Solving Systems of Linear Equations in Three Variables; Applications9.49.4

1. Determine whether an ordered triple is a solution for a system of equations.

2. Understand the types of solution sets for systems of three equations.

3. Solve a system of three linear equations using the elimination method.

Is (2, –1, 3) a solution of the system?

x + y + z = 4

2 + (–1) + 3 = 4

4 = 4

TRUE

3 = 3

TRUE

2x – 2y – z = 3

2(2) – 2(–1) – 3 = 3

– 4x + y + 2z = –3

– 4(2) + (–1) + 2(3) = –3

–3 = –3

TRUE

Because (2, 1, 3) satisfies all three equations in the system, it is a solution for the system.

324

322

4

zyx

zyx

zyxOrdered Ordered

TripleTriple(x, y, z)(x, y, z)

Slide 4- 3Copyright © 2011 Pearson Education, Inc.

Determine if (2, –5, 3) is a solution to the given system.

a) Yes

b) No

5 2 7

3 5 16

5

x y z

y z

z

9.4

Slide 4- 4Copyright © 2011 Pearson Education, Inc.

Determine if (2, –5, 3) is a solution to the given system.

a) Yes

b) No

5 2 7

3 5 16

5

x y z

y z

z

9.4

Copyright © 2011 Pearson Education, Inc.

Types of Solution Sets A Single Solution:

If the planes intersect at a single point, that ordered triple is the solution to the system.

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Types of Solution Sets Infinite Number of Solutions: If the three planes intersect along a line, the system has an infinite number of solutions, which are the coordinates of any point along that line.

Infinite Number of Solutions:

If all three graphs are the same plane, the system has an infinite number of solutions. They are the coordinates of all points in the plane.

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Types of Solution Sets No Solution: If all of the planes are parallel, the system has no solution.

No Solution: Pairs of planes also can intersect, as shown. However, because all three planes do not have a common intersection, the system has no solution.

Solve the system using elimination.

We select any two of the three equations and work to get one equation in two variables.

Add equations (1) and (2):

(1)

(2)

(3)

6

2 2

x y z

x y z

(1)

(2)

(4)2x + 3y = 8Add to eliminate z

83

22

6

zyx

zyx

zyx

Next, we select a different pair of equations and eliminate the same variable.

Use (2) and (3) to again eliminate z.

2 2

3 8

x y z

x y z

(5)

x – y + 3z = 8

Multiply by 3

4x + 5y = 14

3x + 6y – 3z = 6

Now solve the resulting system of equations (4) and (5).

4x + 5y = 14

2x + 3y = 8(5)

(4)

*(1)

*(2)

(3)

83

22

6

zyx

zyx

zyx2x + 3y = 8 (4)

Multiply equation (4) by –2 and then add to equation (5):

Substitute into either equation (4) or (5) to find x.

4x + 5y = 14–4x – 6y = –16

–y = –2 y = 2

4x + 5y = 14

2x + 3y = 8

(5)

(4)

4x + 5y = 14

2x + 3y = 8(5)

(4)Multiply by -2

1

22

862

8232

832

x

x

x

x

yx

(1)

(2)

(3)

83

22

6

zyx

zyx

zyx-2

x + y + z = 6

1 + 2 + z = 6z = 3.

To find z, substitute into any of the original equations.

(1, 2, 3)

Ordered triple

Solve the system using elimination.

(1)

(2)

(3)

2 2

2

x y z

x y z

(2)

(3)

(4)3x + 2y = 4

Eliminate z from equations (2) and (3).

2

22

3693

zyx

zyx

zyx

Eliminate z from equations (1) and (2).

Multiply by 6

(1)

*(2)

*(3)

3 9 6 3

2 2

x y z

x y z

3 9 6 3

12 6 6 12

x y z

x y z

15x + 15y = 15

Eliminate x from equations (4) and (5).

3x + 2y = 415x + 15y = 15

Multiply by 5 15x – 10y = 2015x + 15y = 15

5y = 5y = 1Use y = 1, to find x in equation 4.

3x + 2y = 43x + 2(1) = 4

x = 2

continued3x + 2y = 4 (4)

(5)

2

22

3693

zyx

zyx

zyx

continued

Substitute x = 2 and y = 1 to find z.x + y + z = 22 – 1 + z = 2

1 + z = 2 z = 1

The solution is the ordered triple (2, 1, 1).

y = 1x = 2(1)

(2)

(3)

2

22

3693

zyx

zyx

zyx

Solving Systems of Three Linear Equations Using Elimination

1. Write each equation in the form Ax + By+ Cz = D.

2. Eliminate one variable from one pair of equations using the elimination method.

3. If necessary, eliminate the same variable from another pair of equations.

continued

4. Steps 2 and 3 result in two equations with the same two variables. Solve these equations using the elimination method.

5. To find the third variable, substitute the values of the variables found in step 4 into any of the three original equations that contain the third variable.

6. Check the ordered triple in all three original equations.

Solve the system using elimination.(1)

(2)

(3)

2 6 2 2

2 2 2

x y z

x y z

(1)

(2)

5y 4z = 0

Eliminate x from equations (1) and (2).

132

222

13

zyx

zyx

zyx

13 zyx

222 zyx

Multiply by -2

(4)

Eliminate x from equations (1) and (3).

*(1)

*(2)

(3)

3 1

2 3 1

x y z

x y z

5y + 4z = 2 (5)

Eliminate y from equations (4) and (5).

5y 4z = 05y + 4z = 2

0 = 2

All variables are eliminated and the resulting equation is false. This system has no solution. It is inconsistent.

continued

132

222

13

zyx

zyx

zyx

13 zyxMultiply by -1

132 zyx

5y 4z = 0 (4)

Slide 4- 18Copyright © 2011 Pearson Education, Inc.

Solve the system.

a) (–2, 2, –5)

b) (–5, 2, –2)

c) infinite number of solutions

d) no solution

2 4 1

5 2 9

3 2 14

x y z

x y z

x y z

9.4

Slide 4- 19Copyright © 2011 Pearson Education, Inc.

Solve the system.

a) (–2, 2, –5)

b) (–5, 2, –2)

c) infinite number of solutions

d) no solution

2 4 1

5 2 9

3 2 14

x y z

x y z

x y z

9.4