INSTRUCTORS  Carleen Eaton Grant Fraser  Dr. Carleen Eaton

Slide Duration:

Section 1: Equations and Inequalities
Expressions and Formulas

22m 23s

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
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

20m 15s

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

19m 10s

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
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

17m 31s

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

17m 14s

Intro
0:00
Properties of Inequalities
0:08
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

25m

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
Section 2: Linear Relations and Functions
Relations and Functions

32m 5s

Intro
0:00
Coordinate Plane
0:20
X-Coordinate and Y-Coordinate
0:30
Example: Coordinate Pairs
0:37
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: Continous
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 Continous
23:03
Example 3: Discrete or Continous
25:53
Example 4: Function Notation
30:05
Linear Equations

14m 46s

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

23m 7s

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

23m 5s

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

31m 5s

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

21m 42s

Intro
0:00
Graphing Linear Inequalities
0:07
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
Section 3: Systems of Equations and Inequalities
Solving Systems of Equations by Graphing

17m 13s

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

23m 53s

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

27m 12s

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

28m 53s

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
Section 4: Matrices
Basic Matrix Concepts

11m 34s

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

21m 36s

Intro
0:00
0:18
Same Dimensions
0:25
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
10:24
Example 2: Matrix Subtraction
11:58
Example 3: Scalar Multiplication
14:23
Example 4: Matrix Properties
16:09
Matrix Multiplication

29m 36s

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

33m 13s

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

28m 25s

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

22m 25s

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

22m 32s

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
Section 5: Quadratic Functions and Inequalities

31m 48s

Intro
0:00
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

27m 3s

Intro
0:00
0:16
Standard Form
0:18
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

19m 53s

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

35m 45s

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
10:12
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

27m 11s

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
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

22m 48s

Intro
0:00
0:21
Standard Form
0:29
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
13:50
16:03
19:00
Example 4: Discriminant
21:33
Analyzing the Graphs of Quadratic Functions

30m 7s

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

27m 5s

Intro
0:00
0:11
Test Point
0:18
0:29
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
Section 6: Polynomial Functions
Properties of Exponents

19m 29s

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
Mutliplying 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

13m 27s

Intro
0:00
0:13
Like Terms and Like Monomials
0:23
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

31m 11s

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

22m 30s

Intro
0:00
Polynomial in One Variable
0:13
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
9:23
Odd Degrees
12:51
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

33m 29s

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

21m 10s

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
8:22
8:44
Example 1: Factor Polynomial
12:03
Example 2: Factor Polynomial
13:54
15:33
Example 4: Solve Polynomial Function
17:24
Remainder and Factor Theorems

31m 21s

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

31m 27s

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: Positve
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

31m 16s

Intro
0:00
Equation
0:08
List of Possibilities
0:16
Equation with Constant and Leading Coefficient
1:04
Example: Rational Zero
2:46
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
Section 7: Radical Expressions and Inequalities
Operations on Functions

34m 30s

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

22m 42s

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

30m 4s

Intro
0:00
Square Root Functions
0:07
Examples: Square Root Function
0:16
Example: Not Square Root Function
0:46
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

20m 46s

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
12:23
13:29
16:07
18:18

41m 11s

Intro
0:00
0:16
Quotient Property
0:29
Example: Quotient
1:00
Example: Product Property
1:47
3:24
3:47
6:33
7:16
Rationalizing Denominators
8:27
9:05
11:47
Conjugates
12:07
13:11
16:12
16:20
16:28
19:04
Distributive Property
19:10
19:20
24:11
28:43
32:00
36:34
Rational Exponents

30m 45s

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
19:03
Example 2: Write with Rational Exponents
20:40
Example 3: Complex Fraction
22:09
Example 4: Complex Fraction
26:22

31m 27s

Intro
0:00
0:11
0:22
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
11:27
Restriction: Index is Even
11:53
12:29
15:41
17:44
20:24
24:34
Section 8: Rational Equations and Inequalities
Multiplying and Dividing Rational Expressions

