holt mcdougal algebra 2 3-6 solving linear systems in three variables 3-6 solving linear systems in...

Holt McDougal Algebra 2

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

Holt Algebra 2

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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



Represent solutions to systems of equations in three dimensions graphically.

Solve systems of equations in three dimensions algebraically.

Objectives



Systems of three equations with three variables are often called 3-by-3 systems. In general, to find a single solution to any system of equations, you need as many equations as you have variables.



Recall from Lesson 3-5 that the graph of a linear equation in three variables is a plane. When you graph a system of three linear equations in three dimensions, the result is three planes that may or may not intersect. The solution to the system is the set of points where all three planes intersect. These systems may have one, infinitely many, or no solution.



Identifying the exact solution from a graph of a 3-by-3 system can be very difficult. However, you can use the methods of elimination and substitution to reduce a 3-by-3 system to a 2-by-2 system and then use the methods that you learned in Lesson 3-2.



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



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



11x – 11y = –55

7x – 2y = –25You now have a 2-by-2 system.

4

5

Example 1 Continued



–2(11x – 11y = –55)

55x = –165

11(7x – 2y = –25) –22x + 22y = 110 77x – 22y = –275

4

5

1

1Multiply equation - by –2, and equation - by 11 and add.

4

5

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

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

x = –3 Solve for x.

Example 1 Continued



11x – 11y = –55

11(–3) – 11y = –55

4

1

1

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

y = 2

Substitute –3 for x.

Solve for y.

Example 1 Continued



2x – 3y + z = –16

2(–3) – 3(2) + z = –16

2

1

1

Step 4 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



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.



–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



You now have a 2-by-2 system.


4

5

–5x – 5y = 5

5x + 9y = –1



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



–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



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



You can also use substitution to solve a 3-by-3 system. Again, the first step is to reduce the 3-by-3 system to a 2-by-2 system.



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



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.



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.



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



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




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




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.



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.



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.





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.




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.



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



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.



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.



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



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



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




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.




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



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



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




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.



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



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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