Solving linear systems by linear combinations
This presentation is the property of its rightful owner.
Sponsored Links
1 / 29

Solving Linear Systems by Linear Combinations PowerPoint PPT Presentation


  • 110 Views
  • Uploaded on
  • Presentation posted in: General

Solving Linear Systems by Linear Combinations. AII, 2.0: Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices.

Download Presentation

Solving Linear Systems by Linear Combinations

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Solving linear systems by linear combinations

Solving Linear Systems by Linear Combinations

AII, 2.0: Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices.

LA, 6.0: Students demonstrate an understanding that linear systems are inconsistent (have no solutions), have exactly one solution, or have infinitely many solutions


Solving linear systems by linear combinations1

Solving Linear Systems by Linear Combinations

Objectives

Key Words

Solve a system of linear equations in two variables by the linear combination method

EC: Choosing a Method

Linear combination method


Prerequisite check if you do not know you need to let me know

Prerequisite Check:If you do not know, you need to let me know

Simplify

Evaluate

What do you have if you have twice of a bag with 2 apples and 3 oranges?

What is twice of ?

What is times ?


Using the linear combination method

Steps:

  • Multiply, if necessary, one or both equations by a constant so that the coefficients of one of the variables differ only in sign.

  • Add the revised equations from Step 1. combining like terms will eliminate one variable. Solve for the remaining variable.

  • Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable.

  • Check the solution in each of the original equations.

Using the Linear Combination Method

Step-by-Step


Solving linear systems by linear combinations

Example 1

2x

3y

6

=

4x

5y

8

=

SOLUTION

STEP 1

Multiply the first equation by 2 so that the coefficients of x differ only in sign.

4x

+

6y

12

=

2x

3y

6

=

4x

5y

8

4x

5y

8

=

=

y

4

=

Multiply One Equation

Solve the linear system using the linear combination method.

Equation 1

Equation 2


Solving linear systems by linear combinations

Example 1

STEP 3

Substitute4 for y in one of the original equations and solve for x.

Write Equation 1.

2x

3y

6

=

(

)

Substitute 4 for y.

3

4

2x

6

=

Simplify.

2x

+

12

6

=

Subtract 12 from each side.

2x

6

=

Solve for x.

x

3

=

y

4

=

Multiply One Equation

STEP 2

Add the revised equations

and solve for y.


Solving linear systems by linear combinations

Example 1

The solution is .

ANSWER

(

)

3,

4

STEP 4

Check by substituting 3 for x and 4 for y in the original equations.

Multiply One Equation


Solving linear systems by linear combinations

Example 2

7x

12y

22

=

5x

+

8y

14

=

SOLUTION

STEP 1

Multiply the first equation by 2 and the second equation by 3.

14x

24y

44

=

15x

+

24y

42

=

x

2

=

5x

+

8y

14

7x

12y

=

22

=

Multiply Both Equations

Solve the system using the linear combination method.

Equation 1

Equation 2


Solving linear systems by linear combinations

Example 2

x

2

=

x

2

=

STEP 3

Substitute2 for x in one of the original equations and solve for y.

Write Equation 2.

(

)

5

2

Substitute 2 for x.

+

8y

14

=

Multiply.

10

+

8y

14

=

Solve for y.

y

3

=

5x

+

8y

14

=

Multiply Both Equations

STEP 2

Add the revised equations

and solve for x.


Solving linear systems by linear combinations

Example 2

The solution is (2, 3).

ANSWER

Multiply Both Equations

STEP 4

Check by substituting 2 for x and 3 for y in the original equations.


Solving linear systems by linear combinations

Example 3

2x

4y

7

=

SOLUTION

4x

+

8y

12

=

Multiply the second equation by 2 so that the coefficients of y differ only in sign.

4x

+

8y

12

4x

+

8y

12

=

=

2x

4y

7

4x

8y

14

=

=

Add the revised equations.

0

2

=

A Linear System with No Solution

Solve the system using the linear combination method.

Equation 1

Equation 2


Solving linear systems by linear combinations

Example 3

A Linear System with No Solution

ANSWER

Because the statement 02 is false, there is no solution.

=


Solving linear systems by linear combinations

Checkpoint

(1, 1)

ANSWER

x

4y

5

=

2.

2x

y

4

=

infinitely many solutions

ANSWER

4x

2y

8

=

3.

3x

2y

2

=

(4, 5

)

ANSWER

4x

3y

1

=

Solve a Linear System

Solve the system using the linear combination method.

1.

2x

+

y

1

=


Solving linear systems by linear combinations

Checkpoint

ANSWER

if you get a false equation; if you get a true equation

Solve a Linear System

4. How can you tell when a system has no solution?

infinitely many solutions?


