3.6 Systems with Three Variables

1. Solving Three-Variable Systems by Elimination

2. Solving Three-Variable Systems by Substitution

1) Solving Three-Variable Systems by Elimination

• A three-variable system produces a 3D graph that is a plane

• A three equation system produces three planes

• The planes may never intersect, intersect once, or have an infinite number of intersections

• Use elimination to solve

1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

1

2

3{

1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

1

2

3

You have to work with what you are given to eliminate one variable at a time.

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1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

Step 1: Add and to cancel y.

1

2

3

1 2

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1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

Step 1: Add and to cancel y.

x – 3y + 3z = -4

2x + 3y – z = 15

1

2

3

1 2

1

2

{

1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

Step 1: Add and to cancel y.

x – 3y + 3z = -4

2x + 3y – z = 15

3x + 2z = 11

1

2

3

1 2

1

2

4 New two-variable equation

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1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

Step 1: Add and to cancel y.

1

2

3

2 3

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1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

Step 1: Add and to cancel y.

2x + 3y – z = 15

4x – 3y – z = 19

1

2

3

2

2

3

3

{

1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

Step 1: Add and to cancel y.

2x + 3y – z = 15

4x – 3y – z = 19

6x - 2z = 34

1

2

3

2

2

3

5 New two-variable equation

3

{

1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

Step 2: Add and to solve for x.

1

2

3

4 5

{

1) Solving Three-Variable Systems by Elimination

Example 1:Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 154x – 3y – z = 19

Step 2: Add and to solve for x. 3x + 2z = 11 6x - 2z = 34

1

2

3

4

5

4 5

{

1) Solving Three-Variable Systems by Elimination

Example 1:Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 154x – 3y – z = 19

Step 2: Add and to solve for x. 3x + 2z = 11 6x - 2z = 34 9x = 45 x = 5

1

2

3

4

5

4 5

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1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

Step 3: Sub x in to solve for z.

3(5) + 2z = 11

2z = -4

z = -2

1

2

3

4Sub x = 5 into 4

4

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1) Solving Three-Variable Systems by Elimination

Example 1:Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 154x – 3y – z = 19

Step 4: Sub x and z into one of the original equations. Solve for y.

x – 3y + 3z = -4 5 – 3y + 3(-2) = -4

5 – 3y – 6 = -4 -3y = -4 + 6 – 5 -3y = -3 y = 1

1

2

3

1Sub x = 5, z = -2 in 1

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1) Solving Three-Variable Systems by Elimination

Example 1:

Solve by elimination. x – 3y + 3z = -4

2x + 3y – z = 15

4x – 3y – z = 19

Therefore, the solution is (5, 1, -2).

1

2

3{

1) Solving Three-Variable Systems by Elimination

Example 2:

Solve by elimination. 2x – y + z = 4

x + 3y – z = 11

4x + y – z = 14

1

2

3{

1) Solving Three-Variable Systems by Elimination

Example 2:

Solve by elimination. 2x – y + z = 4

x + 3y – z = 11

4x + y – z = 14

Step 1: Add and to cancel z.

2x – y + z = 4

x + 3y – z = 11

3x + 2y = 15

1

2

3

1 2

1

2

4

{

1) Solving Three-Variable Systems by Elimination

Example 2:

Solve by elimination. 2x – y + z = 4

x + 3y – z = 11

4x + y – z = 14

Step 1: Subtract and to cancel z.

x + 3y – z = 11

4x + y – z = 14

-3x + 2y = -3

1

2

3

2 3

2

3

5

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1) Solving Three-Variable Systems by Elimination

Example 2:

Solve by elimination. 2x – y + z = 4

x + 3y – z = 11

4x + y – z = 14

Step 2: Use equations and to find x and y.

3x + 2y = 15

-3x + 2y = -3

4y = 12

y = 3

1

2

3

54

4

5

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1) Solving Three-Variable Systems by Elimination

Example 2:

Solve by elimination. 2x – y + z = 4

x + 3y – z = 11

4x + y – z = 14

Step 2: Use equations and to find x and y.

3x + 2y = 15

3x + 2(3) = 15

3x + 6 = 15

3x = 9

x = 3

1

2

3

54

Sub y = 3 in 4

4

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1) Solving Three-Variable Systems by Elimination

Example 2:

Solve by elimination. 2x – y + z = 4

x + 3y – z = 11

4x + y – z = 14

Step 3: Solve for the remaining unknown.

2x – y + z = 4

2(3) – 3 + z = 4

6 – 3 + z = 4

z = 4 + 3 – 6

z = 1

1

2

3

Sub x = 3 and y = 3 in 1

1

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1) Solving Three-Variable Systems by Elimination

Example 2:

Solve by elimination. 2x – y + z = 4

x + 3y – z = 11

4x + y – z = 14

Therefore, the solution is (3, 3, 1).

1

2

3{

Example 3:

Solve by elimination. -x + 2z = -9

-x – 3y – 4z = 2

-3x – 2y + 2z = 17

1) Solving Three-Variable Systems by Elimination

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Example 3:

Solve by elimination. -x + 2z = -9

-x – 3y – 4z = 2

-3x – 2y + 2z = 17

Therefore, no unique solution.

1) Solving Three-Variable Systems by Elimination

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Summary of Steps

1) Elimination twice, create equations (4) and (5)

2) Solve for unknown a

3) Substitute a into equation (4) or (5) to find b

4) Substitute a and b into (1), (2) or (3) to find c

Homework

p.157 #1-3, 25, 26