, but it is not actually going to reach it.0654

So, it is going to continue on and approach, but not reach, it; it is going to approach like that.0663

All right, now this is one way to just graph out the asymptotes.0671

You can also find the equation; and for right here, we are working with a hyperbola with a horizontal transverse axis.0676

So, we are going to look at this equation.0686

If I was working with a vertical one, I would look at this equation.0688

Now, what does this mean? Well, y equals ±(B/A)x.0691

What this actually is: this B/A gives the slope of the asymptote.0697

Recall that y = mx + b; since the center is at (0,0), the y-intercept is 0; so here, b = 0, so I am going to have y = mx.0703

The slope, m, is B/A; B/A for this line is increasing to the right (m = B/A);0714

and the slope here equals -B/A, where the line is decreasing as we go towards the right.0724

So again, there are two ways to figure out these asymptotes.0734

You can just sketch it out by drawing this rectangle with vertices at (-A,B), (A,B), (-A,-B)...that is actually (A,B), positive (A,B)...or (-A,-B).0737

Draw that rectangle and extend the diagonals.0754

Or you can use the formula, which will give you the slope for these two asymptotes.0758

If you just started out knowing the A's and B's and drew these, then you could easily sketch the hyperbola,0765

because you know that it is going to approach these asymptotes.0772

OK, so we have talked a lot about graphing.0778

And just to bring it all together: you are going to begin by writing the equation in standard form.0780

And then, for hyperbolas, you are going to graph the two asymptotes, as I just showed.0786

So, let's start out with an example: let's make this x2/9 - y2/4 = 1.0790

Since I have this in the form x2/A2 (this x2 term is positive here),0801

divided by y2/B2 = 1, what I have is a horizontal transverse axis.0809

So, this tells me that there is a horizontal transverse axis.0819

So, that is how this is just roughly sketched out already, showing the transverse axis along here.0830

Since it is in this form, I know that A2 = 9; therefore, A = 3.0841

This has a center right here at (0,0).0853

And this point here is going to be A, which is 3, 0.0860

Right here, I am going to have -A, or -3, 0.0867

So, my goal is to make that rectangle extend out the diagonals.0872

And then, I would be able to graph this correctly.0876

OK, B2 = 4; therefore, B = 2; so right up here at (I'll put that right there) (0,2)...that is going to be B.0882

And then, right down here at (0,-2)...0903

Now, all I have to do is extend the line up here and here; and these are going to meet at (2,3).0907

Extend a line out here; I am going to have a vertex right here at (-3,2).0916

I am going to have another vertex here at (-3,-2), and then finally, one over here at (3,-2).0924

Now, this was already sketched on here for me; but assuming it was not there, I would have started out by drawing this box,0936

and then, drawing these lines extending out--the asymptotes.0944

And what is going to happen is that this hyperbola is actually going to approach, but it is never going to intersect with, the asymptote.0960

So again, write the equation in standard form, which might require completing the square.0972

I gave it to you in standard form already; use that to figure out this rectangle.0977

And you are going to need to know A and B to figure out this rectangle.0982

Draw the asymptotes, and draw then the hyperbola approaching (but not reaching) those asymptotes.0985

So far, we have been talking about hyperbolas with a center at the origin (0,0).0993

However, that is not going to always be the case.1000

If the center is at another point, (h,k), that is not (0,0), then standard form looks like this.1002

It is very similar to what we saw with the origin of the center, except instead of just x2/A2, we now have an h and a k.1008

For a horizontal transverse axis, you are going to have (x - h)2/A2 - (y - k)2/B2.1015

For a vertical transverse axis, this term is going to be first; it will be positive.1027

And then, you are going to subtract (x - h)2/B2.1033

But the k stays associated with the y term.1037

For example, given this equation, (y - 3)2 - (x - 2)2...and we are going to divide that by 16,1041

and divide this by 9, and set it all equal to 1: what this is telling me is that the center is at (2,3), because this is h;1053

that A2 = 16, so A = 4; and that B2 = 9, so B = 3.1065

From that, I can graph out this hyperbola.1072

And this has a vertical transverse axis; something else to be careful of--let's say I had something like this:1076

(y + 5)/10, the quantity squared, plus (x + 4), the quantity squared, divided by 12, equals 1.1087

The center is actually at (that actually should be a negative right here--this is a difference) (-4,-5).1101

And the reason for that is that this is the same as (y - -5)2, and then (x - -4)2.1112

A negative and a negative is a positive.1125

So, you need to be careful: even though it is acceptable to write it like this, it is good practice,1128

if you are trying to figure out what the center is, to maybe write it out like this,1134

so that you have a negative here, so that whatever is in here is already k, or already h.1139

You don't have to say, "Oh, I need to make that a negative; I need to change the sign."1145

So, that is just something to be careful of.1148

Here, I already had negative signs in here; they are completely in standard form--I have h and k here; h and k is (-4,-5).1151

All right, to get some practice, we are going to first find the equation of a hyperbola that I am going to give you some information on.1160

I will give you that one of the vertices is at (0,2); the other vertex is at (0,-2).1168

The other piece of information is that you have a focus at (0,4), and a focus called f2 at (0,-4).1178

So, looking at this, I can see that this is the transverse axis, and then the center is right there.1188

So, I have a horizontal transverse axis.1196

I can also see that the midpoint right here, the center, is at the origin; so the center equals (0,0).1207

So, this is actually vertical--correction--a vertical transverse axis, going up and down: a vertical transverse axis.1217

Since this is actually a vertical transverse axis with a center at (0,0), I am working with this standard form:1227

y2/A2 - x2/B2 = 1.1234

So, the A2 term is with the y2 term, since this is a vertical transverse axis.1240

All right, in order to find the equation, I need to find A2.1247

This distance, from 0 to the vertex, is 2, because this is at (0,2).1251

Therefore, A equals 2; since A = 2, A2 = 22, or 4.1258

I have A2; I need to find B2; I am not given that.1267

But what I am given is an additional piece of information, and that is that there is a focus here and a focus here.1270

This allows me to find C: the distance from the center to either focus (let's look at this one)--from the center, 0, down to -4--1280

the absolute value of that is 4; therefore, C = 4; the distance is 4.1291

C2, therefore, equals 42, or 16.1298

Recall the relationship: C2 = A2 + B2 for a hyperbola.1303

So, I have C2, which is 16, equals A2, which is 4, plus B2.1309

16 - 4 is 12; 12 = B2; therefore, B = √12, which is about 3.5.1315

If you wanted to draw B, then you could, because that is right here at (3.5,0).1330

But what we are just asked to do is write the equation; and we have enough information to do that,1338

because I have that y2 divided by A2; I determined that A2 is 4;1342

minus x2/B2; I determined that that is 12; equals 1.1348

So, this is the equation for this hyperbola, with a vertical transverse axis and a center at (0,0) in standard form.1355

The next example: Find the equation of the hyperbola satisfying vertices at (-5,0) and (5,0) and a conjugate axis that has a length of 12.1368

Just sketching this out to get a general idea of what we are looking at--just a rough sketch--vertices are at (-5,0) and (5,0).1380

That means that the center is going to be right here at (0,0).1400

So, the center is at the origin; since the vertices are here and here, then I have a horizontal transverse axis;1406

this is going to go through like this, and then like this.1424

So, my second piece of information is that I have a horizontal transverse axis.1432

Since I have a horizontal transverse axis, then I am going to have an equation in the form x2/A2 - y2/B2 = 1.1440

The center is at the origin; it has a horizontal transverse axis; this is a standard form that I am working with.1451

I need to find A: well, I know that the center is here, and that A is this length; so from this point to the center,1457

or from this point (the vertex) to the center, is 5: A = 5.1466

Since A = 5, A2 = 52; it equals 25.1473

The other information I have is that the conjugate axis has a length of 12.1486

So, the length of the conjugate axis, recall, is 2B; here they are telling me that that length is 12.1490

Therefore, 12/2 gives me B; B = 6; so, that would be up here and here: (0,6) and then (0,-6).1503

This would be the conjugate axis; so this is B = 6.1517

Since B equals 6, I want B2 that equals 62, which equals 36.1523

Now, I can write this equation: I have (this is my final one) x2/A2, which is 25,1533

minus y2/B2, and I determined that that is 36, equals 1.1545

So, this is a hyperbola with a center at the origin.1551

And A2 is 25; B2 is 36; and it has a horizontal transverse axis.1554

We are asked to graph this equation; and it is not in standard form.1566

But when I look at it, I see that I have a y2 term and an x2 term, and they have opposite signs.1571

So, I am working with the difference between a y2 term1577

and an x2 term, which tells me that this is the equation for a hyperbola.1579

If they were a sum, this would have been an ellipse, since they have different coefficients.1585

But it is a difference, so it is a graph of a hyperbola.1589

What I need to do is complete the square to get this in standard form.1591

OK, so first I am grouping y terms and x terms: y2 + 12y - 6x2 + 12x - 36 = 0.1597

What I am going to do is move this 36 to the other side and get that out of the way for a moment by adding 36 to both sides.1617

The next thing I need to do with completing the square is factor out the leading coefficient, since it is something other than 1.1627

So, from the y terms, I will factor out a 2; that is going to leave me with y2 + 6y.1634

You have to be careful here, because you are factoring out a -6, so I need to make sure that I worry about the signs.1640

And that is going to leave behind an x2 here; here, it is going to leave behind, actually, -2x.1649

So, checking that, -6 times x2 is -6x2--I got that back.1656

-6 times -2x is + 12x; equals 36.1661

Now, to complete the square, I have to add b2/4 in here, which equals...b is 6; 62/4 is 36/4; that is 9.1669

I need to be careful to keep this equation balanced.1686

Now, this is really 9 times 2 that I am adding; 9(2) = 18--I need to add that to the right.1688

Working with the x terms: b2/4 = 22/4, which is 4/4; that is 1, so I am going to add 1 here.1697

-6 times 1 needs to be added to the other side; so I am going to actually subtract 6 from the right to keep it balanced.1714

Now, I am rewriting this as (y + 3)2 - 6(x - 1)2 =...18 - 6 is 12; 36 + 12 gives me 48.1726

The next step, because standard form would have a 1 on this side, is: I need to set all this equal to 1.1747

I need to divide both sides of the equation by 48.1753

This cancels, so it becomes y + 32; the 2 is gone; this becomes a 24; minus...6 cancels out,1769

and that leaves me with (x - 1)2; 6 goes into 48 eight times; and this is a 1.1780

OK, so it is a lot of work just to get this to the point where it is in standard form.1788

But once it is in standard form, we can do the graph, because now I know the center; I know A2 and B2.1792

We have this in standard form; so now we are going to go ahead and graph it.1799

I will rewrite the standard form that we came up with, (y + 3)2/24 - (x - 1)2/8 = 1.1803

Looking at this; since this is positive, I see that I have a vertical transverse axis.1815

The other thing to note is this plus here: recall that, if you have (y + 3)2, this is the same as (y - -3)2.1824

And when we look at standard form, we actually have a negative here.1835

So, you need to be careful to realize that the center is at (1,-3), not at (1,3).1838

Let's make this 2, 4, 6, 8, -2, -4, -6, -8; the center, then, is going to be at (1,-3).1846

The next piece of information: A2 = 24; therefore, A = √24, which is approximately 4.9.1856

B2 = 8; therefore, B = √8; therefore, if you figure that out on your calculator, that is approximately 2.8.1871

Since I have the center, and I have A and B, I can draw the rectangle that will allow me to extend diagonals out to form the asymptotes.1883

The goal is to write this in standard form, find A and B, find the center, make the rectangle, and make the asymptotes;1892

and then, you can finally draw both branches of the hyperbola.1902

All right, so if the center is here at (1,3), then I am going to have 2 vertices.1906

And what is going to happen, since this is a vertical transverse axis, is: one vertex is going to be up here; the other is going to be down here.1912

The center is at (1,3); that means I am going to have a vertex at 1, and then it is going to start at the center,1926

and then it is going to be 4.9 directly above that center; so that means this is going to be at -3 + 4.9.1933

The y-coordinate will be at -3 + 4.9, which equals (1,1.9).1944

Therefore, at (1,1.9) (that is right there)--that is where there is going to be one vertex.1950

And this is A--this is the length of A.1958

The second vertex is going to be at (1,-3); that is the center; and then I am going to go down 4.9--that is the length, again, of A.1963

That is -3 - 4.9 (or + -4.9; you can look at it that way) = (1,-7.9), down here.1979

OK, vertices are at (1,-7.9) and (1,1.9).1999

Now, I need to find where B is--where that endpoint over here is, horizontally--so that I can make this rectangle.2014

I know that B equals approximately 2.8; that means that I am going to have a point over here at 1 + 2.8...-3.2024

Well, 1 + 2.8 is (3.8,-3); so (3.8,-3) is right there.2040

I can reflect across; and I am going to have a point at 1 - 2.8, -3, which is going to give me...1 - 2.8 is (-1.8,-3).2054

That is going to be...this is 2...-2 is right here; so that is going to be right about there.2068

I now have these points; and recall that I can then extend out to make a box.2076

There is going to be a vertex here; I am going to extend across; there is going to be a vertex here.2083

Bring this directly down; there is a vertex here, and then another vertex right here.2090

Again, I got these points by knowing where the center is, knowing where the vertices of the hyperbola are, and then knowing the length of B.2096

This is B; then this length is A.2109

Once I have this rectangle, I can go ahead and draw the asymptotes by extending diagonals out.2113

Another way to approach this, recall, would have been to use the formula for the slope that we discussed, for the slope of the asymptotes.2130

Either method works.2140

I know that I am going to have a hyperbola branch up here; the vertex is right here, and it is going to approach, but never reach, this asymptote.2143

It is going to do the same thing with the other branch: a vertex is here; it is going to approach, but never reach, the asymptote.2156

OK, so this was a difficult problem; we were given an equation in this form.2170

We had to do a lot of work just to get it in standard form.2175

And then, once we did, we were able to find the center and form this rectangle, draw the asymptotes, and then (at last) graph the hyperbola.2178

Example 4: We don't need to do graphing on this one.2189

We are just finding the equation of a hyperbola with the center here, (0,0), and a horizontal transverse axis.2192

I am going to stop right there and think, "OK, I have a center at (0,0) and a horizontal transverse axis."2200

So, the standard form is going to be x2/A2 - y2/B2 = 1.2206

Since the center is at (0,0), I don't have to worry about h and k.2215

The horizontal transverse axis has a length of 12; well, the transverse axis length, recall, is equal to 2A.2219

I am given that that length is 12; if I take 12/2, that is going to give me A = 6.2231

The conjugate axis--recall that the length of the conjugate axis is equal to 2B, which is 6: B = 3.2240

Now, I need to find A2, which is 62, or 36, to put in here.2257

B2 is 32, which is 9.2264

Now, x2/36 - y2/B2 (which is 9) = 1.2267

So, this is the equation for a hyperbola with the center at the origin, a horizontal transverse axis, and a conjugate axis with a length of 6.2279

That concludes this lesson on hyperbolas; thanks for visiting Educator.com!2290

Welcome to Educator.com.0000

In the past four lectures, we have discussed various conic sections: parabolas, circles, ellipses, and hyperbolas.0003

And this lecture is designed to bring that information together and to give you some context about this.0010

First of all, what are conic sections? We know we can name them; we know what they are; but where do they come from?0016

Well, they are literally sections of a cone: when you take a double cone (it is actually a double cone, as follows, with the points together),0023

and you section them (sectioning is slicing)--when you take slices of them, using a plane, you come up with these four types of curves.0034

So, as you can see, when you take a plane and section, or slice, the cone across, you are going to end up with a circle.0043

If you tip that plane at an angle, the result is an ellipse.0054

If you encompass the edge of the come, you end up with a parabola.0065

And if you slice through in such a way that you capture the edges of both cones, then you end up with a hyperbola; and there you can see the two branches.0071

So, this is where conic sections come from; and they have many applications in science.0079

We have talked about the standard form of each conic section (for example, the standard form of a circle, or the standard form of an ellipse).0087

This standard form that I am talking about now is a very general form.0095

It gives you a general equation, ax2 + bxy + cy2 + dx + ey + f = 0.0098

So, what we are going to do in a minute is talk about how you can look at this general form and determine0109

which type of conic section you are working with, so that you can put the equation0115

in the standard form particular to that type of conic section.0119

And as we have been going through, I have mentioned some ways that you can tell, if you just have an equation in the general form,0124

what type of conic section you are working with; and now I am going to bring that all together.0131

OK, if b = 0, we can analyze that standard form of the conic section to determine what type of conic section the equation represents.0137

Looking back at that general standard form again, ax2 + bxy + cy2 + dx + ey + f = 0.0146

Here, we are having the limitation that b = 0.0168

And throughout this course, when we work with conic sections, we have only worked with ones where b is 0.0171

When b is 0, you end up with this.0178

Once you have this standard form, then you can go ahead and analyze it in ways we are going to discuss in a minute0192

to determine which type of conic section you have (what the equation describes).0197

But let's talk for a minute about what the xy tells you.0201

So far, we have worked with shapes such as parabolas; and some were oriented vertically; some, horizontally.0204

We also worked with ellipses (some had a horizontal major axis; some had a vertical major axis); and the same with hyperbolas.0215

So, even though the center may have been shifted, these were all either strictly vertical or strictly horizontal.0230

What this bxy term does is rotates it so that instead of, say, having an ellipse that has0237

a completely vertical major axis or horizontal major axis, you could end up with an ellipse like this--0243

the major axis is at an angle--or a parabola that is like this.0250

And that is definitely more complicated to work with; and it doesn't allow us to complete the square, then,0259

to shift an equation from the general form to a specific standard form.0265

So, later on, if you continue on in math, you may end up working with these shapes.0270

But for this course, we are limiting it to conic sections that are either vertical or horizontal; but they are not tipped at any other type of angle.0274

In order to identify conic sections, you need to look at the coefficients of the x2 and y2 terms.0286

So, let's rewrite this; and again, the assumption is that b = 0.0293

So, I am just going to have ax2 + cy2 + dx + ey + f = 0.0299

Parabola: Recall that, with a parabola, you have an x2 term or a y2 term, but not both.0310

Therefore, either a is 0 (so this drops out) or c is 0 (so this drops out).0325

An example would be something like x = 3y2 + 2y + 6.0331

Or you might have y = 2x2 - 4x + 8.0338

So, neither of these has both an x2 and a y2 term in the same equation.0344

For a circle, recall that what you are going to end up with is an x2 and a y2 term0352

on the same side of the equation, with the same sign; and they are going to have the same coefficients.0361

Therefore, a is going to equal c.0366

An example would be x2 + y2 + 3x - 5y - 10 = 0.0369

Here, a equals 1, and c equals 1; those are the same coefficients; x2 and y20378

have the same sign and the same side of the equation; so it is a circle.0386

If we are working with an ellipse, this time the x2 term and y2 terms are going to be0392

on the same side of the equation, with the same sign (like with the circle), but a and c are different.0399

They are unequal; that tells me that I am working with an ellipse.0405

For example, 12x2 + 9y2 + 25x + 28y + 40 = 0.0409

Here, I have a = 12 and c = 9; so this is the equation describing an ellipse.0423

Finally, with a hyperbola, these are pretty straightforward to recognize, because you are going to have0431

an x2 term and a y2 term, but they are going to have opposite signs.0436

Their coefficients will have opposite signs.0440

For example, 4x2 - 8y2 + 10x + 6y - 34 = 0.0443

So, I have a = 4 and c = -8; since a and c have opposite signs, this is an equation describing a hyperbola.0458

You can use these rules to allow you to identify conic sections when you are given an equation in what we are going to call "general form."0467

It is standard form, but it is a very general standard form for any type of conic section.0476

OK, now we are going to work on identifying the various conic sections by looking at their equations.0481

First, write in standard form, and identify the conic section.0488

OK, so general standard form is what I am talking about right now: it is x2 + 2y2.0495

I need to subtract 4x from both sides, subtract 12, and set everything equal to 0.0503

What this tells me is that I have a = 1 and c = 2.0510

Since a = 1 and c = 2, these have the same sign (the x2 and the y2 terms); but they have different coefficients.0515

And that means that what I am working with is an ellipse.0525

You could go on, then, and write this in the specific standard form for an ellipse.0531

Let's do that by completing the square: start out by grouping...let's rewrite it here.0537

And then, let's group the x and the y terms; so x2 terms group together; y terms group together.0545

And now, add 12 to both sides to move that over, to make completing the square a little bit easier.0555

To complete the square for x2 - 4x, I need to add b2/4.0563

b2/4 is equal to 42/4, is 16/4; it is 4.0571

So, I add x2 - 4x + 4; and it is very important to remember to add the 4 to the right side, as well.0582

There is no factor out here; I don't need to multiply--it is just 1; so 4 times 1 is 4; that gives me 12 + 4.0592

All right, that is x2 - 4x + 4 + 2y2 = 16.0601

This can be rewritten as (x - 2)2 + 2y2 = 16.0609

But recall, in standard form for an ellipse, you need to have a 1 on the right.0616

So, rewrite this up here, and then divide both sides by 16.0622

This is just (x - 2)2/16; this will cancel; this will become 8; and then 16/16 is 1.0633

So, we started out with this equation, put it into the general standard form to identify that this is an ellipse,0645

and then went on to complete the square; and now I have it written in standard form specifically for an ellipse,0652

which is much more useful when you are working with that and trying to graph.0657

This time, without completing the square, all we are going to do is identify the conic section.0664

And this is already in standard form; therefore, a = 2; c = -3.0668

Since a and c have opposite signs, this is the equation for a hyperbola.0678

I have an x2 term and a y2 term, both, so it is not a parabola.0687

They have opposite signs; therefore, it must describe a hyperbola.0691

OK, write in standard form and identify the conic section.0698

Right now, this is not in any type of standard form; so I am going to work with the general standard form.0702

First, I am going to subtract 36x2 from both sides.0708

Then, I am going to subtract 128 from both sides.0717

This means that I have a = -36, and c = 16; since these two are opposite signs, this is an equation describing a hyperbola.0727

OK, now, let's go ahead and put this in standard form specific to a hyperbola.0741

And let's start out by moving this 128 back over to the right; this is actually 32.0747

Next, I do have a common factor of 4, so I am going to divide both sides by that, so that I am working with smaller numbers.0763

That is -9x2 + 4y2 + 8y =...128/4 would give 32.0770

All right, now to make this already move it more towards looking like a hyperbola, I am going to put the positive terms here in front:0786

4y2 + 8y - 9x2, because I am going to have a difference.0793

To complete the square, I first need to factor out that 4; then I need to add b2/4 to this expression.0801

This is going to equal 22/4; that is 4/4, which is 1.0816

Here is where I need to be careful, because I need to make sure I add 4 times 1 to the right, which is 4, to keep the equation balanced.0825

At this point, I am going to rewrite this as (y + 1)2 - 9x2 = 36.0837

The last step is: I want the right side to be 1, so I am going to divide both sides by 36.0844

4 goes into 36 nine times; 9 goes into 36 four times; and then this cancels out to 1.0860

OK, so I started out with an equation that wasn't in any kind of standard form.0872

I put it in general standard form, and then determined it was a hyperbola, completed the square, and ended up0876

with an equation in the standard form for a hyperbola, so that I can use that to graph the hyperbola, if needed.0882

Write in standard form and identify the conic section.0891

So, this is almost in the general standard form, but not quite.0894

I have 4x2; I need to move this -3y2 next, then -16x - 18y - 12 = 0.0897

Now, I can easily see that a = 4 and c = -3; since these have opposite signs, that means that this is an equation describing a hyperbola.0907

OK, the next task is to complete the square.0921

I am going to first add 12 to both sides to remove the constant from the left side.0926

Then, I am going to group the x terms, which is 4x2 - 16x.0937

And then, I have a -3y2 - 18y, and that all equals 12.0946

I have a leading coefficient that is something other than 1, so I am going to factor out the 4, leaving behind x2 - 4x.0955

Here, I need to factor out a -3; that is going to leave behind y2 + 6y.0965

You need to be careful with the signs here; just double-checking: -3 times y2 is -3y2.0969

-3 times positive 6y is -18y, when you factor out with that negative sign; equals 12.0976

Completing the square: b2/4, in this case, is 42/4, is 16/4; that is 4.0987

So, I am going to add 4 here; I am also going to add 4 times 4, or 16, to the right, to keep the equation balanced.0998

For the y expression, I have y2 + 6y; therefore, b2/4 = 62/4, which is 36/4, which is 9.1012

-3 times 9 is -27; so I am going to subtract 27 from the right side, again keeping the equation balanced.1029

I am rewriting this as (x - 2)2 - 3(y + 3)2 = 16 + 12, is 28, minus 27; conveniently, I end up with a 1 on the right.1041

Now, this is almost in standard form; generally, with standard form for a hyperbola, this term will be in the denominator.1059

So, it is possible to rewrite it like this; and it might be easier to look at it that way,1071

so that you can immediately know that this is a2, instead of having to think it out.1077

Putting it in truly standard form is also a good idea, because recall that, if I have the numerator divided by 1/4, that is the same as this times 4.1082

And that tells me that I have a hyperbola with a center at (2,-3); you have to watch out for this positive sign.1096

And it has a horizontal transverse axis.1103

So, today we learned exactly what conic sections are, where they come from,1109

and how to look at an equation and determine what type of conic section it describes.1114

Thanks for visiting Educator.com; see you again soon!1120

Welcome to Educator.com.0000

Today, we are going to talk about solving quadratic systems of equations.0002

In earlier lectures, we discussed talking about linear systems of equations, and used various methods to solve those: for example, substitution and elimination.0006

And we are going to use some similar methods for quadratic systems, although these systems are more complex.0017

There are two types of quadratic systems that we are going to be working with today.0023

The first is a linear-quadratic system: linear-quadratic systems are a set of two equations in x and y,0027

in which one of the equations is linear, and the other is quadratic.0035

We will talk in a second about how to solve these; let's just stop at the definition right now.0041

x2 + 3x + y = 9; x = 2y: these two equations, considered together, would be a linear-quadratic system.0046

I have a linear equation here and a quadratic equation here.0057

And right here, this is a fairly simple linear equation to start out with.0063

And we can use that to apply the method discussed above.0068

So, given this type of system, these can be solved algebraically by isolating0072

one of the variables in the linear equation, and then substituting it into the quadratic equation.0077

OK, so conveniently, x is already isolated.0083

Often, you will be given a linear equation as part of a system where you have to do some manipulation to isolate either x or y.0088

But I already have x = 2y, so that is perfectly set up for me to substitute 2y in for the x variable in this quadratic equation.0096

So, I am rewriting the quadratic equation, and now substituting 2y in for x.0105

That gives me 2y2 + 3(2y) + y = 9.0113

What I am left is a quadratic equation, which I can solve by the usual methods that we have learned.0122

This is 22 is 4y2 + 6y + y = 9.0128

That gives me 4y2 + 7y = 9, or 4y2 + 7y - 9 = 0.0136

And then, you could go on to solve this, using the quadratic formula.0148

Once you find y, you can go ahead and substitute that value(s) in here, and then determine the corresponding value of x.0153

In a linear quadratic system, you may have either 0, 1, or 2 solutions.0170

And the easiest way to understand this is to think about it in terms of a graph.0176

A linear equation is going to give you a line: a quadratic equation might give you a parabola or an ellipse, a hyperbola, a circle...0180

So, let's look at some possibilities: maybe I have a system that ends up graphing out like this.0188

It gives me that line and this parabola.0196

Well, the solutions are going to be where these two intersect; but in this system, they will never intersect.0201

So, this system is going to have 0 solutions.0206

I may have another situation (again, I will use a parabola as my example) where I have a line that goes through right here at one point.0212

And this is going to give me one solution.0224

Perhaps I have an ellipse, and I have a line going through it like this.0228

It intersects at two spots, so I would have two solutions.0237

This helps to illustrate why you may have no solutions, one, or two solutions.0241

More complicated systems involve quadratic-quadratic equations.0251

This would be a quadratic-quadratic system in which there is a set of two quadratic equations in x and y.0255

For example, x2 + y2 = 5 and 2x2 - y = 4:0262

you can see that you have two quadratic equations.0272

You could try to use substitution, but it could get a little bit messy.0276

So, often, elimination is the easiest way to use, so we solve them algebraically by elimination.0281

And you will recall, working with linear equations, that in elimination, what you try to do is get variables to have the same or opposite coefficients.0286

And then, either add or subtract the equations so that a variable drops out.0294

And here I want to get either x2 or y2 to drop out.0298

Well, I have an x2 term in each; so that, I could end up having drop out if I multiplied this top equation by 2.0304

So first, I am going to multiply the first equation by 2.0314

2 times x2 + y2 = 5: and this is going to give me 2x2 + 2y2 = 10.0317

So, I am going to then go ahead and write the other equation just below that: 2x2 - y = 4.0333

This is equation 1; this is equation 2; I multiplied equation 1 by 2, and then I ended up with this; and I have equation 2 right here.0347

Now, what I am going to do is subtract this second equation from the first.0359

Rewriting that: this gives me 2x2 - 2x2; these terms drop out.0366

The x2 terms drop out; a negative and a negative is a positive, so this is actually going to give me, let's see, 3...0375

actually, there are slightly different coefficients, so I can't add those; this is 2y2 - a negative0394

(that becomes a positive) y, and then 10 - 4 is equal to 6.0402

OK, so 2x2 - 2x2--those drop out; I have a 2y2 down here.0407

I have a negative and a negative, gives me a positive y; and then this becomes a -4; so 10 - 4 is 6.0414

So, I end up with 2y2 + y = 6 by using elimination.0422

Therefore, I have gotten rid of one of the variables.0427

From there, I have a regular quadratic equation that I can solve: that gives me 2y2 + y - 6 = 0.0431

I am going to try to solve that by factoring; and this is going to give me (2y + something) (y - something).0440

So, let's look at 6: if we take factors of 6, those are 1 and 6, and 2 and 3.0448

And I want them to be close together, so that I just end up with a coefficient of 1 here; so let's try 3 - 2 = 1.0455

Therefore, I am going to take + 3, - 2.0466

But I have to take this 2y into account; so let's go ahead and see if this works.0478

This becomes 2y2, and then this is minus 4y, plus 3y; so that gives me a -y; so that won't work--let's try it the other way.0482

Let's actually make this a negative and see what happens.0496

This gives me 2y2, and then this is 4y, minus 3y; that gives me a y; and then, -3 times 2 is -6.0500

So, this actually factored out, and it worked.0510

Using the zero product property, let's go back up here and get (2y - 3) (y + 2) = 0.0514

2y - 3 = 0, or y + 2 = 0; if either of these equals 0, the whole quantity equals 0.0522

Solving for y gives me 2y = 3 or y = 3/2; solving for y gives me y = -2.0530

Now, what I need to do is go back and find the corresponding x-value, so I have coordinate pairs as my answers.0541

All right, let's first work with y = -2.0552

And going back to this equation (because this one is simple to work with), I am going to substitute in that value for y2 and see what I get.0558

x2 + (-2)2 = 5: that is x2 + 4 = 5, or x2 = 1.0568

Therefore, x = 1 and -1; now, that is when y is -2.0580

So, my solutions are (1,-2) and (-1,-2)--I have two solutions there.0588

Now, I need to repeat that process for y = 3/2; let y equal 3/2, and then substitute in here.0603

This is going to give me x2 + (3/2)2 = 5, so x2 + 9/4 = 5.0610

So, x2 = 5 - 9/4; that gives me x2 =...a common denominator of 4; it would be 20/4 - 94, so x2 = 11/4.0620

Therefore, x = ±√11/4; now, looking over here, when y is 3/2, x could be √11/4, or it could be -√11/4.0637

So, I have another two members to my solutions: I have that x could be (√11/4,3/2), and (-√11/4,3/2)0656

Here, we have a situation where we actually have four solutions.0673

And I solved this by elimination; and then I had to go back and take each of the solutions I ended up with,0679

-2 and 3/2 for y, and find corresponding solutions for x.0686

And each of those yielded two values for x, so I ended up with 4 solutions altogether.0691

And I can also illustrate that point using graphing--of why you can end up with 0, 1, 2, 3, or 4 solutions.0700

Quadratic-quadratic systems: think about all of the different curves that you could come up with, and the possibilities.0708

You could maybe have a circle and a parabola as a system that never intersect: this is 0 solutions.0714

Perhaps you have an ellipse, and then another ellipse, like this: that gives you 1, 2, 3, 4 solutions.0722

You may have an ellipse, and then a circle, right here; and this gives you two solutions.0735

You could have, say, a parabola and an ellipse here that intersect at just one point.0747

So, you can see how that, with various combinations, you could get 0, 1, 2, 3, or 4 solutions.0754

And we just saw that demonstrated algebraically in the previous example--that we ended up with 4 solutions.0760

Systems of quadratic inequalities: these are systems of inequalities that have two inequalities, and at least one of these is a quadratic inequality.0769

So, for example, x2/9 + y2/4 <1; and then that is considered along with (x - 1)2 + (y - 3)2 < 9.0781

If you look at this, you have two quadratic equations; and you would recognize this0807

as an ellipse in standard form; and this gives the equation for a circle in standard form.0811

We have worked with some systems of inequalities before, and talked about how these can be solved by graphing.0818

And we are going to do the same thing here.0823

We are going to first graph the corresponding equation; and that will give us the boundary for the solution set.0825

Then, we will use a test point to determine on which side of the boundary the solution set lies.0832

We will do the same thing for the second equation that corresponds with the inequality:0840

find the boundary; use a test point; find where the solution set is.0845

And then, the overlap between those two solution sets gives you the solution for the system.0849

Illustrating this with this example: I am going to start out be graphing the boundary, and then finding the test point, for this first inequality.0855

So, the corresponding equation is going to be x2/9 + y2/4 = 1.0866

Since this is an ellipse, looking at it in this form, I know that it has a center at (0,0).0883

I know that A2 = 9, so a = 3; and since the larger term is associated0892

with the x2 term, I also know that this has a horizontal major axis.0901

The major axis is going to be oriented this way.0907

B2, right here, is equal to 4; therefore, B = 2.0911

So, that allows me to at least sketch this out: since A = 3, then I am going to have a vertex here, and the other vertex here, at (-3,0).0917

B is 2, so I am going to go up 2 and down 2; and this allows me to just sketch out the ellipse.0928

All right, the next step is to use a test point, because now I have a boundary.0941

And I actually need to be careful; I need to determine if the boundary should be dotted, or if it should be solid.0948

And looking here, I actually have a strict inequality.0956

What that tells me, recall, is that the boundary is not part of the solution set.0960

And the way we make that known is by using a dotted or dashed line.0966

Clarifying that, so we have the correct type of boundary...these breaks in the boundary indicate that this boundary is not part of the solution set.0976

OK, so I graph the boundary; I checked, and I had a strict inequality.0993

Now, I am going to take a test point; this is a convenient test point,0999

because the boundary has divided this into two regions: the region outside the ellipse, and the region inside it.1004

And I need to determine where the solution set is.1010

So, let's take the test point (0,0): I am going to put these values back into that original inequality.1013

0 is less than 1; this is true; therefore, this test point is part of the solution set.1033

So, the solution set must be inside this boundary.1040

So, I graphed my first boundary, and I determined where the solution set for this inequality is.1050

Looking at the second inequality: the corresponding equation will give me the boundary line for that.1055

Looking at this, this is written in standard form for a circle.1064

So, this is a circle with the center at (1,3), and r2 = 9, so the radius equals 3.1067

And this is a circle; and the center is at (1,3), so the center is right here.1078

And this is also a strict inequality, so I am going to use the dotted line when I draw this boundary line.1086

And the radius is 3: since the radius is 3, then that would be 4, 5, 6: that would go up to here.1099

And then, this is at 1, so then I would have the end right here, here, here.1106

OK, this is enough to get a rough sketch of the circle.1114

So, first I will just draw it as solid, and then go back and make it dashed, since this is a strict inequality.1133

All right, so this is a dashed line; it is a circle; it has a center at (1,3), and a radius of 3.1149

So again, this boundary line is not part of the solution set.1161

I am going to use another test point and insert it into that inequality to figure out where my solution set is.1165

So, for this, let's go ahead and use (1,1) as a test point, right there...the test point is going to equal (1,1).1173

So, that is going to give me 12/9 +...oops, I actually need to go back into that inequality...there is my equation1185

for the circle, (x - 1)2 + (y - 3)2 < 9; my test point is (1,1).1207

So, that is going to give me (1 - 1)2 + (1 - 3)2 < 9.1214

1 - 1 is 0, plus 1 - 3...that is -2, squared is less than 9; so 0...and this is -2 times -2 is 4...so 4 is less than 9.1222

Yes, this is true; therefore, this point is part of the solution set.1235

So, the solution set lies inside the circle.1240

The solution set for the system of inequalities is going to be this area here that is the overlap1251

between the solution set for the circle and the solution set for the inequality involving the ellipse.1260

And you can see, right here, the area where there is both red and black; it is that area of overlap.1266

So again, this is similar to methods we have used before, involving solving systems of inequalities,1270

where we graph the boundary line for one inequality; we graph the boundary line for the other inequality;1277

we use test points to find the solution sets for each inequality; and then, we determine the area of overlap1282

between those two solution sets, and that is the solution set for the system of inequalities.1289

And here, we are doing that with quadratic inequalities involving a circle and an ellipse.1295

Example 1: this is a linear-quadratic system: I have a linear equation here, and a quadratic equation here.1303

Recall that the easiest way to solve for these is by substitution.1310

Therefore, I am going to isolate x; I am going to rewrite this as x = y + 2.1314

Then, I go back to this first equation (this is from equation 2); I go back, and I make sure I substitute this into the other equation.1320

So, wherever there is an x, I am going to then put y + 2 instead; and I need to square that in this case.1331

This equals 36; I am going to write this out as y2 + 4x + 4 + y2 = 36.1339

OK, y2 + y2 gives me 2y2 + 4y (we are working with y here) + 4 = 36.1360

Now, it is a regular quadratic equation that I can just solve as I usually would.1372

And I see here that I have a common factor of 2; let's first go ahead and subtract 4 from both sides.1377

And this is going to give me 32; I am going to divide both sides by 2 to make this simpler, which is going to give me y2 + 2y = 16.1391

And now, I am going to solve it as I would any other quadratic equation: y2 + 2y - 16 = 0.1403

You could try this out, but it actually doesn't really factor out.1411

This is a situation where we have to go back to the quadratic formula, y = -b ±√(b2 - 4ac), divided by 2a.1415