Solving linear systems by linear combinations

Example 4

Use a Linear System as a Model

Catering A customer hires a caterer to prepare food for a party of 30 people. The customer has $80 to spend on food and would like there to be a choice of sandwiches and pasta. A $40 pan of pasta contains 10 servings, and a $10 sandwich tray contains 5 servings. The caterer must prepare enough food so that each person receives one serving of either food. How many pans of pasta and how many sandwich trays should the caterer prepare?


Solving linear systems by linear combinations

Servings per

sandwich tray

Servings

per pan

Sandwich

trays

Servings

needed

Money to spend

on food

Price

per pan

Price

per tray

Sandwich

trays

Example 4

Use a Linear System as a Model

SOLUTION

VERBAL

MODEL

Pans of

pasta

+

=

Pans of

pasta

+

=


Solving linear systems by linear combinations

Example 4

Use a Linear System as a Model

(servings)

Servings per pan of pasta 10

LABELS

=

(pans)

Pans of pasta p

=

(servings)

Servings per sandwich tray 5

=

(trays)

Sandwich trays s

=

(servings)

Servings needed 30

=

(dollars)

Price per pan of pasta 40

=

(dollars)

Price per sandwich tray 10

=

(dollars)

Money to spend on food 80

=


Solving linear systems by linear combinations

+

+

10p

10p

5s

5s

30

30

=

=

+

+

Equation 2 (money to spend on food)

40p

40p

10s

10s

80

80

=

=

Multiply Equation 1 by 2 so that the coefficients of s differ only in sign.

20p

10s

60

=

+

40p

10s

80

=

20p

20

=

Add the revised equations and solve for p.

p

1

=

Example 4

Use a Linear System as a Model

ALGEBRAIC

MODEL

Equation 1 (servings needed)


Solving linear systems by linear combinations

+

10p

5s

30

=

Write Equation 1.

(

)

1

Substitute 1 for p.

+

10

5s

30

=

Subtract 10 from each side.

5s

20

=

Solve for s.

s

4

=

ANSWER

The caterer should make 1 pan of pasta and 4 sandwich trays.

Example 4

Use a Linear System as a Model

Substitute 1 for p in one of the original equations and solve for s.


Solving linear systems by linear combinations

Checkpoint

ANSWER

2 pans of pasta and 4 sandwich trays

Solve a Linear System

Another customer asks the caterer in Example 4 to plan a party for 40 people. This customer also wants both sandwiches and pasta and has $120 to spend. How many pans of pasta and how many sandwich trays should the caterer prepare?

5.


Conclusions

Conclusions

Summary

Assignment

  • How do you solve a system of linear equations algebraically?

    • To use the linear combination method, multiply one or both equations by constants to get opposite coefficients for one variable. Add the revised equations to solve for one variable. Then substitute the value you found into either one of the original equations to find the value of the other variable.

Pg142 #(10,14,26,28,34)

Due by the end of the class.


Choosing a method

Choosing a Method

What method is more convenient? Graphing, Substitution, or Linear Combination.


Choosing a method1

Choosing a Method

Substitution

Linear Combination

If one of the variables has a coefficient of 1 or -1, the substitution method is convenient.

In general, you should solve for the variable.

If neither variable has a coefficient of 1 or -1, the linear combination method is often more convenient, although you can still use substitution.


Choosing a method2

Choosing a Method

Which Method Will You Choose?

Substitution Method:

Step-by-Step

Choose a method to solve the linear system. Explain your choice. Then solve the system.

Solve one equation for one of its variables

Substitute the expression from Step 1 into the other equation and solve for the other variable

Substitute the value from Step 2 into the revised equation from Step 1 and solve

Check the solution in each of the original equations


Choosing a method3

Choosing a Method

Which Method Will You Choose?

Linear Combination Method:

Step-by-Step

Choose a method to solve the linear system. Explain your choice. Then solve the system.

Multiply, if necessary, one or both equations by a constant so that the coefficients of one of the variables differ only in sign.

Add the revised equations from Step 1. combining like terms will eliminate one variable. Solve for the remaining variable.

Substitute the value obtained in Step 2 into either of the original equations and solve for the other variable.

Check the solution in each of the original equations.


Choosing a method4

Choosing a Method

Choose a method to solve the linear system. Explain your choice. Then solve the system.

Answer:

Substitution Method

y has a coefficient of 1


Choosing a method5

Choosing a Method

Choose a method to solve the linear system. Explain your choice. Then solve the system.

Answer:

Linear Combination Method

Neither variable has a coefficient of 1 or -1


Choosing a method6

Choosing a Method

Choose a method to solve the linear system. Explain your choice. Then solve the system.

Answer:

Substitution Method

y has a coefficient of -1 and x has a coefficient of 1


Additional practice problems solve graphically and algebraically

Additional Practice Problems solve graphically and algebraically

http://www.classzone.com/cz/books/algebra_2_cs/resources/applications/animations/html/explore_learning/chapter_3/dswmedia/7_5_special_sys.html


  • Login