40m 54s

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

55m 4s

Intro
0:00
Least Common Multiple (LCM)
0:27
Examples: LCM of Numbers
0:43
Example: LCM of Polynomials
4:02
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
31:44
Example 3: Subtract Rational Expressions
39:18
Example 4: Simplify Rational Expression
38:26
Graphing Rational Functions

57m 13s

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

20m 21s

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

55m 14s

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
Doublecheck
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
Section 9: Exponential and Logarithmic Relations
Exponential Functions

35m 58s

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

45m 54s

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
39:19
Properties of Logarithms

28m 43s

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

25m 23s

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
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
21:52
Base e and Natural Logarithms

21m 14s

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
16:54
18:16
Exponential Growth and Decay

24m 30s

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
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
Section 10: Conic Sections
Midpoint and Distance Formulas

32m 42s

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

41m 27s

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

21m 3s

Intro
0:00
What are Circles?
0:08
Example: Equidistant
0:17
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
11:51
15:08
Example 4: Equation of Circle
16:57
Ellipses

46m 51s

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

38m 15s

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

18m 43s

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

47m 4s

Intro
0:00
0:22
0:45
Solutions
2:49
Graphs of Possible Solutions
3:10
4:10
Example: Elimination
4:21
Solutions
11:39
Example: 0, 1, 2, 3, 4 Solutions
11:50
12:48
13:09
21:42
29:13
35:02
40:29
Section 11: Sequences and Series
Arithmetic Sequences

21m 16s

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

21m 36s

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

23m 3s

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

22m 43s

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

18m 32s

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

14m 34s

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

48m 30s

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
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• ## Related Books 1 answer Last reply by: Dr Carleen EatonSun Jul 8, 2018 1:26 PMPost by John Stedge on June 29, 2018what is an example of 3 solutions?

• In a linear-quadratic system, use substitution to solve.
• When graphing inequalities, remember the conventions about graphing boundaries using either solid or dotted lines.
• If possible, check your solutions to systems of equations by graphing.

