Holt Algebra 2 3-6 Solving Linear Systems in Three Variables 3-6 Solving Linear Systems in Three Variables Holt Algebra 2 Warm Up Warm Up Lesson Presentation

Holt Algebra 2

3-6Solving Linear Systems in Three Variables 3-6Solving Linear Systems in Three Variables

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 2

3-6Solving Linear Systems in Three Variables

Warm UpSolve each system of equations algebraically.

Classify each system and determine the number of solutions.

1. 2.x = 4y + 10

4x + 2y = 4

6x – 5y = 9

2x – y =1(2, –2) (–1,–3)

3. 4.3x – y = 8

6x – 2y = 2

x = 3y – 1

6x – 12y = –4

inconsistent; none consistent, independent; one

Holt Algebra 2


Holt Algebra 2


Holt Algebra 2


Use elimination to solve the system of equations.

Example 1: Solving a Linear System in Three Variables

Step 1 Eliminate one variable.

5x – 2y – 3z = –7

2x – 3y + z = –16

3x + 4y – 2z = 7

In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1 and z is easy to eliminate from the other equations.

1

2

3

Holt Algebra 2


Example 1 Continued

5x – 2y – 3z = –7

11x – 11y = –55

3(2x –3y + z = –16)

5x – 2y – 3z = –7

6x – 9y + 3z = –48

1

2

1

4

3x + 4y – 2z = 7

7x – 2y = –25

2(2x –3y + z = –16)3x + 4y – 2z = 74x – 6y + 2z = –32

3

2

Multiply equation - by 3, and add to equation .1

2

Multiply equation - by 2, and add to equation .3

2

5

Use equations and to create a second equation in x and y.

3 2

Holt Algebra 2


11x – 11y = –55

7x – 2y = –25Step 2: You now have a 2-by-2 system. Solve for x and y.

4

5

Example 1 Continued

Holt Algebra 2


2x – 3y + z = –16

2(–3) – 3(2) + z = –16

2

1

1

Step 3 Substitute for x and y in one of the original equations to solve for z.

z = –4

Substitute –3 for x and 2 for y.

Solve for y.

The solution is (–3, 2, –4).

Example 1 Continued

Holt Algebra 2


Use elimination to solve the system of equations.

Step 1 Eliminate one variable.

–x + y + 2z = 7

2x + 3y + z = 1

–3x – 4y + z = 4

1

2

3

Check It Out! Example 1

In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1.

Holt Algebra 2


–x + y + 2z = 7

–5x – 5y = 5

–2(2x + 3y + z = 1) –4x – 6y – 2z = –2

1

2

1

4

5x + 9y = –1

–2(–3x – 4y + z = 4)–x + y + 2z = 7

6x + 8y – 2z = –81

3

Multiply equation - by –2, and add to equation .1

2

Multiply equation - by –2, and add to equation .1

3

5

Check It Out! Example 1 Continued

–x + y + 2z = 7

–x + y + 2z = 7

Use equations and to create a second equation in x and y.

1 3

Holt Algebra 2


You now have a 2-by-2 system.


4

5

–5x – 5y = 5

5x + 9y = –1

Holt Algebra 2


4y = 4

4

5

1

Add equation to equation .45

Step 2 Eliminate another variable. Then solve for the remaining variable.

You can eliminate x by using methods from Lesson 3-2.

Solve for y.


–5x – 5y = 55x + 9y = –1

y = 1

Holt Algebra 2


–5x – 5(1) = 5

4

1

1

Step 3 Use one of the equations in your 2-by-2 system to solve for x.

x = –2

Substitute 1 for y.

Solve for x.


–5x – 5y = 5

–5x – 5 = 5

–5x = 10

Holt Algebra 2


2(–2) +3(1) + z = 1

2x +3y + z = 12

1

1


z = 2

Substitute –2 for x and 1 for y.

Solve for z.

The solution is (–2, 1, 2).


–4 + 3 + z = 1

Holt Algebra 2


The table shows the number of each type of ticket sold and the total sales amount for each night of the school play. Find the price of each type of ticket.

