Chapter 1 Section 3 Intersection Point of a Pair of Lines Read pages 18 – 21 Look at all the Examples.

Slides:



Advertisements
Similar presentations
Learning Objectives for Section 4.1
Advertisements

Chapter 1 Linear Equations and Graphs
Learning Objectives for Section 4.1
Chapter 4 Systems of Linear Equations; Matrices
7 = 7 SOLUTION EXAMPLE 1 Check the intersection point Use the graph to solve the system. Then check your solution algebraically. x + 2y = 7 Equation 1.
Thinking Mathematically Algebra: Graphs, Functions and Linear Systems 7.3 Systems of Linear Equations In Two Variables.
Solving Systems by Graphing
3.1 Solve Linear Systems by Graphing. Vocabulary System of two linear equations: consists of two equations that can be written in standard or slope intercept.
Lesson 9.7 Solve Systems with Quadratic Equations
Algebra 2 Chapter 3 Notes Systems of Linear Equalities and Inequalities Algebra 2 Chapter 3 Notes Systems of Linear Equalities and Inequalities.
Algebra 2 Chapter 3 Notes Systems of Linear Equalities and Inequalities Algebra 2 Chapter 3 Notes Systems of Linear Equalities and Inequalities.
Unit 1 Test Review Answers
Starter Write an equation for a line that goes through point (-3,3) and (0,3) y = 1/2 x + 3/2.
Substitute 0 for y. Write original equation. To find the x- intercept, substitute 0 for y and solve for x. SOLUTION Find the x- intercept and the y- intercept.
Substitute 0 for y. Write original equation. To find the x- intercept, substitute 0 for y and solve for x. SOLUTION Find the x- intercept and the y- intercept.
Section 7.1 Solving Linear Systems by Graphing. A System is two linear equations: Ax + By = C Dx + Ey = F A Solution of a system of linear equations in.
3.1 Solving equations by Graphing System of equations Consistent vs. Inconsistent Independent vs. Dependent.
3.6 Solving Absolute Value Equations and Inequalities
Chapter 2: Equations and Inequalities
1 © 2010 Pearson Education, Inc. All rights reserved © 2010 Pearson Education, Inc. All rights reserved Chapter 7 Systems of Equations and Inequalities.
Linear Systems of Equations Section 3.1. What is a “system” of equations?
Chapter 4: Systems of Equations and Inequalities Section 4.7: Solving Linear Systems of Inequalities.
Solve for the variable 1. 5x – 4 = 2x (x + 2) + 3x = 2.
Chapter 8 Systems of Linear Equations in Two Variables Section 8.3.
3.3 Linear Programming. Vocabulary Constraints: linear inequalities; boundary lines Objective Function: Equation in standard form used to determine the.
EXAMPLE 1 Solve a system graphically Graph the linear system and estimate the solution. Then check the solution algebraically. 4x + y = 8 2x – 3y = 18.
Algebra 1 Section 7.6 Solve systems of linear inequalities The solution to a system of linear inequalities in two variable is a set of ordered pairs making.
5.1 Solving Systems of Linear Equations by Graphing
Algebra 1 Section 4.2 Graph linear equation using tables The solution to an equation in two variables is a set of ordered pairs that makes it true. Is.
Algebra 2 Chapter 3 Review Sections: 3-1, 3-2 part 1 & 2, 3-3, and 3-5.
UNIT 2 – Linear Functions
Algebra 1 Section 6.4 Solve absolute Value Equations and Inequalities
Do-Now Evaluate the expression when x = –3. –5 ANSWER 1. 3x
X.2 Solving Systems of Linear Equations by Substitution
Algebra 1 Section 6.5 Graph linear inequalities in two variables.
ALGEBRA 1 CHAPTER 7 LESSON 5 SOLVE SPECIAL TYPES OF LINEAR SYSTEMS.
Chapter 1 Linear Equations and Graphs
Chapter 1 Linear Equations and Graphs
Graphing Linear Equations
Chapter 3: Linear Systems
Do Now Solve the following systems by what is stated: Substitution
College Algebra Chapter 2 Functions and Graphs
3-1 Graphing Systems of Equations
Solve Systems of Equations by Graphing
Linear Systems Chapter 3.
Solve a system of linear equation in two variables
Lesson 7.1 How do you solve systems of linear equations by graphing?
Solve Linear Systems by Graphing
Systems of Linear and Quadratic Equations
Graphing Linear Functions
College Algebra Chapter 2 Functions and Graphs
7.2 Solving Systems of Equations Algebraically
College Algebra Chapter 5 Systems of Equations and Inequalities
Chapter 3 Graphs and Functions.
Chapter 3 Section 1 Systems of Linear Equations in Two Variables All graphs need to be done on graph paper. Four, five squares to the inch is the best.
Systems of Equations and Inequalities
of Linear Inequalities
6.4 Slope-Intercept Form of the Equation for a Linear Function
Objectives Identify solutions of linear equations in two variables.
Solving Equations 3x+7 –7 13 –7 =.
Solving Linear Equations by Graphing
Systems of Equations Solve by Graphing.
Chapter 1 Linear Equations and Graphs
Writing Equations of Lines
Section Linear Programming
To Start: Simplify the following: -5(2)(-4) -4(-3)(6) -6(2)(-1) = 40
Integrated Math One Module 5 Test Review.
Writing Equations of Lines
Intersection Method of Solution
Solving Linear Systems by Graphing
Presentation transcript:

