aii, 2.0: students solve systems of linear equations and inequalities (in two or three variables) by...

A I I , 2 . 0 : S T U D E N T S S O LV E S Y S T E M S O F L I N E A R E Q U AT I O N S A N D I N E Q U A L I T I E S ( I N T W O O R T H R E E VA R I A B L E S ) B Y S U B S T I T U T I O N, W I T H G R A P H S, O R

W I T H M AT R I C E S .L A , 6 . 0 : S T U D E N T S D E M O N S T R AT E A N

U N D E R S TA N D I N G T H AT L I N E A R S Y S T E M S A R E I N C O N S I S T E N T ( H AV E N O S O LU T I O N S ) , H AV E

E X A C T LY O N E S O LU T I O N, O R H AV E I N F I N I T E LY M A N Y S O LU T I O N S

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

Solving Linear Systems by Linear Combinations

Simplify Evaluate

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

2. What is twice of ?3. What is times ?

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

Using the Linear Combination Method

Step-by-Step

Steps:1. Multiply, if necessary, one or both

equations by a constant so that the coefficients of one of the variables differ only in sign.

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

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

4. Check the solution in each of the original equations.

Example 1 Multiply One Equation

Solve the linear system using the linear combination method.

Equation 163y2x =–


SOLUTION

STEP 1 Multiply the first equation by 2 so that the coefficients of x differ only in sign.

–

63y2x =–

85y4x =– 85y4x =–

126y4x =+– –

4y = –


STEP 2 Add the revised equations and solve for y.

STEP 3 Substitute 4 for y in one of the original equations and solve for x.

–

Write Equation 1.63y2x =–

Substitute 4 for y.62x =– ( )4–3 –

Simplify.6122x =+

Subtract 12 from each side.62x = –

Solve for x.3x = –

4y = –


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

– –

ANSWER The solution is .( )3,– – 4

Example 2 Multiply Both Equations

Solve the system using the linear combination method.

Equation 1

Equation 2

SOLUTION

STEP 1

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

2212y7x =– –

148y5x =+–

2212y7x =– –

148y5x =+–

4424y14x =– –

4224y15x =+–

2x = ––


STEP 2 Add the revised equations and solve for x.

2x =

2x = ––

STEP 3 Substitute 2 for x in one of the original equations and solve for y.

148y5x =+– Write Equation 2.

148y =+ Substitute 2 for x.– ( )25

148y10 =+– Multiply.

3y = Solve for y.


STEP 4 Check by substituting 2 for x and 3 for y in the original equations.

ANSWER The solution is (2, 3).

Example 3 A Linear System with No Solution


Equation 1


128y4x =+– –

SOLUTION

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

8y

–

124x =+

148y4x =

– –

74y2x =–

128y4x =+– –

20 =Add the revised equations.

Example 3 A Linear System with No Solution

ANSWER

Because the statement 0 2 is false, there is no solution.=

Checkpoint

ANSWER infinitely many solutions

1.


Solve a Linear System

54yx =–

1y2x =+ANSWER (1, 1)–

2. 4y2x =–

82y4x =–

3. 22y3x =–

13y4x =–ANSWER (4, 5 )

Checkpoint

ANSWER

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

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

Solve a Linear System

Use a Linear System as a ModelExample 4

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?


SOLUTION

VERBALMODEL

•Servingsper pan

Pans of pasta

Sandwich trays =+

Servings per

sandwich tray

Servings needed•

•Price

per panPans of pasta

Sandwich trays

=+ Priceper tray

Money to spend

on food•


LABELS Servings per pan of pasta 10 = (servings)

Pans of pasta p = (pans)

Servings per sandwich tray 5 =

(trays) Sandwich trays s=

Servings needed 30 =

Price per pan of pasta 40 = (dollars)

Price per sandwich tray 10 = (dollars)

Money to spend on food 80 = (dollars)

(servings)

(servings)


ALGEBRAICMODEL

Equation 1 (servings needed) 30=10p + 5s

Equation 2 (money to spend on food)

80=40p + 10s

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

30=10p + 5s

80=40p + 10s

=20p 10s– – 60–

80=40p + 10s

20=20pAdd the revised equations and solve for p.

1=p


ANSWER

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

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

Write Equation 1. 30=10p + 5s

Substitute 1 for p. 30=10 + 5s( )1

Subtract 10 from each side. 20=5s

4=s Solve for s.

Checkpoint Solve a Linear System

5. 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?

ANSWER

2 pans of pasta and 4 sandwich trays

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.

Conclusions

WHAT METHOD IS MORE CONVENIENT? GRAPHING, SUBSTITUTION, OR LINEAR

COMBINATION.

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 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.

1. Solve one equation for one of its variables

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

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

4. Check the solution in each of the original equations

Choosing a Method

Which Method Will You Choose?

Linear Combination Method:Step-by-Step


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

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

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

4. Check the solution in each of the original equations.

Choosing a Method

Choosing a Method


Answer:Substitution Methody has a coefficient of 1

Choosing a Method


Answer:Linear Combination MethodNeither variable has a coefficient of 1 or -1

Choosing a Method


Answer:Substitution Methody has a coefficient of -1 and x has a coefficient of 1

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

aii, 2.0: students solve systems of linear equations and inequalities (in two or three variables) by...

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