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### Alg II 3-5 Sytems Three Variables

• 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)
• 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)
• 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

### Editor's Notes

1. Solution: (4, 2, -3)Solution: (2, -1, 4)Solution: (-1, 2, -4)
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