It is a little more time-consuming, but it will get us the answer when factoring doesn't work.1431

So, let's rewrite this up here: y2 + 2y - 16 = 0.1436

y =...well, b is 2, so that is -2 ±√((-2)2 - 4 times a (is 1), and then c is negative 16), all divided by 2 times a, which is 1.1443

Simplifying: y = -2 ±√...-22 is 4; -4 times 1 is -4; -4 times -16 is + 641461

(those negatives, times each other, become positive) all over 2.1478

Therefore, y = -2±√...that is 64; -4 times 1 times -16 is 64, plus 4 is 68, divided by 2.1485

Now, you actually could determine that 68 is equal to 4 times 17; therefore, √68 equals the perfect square of 4, times 17.1507

So, I can then pull this 2 out by taking the square root of 4, which is 2; and it becomes 2√17.1523

So, I am going to do that over here, as well.1531

All right, you can factor out a 2, and those will cancel; I am just going to do that over here.1540

y = 2(-1) ± 1√17/2; that will cancel.1547

What you are left with -1 ± 1√17, divided by 2.1555

OK, so this is not an easy problem, because you ended up having to use the quadratic formula.1567

So, let's look at what we actually have: we have y = -1 + 1√17, and we also have y = -1 - √17.1576

So, we don't really need this 1 here.1591

The next step is to go back and substitute in: fortunately, we have an easy equation here1595

that we can substitute: x - y = 2, which is the same as x = y + 2.1601

I am going to work up here; and I am going to say, "OK, when y is -1 + √17, then x = this (that is y) + 2."1607

Therefore, x = 2 - 1 (is 1)...1 + √17.1624

That gives us an ordered pair: (1 + √17, -1 + √17).1632

That is an ordered pair; then I am going to take this second possibility, where y equals -1 - √17.1643

And I am going to substitute that into this, as well: x = -1 - √17 + 2.1656

This is going to give me x = -1 + 2...that is going to give me positive 1...so 1 - √17.1669

So, my second ordered pair is going to be (let's write it over here) x = 1 - √17, and then the y-value is -1 - √ 17.1677

So, it gets a little confusing with all of those signs; you want to make sure that you are careful1692

to check your work, and that you don't have any of the signs mixed up.1697

But what it came down to is using the quadratic equation to determine that y = -1 + √17, and y = -1 - √17.1701

Then, you take each possibility, starting with the first one; substitute for y in this equation,1711

which is the same as this (just rearranged); and substitute this to determine the corresponding value of x1719

(because remember, the solution is going to be an ordered pair, an x and a corresponding y value).1724

So, I did that for this first one; then I went and took the second one, repeated that process,1730

and got that, when y is -1 - √17, x is 1 - √17.1734

These two are an ordered pair, as well--that is where this came from; and these are the two solutions for this linear quadratic system of equations.1743

Example 2: we are going to solve...this is a quadratic-quadratic system, because I have a second-degree equation here and here.1755

I have two second-degree equations.1761

Recall that the best way to solve these is by elimination.1763

Looking at this, the first thing I can do to make this a little easier is actually to divide this first equation (equation 1 and equation 2) by 2,1768

because right now, it is kind of messy; it is bigger numbers than I need to be working with.1781

So, 2 goes into 8 four times; 2 goes into 2 once; and 2 goes into 40 twenty times.1786

All right, now I am going to go ahead and take equation 2, 4x2 + y2 = 100, and rewrite it down here.1797

And I see that, if I subtract this second equation, the y2 terms will drop out, because they already have the same coefficient.1805

I am going to rewrite this as 4x2 + y2 = 100; plus...I am going to make this a plus,1819

and then apply that as -4x; minus y2; and then I have a -20 here; I am just adding the opposite of each term.1828

So here, I end up with 4x2 - 4x (these are different, so I can't just combine them);1843

y2 - y2...the y2's drop out; 100 - 20 is 80.1857

All right, now I am going to divide both sides by a factor of 4 to simplify this.1863

This gives me x2 - 4 = 20; now it is just a quadratic equation that I need to solve.1867

x2 - 4x - 20 = 0: let's rewrite that up here.1875

x2 - 4x - 20 = 0: and let's hope that we can solve this by factoring,1882

instead of having to go back to the messy quadratic formula.1888

x...the factors of 20 are 1 and 20, 2 and 10...actually, this is just x, because we got rid of that when we divided by 4...1893

this is just x; we divided by 4 to give us x2, x, and 20, which will make this even easier to factor.1905

OK, so since this is -1 for a coefficient, I want some factors that are close together, like 4 and 5.1912

And if I take 4 - 5, I am going to get -1; so I am going to use this combination.1920

This is going to give me (x - 5) (x + 4) = 0; just checking that, x times x is x2, and then outer terms is 4x,1926

minus the inner terms--that is 4x + -5x gives me -x; -5 times 4 is -20.1939

The zero product property tells me that, if x - 5 equals 0, or x + 4 = 0, this whole thing will equal 0.1948

So, solving for x will give me these two solutions: x = 5 and x = -4.1956

OK, I have the x-values: the next thing is to find the y-values.1962

So, I need to go back and substitute into one of these equations.1968

I am just going to select the top one; and what I need to do is determine what y will be when x is 5, and what y will be when x is -4.1973

So, starting out, I am letting x = -4, and then using this 8x + 2y2 = 40.1986

So, this is 8 times -4, plus 2y2, equals 40; that is -32 + 2y2 = 40.1997

Adding 32 to both sides gives me 2y2 = 72; dividing both sides by 2 gives me y2 = 36 (72/2 is 36).2008

Therefore, if I take the square root of 36, I get that y = ±6; so y = 6, and y = -6, when x = -4.2023

OK, let's write some ordered pairs up here as solutions.2035

When x = =4, you could equal 6; when x equals -4, another solution could be that y = -6.2039

I am going to repeat this process with x = 5, substituting into this equation.2047

This is going to give me 8(5) + 2y2 = 40.2053

That is 40 + 2y2 = 40; 40 - 40 is 0; 2y2 = 0; divide both sides by 2; I get y2 = 0.2058

The square root of 0 is 0, so I only get one solution for y here.2072

So, when x is 5, y is 0; I ended up with 3 solutions for this system of quadratic equations,2076

because it turned out that I had two values for x; one of these values of x, substituting in, yielded 2 values for y.2085

The second value for x yielded only one result for y; so I have three solutions here.2094

Another system: this is another quadratic-quadratic system, so I am going to use the approach of elimination.2104

Before I do anything, though, I can simplify these, because there are common factors.2110

I am going to take the first equation; and first, I will just rewrite it so that it is in more of a standard form.2115

So, I am going to subtract 4x2 from both sides; so it is -4x2 + 4y2 = -28.2124

OK, this is still equation 1; I am going to divide both sides by the common factor of 4.2134

That is going to give me -x2 + y2 = -7.2141

For the second equation, I have a common factor of 5; so I will divide both sides by 52147

to get x2 + y2... 125 divided by 5 is 25.2152

Now, all I have to do is add these two together, because I have a -1 for a coefficient here, and a 1 here; these cancel out.2159

y2 + y2 is 2y2; 25 - 7 is 18.2166

Just solve for y2: y2 = 18/2, so I divided both sides by 2; y2 is 9.2175

Therefore, y = ±3, by taking the square root of 9.2183

That means that y = 3, and y could also equal -3.2188

Now, I need to go back and substitute into one of these equations when y = 3, and figure out what x is.2196

Then, I need to see, when y is -3, what x is going to be.2204

Let's see, the easiest one to work with would be this: and I could go back in and use the top one,2211

but since I divided both sides by the same thing, I didn't really change this equation.2221

And it is a lot easier to work with this without these larger coefficients.2228

So, what I am going to do is say, "Let's let y equal 3."2232

And then, I am going to look at this: x2 + y2 =...actually, this first one; the first one is a smaller number, over here.2237

I am going to say -x2 + y2 = -7.2247

And let's rearrange this a bit, because we are looking for x.2251

So, let's move this y2 to the other side; and now I am stuck with a bunch of negatives.2254

And what I can do is just multiply both sides of the equation by -1, and that gives me x2 = y2 + 7.2261

Now, this is the one I am going to substitute back into: again, you could have taken either of these forms;2270

but I just took this and made it easier to work with, and solved for x2.2275

I am going to substitute 3 in wherever this is a y: so x2 = 32 + 7; therefore, x2 = 9 + 7.2281

So, x2 = 16, which means that x = ±4; so x could equal 4, and x could equal -4, when y is 3.2291

So, let's start our solutions up here.2307

When x is 4, y is 3; when x is -4, y is 3; that is two solutions so far.2311

That is when y is 3; but recall, y can also equal -3.2322

So, when y is -3, I am going to go back into that equation and substitute -3 and see what I get for x.2326

(-3)2 is 9; 9 + 7 is 16; so again, I get x = ±4; so x can equal 4, and x can equal -4.2340

But this time, y is -3; so this is two different solutions from what I had up here.2352

So, when x is 4, y is -3; that point is a solution for this system of equations.2358

When x is -4, y is -3; so there are four solutions.2366

If you graphed this out, you would find that these intersected at four points.2372

This is pretty complicated: the initial part actually wasn't that bad, but keeping track of all of the different solutions was a little bit challenging.2377

We started out with these two equations that I simplified by dividing the first by its common factor of 4, and the second by its common factor of 5.2385

Then, you added them together; the x2 terms dropped out, which allowed you to just solve for y.2393

I got two solutions for y: 3 and -3; I took each of those solutions, y = 3 and y = -3, and plugged them into this equation,2399

right here, to find corresponding values for x; that yielded these two solutions; and then y = -3,2411

when I found the x-values that corresponded to that...I got two more solutions, for a total of four solutions.2422

This time, we have a system of quadratic inequalities: I have x2 ≤ y, and then 4x2 + 4y2 < 36.2429

So, remember, we are going to solve these by graphing the corresponding equation to find the boundary line.2440

And then, we are going to use test points to find the solution sets for each, and then the solution set for the system.2446

Starting out with x2 ≤ y: this is x2 = y--that is the corresponding equation.2452

Finding a few points: x and y--recall that y = x2, so when x is 0, 02 gives you 0, so y is 0.2465

When x is 1, then we get 12; y is 1; when x is 2, we get...2 times 2 is 4; y is 4.2482

When x is -1, -1 times -1 is 1; when x is -2, -2 times -2 is 4; all right.2493

You recognize this as a parabola that opens upward; and it has its vertex, which is a minimum, right here at the center: (0,0).2503

I have a point here at (1,1), and a point at (-1,1); I have another point at (2,4), and then a point at (-2,4).2513

I could have also just graphed half, and used reflection symmetry to graph the other part of the parabola.2524

OK, now, this is less than or equal to; so I am actually using a solid line for my boundary,2531

because the boundary line is included as part of the solution set.2540

Now, I have the graph of the boundary; and I need to use a test point to determine where my solution set lies.2544

Does it lie inside the parabola, or outside?2552

And I am going to use the test point right here, (0,2); that would be convenient to work with.2555

x2 ≤ y; and I have a test point at (0,2); so when x is 0, that would give me 02 ≤ 2.2560

Is 0 less than or equal to 2? Yes, this is true; therefore, the test point is part of the solution set.2574

So, this solution set for this first inequality lies inside the parabola.2581

That is shaded in; and it is going to include the boundary.2594

This second equation that I have is 4x2 + 4y2 < 36.2600

So, let's rewrite this down here, and find the corresponding equation, 4x2 + 4y2 = 36.2608

And as you can see, there is a common factor of 4; so I am dividing both sides by that.2621

Looking at this, you can see that this describes a circle; and the circle has a center at (0,0).2630

r2 = 4; therefore...actually, that should be 9, because I divided both sides by 4; so 36/4 is 9: r2 = 9.2643

Therefore, r = 3; so this equation--all I did is divide both sides by 4, and I can see now that I have2658

an x2 and y2 term here on the same side of the equation, and they are equal to 9.2666

Therefore, it is a circle at center (0,0); they have the same coefficient; and r2 = 9, so the radius of the circle is 3.2674

So, the center of the circle is here; the radius is 3.2683

There would be a point there...the edge of the circle there...there...and there.2686

Then, you can fill in; now, this is a strict inequality, so I am just using a dashed, or dotted, line, as the border.2692

I am not going to make this a solid line, because this border is not part of the solution set.2706

The next step is to use a test point to determine, just for this inequality (this is the second inequality), where the solution set is.2711

And I am going to use the test point for the circle, right, test point (0,0), at the origin; that is easy to work with.2721

So, I go back up here to the original, 4x2 + 4y2 < 36.2730

So, 4 times 02 + 4 times 02 < 36.2736

This just gives me 0: 4 times 0 is 0; and this is 02...4 times 0 is also 0.2742

Is 0 less than 36? Yes, so again, I have a solution set that is on the inside of this boundary.2752

OK, so this is the solution set for the second inequality; this is the solution set for the first inequality.2766

And the solution set for the system is going to be right in here, where the blue and the black overlap.2771

This boundary is included in part of the solution set: the boundary of this circle is not.2780

So, we solved this by graphing the boundary for the first inequality, then the boundary for the second inequality,2786

and using test points to find the solution sets for the individual inequalities.2794

And then, the overlap between the two is the solution set for the entire system.2799

So today, we worked on systems involving quadratic equations--systems of equations where one was linear,2806

one was quadratic, or both were quadratic, as well as some systems of quadratic inequalities.2814

And that concludes today's lesson; thanks for visiting Educator.com.2822

Welcome to Educator.com.0000

Today, we are going to start a series of lectures on sequences and series, starting with arithmetic sequences.0002

A sequence is a list of numbers given in a certain number; and each of these numbers is called a term.0010

So, the general form that a sequence is written in is this: a1,0016

or just a, is the first term; a2 is the second term, and on and on.0021

Each term is given by a; and then n is the term number.0026

For example, a typical sequence would be something like 5, 10, 15, 20; and this type of sequence is called a finite sequence.0032

Another type of sequence is one like this: 3, 9, 27, 81...and then it has the three dots, the ellipses, at the end,0052

indicating that it goes on forever: this is called an infinite sequence.0064

Looking back at this second function, looking a little bit more deeply, we could actually rewrite this second function as an = 3n.0074

I will go into more detail about this in a few minutes.0089

But just in general, sequences in general (not just limiting ourselves to arithmetic sequences)...0092

The idea, then, would be: if I was looking for the first term, a1 (here n = 1--it is the first term),0099

I would then substitute in 1 where the n was; so that gives me 3.0106

So, the first term is 3: a2 is 3 to the second power, 9; a3 is 3 to the third power, which is 27; and so on.0112

So, you can develop an equation for this sequence that will tell you what a particular term is--what the value of that term is.0126

If I wanted to know the 17th term, then it would simply be 317, whatever that comes out to.0134

So, this is just sequences in general; but today we are focusing on arithmetic sequences.0142

In an arithmetic sequence, each term after the first one is obtained by adding a constant--0149

not multiplying by a number or anything--it is just by adding; that is limiting this to arithmetic sequences.0154

The constant that you add to obtain the subsequent term is called the common difference.0162

In the previous slide, I showed you a typical sequence: 5, 10, 15, 12 and this is an arithmetic sequence.0171

The common difference, which is often just called d, is 5; so this is the common difference.0181

Therefore, I can start with the first term; and to get the second term, I am going to add 5 to that.0191

So, a2 = 10; a3, 10 + 5, equals 15; and so on.0199

One thing to be aware of is that the common difference can be negative; it also can be a fraction.0211

So, the common difference can be a positive whole number; it could be a fraction; or it could be an integer that is negative.0218

And we are going to see examples of all of those in a few minutes.0227

If you want to find the value of a particular term in an arithmetic sequence, you can use the formula for the nth term.0233

Here, this is called the formula for the general term sometimes (either for the nth term or the general term).0242

an = the first term, plus (n - 1) times the common difference.0249

So, if I know the first term and the common difference, and I am looking for a particular term (let's say the 20th term),0257

then I would know n, so I can find the value of a20.0266

Looking at a little bit different sequence here: 5, 3, 1, -1, -3, and continuing on--looking at this,0273

I can see that the numbers are getting smaller.0286

Since the numbers are getting smaller, I know that the common difference is negative.0289

And I can easily find that common difference by taking one of the terms and subtracting the previous term.0292

So, I will go ahead and take 3; I will subtract the previous term; and that is going to be equal to -2, so the common difference is -2.0299

All right, so if I have this common difference, then if I wanted to look for a particular term, I could find it.0312

For example, I may want to find a10: then I could use this formula.0321

a10 would equal the first term, 5, plus 10 minus 1, times the common difference, which is -2.0336

So, a10 = 5 + (10 - 1)...is 9...times -2 (that is times -2, actually, making that clear).0349

a10 = 5 + -18, or a10 = -13.0365

Therefore, by having this formula, I can find any term in here, without going through the cumbersome of just subtracting 2, subtracting 2, and so on.0376

I knew the first term; I knew n; and I knew the common difference; the rest is just calculating.0386

OK, the equation for the nth term: you can use the formula that I just gave in order to find the equation for the nth term.0397

And I gave an example of applying that general equation.0404

Let's talk about using that formula to find a specific equation again.0409

So, looking at another example: 9, 12, 15, and so on--I have been given some terms in the sequence.0416

And I think back to my formula: an = a1 + (n - 1)d.0431

I want to find an equation for the nth term specific to this sequence.0441

So, I am going to look at what I have: I have a1...I have the first term, often just called a; that equals 9.0446

I need to find the common difference: I can find the common difference by saying 12 - 9 = 3.0453

Therefore, I can write a general equation: the general term, or the nth term, an, equals 90459

(9 is the first term), plus (n - 1), times the common difference, which is 3.0472

So, an = 9 + 3n - 3; an = 6 + 3n.0479

Now, I have a formula that I can use; so if I wanted to find any term, such as a21, I could use this.0493

a21 = 6 + 3(21); so, a21 would then equal 63 + 6, which equals 69.0504

OK, so again, I have my general equation here; and then I said, "All right, I have my general formula; I can write a specific equation for this situation."0532

The first term is 9; and the common difference is 3; that comes out to 9 + 3n - 3; 9 - 3 is 6, so it is 6 + 3n.0544

Now, if I want to find the 21st term, then I will say, "OK, I am going to go ahead and use my specific equation0558

for this scenario, for this sequence," which is a21 = 6 + 3(21).0570

That gives me 63 + 6; a21 is 69.0582

Arithmetic means: arithmetic means does not mean finding the mean of the sequence or anything; this is totally different.0588

Arithmetic means are actually the terms between two non-successive terms of an arithmetic sequence.0595

And you can actually use the formula that we learned for the nth term to find the common difference, d.0603

Once you know the common difference, d, you can use d to find the arithmetic means between terms.0610

For example, if I were given a sequence, -3...and then there were some missing terms: one term is missing,0616

two terms are missing, three, four, and then they give me the last one; I have my first term, a1;0625

and I have 1, 2, 3, 4, 5, 6 terms in this; so a1 = -3, a6 = 17.0634

Then, I look back to that formula for the general term, an = the first term, plus (n - 1) times the common difference.0645

In this case, I am going to be using the formula, not to find an yet; first I need to find the common difference.0654

I then use the common difference to find these terms.0661

Let's go ahead and find the common difference.0664

If a6 is 17, and I put that here; -3 is the first term, and I am working with a6;0666

so in this case, there are 6 terms, a6...6 - 1 times the common difference, which is what I am looking for.0675

Now, it is just a matter of solving for that: this gives me 17 = -3 + 5d; add 3 to both sides; that is going to give me 20 = 5d.0685

And then, I am going to divide 20 by 5 to give me 4 = the common difference.0699

So now that I have a common difference of 4, all I have to do is find a2; that is going to equal -3 + 4, so that is going to equal 1.0706

To find a3, it is going to be 1 + 4 = 5; a4 = 5 + 4; that is 9.0715

And a5 = 9 + 4, which is 13.0726

OK, we have -3; and now, our missing terms: 1, 5, 9, and 13; and then I have 17, which was already given.0732

So, I found the arithmetic means, or the missing terms, by using the formula for the nth term0746

to find the common difference, and then taking that common difference and adding it to each term to find the next term.0750

Example 1: Write an equation for the nth term of this sequence.0758

Recall that the equation for the nth term, just the general formula, is an, the general term,0764

equals the first term, plus (n - 1) times the common difference.0774

Therefore, we need to find the common difference; and you can find that by taking any term and subtracting the one just before it.0779

Therefore, 26 - 19 = 7; so the common difference is 7.0786

an equals the first term; well, I also have the first term--that is -2.0797

So, that is -2 + (n - 1) times 7, so an = -2 + 7n - 7; an =...-2 + -7 is -9, so 7n - 9.0801

So, this is the equation to find any term of this sequence.0822

Example 2: Find the arithmetic means: this time, we need to find the missing terms.0829

And we can do that because we, again, have that equation for the nth term,0834

which is an = the first term + (n - 1) times the common difference.0840

In order to do this, though, I need to find the common difference; and I can do that because I have the first term,0846

which is -7; I also have 1, 2, 3, 4, 5, 6, 7...I also have the seventh term, which is equal to 11.0854

And in this case, n = 7; so I just go ahead and use these values to find d.0863

Therefore, I am going to end up with 11 = the first term, which is -7, plus n, which is 7, minus 1, times d.0872

Therefore, 11 = -7 + 6d; I am going to add 7 to both sides to simplify--that gives me 18 = 6d.0887

Divide both sides by 6; I end up with d = 3.0896

Now that I have this, all I need to do is say, "OK, for the second term" (I have the first term),0900

"I am going to take the first term, -7, and just add 3 to that to get -4."0908

For a3, I am going to take -4 and add 3 to that to get -1.0914

a4 = -1 + 3, so that is going to give me 2.0925

a5 = 2 + 3, which is actually equal to 5; and then, a6, which is also missing, is going to be 5 + 3, or 8.0933

Therefore, the term I was given was -7; and then, I found the missing terms, -4, -1, 2, 5, and 8; these were the missing terms, the arithmetic means.0943

And I was given the last term, 11.0960

And the key thing is to find the common difference, using this general equation, and then to take that common difference and use it to fill in the missing terms.0962

Which term is 763 in the arithmetic sequence (and the arithmetic sequence is given)?0974

So, you have a term, an = 763; and what this is really asking is what place that is.0983

Is it the fifth term? Is it the seventeenth term? What number term is this?0991

I know that its value is 763; and let's look at what else I know.0997

Well, I know that the first term is -7; and I can easily find the common difference.1002

I can just take 15 - 4, for example, which is 11; so knowing this value, the first term, and the common difference,1009

I can go back to my equation for the nth term, an1017

equals the first term, plus n - 1 times the common difference.1022

What I am looking for is n, the term number: for an = 763, what is n?1027

I can put in 763 = -7 +...n is my unknown, so I have (n - 1) times this common difference of 11.1037

So, 763 = -7 + 11n - 11; adding 7 to both sides is going to give me 770 = 11n - 1.1047

Now, adding 11 to both sides gives me 781 = 11n; and the last step is just to divide both sides by 11.1065

And if you figure that out, it comes out to n = 71.1074

Therefore, the term number equals 71, or a71 = 763.1078

So, this time, I was given a term, and was asked to figure out which term it is in the sequence.1090

Where does it land in this sequence?--I could do that because I had the first term; I had the common difference; and I had the value of that term.1097

All right, find an equation for the nth term of the arithmetic sequence with a101 = 100 and a common difference of 7.1106

Well, to figure out this equation for the nth term, I need the first term.1118

So, I have the common difference; but the thing that is missing here is the first term.1125

So, how do I figure that out? Well, I do know another term; I know an an.1138

Since I know an, or a101, and I know that n is 101 in that case, and I know the common difference, I can solve.1143

So, I am going to first solve for the first term.1151

Then I will go back and use that first term to develop an equation for the nth term for this sequence.1156

So, I have a1 = 100, and I don't know my first term; I know my n for this term is 101,1163

and I am going to say minus 1, times the common difference, which is 7: this gives me 100 = a1 + 100 times 7.1176

100 = a1 + 700; subtract 700 from both sides; that is going to give me...actually, let's see:1187

yes, it is going to give me -600 equals this first term.1202

So, I am just rewriting it in a more standard form: the first term equals -600.1207

Now that I know the first term, I go back again and look at that general equation for the nth term.1212

And recall that I am asked to find a specific equation for the nth term for this sequence.1220

And I can do that, because I know that the first term is -600, and the common difference, d, is equal to 7.1226

So, an = -600 + 7n - 7; this equals an = 7n - 607, just simplifying.1236

I had to take an extra step here, because I wasn't given the first term.1251

But since I was given another term, I could find the first term.1254

And then, I went ahead, and I used that to find the equation for the nth term for this sequence, which is an = 7n - 607.1258

This concludes the lesson on arithmetic sequences on Educator.com; thanks for visiting!1269

Welcome to Educator.com.0000

In a previous lesson, we talked about arithmetic sequences; and in this lesson, we will continue on with the discussion, to discuss arithmetic s0ries.0002

First of all, what are arithmetic series? An arithmetic series is the sum of the terms of an arithmetic sequence.0012

So, I will briefly review arithmetic sequences; but if you need a complete review of that,0019

go and check out the previous lesson, and then move on to arithmetic series.0024

Recall that a series is a list of numbers in a particular order; and each term in the series is related to the previous one by a constant called the common difference.0028

A typical arithmetic sequence would be something like 20, 40, 60, 80, 100.0043

Taking 40 - 20 or 60 - 40, I get a common difference of 20, which means that, to go from one term to the next, I add 20.0051

So, this is the sequence: it is a list of numbers; the series is actually a sum of numbers,0063

so I am going to add a + between each number: 20 + 40 + 60 + 80 + 100.0071

So, an arithmetic series is in the general form: first term, a1 + a2 + a3, and on and on until the last term, an.0082

This is a finite arithmetic series, because there is a specific endpoint; there is a finite number of terms.0097

As with arithmetic sequences (those could be finite or infinite), you could have finite or infinite arithmetic series.0106

So, looking at a different sequence: 100, 200, 300, 400, and then the ellipses to tell me that this is an infinite sequence:0115

I could have a corresponding series, 100 + 200 + 300 + 400 + ...on and on; this is an infinite arithmetic series.0127

There is a formula (or actually, two formulas) here to allow you to find the sum of an arithmetic series.0148

Sometimes you will be asked to find the sum of the first n terms of an arithmetic series--maybe the first 17 terms or the first 10 terms.0154

And that would be s17 or s10.0162

There are two formulas: the first one involves taking the number of terms, dividing it by 2, and then multiplying it0166

by 2 times the first term, plus the quantity (n - 1) times the common difference.0173

The second term is the sum of the first n terms: again, it equals the number of terms divided by 2.0179

But this time, you are going to multiply that by the sum of the first and last terms.0185

So, you can see the difference: you use the first formula if you know the common difference, and the second formula if you know the last term.0190

So, you have to choose a formula, depending on what you know.0199

For example, let's say that a series has, as its first term, a 5: 5 is the first term.0202

And it has a common difference of 4, and you are asked to find the sum of the first 10 terms, s10.0211

I am looking, and I see that I have the common difference; so I am going to go ahead and use the first formula.0220

And let's go ahead and do that: s10 = 10, because, since I am finding the sum of the first 10 terms,0229

n = 10, divided by 2, and that is times 2 times 5 (the first term), plus n (which is 10) minus 1, times the common difference (which is 4).0241

So, s10 = 5(10) +...10 - 1 is, of course, 9, times 4.0253

Therefore, s10 = 5 times...9 times 4 is 36; 36 + 10 is 46.0262

And if you multiply that out, or use your calculator, you will find that the sum of the first 10 terms0272

in a series with this first term and this common difference is actually 230.0276

So, that is one example; in another example, perhaps you are asked the sum of the 16 terms in a series,0283

if the first term is 20, and the last term of the ones we are trying to find, a16, is equal to -10.0292

Well, we are going to use the second formula, because we have the first and last terms, but we don't have the common difference.0305

So, s16 = n; in this case, I am looking for the sum of 16 terms, so n is 160311

(I am going to put that right here--n is 16); that is going to give me 16/2, times the first term, which is 20, plus that last term, which is -10.0323

So, this is s16 = 8, times 20 - 10, which is 10; that gives me 80 as the sum of the first 16 terms.0336

You need to learn both formulas and simply know when to use a particular one.0348

Sigma notation: a series can be written in a concise form, using what is called sigma notation.0353

And what sigma refers to is the Greek letter Σ; and in this case, sigma means "sum."0359

So, what this symbol means, in the context in which we are using it, is "sum."0366

And you will see it written something like this: you will see a letter here, called the index (that variable is called the index).0371

And we have been working with n, but often i is used; it could be n; it could be i; it could be k; it could be something else.0378

i = 1; I am just giving an example; and let's say, up here, the upper index is going to be 10.0385

And then, there will be the formula for the general term, which we talked about earlier on,0393

when we talked about arithmetic sequences and the formula for an that is particular to a sequence.0399

So, if you need to go ahead and review that, it is in the previous lecture.0406

This is the formula for the general term of the sequence.0409

And this is read as "the sum of an as i goes from 1 to 10."0412

So, it is the sum of the terms that are found, using this particular formula, when you plug in 1, then 2, then 3, and up through 10.0420

That is in general; let's talk about a specific example with a specific formula.0429

You could have another arithmetic series, written in sigma notation, where n goes from 5 to 12.0435

So, n goes from 5 to 12; so I am going to start with 5, and I am going to end with 12.0448

That means that there are actually 8 terms; there are 8 terms in the series, because it is 5 through 12, inclusive.0454

And the formula, we are going to say, is n + 3: so the formula to find a particular term, an, is n + 3.0460

I could find the terms, then, by saying that for the first term, a1, I am going to use 5 as my n, so it equals 5 + 3; therefore, it is 8.0472

For a2, I am then going to go to 6: so that is going to be 6 + 3; that is 9.0485

a3 =...then I am going to go to the next value, which is 7: 7 + 3 = 10; and on up.0497

So, if I wanted to find the last term, a8 (because there are 8 terms), then I would put in 12; and 12 + 3 is 15.0512

There would be terms in between here, of course.0524

This is just a concise way of writing a series; and we have already talked about how to work with these series, and what the different terms mean, and formulas.0527

But now, this is just a different way of writing them.0539

We need to find the first term of the arithmetic series with a common difference of 3.5 and equal to 20,0541

and the sum of the terms, s20, equaling 1005.0547

Since we know the common difference, we can use this formula: sn = n/2, times 2 times the first term, plus (n - 1)d.0553

Except, in this case, I am not looking for the sum: I have the sum; I am looking for the first term.0566

Therefore, 1005 = 20/2, times 2 times a1, plus n (n is given as 20), minus 1, times the common difference of 3.5.0571

This gives me 1005 = 10(2a1) + 19(3.5).0589

19 times 3.5 is actually 66.5; therefore, 1005 equals 10 times 2a1, which is 20a1, plus 10(66.5), which is 665.0601

1005; subtract 665 from both sides to get 340 = 20 times that first term.0622

Divide both terms by 20, and you will get that the first term is equal to 17.0631

So, we use this formula for the sum of the series; but in this case, we were looking for the first term.0636

We had the sum; we were looking for the first term.0646

And the solution is that the first term is 17.0647

In the second problem, we are asked to find the first three terms of the arithmetic series with n = 17, an = 103, sn = 1102.0653

In order to find the first three terms, I need to find the common difference.0666

And I also need to find the first term; I need the first term, and then I need the common difference, to find the second and third terms.0670

I can use the formula sn = n/2 times the first term plus the last term, because I don't have the common difference--I am looking for it.0680

But what I do have is an, so my first step is going to be to find the first term.0691

The sum is 1102; n is 19; I don't know the first term; and I know that an is 103.0698

I am going to multiply both sides by 2 to get 2204 = 19 times this, a1 + 103.0710

I am then going to divide both sides by 19, and that comes out to 116 = the first term, plus 103.0721

And then, I just subtract 103 from both sides; and now I have my first term.0729

So, I am asked to find the first three terms: the first term is 13.0734

To find the next two terms, I find the common difference.0737

What I am going to do is switch to the other formula: that other formula, sn,0740

equals n/2 times 2a1, the first term, plus n - 1 times the common difference.0744

I found the first term; now that I have that, I can find the common difference, because I can put the first term in here.0754

So again, sn is 1102; n is 19; and this gives me 2 times 13, plus I have an n of 19 - 1, and I am looking for the common difference.0761

I am going to multiply, again, 1102 times 2 to get 2204 = 19...2 times 13 is 26, plus 19 - 1...that gives me 18d.0775

I am going to divide both sides by 19 to get 116 = 26 + 18d.0790

Subtracting 26 from both sides gives me 90 = 18d; the final step is to divide both sides by 18 to get a common difference of 5.0798

I have a1 is 13; I am going to take 13 + the common difference of 5 to get the second term (that is 18).0810

So, a2 is 18; the third term--I am going to take 18, and I am going to add 5 to that to yield 23.0823

So, I was asked to find the first three terms, and I did that by first using this formula to find the first term,0836

then going to the other formula and finding the common difference to get 13, 18, and 23 as my solutions.0843

The third example: I am asked to find the sum of the series 6 + 11 + 16 + 21, and on and on, with the last term of 126.0854

That means that what I have is the first term and the last term.0867

I am going to find the sum using this formula, because I can figure out my common difference.0874

So, I am going to use the formula n/2, times 2a1, plus a - 1, times d.0884

I could easily find the common difference, because I know I will just take 11 - 6, so I have a common difference of 5.0896

Looking at this formula, the only issue is going to be that I don't know n.0904

But I can figure out n; and that is because I have another formula.0910

Recall the formula for the general term: we discussed this in the lecture on arithmetic sequences.0917

an equals the first term, plus n - 1, times the common difference.0926

an is 126; so that shouldn't be 16--that is 126: an is 126, and I have the first term equal to 6; and I am solving for n.0937

I know that the common difference is 5.0956

126...and then I am subtracting 6 from both sides: that gives me 120 = (n - 1)5, so 120 = 5n - 5.0959

Adding 5 to both sides gives me 125 = 5n; 125/5 is 25, so I have n = 25.0970

Now, again, I am asked to find the sum of the series; and I can use this formula, because I now have n; I have the first term; and I have the common difference.0982

So, let's go ahead and use that: it is actually s25 = n, which is 25, divided by 2, times 2 times that first term0990

(which is 6), plus (n - 1) (n is 25, minus 1), times the common difference of 5.1000

Now, it is just a matter of simplifying: this is going to give me 25/2 times 12, plus 24 times 5.1009

So, the sum equals 25/2, times 12; and then if you multiply out 24 times 5, you will get 120.1022

Therefore, the sum equals 25/2; 120 + 12 is going to give you 132.1033

Now that I have gotten it down to this point, I can simplify,1050

because this is going to give me 132/2, times 25; so that is s25 = 25 times 66.1054

You can multiply it out; or it is a good time to use your calculator to find that the sum equals 1650.1065

So again, in order to use this formula, I had my common difference.1074

I didn't have n; I solved for n; n equals 25.1077

I went back in, substituted those values in, and then came out with s25; the sum of this series is 1650.1081

Example 4: we are working with sigma notation, so you need to know how to read this notation.1093

And I am going to start with 4 and end with 14; and I can find the first term, because I am also given the formula for a general term in this series.1098

So, to find the sum of the series, let's just start out by finding the first term, because we know we are going to need that.1116

a1...we are going to begin with 4, so for the first term, n is going to equal 4.1123

It is 2 times 4, minus 3; a1 = 8 - 3, so the first term is going to be equal to 5.1131

Now, recall that we have two formulas that we can use to find the sum.1143

We have one formula that involves knowing the common difference.1146

We have another formula that requires us to know the first term and the last term.1150

I found the first term; since I know that, for the last term, n = 14, I can find that, as well.1156

And what that means is that I can use this formula: that the sum of the series is going to be equal to n/2, times the first term, plus the last term.1162

Therefore, let me find the last term: an = 2(14) - 3; this is going to give me an = 28 - 3, so the last term equals 25.1172

Now, what is n? Well, this is telling me that the number of terms...I would have to take each number from 4 through 14, inclusive.1196

And if you figure that out, that is actually 11 terms, because you are including 14.1206

So, starting with 4 and going up through 14, there are actually 11 terms, so n = 11.1209

It is really a11 = 15 I am asked to find the sum of these 11 terms.1215

And I can do that now, because I know that I have n = 11, divided by 2, and then I am going to get the first term1222

(that is 5), plus the last term, which is 25; so the sum is 11/2, times 5 + 25 (is 30).1228

Therefore, this cancels; I am going to get 11 times 15, and that is simply 165.1239

OK, so in sigma notation, this gives me a lot of information, because I saw that I knew the formula to find a particular term.1251

And I knew the n for the first term and the n for the last term.1260

So, I knew I could use this formula, because I could find the first and last terms.1264

So, I made n equal to 4 to find the first term, which is 5; I made n equal to 14, which is to help me find the last term, which is 25.1268

And then, I knew that, since it was going from 4 to 14, that n is equal to 11.1277

Once I had first term, last term, and n, it was just a matter of calculating the sum, which was 165.1282

That finishes up today's lesson on arithmetic series; thanks for visiting Educator.com!1290

Welcome to Educator.com.0000

In the previous lessons, we talked about arithmetic sequences and series.0002

So, we are going to go on to discuss geometric sequences.0007

What are geometric sequences? Recall that a sequence in general is a list of numbers in a certain order.0012

So, it is in the general form first term, second term, and on...and it could end at a particular term, an, or it may go on indefinitely.0020

In previous lessons, we talked about arithmetic sequences: for example, 5, 10, 15, 20.0034

And each term was related to the previous one by a common difference, d (here d = 5).0043

What we did is added whatever the common difference was to a term to get the next, to get the next, and so on.0051

So, if you need to review arithmetic sequences or sequences in general, it might be a good idea to go back and start with that lecture.0057

And now, we are going to continue on and learn about geometric sequences.0064

So again, a geometric sequence is a list of numbers; so a geometric sequence is a sequence0067

in which each term after the first is found by multiply the previous term by a non-zero constant r.0073

When we talked about arithmetic sequences, we added the common difference.0081

Here we are going to multiply a term by the common ratio r to get the next term.0087

Therefore, an equals the previous term, which is an - 1, times the common ratio.0092

So, if I were to look for a4 in a particular series, I would find it by saying,0103

"OK, the previous term, a4 - 1, times the common ratio..." so the term a4 would equal r times the third term.0109

