Solve a Linear System in Three Variables Objectives: 1.To geometrically interpret the solution to a...

Solve a Linear System in Three VariablesSolve a Linear System in Three Variables

Objectives:

1.To geometrically interpret the solution to a linear system in three variables

2.To solve a linear system in three variables using substitution and elimination

Activity 1: Graphing in 3DActivity 1: Graphing in 3D

A linear equation in three variableslinear equation in three variables x, y, and z can be written

ax + by + cz = d,

where a, b, c, and d are real numbers, all of which are not zero.

Activity 1: Graphing in 3DActivity 1: Graphing in 3D

We graph this equation in 3-D, on a coordinate system with an x-, y-, and a z-axis, dividing space into eight octants.

Points in space are located with an ordered tripleordered triple (x, y, z).

Activity 1: Graphing in 3DActivity 1: Graphing in 3D

The solution to a linear equation in three variables is the set of all points (x, y, z) that satisfy the equation.

In this activity, we will discover the shape of the graph of a linear equation in 3 variables.

Activity 1: Graphing in 3DActivity 1: Graphing in 3D

We are going to use a three-dimensional coordinate system to graph the equation 3x + 4y + 6z = 12.

Step 1: Start by finding the x-intercept. Substitute 0 in for y and z and solve for x. Plot this point.

Activity 1: Graphing in 3DActivity 1: Graphing in 3D

Step 2: Next find the y-intercept by substituting 0 in for x and z and solving for y. Plot this point.

Step 3: Finally find the z-intercept by substituting 0 in for x and y and solving for z. Plot this point.

Activity 1: Graphing in 3DActivity 1: Graphing in 3D

Step 4: Connect your three points: x-intercept to y-intercept, y-intercept to z-intercept, and z-intercept to x-intercept.

What shape is the graph of a linear equation in 3 variables?

Activity 1: Graphing in 3DActivity 1: Graphing in 3D

Recall a postulate from geometry which states: Through any 3 noncollinear points, there exists

exactly one plane.

Thus, we can conclude that the graph of a linear equation in 3 variables is a plane.

Activity I: Graphing in 3DActivity I: Graphing in 3D

Microsoft Mathematics 4.0:

• Macs

• Click the Graphing tab

• Choose 3D from the drop down menu

• Type in the equation and click Graph

Free Graph PaperFree Graph Paper

http://www1.itlnet.net/users/smatheson/graphpaper/

Exercise 1Exercise 1

Sketch the graph of the equation.

3x + 9y – 3z = -18

Linear System in 3 VariablesLinear System in 3 Variables

A linear equation in three variableslinear equation in three variables x, y, and z can be written

ax + by + cz = d,

where a, b, c, and d are real numbers, all of which are not zero.

A linear system of equations in three variables has 3 such equations.

Linear System in 3 VariablesLinear System in 3 Variables

A linear equation in three variableslinear equation in three variables x, y, and z can be written

ax + by + cz = d,

where a, b, c, and d are real numbers, all of which are not zero.

Linear System in 3 VariablesLinear System in 3 Variables

A linear equation in three variableslinear equation in three variables x, y, and z can be written

ax + by + cz = d,

where a, b, c, and d are real numbers, all of which are not zero.

The solutionsolution to such a system is the ordered tripleordered triple (x, y, z) that satisfies all the equations.

Graphs of 3D SystemsGraphs of 3D Systems

Recall that a system of linear equations in two variables can be either consistentconsistent or inconsistentinconsistent, and that consistent systems can be either independentindependent or dependentdependent.

Possible SolutionsPossible Solutions

Geometrically, the solution to any system of equations is the point or points of intersection.

Possible SolutionsPossible Solutions

Geometrically, the solution to any system of equations is the point or points of intersection.

Solving AlgebraicallySolving Algebraically

We’d probably not want to solve a linear system in 3 variables by graphing. Instead, there would probably be far less bloodshed if we solved such a system algebraically, using either elimination or substitution.

For the elimination method, you first eliminate one of your variables so that you have 2 equations with 2 variables. Easy.

Exercise 2Exercise 2

Solve the system.

2x – y + 6z = -4

6x + 4y – 5z = -7

-4x – 2y + 5z = 9

Elimination MethodElimination Method

In Step 1, you’ll have to eliminate the same variable from 2 different sets of the equations.

