1. 3-5 Systems with
Three Variables
Algebra II Unit3 Linear Systems
© Tentinger

2. 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

3. 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.

4. 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

5. Solving a system using
Elimination
• Step 1: Pair the equations to eliminate one variable, z. Then
you will have two equations with two unknowns.
• AddSubtract

6. 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

7. 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)

8. Solve using Elimination

9. Solving Equivalent Systems

10. Solving a System using
Substitution:
• Step 1: choose 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

11. 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)

12. Solve by substitution

13. 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

14. 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)

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