Looking at an example of this, this, again, was an arithmetic sequence.0124

Now, we are going on and talking about an example that gives you a geometric sequence: 3, 12, 48, 192.0150

Working with these, it is important to find a common ratio: the common ratio is given by taking a term0166

(any term--I am going to take 12) and dividing it by the previous term.0173

So, this is a2; the first term is a1; therefore, if I take a2/a2 - 1, that is just a2/a1.0178

So, take a term; divide it by the previous term; and this is going to give me the common ratio of 4.0191

So, you can see where these two equations come from.0206

To find the next term, you multiply the term previous to it by the common ratio.0209

To find the common ratio, you take a term, and you divide it by the term that came just before.0215

Now, as we talked about arithmetic sequences, I said that the common difference could be a negative number; it could also be a fraction.0223

And that is true, as well, with the geometric series.0231

For example, I could have 2, -6, 18, -54: this is another geometric sequence, and I want to find the common ratio.0237

So, I will take a term; I am going to go for -6, and I am going to divide it by the previous term, which is 2; and it is going to give me -3.0250

The thing that you will notice is: when you have a negative common ratio, the terms are going to alternate their signs.0257

So, I am going to have a positive, a negative, a positive, a negative.0265

If it is a positive common ratio, then the terms will just all be positive.0269

You can also have a common ratio that is a fraction; and we will work with those examples later on in the lesson.0274

For a geometric sequence, there is a formula for the nth term.0282

So here, the formula for the nth term is that the general term, an,0286

equals the first term, times the common ratio, taken to the power n - 1.0290

Looking at a geometric sequence: 2, 6, 18, and we will make this an infinite sequence...0296

it is going to go on and on...perhaps I was asked to find the sixth term, a6.0302

What is that? Well, I can use this formula.0309

a6 equals the first term (I have the first term--it is 2), and I also need to find the common ratio.0312

The common ratio is going to be any of the terms (I will take 18), divided by the previous term (6), which is 3.0320

So, I have the common ratio; I have the first term; and I have n; here, since I am looking for the sixth term, n will equal 6.0329

So, the sixth term is equal to the first term (which is 2), time the common ratio (which is 3), raised to the power of 6 - 1.0339

The sixth term is equal to 2, times 35.0354

So, recalling powers of 3: 3 times 3 is 9, times 3 is 27; so 33 is 27; 34 is 81; and 35 is 243, 81 times 3.0359

So, a6 = 2 times 243, or simply 486.0381

So again, this is the formula for the nth term of a geometric sequence.0389

Given the sequence, I could find the sixth term, because I know the first term, 2; I know the common ratio;0394

I was able to figure out that it is 3; and I know that n is equal to 6; so I got that a6 is 486.0404

Geometric means: again, thinking back to arithmetic sequences, we said that arithmetic means are missing terms in an arithmetic series.0412

And that is analogous to this situation: geometric means are missing terms between two non-successive terms of a geometric sequence.0421

2048 is the first term; then I have three missing terms--those are the geometric means; and then, I have my last term, 8.0429

Use the common ratio to find the geometric means.0442

If I find the common ratio, then all I need to do is take a term and multiply it by the common ratio to find the next term;0444

multiply that by the common ratio to find the next term; and so on.0450

If I want to find the missing terms here, I am going to use my formula for the nth term,0454

an = the first term, times rn - 1.0458

Let's look at what I am given: the first term is 2048: 1, 2, 3, 4, 5...the fifth term is 8; OK.0463

Using this formula: a...I need to find the common ratio, and I can do that because I have the first and last terms, and I have n.0476

1, 2, 3, 4, 5; n = 5, right here; so an is 8; that is equal to the first term, times the common ratio, raised to the power 5 - 1.0485

8 = 2048 times r4; so if I take 8/2048, equals r4, this simplifies to 1/256 = r4.0501

If you think about your roots and your powers, the fourth power is actually plus or minus...0524

I am going to take the fourth power of 1/256; it is actually ± 1/4.0535

And if you multiply this out, you will find that 42 is 16, times 2 is 32, times 2 is 64...0540

Excuse me, 1/4 times 1/4 is 1/16, and continuing on, you will find that the fourth power of 1/4 is 1/256.0551

OK, the important thing to note, though, is that it is not just that the fourth power is 1/256; it is actually plus or minus,0568

because I could take -1/4 times -1/4 times -1/4 times -1/4, and I would also get 1/256.0579

Therefore, with geometric means, you may end up with two sets of answers.0591

All right, so I found my common ratios, which could be ± 1/4; and I have my first term, 2048.0599

So, to find my second term, I just take 2048 times 1/4; that is 5/12.0606

To find my third term, I just take 512, times 1/4; and I am going to get 128.0614

To find the fourth term, I am going to take 128 times 1/4, and I am going to get 32.0626

So, that is one set of answers; I actually have two sets of answers.0635

If r = 1/4, then the missing terms (the geometric means) are 512, 128, and 32.0639

If r is actually equal to -1/4, then the signs will alternate, so what I am going to get is 2048, then -512, then positive 128, then -32.0665

So, I have two possible sets of geometric means.0682

Again, to find the geometric means, I am just going to find the common ratio.0686

And I was able to do that using this formula; I got two answers--the common ratio is either plus or minus 1/4.0691

I took the first term; I multiplied it by 1/4, and then multiplied that by 1/4, and on to get this set of geometric means.0698

I took my other possible solution, r = -1/4, and multiplied 2048 by that to get -512, times -1/4 is 128, times -1/4 is -32.0705

So, that is something to keep in mind: that, with geometric means, you can get two sets of solutions.0718

Example 1: Find the ninth term of the geometric sequence with a fifth term of 80 and a common ratio of 2.0726

The formula for the nth term is first term, times rn - 1, the common ratio to the n - 1 power.0734

Here, n = 9; and I am given a5, and I am already given r.0744

So, let's go ahead and work on this.0755

I have my common ratio of 2, and n is 9, so that is 9 - 1; so a9(which is what I am looking for) equals0760

the first term, times 28; I am stuck.0776

I can't go any farther; I need the first term--I need a1.0780

But there is something else I haven't used yet, and that is the fact that I know a5.0786

In order to go over here and find the first term, I can do that by saying, "OK, let's look at this formula again."0791

I know what a5 is; so let's look at this formula again and use it to solve for the first term.0800

a5 equals the first term, times r; in this case, n will be 5; and that is to the 5 - 1.0809

So, I can substitute in; I know that 80 equals the first term, times r4; and I already do know r, so let's put that in, as well.0818

The first term is 2 to the fourth; 2 times 2 is 4, times 2 is 8, times 2 is 16.0827

So, 24 is actually 16; if I divide both sides by 16, I am going to get 80/16, and the first term is 5.0838

Now, I can go back and finish my problem.0849

I know that a1 is 5; since a1 = 5, let's finish this out.0852

a9 = 5 times 28; when you continue on with powers of 2, we know that 24 is 16.0861

If you continue on up, you are going to find that 28 is 256.0870

So, a9 = 5(256); 5 times 256 is 1280, and that is what we were looking for.0875

All right, again, it looked straightforward, but we had to take a detour, because when we started out0886

using that formula to find the ninth term, we discovered that we got stuck at this step, because we didn't have the first term.0893

But, since they gave us another term and the common ratio, I was able to go back,0902

substitute in 80 here, put in my common ratio of 2, and solve for the first term.0906

Then, I finish out the problem to find that the ninth term is equal to 1280.0914

Find the geometric means: so we need to find the three missing terms.0919

I always look at what I am given first.0923

Well, I am given the first term, and I am given 1, 2, 3, 4, 5...the fifth term.0925

As always, I am going to use my formula here, that the nth term is equal to the first term, times the common ratio raised to the n - 1 power.0934

To find the geometric means, I need the common ratio--what is r?0944

If I have r, I multiply it by 4 to get the second term, then the second term by the common ratio to get the third, and so on.0948

But I don't know r: what I do know are these two things, so I can find r.0957

I have that a5 is 324; and my first term is 4; in this case, n is going to be 5; now I can find the common ratio.0961

This gives me 324 = 4r4; divide both sides by 4--that gives me 81 = r4.0978

The fourth root of 81 is 3; 3 to the fourth power is 81.0988

But there is something I have to remember: the other fourth root of 81 is -3.0995

If I take -3 times -3 times -3 times -3, that is 9, -27, times -3 is also 81.1002

So, I have two possibilities here: r can equal plus or minus 3.1013

I am going to have two sets of results here.1023

Let's let r equal 3; if r equals 3, then I am going to end up with 4 as my first term; I add 3 to that--I am going to get 7.1027

I am going to go ahead and (let's see) add 3 to that...1039

Actually, a correction: I was thinking of arithmetic series; this is a geometric series--I am going to multiply.1051

I need to multiply each term, so 4 times 3 is going to give me 12, times 3 is going to give me 36, times 3 will give me 108, times 3 is 324.1057

So, make sure, when you are working with geometric series, that you are multiplying, not adding.1076

So again, if r = 3, I am going to get geometric means of 12, 36, and 108.1080

If r equals -3, I am going to get 4 times -3 is -12; -12 times -3 is going to give me positive 36, times -3 is -108.1087

They are alternating signs; so there are two possible solutions for the geometric means: 12, 36, 108; or -12, 36, -108.1100

Write an equation for the nth term of the geometric sequence -2, 1/2, and -1/8.1112

The formula for the general term is the first term, times the common ratio raised to the n - 1 power.1123

So, if I am looking for an equation for the nth term here, I am going to need the common ratio.1133

To find the common ratio, I will just take a term and divide it by the previous term.1140

The common ratio...I could take 1/2, and I am going to divide that by -2.1145

Recall that I could just rewrite this, to make it a little clearer, as 1/2 divided by -2--just write it out.1148

And that is the same as multiplying 1/2 by the inverse of -2, and the inverse of -2 is -1/2.1156

So, that is -1/4; the common ratio is -1/4.1171

Now, I can go ahead and write my equation: an = the first term, which is -2, times (-1/4)n - 1.1178

And I only have three terms here; but just writing it in a more general form...n - 1, because it is just asking me for the nth term.1194

-2 times -1/4...a negative times a negative is going to be a positive, so that is just going to be 2/4, or 1/2.1210

Oh, actually, I cannot simplify that any further--correction.1225

I can't simplify that any further, because it is (-1/4)n - 1; we are actually done at this step.1228

We are done right here with the general formula, because I don't have n.1234

All right, so the equation for the nth term is simply going to be an = -2(-1/4)n - 1.1238

So, I could find any term from this geometric sequence, using this equation.1252

All right, write the next three terms of the geometric sequence: -1/3, 1/2, -3/4.1260

In order to find a term, I need to have the common ratio.1268

So, let's find that common ratio by taking 1/2 and dividing it by the previous term, which is -1/3.1272

This is the same as 1/2 divided by -1/3; and remember, I can always rewrite that as 1/2 times the inverse, which is -3, or -3/2.1279

Therefore, r = -3/2; now that I have the common ratio, I can find the next three terms.1290

So, we stopped with the third term--I am looking for the fourth term, the fifth, and the sixth.1297

So, the fourth term is going to be equal to -3/4, times -3/2; this is just going to be 9/8.1306

The fifth term is going to be equal to 9/8, times that common ratio of -3/2.1317

-3/2 times 9/8 is going to give me -27/16: a6 (the sixth term) is going to be -27/16 times -3/2.1326

A negative and a negative is going to give me a positive, and 27 times 3 is actually 81.1343

16 times 2 is 32; and you could leave these as fractions, or you could rewrite them as mixed numbers.1349

I am just going to leave them as they are; so the next three terms are 9/8, -27/16, and 81/32.1357

And I could have looked here and just predicted that the common ratio is negative, because I have these alternating signs: negative, positive, negative.1366

That concludes this lesson on Educator.com, covering geometric sequences; thanks for visiting!1376

Welcome to Educator.com.0000

Today we are going to talk about geometric series.0002

In the previous lesson, I introduced the concept of geometric sequences; so this continues on with that knowledge.0004

So, what are geometric series? A geometric series is the sum of the terms in a geometric sequence.0012

Again, make sure that you have geometric sequences learned, that you understand that well, before going on to geometric series.0017

But just briefly, recall that a geometric sequence is a list of numbers.0025

And what is unique about this list is that you find one term by multiplying the previous term by a number r, which is the common ratio.0030

For example, here the common ratio is 2: so I multiply 8 times 2 to get 16, times 2 is 32, times 2 is 64.0042

You can always find that common ratio by taking a term and dividing it by the previous term.0053

This is a geometric sequence: today we are going to move on to talk about geometric series.0062

And a geometric series is the sum of the terms; so from this, we could get a geometric series, 8 + 16 + 32 + 64.0072

This is the geometric series: first term + second term + third term, and on and on, until we get to that last term.0083

Now, this is a finite series, but you could also have an infinite series, where it just continues on indefinitely.0100

So, what we want to find, often, is the sum of a particular number of terms in the series.0112

Now, I could look up here and say, "OK, I want to find the first three terms: that is 8 + 16 + 32."0121

And then, I could just add that up and figure out what it is.0128

But that is going to get really cumbersome to add manually; so we have a formula for the sum of a geometric series.0132

The sum of the first n terms of a geometric series is given by this formula.0139

And we have the limitation that r does not equal 1, because if r equaled 1, if I had, say, 3,0143

and I just multiplied it by the common ratio of 1, I would just get 3 again and again and again, and it wouldn't really change.0151

So, the limitation is that the common ratio cannot equal 1.0158

Looking at an example, 6 + 18 + 54 + 162, let's say I wanted to find the sum of this entire series--the sum of all the four terms.0162

I could use this formula: my first term here is 6--what is my common ratio?0174

Well, I can say 18/6 is 3; so the common ratio is 3.0179

Therefore, the sum of these four terms would be 6, times (1 - 3n) (n = 4 in this case),0186

divided by (1 - 3); this is going to give me 6(1 - 34).0197

Well, recall that 3 times 3 is 9, times 3 is 27, times 3 is 81; so this is 81, divided by 1 - 3.0208

So, the sum is 6 times...1 - 81 would give me -80, divided by...1 - 3 is -2.0219

Let's go up here to continue on: 6 times -80 is going to give me -480, because 6 times 8 is 48; add a 0;0231

divided by -2; the negatives cancel out; 480/2 is 240.0241

So, I was able to find this sum by using a formula, rather than just adding each number.0248

And the formula requires that I know the common ratio, the first term, and n.0255

Sigma notation: as with arithmetic series, we can also use sigma notation as a concise way to express a geometric series.0261

Again, the Greek letter Σ means sum; and the variable that we are going to use is called the index.0270

So here, I have a lower index; and then the upper index tells me how high to go.0278

So, here I have n going from 1 to 5; and then I am going to have the formula for the general term0286

written right here, so I know how to find each term in this geometric series.0292

Looking at an example that is specific: as I said, they often use the letter i for the index in sigma notation, so I am going to use i.0297

i going from 1 to 6...and the formula to find each term is going to be 3 raised to the i power.0308

Therefore, I can find this series; it is going to be 3 raised to the first power, plus 3...0316

I started out with 1, and I inserted that here; next I am going to go to 2.0328

Then, I am going to go to 3, then 4, 5, and 6.0333

And then, you could, of course, figure this out; this is 3 + 9 + 27 + 81 + 729 + 2187.0343

So, this notation means this; and again, this is just a different way of writing a geometric series.0355

But the concepts that we have discussed remain the same.0365

All right, we learned one sum formula: and there is a second formula.0368

This second formula is very valuable when you know the first and last terms, and you know the common ratio, but you don't know the number of terms.0374

If n is not known, use this formula.0382

For example, maybe I have a series, and I know that the first term is 128; that the last term is 4; and that the common ratio is 1/2.0388

But I don't know n--n is not known--and I am asked to find the sum.0400

I can do that using this formula: the sum is going to be the first term, minus this last term, times r, divided by (1 - r).0410

which is going to be equal to 128 minus...4 times 1/2 is 2...divided by...1 minus 1/2 is just 1/2.0422

This is going to be equal to 126 divided by 1/2, and that is the same as 126 multiplied by the reciprocal of 1/2, which is 2.0432

And that is 252; so the sum is going to be 252, and I was able to find that because I knew the first term; I knew the last term; and I knew the common ratio.0442

And I had a formula that did not require me to know n.0457

All right, the formula for sn can be used to find a specific term in the series.0462

And a very important term is the first term: so we often use sn to find the first term in a series.0468

So, looking at one of the sum formulas that we just discussed, that would be the first term, times (1 - rn), divided by (1 - r).0477

Let's say that you are given that the sum is 62.0489

And you are also given that the common ratio is 2, and that the number of terms is actually 5.0496

And you want to find the first term; you can do that with this formula.0504

I know that 62 equals the first term, times 1 - 2, raised to the n power (here, n is 5), times 1 - 2.0509

So, 62 = a1...2 to the fifth...2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32; so that is 32.0521

1 - 32, divided by -1...therefore, this equals...rewriting this as 62 =...the first term...1 - 32 would give me -31,0533

divided by -1; if you just pull the negative out front, then those are going to end up canceling; -31a/-1 is just going to give you a positive.0551

So, I am going to rewrite this as 62 = 31 times the first term, because this is just a -1 down here, and I can rewrite this much more simply.0570

OK, dividing both sides by 31 gives me that the first term equals 2.0585

So, I was able to find the first term by being given the sum and the common ratio and the number of terms.0590

Frequently, you will use one of your sum formulas to find the first term in a geometric series.0597

All right, on to Example 1: Find the sum, sn, for the geometric series with a first term of 162,0603

with a common ratio of 1/3, and with a number of terms equal to 6.0613

All right, the formula for the sum: a1(1 - rn)...and since I know n, I can use this formula,0619

rather than the other one...divided by (1 - r); and I want to find sn.0632

sn equals the first term, which is 162, times (1 - 1/36), all of that divided by (1 - 1/3).0639

Therefore, the sum equals...162 times 1, minus 162 times 1/3 to the sixth power, divided by...1 - 1/3 is going to leave me with 2/3 in the denominator.0655

Now, this looks like it might be complicated to work with; but there is a shortcut.0672

If you think of...I have my 162; I can think of 162 as 2 times 81, and 81 is a multiple of 3, so you might be able to see where I am going with this.0680

2 times 81...as I said, that is a multiple of 3; 3 times 3 is 9, times 3 is 27, times 3 is 81.0698

Therefore, I could rewrite this as 162 - 2...let's move this out of the way a bit...times 3 to the fourth power, times 1/3 to the sixth power.0708

I can use rules governing how I work with powers, and say, "OK, if this gives me 34/36, 162 -2 times...0725

if I look at this as 34/36, they have the same base; if I take 6 - 4, this is going to give me 1/32."0742

So, this is 2 times (1/3)2, divided by 2/3--it is much easier to work with now.0762

This is simply going to be 162 minus 2 times 1/9, or 2/9; all divided by 2/3.0772

Coming up here to finish this out: the sum, therefore, equals 162 - 2/9; that is going to give me 161 and 7/9, all divided by 2/3.0785

Recall that, then, dividing by 2/3 is the same as multiplying by 3/2.0799

That is not a very pretty answer, but you can simplify this, calculate it out, use a calculator...but this does give the sum.0810

So again, I was given the first term, the common ratio, and the number of terms.0818

I can use this formula, since I have the number of terms.0822

This looked like it was going to be very messy to work with: 162 times (1/3)6.0825

But by recognizing that I could break 162 into 2 times a multiple of 3, I was able to get this into a base with 3, 34.0830

Dividing 34 by 36 canceled out, and I got (1/3)2;0842

and that simplified things a lot to give me this answer that I have in the upper left.0848

Find the sum of the first 8 terms of the geometric series.0855

So, here we are asked to find the sum of the first 8 terms of this series, and that would be s8.0859

Thinking about which formula I am going to use: I know n; since I am looking for the first 8 terms, I know n.0867

I can also easily find r, because I can just take 1 divided by 1/4; 1 divided by 1/4 is the same as 1 times 4, or 4.0877

So, I have r; I have n; the other thing is the first term, and I have that.0888

With these three pieces of information, I know that I can use the formula0895

that the sum equals the first term, times (1 - rn), divided by (1 - r).0898

So, the first 8 terms...that is going to be 1/16, times 1 minus...r is 4...raised to the n power, where n is 8, divided by (1 - 8).0904

OK, let's look at what I have: I have 1/16 times 1, minus 1/16 times 48.0932

Actually, this is not 1 - n; it is 1 - r; so let's go ahead and change that to a 4.0955

All right, and then I have 1/16 times 48; as with the previous problem,0960

you might recognize that there is something to make this a lot easier than taking 4 to the eighth power and dividing it by 16.0965

You are going to recognize that you can rewrite this 1/16 as (1/4)2.0971

So, 1/4 times 1/4 would give me 1/6; that times 48 allows me to do some canceling to make things simpler.0985

1 - 4 just gives me -3; let's go up here and work this part out.0998

This is (1/4)2 times 48; this is the same as taking 48 and dividing it by 42.1003

Using my rules governing exponents, 48/42...if I want to divide, and I have like bases,1015

I just subtract the exponents; so this is going to give me 4 raised to the sixth power.1028

So, knowing your rules of exponents is important to solve when you are working with this type of problem.1032

Therefore, I am going to get 1/16 - (1/4)2 times 481038

(that is the same as 46, so I am going to go ahead and rewrite it that way) divided by -3.1044

At this point, you are going to have to do a lot of multiplying, or else use your calculator to determine that 46 is actually 4096.1052

And we are dividing by -3; so the sum is 1/16 - 4096; that comes out to -4095 and 15/16, all divided by -3.1059

And a negative and a negative will give me a positive, so that would give me 4095 and 15/16, divided by 3.1077

You can work this out with your calculator and see that this sum is approximately equal to 1365.3.1086

This does give us our answer; but we can estimate to put it in a neater decimal form.1094

Finding the sum: this time, we are given the first term; we are given the last term; and we are given the common ratio.1104

But we don't know the number of terms--we don't know what n is.1113

But that is OK, because we have that other formula that allows us to find the sum of a geometric series when we know the first term;1117

we know the last term; and we know the common ratio (that second formula that we worked with).1125

All right, looking for the sum: I have the first term; this is 4 - an (which is 8748), times the common ratio of 3, all of that divided by 1 - 3.1140

That is going to give me that the sum equals 4 minus...multiply this out, or use your calculator, to get 26244, divided by 1 - 3 (gives me -2).1159

Coming up here to the second column: the sum is going to be equal to 4 - 26244, or -26240, divided by -2.1174

A negative divided by a negative gives me a positive; and 26240 is even--I divide it by 2, and I get a nice whole number, 13120.1192

So, the sum of this geometric series is 13120; and I found that using the formula1201

that only requires that I know the first term and the last term, and the common ratio.1207

I didn't have to know the number of terms in the series, and I could still find the sum of the terms.1211

Example 4: We are asked to find the first term.1217

As I mentioned, it is important to have the first term frequently when you are working with these series.1219

And you can use the sum formulas to find that first term.1224

I am going to look at the information I have: I have the sum; I have the common ratio; and I have the number; but I don't have the last term.1227

So, I use the formula that involves the number of terms, not the last term.1233

And what I am looking for is the first term; therefore, the sum is 1020, equals the first term,1244

times 1 - r (is 2), and it is 2 raised to the eighth power, divided by 1 - 2.1254

And it is 2 raised to the eighth power, divided by (1 - 2); this is going to give me 1020 equals the first term, times 1 minus...1258

if you go through your powers of 2, you will find that 2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32.1270

So, 25 - 32; then, we are going to get 26 is 64; 27 is 128; and then, 28 is 256.1283

That is 1 - 256, divided by 1 - 2 (is -1); therefore, 1020 = a1...1 - 256 is -255; divided by -1.1297

Therefore, 1020 = -a1 times 255, divided by -1.1316

Well, a negative and a negative gives me a positive, so that is 1020 = a1 times 255.1325

I just take 1020 and divide by 255; so I have divided both sides by 255.1335

And I am going to get that the first term equals 4.1340

I was asked to find the first term, and I found that using the sum formula requiring the first term, the common ratio, and the number of terms.1345

And I determined that the first term in this geometric series is 4.1353

That finishes up this lesson on geometric series on Educator.com; thank you for visiting!1358

Welcome to Educator.com.0000

Today, we continue on our discussion of sequences and series with infinite geometric series.0002

What are infinite geometric series? Well, this is a type of series in which there is an infinite number of terms.0010

Earlier on, I mentioned that, for either arithmetic or geometric series, you may have a limited number of terms,0018

which is a finite series; or it may go on indefinitely, which is an infinite series.0024

For example, the geometric series 1 + 1/4 + 1/16 + 1/64 is a geometric series with 4 terms; and it has a common ratio r = 1/4.0030

This is a finite series: it has a limited number of terms.0047

Consider another geometric series: 3 + 6 + 12 + 24, and then, when you see the ellipses (the three dots),0052

it indicates that it goes on indefinitely; so this is an infinite series.0063

The sums, sn, are called partial sums of the infinite series.0070

For example, I may decide that I want to find the first 7 terms of this series, or the first 5 terms of an infinite geometric series.0074

That would just be a partial sum of this series.0084

And we will talk, in a minute, about special cases, when you actually can find the sum overall of an infinite geometric series.0087

So, recall from the previous lesson the formula for the sum of a geometric series,0097

sn = the first term, times (1 - rn), divided by (1 - r).0103

So, if I were to try to find the first seven terms of this series, s7, I could use this formula.0114

I have the first term; I need to find r; recall that I can find r by taking a term and dividing it by the one that goes just before.0120

So here, r = 2; so this is 3(1 - 2n), so that is 7; and then, divided by 1 - 2.0130

So here, I found the partial sum for this infinite geometric series.0149

You can actually find the sum of an infinite geometric series (not just the partial sum) in some cases.0160

And I know that this sounds counterintuitive--how can you find the sum of something that goes on forever?0166

But if you look, you can see why.0172

All right, first of all, it is very important to know that this is limited to cases in which the series is convergent.0178

And the word "convergent" indicates that the sum converges toward a particular number.0184

A series is convergent if and only if the absolute value of r is less than 1.0191

So, if you are working with a geometric series in which the value of the common ratio is either greater than -1 or less than 1,0196

such as 3/4, for example--if r is 3/4 (the absolute value of that is 3/4), or if r is -1/2 (I take the absolute value of that--it would be 1/2),0205

both of these are convergent; I could actually find the sum of those.0219

Consider the series 1/3 + 1/9 + 1/27, going on indefinitely.0222

Remember, to find r (let's go up here and find r), we are going to take 1/9, divided by 1/3.0234

This is the same as 1/9 times 3, which equals 3/9, which equals 1/3; therefore, r = 1/3.0241

If I am asked to find the sum of this, I go ahead and use this formula, x equals the first term, which is 1/3, divided by 1 - 1/3, equals (1/3)/(2/3).0258

1/3 divided by 2/3 is the same as 1/3 times 3/2, or 3/6.0273

The sum of this infinite geometric series is 1/2.0283

Let's look at this another way: just go ahead and add up some of the terms and see what happens.0288

s1 for this term is 1/3; that is all you have: 1/3.0297

So, s2 would be adding 1/3 + 1/9; 1/3 + 1/9 equals...giving this a common denominator,0304

I multiply both the numerator and the denominator by 3 to get 3/9 + 1/9 is 4/9; that is s2.0313

s3 = 1/3 + 1/9 + 1/27; well, I know that these two are equal to 4/9, so that is 4/9 + 1/27.0322

So again, I need to get a common denominator; and if you work that out, you will find that s3 is 13/27.0333

s4...I would continue on: 1/3 + 1/9 + 1/27 + 1/81...and if you figure that out, it becomes 40/81.0343

I won't work out the rest of these right here; but I will just tell you that s5 is 121/243.0360

s6 is 364/729; and s7 is 1093/2187.0370

Let's look at the pattern here: it started out as 1/3; then it became 4/9, 13/27, 40/81, 121/243, 364/729, 1093/2187.0383

What is happening is: this sum is converging upon 1/2.0400

Another way that we say this is that the limit is 1/2; and that is terminology you will hear later on in higher math courses.0414

But for right now, just be aware that you can only find the sum of an infinite geometric series if it is convergent,0422

meaning that, as you take more and more terms and add them to the series, add them, and get their sum,0428

you will see that the sum of the series is converging upon a particular number.0435

And so, we just use this formula as a great shortcut to find the sum.0440

Sigma notation is something we discussed earlier on.0452

Go back and look at the lectures on geometric sequences and geometric series, if any of these concepts are new.0456

But sigma notation--you will recall that the Greek letter sigma means sum, and we used it with other geometric series.0462

For an infinite series, you are going to have something like this.0470

We have our lower index, i = 1; and this is the series as i goes from 1 to infinity.0476

The difference here is that, instead of stopping at a specific value up here, it goes on through infinity.0484

And again, we have the formula for the sequence right here.0490

Another example, or a specific example, would be a series, again, where i goes from 1 to infinity,0494

but we have a formula over here, 1/4 times 1/2, raised to the n - 1 power.0501

So, what you could do, then, is put 1 in here and find your first term, a1.0507

Then, put 2 here; find your second term; and go on infinitely, because of the type of series that this is.0513

We can use the concepts that we just learned to actually write a repeating decimal as a fraction.0523

The sum formula I just described can be used in this way.0529

A repeating decimal would be something like this: .44444...and it just goes on and on that way.0533

What we do is rewrite this as a geometric series.0542

1: First step--how do you do that?0547

Well, look at what this really means: it really means 0.4 + 0.04 + 0.004 + 0.0004, and so on.0556

So, I rewrote this, but just as a series; and it is an infinite geometric series.0573

Then, find the sum of the series: recall that you can only find the sum of an infinite geometric series0580

if the absolute value of r is less than 1; this is required, or you can't find the sum.0590

Well, let's show here that we are OK, because the absolute value of r is actually less than 1.0604

So, in order to find the common ratio, r, I am going to take .04, and I am going to divide it by the term before it, which is .4.0609

So, I move the decimal over one place; that is going to give me .4/4, which is .1; so the common ratio, r, equals .1.0618

.1 is less than 1, so I am fine: I can use the sum formula, the first term divided by (1 - r).0626

So, the sum equals the first term, divided by (1 - .1), equals .4/.9.0635

Dividing that, move the decimal over; you can get rid of that decimal; that gives me 4/9.0645

So, I found that the sum of this...0652

I started out; I have this decimal that is a repeating decimal (it goes on forever); but I saw that I could rewrite this as a series.0657

So, this, therefore, is equivalent to the sum of the series.0665

My next step was to figure out what the sum of this series is, and it is actually 4/9.0670

Therefore, .444 repeating can be rewritten as 4/9; those are equivalent.0674

Recall that you can have a repeating decimal that doesn't just have one number repeat and repeat; it could be multiple numbers.0683

It could be .383838 repeating, and you could do the same thing.0690

You could rewrite this as .38 + .0038 + .000038, and so on; and then repeat as above, by finding the sum of the series.0698

So, this could also be used for repeating decimals where there is a longer repeat.0717

Also recall that we can write these as .4 with a bar over it, or .38 with a bar over it--that is just a different notation.0722

All right, let's find the sum of this infinite geometric series.0736

First, though, I am going to verify that I actually can find the sum of this by figuring out what r is.0741

So, I am going to take 24, the common ratio, divided by 32; one term divided by the previous one gives me the common ratio.0746

This simplifies out to 3/4: 3/4 is less than 1, so yes, I can find the sum, using this formula.0753

The formula: I need to use the first term: 32/(1 - 3/4) = 32/(1/4).0769

We can rewrite this as 32 times 4; and 32 times 4...4 times 2 is 8; 4 times 3 is 120; so that gives me 128.0778

The sum of this infinite geometric series is 128, and I was able to find that using this formula, because I had an absolute value of r that is less than 1.0791

Write as a fraction: recall that this notation means the same thing as .36363636, and so on.0806

The first step is to write this as a geometric series, so I am going to rewrite this as .36 + .0036 + .000036, and so on.0821

And I am going to use my formula for the sum of an infinite geometric series, which is the first term, divided by (1 - r).0844

What is r? Well, as usual, I can find the common ratio, r, by taking a term, .0036, and dividing by the previous term, .36.0853

Move the decimal over two places to get .36/36; and that is going to give me .01.0864

.01 is less than 1, so I can find the sum.0871

Take the first term; divide it by (1 - .01); this gives me .36/.99.0875

I can move the decimal over two places, to give me 36/99; simplify that, because there is a common factor of 3; this is actually 12/33.0885

I can see, again, that I have another common factor of 3; so this is going to give me 4/11.0896

Therefore, this 0.36 repeating decimal can be rewritten as 4/11; so I wrote this repeating decimal as a fraction.0901

Find the sum for this infinite geometric series: before I proceed, I check that the absolute value of the common ratio is actually less than 1.0918

10/12...to find the common ratio, divide that by the term that came just before, which is 5/4.0927

And this is going to give me 10/12 times 4/5 equals 40/60; so this is going to give me 2/3.0933

And since 2/3 is less than 1, I can find the sum of this series using my formula for the sum of an infinite geometric series.0949

OK, the first term is 5/4; I am dividing that by 1 - 2/3 to get 5/4 divided by 1/3.0961

This is the same as 5/4; and then I just take the inverse of 1/3, and 5/4 times the inverse of 1/3; that is 3, so this is going to give me 15/4.0977

And you can keep this as an improper fraction, or write it as a mixed number.0989

The sum of this infinite geometric series is 15/4.0993

Again, I was able to find that because the common ratio had an absolute value that was less than 1.0999

Write as a fraction: we have another decimal that is repeating--goes on infinitely: 0.99999, and so on.1009

Start out by rewriting this as a geometric series.1021

I have 0.9 + 0.09 + 0.009, and so on; it goes on infinitely.1027

Find the common ratio, r: r =...I am going to take .09, and divide that by .9.1041

Move the decimal over one place to get .9/9; therefore, the common ratio is .1.1048

Next, I use my formula for the sum of an infinite geometric series, which is the first term, divided by 1 - r.1055

And this is going to give me the first term, .9, divided by 1 - .1; that is .9/.9; this is simply 1.1063

So, this asked me to write it as a fraction; and .999 = 1, or you could say 1/1.1081

This is still a fraction, because you could just write it as 1/1.1090

And it is counterintuitive, if you think ".999 repeating is actually 1"; it doesn't seem like it is, but it is actually correct.1093

That concludes this lesson of Educator.com on infinite geometric series; thanks for visiting!1105

Welcome to Educator.com.0000

Today, we are going to be covering recursion and special sequences.0002

The Fibonacci sequence is a particular sequence that states that one term, an, is equal to the previous two terms,0006

an - 1 + an - 2--the term that came just before it and the one before that.0016

Fibonacci is an Italian mathematician who lived in the 13th century.0023

And he actually brought the knowledge of this type of sequence to Europe; and that is why the sequence is called the Fibonacci sequence.0029

This type of formula that you see right here is called a recursive formula.0037

Recursive formula means that the value of a term depends on previous terms.0044

In order to use this type of formula, you would have to start out with a couple of values.0052

Looking at the Fibonacci sequence, you need to start out with your first term, 1, and your second term,0058

which actually is also 1, because if you don't have those two terms, you can't find the next term.0065

So, a3 is going to be equal to the term that came just before it, an - 1 (and 3 - 10070

is going to give me a2), and then the term just before that, which (in this case) is actually the first term.0086

a3 = 1 + 1, or 2; a4...now I am going to add the previous two terms here, 2 and 1, to get 3.0092

The previous two terms, 3 and 2, get me 5.0102

I can use this to form a sequence; so I take these terms, and I write them as a sequence:0106

1, 1, 2, 3, 5, and you could go on very easily: 5 + 3 is 8; and so on.0111

This gives you one recursive formula, which will provide a special sequence called the Fibonacci sequence.0128

But just to give you an example of a recursive formula, more generally, remember that a recursive formula0135

is one in which the value of a term depends upon previous terms.0143

I might say, "OK, an = 4 times the previous term, plus 2 times the term before that."0148

So, if I wanted to find a particular term in this series, and I was told the first term is equal to 1 and the second term is equal to 2,0159

and I was asked to find the third term--"What is a3?"--well, a3 = 4 times the term0169

that came just before, which is 2, plus 2 times the term before that, which is equal to 1.0178

So here, a3 = 8 + 2, which is actually going to be 10.0188

So, in this case, I was given a formula, an equals 4 times the previous term, 4 times 2,0203

plus 2 times the term before that, giving me that the third term is going to be equal to 10.0212

Iteration is the process of composing a function with itself again and again; what does that mean?0229

Iteration is something that can be used to generate a sequence recursively.0238

So, if we are composing a function with itself again and again, what the next iteration is depends on the previous one.0242

And that goes back to what we talked about with recursion.0249

Again, you need a starting point, though; so here, we need to start out with a function and a first value that we are going to use with the function.0252

And that is x0, or "x naught."0264

The easiest way to understand this is through an example.0268

f(x) = 3x - 1; x0 = 1; so I have been given my function, and I am given this x0.0272

So, I could be asked to find the first iterate; I am going to find f(x0), which here equals f(1).0283

So, as usual, to find the value of a function when you are given what you want x to be, you just substitute 1 for x.0299

3 times 1, minus 1, is 3 minus 1, which equals 2.0309

So, I found the first iterate; to find the next iterate, I am actually going to take f(f(x0)), which is f(f(1).0318

And we determined that f(1) is 2, so I just find f(2).0335

This gives me 3 times 2, minus 1 equals 6, minus 1, which equals 5.0340

If I want to find another iterate, then I am going to take f(f(f(x0))).0348

Here, I know that this in here is 5; so I am taking f(5); this is 3(5) - 1 = 15 - 1 = 14.0359

And you can continue on like that: if I wanted to find the next term, it would just be f(f(f(f(x0)))).0379

And you go on that way; so again, this is the process of composing a function with itself again and again and again.0388

I could use these numbers to form a sequence; so I have generated a sequence recursively.0396

My first term, a1, is going to be right up here; it equals 1.0403

My second term is going to be 2; my third term is 5; and then, my fourth term is going to be 14.0410

And I could write this out as a sequence: 1, 2, 5, 14.0421

In the first example, I am asked to find the first 5 terms; and I am given that the first term is 3.0429

And when I look at this, this is a recursive formula, because the term that I am looking for depends on the term before it.0435

So, if I am looking for an + 1, it is going to be equal to 2 times the previous term, minus 3.0448