Solve:
x2 + y2 + 5x − y − 6 = 0
x − y = 2
• This is a linear\ quadratic system. Solve by substitution.
• Solve for x for the second equation
• x = y + 2
• Substitute x into the first equation
• x2 + y2 + 5x − y − 6 = 0
• (y + 2)2 + y2 + 5(y + 2) − y − 6 = 0
• y2 + 4y + 4 + y2 + 5y + 10 − y − 6 = 0
• Combine like terms
• 2y2 + 8y + 8 = 0
• Divide entire equation by 2
• y2 + 4y + 4 = 0
• Notice how this is a perfect square trinomial, solve for y
• (y + 2)2 = 0
• y = − 2
• Solve for x, using y = − 2
• x = y + 2
• x = − 2 + 2
• x = 0
Solution: (0, − 2)
Solve:
− 2y2 + 6x + 3y + 153 = 0
2x + y = 3
• This is a linear\ quadratic system. Solve by substitution.
• Solve for y for the second equation
• y = − 2x + 3
• Substitute y into the first equation
• − 2y2 + 6x + 3y + 153 = 0
• − 2( − 2x + 3)2 + 6x + 3( − 2x + 3) + 153 = 0
• − 2(4x2 − 12x + 9) + 6x − 6x + 9 + 153 = 0
• − 8x2 + 24x − 18 + 6x − 6x + 9 + 153 = 0
• Combine like terms
• − 8x2 + 24x + 144 = 0
• Divide entire equation by − 8
• x2 − 3x − 18 = 0
• Factor and solve using the zero - product property
• (x − 6)(x + 3) = 0
• x − 6 = 0 and x + 3 = 0
• x = 6 and x = − 3
• Solve for y, using x = 6;x = − 3
• y = − 2x + 3
• y = − 2(6) + 3 = − 9
• y = − 2( − 3) + 3 = 9
Solution: ( − 3,9) and (6, − 9)
Solve:
2x2 − x + 18y − 28 = 0
x − 2y = 4
• This is a linear\ quadratic system. Solve by substitution.
• Solve for x for the second equation
• x = 2y + 4
• Substitute x into the first equation
• 2x2 − x + 18y − 28 = 0
• 2(2y + 4)2 − (2y + 4) + 18y − 28 = 0
• 2(4y2 + 16y + 16) − 2y − 4 + 18y − 28 = 0
• 8y2 + 32y + 32 − 2y − 4 + 18y − 28 = 0
• Combine like terms
• 8y2 + 48y = 0
• Divide entire equation by 8
• y2 + 6y = 0
• Factor and solve using the zero - product property
• y = 0 and y + 6 = 0
• y = 0 and y = − 6
• Solve for x, using y = 0; y = − 6
• x = 2y + 4
• x = 2(0) + 4 = 4
• x = 2( − 6) + 4 = − 8
Solution: ( − 8, − 6) and (4,0)
Solve:
− 2x2 − 37x + 6y − 188 = 0
x + 2y + 4 = 0
• This is a linear\ quadratic system. Solve by substitution.
• Solve for x for the second equation
• x = − 2y − 4
• Substitute x into the first equation
• − 2x2 − 37x + 6y − 188 = 0
• − 2( − 2y − 4)2 − 37( − 2y − 4) + 6y − 188 = 0
• − 2(4y2 + 16y + 16) + 74y + 148 + 6y − 188 = 0
• − 8y2 − 32y − 32 + 74y + 148 + 6y − 188 = 0
• Combine like terms
• − 8y2 + 48y − 72 = 0
• Divide entire equation by − 8
• y2 − 6y + 9 = 0
• Factor and solve using the zero - product property
• (y − 3)2 = 0
• y − 3 = 0
• y = 3
• Solve for x, using y = 3;
• x = − 2y − 4
• x = − 2(3) − 4 = − 10
Solution: ( − 10,3)
Solve:
x2 + y2 − 6x + 6y − 19 = 0
x2 + y2 − 6x − 7y + 20 = 0
• Multiply the first equation by − 1 to create an equivalent system.
• -1* (x2 + y2 − 6x + 6y − 19 = 0)
• − x2 − y2 + 6x − 6y + 19 = 0
• Add the system of equation
• (− x2 − y2 + 6x − 6y + 19 = 0) + (x2 + y2 − 6x − 7y + 20 = 0)
• − 13y + 39 = 0
• Solve for y
• − 13y = − 39
• y = 3
• Solve for x using either the first or second equation form the system.
• x2 + y2 − 6x − 7y + 20 = 0
• x2 + (3)2 − 6x − 7(3) + 20 = 0
• x2 + 9 − 6x − 21 + 20 = 0
• x2 − 6x + 8 = 0
• Factor and solve using the zero - product property
• (x − 4)(x − 2) = 0
• x − 4 = 0 and x − 2 = 0
• x = 4 and x = 2
Solution: (4,3) and (2,3)
Solve:
− x2 + y2 + 14x − 13y − 8 = 0
x2 + y2 − 14x − 13y + 88 = 0
• Add the system of equatios
• (− x2 + y2 + 14x − 13y − 8 = 0) + (x2 + y2 − 14x − 13y + 88 = 0)
• 2y2 − 26y + 80 = 0
• Solve for y, factor but divide entire equation by 2 first
• y2 − 13y + 40 = 0
• (y − 5)(y − 8) = 0
• y = 5 and y = 8
