3-5 Systems with Three VariablesAlgebra II Unit3 Linear Systems© Tentinger

Essential Understanding and Objectives

• Essential Understanding: To solve systems of three equations in three variables, you can use some of the same algebraic methods you used to solve systems of two equations in two variable.

• Objectives:• Students will be able to:• Solve systems of three variables using elimination• Solve systems of three variable using substitution

Iowa Core Curriculum• Algebra• Extends A.REI.6 Solve systems of linear equations exactly and

approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.

Three Variable Equations• Two variable equations represent lines• Three variable equations represent planes• Like two variable equations, you can have no solution, one

solution, or infinitely many solutions

• Graphs of solutions• http://www.mathwarehouse.com/algebra/planes/systems/three-

variable-equations.php

• No solution: no point lies in all three planes• One Solution: the planes intersect at one common point• Infinitely Many Solutions: The planes intersect at a line

Solving a system using Elimination

• Step 1: Pair the equations to eliminate one variable, z. Then you will have two equations with two unknowns.

• Add Subtract

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2x −y + z = 4

x + 3y −z =11

4x + y −z =14

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2x −y + z = 4

x + 3y −z =11

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3x + 2y + 0z =15

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x + 3y − z =11

4x + y − z =14

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−3x + 2y + 0z = −3

Solving a system using Elimination

• 2: Write the new equations as a system. Solve for x and y• Add and solve for y.• Substitute your answer and solve for x

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2x −y + z = 4

x + 3y −z =11

4x + y −z =14

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3x + 2y =15

−3x + 2y = −3

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Solving a system using Elimination

• Step 3: Solve for remaining variable, z. Substitute in answers for x and y into the original equations

• Step 4: Write the solution as an ordered triple: (3, 3, 1)€

2x −y + z = 4

x + 3y −z =11

4x + y −z =14

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Solve using Elimination

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x −y + z = −1

x + y + 3z = −3

2x −y + 2z = 0

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3x + y − z =1

x + 2y + z = 4

3x − y − z = 3

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x + y + 2z = −7

3x + y − 2z = 7

−x − 3y + z = −9

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Solving Equivalent Systems

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x + y + 2z = 3

2x + y + 3z = 7

−x − 2y + z =10

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x −2y + 3z =12

2x + 2y −z = 5

2x + 2y −z = 4

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Solving a System using Substitution:

• Step 1: chose the equation whose variable is easy to isolate.• X+5y=9 x = -5y+9• Step 2: Substitute the expression into the other two remaining

equations and simplify• 2(-5y+9) + 3y – 2z = -1 4z – 5(-5y+9) = 4• -7y -2z = -19 25y +4z = 49

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2x + 3y − 2z = −1

x + 5y = 9

4z − 5x = 4

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Solving a System using Substitution:

• Step 3: Write the two new equations as a system and solve for the remaining variables

• use elimination to solve for y then substitute to solve for z

• y = 1, z = 6

• Step 4: Use the original equation to solve for x• Solution (4, 1, 6)

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2x + 3y − 2z = −1

x + 5y = 9

4z − 5x = 4

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−7y −2z = −19

25y + 4z = 49

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Solve by substitution

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x −2y + z =1

2x + z = 9

−3x + y = −3

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x −2y + z = −4

−4x + y −2z =1

2x + 2y −z =10

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Application• You manage a clothing store and budget $5400 to restock 200

shirts. You can buy T-shirts for $12 each, polo shirts for $24 each, and rugby shirts for #36 dollars each. If you want to have the same number of T-shirts as polo shirts, how many of each shirt should you buy?

• Relate:• T-shirts + polo shirts + rugby shirts = 200• T-shirts = polo shirts• 12 * Tshirts + 24*polo shirts + 36*rugby shirts = 5400

• Define:• X = tshirts• Y = polo• Z = rugby

Application• You manage a clothing store and budget $5400 to restock 200

shirts. You can buy T-shirts for $12 each, polo shirts for $24 each, and rugby shirts for #36 dollars each. If you want to have the same number of T-shirts as polo shirts, how many of each shirt should you buy?

• Write:

• Solve:• Substitute x in for equations 1 and 3 then simplify• Write the new equations as a system then solve for y and z• Substitute y and z back into one of the original equations to get x

• Solution: (50, 50, 100)

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x + y + z = 200

x = y

12x + 24y + 36z = 5400

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Homework• Pg. 171-172• #14-16, 24-26, 32, 34-37