And I need to find the first 5 terms.0456

Well, the first term is given (the first term is 3); and I need to find a2.0458

a2 is going to equal 2, times the previous term; so, given whatever n + 1 is, I am going to take away 1 from that and get the previous term.0464

So, that is going to give me 2 times 3 minus 3, equals 6 - 3, which equals 3.0481

I found the second term; I need to find the third term.0489

So, I am going to take 2 times the previous term (the value of that is 3), minus 3.0493

You are starting to see a pattern here: again, we have 3.0499

Well, I know what is going to happen if I put 3 back into this formula, but I will go ahead and do it anyway.0503

2 times 3, minus 6, equals 6 - 3, is 3; I have one more term to find...again, 2 times 3, minus 3, equals 6 - 3, equals 3.0508

So, this is actually a sequence where this formula generates the same value over and over and over.0521

So, I am just going to end up with a sequence that looks like this; and that is actually going to go on forever.0526

This is kind of an interesting formula where we ended up with the same value over and over and over in this sequence.0532

Example 2: this time we are going to work with iterates--we are asked to find the first three iterates of this function.0543

And the function is f(x) = 4x - 7, and x0 = 3.0549

To find the first iterate, I am going to go ahead and find f(x0, which equals 4.0555

And then, I am going to insert 3 here.0567

Actually, just to clarify: this is equal to f(3), because x0 is 3.0569

So, we are going to write this as 4(3) - 7 = 12 - 7; that equals 5.0575

That is the first iterate; the second iterate is going to be f(f(x0)), which, in this case, is going to be f(5).0582

That equals 4(5) - 7; that is going to give me 20 - 7, which is 13.0592

Next, f(f(f(x0))): I know that this, in here, is 13, so I just take f(13) = 4(13) - 7.0600

4 times 13 is 52, minus 7 gives me 45; so I found the first three iterates, and those are 5, 13, and 45.0615

Example 3: Find the first five terms--they give me the first term; they give me the second term;0639

and they give me a formula that is a recursive formula, because it depends on the previous two terms.0645

So, an = 2(an - 1) + 3(an - 2).0653

I need to find the first five terms: the first term is given; the second term is given; I just need to find a3, a4, and a5.0660

Let's start with a3: a3 is going to equal 2 times the term that came just before,0672

or a2, plus 3 times the term before that, which is a1.0682

This is going to be equal to 2 times 3, plus 3 times 2, equals 6 + 6; that is going to give me 12.0688

So, a3 is 12; a4 is going to be equal to 2 times a3, plus 3 times a2,0699

which equals 2 times...a3 is 12, plus 3 times a2, which is 3.0708

Of course, 2 times 12 is 24, plus 3 times 3 (is 9)...24 + 9 is 33: a4 is 33.0717

a5 is equal to 2 times a4, plus 3 times a3; this is going to be equal to 2 times 33, plus 3 times 12.0730

That is going to give me 66, plus 3 times 12 (is 36), and that adds up to 102.0742

So, I found the first 5 terms; I was actually given these 2; and then, I used a recursive formula to find the third, fourth, and fifth terms.0751

Find the first three iterates: f(x) = x2 + 3x - 4; and then, x0, or "x naught," is equal to 1.0763

The first iterate will be f(x0), which equals f(1); that gives me 12 + 3(1) - 4, equals 1 - 3 - 4.0772

So, I have, let me see...actually, that is plus right here...so that is 12 + 3 - 4, so 1 + 3 - 4; that is 4 - 4; the first iterate is 0.0790

The next iterate: f(f(x0)): this is going to be equal to...since f(x0) is 0, it is just f(0).0805

02 + 3(0) - 4 is 0 + 0 - 4; so that is going to give me -4.0817

Next, I want f(f(f(x0))); this is going to be equal to f(-4) = (-4)2 + 3(-4) - 4.0829

This equals 16 - 12 - 4; well, -12 and -4 are -16, so this is 16 - 16 = 0; so the first three iterates are 0, -4, and 0.0847

That concludes this lesson of Educator.com; thanks for visiting, and see you soon!0869

Welcome to Educator.com.0000

In today's lesson, we are going to be covering the binomial theorem.0002

We are going to start out by talking about what we mean when we say we are going to expand a binomial.0007

So, if you expand a binomial, this is what happens.0013

I have (a + b)n; and I am going to expand that for n = 1, 2, 3, 4.0019

I am going to let n equal 1; and I am going to then get (a + b)1, which is just a + b.0027

If I let n equal 2, then I have(a + b)2; that is going to give me, if you will recall, a2 + 2ab + b2.0041

Let n equal 3: you get (a + b)3, which is a little bit more complicated to work out.0057

But if you figure that all out, (a + b)(a + b)(a + b), you would end up with a3 + 3a2b + 3ab2 + b3.0064

Letting n equal 4, (a + b)4 is going to give you a4 + 4a3b +0079

6a2b2 + 4ab3 + b4.0091

One more: let n equal 5: this is going to give me (a + b)5.0099

This equals a5 + 5a4b + 10a3b2 + 10a2b...0108

actually, this is going to be b3...+ 5ab4 + b5.0127

All right, before we go on to talk about Pascal's triangle, let's note a few things about this expansion.0136

And these will be summed up on the next slide.0144

A few things that you will notice are that the first term is an, and the last term is bn.0146

The second thing is that the number of terms equals n + 1.0157

You will also notice a pattern: let's look at n = 4, (a + b)4.0179

The first term is an, so it is a4; in each subsequent term, the power that a is raised to (the exponent) decreases by 1.0185

So, I started out with a4; then here it's a3; a2; a; and then a is gone.0195

So, the exponent for a decreases by 1 for each term; b does the opposite.0203

There is no b here; this would be b0...it is 1; and then in my next term, I get b to the first power, b2, b3, b4.0213

So, with each term, exponents for a are decreasing; exponents for b are increasing.0223

The next thing that you might notice is the coefficients; let's look at the coefficients.0230

The coefficient of the second term is equal to n; again, this is all summed up on the next slide, but this is just to introduce it here.0234

So, if I am looking at n = 3 (that is (a + b)3), and I expand that; then I get that the second term has a coefficient of 3.0242

The second term here has a coefficient of 4; the second term here has a coefficient of 5.0252

Pascal's triangle--we are getting to that now.0258

I can take the coefficient of these terms and use them to create an array that is called Pascal's triangle.0260

Looking at it, it actually starts out with 1; so if we let n equal 0, that is just going to give me (a + b)0, which is going to be 1.0266

Then, in my next row, I am going to have 1, 1, and 1; those are my coefficients.0277

In the next row, they are 1, 2, 1; in the next row, 1, 3, 3, 1.0289

Now, I can look up here to get the next row; but I don't even need to, because each number is the sum of the two numbers just above it.0302

So, 1 + 1 forms 2; 1 + 2 gives me 3; 2 + 1 gives me 3.0314

So, all I have to do, actually--these outside numbers are always 1; but I could just say 1 + 3...this is going to be 4.0325

3 + 3 is equal to 6; 3 + 1...that is equal to 4.0334

So, I look up here to verify that; and indeed, it is 1, 4, 6, 4 1.0340

In the next row, I would add 1, and I add 1 + 4; I am going to get 5; 4 + 6 gives me 10;0345

6 + 4 gives me 10; 4 + 1 gives me 5; and again, I have a 1 out here.0352

And that matches up with what I have for the expansion of (a + b)5.0357

So, Pascal's triangle is an interesting array that comes from the coefficients of the expansion of this binomial.0363

Something else to note is the symmetry around the middle term; you might have seen that up here, but it is even more obvious right here.0371

The middle term is 2; it is flanked by 1's; here I have two middle terms, because there is an even number of terms;0377

so the two middle terms are 3; they are flanked by 1's; now I have a middle term of 6; next to it are 4's; outside of that, 1's.0384

Two middle terms are 10; the next terms are 5; the next terms are 1.0392

And that is helpful, because if you are looking for the coefficients, you don't need to find all of them; you only need to find half of them.0397

And then, you can use this symmetry to know what the other half are.0403

Summing up these properties of the expansion of (a + b)n: there are n + 1 terms.0409

Let's use an example that we just talked about, (a + b)5, which gave me a5 +0418

5a4b + 10a3b2 + 10a2b3 +0427

5ab4 + b4, to illustrate this.0437

There are n + 1 terms; so here, n = 5; therefore, I am going to have n + 1 terms, equals 5 + 1, so I am going to have 6 terms.0442

And I do have 1, 2, 3, 4, 5, 6 terms.0455

The first term is going to be an (that is a5), and the last term is bn, which should be b5 right here.0458

In successive terms, the exponent of a decreases by 1, while that of b increases by 1.0470

I start out with a5, then a4, to the third, to the second, and then a.0475

b goes the opposite way: I start with b, b2, b3, b4, and b5.0482

I pointed out before that the second term has a coefficient that is equal to n.0490

Something I also pointed out in the last slide is that there is symmetry around the middle term or terms,0498

which tells me that, if I figure out these middle coefficients, well, I already know that the first and last coefficients are 1.0504

And I know that the second (and, by symmetry, second-to-last) coefficients are equal to n.0511

So then, all I would have to figure out is one of these; and I know that the other one is the same.0519

Something else to be aware of is that the sum of the exponents in each term is n.0523

Let's look right here: for this term, 10a3b2, if I add up 3 and 2, that is going to give me 5.0528

So, the sum of the exponents is equal to 5.0537

Or looking at this, 10a2b3, I get the same thing.0540

The sum of the exponents of each term is equal to n.0544

The next concept that we will learn requires that you remember how to work with factorials.0552

So, I am going to go ahead and review factorials: factorials are products, and the symbol that looks like an exclamation point is actually read as "factorial."0557

This is a product: n! is equal to n times n - 1 times n - 2 times n -3...and you continue on that way until you get all the way to 2, and then finally 1.0567

For example, 4 factorial: here, n = 4; this is going to be equal to 4, so here n = 4; n - 1...4 - 1 is 3, so that is 4 times 3.0586

4 - 2 gives me 2; 4 - 3 gives me 1; and this equals...4 times 3 is 12, times 2, times 1, is just 24.0599

So, 4! is just the product of 4 times 3 times 2 times 1.0613

If you are working with fractions that involve factorials in the numerator and denominator, you can often do a lot of canceling out to make things easier.0618

If you were working with 8!/6!,that gives you 8 times 7 times 6 times 5 times 4, and then on down to 1.0625

In the denominator, you have 6 times 5 times 4 times 3 times 2 times 1.0637

So, instead of multiplying all of this out and dividing, it makes a lot more sense to start canceling out all of these common terms.0644

1 through 6; 1 through 6; this just leaves me with 8 times 7, or 56.0657

Now, the reason we reviewed it is to work with this formula.0665

First, we are talking about the binomial theorem, and then the binomial formula.0668

The binomial theorem is really what we just discussed; and it is the idea that, when you expand (a + b)n,0672

you are going to get an equation with the properties we just discussed.0682

And those properties are that the first term is going to be an;0688

the second term is going to have some coefficient; and it is going to be raised to the power an - 1.0692

b is going to be to the first power; then you are going to get another term; and you are going to have a total of n terms.0699

You are going to have another term; it is going to have a coefficient that is going to be an - 2.0707

b is going b2; it is going to increase by 1.0711

Then, you are going to get another coefficient, an - 3; b is going to become b3.0714

And that is going to go on and continue on until you get abn - 1, bn, and so on.0721

Excuse me, plus...that is an - 1, times b, plus bn; the last term will be bn; the first term is an.0731

Each of these terms can be given in the general form an - kbk.0751

And we will look at some examples in a few minutes, and talk about what this means, as we relate it to the binomial formula.0760

This is the binomial formula; this is pronounced "n choose k."0769

Now, why are we even looking at this--what is this for?0781

Think about how we can find these coefficients.0785

We can use Pascal's triangle, but that could be really impractical and quite a bit of work,0788

if you have to go through that whole large array, if you have a really big expansion.0793

And if you just find the coefficient for a particular term, you are going to have to go through a lot of work to get there.0799

So, as usual, we have a formula that gets us directly there, without having to go through all that work.0804

This formula will give me the coefficients for a term that has the exponent an - kbk.0812

So, the binomial formula allows me to find this coefficient.0823

Let's use an example of (a + b)7: I can find the coefficient, using this, of a particular term;0829

or I can expand the whole thing and find all of the terms.0839

Let's look at what this expansion would look like: it would look like this.0841

I know that the first term is going to be a7; recall that the second coefficient is going to be equal to n.0845

So this one I also know; it is 7; and then, I know that a...the exponent is going to decrease by 1; and b is going to appear, raised to the first power.0851

I also know that I am going to have 8 terms, because I am going to have n + 1 terms, or 7 + 1.0866

So, I have another term here; I don't know the coefficient, but I do know that it is going to be a5b2.0872

I have another term with an unknown coefficient: a4b3.0878

I have another term, again with an unknown coefficient: a3b4.0884

Another term is a2b5; another term is ab6; plus b7.0890

I have 8 terms, so I am going to have these two middle terms the same; these two terms will be the same;0900

the second-to-last terms will be the same; so I know this is 7; and these two outer terms will be the same, 1 and 1.0907

I may be asked to find this expansion; I found this much; now, in order to finish it out, I need to find this coefficient and this coefficient.0917

If I have those, I have the rest, because I would have the first half, and then I just reflect for the second half.0927

Let's use this formula to find the coefficient for this third term.0935

In order to use this formula, I need to know two things: I need to know n, and I need to know k.0950

n is easy--we know n equals 7; k...if you look up here, k is the exponent of b for that term.0956

So, I am looking for the coefficient that goes right here, where n is 7 and k is 2.0969

Once I have that, it is a matter of using the formula: n!...this is going to be 7!/k!, which is 2!...times (7 - 2)!;0978

this equals 7! divided by 2!, times 5!.0995

I am going down here to complete this: 7 times 6 times 5 times 4 times 3, and so on.1002

In the denominator, I have 2!, 2 times 1, times 5 times 4 times 3 times 2 times 1 (factorials).1010

Again, I am going to cancel out common factors to make this much easier to work with.1023

This leaves me with 7 times 6, divided by 2; that is simply 42/2, or 21.1030

Therefore, the coefficient for the third term is actually 21.1043

Since I know that, and I know that we are going 1, 7, 21, starting from this end1052

(we are also going 1, 7, 21), I could do the same thing to find the fourth term coefficient.1057

The fourth term coefficient would have, again, an n equal to 7; but this time, k would be equal to 3.1067

Note one thing: instead of having to write all this out, to figure out what k is and everything,1077

you can just be aware that the term number equals k + 1.1082

Let's look at the fourth term: the term number is 4--that equals k + 1; k = 3.1092

For the fourth term, k = 3; so that is 1, 2, 3, 4; I know n = 7; and since this is the fourth term, I quickly know that k = 3.1101

So, you can use the binomial theorem, and specifically the binomial formula, to find a specific term of the expanded form of (a + b)n.1117

And I talked, in the last section, about how you can use that formula to find the whole expansion.1126

Or you may just need to find a certain coefficient within the expansion.1133

Recall that the binomial formula is equal to n!, divided by k!, times (n - k)!, where n is this n,1140

and k is the value of the exponent for the b in the term that you are looking for the coefficient of.1158

As an example, we are going to do something slightly more complex.1167

Instead of just a + b, we are actually going to say a - 2b; and we are going to expand that--it is going to be to the fifth.1169

So, we have to account for this -2--that the second term here is not just b; it is -2b.1178

I know that the first term is going to be a5.1188

I know that the second term is going to have a coefficient of n, so it is a coefficient of 5, and that a is going to decrease by 1.1193

I have to be very careful and not just write b; I have to write -2b.1200

Then, I have an unknown coefficient; but I do know that I am going to get a3, because the exponent that a is raised to will decrease by 1.1206

And here, I just have to say that b is really -2b; that is going to be squared; it is going to increase by 1.1216

I have another unknown coefficient, and it becomes a3 and goes down to a2; -2b squared goes up to -2b cubed.1226

Then, there is another coefficient here, and I get a3, -2b to the fourth, and then finally...1237

actually, this is just a; it goes from a2 to a; and then finally, a is gone, and I have -2b5.1247

By symmetry, I know the coefficient of these outer terms, right here, is actually 1; you can think of that as 1.1256

The two outer terms, then, have coefficients of 5.1266

All right, so back to finding a specific term: let's say that I want to find the coefficient for this third term.1270

n is going to equal 5; so in order to find the coefficient for this third term, I need to use the binomial formula; I need to have n, and I need to have k.1298

k...I could just say it is one less than the term number, so k is 2.1307

Or I could look up here and say it is whatever power this is raised to, which is 2.1311

Next, I insert these values into the formula; that is going to give me 5!, divided by 2!, times (n - 2)!.1318

This is equal to 5! divided by 2!, times 3!, equals 5 times 4 times 3 times 2 times 1, divided by 2! (2 times 1), times 3 times 2 times 1.1334

Cancel out common factors; that leaves me with 5 times 4, divided by 2 times 1, or 20 divided by 2, is 10.1354

Therefore, coefficient for the third term is equal to 10.1368

I can go back up and put this in here, but there is some simplifying that needs to be done.1374

So, the third term equals 10a3, (-2b)2; this equals 10a3...-2 times -2 is 4b2.1379

This equals 10 times 4, a3, b2, which equals 40a3b2.1398

It is a little bit more complicated than just putting the 10 in there.1408

And then, something to be careful of, as well: then I can say that by symmetry, this is also 10.1411

But I need to work with what is in here, which is actually a2(-2b)3, so it is somewhat different.1416

So, let's look at what this fourth term is: 1, 2, 3,...the fourth term.1425

Again, I am going to use that coefficient, 10, but here I have a2(-2b)3.1431

So, it is 10a2...-2 times -2 is 4, times -2 is -8; so it is times -8b3.1439

I can rewrite this as 10 times -8a2b3 = -80a2b3.1454

When I simplify that fourth term, this is what it is going to look like.1466

That was using the binomial formula to find a specific term of the expanded form.1471

In the first example, we need to expand (x + 2)4.1481

The general form is (a + b)n; here, I do the same thing, just realizing that a is equal to x, and b is equal to 2.1486

I am going to get something that looks like this: (a + 2)4; I am going to have five terms.1497

I am going to have n equal 4; so, there are going to be 5 terms.1502

The first term is going to be a2; in this case, it is x, so it is x2.1508

Excuse me, not squared; to the n--it is going to be a to the n, which is 4, so it is actually going to be x4.1517

In the next term, I have the coefficient; and the coefficient of the second term is equal to n.1525

So, that is going to be 4, times x...this exponent is going to decrease by 1...times 2.1531

The next term: I am going to have some unknown coefficient; x is going to go from cubed to squared;1543

2 is going to go from being the first power to the second power.1555

I am going to have another term, and then x2 is just going to become x; 2 is going to go from being squared to being cubed;1560

and then, finally, x drops out, and I get, for my last term, 2n, or 24.1572

By symmetry, I know that I am going to have a middle term (I have an odd number of terms--I have one middle term);1581

the two next to it are going to have the same coefficient--it is going to be 4.1587

I can simplify this later on, because I am going to have to deal with this 23 and this 2, and multiply that times the coefficient.1591

But right now, let's just leave it like this.1600

I am almost done with my expansion; for this middle term, there are two ways I could go, because this is not that large of an expansion.1603

I could use the binomial formula, and we are going to get more practice with that in a minute.1612

But this time, I am actually going to use Pascal's triangle.1615

Remember that it starts out with (a + b)0, and you end up with 1 as the only coefficient, because this equals 1.1617

Then, I get (a + b)1 = (a + b); that gives me 1 and 1 as coefficients.1632

Once I have those two, I can find the rest, because I know I have 1's out here.1639

And to find the other terms, I add the two above: 1 + 1 gives me 2.1645

So, for (a + b)2, these are my coefficients; then I want (a + b)3.1651

And what I am finally looking for, to find this coefficient, is when (a + b) is expanded to the fourth.1659

So, I have a 1 out here and a 1 out here; 1 + 2 gives me 3; 2 + 1 gives me 3;1667

so the coefficients for this expansion (that is going to have 4 terms) are going to be 1, 3, 3, 1.1674

Finally, the one I am looking for: I have a 1 and a 1, which I have up here;1680

1 + 3 is 4; 1 + 3 is 4, which I already knew; this gives me my middle coefficient of 6.1685

Now, I could have used the binomial formula to find this; but I wanted1695

to just do it a little bit differently this time, since it was a fairly small expansion, and use Pascal's triangle.1699

Now, I have the expansion, but it is not really simplified; we need to simplify these various terms.1706

So, let's take...the first term is already simplified; the second term equals (here is the second term) 4 times x3 times 2; that equals 8x3.1712

The third term equals (that is 1, 2...the third term) 6x2, 22, equals 6x2 times 4, equals 24x2.1733

That is the third term; the fourth term equals 4x times 2 cubed; that is 4x times 8; 8 times 4 is 32, so that is 32x.1750

The fifth term equals 24; 23 is 8, times 2 again gives me 16.1764

Rewriting it all here, the expansion of (x + 2)4 is1774

x4 + (second term) 8x3 + (third term) 24x2 + (fourth term) 42x + (the final term) 16.1779

Again, I was able to expand this using my properties that I am aware of for the expansion of a binomial1794

to find what these exponents are, how many terms there are going to be, and what this coefficient and this coefficient are.1802

I was left with just finding one coefficient that I could have used Pascal's triangle for.1810

That is what I did; I found the 6 (or I could have used the binomial formula).1814

Once I had these coefficients, all I did was to simplify; and this is the expansion, right down here.1818

Here, I am asked to find the fourth term of the expansion of (a + 2b)8.1829

So, I am asked to find this specific term; and we know that we can use the binomial formula, n!/k! times (n - k)!.1834

I know that I have n; n = 4; what is k?1847

Well, the term number equals k + 1; the term number is 4, so this means 4 = k + 1.1853

This gives me k = 3; so I have n = 4; k = 3; I can find the coefficient.1866

I need to find the coefficient of this fourth term; it is going to have a coefficient.1875

For the first term, it is going to be a raised to some power; for the second component of it, it is going to be 2b (not just b), raised to some power.1884

How do I know what these powers are?1902

Well, recall that it is going to be equal to (let's give ourselves some more room)...1904

when you find a term used in a binomial formula, you also know that it is in the form n - k, 1910

and then the second part is raised to the k power, because that is what k is--1919

it is whatever you are raising b to; and in this case, b = 2b.1926

I also know that the sum of the exponents has to equal n; the sum of the exponents of a term equals n.1934

This makes sense, because n - k + k needs to equal n; the sum of the exponents must equal n, and that works out; so this makes sense.1945

So, the fourth term: I need to find the coefficient using the binomial formula.1954

I know what the power a is going to be raised to is, n - k; and I know that the power that 2b is going to be raised to is k.1960

Let's go ahead and find the coefficient: the coefficient of the fourth term is going to be equal to n!, which is 4!...1967

excuse me, n is actually 8--a correction on that: we are finding the fourth term, but n is right here, so n is actually 8.1981

So, it is n! (that is 8!), divided by k! (that is 3!); n (which is 8), minus 3, factorial, equals 8!/3! times 5!.1993

That gives me 8! divided by 3 times 2 times 1, times 5!.2013

I do my canceling to make things easier to work with.2026

This leaves me with 8 times 7 times 6, divided by 3 times 2 times 1.2032

6 is equal to 3 times 2, so I have more common factors and more canceling.2040

That just leaves me with 8 times 7, divided by 1, which is 56; that was the hardest part here--finding the coefficient.2045

The coefficient of this fourth term is equal to 56.2053

To write that out completely, I am going to write it out as 56an - k; that is 8 - 3;2058

times (2b) raised to the k power (k is 3); this equals 56a5(2b)3.2071

Simplifying further: 56a5...2 cubed is 8, times b3.2087

If you multiply out 56 times 8, you will get 448; a5, b3.2095

This is the fourth term of this expanded form of (a + 2b)8.2103

It is a little bit complicated: the first thing I had to do is find the coefficient.2111

And I did that by using my binomial formula and knowing that n = 8 and k = the term number, plus 1.2115

The term number plus 1 is k, so k is one less than the term number: k is 3.2128

I was able to find, then, using this formula, that the coefficient is 56.2133

The second thing I had to do is figure out what power I should raise a to, and what power I should raise 2b to.2139

Well, when I am using the binomial formula, I am finding the coefficient for a term in this form, which is an - k2bk.2145

So, I knew my n and my k; so I put my coefficient here and said, "OK, I have a8 - 3(2b)3," using this.2153

Simplify, and then multiply to get 448a5b3.2163

We are asked to expand (2x + 3y)5...this binomial to the fifth.2172

And the general form is (a + b)n.2178

So, here we actually have that a is equal to 2x, and b is equal to 3y; so that makes things a little bit more complicated at the end.2181

We have to take these into account when we simplify our final answer.2190

Since this is the fifth power, I have an n equal to 5.2197

I am going to have a total of 6 terms.2202

It is going to look something like this at the end: (2x + 3y)5 =...my first term is 2x raised to the n power,2205

which is the fifth; plus the coefficient, and then 2x is going to decrease to the fourth power, and I am going to have my 3y here.2214

Plus...some other coefficient; 2x becomes cubed, and 3y increases from an exponent of 1 to an exponent of 2.2225

Another coefficient is going to bring me to 2x23y3;2241

another coefficient...now I just have 2x, and 3y4; and then, finally, 3y5.2250

And I expect to have 6 terms--let's verify that I am not missing anything: 1, 2, 3, 4, 5, 6.2261

Down here at the bottom, I am going to start building up what our final answer is going to look like, because I have some simplifying that I need to do.2270

The first term, I can figure out: I already know that it is 2x5.2281

23 is 8, times 2 is 16, times 2 is 32; so that is 32x5.2286

The second term: recall that the coefficient for the second term is equal to n--that is 5.2293

Also, remember that symmetry is involved: since there are 6 terms (that is an even number of terms), I am going to have two middle terms.2301

These two are going to have the same coefficient; the two next to those will have the same coefficients; and the two outer have the same coefficients.2309

It is not the same value, once I simplify them; but the same coefficient.2319

So, since these two are the same, and these two are the same, this term also has a coefficient of 5.2323

Let's work this one out up here; this is the second term, and I know that it is equal to 5 times (2x)4 times 3y.2332

That is...2 times 2 is 4; 2 cubed is 8, times 2 is 16; so that is times 16x4, times 3y.2346

Multiplying 5 times 16 times 3 is going to give you 240x4y; I am going to put that in here; that is 240x4y.2358

I don't know what I have for these two middle ones, but it is going to be something here, and it is going to be x3y2.2373

Something is here, then x2y3; I can figure this one out:2384

right here, this is the third term; it equals 5 times 2x, times (3y)4.2394

5 times 2x is just 10x; 3 times 3 is 9, times 3 is 27, times 3 is going to give me 81.2406

So, that is times 81y4; so this one is easy, because it is just 81 times 10, is 810xy4.2419

So, that was...actually not the third term: correction--we don't know the third term; the first, second, third, and fourth...2436

the two middle terms we don't know; we do know the fifth term--this is the fifth term.2448

All right, so the fifth term I can fill in: 810xy4.2456

The sixth term, right here, I can also fill in: the sixth term equals (3y)5.2465

33 is 27; 34 is 81; 35, if you figure it out, is 243; 243y5.2475

Just using the properties of binomial expansions, I got most of the terms.2489

And since there is symmetry, the two middle terms will have the same coefficient inserted up here.2496

And then, I can work out the rest and finalize it.2505

So, I only need to find one coefficient, using the binomial formula.2507

The binomial formula, recall, is n!/k!, times (n - k)!.2511

I know that n = 5; for the first, second, third term, k is equal to 2; you could just say "Third term: 3 - 1 is 2";2521

or you could look up here and say, "OK, 3y is raised to the second power, so k = 2."2532

Therefore, I am going to have 5!/2!, times (5 - 2)!; this equals 5!/2! times 3!,2537

which equals 5 times 4 times 3 times 2 times 1; 2 times 1; 3 times 2 times 1.2559

1, 2, and 3 all cancel; this gives me 5 times 4, divided by 2; that is 20/2, or 10.2568

So, the missing coefficient here is 10; by symmetry, the missing coefficient right here is also 10; I am almost done.2578

So, for the first, second, third term, I have my coefficient; but I need to simplify that a bit.2588

My third term equals 10 times (2x)3 times (3y)2.2595

This is going to be equal to 10 times...2 cubed is 8x3; 3 squared is 9y2.2605

This is going to give me 80 times 9x3y2, which is just 720x3y2.2617

That is my third term, so I am going to put a 720 right here.2625

The fourth term: I also have a 10 right here.2630

What is slightly different, though: it is 10 times (2x)2(3y)3.2635

This is going to give me 10 times 4x2; 3 cubed is 27, so it is y3.2642

This is 40 times 27...let's put all of the constants together...x2y3.2651

40 times 27 actually comes out to 1080; you can multiply that out for yourself; x2y3.2663

The fourth term, therefore: I am going to put a 1080 right here.2673

So, this is quite a bit of work; but we actually only needed to find one coefficient using the binomial formula,2676

because by symmetry, these two middle ones were the same (the two middle coefficients that we will put in the blank);2682

and I knew that the second and the second-to-the-last had coefficients equal to n.2690

Once I found that a 10 went into these blanks, all I needed to do was work out what each of these terms would simplify to,2695

to end up with the expansion that I found down here.2703

We are asked to find a specific term this time, not the entire expansion.2710

I just want to find the fourth term of (x - 3y)7, using the binomial formula, n!/k!, times (n - k)!.2713

All right, n = 7; what does k equal? Well, it is the fourth term; k is going to be one less than that; 4 - 1...I am going to have k = 3.2725

This is going to give me the coefficient for a term in this form.2737

It is going to have a coefficient, and then it is going to have x raised to the n - k power, times -3y, raised to the k power.2741

This is going to become x7 - 3, (-3y)3.2757

Therefore, I am going to get some coefficient (that I am going to find in a moment), times x4, times (-3y)3.2768

Let's go ahead and find the coefficient: this missing coefficient equals n!, 7!, divided by k!, times (7 - 3)!;2775

that equals 7!/3!, times 4!, equals (let's write this out)...divided by 3 times 2 times 1, times 4 times 3 times 2 times 1.2793

Do some canceling out: 1's, 2's, 3's, and 4's are the same.2812

This leaves me with 7 times 6 times 5, divided by 3 times 2 times 1.2819

6 has common factors here of 3 and 2; I can just cancel those out.2825

So, this is 7 times 5, divided by 1; so just 7 times 5 equals 35.2832

I know that my coefficient is, then, 21 that is what is going to go in this blank.2838

So, I am going to end up with...the fourth term is going to be 35 (I am going to put 35 right here in this blank)2842

x4(-3y)3, which equals 35x4...-3 squared is 9, times -3 again is -27.2849

So, it is -27y3; so it is time to get out your calculators and figure out 35 times -27, or multiply it out, if necessary.2864

And you will find that that equals -945x4y3.2875

OK, we found the fourth term of this expansion by using the binomial formula to find the coefficient,2881

and also, an awareness that the exponent of this x is going to be n - k; the exponent for the -3y is going to be k.2887

We figured out the coefficient (it is 35), and then simplified, using our rules for working with exponents.2896

That concludes this session on the binomial theorem here at Educator.com; thanks for visiting!2905

Welcome to Educator.com.0000

For today's Algebra II lesson, we are going to be discussing relations and functions.0002

And recall that some of these concepts were discussed in Algebra I, so this is a review.0008

And if you need a more detailed review, check out the Algebra I lectures here at Educator.0013

Beginning with the concept of the coordinate plane: the coordinate plane describes each point as an ordered pair of numbers (x,y).0021

The first number is the x-coordinate, and the second is the y-coordinate.0031

For example, consider the ordered pair (-4,-2): this is describing a point on the coordinate plane with an x-coordinate of -4 and a y-coordinate of -2.0036

Or the pair (0,2): the x-value would be 0, and the y-value would be 2; this is the point (0,2) on the coordinate plane.0057

Or (3,5): x is 3; y is 5.0073

Also, recall that the quadrants are labeled with the Roman numerals: I, and then (going counterclockwise) quadrant II, quadrant III, and quadrant IV.0080

In the coordinate pairs in the first quadrant, the x is positive, as is the y.0097

In the second quadrant, you will have a negative value for x and a positive value for y, such as (-2,4)--that would be an example.0103

In the third quadrant, both x and y are negative; and then, in the fourth quadrant, x is positive; y is negative.0114

And we will be using the coordinate plane frequently in these lessons, in order to graph various equations.0127

Recall that a relation is a set of ordered pairs.0135

The domain of the relation is the set of all the first coordinates, and the range is the set of all the second coordinates.0140

A relation is often written as a set of ordered pairs, using braces to denote that this is a set,0149

and then the ordered pairs, each in parentheses, separated by a comma.0158

Sometimes, the relation is represented as a table; so, -2, 1; -1, 0; 0, 1; and 1, 2.0172

We will be doing some graphing of relations also, in just a little bit.0192

So, as discussed up here, the domain is the set of all first coordinates.0196

And the set of all first coordinates here would be {-2, -1, 0, 1}.0207

The range is the set of the second coordinates; so the range right here--all the y-values--is {1, 0, 1, 2}.0219

However, you don't actually need to write the 1 twice; so in actuality, it would be written as such.0232

It is OK to repeat the values if you want, but usually, we just write each value in the domain or range once; each is represented once.0245

Functions are a certain type of relation: so, all functions are relations, but not all relations are functions.0261

A function is a relation in which each element of the domain is paired with exactly one element of the range.0272

For example, consider the relation shown: {(1,4), (2,5), (3,8), (4,10)}.0280

Each member of the domain corresponds to exactly one element of the range.0298

We don't have a situation where it is saying {(1,4), (1,6), (1,8)}, where that member of the domain is paired with multiple members of the range.0305

Another way, again, to represent this as a table--another method that can be used--is mapping.0314

And mapping is a visual device that can help you to determine if you have a function or not,0321

by showing how each element of the domain is paired with an element of the range.0327

A map would look something like this: over here, I am going to put the elements of the domain, 1, 2, 3, and 8;0333

over here, the elements of the range: 4, 5, 8, and 10.0347

And then, using arrows, I am going to show the relationship between the two.0355

So, 1 corresponds to 4 (or is paired with 4); 2 to 5; 3 to 8; and (this should actually be 4) 4 to 10.0360

OK, so as you can see, there is only one arrow going from each element of the domain to each element of the range.0374

And that tells me that I do have a function.0383

Let's look at a different situation, using a table form: let's look at a second relation.0386

In this one, I am going to have (-2,2), (-3,2), (-4,5), and (-6,7).0392

And I am going to go ahead and map this: -2, -3, -4, and -6: these are my elements of the domain.0403

For the range, I don't have to write 2 twice; I am just going to write it once; 5, and 7.0416

OK, -2 corresponds to 2; -3 also corresponds to 2; -4 corresponds to 5; and -6 corresponds to 7.0422

This is also a function, so both of these are relations, and they are also functions.0436

It is OK for two elements of the domain to be paired with the same element of the range; this is allowed.0445

What is not allowed is if I were to have a situation where I had {1, 2, 3}, {4, 5, 6}; and I had 1 paired with 4, and 1 paired with 5.0452

So, if you have two arrows coming off an element of the domain, then this is not a function.0468

Here are two examples of relations that are also functions.0476

There is a specific type of function that is called a one-to-one function.0482

And a function is one-to-one if distinct elements of the domain are paired with distinct elements of the range.0486

In the previous example, we saw a situation where we did have a one-to-one function, and another situation where we did not.0494

OK, so to review: the ordered pairs in that first function that we just discussed were (1,4), (2,5), (3,8), and (4,10).0501

OK, and we can use mapping, again, to determine what the situation is with this relation (which is also a function).0519

The domain is {1, 2, 3, 4}; and the range is {4, 5, 8, 10}.0529

When I put my arrows to show this relationship, you see that distinct elements of the domain are paired with distinct elements in the range.0540

1 is paired with 4; they are each unique--each pair is unique.0552

Looking at the other function that we discussed: the pairs are (-2,3), (-3,2), (-4,3)...slightly different, but the same general concept...slightly different, though.0558

OK, here I have -2, -3, -4, and -6; over here, in the range, I have 3, 2...I am not going to repeat the 3--I already have that...and then 7.0587

-2 corresponds to 3; -3 corresponds to 2; -4 also corresponds to 3; -6 corresponds to 7.0605

This is still a function; OK, so these are both functions: function, function.0620

However, this is a one-to-one function; this is not one-to-one.0628

They are both functions, since each element of the domain is paired only with one element of the range.0639

But in this case, it is not a unique element of the range: these two, -2 and -4, actually share an element of the range.0648

In other words, this is unique; it is a one-to-one correspondence.0656

OK, we can graph relations and functions by plotting the ordered pairs as points in the coordinate plane, as discussed a little while ago.0662

There are a couple of types of graphs that you can end up with.0673

The first is discreet, and the second is continuous; let's look at those two different types.0676

Consider this relation: OK, so if I am asked to graph this relation, I am going to graph each point:0682

(-4,-2): that is going to be right here; (-2,1)--right here; here, (0,2), 0 on the x, 2 on the y.0698

This is a discrete function--discrete graph--discrete relation.0714

This actually is both a relation and a function; so it is a discrete relation or a discrete function.0721

And the reason is because I have a set of discrete points; they are not connected.0727

And I can't connect them, because I haven't been given anything in between, or a way to know if or what lies in between these.0734

I can't just connect them when I don't know; there could be a point up here, or actually this is just the entire relation.0741

So, I can just work with what is given.0747

OK, a different scenario would be if I am given a relation y=x+1.0749

And I can go ahead and plot this out, if I say, "OK, when x is -1, -1+1 is 0; when x is 0, y is 1; when x is 1, 1+1 is 2; when x is 2, y is 3."0759

OK, so I am going to go ahead and plot this out.0779

When x is -1, y is 0; when x is 0, y is 1; when x is 1, y is 2.0781

Let's remove this out of the way.0790

When x is 1, y is 2; when x is 2, y is 3.0793

Now, I have a set of points, because these are the points I chose.0798

But because I am given this equation, there is an infinite number of points in between.0801

I could have chosen an x of .5 to get the value 1.5 here, to fill that in--and on and on, until this becomes continuous and forms a line.0807

So, this is a continuous function: the graph is a connected set of points,0821

so the relation (or the function in this case, since we do have a function) is continuous relation or continuous function,0834

because I have a line; whereas this, which is a set of points, is a discrete relation.0841

