Solve Systems of Linear Equations in Three Variables. Section 3.4.

Post on 23-Dec-2015

232 views 1 download

Tags:

Transcript of Solve Systems of Linear Equations in Three Variables. Section 3.4.

Solve Systems of Linear Equations in Three Variables.

Section 3.4

A linear equation in three variables x, y, and z is an equation of the form ax + by + cz = d where a, b, and c are not all zero.

Linear Equation in Three Variables

The following is an example of a system of three linear equations.

The solution of such a system is an ordered triple (x, y, z) whose coordinates make each equation true.

System of Three Linear Equations

2 5

3 2 16

4 3 5 3

x y z

x y z

x y z

The graph of a linear equation in three variables is a plane in three-dimensional space.

The graphs of three such equations that form a system are three planes whose intersection determines the number of solutions.

Exactly one solutionThe planes intersect in a single point.

Infinitely many solutionsThe planes intersect in a line or the same plane.

No solutionThe planes have no common point of intersection.

1. Rewrite the linear system in three variables as a linear system in two variables. by using the elimination method.

2. Solve the new linear system for both of its variables.

3. Substitute the values found in #2 into one of the original equations and solve for the remaining variable.

The Elimination Method for a Three-Variable System

If you obtain a false statement such as 0 = 1, in any of the steps, then the system has not solution.

If you do not obtain a false equation, but obtain an identity such as 0 = 0, then the system has infinitely many solutions.

Solve the system.

1. Eliminate the y since it has a coefficient of -1 in the 1st equation.

Example 1

2 6 4 1

6 4 5 7 2

4 2 5 9 3

x y z

x y z

x y z

2 6 4 1

6 4 5 7 2

x y z

x y z

4 2 6 4

6 4 5 7

x y z

x y z

8 4 24 16

6 4 5 7

x y z

x y z

14 19 23x z

2 6 4 1

4 2 5 9 3

x y z

x y z

2 2 6 4

4 2 5 9

x y z

x y z

4 2 12 8

4 2 5 9

x y z

x y z

8 7 17x z

2. Solve the new two variable linear system.

14 19 23

8 7 17

x z

x z

8 14 19 23

14 8 7 17

x z

x z

112 152 184

112 98 238

x z

x z

54 54z 1z

8 7 1 17x

8 7 17x z

3x

2 6 4x y z

2 3 6 1 4y

4y

The solution is (-3, 4, 1).

3. Substitute −3 for x and 1 for z in one of the three equations and solve for y.

Solve the system.

Example 2

2 1

3 3 3 8 2

2 4 7 3

x y z

x y z

x y z

2 1

3 3 3 8 2

x y z

x y z

3 2

3 3 3 8

x y z

x y z

3 3 3 6

3 3 3 8

x y z

x y z

0 2 False

No solution

Solve the system.

Example 3

6 1

6 2

4 4 24 3

x y z

x y z

x y z

6 1

6 2

x y z

x y z

2 2 12 or 6 x z x z

6 2

4 4 24 3

x y z

x y z5 5 30 or 6 x z x z

The solution is the line x + z = 6

So there are infinitely many solutions.

Solve the system.

Example 4

3 2 10 1

6 2 2 2

4 3 7 3

x y z

x y z

x y z

3 2 10 1

6 2 2 2

x y z

x y z

2 3 2 10

6 2 2

x y z

x y z

6 2 4 20

6 2 2

x y z

x y z

12 3 18 x z

6 2 2 2

4 3 7 3

x y z

x y z

2 6 2 2

4 3 7

x y z

x y z

12 4 2 4

4 3 7

x y z

x y z

13 5 3 x z

12 3 18 x z

4 6

13 5 3

x z

x z

4 6 x z

5 4 6

13 5 3

x z

x z

20 5 30

13 5 3

x z

x z

33 33x 1x

13 5 3 x z

13 1 5 3 z

5 10z

2z

3 1 2 2 10 y

3y

(1, 3, −2)

a) Define the unknowns.

b) Set up the system of equations.

c) Solve the system of equations.

d) Write a sentence to answer the question.

System of Three Linear Equations Application Problems

A coin bank holds nickels, dimes, and quarters. There are 45 coins in the bank and the value of the coins is $4.75. If there are five more nickels than quarters, find the number of each type of coin in the bank.

Example 1

a)

N = # of nickels

D = # of dimes

Q = # of quarters

b)

We will use substitution to solve the 1st part of this problem.

N + D + Q = 45.05N + .10D + .25Q = 4.75

N = Q + 5

Q + 5 + D + Q = 45

.05 Q + 5 + .10D + .25Q = 4.75

N + D + Q = 45

.05N + .10D + .25Q = 4.75

N = Q + 5

2Q + D = 40

.30Q + .10D = 4.50

c)

2Q + D = 40

.30Q + .10D = 4.50

2Q + D = 40 .10

.30Q + .10D = 4.50

.20Q .10 D = 4

.30Q + .10D = 4.50

.10Q = .50Q = 5

2Q + D = 40

2 5 + D = 40D = 30

N + 30 + 5 = 45N = 10

d) There are 10 nickels, 30 dimes, and 5 quarters.

John invested $6500 in three different mutual funds for one year. He earned a total of $560 in simple interest on the three investments. The first fund paid 5% interest, and the second fund paid 8% interest, and the third fund paid 10% interest. If the sum of the first two investments was $500 less than the amount of the third investment, find the amount he invested at each rate.

Example 2

x = amount invested in 5% fund

y = amount invested in 8% fund

z = amount invested in 10% fund

6500 x y z

6500

5 8 10 56000

500

x y z

x y z

x y z

.05 .08 .10 560 x y z500 x y z

6500

5 8 10 56000

500

x y z

x y z

x y z

6500

500

x y z

x y z

2 2 6000x y

6500

5 8 10 56000

x y z

x y z

6500 10

5 8 10 56000

x y z

x y z

10 10 10 65000

5 8 10 56000

x y z

x y z

5 2 9000x y

2 2 6000

5 2 9000

x y

x y

3 3000x 1000x

2 1000 2 6000y 2 4000y

2000y 6500 1000 2000z 3500z

John invested $1000 at 5%, $2000 at 8%, and $3500 at 10%.

The sum of the digits of a three digit number is 12. Five times the units digit plus 6 times the tens digit is 28. If 2 times the tens digit is subtracted from 3 times the hundreds digit, the result is 15. Find the number.

Example 3

U = the units digit

T = the tens digit

H = the hundred digit

12 U T H5 6 28 U T3 2 15 H T

12

5 6 0 28

0 2 3 15

U T H

U T H

U T H

6 6 6 72

5 6 0 28

U T H

U T H

12

5 6 0 28

0 2 3 15

U T H

U T H

U T H

6 44U H 2 2 2 24

0 2 3 15

U T H

U T H

2 5 39U H

6 44

2 5 39

U H

U H

6 44 2

2 5 39

U H

U H

2 12 88

2 5 39

U H

U H

7 49H 7H

6 44U H

6 7 44U 2U

2 7 12T 3T

The number is 732.