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Solving Systems of Equations in 3 Variables

Learn how to solve systems of equations in 3 variables using elimination and substitution methods. Understand the relationship between the solution and the graph of each system.

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Solving Systems of Equations in 3 Variables

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  1. pencil, red pen, highlighter, notebook, calculator total: 4/15/13 Have out: Bellwork: 1) Find the solution to each system of equations. a) 2x + 3y = –17 b) 3x + 2y = 11 y = x – 4 4x + 3y = 14 2) Explain how your solution relates to the graph of each system.

  2. + –15 –15 total: 1. Find the solution to each system of equations. 3x + 2y = 11 3(3x + 2y = 11) a) 2x + 3y = –17 b) 4x + 3y = 14 –2(4x + 3y = 14) y = x – 4 9x + 6y = 33 +1 +1 2x + 3(x – 4) = –17 –8x – 6y = –28 2x + 3x – 12 = –17 x = 5 +1 5x – 12 = –17 y = x – 4 3(5) + 2y = 11 +12 +12 +1 +1 y = (–1) – 4 15 + 2y = 11 5x = –5 y = –5 +1 5 5 +1 2y = –4 x = –1 (–1, –5) 2 2 (5, –2) +1 +1 +1 y = –2 2) Explain how your solution relates to the graph of each system. The solutions represent the point of intersection for each system. +2

  3. z y x Today we are going to solve systems of equations in 3 variables. ax + by + cz = d Any equation that can be written in the form _______________ is a linear equation in 3 variables (where a, b, c, and d are constants and a, b, c are not all zero). ordered triple (x, y, z) The solution to such an equation is an __________________. plane The graph of a linear equation in 3 variables is a _____. plane

  4. Given a systems of 3 linear equations in 3 variables, there can be one solution, infinitely many solutions, or no solution. One Solution Infinitely Many Solutions  planes intersect in exactly one point  planes intersect in a line  planes intersect in the same plane No Solution  planes have no point in common

  5. Solving a system of equations in 3 variables is similar to solving a system of equations in 2 variables using substitution and/or elimination methods. Example #1: Solve the system of equations. A x + y + z = 6 Sometimes it helps to number or letter each equation. B 2x – y + 3z = 9 –x + 2y + 2z = 9 C Dealing with 3 variables can be complicated, so let’s try to eliminate one of the variables right away. Which two equations could we pick to eliminate a variable? A and C

  6. Example #1: Solve the system of equations. A x + y + z = 6 B 2x – y + 3z = 9 –x + 2y + 2z = 9 C A x + y + z = 6 Begin by adding A and C together.  This will eliminate an x–term. –x + 2y + 2z = 9 C Label the new equation D. D 3y + 3z = 15 Let’s pair up two more equations to eliminate x. Which ones could we pick? A and B Multiply A by –2 to help eliminate x. –2x – 2y – 2z = –12 A –2(x + y + z) = –2(6) x + y + z = 6 2x – y + 3z = 9 2x – y + 3z = 9 B – 3y + z = –3 E Label the new equation E.

  7. Example #1: Solve the system of equations. A x + y + z = 6 B 2x – y + 3z = 9 –x + 2y + 2z = 9 C Now we have a couple 2-variable linear equations, D and E. Solve for y and z. D 3y + 3z = 15 – 3y + (3) = –3 – 3y + z = –3 E – 3y = –6 4z = 12 y = 2 z = 3 Pick one of the original 3–variable equations to solve for x. x + y + z = 6 A The solution is the ordered triple x + (2) + (3) = 6 (1, 2, 3) x + 5 = 6 x = 1

  8. Example #2: Solve the system of equations. A x + 2y + z = 10 It helps to number or letter each equation. B 2x – y + 3z = –5 2x – 3y – 5z = 27 C Pick any two equations to eliminate a variable. Let’s eliminate y for the fun of it.  Choose B and C B –3(2x – y + 3z) = –3(–5) 2x – y + 3z = –5 –6x + 3y – 9z = 15 2x – 3y – 5z = 27 C 2x – 3y – 5z = 27 – 4x – 14z = 42 D Label the new equation D. Pick another pair of equations to eliminate the y–term. A and B x + 2y + z = 10 A x + 2y + z = 10 4x – 2y + 6z = –10 2(2x – y + 3z) = 2(–5) 2x – y + 3z = –5 B E 5x + 7z = 0 Label the new equation E.

  9. Now we have a couple 2-variable linear equations. Solve for x and z. – 4x – 14z = 42 D – 4x – 14z = 42 5x + 7z = 0 2(5x + 7z) = 2(0) 5x + 7z = 0 E 10x + 14z = 0 5(7) + 7z = 0 35 + 7z = 0 6x = 42 7z = –35 x = 7 z = –5 Pick one of the original 3–variable equations to solve for y. x + 2y + z = 10 (7) + 2y + (–5) = 10 The solution is the ordered triple 2 + 2y = 10 (7, 4, –5) 2y = 8 y = 4

  10. Summary: Ax+By+Cz = D 1. Simplify and put all three equations in the form ___________ if needed. variables eliminate 2. Choose to _________ any one of the _________ from any ______________. pair of equations 3. Eliminate the ______ variable chosen in step 2 from any ___________________, creating a system of two equations and 2 unknowns. SAME other pair of equations system of 2 equations 2 4. Solve the remaining __________________ (with _ unknowns) found in step 2 and 3. third variable 5. Solve for the ___________. Write the answer as an __________________. ordered triple (x, y, z). 6. Check.

  11. Complete the worksheets and the CST practice.

  12. References & Resources http://www.jcoffman.com/Algebra2/ch3_6.htm http://www.analyzemath.com/Calculators/Calculator_syst_eq.html http://fp.academic.venturacollege.edu/rbrunner/3e_3s.htm http://www.mathwarehouse.com/algebra/planes/systems/three-variable-equations.php

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