OK, one visual way to tell if a relation is a function is using the vertical line test.0849

And a relation is a function if and only if no vertical line intersects its graph at more than one point.0856

This is most easily understood through just working through an example.0866

Consider if you were given the following graph.0870

OK, the vertical line test: what you are seeing is, "Can you put a vertical line0874

somewhere on the graph so that it intersects the graph at more than one point?"0880

And I can: I put a vertical line here, and it intersects this graph at 1, 2, 3 places.0887

Over here, it only intersects at one place; that is fine; but if I can draw a vertical line anywhere on the graph0897

that intersects at more than one place, then we say that this failed the vertical line test.0903

And when something fails the vertical line test, it means that it is not a function.0912

The reason this works is that, if two or three or more points share the same x-value,0919

then they are going to lie directly above or below each other on the coordinate plane.0926

For example, looking right here, I have x = 3; x is 3; y is 0.0931

Then, I look right above it, up here: again, x is 3, and y is...say 2.1--pretty close.0940

Then, I look up here; again, x is 3, and y is about 4.6, approximately.0950

So, when x-values are the same, but then the y-values are different, that is telling me0959

that members of the domain are paired with more than one member of the range; by definition, that is not a function.0968

OK, consider a different graph--consider a graph like this of a line--a straight line.0975

OK, now, anywhere that I pass a vertical line through--anywhere on this graph--it is only going to intersect at one point.0984

So, this passed the vertical line test.0996

Therefore, this line, this graph, represents a function.1006

OK, so the vertical line test is a visual way of determining if a relation is a function.1014

Working with equations: an equation can represent either a relation or a function.1024

If an equation represents a function, then there is some terminology we use.1031

And let's start out by just looking at an equation that represents a function.1036

The variable corresponding to the domain is called the independent variable, and the other variable is the dependent variable.1043

So, here I have x and y; and let's look at some values--let's let x be -1.1050

Well, -1 times 2 is -2, minus 1--that is going to give me -3.1057

When x is 0, 0 times 2 is 0, minus 1 is -1.1063

When x is 1, 1 times 2 is 2, minus 1--y is 1.1071

When x is 2, 2 times 2 is 4, minus 1 gives me 3.1077

So, looking at how this worked, x is the independent variable.1084

The value of x is independent of y; I am just picking x's, and here it could be any real number.1099

We sometimes also say that this is the input; and the reason is that I pick a value for x (say 0),1109

and I put it in--I input it into the equation; then, I do my calculation, and out comes a y-value.1116

So, the value of y is dependent on x; therefore, it is the dependent variable; and we also sometimes say that it is the output.1127

You put x in and do the calculation; out comes the value of y; so x is independent, and y is dependent.1141

The notation that you will see frequently in algebra is function notation.1151

We have been writing functions like this: y = 4x + 3; but you will often see...1157

instead of an equation written like this, if it is a function, you will see it written as such.1165

And when we say this out loud, we pronounce it "f of x equals 4x plus 3."1170

And we are talking about the value of a function for a particular value of x.1179

So, we say, "The function of f at a particular x."1186

Let's let x equal 3; then, we can talk about f of 3--the value of the function, the value of y,1193

of the dependent variable, when the independent variable, x, is 3.1204

And in that case, since it is telling us that x is 3, I am going to substitute in 3 wherever there is an x.1209

And I could calculate that out to tell me that f(3) is...4 times 3 is 12, plus 3...so f(3) is 15.1217

Here, x is an element of the domain, of the independent variable; f(x) is an element of the range.1231

So again, we are going to be using this function notation throughout the remainder of the course.1246

Looking at the first example: the relation R is given by this set of coordinate pairs.1253

Give the domain and range, and determine of R is a function.1260

Well, recall that the domain is comprised of the first element of each of these coordinate pairs.1267

So in this case, the domain would be {1, 2, 6, 5, 7}.1275

The range: the range is comprised of the second element of each ordered pairs, so I have 4, 3...4 again;1291

I don't need to write that again; 3--I have 3 already; and 5; I am just writing down the unique elements.1303

This is the domain, and this is the range.1309

Now, is R a function? Well, I can always use mapping to just help me determine that.1311

And I am going to write down my members of the domain, and my elements of the range.1320

And then, I am going to use arrows to show the correspondence between each:1330

1 and 4--1 is paired with 4; 2 is paired with 3; 6 is also paired with 4; 5 is paired with 3; and 7 is paired with 5.1333

Now, I am looking, and I only see one arrow leading from each element of the domain.1350

There is no element of the domain that is paired with two elements of the range.1354

So, in this case, this is a function; so, is R a function? Yes, this relation is a function--R is a function.1359

So, always double-check and make sure you have answered each part.1373

I found the domain; I found the range; and I determined that R is a function.1375

OK, the relation R is given by the equation y=2x2+4; is R a function?1384

What are the domain and range? Is R discrete or continuous?1393

Let's just look at some values for x and y to help us determine if this relation is a function.1400

If I let x equal -1, -1 times -1 is 1, times 2 is 2, plus 4 is 6.1412

OK, if x is 0, this is 0, plus 4--that gives me 4.1422

If x is 3, 3 squared is 9, times 2 is 18, plus 4 is 22.1428

So, as you are going along, you can see that, for any value of x, there is only one value of y; therefore, R is a function.1434

What is the domain? Well, I could pick any real number for an x-value that I wanted, so the domain is all real numbers.1450

You might, at first glance, say, "Oh, the range is all real numbers, as well"; but that is not correct, because look at what happens.1467

Because this is x2, whenever I have a negative number, it becomes positive; if I have a positive number, it stays positive, of course.1475

Therefore, if I have, say, -1, that becomes 1; this becomes 6.1486

So, I am not going to get any value lower than...for y, the smallest value I will get is for when x is 0.1493

OK, so if x is 0, y is 4; because -1 is going to give me a bigger value--it is going to give me 6.1505

If I do -2, that is going to be 4 times 2 is 8, plus 4 is 12.1510

So, the lowest value that I will be able to get for y will occur when x is 0.1516

And that is going to give me a y-value of 4.1522

Therefore, the range is that y is greater than or equal to 4.1525

So, the most difficult part of this was just realizing that the range is not as broad as it looked initially.1530

Because this involves squaring a number, there is a limit on how low you are going to go with the y-value.1538

So, this is a range with a domain of all real numbers, and a range of greater than or equal to 4.1545

OK, in Example 3, graph the relation R given by 2x - 4y = 8.1559

Is R a function? Find its domain and range. Is R discrete or continuous?1572

OK, so graph the relation given by 2x - 4y = 8.1583

Let's go ahead and find some x and y values, so that we can graph this.1590

When x is 0, we need to be able to solve for y; when x is 0, let's figure out what y is.1598

0 - 4y equals 8; therefore, y equals -2 (dividing both sides by -4).1607

OK, when x is 2, 2 times 2 minus 4y equals 8; that is 4 minus 4y equals 8; that is -4y equals 4; y = -1.1615

And let's do one more: when x is -2, this is going to give me -4 - 4y = 8.1635

That is going to then give me, adding 4 to both sides, -4y = 12, or y = -3; that is good.1646

All right, so when x is 0, y is -2; when x is 2, y is -1; when x is -2, y is -3, right here.1657

I am asked to graph it; and I have some points here that I generated,1686

but I also realize that I could have picked points in between these, which would actually end up connecting this as a line.1691

So, I am not just given a set of ordered pairs; I am given an equation that could have an infinite number of values for x,1702

which would allow me to graph this as a continuous line.1707

Therefore, I graphed the relation...is R a function?1713

Is R discrete or continuous? Well, I have already answered that--seeing the graph of this, I know that this is continuous.1720

And let's see, the next step: is R a function?1731

Yes, it is a function, because if I look, for every value of x (for every value of the domain), there is one value only of the range.1746

So, every element of the domain is paired with only one element of the range.1758

It is continuous, and it is a function.1762

Find the domain and range: well, this is another case where I could choose x to be any real number, so it would be all real numbers--any real number.1765

Here, the situation is the same for the range--all real numbers.1778

Depending on my x-value, I could come up with infinite possibilities for what the range would be, what the y-value would be.1784

R is a function; its domain and range are all real numbers; and this is a continuous function.1792

OK, in Example 4, we are given f(x) = 3x2 - 4, and asked to find f(2), f(6), and f(2k).1805

First, f(2): recall that, when you are asked to find a function for a particular value of x,1824

you simply substitute that value for x in the equation; so f(2) equals 3(4) - 4, so that is 12 - 4; so f(2) = 8.1834

Next, I am asked to find f(6), and that is going to equal 3(62) - 4.1852

f(6) = 3(36) -4, and that turns out to be 108 - 4, so f(6) is 104.1860

Now, at first, this f(2k) might look kind of difficult; but you treat it just the same as you did with the numbers, when x is a numerical value.1877

Everywhere I see an x, I am going to insert 2k.1887

And figuring this out, 2 times 2, 2 squared, is 4; k times k is k2.1891

3 times 4 is 12, so I have 12k2 - 4; so f(2k) = 12k2 - 4.1902

So again, if you are asked to find the function of a particular value of x, you simply substitute whatever is given, including variables, for x.1910

That concludes this lesson of Educator.com; I will see you back here soon!1919

Welcome to Educator.com.0000

In today's lesson, we will be covering linear equations.0002

Again, this is some review from Algebra I, which we will discuss here.0008

And if you need further detail, go back to the Algebra I lectures and check those out for this subject.0011

So, a linear equation is an equation of the form ax + by = c, where, a, b, and c are constants.0018

For example, 2x + 5y = 3, and here a = 2, b = 5, and c = 3.0028

Another example could be x - 7y = 4; here, even though it is not written, there actually is a 1 in front of the x.0041

Just by convention, we don't write it out when the coefficient is 1.0052

So, a = 1, b = -7, and c = 4.0055

We will talk a little bit more about what this form is.0062

A linear equation represents a function, which is known as a linear function; and that can be written in the form f(x) = mx + b.0067

And this is a very useful form of the equation that will be discussed in another segment.0078

And an example of this would be something such as f(x) = 4x + 2; you may also see it written as y = 4x + 2.0084

And here, m = 4; b = 2; and these two numbers represent certain elements that we will talk about in a few minutes.0107

OK, so the first form of the linear equation that we talked about just a minute ago is also called standard form.0118

So, a linear equation is in standard form if it is written in the form ax + by = c,0129

where a, b, and c are integer constants (and this is important) with no common factor.0135

If a common factor remains, it is not in standard form.0140

For example, if I have an equation 9x + 4y = 10, this is in standard form.0144

I have ax (9 is a), plus by (b is 4), equals c (c is 10), and there is no common factor.0156

Consider another possible equation: 6x + 8y = 12; this is not in standard form, because 6, 8, and 12 have a common factor.0168

What I need to do is divide both sides of this equation by 2 to get 3x + 4y = 6; now, it is in standard form.0189

Another example would be if I had 7y + 5 = -x; again, it is not in standard form.0207

I can add x to both sides to get x + 7y + 5 = 0, and then subtract 5 from both sides to get x + 7y = -5; this is standard form.0218

We can graph linear equations using the intercept method.0246

Recall that the point where a line crosses the x-axis is its x-intercept.0250

And the point where the graph crosses the y-axis is the y-intercept.0259

For example, if I had a line such as this, first the x-intercept is right here; the x-intercept is -2.0265

And at the x-intercept, y is always going to be 0; so, the coordinate pair would be (-2,0), because it crosses the x-axis, and so y is 0.0283

The y-intercept is right here: y equals 4, and at this point, x is going to equal 0, so (0,4) is the y-intercept.0294

And we can use this knowledge in order to graph a linear equation.0308

For example, if I have a linear equation 2x + 3y = 6, all I need in order to plot this are two points on the line.0314

Now, I can easily get two points by finding the x- and y-intercepts.0328

I know that, with the y-intercept, x will be 0; so if I let x equal 0, that is going to give me 0 times 2 is 0,0334

so that will be 3y = 6; 6 divided by 3--y equals 2.0343

OK, to find my second point, I am going to find the y-intercept (this is actually the x-intercept right here)...0.0354

Excuse me, this is the y-intercept right here; now I am looking for the x-intercept.0368

The x-intercept is the point at which y equals 0, and the line of the graph crosses the x-axis.0375

So, this time, I am going to let y equal 0; so I am going to have 2x, plus 3 times 0, equals 6, or 2x + 0 = 6.0383

So then, I end up with x = 3; this is the x-intercept.0395

Now, I have two points; I can plot the line.0400

At the y-intercept, x is 0; y is 2; the y-intercept is right here, at (0,2); the x-intercept is right here at (3,0).0405

Now, I have two points; and when I have two points, that means I can connect them to form a line.0422

And I did that by finding my x-intercept at (3,0) and my y-intercept at (0,2).0431

The technique is to find the x- and y- intercepts by letting x equal 0, and putting that into the equation,0449

and solving for y to get the y-intercept, and letting y equal 00458

to find the x-intercept, and then using those two points to plot the line.0467

In the first example, is the function linear? Well, recall that a linear function can be written in standard form.0475

And standard form is ax + by = c: I look at what I have here, and I have ax (so it seems like I am going along OK),0482

minus by (or that would be the same as plus -by, so that is fine), equals c; but then I have this over here.0492

And this is not part of standard form.0500

And you could attempt to get rid of that; maybe you say, "OK, I will multiply both sides by x to get rid of it."0502

But then, see what happens: all right, let's see what happens if I try to multiply both sides by x to get rid of this x in the denominator.0508

OK, I end up with 2x2 - 3xy = 4x + 1; so I am no better off.0518

I still don't have it in standard form; there is no way for me to get this in standard form; therefore, this is not a linear function.0530

It is not in standard form; I can't get it into standard form.0546

OK, is this function linear? Well, recall that we talked earlier about the form g(x) = mx + b.0550

And remember, g(x) is the same idea as f(x): you can use different letters, so it is still just that we are looking for this form for a linear function.0566

Well, I am looking at this, and I have what looks like an mx and a b, but the problem is that I also have this.0580

And this is not a part of the form for a linear function--this x2 term--therefore, this is not a linear function.0588

Once again, I can't get it into the right form for a linear function.0598

OK, in Example 3, we are asked to write the equation in standard form.0604

And recall that standard form is ax + by = c.0608

And I am dealing with some fractions, and I need to get rid of those in order to get this in standard form.0613

I can do that by multiplying first both sides of the equation by 3; this is going to give me 3y/6 =...here, the 3's will cancel out; -7 times 3--that is -21.0620

I can simplify this to y/2 = x - 21; so, I am still left with a fraction.0637

I am going to multiply both sides of the equation by 2 to eliminate that fraction.0644

The 2's cancel out, so I have y = 2x...-21 times 2; that is -42.0652

Next, I am going to subtract 2x from both sides to get the x on the left side of the equation.0665

And just switching around the order of the x and y to match this: -2x + y = -42.0677

And you could stop here, or you could multiply both sides by -1; and that is more conventional, to have the a term be positive.0688

So, that is going to leave me with 2x - y = 42.0698

I started out with this equation, and I wanted to get it into this form.0707

And I could achieve that by first multiplying by 3 to get rid of this fraction, then (I ended up with this)0711

multiplying by 2 to get rid of the y fraction, then simply subtracting 2x from both sides to get the 2x (the ax)0721

on the left side of the equation, and finally just a little more work to clean this up, multiplying both sides by -1.0730

This is the equation in standard form, where a equals 2, b equals -1, and c equals 42.0738

Example 4: Find the intercepts and graph the equation: 4x - 5y = 20.0748

We already talked about how, with the x- and y-intercepts, you can find those, and then you will have two points, and you can graph an equation.0756

So, first, in order to find the y-intercept, I am going to let x equal 0; this will give me the y-intercept, the point at which the line crosses the y-axis.0766

So, to find the y-intercept, I am going to substitute 0 for x.0779

This is going to give me 0 - 5y = 20, or -5y = 20; divide both sides by -5 to get y = -4.0788

OK, next I want to find the x-intercept: to find that, I am going to let y equal 0.0801

I am going to go back and do this again, this time letting y equal 0.0810

This is for the y-intercept, and this is for the x-intercept.0821

This is going to give me 4x - 0 = 20, or 4x = 20, which is x = 5.0824

Now that I have my x- and y-intercepts, I actually have two points on the line.0834

My first point is (0,-4), the y-intercept; my second point is at (5,0), and that is the x-intercept.0839

OK, connecting these two points gives me the graph of this equation.0859

We use the intercept method in order to graph, finding the x- and y-intercepts,0869

plotting those out on the coordinate plane, and then using those to form a line.0874

That concludes this lesson for Educator.com; and I will see you here again soon.0880

Welcome to Educator.com.0000

In today's lesson, we will be discussing slope.0002

Recall the definition of slope: the slope, m, of the line passing through a point (x1,y1),0007

and another point (x2,y2), is given by:0015

the slope equals the change in y, over the change in x (the change in the y coordinates, over the change in the x coordinates).0020

Therefore, if you have any two points on a line, you can use that to find the slope.0030

For example, imagine that you are given the graph of a line that looks like this, and you are asked to find the slope.0037

Well, I have a point right here at x is 0, so my (x1,y1) point is going to be x is 0, y is 2.0053

I have a second point over here, and I am going to call this (x2,y2); and here, x is -2, y is 0.0065

So, I am using the intercepts; but I could have used any other point on this line.0076

OK, an important point is that I could have assigned either one of these as...0081

I could have said this is (x1,y1), and this is (x2,y2).0086

It doesn't matter, as long as you are consistent (I can't say (x1,y2), (x2,y1)).0090

As long as I am consistent, and I follow that consistency when I am subtracting, I am fine.0096

So, I arbitrarily assigned this as (x1,y1), and this point as (x2,y2).0103

So, I found two points on the line; now, my y2 value is 0, and I am going to subtract my y1 value (which is 2) from that.0107

The x2 value is -2; my x1 value is 0, so minus 0; so here, the slope equals -2 over -2, or 1.0120

So, I could have used any two points on this line, and found the difference between those x and y coordinates--the change in y over the change in x.0138

So, what this is telling me is that...we say m = 1, but what we really mean is m = 1/1; think about it that way, and say that.0147

For every increase in y by 1, I am going to increase x by 1; increase y by 1, increase x by 1.0156

So, I may be given a line and asked to find the slope; or I may directly be given a set of points and asked to find the slope.0167

So, if I were given two points on the line, but not given the line itself, or a graph, or anything, I could easily find the slope.0175

This is telling us the vertical change over the horizontal change.0182

OK, some cases to keep in mind when you are dealing with slope: a horizontal line has a slope of 0.0187

And let's think about why this is so: consider a horizontal line up here at, say, 4; OK, so this is 4.0198

And if I look at any point--for example, right here--here, x is 1; y is 4; and then maybe I look over here at this point; here x is 3; y is still 4.0212

Now, recall that the slope formula is the change in y over the change in x.0229

So, I have my two points; and I am going to call this (x1,y1), (x2,y2).0239

Again, I could have done it the other way around.0246

All right, so I have y2 (that is 4), minus y1;0249

and then I have my x2 (which is 3), minus x1 (which is 1).0255

It actually doesn't matter what I have down here, because, since I have a horizontal line, the y-values everywhere here are going to equal 4.0261

So, 4 - 4 is going to be 0; therefore, the slope is 0.0273

So, a horizontal line has a slope of 0, because the y-values are the same at every point, so the difference in y is going to be 0.0281

OK, vertical lines are also a special case: and they have an undefined slope.0291

For example, let's look at a vertical line right here at x = -3.0298

OK, if I look right here at (I could pick any point)...I am going to pick this point, x is -3; y is 1.0306

Then, I will pick another point right here; let's say x is -3; y is -2.0315

So, I have my (x1,y1), and my (x2,y2).0322

Again, slope is change in y over change in x; change in y is y2 (that is -2), minus 1, over change in x (-3 - -3).0330

This is going to give me -3 over...-3 minus -3 is -3 plus 3; that is over 0.0349

And this is not allowed; this is undefined.0359

We say that a vertical line has an undefined slope because, since the change in x is 0,0367

because x is the same at every point, what you end up with is a denominator that is 0, and that is undefined; that is not allowed.0372

So, for a horizontal line, the slope is 0; for a vertical line, slope is undefined.0379

A couple of other things to keep in mind regarding slope: one is that in a line that rises to the right (such as this one), m is a positive number.0393

So, it is increasing to the right; the values are increasing.0407

If you have a line that falls to the right, such as this one, here the slope is positive; in this case, the slope is negative.0413

Rises to the right--slope is positive; falls to the right--slope is negative.0428

A horizontal line has a slope of 0; a vertical line has an undefined slope.0433

Parallel lines: two lines are parallel if and only if they are vertical or have the same slope.0439

Let's first check out the case where the lines are not vertical, but they are still parallel.0446

If I have two lines that are parallel, it is going to look like this, for example.0451

These are going to have the same slope; so if this is my line 1 and line 2, m1 will equal m2--the slopes will be equal.0459

And that is because they are changing at the same rate--these two lines are changing at the same rate, so they never intersect.0468

Now, we set vertical lines aside; we put them separately; we don't say they have the same slope.0476

And remember that the reason is because a vertical line has an undefined slope.0483

So, I can't say these two have the same slope, because their slope is undefined.0492

However, we recognize that they are still parallel; so we say that two lines are parallel0496

if they are vertical (that is a separate case), or if they have the same slope; but both of these represent parallel lines.0501

Perpendicular lines: two non-vertical lines are perpendicular if and only if the product of their slopes is -1.0512

So, "perpendicular" would tell me that these two lines intersect at right angles.0522

And if this is my line 1 and line 2, and then I have a slope m1 and m2, what this is stating0535

is that the product of m1 and m2 is -1.0544

Let's look at an example: let's say I have an m1 that is equal to 4, and I say that it is perpendicular to a line L2.0552

So, this is line 1, and this is line 2--the perpendicular line.0569

I can find the slope of the perpendicular line, because I know that m1 times m20575

is -1 (since these are perpendicular lines), and I am given m1.0584

So, 4 times m2 equals -1; therefore, the perpendicular line would have a slope of -1/4.0591

So, the perpendicular line has -1/4 as the slope, and that is the negative reciprocal of m1.0606

Knowing the relationship between parallel lines and the slopes of parallel lines and the slopes of perpendicular lines0618

can be helpful in answering problems and graphing lines; and we will see that right now with the examples.0624

OK, in this first example, find the slope of the line passing through the points (-9,-7) and (-6,-3).0633

Well, the definition of slope is the change in y over the change in x.0643

And I could choose to assign this either way; but I am just going to go ahead0651

and say this is (x1,y1), and this is going to be my (x2,y2).0654

And it doesn't matter which way you assign, as long as you are consistent.0660

So, I could have called this (x2,y2), and it would have been fine; we would have gotten the same answer.0663

OK, y2 is -3, and y1 is -7, so minus -7; over x2 (which is -6), minus x1 (which is -9).0668

Simplifying this: this is -3; a negative and a negative gives me a positive; and that gives me 4/3.0686

So, the slope of a line passing through these points is 4/3, just using my slope formula.0697

OK, Example 2: Graph the line passing through the point (-2,-1) with a slope of -2/3.0705

Well, I am given the slope, and I am given a point on the line.0717

So, let's start out with the point that I am given, which is (-2,-1); that is right here.0722

Remember that the slope is the change in y over the change in x.0730

So, if the change in y is -2, then I am starting out here at -1; I am going to go to -2, -3.0735

For every 2 that y is decreased in value, x is going to increase by 3: 1, 2, 3; so, I am going to end up with this point right here.0744

You could have also looked at this as this, and I could say that for every 2 y is increased, x is decreased by 3: 1, 2, 3, right here.0756

The same thing--you could look at it either way.0771

OK, so now, I am able to graph this line because I had a starting point right here, and I have the slope.0777

So, I know, starting from here, how much the x is going to change and how much the y is going to change.0783

So then, I simply connect these lines to give my graph.0789

And notice that this line is decreasing to the right: and that fits with what I have, which is a negative slope.0800

So, simply plot the point you are given; and then use the change in y over the change in x to find additional points on the line.0806

OK, graph the line passing through (1,-3) and parallel to the graph of this equation.0820

Well, I have my point; x is 1; y is -3; so, I have my starting point.0829

I need my slope; and I am not directly given the slope; however, I am given a parallel line.0835

And recall that parallel lines have the same slope.0843

That means that, if I find the slope of this line, I am going to have the slope of the line that I am looking for.0853

In order to find the slope, I need to be able to find the change in y over the change in x; so I need two points on this line.0860

So, the slope is (y2-y1)/(x2-x1).0870

Thinking of an easy way to find a couple points on this line, I am going to use the intercept method.0876

And I could find these points and graph, or I could just find these points and plug them into my slope equation.0883

First, I am going to let x equal 0 to find the y-intercept.0890

OK, so this gives me -3y = 6, or y = -2.0901

When x is 0, y is -2; when y is 0, let's go ahead and plug that in; 2x - 3(0) = 6, so this is 2x - 0 = 6; 2x = 6.0913

Divide both sides by two; that gives me x = 3.0933

OK, so when x is 0, y is -2; when x is 3, y is 0.0937

Now, I can find the slope of the sign, because this can be my x1;0946

this can be my y1; this can be my x2; this can be my y2; so let me find the slope.0953

This is y2, which is 0, minus y1, which is -2.0959

This is x2, which is 3, minus 0, which is x1.0967

This is going to give me 0 minus -2 (0 plus 2), over 3 minus 0; so the slope is 2/3.0975

Therefore, the slope of this line is 2/3, and since parallel lines have the same slope, the slope of the line I am looking for is also 2/3.0988

Now, I was given this point, (1,-3); now I know that the vertical change is going to be an increase in y by 2;1003

and there is going to be a change in x--an increase by 3--1, 2, 3; so I have another point.1011

Now that I have this second point, I can go ahead and graph this line.1019

And this is the line I was asked to graph, so I have completed what I have been asked to do.1027

But just to look at the idea that this is indeed a parallel line, I am going to go ahead and plot this other line, as well.1032

This is the line we were asked to find; this other line--when x is 0, y is -2; when x is 3, y is 0.1038

And you can plot that out and see that, visually, this does look like a parallel line.1047

And these two are changing at the same rate: the slope of both of these is 2/3.1052

I was able to graph the line passing through this point and parallel to the graph of this line1059

by finding the slope of this line, and then realizing that my parallel line is going to have the same slope, which is 2/3.1068

Example 4: Graph the line passing through the point (-2,1) and perpendicular to the graph of this equation: 3x + 4y = 12.1077

OK, I am given a point; and let's see, let's call this -2, -4, -6, -2, -4, -6, 2, 4, 6, 8; OK.1090

-2 would be right here; and then 1 would be right about there; so that is my point.1110

But I also need the slope: recall that the slopes of perpendicular lines--the product of those slopes is equal to -1.1116

So, the product of perpendicular lines' m1 and m2 (I have two perpendicular lines;1127

the slope of one is m1, and the other is m2)--their product is -1.1133

So, I am going to call this first line line 1, and its slope is going to be m1; I am going to find m1.1137

My line I am going to call line 2; and its slope is going to be m2.1147

I need to find the slope of this line, recalling that slope is the change in y over the change in x.1153

So, in order to find this change, I need to find two points on this line.1160

I am going to go for something easy, so I am going to let x equal 0, and I am going to plug that in here.1167

I am just going to find the intercepts as my two points; but I could have found any two points on that line to get the slope.1172

This is going to give me y = 3; when x is 0, y is 3.1185

For my second point, I am going to let y equal 0; that is 3x + 4(0) = 12; so, if 3x = 12, x = 4.1189

So now, I have two points; two points means I can find the slope.1206

I am going to say that this is my (x1,y1), (x2,y2), so I can keep track of everything.1210

y2 is 0; y1 is 3; so 0 - 3, over x2 (which is 4), minus 0.1219

OK, I have found m1: I have found the slope of this line.1233

And I know that the line I am looking for, which is going to have a slope m2, is perpendicular to this line.1239

So, I am going to go right here and say that m1...I know that is equal to -3/4; and I am looking for m2.1246

m1 times m2 equals -1; so now, all I have to do is substitute this in.1258

-3/4 times m2 equals -1; I am going to multiply both sides by -4/3 to move this over to the right.1265

And that is going to give me -1, times -4/3, or m2 = 4/3.1277

OK, now I found this slope, and I need to graph the line.1285

I have my starting point here at (-2,1); and the change in y is going to be 4, so that is going to be 1, 2, 3, 4,1291

for an increase in x of 3; so that is 1, 2, 3, right there.1300

Increase in y by 4: 1, 2, 3, 4; increase x by 3: 1, 2, 3--so right about there; OK.1308

I can go ahead and plot this line out.1325

Reviewing what we did: we were given a point that the line passes through, and we needed the slope in order to graph it.1329

We were told that this line is perpendicular to the line described by this equation.1335

Therefore, we found the slope of this line; and we did that by finding two points in the line (the x- and y-intercepts) and plugging that into the slope formula.1340

Once I found the slope of this perpendicular line, I wanted to find the slope of my line.1351

And I did that by recalling that the product of the slopes of perpendicular lines equals -1.1356

So, -3/4 times the slope of the perpendicular line is -1.1363

I figured out that the line I am looking for has a slope of 4/3, and then I just took my point that I was given and increased y by 4, and then x by 3.1368

That concludes this lecture about slope on Educator.com.1382

Welcome to Educator.com.0000

In today's lesson, we are going to talk about writing linear functions.0002

And in particular, we are going to discuss two forms of these functions.0006

The first form is the slope-intercept form; and this is a very useful form of the equation, because it can help you to graph a linear equation.0012

The slope-intercept form of a line is y = mx + b, where m is the slope, and b is the y-intercept.0022

Recall that the y-intercept is the point at which the graph of the equation (the line) intersects the y-axis.0033

For example, if we look at an equation in this form, and it is given as y = -2x + 1, then the slope is -2, and the y-intercept, or b, is 1.0043

This information alone allows me to graph the line.0060

Before, we talked about graphing the line by finding a couple of points.0063

And we, in particular, used the intercept method, where we found the x-intercept and the y-intercept, and graphed that.0067

This time, I am going to use a slightly different method.0073

So, here I have the y-intercepts: this is the y-coordinate where the graph of the line is going to cross the y-axis.0076

That is going to be at y = 1; x will be 0, and y is 1.0088

Now, I also have the slope; the slope is the change in y, over the change in x.0094

And this is written as -2; but you can think of it in your mind as -2/1.0100

For every 2 that y is decreased, x is increased by 1.0105

And I am already thinking of what I expect this graph to look like.0110

Recall that, if the slope is negative, the line is going to decrease going from left to right--it is going to go this way.0113

If the slope is positive, the line is going to increase going from left to right.0120

I am starting right here, and I am going to decrease y by 2--1, 2--for every increase in x by 1.0125

So now, I have another point: 1, 2, and then increase x by 1.0133

And I now have plenty to go ahead and graph this with.0139

So, you can see how this form of the equation is very helpful in getting information about the line and actually graphing the line.0152

The second form of these linear equations that we are going to look at today is point-slope form.0161

The point-slope form of the equation of a line is y - y1 = m (x - x1).0169

And you will see that this is related to slope and the slope formula.0177

So, since the slope is (y2 - y1)/(x2 - x1), you can look and see that that is pretty familiar.0182

Now, previously we talked about having two points: (x1,y1), and the other (x2,y2).0191

Well, when we are working with the point-slope form, we have one point (x1,y1),0199

and then the other point could be anywhere on the line; it is just another point (x,y) that isn't specified.0205

So, look at how you could manipulate this to be similar to this,0212

if I said, "OK, the slope is some point on the line, minus a given point, over some point on the line, minus the given point."0215

Now, I am going to multiply both sides of this by (x - x1)--both sides of the equation.0226

These cancel out; and that is going to give me (x - x1) times m equals (y - y1).0243

A little bit of rearranging: I am going to put the y portions on the left side of the equation, and the x and the slope on the right side of the equation.0255

And you see, that gives me the point-slope form.0267

And this is useful, because imagine if I am given some facts about a line, and I am told that the slope equals 4, and that this line passes through a certain point.0271

And I am told that it passes through the point (-2,6).0287

Knowing this, I can write the equation for this line in point-slope form.0290

So, point-slope form: y - y1 = the slope times (x - x1).0295

So, y is any other point on this line.0301

y1 is 6; slope is 4; x is another x point on this line that goes along with this y coordinate, minus x1, which is -2.0306

Simplifying this: a negative and a negative is a positive.0319

What is helpful about this point-slope form is: with a little bit of work, I can put this into the slope-intercept form.0329

Recall that the slope-intercept form (this is the point-slope form of the equation) is y = mx + b.0338

I am going to multiply this out; this is 4x + 8; and I want to isolate y by adding 6 to both sides.0358

So now, I have it in slope-intercept form; this is a very useful form of the equation.0374

And again, this can allow us to graph the equation.0384

Parallel and perpendicular lines: first, recall that parallel lines have the same slope.0388

So, if I have two parallel lines, line 1 and line 2, and the first one has a slope of m1, and the second one has a slope of m2, those are equal.0402

Perpendicular lines: the product of the slope of two perpendicular lines is equal to -1.0415

The slope-intercept form and point-slope form can be used to solve problems involving parallel and perpendicular lines.0429

And the reason is: if I am told that a line is parallel to another line, I have the slope.0435

With a little bit more information, I can write the equation in slope-intercept form.0442

Or I may be given the equation in slope-intercept form and told that a line is parallel to that line.0446

The same with perpendicular lines: by having these forms of the equation, which involve slope,0453

and knowing the relationship between two lines and their slopes, I can write the equation for the second line.0458

I can graph the lines, and it is all about knowing the relationships between these two lines.0464

For example, if I am told that the graph of a line is parallel to the line described by the equation y = 1/2x - 4,0471

and I am also told that the line I am looking for passes through a point (4,-3),0499

I can graph the line I am looking for; I can also write an equation for the line I am looking for.0514

So, the graph of a line is parallel to the line y = 1/2x -4, and the line I am looking for passes through a certain point.0521

Well, since parallel lines have the same slope, and I am looking at this, and it is in slope-intercept form, which is y = mx + b, I now have the slope.0529

So, the slope equals 1/2; since these lines are parallel, I also have the slope of the line I am looking for, and it is 1/2.0540

The line I am looking for passes through (4,-3); so that is (4,-3), and I am going to plot that out: (4,-3).0549

And I have the slope, so I know that when I increase y by 1, I increase x by 2; therefore, I can plot the line.0564

OK, and this is as expected, because it has a positive slope, so it is increasing as it goes to the right.0577

The same holds true for perpendicular lines: for example, if I am told that the graph of a line is0598

(this stands for perpendicular) perpendicular to the graph of the line defined by the equation y = 1/4x + 6,0612

and the line passes through some point (say (1,2)), I can graph this line.0627

And the reason I can graph it is that I know that this slope is 1/4, and I recall that the two slopes are related by this formula.0638

So, if I am given a point on a line, as well as the knowledge and relationship between that line and a parallel or perpendicular line,0648

I have the point; I can find the slope; I can graph the line.0659

OK, first example: Find the equation in slope-intercept form of the line with the slope 2/3 and passing through (2,-4).0665

Slope-intercept form is y = mx + b; and I am given slope, so I am given m = 2/3; so let's start from there: y = 2/3 x + b.0678

Well, in order to write this out, I also need to find b; and b is unknown.0694

However, I am given an x value and a y value; and since I have m, x, and y, I can solve for b.0699

So, substituting in -4 for y, and 2 for x, now I can solve for b.0707

So, -4 = 2 times 2...that is 4, so that is 4/3, plus b.0718

Subtract 4/3 from both sides...equals b...and you could really think of this as -1 and 1/3; it might be easier to look at it that way.0734

This is -5 and 1/3...equals b.0746

OK, going back to the beginning here: y = mx + b, so y equals...m is given as 2/3; x; and then b is -5 and 1/3.0749

Using the facts that I was given, I am able to write this in slope-intercept form.0765

I was given the slope, and I was given a point on a line.0773

The point on the line, and the slope: by plugging those into this equation, I could find the y-intercept; and therefore, I could write this out in slope-intercept form.0776

Example 2: find the equation in slope-intercept form of the line passing through these two points.0786

Slope-intercept form is y = mx + b, so I need to have the slope, and I need to have the y-intercept, in order to write this.0793

The slope is the change in y, over the change in x; and I am given two points, so I can find the change in y over the change in x.0802

I am going to call this (x1,y1), and this (x2,y2).0813

OK, so m = y2 (which is -3) minus -7, over -6 minus -2.0819

-3...and that negative and negative becomes a positive, so plus 7, minus 6--a negative and a negative--that is plus 2.0833

-3 + 7 is 4; -6 + 2 is -4; I have the slope now--the slope is -1.0841

My next thing is to find b, which is the y-intercept; so, I have y = mx (-1x; we usually just write this as -x,0853

but we are writing it out right now as -1x) + b.0864

I need to solve for b, and I can do that, because I have some x and y values I can substitute in.0868

You could choose either set of coordinates, either ordered pair.0874

I am going to go ahead and choose this first one: y is -7; it equals (instead of writing -1, I am just going to write) -x (which is -2) + b.0880

Solving for b: -7 equals...a negative and a negative is a positive, so that is 2 + b.0898

Subtracting two from both sides, b equals -9.0905

In order to write this equation, I needed to have my slope, which I do.0910

And so, I have that m equals -1, and b equals -9.0916

So now, I can go ahead and write this as y = -x (that is mx, where m is -1) + b (and that is -9): y = -x - 9.0922

This is the equation for this line, written in slope-intercept form.0942

Example 3: Find the equation in slope-intercept form of the line passing through the point (-2,-3), and parallel to the graph of y = 3x - 7.0951

Slope-intercept form is y = mx + b: the first thing I need to do is to find the slope.0964

I am not directly given the slope; however, I am told that this line is parallel to the line described by this equation.0972

Parallel lines have the same slope, so if I know this slope, I know the slope for the line I am looking for an equation of.0981

Well, this is in slope-intercept form; therefore, y = mx + b, so I have the slope.0994

The slope of this line is 3, and the slope of the parallel line (which is my line) is also 3.1005

So now, I have y = 3x + b; and in order to fully have this in slope-intercept form, I need to have b.1012

Well, I have a point on this line, (-2,-3); so I am going to substitute in; I am going to let x equal -2 and y equal -3, and then solve for b.1021

OK, so y is -3; x is -2; and I have the slope: so 3 times -2 plus b; that gives me -3; 3 times -2 is -6; plus b.1041