Example 2: Business Application

Orchestra Mezzanine Balcony Total Sales

Fri 200 30 40 $1470

Sat 250 60 50 $1950

Sun 150 30 0 $1050

Holt Algebra 2


Example 2 ContinuedStep 1 Let x represent the price of an orchestra seat,

y represent the price of a mezzanine seat, and z represent the present of a balcony seat.

Write a system of equations to represent the data in the table.

200x + 30y + 40z = 1470

250x + 60y + 50z = 1950

150x + 30y = 1050

1

2

3

Friday’s sales.

Saturday’s sales.

Sunday’s sales.

A variable is “missing” in the last equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward.

Holt Algebra 2


5(200x + 30y + 40z = 1470)

–4(250x + 60y + 50z = 1950)

1

Step 2 Eliminate z.

Multiply equation by 5 and equation by –4 and add.1

2

2

1000x + 150y + 200z = 7350

–1000x – 240y – 200z = –7800

y = 5

Example 2 Continued

By eliminating z, due to the coefficients of x, you also eliminated x providing a solution for y.

Holt Algebra 2


150x + 30y = 1050

150x + 30(5) = 1050

3 Substitute 5 for y.

x = 6

Solve for x.

Step 3 Use equation to solve for x.3

Example 2 Continued

Holt Algebra 2


200x + 30y + 40z = 14701 Substitute 6 for x and 5 for y.

1

z = 3

Solve for x.

Step 4 Use equations or to solve for z.21

200(6) + 30(5) + 40z = 1470

The solution to the system is (6, 5, 3). So, the cost of an orchestra seat is $6, the cost of a mezzanine seat is $5, and the cost of a balcony seat is $3.

Example 2 Continued

Holt Algebra 2



Jada’s chili won first place at the winter fair. The table shows the results of the voting.

How many points are first-, second-, and third-place votes worth?

Name

1st Place

2nd Place

3rd Place

TotalPoints

Jada 3 1 4 15

Maria 2 4 0 14

Al 2 2 3 13

Winter Fair Chili Cook-off

Holt Algebra 2



Step 1 Let x represent first-place points, y represent second-place points, and z represent third- place points.

Write a system of equations to represent the data in the table.

3x + y + 4z = 15

2x + 4y = 14

2x + 2y + 3z = 13

1

2

3

Jada’s points.

Maria’s points.

Al’s points.

A variable is “missing” in one equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward.

Holt Algebra 2


3(3x + y + 4z = 15)

–4(2x + 2y + 3z = 13)

1

Step 2 Eliminate z.

Multiply equation by 3 and equation by –4 and add.3

1

3

9x + 3y + 12z = 45

–8x – 8y – 12z = –52

x – 5y = –7 4


2

–2(x – 5y = –7)4

2x + 4y = 14–2x + 10y = 14

2x + 4y = 14

y = 2

Multiply equation by –2 and add to equation .2

4

Solve for y.

Holt Algebra 2


2x + 4y = 14

Step 3 Use equation to solve for x.2

2

2x + 4(2) = 14

x = 3

Solve for x.

Substitute 2 for y.


Holt Algebra 2



z = 1 Solve for z.

2x + 2y + 3z = 133

2(3) + 2(2) + 3z = 13

6 + 4 + 3z = 13

The solution to the system is (3, 2, 1). The points for first-place is 3, the points for second-place is 2, and 1 point for third-place.


Holt Algebra 2


Consistent means that the system of equations has at least one solution.

Remember!

The systems in Examples 1 and 2 have unique solutions. However, 3-by-3 systems may have no solution or an infinite number of solutions.

Holt Algebra 2


Classify the system as consistent or inconsistent, and determine the number of solutions.

Example 3: Classifying Systems with Infinite Many Solutions or No Solutions

2x – 6y + 4z = 2

–3x + 9y – 6z = –3

5x – 15y + 10z = 5

1

2

3

Holt Algebra 2


Example 3 Continued

3(2x – 6y + 4z = 2)

2(–3x + 9y – 6z = –3)

First, eliminate x.

