You sell tickets for admission to your school play and collect a total of $104. Admission prices are $6 for adults and $4 for children. You sold 21 tickets.

Slides:



Advertisements
Similar presentations
The student will be able to:
Advertisements

4.3 Systems of Equations - Elimination Objective: The student will be able to: Solve systems of equations using elimination with addition and subtraction.
Bell Work2/12/15 Solve the system by elimination..
The student will be able to:
Systems of Linear Equations
Warm up: Solve using a method of your choice.
Ch 5.4 (part 2) Elimination Method (multiplying both equations)
Solving Systems of Linear Equations in Two Variables
Warm Up 12/5 1) Is (-2, 3) a solution? 3x + y = -3 3x + y = -3 2x – 4y = 6 2x – 4y = 6 2) Find the solution by graphing y = -4 + x x + y = 6 3) Solve:
Solving Systems of Equations. Solve systems of equations using addition and subtraction.
5.2: Solving Systems of Equations using Substitution
Solving Systems of Equations using Elimination. Solving a system of equations by elimination using multiplication. Step 1: Put the equations in Standard.
Solving by Substitution Method or Elimination (Addition) Method
Solving Systems Using Elimination
3-2 Solving Linear Systems Algebraically Objective: CA 2.0: Students solve system of linear equations in two variables algebraically.
Lesson 7.4A Solving Linear Systems Using Elimination.
What is a system of equations? A system of equations is when you have two or more equations using the same variables. The solution to the system.
SOLVING SYSTEMS USING ELIMINATION 6-3. Solve the linear system using elimination. 5x – 6y = -32 3x + 6y = 48 (2, 7)
3-2 Solving Systems Algebraically. In addition to graphing, which we looked at earlier, we will explore two other methods of solving systems of equations.
Solving Systems by Elimination 5.4 NOTES, DATE ____________.
Systems of Equations By Substitution and Elimination.
7.2 Solving Linear Systems by Substitution. Steps: 1. Solve one of the equations for one of the variables. 2.Substitute that expression into the other.
5.4 Elimination Using Multiplication Algebra 1 Objective: Each student will understand if addition and subtraction does not eliminate a variable – how.
Solving Systems Using Elimination Section 6-3 Part 2.
Solving Systems of Linear Equations in 2 Variables Section 4.1.
Objective The student will be able to: solve systems of equations using elimination with addition and subtraction.
Linear Systems Systems of Equations Solve by Graphing Solve by Substitution Solve by Elimination Applications of Systems.
Algebra 2 Solving Systems Algebraically Lesson 3-2 Part 2.
The student will be able to:
The student will be able to:
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Solving Systems of Equations
Solving Systems of Equations with Substitution
The student will be able to:
Solving Systems of Linear Equations
SYSTEMS OF LINEAR EQUATIONS
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
The student will be able to:
The student will be able to:
Objective The student will be able to: solve systems of equations using elimination with multiplication.
Solving a system of equations by elimination using multiplication.
The student will be able to:
REVIEW: Solving Linear Systems by Elimination
The student will be able to:
The student will be able to:
The student will be able to:
Solving systems of equations
Notes Solving a System by Elimination
The student will be able to:
Solving Systems of Equations
The student will be able to:
The student will be able to:
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
The student will be able to:
Elimination Using Multiplication.
Solving Systems of Equations
The student will be able to:
The student will be able to:
Solving Systems of Equations
The student will be able to:
The student will be able to:
The student will be able to:
The student will be able to:
The student will be able to:
Solving Systems by ELIMINATION
The Substitution Method
The student will be able to:
The student will be able to:
Step 1: Put the equations in Standard Form. Standard Form: Ax + By = C
The student will be able to:
Presentation transcript:

You sell tickets for admission to your school play and collect a total of $104. Admission prices are $6 for adults and $4 for children. You sold 21 tickets. How many adult tickets and how many children tickets did you sell? 11:30

Objective The student will be able to: Set up a system of equations Use the “substitution” method and the “elimination” effectively to solve for all unknown variables. Manipulate- within reason

Manipulating 6x + 4y = 104 x + y = 21

Steps for solving word problems: 1. Indicate what each variable stands for. Write a system of 2 equations that can be used to solve the problem. Solve the system- substitution or column elimination 4. Check your work in both original equations.

Substitution: 1. Pick the simpler of the two equations Pick a variable and solve for it Go to “the other” equation and kick out that variable you solved for. Sub in the expression to which the variable is equal (step 2) Solve for the single variable equation

Consider y – x = 4 y – 2x = -2

Elimination using multiplication. Step 1: Put the equations in Standard Form. Standard Form: Ax + By = C Step 2: Determine which variable to eliminate based off of LCM. Look for variables that have the same coefficient. Step 3: Multiply each equation fully to get opposite LCM coefficients Solve for the variable. Step 4: Add both equations column by column Step 5: Solve the one variable result equation

1) Solve the system using elimination. 2x + 2y = 6 3x – y = 5 Step 1: Put the equations in Standard Form. They already are! None of the coefficients are the same! Find the least common multiple of each variable. LCM = 6x, LCM = 2y Which is easier to obtain? 2y (you only have to multiply the bottom equation by 2) Step 2: Determine which variable to eliminate.

1) Solve the system using elimination. 2x + 2y = 6 3x – y = 5 Multiply the bottom equation by 2 2x + 2y = 6 (2)(3x – y = 5) 8x = 16 x = 2 2x + 2y = 6 (+) 6x – 2y = 10 Step 3: Multiply the equations and solve. 2(2) + 2y = 6 4 + 2y = 6 2y = 2 y = 1 Step 4: Plug back in to find the other variable. (2, 1)

1) Solve the system using elimination. 2x + 2y = 6 3x – y = 5 (2, 1) (2, 1) 2(2) + 2(1) = 6 3(2) - (1) = 5 Step 5: Check your solution.

  There are three possible outcomes that you may encounter when working with these systems: 1.) one solution 2.) no solution infinite solutions  

  One Solution If the system has one solution, it is an ordered pair (x, y) that makes BOTH equations true.  In other words, when you plug in the values of the ordered pairs, it makes ALL equations TRUE.   

No Solution If the equations are parallel to each other, they will never intersect.  This means they do not have any points in common.  In this situation, no pair values will make both equations true. Example 5: pg 491

Infinite Solutions If the two equations are equivalent to each other, then there is an infinite number of solutions.  Example 6 Pg. 491    

7.1 Pg. 483 #1;5-7; 16; 67; 68 7.2 Pg. 495 # 2-6(even); 12-14; 48; 49