Adding 6 to both sides, b equals 3.1061

I have the slope; I have the y-intercept; I can write this out in slope-intercept form: y = (slope is 3) 3x...and b is 3.1068

OK, so this one was a little bit more complicated than before, because they didn't directly give us the slope or two points on the line.1083

But what they did give us is the fact that this line is parallel to the line described by this equation.1089

Knowing that means that I have this slope (which is 3), and since my line is parallel, it is the same slope.1096

Once I have the slope, I just take the point on that line, substitute that in for x and y, and solve for b.1104

So, to write it in this form, I am essentially given m; I can figure out b; and this is the equation in slope-intercept form.1111

Find the equation in slope-intercept form of the line passing through (2,-1) and perpendicular to the graph of 2x - 3y = 6.1122

In order to write this in slope-intercept form, y = mx + b, I need to have the slope of this line.1132

I am not given the slope directly; however, I am told that it is perpendicular to the graph of this line.1139

Recall that the slopes of perpendicular lines are related by this equation.1146

So, if you have the slope of two lines that are perpendicular, and you take their product, it is equal to -1.1152

So, let's let the slope of this line equal m1; and m2 is the slope of my line, the line I am looking for.1158

Let's go ahead and figure out m1: what I need to do is write this equation in slope-intercept form,1172

and that will give me the slope of this line, which in turn will give me the slope of the line I am looking for.1180

First, I am going to subtract 2x from both sides; then, I am going to divide both sides by -3.1190

That is going to give me y = -2x/-3 + 6/-3.1201

Simplify: the negatives cancel out, and I will get 2/3x - 2.1209

OK, since this is in slope-intercept form, this is m; this is the slope; so m1 equals 2/3.1217

I now want to find m2, which is the slope of the perpendicular line, or the line that I am looking for.1226

m1 times m2 equals -1; and I am given m1--I am given that m1 is 2/3.1235

So, it is 2/3 times m2 equals -1.1243

If I multiply both sides by 3/2, I can isolate m2, and it is -1 times 3/2.1248

Therefore, the slope of the line that I am looking for is -3/2.1257

So now, I have the slope: I have that, for my line, y = -3/2x + b.1265

I have slope; I need the y-intercept; well, I am also given a point on this line, which means I have an x and y value to substitute here.1274

So, y is -1; x is 2; the 2's cancel out, and that is going to give me -3 plus b.1282

I am going to add 3 to both sides, which is going to give me b = 2.1298

So now, I have the slope; I have the y-intercept; so I can write the equation for this line in slope-intercept form: slope is -3/2; b is 2.1304

Again, we approached this by realizing that the slope of the line we are looking for is related to the perpendicular line1322

by the equation m1 times m2 (the product of the slopes of the perpendicular lines) = -1.1331

So then, I went about looking for the slope of this line; rewriting it in slope-intercept form gave me y = 2/3x - 2.1338

So, I know that this slope is 2/3; once I have this slope, I can find my slope: m1 times m2 equals -1.1347

So, that is 2/3 times m2 equals -1; m2 equals -3/2.1357

I go back to this form y = mx + b, now knowing the slope of my line.1363

Substitute in x and y values and the slope to solve for b; b equals 2.1369

I have b; I have the slope; that allows me to write this equation in slope-intercept form.1376

That concludes this lesson of Educator.com.1383

Welcome to Educator.com.0000

In today's session, we are going to talk about several special functions.0002

The first one we are going to discuss is the step function; and the step function makes a unique-looking graph.0007

It is a function that is constant for different intervals of real numbers.0013

And the result is a graph that is a series of horizontal line segments, so they look like steps; and that is where the name "step function" comes from.0018

The best way to understand this is through an example.0028

So, for example, if apples were sold at a price of a dollar per pound, and the price is such that0030

you are charged $1 for each pound, or any part of a pound--in other words, they round up in order to determine the price;0053

if you had a pound and a half of apples, they are going to charge you $1 for the pound, and then another $1 for the half pound.0070

So, they charged you a dollar for a pound, and then any part of a pound is considered a full pound in terms of pricing.0079

So, if x is the number of pounds, and y is the cost, then let's see what kind of values we get and what the graph looks like.0088

OK, if I have .8 pounds, I am going to get charged $1: they are going to round up.0104

If I have a full pound, I am going to get charged $1; if I have a little bit over a pound--I have 1.2 pounds--0113

I am going to get charged $1 for the first pound, and then that .2 is going to be another dollar, so $2.0124

1.4 pounds--again, $2, and on up...2 pounds--a dollar for the first pound and a dollar for the second pound.0135

2.5--$2, and then the .5 is another $1, so that bumps me up to $3; until I hit 3 pounds, it is $3, also.0145

3.8: $1 for the first pound, $1 for the second pound, $1 for the third pound, and another $1 for that .8, so $4.0155

So, you can see that the function is constant for different intervals.0166

So, for this first interval, from a little bit above 0, all the way to 1, the y-value is constant; it is $1.0170

For the next interval, which is just above 1, all the way to 2, including 2, it is going to be $2.0180

Once I get above 2, up to and including 3, it is $3; and so on.0188

So, this is constant for different intervals of real numbers.0193

Looking at what the graph looks like: any part of a pound up to and including a dollar for that first pound is a dollar.0198

Now, 0 pounds of apples are going to be $0; so I am not going to include 0.0207

But even .1...the slightest part of a pound is going to be $1, so I put an open circle to indicate that 0 is not included;0213

but just above that is, all the way up to and including 1; since 1 is included (1 is also a dollar), I am going to put that as a closed circle.0224

So here, I have pounds; and here, I have the cost on the y-axis in dollars.0233

Now, once I get just above 1 (say 1.1 pounds), they are going to charge me $2--not including 1, but just above it--open circle.0243

All the way up to 2...at 2 pounds, I will also be charged $2.0254

Once I hit just above 2, I am going to be charged $3, all the way to 3 and including 3.0260

And on...so, you can see how this looks like a series of steps, and how this is a result of the fact that the function is constant for different intervals of x.0270

This function is the same for this entire interval; then it is the same for the second interval; and on.0287

So, this is a step function.0292

A second type of function that you will be working with is an absolute value function.0295

And these functions have special properties: looking first at f(x) = |x|, just the simplest case, here f(x) is just going to be the absolute value of x.0301

When x is 0, f(x) is also 0; when x is 1, the absolute value is 1; and so on for positive numbers.0317

Now, let's look at negative numbers: for -1, the absolute value is 1; -2--the absolute value of x is 2; and on.0329

The result is a certain shape of graph: when x is 0, f(x) is 0; x is 1, f(x) is 1; x is 2, f(x) is 2; and on.0343

Now, for negative numbers: -1, f(x) is 1; -2, f(x) is 2; -3, f(x) is 3; and it is going to continue on like that.0358

So, absolute value graphs are v-shaped; so we are going to end up with a v-shaped graph.0372

Depending on the function, the graph can be shifted up, or it can be shifted to the right or to the left; and let's see how that could happen.0385

Let's now let f(x) equal (here f(x) equals |x|)...let's say f(x) equals the absolute value of x, plus 1.0393

OK, so we are given x, and the absolute value of x is 0; we are adding 1 to them, so this is going to become 1.0404

The absolute value of x is 1; 1 + 1 is 2; 3; 4; the absolute value of -1 is 1; 1 + 1 is 2.0412

The absolute value of -2 is 2; add one to that--it is 3; and add 1 to 3 to get 4.0425

OK, so it is the same as this, except increased by 1: each value of the function has been increased by 1.0432

So now, let's see what my graph is going to look like.0439

Right here, I have the graph for f(x) = |x|; now, I am going to look at this graph.0443

When x is 0, f(x) is 1; when x is 1, f(x) is 2; when x is 2, f(x) is 3; so you can see what is happening.0451

And then here, I have x is 3, f(x) is 4; negative values--when x is -1, f(x) is 2; when x is -2, f(x) is 3; when x is -3, f(x) is 4.0465

OK, I am drawing the line through this: this is the graph of f(x) = |x| + 1.0492

So you see that this graph, the v-shaped graph, is simply shifted up by 1.0509

And again, you can also shift this from side to side; and we will see an example of that later on.0513

So, in the absolute value function, it is very important to find both negative and positive values of the function.0519

So, assign x 0; assign it some positive values; and it is very important to find what f(x) will be when x is negative,0526

because if I didn't--if I picked only positive values--I would end up with half of a graph.0537

So, to get the entire v-shape, choose negative and positive values for x.0541

The third special function that we are going to discuss is called a piecewise function.0548

And a piecewise function is a function that is described using two or more different expressions.0553

The result is a graph that consists of two or more pieces.0560

Just starting out with one that consists of two pieces: the notation is usually like this--one large brace on the left:0564

f(x) equals x + 2 for values of x that are less than 3; and f(x) equals 2x - 3 for values of x that are greater than or equal to 3.0573

So you see that, for different intervals of the domain, the function is defined differently.0590

So, it is a function that is described using two or more different expressions.0597

So, for the part of the domain where x is less than 3, this is the function.0600

For the interval of the domain that is greater than or equal to 3, this is the function that I am going to use.0605

Let's see what happens: let's first use this part of the function where f(x), or y, equals x + 2.0613

And for the domain, remember that x is going to be less than 3.0625

So, I will go ahead and start out with 2: when x is 2, f(x) (or y) is 4.0630

When x is 1 (I have to remain at x-values less than 3), f(x) is 3; when x is 0, f(x) is 2.0636

Just picking a negative number: when x is -4, f(x) is -2.0648

I am going to graph that here; OK, when x is 2, f(x), or y, is 4; x is 1, y is 3; x is 0, y is 2; x is -4, y is -2.0656

OK, now, this is for values of x that are less than 3; 3 is about here.0689

Therefore, anything just below 3, but not including 3, is going to be part of this graph.0698

So, we are going to use an open circle here to indicate that 3 is not going to be included as part of this function--the domain of this function.0703

So, it is going to begin at just below 3 and continue on indefinitely; that is the first piece of the graph.0716

The second piece of the graph is for x such that x is greater than or equal to 3; and here, y is going to be 2x - 3.0723

So, it is greater than or equal to 3, so I am first going to let x equal 3.0735

3 times 2 is 6, minus 3 is 3; getting larger--when x is 4, it is 2 times 4 is 8, minus 3 is 5; when x is 5, 5 times 2 is 10, minus 3 is 7.0738

Now, this does include 3--this section of the graph--this piece; so, when x is 3, y is 3, right here.0755

When x is 4, y is 5, right here; when x is 5 (let me shift that over just a bit), y is 7; OK.0768

Now, if you look, this actually did end up including all possible values of x (all real numbers),0801

because when x is less than 3, I use this function; and then, as soon as x becomes 3 or greater, I shift to this other function.0807

So, you can see how there are two pieces to the graph; and you actually can have situations where there are more than 2.0816

You could be given, say, f(x) is 4x + 3, 2x + 7, and x - 1, and then given limits on the domain for each of those.0822

So, there are at least two pieces; however, there can be more.0836

In Example 1, we have a greatest integer function.0840

Before we start working in this, let's just review what we mean by the greatest integer function.0845

So, when you see this notation with the brackets, let's say that you have a number in here, such as 4.7.0848

What this is saying is that this value is equal to the greatest integer that is less than or equal to 4.7; so that is 4.0855

It is the greatest integer less than or equal to whatever is in here; if it was 2.8, it would be 2.0866

Be careful with negative numbers: let's say I have -3.2--the temptation is to say, "Oh, that is equal to -3";0875

but if you look at it on the number line, -3.2 is right about here; OK, so if I have -3.2,0882

and I am trying to find the greatest integer that is less than or equal to 3.2, it is going to have to be something over here--smaller.0893

So, it is actually going to be -4; so just be careful when you are working with negative numbers.0903

Whatever is inside here--whatever that value is--the function is equal to the greatest integer that is less than or equal to this value in here.0907

Understanding that, you can then find the graph; so let's find a bunch of points for this,0920

so we make sure we know what is going to happen with various situations.0927

When x is 0, this inside here is going to be 2; the greatest integer less than or equal to 2 is 2.0935

When x is .6, then you are going to get 2.6 in here; the greatest integer less than or equal to 2.6 is 2.0945

.8--I get 2.8; again, I round down to 2.0953

All right, so when I hit 1, 1 plus 2 is 3, and the greatest integer less than or equal to 3 is 3.0958

Slightly above 1: that is going to give me 1.2 + 2 is 3.2; the greatest integer less than or equal to 3.2 is also 3.0967

OK, so you can get the idea of what this is going to look like.0976

And that continues on; and then, when we hit 2, 2 + 2 is 4; the greatest integer is going to be 4.0980

For negative numbers: let's take -.5: -.5 and 2 is 1.5; the greatest integer less than or equal to 1.5 is going to be 1.0988

Now, notice: I have a negative number for x, but this did not come out to be a negative number; so that is different from the case I was discussing there.1005

Let's go a little bit bigger--let's say -3: -3 and 2 is -1, and that is going to be -1.1014

Let's say I take -3.5: -3.5 and 2 is going to equal -1.5: again, just thinking about that to make sure you have it straight,1024

I have -1.5; so I have 0; I have -1; I have -1.5; I have -2; the greatest integer less than or equal to this is actually -2.1037

OK, now plotting this out: when x is 0, f(x) is 2; when x is slightly above 0 (it's .6), f(x) is 2; .8--it is 2, all the way up until I hit 1.1053

At 1, f(x) becomes 3; therefore, 1 is not included in this interval.1075

So, you can already see that this is going to be a step function, because we have intervals.1082

For different intervals of the domain, we have that same value for the range.1089

All right, for values between 1 and 2, f(x) will be 3; once we hit 2, I have to do an open circle, because at 2, the value for f(x) jumps up to 4.1095

OK, so you can see what this is going to look like; and that pattern is just going to continue.1112

Let's look over here at negative numbers: when x is slightly less than 0, then you are going to end up with an f(x) that is 1.1116

So, for values slightly less than 0, but not including 0, this is what you are going to end up with.1137

OK, looking, say, when x is -3: when x is -3, f(x) will be -1.1145

But when we go slightly more negative than that, when x is -3.5, f(x) is going to be -2; it is going to be down here.1161

So, the steps on this side are going to have the open circle on the right.1175

And I am going to jump down, and it is not going to include -2, because -2 and 2 is 0;1184

so -2 is going to be right here for the x-value, and the f(x) will be 0.1193

But as soon as I get to a little bit bigger than -2, the greatest integer is going to be down here.1199

OK, and so, we continue on like that with the steps; and you can see how this is a step function.1207

You just have to be very careful and pick multiple points until you can see the pattern1220

where for a certain interval of the domain, the range is a particular value.1225

OK, so that was a step function, and it involved the greatest integer function.1234

Example 2: now we are working with absolute value.1240

g(x) equals the absolute value of x, minus 3.1244

And we already know that the shape of this graph is going to be in a v.1248

But we don't know exactly where that v is going to land, so let's plot it out.1252

When x is 0, the absolute value of x is 0; minus 3--that gives me -3.1258

When x is 1, the absolute value is 1; minus 3...g(x) is -2.1264

When x is 2, the absolute value is 2; minus 3 is going to give me -1.1269

Now, let's pick some negative numbers for x, because that is really important to do with an absolute value graph.1277

When x is -1, the absolute value is 1, minus 3 gives me -2; you can already see that my v shape is going to occur.1283

When x is -2, the absolute value is 2; minus 3 is -1.1291

The absolute value of -3 is 3; minus 3 is 0; so this is enough to go ahead and plot.1299

x is 0; g(x) is -3; x is 1, g(x) is -2; x is 2, g(x) is -1; over here with the negative values,1305

when x is -1, g(x) is -2; when x is -2, g(x) is -3; when x is -3, g(x) is 0.1316

So, you can see that I have a v-shaped graph, and compared with my graph that would look like this,1325

that would have the v starting right here, it has actually shifted down by 3; that is an absolute value function.1338

Here you can see that you are given a piecewise function, because there are two different pieces.1348

And this could also be written in this notation.1357

There are two different sections to the graph; and we see that the function is defined differently for different intervals of the domain.1360

Starting with if x is greater than 2 (this is going to be for x-values where x is greater than 2): f(x) is going to be x + 1.1371

When x is 3, f(x) is 4; when x is 4, f(x) is 5; when x is 5, f(x) is 6; OK.1386

When x is 3, f(x) is 4; when x is 4, f(x) is 5; and it is going to go on up.1404

And that is going to go all the way, until just greater than 2.1416

2 is not going to be included on this graph, because it is a strict inequality; so I am going to use an open circle, and this is going to continue on.1422

Now, for x less than or equal to 2, I have a different situation: I am looking at f(x) is -2x.1430

OK, so when x is 2, 2 time -2 is -4; when x is 1, 1 times -2 is -2; when x is 0, f(x) is 0; when x is -2, that is -2 times -2, which is positive 4.1443

So, starting with x is 2: when x is 2, f(x) is -4; and that is including the 2.1464

When x is 1 (these are values less than or equal to 2, so I am getting smaller), f(x) is -2.1476

x is 0; f(x) is also 0; when x is -2, f(x) is up here at 4; OK, so I have a steep line going right through here.1494

So, you can see: this is a piecewise function consisting of two pieces; and here, one picks up where the other leaves off.1509

For values greater than 2, this is my graph; for values of x less than or equal to 2, this is my graph; so this is a piecewise function.1515

OK, this time, in Example 4, we have both greatest integer and absolute value in this function.1525

Recall that, for the greatest integer function, what that is saying is that whatever is inside this bracket--let's say it's 1.2--1532

it is asking for the greatest integer less than or equal to 1.2; in that case, this would be 1.1540

Or if I had 4.8, it would be 4.1545

For negative numbers, like -3.2, the greatest integer less than or equal to -3.2 is -4.1551

OK, now, since this is a bit complicated, it is helpful just to take it in stages.1560

So, I am going to look first at what the greatest integer of x is; and then, I am going to look for the absolute value of what that is.1567

If x is .2, the greatest integer less than or equal to .2 is 0; the absolute value of 0 is 0; so this is the function that we are looking for.1578

And the same would hold true of .5: round down to 0; the absolute value is 0.1590

When we hit 1, the greatest integer less than or equal to 1 is 1, and the absolute value of that is 1.1596

1.2: again, we are going to go down to the greatest integer that is less than or equal to 1.2, which is 1; and the absolute value is 1.1608

The same for 1.8, and all the way up until 2; once we hit 2, the greatest integer less than or equal to 2 is 2; the absolute value is 2.1621

So, that is working with positive numbers, greatest integer, and the absolute value; it is the same; OK.1629

So, let's go to negative: for -.4, the greatest integer that is less than or equal to -.4...I am looking, and I have 0, and 1,1636

and -1, and -.4 is about here; so I am going to go down to -1; the absolute value of that is 1.1650

Here you can see that the greatest integer is not the same as the absolute value.1659

Or for -1, the greatest integer less than or equal to -1 is -1; the absolute value is 1.1664

-1.8: the greatest integer that is less than or equal to -1.8...I am going to go down to -2; and the absolute value is 2.1674

For -2, the greatest integer less than or equal to -2 is -2; the absolute value is 2.1686

So, you see that there are intervals here--intervals of the domain end up with the same value for the function.1694

So, I am going to have a step function.1704

But remember that absolute value graphs are v-shaped, so I am going to end up with a v-shaped step function.1709

Let's plot these out: for 0, the greatest integer of 0 would be 0, and then the absolute value would be 0.1715

So, with 0, we are going to include it; and for all values up to but not including 1, the function is going to equal 0.1726

Once we get to 1, I have an open circle, because it is not included.1736

When x is 1, f(x) is 1; so I am going to jump up here.1740

All the values between 1 and 2, but not including 2, will have an f(x), or a y-value, that is 1.1746

As soon as I hit 2, open circle: I am going to jump up, and once I hit 2, f(x) is 2, all the way up to, but not including, 3.1754

And it is going to go on that way: and you see now, we have the step function, and it is v-shaped like absolute value.1771

Let's look over here on the negative side of things.1778

For -.4, somewhere in here, it is going to equal 1; -1 is also equal to 1; so here, on the left side, I have a closed circle, and an open circle on the right.1782

It is the opposite of what I had over here.1799

When I get to less than -1, my value for f(x) is going to jump up to 2; this is a closed circle;1802

I get slightly less than, but not including, -1; it is going to jump up to 2.1815

-2: my value is also 2, and everything in between; and then, when I get to just slightly more negative than -2, like -2.1, it is going to jump up to 3.1822

You can see how this is v-shaped, and it is a step function.1841

The step function comes from it being the greatest integer function; the v shape comes from that absolute value.1847

And you also just had to be careful how you are doing the open and the closed circles; OK.1853

Welcome to Educator.com.0000

In today's lesson, we are going to be graphing inequalities.0002

Recall that, in order to graph a linear inequality, you have to take a series of steps.0008

And the graph of a linear inequality is a shaded region in the coordinate plane.0014

And those shaded regions are known as half-planes; the boundary of the region, or the half-plane, is a straight line,0019

which is the graph of the corresponding linear function.0027

The region, or the half-plane, containing the solution set can be determined by using a test point.0032

OK, so the steps in order to graph a linear inequality are to first graph the corresponding linear function; and that is a straight line.0040

And that line is dashed if the inequality you are working with is a strict inequality (less than or greater than).0058

The line is solid if you are dealing with less than or equal to, or greater than or equal to; and we will talk in a minute about what that means.0071

After you have graphed the line, then you end up with the coordinate plane divided into two half-planes.0079

You use a test point to determine the region containing the solution set.0097

So, I am going to go ahead and do an example now: but if you need more review on this, check out the Algebra I video on this topic.0104

All right, the easiest way to understand is to look at an example: let's look at the inequality y - 3x < -2.0112

My first step is to graph the corresponding linear function, which would be y - 3x = -2.0123

So, I am going to go ahead and find some x-values and some y-values--some coordinate pairs on this line.0132

When x is 0, y is -2; so that gives me the y-intercept.0140

When x is 1, this is going to give me -3; add 3 to both sides--y will be 1.0145

When x is 2, -3 times 2 is -6; add that to both sides, and y is 4.0155

All right, I have enough to graph the line: x is 0, y is -2--the y-intercept; then I have another point at (1,1) and another point at (2,4).0164

Because this is a strict inequality, I am going to use a dashed line.0177

This line divides the coordinate plane into two half-planes: an upper and lower half-plane.0187

One of these half-planes contains the solution set for the inequality; the way I figure out which one is to use a test point.0193

Pick a point that is not on the boundary line, and that is easy to work with.0201

(0,0) is very easy to work with; so I am going to take my test point, and I am going to go back to my inequality and plug it in.0206

I am going to see what happens: this gives me 0 - 0 < -2; 0 is less than -2--that is not true.0216

What this means is that this is not part of the solution set.0230

If this is not part of the solution set, it is in the half-plan that is not contained in the solution set.0234

Therefore, the solution set is actually in the lower half-plane.0240

And you could verify that by checking a point down here.0245

The idea is to graph the corresponding linear function to find the boundary line.0249

Now, I mentioned that this is a strict inequality, so I am using a dashed line.0253

And what that means is that the boundary line is not part of the solution set.0256

If it were a solid line (because I had less than or equal to), then the boundary line would be part of the solution set.0262

But actually, in this, the solution set is just below that boundary line.0268

I determined which half-plane was correct by using a test point.0272

Since the test point ended up giving me an inequality that is not true, that is not valid,0276

I knew that that test point did not lie in the half-plane of the solution set, so I shaded in the lower half-plane.0281

Graphing absolute value inequalities: the procedure for these inequalities is similar to the procedure for linear inequalities.0290

So, we are going to use the same technique.0298

Recall that the technique involved first graphing the line for the corresponding equation.0302

Graph the corresponding equation: in this case, it is going to be an absolute value equation.0309

And then, use a test point to determine the half-plane containing the solution set for the inequality.0318

All right, let's take an example: y ≤ |x-3|.0338

The corresponding equation would be y = |x-3|, so I need to graph that to find my boundary.0347

So, some values for x and y: when x is -2, -2 minus 3 is -5; the absolute value is 5.0357

When x is -1, this becomes -4; the absolute value is 4; when x is 0, this becomes -3; the absolute value is 3.0368

Let's jump up to 3: 3 minus 3 is 0--the absolute value is 0.0381

4 minus 3 is 1; the absolute value is 1; 5 minus 3 is 2--the absolute value is 2.0389

OK, plotting these points: first, (-2,5), (-1,4), (0,3), (3,0), (4,1), (5,2); OK, I have my typical v-shaped graph of absolute value.0399

And you notice that I made this a solid line; and the reason that is a solid line is because I had less than or equal to.0439

So, this boundary is actually going to be part of the solution set.0447

OK, so I graphed the corresponding equation, and I determined that it is a solid line.0451

Now, I need to determine if the solution set is down here, or if it is above the boundary; is it above or below the boundary line?0455

So, I am going to use a test point of (0,0): that is going to be my test point.0462

I go back to the inequality, and I see what happens when I put my test point values in there.0468

This gives me 0 ≤ |-3|; that would say that 0 is less than or equal to 3; that is true.0477

Since that is true, what that tells me is that this is part of the solution set.0489

(0,0) is part of the solution set, and that means that this lower half-plane where (0,0) is, my test point, contains the solution set.0495

So, as I mentioned, this is very similar to the technique for graphing linear inequalities.0505

First, graph the corresponding equation; and use a solid line if it is less than or greater than or equal to.0510

Use a dashed line if it is a strict inequality (just greater than or less than).0520

So then, once you have your boundary line, find a test point that is off the boundary line, and not so close to it as to cause any confusion.0525

And choose one that is simple: (0,0), (1,1)--very easy to work with.0533

Plug that into the absolute value inequality; and if you come out with a true inequality,0538

like this one, you know that you are in the correct area with the solution set.0544

If you come out with a statement or inequality that is not valid, you shade in the other region.0548

OK, the first example: we have an inequality that we are asked to graph: y ≤ 2x - 3.0557

My first step is going to be to graph the corresponding linear equation, in order to determine the boundary.0567

Some values for x and y: when x is 0, y is -3; when x is 1, 1 times 2 is 2, minus 3 is -1.0581

When x is 2, 2 times 2 is 4, minus 3 is 1.0594

OK, so plotting this out: (0,-3), (1,-1), (2,1); I have plotted this boundary line, and I look up here,0599

and I see that it is less than or equal to; so I am going to make this a solid line.0615

It is going to be a solid line: this line will be part of the solution set.0621

So, first I graph the boundary; next, I am going to look at a test point.0626

OK, I can again use this very simple test point of (0,0), the origin.0638

Going back to the inequality to look at my test point, y ≤ 2x -3; the test point is (0,0), so y is 0; x is 0.0646

This is saying that 0 is less than or equal to -3; that is not true--this is not in the solution set.0664

Well, the origin is up here; this means that this whole area is not part of the solution set.0678

So, I am going to go to the lower half-plane and shade that in to indicate that this shaded region, including the boundary line, is the solution set.0684

Again, the first step is to take the corresponding linear equation and graph it.0692

The second step: find a test point that is easy to work with and away from the boundary line, and plug that in.0698

If you come up with an inequality that is not true, like this, shade the other half.0705

If it is true, shade the half containing the test point.0711

Example 2: We are graphing another inequality, 2x + y > 6.0716

First, I am going to find the corresponding equation: 2x + y = 6.0724

I am going to keep this simple: I am going to use the intercept method in order to graph this.0733

I am going to let x equal 0 to find the y-intercept: when x is 0, y is 6.0738

Then, I am going to let y equal 0 to find the x-intercept: when y is 0, then I am going to end up with 2x + 0 = 6, or 2x = 6; x = 3.0745

So, I have my x and y intercepts; it is going to give me...when x is 0, y is 6, right about there; when x is 3, y is 0.0759

OK, then I look, and I have a strict inequality, so I am going to have a dashed line;0771

this line is going to be dashed, and is not going to be part of the solution set.0777

So, I found my boundary line; and that is my first step; my second step is going to be to work with my test point.0788

The test point is right here: it is the origin, (0,0).0798

Going back and looking at the inequality: 2 times 0, plus 0, is greater than 6; 0 plus 0 is greater than 6.0803

Well, 0 is not greater than 6, so this is not true--not in the solution set.0817

The origin is not part of the solution set, so the lower half-plane is not the correct half-plane; I am going to shade, instead, this upper half-plane.0830

Again, graph the linear equation corresponding to the inequality.0839

Since this is strict inequality, plot that line using a dashed line, since this boundary is not part of the solution set.0844

Find a test point; the origin is a really good one, as long as it is not on the boundary line or very close to the boundary line.0853

Substitute (0,0) into the inequality; and you come up with an inequality that is not true.0859

Example 3: Graph 3x - 4 > y.0870

First, graph the boundary line: the corresponding equation is 3x - 4 = y.0876

To graph that, I am going to need some x and y values; when x is 0, y is -4; when x is 1, that gives me 3 - 4 = y, so y equals -1.0889

One more value: when x is 2, y is 2; OK.0908

So, I have enough information to plot my line: when x is 0, y is -4; (1,-1), (2,2).0918

The next issue: strict inequality--that means this is a dashed line.0935

OK, so now, I have this boundary; I have an upper and a lower half-plane; and I need to figure out where my solution set is.0948

The origin is well away from the boundary line, so I can use that as my test point.0956

So, test point is (0,0): the inequality is 3x - 4 > y; and this gives me 0 - 4 > 0, which comes out to -4 > 0.0963

And that is, of course, not true; this is not in the solution set, because this inequality did not hold true.0986

Therefore, I am going to go to the other half-plane, and I am going to shade that in.0997

I graphed the boundary line using a dashed line, since it is a strict inequality.1007

And then, I took a test point, substituted those values into the inequality, and determined that the origin (0,0) is not part of the solution set.1011

So, my other half-plane contains the solution set.1022

Example 4: we are working with an absolute value inequality where y is less than the absolute value of x - 2, plus 1.1027

I am starting out with just thinking that I expect this to look like a v, because absolute value graphs are v-shaped.1036

I am graphing the corresponding equation, y = |x - 2| + 1.1046

And with absolute value inequalities or equations, I like to get more values,1051

because I want to make sure that I don't end up missing half of my v.1055

So, when x is -2, that is going to give me -4; the absolute value of that is 4; plus one is 5.1061

-1 minus 2 is -3; the absolute value is 3, plus one is 4.1070

0 minus 2 is -2; the absolute value is 2, plus one is 3.1077

Now, some positive numbers: 1 minus 2 is -1; the absolute value is 1; plus one is 2.1085

2 minus 2 is 0, plus one is 1; 3 minus 2 is 1; the absolute value is 1, plus one is 2.1094

4 minus 2 is 2; the absolute value is 2, plus one is 3; and now I have enough values to work with.1111

So, (-2,5), (-1,4), (0,3), (1,2), (2,1), (3,2)...so you see the v forming, so I know I have enough points,1119

because I have both parts of my v...(4,3).1145

It's a strict inequality, so I am using a dashed line.1150

And you see that it is the typical v shape, but it is shifted over.1161

And I have a -2 here, and so this is actually shifted over to the right by 2; and plus 1, so it is shifted up by 1.1165

I found the boundary; again, I am fortunate--I can easily use (0,0) as the test point--the origin.1173

Going back to the inequality: this is going to be 0 < |0 - 2| + 1.1187

So, that is 0 < |-2| (the absolute value of -2 is 2) + 1, so this gives me 0 < 3.1201

0 actually is less than 3; therefore, the origin is part of the solution set, so this is a true statement, and the origin is part of the solution set.1219

(0,0) is part of the solution set, and that means that I am in the correct half-plane, and I am going to go ahead and shade that in.1240

OK, so we handled this just with a very similar technique to linear inequalities.1257

Graphing out the absolute value equation, I got my v-shaped graph with a dashed line for the strict inequality.1263

I used a test point, (0,0), substituted that into the inequality, and found that I had a true statement--a true inequality, 0 < 3.1270

Since that is true, I knew that the origin lay within the region that contains the solution set.1279

So, I went ahead and shaded that region to note that my solution set is located there.1289

That concludes this lesson of Educator.com; see you next lesson!1297

Welcome to Educator.com.0000

In today's Algebra II lesson, we will be discussing solving systems of equations by graphing.0002

Recall that a system of equations, for our purposes, is defined as two equations and two variables.0009

Later on, we will be looking at systems of equations involving more than two variables.0016

But right now, we are just going to stick to this definition.0021

For example, a system of equations could be something such as 2x - y = 7, when it is considered along with another equation, such as x + 3y = -4.0025

And a solution to a system of equations would be a value for x and a value for y--a set of values that satisfies both of the equations.0040

The first technique we are going to discuss in solving systems of equations is solving by graphing.0053

In addition, later on, we will be talking about how these systems can be solved algebraically.0059

So, one way to solve is to graph each equation: the solution for the system is the point of intersection of the two graphs.0065

And we will do some examples in a minute; but for now, imagine that you are given an equation,0075

and you figure out some points, or one of the points in the slope, and it turns out that this is the graph of the line described by the equation.0081

And you are given another equation as part of that system; and you go ahead and graph that one out.0098

And it comes out to a line that looks like this.0105

The solution for the system is the point of intersection, so this is the solution.0110

The x- and y-coordinate for this point are the solution.0117

Now, here you can see a weakness in this method: and that is that, if the solution doesn't land right on an integer, it is difficult to get an exact value.0121

And that is one advantage to the algebraic methods.0136

But if you graph carefully, and the solution does land on, say, (3,4), then you could get an exact solution.0138

There are several types of systems of equations; we call a system independent if it has exactly one solution.0150

For example, just as I showed you in the previous slide, sometimes you will graph out a system, and usually,0159

in the problems you will be doing, you will see a single point of intersection, which is the solution--the single solution.0167

And this set of lines describes an independent system.0175

Other times, you will go along, and you will graph out the system of equations.0185

You have graphed the first equation, and perhaps the graph looks like this.0191

Then, you go to graph the second equation, and it turns out that it describes the same line.0196

So, if the system of equations--both equations within that system describe the same line, then there would be an infinite number of solutions.0202

And the reason is that every single point along this line, every single set of (x,y) coordinates is an intersection of the two systems.0216

The line...every single point along this line (and there is an infinite number of points along the line)0226

is the set of solutions, the set of (x,y) values where the two lines intersect, since they are the same line.0233

Here, we call this a dependent system; here we have one solution; here an infinite number of solutions; and this is a dependent system.0239

The third possibility is that there are no solutions; and we call this an inconsistent system.0254

So, thinking about a situation where there would not be any solution, where there is no point of intersection: it would be a set of parallel lines.0263

Drawing this over here for clarity: here is my x and y axis; and if I graphed the first line, the first linear equation,0273

and then I went ahead and graphed the second one, and those two turned out to have the same slope, those would be parallel lines.0284

They are never going to intersect; therefore, we can see on the graph: there is no solution.0292

So, no solution--that is an inconsistent system.0298

So, there are three possibilities: one solution--intersection at one point; an infinite number of solutions--0306

all the points along the line; or no solution, because it is a set of parallel lines that do not intersect.0312

The first example gives us a system of equations with two variables: x + y = 3 and x - y = 1.0322

So, graphing this first line, I am going to start with x + y = 3; and I find a few x and y values, some ordered pairs, so I can do the graphing.0331

First, I am going to let x equal 0; well, if x is 0, I can see easily that y is 3.0343

So, that gives me the y-intercept; next, I am going to let y equal 0 to find the x-intercept.0351

So, this would be 0; x would be 3; and just one more point to make it a better graph0361

(because it is especially important when you are not just graphing one line--you are actually looking0367

for a solution to a system of equations--to have a really good graph, so you can find a point of intersection accurately):0372

so, when x is 1, x plus y equals 3; so when x is 1, y equals 2.0378

So, I am going to graph this line: first, I have the y-intercept at 3; I have the x-intercept at 3; and another point in between those--x is 1; y is 2.0387

And this gives me a better sense of the slope of the line than if I had just done two points.0399

OK, the second equation--graphing that: x - y =1: again, I am finding a few values for x and y--a few sets of values.0406

When x is 0, that would give me 0 - y = 1, so - y equals 1; therefore, y is -1; the y-intercept is -1.0422

Now, finding the x-intercept, let y equal 0; if y is 0, x is 1.0434

One final point: I am going to just let x equal 2; so that would be 2 - y = 1; and that is going to give me -y = -1, so y = 1.0442

Again, I have three sets of points to graph.0457

So, the y-intercept is at -1; the x-intercept is at 1; and one more point: when x is 2, y is 1.0460

OK, drawing this line, I can immediately see that I have a single point of intersection, right here.0476

This is the solution; and this occurs at (2,1).0487

So, the solution for this is x = 2, y = 1.0493

And therefore, this is an independent system, since it has one solution.0501

OK, and you can always check your work by substituting x and y values back into these equations.0515

I have x + y = 3, so that is 2 + 1 = 3; and that checks out.0528

I have x - y = 1, so 2 - 1 = 1, and that checks out; so you can easily check and make sure that both of these are valid solutions for both of the equations.0534

The second example is a little bit more complicated, but using the same system.0551

And this time, instead of just finding (x,y) values, I am going to graph these out by putting both equations into slope-intercept form.0557

So, the first one is 2x - 4y = 12, so I am going to subtract 2x from both sides, and then I am going to divide both sides by -4.0565

And that is going to give me y = 1/2x - 3.0577

Slope-intercept form is very helpful when you are graphing: this is -2, -4, -6, -8, -10, 2, 4, 6, 8, 10.0582

OK, here I have a y-intercept at -3; and I know the slope--the slope is 1/2.0597

That tells me, when I increase y by 1, I am going to increase x by 2; if I increase y by another 1, I am going to increase x by 2.0606

So, you can see the slope of this line right here; that gives me enough to work with.0620

I am going to do the same with the second equation, 4x - y = 10, putting this into slope-intercept form, y = mx + b,0638

subtracting 4x from both sides, and then dividing both sides by -1 to give me y = 4x - 10.0646

So here, I have a y-intercept at -10 and a slope of 4; I am going to increase y by 4 (that is 2, 4), and increase x by 1.0656

Increase y by 2, 4; increase x by 1; increase y by 4 (2, 4); increase x by 1; OK.0669

So, this is a much steeper line; the slope is much steeper--the slope of 4; so I am going to go ahead and draw it like that.0687

And the point of intersection is right here, and again, we have an independent system with a single solution.0695