1

2

6x – 18y + 12z = 6

–6x + 18y – 12z = –6

0 = 0

Multiply equation by 3 and equation by 2 and add.2

1

The elimination method is convenient because the numbers you need to multiply the equations are small.

Holt Algebra 2


Example 3 Continued

5(2x – 6y + 4z = 2)

–2(5x – 15y + 10z = 5)

1

3

10x – 30y + 20z = 10

–10x + 30y – 20z = –10

0 = 0

Multiply equation by 5 and equation by –2 and add.

3

1

Because 0 is always equal to 0, the equation is an identity. Therefore, the system is consistent, dependent and has an infinite number of solutions.

Holt Algebra 2


Check It Out! Example 3a

Classify the system, and determine the number of solutions.

3x – y + 2z = 4

2x – y + 3z = 7

–9x + 3y – 6z = –12

1

2

3

Holt Algebra 2


3x – y + 2z = 4–1(2x – y + 3z = 7)

First, eliminate y.

1

3

3x – y + 2z = 4

–2x + y – 3z = –7

x – z = –3

Multiply equation by –1 and add to equation . 1

2

The elimination method is convenient because the numbers you need to multiply the equations by are small.

Check It Out! Example 3a Continued

4

Holt Algebra 2


3(2x – y + 3z = 7)–9x + 3y – 6z = –12

2

3

6x – 3y + 9z = 21

–9x + 3y – 6z = –12

–3x + 3z = 9

Multiply equation by 3 and add to equation . 3

2

Now you have a 2-by-2 system.

x – z = –3

–3x + 3z = 9 5

4

5


Holt Algebra 2


3(x – z = –3)

–3x + 3z = 9 5

4 3x – 3z = –9

–3x + 3z = 9

0 = 0

Because 0 is always equal to 0, the equation is an identity. Therefore, the system is consistent, dependent, and has an infinite number of solutions.

Eliminate x.


Holt Algebra 2


Check It Out! Example 3b

Classify the system, and determine the number of solutions.

2x – y + 3z = 6

2x – 4y + 6z = 10

y – z = –2

1

2

3

Holt Algebra 2


y – z = –2y = z – 2

3Solve for y.

Use the substitution method. Solve for y in equation 3.

Check It Out! Example 3b Continued

Substitute equation in for y in equation .4 1

4

2x – y + 3z = 6

2x – (z – 2) + 3z = 6

2x – z + 2 + 3z = 6

2x + 2z = 4 5

Holt Algebra 2


Substitute equation in for y in equation .4 2

2x – 4y + 6z = 10

2x – 4(z – 2) + 6z = 102x – 4z + 8 + 6z = 10

2x + 2z = 2 6

Now you have a 2-by-2 system.

2x + 2z = 4

2x + 2z = 2 6

5


Holt Algebra 2


2x + 2z = 4

–1(2x + 2z = 2)6

5

Eliminate z.

0 2


Because 0 is never equal to 2, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions.

Holt Algebra 2


Lesson Quiz: Part I

At the library book sale, each type of book is priced differently. The table shows the number of books Joy and her friends each bought, and the amount each person spent. Find the price of each type of book.

paperback: $1;

Hard-cover

Paper- back

Audio Books

Total Spent

Hal 3 4 1 $17

Ina 2 5 1 $15

Joy 3 3 2 $20

1.

hardcover: $3;

audio books: $4

Holt Algebra 2


2.

3.

2x – y + 2z = 5

–3x +y – z = –1

x – y + 3z = 2

9x – 3y + 6z = 3

12x – 4y + 8z = 4

–6x + 2y – 4z = 5

inconsistent; none

consistent; dependent; infinite

Lesson Quiz: Part II

Classify each system and determine the number of solutions.

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Holt Algebra 2 3-6 Solving Linear Systems in Three Variables 3-6 Solving Linear Systems in Three Variables Holt Algebra 2 Warm Up Warm Up Lesson Presentation