• Solve for x using either the first or second equation form the system.
• y=5
x2 + y2 − 14x − 13y + 88 = 0
x2 + (5)2 − 14x − 13(5) + 88 = 0
x2 − 14x + 48 = 0
(x − 6)(x − 8) = 0
x = 6 and x = 8
• y=8
x2 + y2 − 14x − 13y + 88 = 0
x2 + 82 − 14x − 13(8) + 88 = 0
x2 − 14x + 48 = 0
(x − 6)(x − 8) = 0
x = 6 and x = 8
Solution: (6,5),(8,5),(6,8),(8,8)
Solve:
8x2 + 4y2 − 9x − 8y − 95 = 0
10x2 − 4y2 + 45x + 8y + 41 = 0
• Add the system of equatios
• (8x2 + 4y2 − 9x − 8y − 95 = 0) + (10x2 − 4y2 + 45x + 8y + 41 = 0)
• 18x2 + 36x − 54 = 0
• Solve for x, factor but divide entire equation by 18 first
• x2 + 2x − 3 = 0
• (x + 3)(x − 1) = 0
• x = − 3 and x = 1
• Solve for y using either the first or second equation form the system.
• x = − 3
8x2 + 4y2 − 9x − 8y − 95 = 0
8( − 3)2 + 4y2 − 9( − 3) − 8y − 95 = 0
4y2 − 8y + 4 = 0
y2 − 2y + 1 = 0
(y − 1)(y − 1) = 0
y = 1
• x=1 8x2 + 4y2 − 9x − 8y − 95 = 0
8(1)2 + 4y2 − 9(1) − 8y − 95 = 0
4y2 − 8y − 96 = 0
y2 − 2y − 24 = 0
(y − 6)(y + 4) = 0
y = 6 and y = − 4
Solution: ( − 3,1),(1,6),(1, − 4)
Solve:
2x2 − 24x − y + 69 = 0
− 2x2 + 5y2 + 24x − 9y − 144 = 0
• Add the system of equatios
• (2x2 − 24x − y + 69 = 0) + ( − 2x2 + 5y2 + 24x − 9y − 144 = 0)
• 5y2 − 10y − 75 = 0
• Solve for y, factor but divide entire equation by 5 first
• y2 − 2y − 15 = 0
• (y − 5)(y + 3) = 0
• y = 5 and y = − 3
• Solve for x using either the first or second equation form the system.
• y=−3 2x2 − 24x − y + 69 = 0
2x2 − 24x − ( − 3) + 69 = 0
2x2 − 24x + 72 = 0
x2 − 12x + 36 = 0
(x − 6)(x − 6) = 0
x = 6
• y=5
2x2 − 24x − y + 69 = 0
2x2 − 24x − (5) + 69 = 0
2x2 − 24x + 64 = 0
x2 − 12x + 32 = 0
(x − 4)(x − 8) = 0 x = 4 and x = 8
Solution: (6, − 3),(8,5),(4,5)
Solve:
x2 + y2< 9
(x − 4)2 + y2< 16
• Notice how both equations are circles. Draw both circles
• Circle 1 has a r = 3 and is centered at (0,0)
• Circle 2 has a r = 4 and is centered at (4,0)
• Circle 1 and 2 need to be dashed becuase both have the " < " sign, their circumference is not part of the solution set
• Check which way to shade using a test point.
• Circle One: Test (0,0)
x2 + y2< 9
02 + 02< 9
0 < 9
True
• Circle Two: Test (5,0)
(x − 4)2 + y2< 16
(5 − 4)2 + 02< 16
1 < 16
True
• For Circles 1 and 2, you must shade inside the boundary line of each circle
• The solution is whatever they have in common Solve:
(x − 5)2 + y2 ≤ 36
(x − 5)2 + y2≥ 9
• Notice how both equations are circles. Draw both circles
• Circle 1 has a r = 6 and is centered at (5,0)
• Circle 2 has a r = 3 and is centered at (5,0)
• Circle 1 and 2 need be solid becuase both have the " ≤≥ " sign, their circumference is part of the solution set.
• Check which way to shade using a test point.
• Circle One: Test (5,0)
(x − 5)2 + y2 ≤ 36
02 + 02< 9
0 < 9
True
• Circle Two: Test (5,0)
(x − 5)2 + y2≥ 9
02 + 02≥ 16
0 ≥ 16
NotTrue
• For Circles 1, you must shade inside the circle.
• For circle 2, you must shade outside the circle.
• The solution is the shaded area they have in common. The solution set forms a doughnut shape. *These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Solutions 2:49
• Graphs of Possible Solutions
• Example: Elimination
• Solutions 11:39
• Example: 0, 1, 2, 3, 4 Solutions
• Systems of Quadratic Inequalities 12:48
• 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

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

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

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

And that concludes today's lesson; thanks for visiting Educator.com.2822

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