And it is at (2,-2), so that is my point of intersection: x is 2; y is -2.0708

Again, you could always check your answers by substituting either or both equations with these x and y values, and ensuring that the solutions are valid.0723

So again, I have an independent system; it has one solution--one set of valid solutions.0733

OK, this next example: again, solve by graphing; and my approach is going to be to put these into y-intercept form and use that to graph.0748

First, 2x - 3y = 6, so -3y = -2x + 6; divide both sides by -3; that is going to give me...-2/-3 is 2/3x, and then 6/-3 is -2.0758

So, this line has a y-intercept of -2 and a slope of 2/3; so increase y by 2; increase x by 3.0781

OK, that is my first equation; the second equation--again, putting it into the slope-intercept form, which is an easy way of graphing:0799

This gives me 6y = 4x - 12; dividing both sides by 6 gives me 4/6x - 2.0808

And I am going to go ahead and simplify this to 2/3x - 2.0823

So, you may already see that these are the same equation--these are going to describe the same line.0828

Even if you didn't notice it right away, when you start graphing, you are going to see: the y-intercept is -2; the slope is 2/3.0835

You are going to end up with the same line.0842

Because this is the same line, they intersect at every single point; these two equations--their graphs intersect at every single point,0844

which is an infinite number of points, along the line; they don't intersect everywhere, but everywhere on this line.0852

Therefore, this is a dependent system, and there is an infinite number of solutions; all points along this line are solutions.0857

And this is known as a dependent system.0871

So initially, you looked at this; you might not have recognized that these are actually describing the same line.0877

But once you started to plot it, either by putting it in the y-intercept form or by finding points along these lines,0882

you would have quickly realized that this is the same line.0890

In this last example, again, there is a system of equations that we need to solve by graphing.0896

Again, using the method of slope-intercept form for graphing: 4x - 2y = 8; subtract 4x from both sides, and then divide both sides by -2 to give me y = 2x - 4.0905

OK, the y-intercept is -4; the slope is 2; increase y by 2; increase x by 1; increase y by 2; increase x by 1; and so on.0925

I have three points; that is plenty to graph this line...so that is my first line.0942

The second line: again, slope-intercept form: add 6x to both sides--it gives you 3y = 6x + 12.0947

Now, divide both sides by 3 to get y = 2x + 4.0961

So here, I have a y-intercept of 4 and a slope of 2; so increase x by 2; increase y by 1.0968

So, I am going to get a line looking like this; and what you may quickly realize is that these are parallel lines.0978

One thing you could have noted is that the slope, m, equals 2 for this line, and it equals 2 for this line.0991

Since these are parallel lines, they are never going to intersect; and this is an inconsistent system, and there is no solution.0998

And what we say here is: you could say the solution is just the empty set; there is no solution.1008

This is described as an inconsistent system.1016

OK, that concludes this lesson of Educator.com, describing graphing to solve systems of equations.1024

And I will see you next time.1032

Welcome to Educator.com.0000

In today's lesson, we will be talking about solving systems of equations algebraically.0002

In the previous lesson, we talked about solving systems of equations by graphing.0008

However, that method has some limitations; therefore, we are going to talk about a few other methods of solving.0013

And this is some review from Algebra I; so again, if you need more detail--more review on these concepts--check out our Educator.com Algebra I series.0020

First, we are going to review solving by substitution, first jotting down a system of equations:0030

2x + 3y = 12, and the second equation in the system is x + 4y = 1.0036

In this method, you are going to solve one equation for one variable in terms of the other variable.0047

Then, substitute the expression for the variable in the other equation.0053

Let's look at what this means: first step: solve one equation for one variable in terms of the other one.0059

The easiest thing to do is to find a situation where you have a variable that has a coefficient of 1; and I have that right here.0065

So then, I can solve for x in terms of y pretty easily.0071

So, x + 4y = 1, so I am going to solve for x in terms of y; this gives me x = -4y + 1.0075

The second step is to substitute this expression for the variable in the other equation.0088

So, I am going to substitute -4y + 1 for x in the first equation; and that is going to create0094

an equation that has only one variable, and then I will be able to solve that.0101

It is important that you go back into the other equation to substitute in.0105

This is going to give me 2(-4y + 1) + 3y = 12; then, I can just solve for y.0111

2 times -4y; that is -8y, plus 2 times 1--that is just 2--plus 3y, equals 12.0124

Combine like terms: -8y and 3y gives me -5y, plus 2 equals 12; subtract 2 from both sides to get -5y = 10.0135

Dividing both sides by -5, I get y = -2.0150

Once I have one variable solved, I can easily solve for the other.0155

So, I just go back to either one of these (and this one is easier to work with), and I am going to substitute in...I am going to let y equal -2.0159

x + 4(-2) = 1; this is going to give me x + -8 = 1, or x - 8 = 1; adding 8 to both sides, I get x = 9.0170

Here, y equals -2 and x equals 9.0192

This method of substitution works really well when the coefficient of one of the variables is 1.0197

So, use substitution when a variable in one of the equations has a coefficient of 1.0203

So again, we looked at this system of equations, saw that this variable had a coefficient of 1,0232

then solved for this variable x in terms of the other variable y to get x = -4y + 1.0240

And then, I went ahead and substituted this 4x in the other equation; that gave me one equation with one variable, and I can solve for y.0247

Once you have y, you can substitute that value into either equation and then solve for x.0260

The second method is solving by elimination: in elimination, you add or subtract the two equations to eliminate one of the variables from the resulting equation.0267

And this system works well when you have variables, either the two x's or the two y's, that have either the same coefficient or opposite coefficients.0277

By "opposite coefficients," I mean the same number with opposite signs, such as 2 and -2 or 3 and -3.0286

For example, if you were asked to solve this system, you look and see that there is no variable with a coefficient of 1.0293

Therefore, substitution is not the ideal--it is not the easiest way to go.0306

But what you see is that you have one variable--you have y--that has the same coefficient.0312

So, adding is certainly not going to help me; if I add these together, I will get 6x + 10y = 8.0318

What I need to do is subtract, because my goal is to get one variable to drop out.0324

So, I am going to subtract the second equation from the first.0328

And just to help me keep my signs straight, I rewrite that as adding the opposite; so I am adding -4x, -5y, and -3.0333

OK, this is going to give me 2x - 4x is -2x; the y's are going to drop out: 5y - 5y gives me 0, so I can write + 0, but they just drop out.0352

And then, 5 minus 3 equals 2; this is -2x = 2; I can easily solve for x, and x equals -1.0365

Once I have one variable, I can go to either original equation, and then substitute in order to solve for the other variable.0376

I know that x equals -1, so I am going to substitute that up here.0385

Add 2 to both sides (that is going to give me 5y = 7), and divide both sides by 5 to get y = 7/5.0395

So, solving by elimination, I was able to come up with the solution that x equals -1 and y = 7/5, which will satisfy both of these equations.0403

Again, this works well; use elimination when the same variable (meaning both x's or both y's)0412

have the same or opposite coefficients (the same coefficient, but opposite signs).0426

Sometimes, you look, and you see that there is no variable with a coefficient of 1; and then you say, "OK, I will use elimination."0442

But then, you realize that none of the variables have the same or opposite coefficients.0449

In that case, you can still use elimination, but you are going to have to take an extra step before you do the elimination.0453

And the extra step is to multiply one or both equations so that one of the variables has the same or opposite coefficient in the two new equations.0459

So, your goal is to create a situation where one of the variables has the same or opposite coefficients.0469

Then, you just use elimination, as we did previously.0476

If you had a system of equations, 5x + 4y = 1, and 6x + 3y = 3, you see that neither of these0480

has a variable with a coefficient of 1, and neither set of variables has the same or opposite coefficients.0489

But if I look at these two, 4y and 3y, the least common multiple is 12; so what I want to do0497

is multiply each of these by something, in order to end up with a coefficient of 12 in front of the y.0505

Sometimes you will be lucky, and all you will have to do is multiply one of the equations by a number to get the opposite or same coefficient.0511

Other times, like this, you are going to have to multiply both.0519

So, for the first equation, I am going to multiply by 3; and this is going to give me 15x + 12y = 3.0522

The second equation I am going to multiply by 4: this is going to give me...4 times 6x is 24x + 12 y = 12.0542

OK, I now see that I have the same coefficient for y; so rewriting this over here, I need to subtract.0559

So, I am going to subtract the second equation from the first.0576

And to keep everything straight, I like to go about this by adding the opposite to make sure that I keep my signs straight.0581

So, I am going to add -24x - 12y, and then that is going to be a -12.0589

OK, -24x - 12y - 12: add these together: I get 15 - 24x--that is going to give me -9x; the y's drop out, and then I have 3 - 12; that is -9.0603

Divide both sides by -9 to get x = 1.0618

From here, I substitute back; I will choose this top equation, 5x + 4y = 1, and I know that x equals 1.0622

This is 5(1) + 4y = 1; so that is 5 + 4y = 1, or 4y = -4, so y = -1.0631

So, x equals 1; y equals -1.0645

Again, this is just the first step in elimination; and you use it in a situation like this,0648

where you want to use elimination, but you don't already have a set of variables with the same or opposite coefficients.0653

So, you figure out what the least common multiple is, and multiply one or both equations in order to achieve that.0660

From there, you just proceed as we did before, solving by elimination.0667

There are several different possibilities that can occur when you are solving systems of equations algebraically.0675

Usually, in the problems that you will see, what will happen is what just happened previously,0683

that I showed you, where you will end up with a value for x and a value for y that satisfy both equations.0688

However, there are times when that doesn't occur.0695

You can be going along, doing substitution, doing elimination, and things are going fine;0699

and then you end up with something like this: c = d.0705

Your variables drop out, and you end up with an equation that is saying that a constant is equal to a different constant--for example, 4 = 5.0710

Well, that is not true; so when you see this, this is an equation that is never true.0719

And what this tells me is that the system of equations is inconsistent.0727

So, this is a situation where there is no solution.0742

If you start seeing a situation where the constants drop out, and then you see something like 4 = 50752

or 9 = 10--something that is not true--then you know you have an inconsistent system.0758

The other possibility is that you could have a system that is dependent, or always true.0764

In that case, you are going along; you are doing elimination; you are doing substitution;0773

and then you see that you end up with variables dropping out, and a constant that equals a constant--the same constant--for example, 3 = 3.0778

Well, that is always true; so if you end up with an equation--the sum or difference of the two equations--0788

the system of equations--that is always true, this system is dependent, and it has an infinite number of solutions.0795

Recall, when we were solving systems of equations by graphing: this is analogous to the situation where you would end up with the same line.0812

So, if you have two equations (a system of equations), you graph both equations,0819

and you find out that they are the same line, well, then, there is an infinite number of solutions--all points along those lines.0823

Here, no solution would be like parallel lines, where they never intersect; there is no solution to that system.0829

OK, the first example is: Solve algebraically for the system of equations.0839

As soon as I see that I have variables (or even one variable) with a coefficient of 1, I recognize that substitution would be a really good method to use.0844

So, use substitution when a variable has a coefficient of 1.0853

So, I am going to solve for x in terms of y: x + y = 5, so x = 5 - y.0856

Now, I have solved for x in terms of y; and now I am going to substitute this value into the other equation.0867

2x + 3y = 13: I am going to substitute this expression for x, so 2 times (5 - y), plus 3y, equals 13.0875

Or, 10 - 2y + 3y = 13; this is 10 + y = 13; then subtract 10 from both sides to get y = 3.0889

Once I have this, I can plug y into this simple equation up here: x + y = 5.0906

Substitute in; let y equal 3; and I get x = 2.0916

And this is very easy to check; you can see that, if x is 2 and y is 3, this equation holds true.0924

And you could do the same--plug it back into that first equation, put these values in for x and y, and just verify that these solutions are correct.0929

OK, so again, we are solving by substitution because you had a situation where you had a variable (actually, two variables) with a coefficient of 1.0942

That is the easiest method to use.0950

Here, I am solving this system of equations algebraically; and I don't have any variables with a coefficient of 1, so I am not going to use substitution.0954

But if I look; I have variables (y) that have the opposite coefficient, -3 and 3.0963

What that means is that, if I add these equations, the y's are going to drop out; then, I can just solve for x.0970

Rewrite this right here and add: I am going to add these two.0977

2x + 4x: that gives me 6x; the y's drop out: -3y + 3y is 0; and then 0 + 0 is 9.0988

I am going to divide both sides by 6; this is going to give me 9/6, and that simplifies to 3/2.1004

I have x = 3/2; I am going to pick either one--I will go ahead and pick the top one: 2x -3y = 0.1011

And let x equal 3/2; substitute that for x.1018

The 2's cancel out; this gives me 3 - 3y = 0, or -3y = -3; if I divide both sides of the equation by -3, I will get y = 1.1028

So, the solution is x = 3/2, y = 1; again, I could always check these solutions1042

by substituting the values in here and making sure that this equation holds true.1051

And this was a perfect setup to use elimination, because I already had variables that had the opposite coefficients.1056

I simply added these equations together; the y dropped out, allowing me to solve for x, and then substitute 3/2 in for x.1063

OK, Solve algebraically: again, there are no variables with a coefficient of 1, so I am going to go to elimination.1075

But these do not have the same or opposite coefficients, and the y's do not have the same or opposite coefficients.1083

So, this time I am going to need to use that extra step; I am going to need to use multiplication1090

in order to create a situation where I have variables with the same or opposite coefficients.1095

And I see that, if I multiply the top equation by 3, I will get 6x; and if I multiply the bottom equation by 2, I will get 6x.1102

So, I am going to go ahead and do that--multiply the top equation by 3: that is going to give me 3 times 2x - 3y = -16.1112

And this comes out to 3(2x) is 6x; 3 times -3...that is -9y; and 3 times -16 is -48.1129

The second equation I am going to multiply by 2.1144

So, 2 times 3x, plus 5y, equals 14; OK, 2 times 3x gives me 6x; 2 times 5y is 10y; and 2 times 14 is 28.1147

I have the same coefficient here; so what I need to do is subtract.1165

I am going to subtract, and again, I am going to convert this so that I am adding the opposite.1171

Rewrite the top equation the same, and the bottom as adding the opposite (adding -6x - 10y - 28).1185

OK, when I do that, the 6x's drop out, and that is going to give me -19y equals...-48 and -28 comes out to -76.1196

Divide both sides by -19, and -76 divided by -19 actually equals 4, so it divided very nicely and evenly; so I ended up with y = 4.1215

Now that I have that, I am going to substitute y = 4 into either equation; I am going to choose the top one.1227

So, 2x - 3y = 16; let y equal 4; so, 2x - 3(4) = -16, or 2x - 12 = 16.1233

Add 12 to both sides to get 2x = -4, and divide both sides by 2 to get x = -2.1251

So, the solution is that x equals -2 and y equals 4.1259

So, when I looked at this, I saw that I could use elimination, if I got the x's to have the opposite coefficient1264

by multiplying the top equation by 3 to give me 6x, and the bottom equation by 2 to give me 6x here, as well.1274

So, I did that; and then, these have the same coefficient, so I subtracted and solved for y.1283

Once I had a value for y, I substituted that value into one of the equations and then solved for x to get the solution.1291

Here, I have 2x + 3y = 8, and -4x - 6y = -16.1303

Since there is no variable with a coefficient of 1, I am going to use elimination.1311

But again, I am going to have to do a little work to get the opposite coefficients.1314

And I see here that all I have to do is multiply the top by 2, and that will give me 4x; that is opposite coefficients.1319

I don't have to do anything to the bottom equation.1329

So, that is 2 times 2x, plus 3y, equals 8, and this is going to give me 4x + 6y = 16.1330

That is the opposite coefficient, so I am going to add this new equation and this equation: + -4x - 6y - 16.1343

And you may have already realized that each one of these terms is opposite.1354

And so, here is what is going to happen: 4x - 4x is 0; 6y - 6y is 0; 16 - 16 is 0; so I end up with 0 = 0.1360

And I didn't make any mistakes--I did everything correctly--but then all my variables went away.1373

And what this tells me is that I have a dependent system, and it has an infinite number of solutions.1377

If I were to graph this, I would see that I have an infinite number of solutions--that these are intersecting at every point along the line.1386

So, we call this a dependent system.1399

If I had been going along, and all of the variables dropped out, and then I got something that wasn't true,1405

like 4 = 2 or 4 = 0, then I would have a situation where it is an inconsistent system and there are just no solutions.1410

But here, this is always true; so I have an infinite number of solutions.1418

So today, we covered solving systems of equations algebraically.1425

And that concludes today's lesson for Educator.com; I will see you again.1430

Welcome to Educator.com.0000

Today, we will be covering solving systems of inequalities by graphing.0002

In order to solve a system of inequalities by graphing, you need to graph the solution set of each inequality.0009

So, previously we discussed techniques; and you can review that lecture about how to find the solution set of an inequality by graphing.0015

Here, all you are going to do is graph each inequality.0024

And the solution of the system of the inequalities put together is the intersection of the solution sets of the inequalities.0027

So, what this is, really, is the overlap--the area of overlap--the points that are in common between each of the solution sets.0035

For example, if you are given a system of inequalities: y ≤ x - 3, and y > -2x + 1,0044

recall the techniques for graphing an inequality to determine the solution.0056

First, you want to graph the corresponding linear equation to find the boundary line of the solution set.0063

Then, use a test point to determine the half-plane containing the solution set.0082

Let's go ahead and do that, and then talk about how to find, then, the solution set.0099

This is just graphing each one--finding the solution set for each one; and then we have to talk about how to find the solution set for the system.0105

So, I am going to start with this one; and that is y ≤ x - 3; and the corresponding linear equation would be y = x - 3.0111

So, this is in slope-intercept form, y = mx + b, so I can easily graph this, because I know that the y-intercept is -3 and the slope is 1.0126

Since the slope is 1, increase y by 1; increase x by 1; increase y by 1; increase x by 1; and on.0139

Now, looking back here, this says y is less than or equal to -3; and what that tells me is that I am going0148

to use a solid line here for the boundary line, because the line is included in the solution set; so use a solid line to graph this.0158

The next thing is to determine which half-plane--the upper or lower half-plane--this solution set is in.0174

And I want to use an easy test point, and that is the origin (0,0), as my test point.0179

And it is well away from the boundary line, so I can use that.0185

If it were near or on the boundary, I would want to pick a different test point.0189

I am going to take my test point, (0,0), and substitute those values in to the inequality.0194

So, 0 is less than or equal to 0 minus 3; 0 is less than or equal to -3.0200

Well, that is not true; since that is not true, that means that the solution set is not in this upper half-plane.0208

(0,0) is not part of the solution set; therefore, the solution set is in this lower half-plane.0216

OK, that is the solution set for this inequality; now let's work on the other one.0223

Here, y is greater than -2x + 1; the corresponding linear equation is y = -2x + 1.0229

So, here I have the y-intercept at 1 and a slope of -2.0240

So, decrease y by 2...1, 2; increase x by 1; decrease y by 2; increase x by 1; and so on.0246

Now, since this is a strict inequality, I am going to use a dashed line, because the boundary line here is not part of the solution set.0256

OK, so here is my dashed line; next, I need to find a test point.0267

And again, I am going to use (0,0); I am going to use that test point.0281

Actually, let's use a different one, because that is a bit close; so let's go ahead and pick one0293

that is farther away, so we don't have any chance of causing confusion.0297

Let's use something up here...(2,1); (2,1) would be good for the test point.0301

So, my test point is going to be (2,1): I am going to insert (2,1) into that inequality.0308

1 is greater than -2; x is 2; plus 1; this is my test point; and 1 is greater than -4 plus 1; 1 is greater than -3.0316

And this is true; since this is true, that means that the test point is in the correct half-plane where the solution set lies.0331

So, for this line, this is the upper half-plane, and this is part of the solution set; so I am going to go ahead and shade that in.0340

OK, so far, I had just been going along, doing what I usually do when I would find the solution set for an inequality.0356

But remember, we are looking for the solution set for a system of inequalities.0364

So, this lower half-plane is the solution set for this inequality; this half-plane is the solution set for this inequality.0369

Now, I am looking for the intersection between the two solution sets, and that is right here, in this quadrant right here.0376

Darken that in even more; I am going to go ahead and use another color to emphasize that.0382

OK, so the area bounded by this line and including that line--that is the boundary line for the system of equations.0396

And then, the boundary line over here is the dashed line; so this line is not part of the solution set.0406

The technique: graph each of the inequalities; find their solution sets; and then find the area that is the intersection of the two solution sets.0414

You may come across a situation where there is no solution to a system of inequalities.0427

And this is because the two inequalities may not have any points in common.0433

Remember: the common points in each of the solution sets comprise the solution set for the system of inequalities.0438

If the two inequalities have no points in common, then the solution set of the system is the empty set.0446

For example, let's say I am given y < x - 4 and y ≥ x - 2.0454

So, I am going to go ahead and graph those, starting with y < x - 4.0462

I need to find the boundary line, so the corresponding linear equation is y = x - 4.0467

So, - 4 is the y-intercept, and the slope is 1.0474

So, increase x by 1, and increase y by 1, with each step.0480

Since this is a strict inequality, I am going to use a dashed line.0487

I have my boundary line; now I need to use a test point, and I am going to use this (the origin) as the test point.0495

And I am going to substitute (0,0) into the inequality and see what happens.0503

This tells me that 0 is less than -4; and that is not true.0510

Since that is not true, this is not part of the solution set; the solution set is actually below this line--the lower half-plane.0516

OK, that was my first inequality; now, my second inequality is y ≥ x - 2.0525

The corresponding linear equation is y = x - 2.0532

Here, I have a y-intercept of -2 and a slope of 1; so increase x by 1 and y by 1 each time.0538

Now, here I am going to use a solid line; this line is part of the solution set.0551

And a test point: I can use (0,0) again--that is well away from this boundary line.0561

Substituting these values into this inequality gives me 0 ≥ -2; and that is true: 0 is greater than -2.0567

So, this upper half-plane describes the solution set.0578

Now, you can see what happened here; and you may have already noticed that these two lines have the same slope.0585

Since they have the slope, and they are parallel lines, that means they are never going to intersect.0592

Since this solution set is above this line, and this one is below the line, there are not going to be any points in common.0598

These two lines will go along forever, and the points above and below them will never intersect.0605

So, this is a situation where there is no solution; we just say it is the empty set.0611

So here, we saw a situation where there is no solution, because there are no points in common.0622

OK, the first example is very straightforward: x ≥ 2; y > 3.0628

Starting with the x ≥ 2: the corresponding linear equation would be x = 2, and that just tells me that, if x equals 2,0635

no matter what the y-value is, it is just saying x is 2; so that is going to be a vertical line.0650

And since this is greater than or equal to, I am going to make this a solid line.0657

Now, you could certainly use a test point; or you could just look at this and say,0674

"Well, it is saying that x is greater than or equal to 2, which means that it is going to be values to the right."0678

You could certainly say, "OK, I want to do a test point at (0,0); 0 is greater than or equal to 2--not true, so this is not part of the solution set."0685

Or, like I said, you could just look back here and say, "OK, it is telling me that the values of x are greater than or equal to 2."0697

So, this half-plane contains the solution set.0704

The second inequality, y > 3, has a corresponding linear equation of y = 3.0711

So, if y equals 3, that is going to be a horizontal line right here.0720

And it is a strict inequality, so I am going to make that a dashed line, making it a different color so it shows up.0725

So, this blue line is that y = 3; OK.0735

Again, I could either just go back and say, "All right, this is y > 3, so those points would be up here";0743

or I can always use my test point, (0,0), and say 0 is greater than 3--that is not true, so I know this is not part of the solution set.0750

So, I need to go up here; OK.0761

So again, the solution set for the system of inequalities is the area of intersection of the two solution sets.0774

For this first inequality, the solution set is over here.0782

For the second inequality, the solution set is up here; and I can see the area of intersection is right up here.0786

And it is to the right of this solid vertical line, and it includes the points on the line.0793

And it is above the dashed blue line, and it does not include the points on the line.0799

We are just graphing each inequality and finding this area of intersection.0805

Here it is slightly more complicated, but the same technique.0811

Again, this is a system of inequalities, starting with the first one: y > x + 1.0815

The corresponding linear equation is y = x + 1.0822

The y-intercept is 1, and the slope is 1.0828

So, increase y by 1; increase x by 1; increase y by 1; increase x by 1.0835

And this is also going to be a dashed line, since it is a strict inequality; this boundary line is not part of the solution set.0842

Take a test point, (0,0); substitute back into that inequality to give 0 is greater than 0 + 1; so that says 0 is greater than 1.0858

Well, that is, of course, not true--which means that this half-plane where the test point is, is not the solution set.0874

Instead, it is the upper half-plane; so I am going to shade that in to indicate that this is the solution set for the first inequality.0884

The second inequality is y = -x + 2; well, the inequality is y > -x + 2; the corresponding linear equation is y = -x + 2.0893

So, that gives me a y-intercept of 2 and a slope of -1.0907

Decrease y by 1; increase x by 1; decrease y by 1; increase x by 1; and on down.0914

Again, it is a strict inequality, so we have another dashed line; I am making this a different color so it stands out as a separate line.0923

This blue line is the boundary line for the solution set of the second inequality.0941

OK, again, I am using (0,0) as my test point.0949

I am substituting into that inequality: 0 is greater than -0 + 2, so 0 is greater than 2.0956

And again, this is not true; so looking at this line, this point is not part of the solution set.0966

So, this lower half-plane is not the solution set.0976

I am going to go to the upper half-plane and shade that in.0979

OK, therefore, the solution set for the system of inequalities is right up here in this corner.0996

It is the points above the black line (but not including this line) and above the blue line (but not including that boundary line).1004

The technique, again: graph the first inequality (which was right here); we found the solution set in the upper half-plane.1014

We graphed the second inequality and found its solution set in the upper half-plane.1022

And then, we noted that this area up here is the overlap between the two inequalities; and that contains the solution set for the system.1027

OK, Example 3: 2x - y > 4 is my first inequality.1041

And one thing that I see right away is that this is not in the standard form we usually use.1050

So, I am going to work with this to put it in a standard form, where y is isolated on the left; and that is going to give me -y > -2x + 4.1055

Now, I have to divide by -1; and recall that when you divide an inequality by a negative number,1067

then you have to reverse the direction of the inequality symbol.1073

So, this is going to become y < 2x - 4; and it is absolutely crucial that, if you are multiplying1079

or dividing by a negative number, you immediately reverse the inequality symbol; or you won't end up with the correct solution set.1089

OK, so that is my first inequality; I am going to go ahead and do that same thing with the second inequality before I work with either one of these.1096

So, here I have 4x + 2y ≥ -2, so I am going to subtract 4x from both sides.1107

And then, I am going to divide both sides by 2; and since 2 is a positive number, I can just keep that inequality symbol as it was.1118

This is -4 divided by 2, so that gives me -2x - 1.1125

OK, looking at this first inequality: the corresponding linear equation, now in standard slope-intercept form, is y = 2x -4.1138

So, by putting it in that form, we made it much easier to graph.1153

The y-intercept is at -4, and the slope is 2.1158

So, increase y by 2; increase x by 1 for the slope.1162

Looking back, this is a strict inequality, so I need to use a dashed line, indicating that the boundary is not part of the solution set.1169

And then, a test point--I need to determine where the solution set lies--in the upper or lower half-plane.1189

I am using the origin as my test point, substituting (0,0) in for x and y.1198

This gives me 0 - 0 > 4, or 0 is greater than 4; and that is not true.1208

Since that is not valid, that is not part of the solution set; so this is not the correct half-plane.1216

I am going to shade the lower half-plane instead.1221

OK, over here, the corresponding linear equation is y = -2x - 1.1228

So, the y-intercept is -1; the slope is -2x.1237

So, when I decrease y by 2, I am going to increase x by 1; decrease y by 2; increase x by 1.1243

Again, I have to check here, and I see that I am going to use a solid line for this boundary line,1258

because the line is part of the solution set.1264

And then, I need a test point here; and I want to pick a test point that is not quite so close to this boundary line,1278

just in case my graphing wasn't perfect, so I am going to select (1,2) right up here as my test point, just to be safe.1285

Substitute those values into this inequality right here, or back up here--either way.1296

I am going to go ahead and use this one, because it is simpler: 2 ≥ -2(2) - 1.1308

2 is greater than or equal to -4 minus 1; so, 2 is greater than or equal to -5; and that is true.1319

So, my test point is in the half-plane containing the solution set.1333

I am going to shade this upper half-plane.1340

Once I have done that, I can see that the solution set for the set of inequalities, the system of inequalities, is right here.1348

It is the area bounded by this line and including the line, and then the area bounded by the dashed line,1356

right here, but not including it; so, this lower right section of the graph.1363

For this one, we had to take an extra step, just to make it easier--put this in standard form.1372

And then, we graphed the corresponding linear equation to find the boundary line.1376

We used the test point to find that the lower half-plane was the solution set for this first inequality.1381

The same technique for the other inequality: and the upper half-plane turned out to contain the solution set.1388

So, my area of intersection is right here; that is the solution set for the system.1395

OK, Example 4: y < 2x + 1; y ≥ 2x + 3.1403

We are going ahead and starting out with the first inequality, y < 2x + 1.1412

I am finding my boundary line with the corresponding linear equation, y = 2x + 1.1424

Here, the y-intercept is 1; the slope is 2; so increasing y by 2 means increasing x by 1, and continuing on.1429

I am checking back and seeing that I need a dashed line, because this is a strict inequality.1446

So, I am graphing out this line with a dashed line; so this line will not be part of the solution set--the points on the line.1452

The second inequality is y ≥ 2x + 3; the linear equation is y = 2x + 3.1463

I am going to go ahead and graph this out; and this tells me that the y-intercept here is 3.1475

And the slope is 2: increase y by 2, increase x by 1; or decrease y by 2, decrease x by 1; and so on.1484

So, this is going to give me another line, right next to this one.1499

But this time, I am actually going to use a solid line.1504

So, this is a solid line, because this is greater than or equal to.1509

Now, let's look at some test points for each.1517

For this first one, let's use a test point of (0,0) right here for this line and substitute in.1520

0 is less than 2 times 0 plus 1; that gives me 0 < 0 + 1, or 0 is less than 1.1528

And that is true; so I have this as part of my solution set right here, so it is the lower half-plane.1537

I found the solution set for this inequality.1550

For this one, I am also going to use (0,0); that is well away from that line--that is a good test point.1552

Substitute in: 0 is greater than or equal to 2 times 0, plus 3; 0 ≥ 0 + 3; 0 is greater than or equal to 3.1560

That is not true; so, for this line, (0,0) is not part of the solution set; so it is the upper half-plane.1572

Now, it is a little tough to draw: but if you look here, one thing that you will see is that these lines have the same slope.1585

Because of that, I know that these are parallel lines, so I know that these two lines are parallel and that they will never intersect.1593

Since the points are all below this parallel line in that half-plane, and above this line (they are parallel to it), these solution sets are never going to intersect.1602

In this case, the solution is the empty set; there are no points in common, since these are parallel lines.1614

That concludes this session on solving systems of inequalities by graphing at Educator.com; see you again!1625

Welcome to Educator.com.0000

Today we are going to be going on to talk about solving systems of equations in three variables.0002

In previous lessons, we talked about how to solve systems with two variables.0007

So now, we are going to go up to systems involving three variables.0012

In order to solve these systems, you are actually going to use the same strategies you used for solving systems with two variables.0017

Recall those techniques: substitution, elimination, and multiplication.0027

This time, though, a solution is an ordered triple of values.0031

So, you will end up having three variables: for example, (x,y,z).0035

And the solution would be something like (5,3,-2), where x is 5, y is 3, and z is -2.0040

Now, just to work out an example to show you how to approach these using the same techniques that you already know:0056

looking at this system of equations, I have three equations, and I have three total variables.0082

And the idea is to work with the three equations so that you get one of the variables to drop out.0088

Once you get one of the variables to drop out, you will be left with a system of two equations0096

with two variables, and you already know how to work with that.0100

So, the technique would be to first just consider two of the equations together; I am going to call these equations 1, 2, and 3, so I can keep track of them.0103

I am going to first consider equations 1 and 2.0113

And when I look at these, I see that the two z's have opposite coefficients.0116

And you will recall that elimination works really well in that situation.0125

So, I am going to take 2x + 3y - z = 5 (that is equation 1) and equation 2: 3x - 2y + z = 4, and I am going to add those.0129

This will give me 5x, and then 3y - 2y is going to give me y; the z's drop out; 5 and 4 is 9.0146

So now, I have a new equation, and I will just mark this out so I can keep track of it.0158

So, once you have gotten a variable to drop out, work with two different equations to get the same variable to drop out.0167

So, I got z to drop out; and what I want to do is work with two different equations--I worked with 1 and 2.0175

I could work with 1 and 2, or I could work with 2 and 3.0182

And I am actually going to work with 2 and 3; and I want to get z to drop out.0187

Equations 2 and 3: looking at this, how am I going to get z to drop out?0193

Well, in order to do that, I could use elimination; but I am first going to have to multiply this second equation by 2.0197

So, this is equation 2; and it is 3x - 2y + z = 4; and I am going to multiply that by 2 to give me 6x - 4y + 2z = 8.0207

Now, I am going to take equation 3 (this is equation 2, and I am going to take equation 3): I want to make sure that I am working with two different equations.0229

So, I have 1 and 2, and 2 and 3 (or I could have done 1 and 3--either way).0237

-4x + y + 2z = 3: my goal is to get the z's to drop out.0243

In order to do that, I am going to have to subtract: I need to subtract 3 from 2.0256

I want to be very careful with my signs here, so I am going to change this to adding the opposite.0267

This is going to give me 6x + 4x, which is 10x; -4y and -y is -5y; 2z and -2z drops out 8 minus 3 is 5; OK.0276

At this point, what I am left with is a system of two equations with two unknowns.0297

Once you get that far, you proceed using the techniques that we learned previously (again, substitution, elimination, and multiplication).0303

But since you are only working with two equations with two unknowns, you are on familiar territory.0312

And you can solve for one of the variables and then find that value; substitute in for the other variable0316

and find that value; and then you can go back and find z.0325

And we are going to work more examples on this; but the basic technique is to work with two equations0329

to eliminate a variable, using either elimination or substitution, then work with two other equations0335

to eliminate that same variable, resulting in two equations with two variables,0342

allowing you to solve for one variable, then the other, and then the third.0349

Just as in systems with two variables, a system with three variables may have one solution, no solutions, or an infinite number of solutions.0357

So, recall: the solution set here, if I had three variables (x, y, and z) would be a value for x0368

(such as 2), a value for y (such as -4), and a value for z (such as 1) that would satisfy all three equations.0375

The other possibility is that there may be no solutions; recall from working with systems of equations with two variables--0384

you know that you are in this situation when you are using elimination, or you are using substitution;0393

you are going along; and then you see variables start to drop out, and you end up with an equation0399

where you have a constant equaling another constant, which is never true.0406

So, if you start seeing variables drop out, and you end up with something such as 4 = 7 (which is never true),0413

this tells you that there is no solution to this system of equations.0420

There can be an infinite number of solutions; when you are working with your equations;0426

you are eliminating; you are substituting; you are using your techniques; you are being careful;0434

you are doing everything right, and then you see variables drop out, and you get a constant equaling a constant, like 2 = 2.0438

Well, that is always true; and this means that that system of equations has an infinite number of solutions.0447

So, typically, you will get one solution: a value for x, y, and z that is the set of values that makes the equations true--that satisfies the equations.0457

You may end up, though, with no solutions (there are no solutions to this equation) or an infinite number of solutions for this system of equations.0467

OK, in the first example, we are given a system of three equations with three variables.0479

So again, I am going to work with two of the equations; I will number them 1, 2, and 3.0486

And I will work with two of them, and then a different two.0493

So, first, I am going to look at the second two equations; and they are very easy to work with, because I have opposite coefficients.0500

I could also use substitution, because I have coefficients of 1; but I am just going to use elimination.0509

So first, I am going to work with 2 and 3: y - z = 2, x (let me move that, so it doesn't create confusion) + 2y + z = 2.0515

OK, here I end up with x; all I am doing is adding these together using elimination.0533

x + 3y; the z's drop out; 2 + 2 is 4.0543

OK, so I eliminated z from this first set of equations; now, I need to work with a different set of equations.0550

And there are actually multiple different ways to approach this, and I am going to work with 1 and 2.0567

I already worked with 2 and 3, so I eliminated z; and I want to...actually, I already have...this does not have z in it, so I don't even need to proceed.0579

That way, I already have two equations with two unknowns; this is a particularly easy situation, compared with when all three have 3 variables.0598

OK, so I look up here, and I have my new equation; I can call it equation 4.0607

And then, I have equation 1; and these just have x and y.0613

So, I am just going to proceed, like I usually do, with two equations with two unknowns.0616

Let me rewrite these right here: 2x + y = 3, and this is x + 3y = 4.0620

I can use substitution; that would be fine, because I have a coefficient of 1.0634

So here, I am going to solve for y; and this will give me y = -2x + 3.0640

And then, I am going to substitute into this equation; so I have x, and I am going to substitute for y: plus 3, times -2x, plus 3, equals 4.0651

Working this out: x...3 times -2x is -6x; 3 times 3 is 9; it equals 4.0668

Here, now, I just have an equation with a single variable, so I can solve that.0679

First, I am combining like terms: x - 6x is -5x; plus 9 equals 4.0684

Subtract 9 from both sides: -5x = -5; x = 1.0692

OK, so the first thing I wanted to do is just get rid of one of the variables; I am just working with two variables.0700

And I did that by just adding these two; in the second and third equations, the z dropped out.0705

I was lucky, because the first equation already didn't have a z; so I had two equations with two unknowns.0711

And then, I just used those two; I solved by substitution, and I came up with x = 1.0717

Since I know that x equals 1, I can go ahead and substitute this into this equation to find y.0726

So, looking at equation 1: 2x + y = 3; I know that x equals 1, so that is 2(1) + y = 3, or 2 + y = 3.0734

Subtracting 2 from both sides, I get y = 1; so now, I have x, and I have y.0754

I need to find z: well, this will easily tell me what z is.0759

That is y - z = 2, and I know y: y equals 1, so 1 - z = 2; -z = 1, therefore z = -1.0765

So, the solution to this set of equations is that x equals 1, y equals 1, and z equals -1.0780

The hardest step was just getting rid of that third unknown (the third variable).0791

I did that by adding these two together: then, working with two equations with two variables, I was able to solve for x.0797

Once I am there, all I have to do is start substituting.0805

Here, I substituted x into the first equation and solved for y.0807

Once I got y, then I was able to substitute y into the second equation to solve for z.0813