Chapter 1 Section 3 Intersection Point of a Pair of Lines Read pages 18 – 21 Look at all the Examples

Note: Authors present one method of finding point of the intersection of two lines. There are other algebraic methods that you can use to find the point, so use the algebraic method that you are most comfortable with!

Vertex A vertex is a point formed by the intersection of two lines. Plural form: Vertices

Authors Method of Finding a Vertex 1.Convert each of the two linear equations into standard form 2.Equate the two expressions that are to the right of each ‘y =‘ 3.Solve for x 4.Solve for y, by substituting the value of x that you have found into one of the original equations (it doesn’t matter which one). 5.The point ( x, y ) is the vertex of the two lines.

Exercise 21 (page 22) Graph and find the vertices of the feasible set formed by the given system of inequalities: Solution: Given x and y intercepts 4 x + y > 8y > – 4 x + 8 ( 2, 0 ), ( 0, 8 ) x + y > 5y > – x + 5 ( 5, 0 ), ( 0, 5 ) x + 3 y > 9y > – 1/3 x + 3 ( 9, 0 ), ( 0, 3 ) x > 0x > 0 y > 0y > 0

Exercise 21 Graph x = 0 y = 0 y = – (1/3) x + 3 Not to scale ( 9, 0 ) y = – x + 5 y = – 4 x + 8 I II III IV ( 0, 8 ) ( 2, 0 ) ( 0, 5 ) ( 0, 3 ) ( 5,0 ) Feasible Set

Exercise 21 Graph II x = 0 I Feasible Set y = – 4 x + 8 II y = – x + 5 III IV y = – (1/3) x + 3 y = 0

Exercise 21 : Find the Vertices Vertex I x = 0 y = – 4 x + 8 y = – 4 ( 0 ) + 8 = 8 Vertex I : ( 0, 8 ) Vertex II y = – 4 x + 8 y = – x + 5 – 4 x + 8 = – x + 5 – 3 x = – 3 x = 1 y = – ( 1 ) + 5 = 4 Vertex II : ( 1, 4 )

Exercise 21 : Find the Vertices Vertex III y = – x + 5 y = – 1/3 x + 3 – x + 5 = – 1/3 x + 3 – 2/3 x = – 2 x = 3 y = – ( 3 ) + 5 = 2 Vertex III : ( 3, 2 ) Vertex IV y = – 1/3 x + 3 y = 0 0 = – 1/3 x + 3 x = 9 Vertex IV : ( 9, 0 )

Supply vs Demand Application Exercise 25 (page 23) Supply Curve: p = q Demand Curve: p = – q Determine the quantity that will be produced and the selling price. Solution: q represent the quantity produced p represent the selling price (in dollars) q = – q q = q = 29,500 which rep p = (29500) = 3 ( $3.00, 29,500)

Solution continued q = – q q = q = 29,500 29,500 units Now take one of the equation and substitute 29,500 in for q p = q p = (29,500) p = 3$3.00

Answer Twenty-nine thousand five hundred units are produced that sell for $3.00 per unit.