Elimination MethodElimination Method

The system has no solution no solution if you obtain a contradiction (ex. 0 = 1) while solving the system.

Elimination MethodElimination Method

The system has infinitely many solutions infinitely many solutions if you obtain an identity (ex. 0 = 0) while solving the system.

Protip #1: Letter EquationsProtip #1: Letter Equations

To help you through the often labyrinthine process of solving a 3-variable system, letter each of your equations.

2x – y + 6z = -4 6x + 4y – 5z = -7-4x – 2y + 5z = 9

A

B

C

Protip #1: Letter EquationsProtip #1: Letter Equations

In terms of these letters, write a simple expression that tells you how to add/subtract multiples of each equation.

Label the new equation with a new letter.

2x – y + 6z = -4 6x + 4y – 5z = -7-4x – 2y + 5z = 9

A

B

C

B C+ 6x + 4y – 5z = -7-4x – 2y + 5z = 9+

2x + 2y = 2 D

D1

2 x + y = 1 E

Protip #1: Letter EquationsProtip #1: Letter Equations

Continue this process until the system is solved.

2x – y + 6z = -4 6x + 4y – 5z = -7-4x – 2y + 5z = 9

A

B

C

+

46x + 19y = -62 F

+A B5 6 10x – 5y + 30z = -20 36x + 24y – 30z = -42

Protip #1: Letter EquationsProtip #1: Letter Equations

Continue this process until the system is solved.

2x – y + 6z = -4 6x + 4y – 5z = -7-4x – 2y + 5z = 9

A

B

C

+

27x = -81

-19x – 19y = -19 46x + 19y = -42

+-19 E F

x = -3

Exercise 3Exercise 3

Solve the system.

x + y – z = 2

3x + 3y – 3z = 8

2x – y + 4z = 7

Exercise 4Exercise 4

Solve the system.

x + y + z = 6

x – y + z = 6

4x + y + 4z = 24

Protip #2: Multiple SolutionsProtip #2: Multiple Solutions

When you discover that you have a consistent, dependent system of equations, how do you write your answer?

Graphically, the equations in this system intersect in a line, so you could just write the equation of that line.

But what if you want specific solutions, in the form of ordered triples?

Protip #2: Multiple SolutionsProtip #2: Multiple Solutions

To write your answers as a set of ordered pairs, set one of the variables in your equation equal to a. Now re-write the other variable in terms of a.

x + y + z = 6x + z = 6

Let x = a

Then by substitution in the 2nd equation:

a + z = 6 z = 6 – a

Then by substitution in the 1st equation: a + y + (6 – a) = 6

y = 0

Protip #2: Multiple SolutionsProtip #2: Multiple Solutions

Finally, use your new expressions to write an ordered triple. Substitute values in for a to get a specific solution points.

x + y + z = 6x + z = 6

x = a z = 6 – a y = 0

(a, 0, 6 – a)

Let a = 0: (0, 0, 6)Let a = 1: (1, 0, 5)

Let a = -1: (-1, 0, 7)

Exercise 5Exercise 5

Solve each system.

1. 3x + y – 2z = 10 6x – 2y + z = -2 x + 4y + 3z = 7

2. x + y – z = 2 2x + 2y – 2z = 6 5x + y – 3z = 8

3. x + y + z = 3 x + y – z = 3 2x + 2y + z = 6

Exercise 6Exercise 6

At a carry-out pizza restaurant, an order of 3 slices of pizza, 4 breadsticks, and 2 soft drinks cost $13.35. A second order of 5 slices of pizza, 2 breadsticks, and 3 soft drinks cost $19.50. If four bread sticks and a can of soda cost $.30 more than a slice of pizza, what is the cost of each item?

Substitution MethodSubstitution Method

If it is convenient, you could use substitutionsubstitution to help solve a linear system in three variables.

1.Solve one of the equations for one of the variables.

2.Substitute the expression from Step 1 into both of the other equations.

3.Solve the remaining 2 variable system.

Exercise 7: SATExercise 7: SAT

If 5 sips + 4 gulps = 1 glass and 13 sips + 7 gulps = 2 glasses, how many sips equal a gulp?