So, 1, 1, -1 is the solution for this system of equations with three variables.0822

OK, Example 2: again, my goal is going to be to get a variable to drop out, so I am just left with two variables.0833

Looking at this first and second equation, considering these together, y and -y...if I add those together,0843

the y's will drop out, because they have opposite coefficients (1 and -1).0852

So, I am going to start off by adding the first two equations: this is 1 and 2.0857

x + y + z = 2; and then, I am going to add x - y + 2z = -1; and I am going to come up with a new equation.0861

This is 2x; the y's drop out; z + 2z is 3z; 2 - 1 is 1; OK, I have this.0874

And I worked with these first two; I now need to work with two different equations to get the same variable to drop out.0886

This time, I am going to pick equations 1 and 3, and I want y to drop out.0894

So, I have equations 1 and 3: that is x + y + z = 2, and (equation 3) that is 2x + y + 2z = 2.0900

Now, I need to subtract in order to get the y to drop out.0919

To keep everything straight, as far as my signs go, I am going to keep the first equation the same;0925

but for the second one, I am going to change it to adding the opposite: add -2x, -y, -2z, and -2.0931

OK, x - 2x is -x; the y's drop out, which is just what I wanted; z - 2z is -z; 2 - 2 is 0.0944

Now, I have two equations and two variables; I am rewriting these two up here to see what I have to work with.0956

2x + 3z = 1; now, I just use my usual methods of solving a system of equations with two variables.0965

Since I have coefficients here of -1, it is pretty easy to use substitution, so I am going to solve for x in this second equation,0976

and then substitute that value up in the first equation.0984

I have -x - z = 0, which would give me -x = z, or x = -z.0989

So, I am going to take this -z and substitute it in right here; OK, that gives me 2x + 3z = 1, and let x equal -z.0996

So, 2 times -z, plus 3z, equals 1; that is -2z + 3z = 1.1014

Combine these two like terms to get z = 1; now, I have my first value.1025

OK, so I know that z equals 1, so I am on my way.1034

And I look up here, and I see, "Well, I know that x equals -z, so that makes it very easy to solve for x."1038

If x equals -z, and z equals 1, then x equals -1.1047

So now, I have x = -1, z = 1; I am just missing y.1059

Well, look at that first equation: it tells me that x + y + z = 2.1063

The x is -1; I don't know y; and I know that z is 1; these two cancel, and that gives me y = 2.1074

So, -1 + 1 is 0, so I end up with y = 2.1088

Putting all this together up here as my solution, I end up with x = -1, y = 2, and z = 1.1092

That was a lot of steps; it is really important to keep track of what you are working with--especially, in the beginning,1106

that you work with two equations (I worked with 1 and 2) to get a variable to drop out.1112

I added those, and the y's dropped out; then I want to work with either 1 and 3 or 2 and 3 (two different equations) to get the y to drop out.1118

I chose 1 and 3; and I saw that I could get the y to drop out of 1 and 3 if I just subtracted 3 from 1; that is what I did right here.1128

At that point, I clearly mark out what I ended up with, which is two equations with two variables.1140

We wrote those up here; and I decided I was going to use substitution.1146

I solved for x in this second equation: x equals -z; I substituted that in right here, into the first equation.1151

That allowed me to have one equation with one variable, z; and I determined that z equals 1.1162

From there, it was much easier, because I saw that x equals -z, and I knew z; so x equals -1.1170

I had x; I had z; and I had three equations that I could have used,1180

but I picked the easiest one to substitute in x and z and solve for y to get my set of solutions.1184

Again, this is a set of three equations with three variables that I need to approach systematically.1198

And my first goal is to eliminate the same variable, so I am working with two equations with two variables.1204

And I see several possibilities; you could approach it differently, and you will come up with the same answer, as long as you follow the rules and the steps.1216

I am seeing that -y and y are opposite in terms of coefficients (-1 and 1), so I am going to add those two.1227

I am going to add 1 and 3; that is going to give me 5x; the y's drop out, so 5x - z; 1 - 3 is -2.1235

I just marked that, so I can keep track of it, because I am going to need to use it in a minute,1260

once I generate another equation in which y has been eliminated.1265

OK, I worked with the first and the third; now, I need to work with two different equations;1270

and I am going to work with the first and the second, and I want to eliminate y.1276

In order to eliminate y, I need to multiply the first equation by 2; so I will do that up here.1282

This is going to give me 4x - 2y + 2z = 2; that is the first equation.1294

Now, I am going to add it to the second equation: + x + 2y - z = 0.1305

Adding these together, I am going to get 5x; the y's drop out; 2z - z is z; 2 and 0 is 2.1319

OK, I first worked with the first and the third, and then I worked with the first and the second, to get the y's to drop out.1330

So now, I have two equations and two variables: x and z.1338

So, put these together so I can see what is going on with them: 5x - 2 = -z; 5x + z = 2.1345

Well, I can see that, if I add these, z will drop out; and I am just back to my usual two equations with two variables--usual techniques.1355

5x and 5x is 10x; the z's drop out; -2 and 2 is 0.1365

Divide both sides by 10; it gives me x = 0.1373

So, I have my first value, which makes things much easier.1377

Since I know that x equals 0, I can substitute into either of these to solve for z.1384

So, 5x - z = -2; so 5(0) - z = -2; 0 - z = -2; -z = -2; divide both sides by -1 to get z = 2.1389

I have my second value; OK, so I know x; I know z; I just need y.1411

I am going to solve for y; I could use any of these--I am going to pick the top one and solve for y, knowing that x equals 0 and z equals 2.1417

OK, that is 0 - y + 2 = 1, which gives me -y + 2 = 1.1437

Subtract 2 from both sides to get -y; if I say -1 - 2, that is going to give me -1.1447

Multiply all of that by -1 to give me y = 1.1455

OK, so the solution here is x = 0, y = 1, and z = 2.1461

And I approached that by seeing that I could add 1 and 3 because of -y and y, and those would drop out.1470

I could have added the first two and had the z drop out, and had that be my variable to eliminate; I happened to choose y.1479

I added those together and got this equation.1486

Then, I worked with the first and the second equation--a little more complicated, because to get opposite coefficients,1490

I had to multiply the first equation by 2.1496

I did that to generate this equation, which I added to the second; this is the resulting equation.1500

I then had two equations with two variables; I looked at those two and saw I had opposite coefficients with z.1507

So, when I added them together, z dropped out, and I could solve for x.1514

Once I determined that x is 0, I substituted 0 into this top equation for x to solve for z, and determined that z equals 2.1519

At that point, I just needed to solve for y, so I took this equation and substituted my value for x and my value for z, and determined that y equals 1.1529

So again, we are using the same techniques that we used previously, only you are working with more equations, and there is more to keep track of.1543

OK, in this system of equations (three equations with three unknowns), it is a little bit more complicated.1551

I do have one equation that has a coefficient of 1, but the rest of them have larger coefficients.1559

So, I am going to work with the first and second equations; and what I want to do is eliminate the z.1566

So, in order to do that, I am going to need to multiply this first equation by 2.1577

I want to work with the first and second equations; I want to get the z to drop out.1583

But I need to multiply this by 2 first; so let me do that right over here.1589

That is going to give me 6x - 4y + 2z = 8; so I am going to rewrite that right here: 6x - 4y + 2z = 8.1600

And that came from that first equation, multiplied by 2; and I am going to add it to the second equation.1615

So, + -6x + 4y - 2z = 2: now, you might have already seen what happened.1621

I was really just focusing on "OK, I want to get the z's to be the same or opposite coefficients," so that they would cancel out.1635

But what happened is: everything ended up with opposite coefficients: 6 and -6; -4 and 4; 2 and -2; OK.1640

So, I have 6x - 6x; that is 0; -4y and 4y--0; 2z and -2z--0; 8 and 2--10.1654

0 = 10: well, we know that 0 does not equal 10, so this is not true; it is never true that 0 equals 10.1669

So, there is no solution to this set of equations.1678

I could have had a situation where I got a solution.1682

I could have also had a situation where maybe I got 10 = 10 with a different system of equations.1686

If I had come up with something like 0 = 0 (that is always true) or 10 = 10, then I would have had an infinite number of solutions.1694

But instead, what happened is that my variables drop out; I got c = d, which is saying1701

that I have a constant that is equal to another constant, which is never true; so there is no solution.1706

So, this one turned out to actually be less work than the others.1712

But when this happens, you just want to be really careful that you were doing everything correctly;1714

you didn't make a mistake; but actually, it can end up that there simply is no solution to the system of equations.1718

That concludes this lesson on Educator.com on solving systems of equations with three variables; and I will see you next lesson!1725

Welcome to Educator.com.0000

Today's lesson introduces the concept of matrices.0002

And matrices are used throughout math and science as an approach to problem solving.0006

In this course, we are going to use them to solve systems of equations.0013

However, they are also used in fields such as physics, computers, and genetics.0018

First of all, defining what a matrix is: a matrix is a rectangular array containing variables or constants, which is enclosed by brackets.0026

And the plural form of the word "matrix," which I just used, is matrices; so you will hear me say that in the course.0037

And to give you an example of what a matrix looks like, as I said, a matrix is enclosed in brackets, and it contains variables or constants.0045

And starting out, we are going to be working with matrices that contain constants.0062

And then, towards the end of this series of lectures, we will see a matrix that involves variables.0066

This is an example of a matrix; and these are usually designated by capital letters, so I could call this matrix A.0079

Another matrix, B, might be smaller, perhaps containing only two numbers.0087

Another matrix, C, could have a different set of numbers.0097

Each variable or constant within a matrix is called an element; so, 2 is an element; 1 is an element; each of these is an element.0108

And later on, we will also see matrix equations, because you can perform operations with matrices.0118

For example, I could have a matrix equation that says A + B = C.0126

So, I could have some matrix, A, that when added to B equals another matrix C.0134

So, just as we performed operations on constants and terms, we can perform operations on matrices.0138

OK, starting out, it is important to understand the concept of dimensions when referring to a matrix.0148

And with a matrix, dimensions refers to the number of rows and columns in the matrix.0154

For example, let's say I have a matrix, and you can see, of course, that it has a certain number of rows and a certain number of columns.0161

Well, m refers to the number of rows; and here, that is 1, 2, 3, 4--I have 4 rows.0179

For this matrix, looking at columns (and we will say n here is the number of columns), I have 1, 2 columns.0191

Therefore, I would call the dimensions of this matrix 4 by 2; this is a 4x2 matrix.0200

And this is important, because the dimensions often limit what you are able to do--0207

which operations you can perform on a set of matrices.0214

So, looking at another matrix: this matrix has 1, 2 rows and 1, 2, 3, 4 columns; so this is a 2x4 matrix.0219

It is important--the order of the numbers is very important: here it is 4x2 (4 rows, 2 columns); here it is 2x4, because it's 2 rows, 4 columns.0243

OK, there are certain matrices that are special cases: a 1xn matrix has one row, and is called a row matrix.0262

For example, this is a row matrix, because it has one row (in this first term here, it tells the number of rows).0272

And looking at the number of columns, I have 1, 2, 3, 4, 5, 6; so this is a 1x6 matrix, and it is a row matrix, because it has only one row.0286

A second type of special matrix is a column matrix; here, it is an mx1 matrix--it has one column.0300

So here, it is 1 row; and here, 1 column.0309

For example (let me make this a little bit more spread out)...2, 3, 8, 4, 2: OK, so here, I have 1, 2, 3, 4, 5 rows, but I have only 1 column.0312

So, this would be a 5x1 matrix, and since it has a 1 right here, this is a column matrix.0346

A third type of special matrix is called a zero matrix; and this zero matrix has all its elements equal to 0.0358

For example, this would be an example of a zero matrix.0365

So, these are three special types of matrices: a row matrix with one row, a column matrix that has one column,0375

and a zero matrix, which contains elements that are all 0.0382

We say that two matrices are equal if they have the same dimensions, and their corresponding elements are equal.0389

Let's talk about what corresponding elements are, using this set of matrices as an example.0396

OK, so I am going to call this A, and this matrix B; and I can say that A = B, because the corresponding elements of A and B are equal.0415

Elements are corresponding if they have the same position within a matrix.0424

By "position," I mean the same row number and the same column number.0428

So, right here, this element is in row 2, column 1; its corresponding element is also going to be in row 2, column 1.0432

And these are the same; and my row 1, column 1 elements are the same; and so on.0450

For each position, all of the elements are the same--the corresponding elements; therefore, these two matrices are equal.0456

If I had the same numbers, but they weren't in the same positions, the two matrices would not be equal.0463

The same dimensions (and I do have the same dimensions--1, 2, 3 rows, 2 columns; this is a 3x2 matrix,0471

and the same here--1, 2, 3 by 2 matrix) and corresponding elements are all equal: 1 equals 1, 3 equals 3, and so on.0477

I can say that these two matrices are equal.0487

OK, doing some examples: first, we are asked, "What are the dimensions of this matrix?"0493

And remember that when we do dimensions, we say mxn, so we look first at the number of rows, times the number of columns.0500

So, all I need to do is count the number of rows (that is 1, 2, so m = 2, and that is the number of rows),0513

and columns (n = 1, 2, 3--there are three columns); therefore, this is a 2x3 matrix; the dimensions of this matrix are 2x3, rows by columns.0523

Example 2 (a different matrix here): What are the dimensions of this matrix?0544

Again, we are looking at rows by columns; I have 1, 2 rows; 1, 2, 3, 4, 5 columns.0549

So, this is a 2x5 matrix, because it contains 2 rows and 5 columns; always have rows first, then columns.0568

Write the 3x3 zero matrix: this is telling me that I am going to have 3 rows and 3 columns.0579

And since it is a zero matrix, all elements are zero; OK, so that is going to give me 1, 2, 3 columns and 1, 2, 3 rows.0588

I just need to fill in, and all elements are 0; and again, a matrix is contained within brackets.0606

This is a 3x3 matrix (3 rows, 3 columns); it is a zero matrix; and I have my brackets around it to indicate that it is a matrix.0613

This is a 3x3 zero matrix.0622

Example 4: Write a matrix that is both a row matrix and a column matrix.0628

Recall that a row matrix has one row, and then some number of columns.0632

A column matrix has some number of rows, but only one column.0643

Putting this together, this is telling me that I am going to have...0649

if it is both a row and a column matrix, that means that in this m position,0653

the rows are going to be 1; and the columns are also going to be 1.0659

This is a 1x1 matrix that they are asking for; and I could use any constant or variable.0662

This would be both a row matrix and a column matrix; it has one row, and it has one column.0670

Or it could be 5, or I could use a variable (such as y or x).0675

So, any of these examples would be a row matrix and a column matrix at the same time.0680

That concludes this lesson from Educator.com; and I will see you next time, when we talk more about matrices.0689

Welcome to Educator.com.0000

In today's lesson, we are going to continue on with matrices, and this time doing operations on matrices.0002

Just as you can perform mathematical operations on numbers (such as addition, subtraction, and multiplication), you can do the same thing with matrices.0009

Starting out with addition of matrices: addition is defined only for matrices with the same dimensions.0018

So, just to review: when we talk about dimensions, the dimensions of a matrix are m times n,0026

where m is the number of rows and n is the number of columns.0033

So, two matrices must have the same number of rows and columns in order for addition to occur.0040

If that specification is met, then you add the corresponding elements of the two matrices.0049

The sum matrix will have the same dimensions as the original matrices.0055

To illustrate this, let's say that I have two matrices, A, and then I have a matrix B; and I want to add those together.0062

Well, let's look at my first matrix, A; and it will be 4, 0, 3, 1, -2, -4; OK.0074

And I have a second matrix, B, that I am going to add (A + B).0085

OK, so first looking at the dimensions to make sure that I am allowed to even add these: there are 2 rows on this one and 1, 2, 3 columns.0101

Therefore, this is a 2x3 matrix; matrix B has 2 rows and 1, 2, 3 columns; this is a 2x3 matrix.0109

Since these have the same dimensions, they can be added.0119

The sum matrix (the resulting matrix) is also going to have the dimensions of 2x3: it has the same dimensions as these two original matrices.0123

Now, to add these, I am going to add corresponding elements.0133

Recall that corresponding elements occupy the same position in the matrix.0136

So, if I am looking at this position, this is 1, 2; so it is row 2, and in columns, 1, 2; so it is row 2, column 2.0140

The corresponding element over here is also going to be in this position of row 2, column 2.0149

So, all I am going to do is add corresponding elements: 4 + 5 right here, and then I am going to have 0 and -1;0155

over here, 3 + 2; and down here, 1 + 0; in this position, -2 + 3, and in this position, -4 + -5.0169

OK, if I work out the addition on that, that is going to give me 9, -1, 5, 1, (-2 + 3) is 1, (-4 + -5) is -9.0185

So again, first always verify that the two matrices you are being asked to add have the same dimensions.0202

If they do, then you just add the corresponding element and put it in that same position0208

to get the sum matrix, which will have the same dimensions as the two originals.0216

Matrix subtraction is very similar: again, matrix subtraction is defined only for matrices with the same dimensions.0222

To subtract two matrices, subtract the corresponding elements of the two matrices.0230

The result--the difference matrix--will have the same dimensions as the original matrix.0237

So, it's the same concept as with matrix addition.0241

So, looking at that situation here: if I have 3, 2, 4, -1, 0, 6, -2, 0, and this is A;0244

and I have a second matrix that I am going to call B, and I am asked to find A - B;0267

OK, this is the difference matrix, which will be A - B, over here.0276

OK, so first verify that these have the same dimensions: I have 1, 2, 3, 4 rows, and I have 2 columns.0283

Over here, I have 1, 2, 3, 4 rows and 2 columns.0292

My difference matrix is also going to have 4 rows and 2 columns.0299

OK, so I simply subtract: 3 - 6 is going to give me -3; the corresponding elements 2 - 0 is 2; 4 - -5...0303

if I have 4 minus -5, that is going to give me 9; -1 - 6 is going to give me -7; 0 - 2 is going to give me -2;0317

6 - -1 is going to give me 7 (a negative and a negative are going to give me a positive); -2 - 0 is -2, and 0 - 1 is -1.0341

So again, verify that the two matrices have the same dimensions, and then simply0354

subtract corresponding elements; and since we are doing subtraction, you need to be very careful with the signs.0360

OK, now we are going to talk about scalar multiplication.0368

And scalar multiplication is not the multiplication of one matrix times another.0370

Matrix multiplication, we will actually cover in a separate lecture.0376

This is called scalar multiplication, because what you are going to be doing is multiplying a matrix by a constant.0380

And that constant is referred to as a scalar.0387

So, for example, if I am given a matrix that looks like this, and I am asked to multiply it by 2;0391

well, 2 is a constant called a scalar; and let's say this matrix is called A.0402

If I am asked to find 2A, then I am going to multiply the scalar by the matrix.0408

In order to do that, you multiply each element of the matrix by the scalar.0414

So, each element of the matrix, you multiply by the scalar to get the result.0419

OK, therefore, I am going to say 2 times 2 for this position; for this second position in row 1, column 2, it is 2 times 3.0425

Here, it is 2 times -1; 2 times 0; 2 times 4; and finally, 2 times 1.0437

Doing the multiplication out will give me 4; 2 times 3 is 6; 2 times -1 is -2; 2 times 0 is 0; 2 times 4 is 8; and 2 times 1 is 2.0451

And as you can see up here, the scalar product matrix has the same dimensions as the original.0468

This original matrix had 1, 2, 3 rows and 2 columns.0473

The scalar product over here, 2A, has 1, 2, 3 rows and 2 columns--the same dimensions as the original.0480

Again, for scalar multiplication, simply take the scalar (the constant) and multiply it by each element in the original matrix.0493

Just as we have certain properties when we are working with operations on regular numbers and variables,0503

there are also properties that regulate matrix operations.0510

So, if A, B, and C are matrices with the same dimensions, and k is a scalar, then the following hold.0514

This first one, you will recognize as the commutative property of addition.0522

Recall that with numbers, we have the commutative property, where, if you want to add 3 + 2,0530

you can add it in the other order--2 + 3--and get the same result.0535

And the same is true if you are adding two matrices.0539

So, matrix addition follows the commutative property.0542

The next property you may recognize as the associative property for addition of matrices.0548

What this is stating is that I can do these operations in either order.0562

I can either add the two matrices A + B together, then add C to those; or I can add B + C together first, and then add A to those two.0567

So, I can perform these operations in either order; and the result will be the same.0578

Now, looking at scalar multiplication, combined with the addition of two matrices here:0584

this follows the distributive property, and what this is stating is that I can add matrix A and matrix B,0591

and then multiply this scalar by that result; or I can multiply the scalar times one matrix, multiply the scalar by the other matrix,0604

and then add those two together; and I will get the same result.0615

So, this is the distributive property, which you have seen previously.0618

First example: we are going to add two matrices.0625

First, verify that addition is allowed: do these have the same dimensions? two rows, three columns--OK, so far, so good.0628

1, 2 rows; 3 columns: since these have the same dimensions, then addition is allowed.0637

OK, so I am rewriting these down here so we can work with them more easily.0646

Recall that, for addition, you are going to add corresponding elements; you are going to add the elements occupying the same row and column number.0657

So, 2 + -4 is going to give me -2; -1 + 6 is going to give me 5; 3 + -2 is going to give me 1.0671

0 and -3 is going to give me -3; 6 and 0--I am going to get 6; and then, 4 + 4 is going to give me 8.0692

So, all I have to do to find the sum matrix is to add the corresponding elements of these two matrices to get the result.0705

Example 2 involves subtraction: first, verify the dimensions--1, 2, 3 rows, 1, 2, 3 columns.0718

So, this is a matrix with three rows and three columns, so it is a square matrix, since it has the same dimensions in both directions.0727

Here, I have 1, 2, 3 rows and 1, 2, 3 columns; so this is also a 3x3 matrix; so it is a square matrix, since it has the same dimensions in both directions.0740

Therefore, I can subtract these; when I go ahead and subtract, I am going to get a difference matrix that is also 3x3.0755

Beginning with these two corresponding elements: 2 - -1 (I have to be very careful, when I am working with negatives,0772

that this becomes 2 + 1, which is 3, so I get 3 right here); -1 - -2 is going to give me -1 + 2, which is 1.0779

Here, I just have 6 - 3; that is 3; 3 - 0--that is 3; 2 - 4--and that is just going to be...2 - 4 is going to give me -2.0801

0 - -8 (working down here) is going to be 0 + 8, which equals 8 in this position.0818

1 - 6 is -5; 4 - 0 is 4; and then, 3 - -1 gives me 3 + 1, which equals 4.0831

So again, with subtraction, just be really careful with the signs.0845

And I end up with a difference matrix right here that resulted from taking an element0848

in the first matrix and subtracting the corresponding element of the other matrix from that.0856

Example 3: Find the product--and it is the product of a scalar and a matrix, so this is scalar multiplication.0864

And here, the scalar is -2.0873

I am rewriting this down here.0876

Recall that, in scalar multiplication, you are going to multiply the scalar times each element in the matrix,0884

and the result is going to be a scalar product that has the same dimensions as the original.0890

My original here is a 3x3 matrix; so I am going to take -2 times 2 for the first position; -2 times -1;0895

-2 times 6; -2 times 3; and continue on, multiplying each one...-2 times 0; the third row:0909

-2 times 1; -2 times 4; and finally, -2 times 3.0924

Then, when I do my multiplication, I am going to end up with -4, 2, -12, -6, here is -4, 0, -2, -8, and then -6.0932

So again, in scalar multiplication, simply multiply each element in the matrix by the scalar0956

to get a scalar product matrix with the same dimensions as the original.0963

Example 4 is slightly more complicated: we are asked to find -3A + 6B.0970

OK, so what I need to do is multiply matrix A by -3; multiply matrix B by 6; and then add those together.0978

Just to make sure I can add them eventually, I verify the dimensions as 3 rows, 2 columns and 3 rows, 2 columns.0989

So, I will be able to add them.1000

OK, just to note, looking at the distributive property, it says that k, a scalar, times A + B (these two matrices) equals KA + KB.1002

So, there is actually another way I could do this: I could say, "All right, I am going to factor out the -3, and then I am going to end up with A - 2B."1015

This would be another way to do that; however, it is debatable which way is easier.1038

I am going to go ahead and just follow this original.1044

But you actually, if you felt like this way was easier, could have done it this way.1046

OK, so starting out, the first thing we are going to need to do is multiply the A by the scalar -3.1053

OK, and from that I can find 3A; I want to find 3A.1082

So, this is -3 and A, and I want to find -3A.1089

OK, so recall that all we are going to do is multiply each element of the matrix by the scalar.1097

And that will give me 3 here; -3 times 0 is 0; -3 times 2 is -6; -3 times -1 is 3; -3 times 3 is 9.1105

And let's see, then I have -3 times 4, which is -12; OK.1122

Now, what I also need to do is find 6B; so here, I have 3A--I need to find 6B.1128

So, I have B over here; 0, -3, -4, 6, 4, and 9; so, I have B, and I am going to multiply it by 6, and that is going to give me 6B.1139

So, figuring this out, it is: 6 times 0 is 0; 6 times -3 is -18; 6 times -4 is -24.1157

6 times 6 is 36; 6 times 4 is 24; and 6 times 9--that is 54.1171

All right, so here I have 3A; here I have 6B; now, I need to add those.1178

So, let me go ahead and copy 6B right over here, so I can add it.1184

OK, 0, -18, -24, 36, 24, and 54: now, I am erasing that; these two are no longer equal.1200

OK, all we need to do with matrix addition is to add the corresponding elements.1215

So, I am going to add 3 and 0; and this is going to give me 3.1222

I am going to add 0 and -18 to get -18; -6 and -24 is -30; 3 and 36 is 39;1230

9 and 24...or excuse me, this actually should be -9; correct that--that is -3 times 3; that is -9 + 24 is 15;1241

and then, I have -12 and 54, to give me 42.1251

So, this here is -3A + 6B: so right here is my solution.1256

And what I did is took A; I multiplied it by the scalar -3 to get this -3A.1264

I took 6, and I multiplied it by the second matrix, B, to get 6B.1272

Then, I added -3A + 6B to get -3A + 6B as my solution.1279

That concludes this lesson on matrix operations at Educator.com; I will see you next lesson.1291

Welcome to Educator.com.0000

In today's lesson, we will be covering matrix multiplication.0002

In a previous lesson, we discussed scalar multiplication, which is multiplying a constant by a matrix.0005

This time, we will be talking about multiplying one matrix by another.0012

Before you proceed with matrix multiplication, you need to verify that the dimension requirement has been met.0018

So, suppose that matrix A is m by n, and matrix B has dimensions of p by q.0024

The product of these two matrices can be obtained only if n = p.0031

Now, what is that saying? What that is saying is that the number of columns of the first matrix must equal the number of rows of the second matrix.0036

So, in order to find AB, to find that product, to be allowed to do that kind of multiplication,0045

the number of columns of the first matrix must equal the number of rows of the second matrix.0052

OK, once you find that product, the resulting product matrix, AB, will have dimensions of m times q.0099

So, the product matrix dimensions will have the number of rows of the first matrix and the same number of columns as the second matrix.0109

To make this more concrete, let's look at an example.0115

If matrix A equals 2, -1, 4, 0, and matrix B equals 3, 4, 2, 1, 0, -3, 0, -2; my first question is, "Can I multiply them--can I find AB?"0118

Well, this has one row and four columns; it is a 1x4 matrix.0137

This has four rows and two columns; it is a 4x2 matrix; so this gives me 1x4 and 4x2.0145

So, all I have to do is see, "OK, does this second number equal this first number?" and yes, the second number does equal the first number.0152

Therefore, I can multiply those.0162

Now, what are going to be the dimensions of the product matrix?0164

Well, the dimensions of the product matrix...this is a 1x4, multiplied by a 4x2 matrix; and AB is going to have0168

the same number of rows as this first matrix (which is 1), and the same number of columns as the second matrix (which is 2).0175

So, the product matrix is going to be a 1x2 matrix.0184

So, the important thing is: before you multiply, make sure that you verify that the dimension requirement is met--0188

that the second number of the first matrix is equal to the first number of the second matrix.0193

And to predict the dimensions of the product matrix, you take this first number and this second number; and that will give you the product matrix dimensions, 1x2.0200

OK, multiplying by matrices is not exactly what you would expect.0217

It is not like addition and subtraction of matrices, where (in addition and subtraction) we just took the first matrix0224

and added the corresponding element of that to the corresponding element of the second matrix (and the same in subtraction).0231

You might think, "OK, I am just going to multiply the corresponding elements by each other."0238

But it is actually more complicated than that; and you need to take it step-by-step.0243

So, the best way to understand this is to go through an example: so let's look at a pair of matrices and multiply them out.0247

OK, this is my first matrix; I am going to call it A; and then, I want to multiply it by another matrix, which I am going to call B.0264

So, before I multiply, I have to make sure that they meet the dimension requirements.0277

And A has two rows, and it has four columns; B has four rows and two columns.0284

So, I am allowed to multiply these, because this second number equals the first number.0292

My product matrix, AB, is going to have two rows and two columns; it is going to be a square matrix with dimensions 2x2.0299

Now, as you read up here, it says that the element in row I and column J of the product...0311

so the element in a certain row and column of the product of the matrices A and B...0317

is obtained by forming the sum of the products of the corresponding elements in row I0323

(in a certain row of matrix A) and the corresponding column, J, of matrix B.0329

What does this mean? Well, instead of just saying row I and column 1, let's start with row 1, column 1.0335

I want to find the element that goes right here, in row 1, column 1.0345

And the way I am going to do that is: I am going to go over here to row 1, and I am going to go here in column 1;0350

and I am going to multiply 3 by 4 and find that product, and then I am going to add that to the next product, 2 by 0.0361

Then, I am going to add that to the next product, 0 by 1, to 0 by 3, and so on.0369

So, row 1, column 1; then I am going to go to the corresponding row in A and the corresponding column in B.0375

And working this out, this gives me 3 times 4, plus 2 times 0, plus 0 times 3, plus 1 times -2.0380

Working that out, that is going to give me 12 + 0 + 0 - 2; 12 - 2 is 10.0399

Therefore, my row 1, column 1 element is 10.0410

OK, next let's work on row 1, column 2; that means I am going to go to row 1 here in A, and column 2 here, in B, and do the same operations.0414

So, row 1, column 2: that is 3 times -1, plus 2 times 2, plus 0 times 1, plus 1 times 0.0429

OK, working that out, this gives me -3 + 4 + 0 + 0; that is 4 and -3, so that is 1.0446

Therefore, my row 1, column 2 element for the product is 1.0461

All right, now I need to work with this second row; and I am going to think about what position this is right here.0468

This is row 2, column 1; that means I am going to go to row 2 here and column 1 in B and do the same thing.0472

OK, row 2, column 1: -1 times 4, plus 0 times 0, plus 4 times 3 (so the third element here and the third element here),0487

plus the fourth element here and the fourth element there.0507

All right, figuring this out, this is going to give me -4 + 0 + 12 + -2, or - 2; look at it either way.0512

This is going to give me -4 and 12, which is 8, minus 2...actually, let's make this clearer...plus 12, minus 2...it is going to give me...0526

let's see...that is -4, 0, 4 and 3, 2 and -2; OK, so this is going to give me 12 - 6, or 6.0546

Now, the next row and column position: I have row 2, column 2.0562

Here is row 2; here is column 2; it is going to give me -1 times -1, plus 0 times 2, plus 4 times 1, plus 2 times 0.0577

OK, so that is going to come out to...-1 times -1 is 1, plus 0, plus 4, plus 0; 4 + 1 is 5, so I am going to get 5 right here.0601

So again, if I just looked here, and I said, "OK, that is row 1, column 2," then I would get that0618

by going to row 1 here and column 2 here and multiplying each of those, and then finding the sum of their products.0626

So, even with a bigger matrix, you can always find a particular position by using this method.0638

OK, there are some properties that govern matrix multiplication that you need to be familiar with.0646

If A, B, and C are matrices for which products are defined, and k is any scalar, then these properties hold.0653

The first property you will recognize as the associative property for multiplication.0660

And what it says is that I can either multiply A times B, find that product, and then multiply by C;0667

or I can multiply B times C, find that product, and then multiply A; and I will get the same result.0679

OK, next I see the same thing here (it is the associative property), except this time, it also involves multiplying a scalar.0687

So, here I have matrix multiplication, and then I also have a scalar.0695

And I can either multiply the two matrices and then multiply the scalar times that product;0698

or I can multiply the scalar times the first matrix, then multiply that product by the second matrix;0704

or the scalar and the second matrix, and the product by the first matrix.0712

So, it doesn't matter which order I do the multiplication in, in this situation; these two first, then the other two.0716

Next, you are going to recognize the distributive property.0726

And the distributive property, as usual, says that, if I have these two matrices, B + C,0732

and I am multiplying A by them, another approach that would give the equal result0739

is to first multiply A and B, and then add that product to the product of A and C--the distributive property.0743

And again, the distributive property would work if I placed these as follows, added A and B, and then multiplied C.0751

The same thing: I could say the product of A and C, plus the product of B and C.0763

What is very important to realize is that the order that you multiply matrices in matters a lot.0768

So, if I say, "Oh, I am going to do AB; does this equal BA?" no, it does not always equal that.0775

Sometimes it does; but very often it does not--these are two different things.0783

Therefore, matrix multiplication is not commutative; it does not follow the commutative property,0788

and so you can't simply say, "A times B is the same as B times A"; you can't just change the order of those two.0797

The first example: suppose A is 3x4; B is a matrix that is 4x3; C is 3x3; and D is 4x4.0812

What are possible defined products of the four matrices?0821

Recall that, for a product to be defined, the second number of the first matrix has to equal the first number of the other matrix.0824

So, let's first look at AA; is that defined?0834

Well, if I have 3x4, and I am trying to multiply it by 3x4; these two are not equal, so this is not defined.0840

OK, and let's make a list here of the ones that are defined, because that is what they are asking me for--where the product is defined.0852

If I take A times B, well, the second number of A is equal to the first number of B; so that is defined.0863

Now, let's do it in the opposite order, BA: if I have 4x3 and 3x4, that is also defined.0871

OK, now AC: 3x4 and 3x3--that is not defined.0882

Now, if I put C first--if I say CA--it is 3x3 and 3x4, and these two are equal, so CA would have a defined product.0888

OK, now 3x4 and 4x4; that is AD; that is defined, because of the 4 and the 4.0903

However, if I put D first--if I say I am going to do DA--that is going to give me the D first, which is 4x4, and then A, 3x4; that is not defined.0910

OK, moving on to B: B times B (BB) is 4x3 and 4x3; that is not defined.0923

BC is 4x3 and 3x3; these two are equal, so BC is defined.0934

CB is 3x3 and 4x3--not defined; OK, BD--4x3 and 4x4--not defined.0940

But if I put the D first, then I would get 4x4 and 4x3; that is defined, so DB is defined.0953

OK, now I am up to C; 3x3 and 3x3--CC--that is a square matrix, so that is defined; the multiplication is defined for that.0965

OK, now, CD--3x3 and 4x4--is not defined; then I try the D first--it is not going to matter: 4x4 and 3x3 is still not defined.0975

OK, let's go to D: I can multiply DD, because I have 4x4 and 4x4, and that means that, of course,0988

since this is a square matrix, this second number and the first number there are going to be equal.1000

So, these are my defined products; I have 3, 6, 8 defined products.1005

So, these are all the ones where the second number of the first matrix--1010

the number of columns--was equal to the first number of the second matrix--the number of rows.1016

And I can perform multiplication of these sets of matrices.1021

OK, now doing some matrix multiplication: before I proceed, I am going to make sure that the product is defined--that I can do it.1029

So, I have a 2x2 matrix here, and I have a 2x2 matrix here; therefore, multiplication is allowed.1038

I am rewriting this down here, since this second number is equal to this first number (so multiplication is allowed).1052

Using the method we discussed: my product matrix is going to be 2x2 also--the first number here and the second number there.1066

I am expecting a 2x2 matrix; so let's first look for this position--row 1, column 1.1076

I am going to go to row 1 here and column 1 here and work with those two.1086

I am going to say 3 times 0 (that product), and I am going to add it to the product of 2 times 4.1091

That is going to give me 0 + 8, which is 8; so the element in row 1, column 1 is 8.1099

Now, row 1, column 2--I am going to go to row 1 here--row 1 of this first matrix--and column 2 of the second matrix.1111

And that is going to give me...row 1 is 3, times 6; so row 1, column 2 is 2 and -1, so plus 2 times -1.1124

It is going to give me 18 - 2, which equals 16.1140

OK, now the next position is row 2, column 1; working with that, that is going to give me a -1 times 0, plus 4 times 4,1150

which is going to equal 0 plus 16, which is 16.1170

Next, I want to find...this is row 2, but it is column 2 this time.1177

So, I go to row 2 here and column 2 here: -1 times 6, plus 4 times -1--that is going to give me -6 - 4, or -10.1185

As expected, my product matrix is also 2x2; and again, if I picked any element in here--1208

let's say I picked this 16--I could simply say, "OK, that is row 2, column 1."1215

And I would get that by multiplying and finding the products of row 2 of the first matrix1223

and column 1 of the second matrix, and adding those up--finding the sum of those.1230

This is the result of the multiplication of these two matrices.1236

Example 3: I have 1, 2 rows and 3 columns, so I have a 2x3 matrix and a 3x2 matrix.1244

These two are equal, so I am allowed to multiply them.1255

The product of these two is going to have this number of rows (2) and this number of columns.1258

So, I am going to get a matrix that is going to be 2x2 for my product here.1267

OK, I am setting up the product matrix right here, and first looking for row1 , column 1; so I am going to use row 1 here, column 1 here.1275

2 times 0 is the product; then I am going to add that to the next product, which is -1 times 3, and the third product, which is 0 times -2,1293

which equals...2 times 0 gives me 0, minus 3, plus 0; so it is -3--row 1, column 1, right here, is -3.1314

Row 1 right here, but column 2: I am going to take 2 times 1; OK, in row 1, column 2, the next set of elements is -1 and 6,1331

so plus -1 times 6; then I still have row 1, column 2; I have 0 times -1; figure that out--1355

that is 2 - 6 + 0, so it is just 2 - 6, and that gives me -4, so row 1, column 2 is -4.1368

OK, moving on to the second row: the first thing I need to find is row 2, column 1--row 2 here, column 1 here.1380

0 times 0, plus 3 times 3, plus -2 times -2; OK, that is going to give me 0