1. MATHPOWER TM 11, WESTERN EDITION 1.6.1 1.6 Chapter 1 Systems of Equations

2. Solving Systems of Three Equations in Three Variables- Principles To solve a linear system of three equations in three variables, use the same principles as for two equations in two variables: l Multiply any equation by a constant without changing the equation. l Add or subtract any two equations without changing the solution. 1.6.2

3. Solving Systems of Three Equations in Three Variables- Principles [cont’d] Although a three-variable linear system has three equations, only two equations can be combined at a time. Therefore, to solve a system of three equations, use elimination to reduce the system to a system of two equations in two variables. Then, solve the new system, and find the third variable by substitution. 1.6.3

4. Solve: x + 4y + 3z = 5 (1) x + 3y + 2z = 4 (2) x + y - z = -1 (3) 1.6.4 Solving a System of Three Equations

5. Choose two pairs of equations, and eliminate the same variable from each pair. NOTE: It does not matter which variable is eliminated first, but one choice may be more convenient than others. 1.6.5 Solving a System of Three Equations [cont’d]

6. Group Equations 1 and 2 and use elimination: x + 4y + 3z = 5 x + 3y + 2z = 4 y + z = 1 Group Equations 2 and 3 and use elimination: x + 3y + 2z = 4 x + y - z = -1 2y + 3z = 5 This results in a system of two equations in two variables that can be solved by elimination. Solving a System of Three Equations [cont’d] 1.6.6

7. y + z = 1 2y + 3z = 5 Solving a System of Three Equations [cont’d] Solve for x : x + y - z = -1 x + (-2) - 3 = -1 x = 4 Therefore, the solution is (4, -2, 3). 2y + 2z = 2 2y + 3z = 5 -z = -3 z = 3 Solve for y: y + z = 1 y + 3 = 1 y = -2 Multiply by 2. Verify: x + 4y + 3z = 5 4 - 8 + 9 = 5 x + 3 y + 2z = 4 4 - 6 + 6 = 4 1.6.7

8. 1.6.8

9. Note: It doesn’t matter how the equations are paired as long as the same variable is eliminated from each pair. l You could combine –Equations 1 and 3 –Then 1 and 2 l Or, you could combine –Equations 1 and 2 –Then 2 and 3 l Or, you could combine –Equations 1 and 3 –Then 2 and 3 1.6.9 Solving Systems of Three Equations-Principles

10. Note: It may be easier to solve a system by substitution. Solve: x + y + z = 25 (1) y = x + 2 (2) z = y - 3 (3) 1. Solve Equation 2 for x. 2. Substitute for x (Equation 2) and z (Equation 3) in Equation 1. 3. Simplify Equation 1 to determine y. 1.6.10 Solving Another System of Three Equations

11. x + y + z = 25 1 y = x + 2 2 z = y - 3 3 Rewrite Equation 2 as x = y – 2. Now both y and z can be replaced in Equation 1 leaving y as the only variable. x + y + z = 25 y - 2 + y + y - 3 = 25 3 y = 30 y = 10 Solve for z: z = y - 3 z = 10 - 3 z = 7 Solve for x: x = y - 2 x = 10 - 2 x = 8 The solution is (8, 10, 7). Verify Your Solution! Solving Another Systems of Three Equations [cont’d] 1.6.11

12. x + y + z = 25 (1) 8 + 10 + 7 = 25 25 = 25 y = x + 2 (2) 10 = 8 + 2 10 = 10 z = y - 3 (3) 7 = 10 - 3 7 = 7 Verify: Therefore, the solution is (8, 10, 7). 1.6.12 Solving Another System of Three Equations [cont’d]

13. Solve: x + 5y + 3z = 4 (1) 2x + y + 4z = 1 (2) 2x - y + 2z = 1 (3) 2x - 1y + 2z = 1 (3) 2y + 2z = 0 2x - 1y + 2z = 1 (3) 2x + 10y + 6z = 8 (1) 11y + 4z = 7 4y + 4z = 0 7y = 7 y = 1 11(1) + 4z = 7 4z = -4 z = -1 x + 5y + 3z = 4 (1) x + 5(1) + 3(-1) = 4 x = 2 The solution is (2, 1, -1). Solving Systems of Three Equations- Practice 1.6.13 2x + 1y + 4z = 1 (2)

14. Solve: x + 3y + 4z = 19 (1) x + 2y + z = 12 (2) x + y + z = 8 (3) x + 2y + z = 12 (2) x + y + z = 8 (3) y = 4 x + y + z = 8 (3) x + 3y + 4z = 19 (1) -2y - 3z = -11 -2(4) - 3z = -11 -3z = -3 z = 1 x + y + z = 8 (3) x + 4 + 1 = 8 x + 5 = 8 x = 3 The solution is (3, 4, 1). 1.6.14 Solving Systems of Three Equations- More Practice

15. The operator of a ski resort sells three types of tickets: full-day skiing at $40, half-day skiing at $25, and rental of ski equipment at $15. At the end of the day, he finds that he has sold a total of 517 tickets. The total number of people skiing during the day was counted at 425. He also has a total cash intake of $16 505. Determine the sales of each type of ticket. Let x = full-day ticket sales. Let y = half-day ticket sales. Let z = rental of equipment. A Problem Involving a System 1.6.15

16. x + y + z = 517 1 x + y = 425 2 40x + 25y + 15z = 16 505 3 Since x + y = 425 425 + z = 517 z = 92 From Equation 2: y = 425 - x Substitute into Equation 3: 40x + 25(425 - x) + 15(92) = 16 505 40x + 10 625 - 25x + 1380 = 16 505 15x +12 005 = 16 505 15x = 4 500 x = 300 y = 125 A Problem Involving a System [cont’d] 1.6.16

17. A Problem Involving a System [cont’d] Verify the Solution: x + y + z = 517 (1) 300 + 125 + 92 = 517 517 = 517 x + y = 425 (2) 300 + 125 = 425 425 = 425 40x + 25y + 15z = 16 505 (3) 40(300) + 25(125) + 15(92) = 16 505 12000 + 3125 + 1380 = 16 505 16 505 = 16 505 Therefore, the operator sold 300 full-day, 125 half-day, and 92 ski rental tickets. 1.6.17

18. Homework Questions: Pages 44 #1(odds), 3, 4(odds), 6, 9, 12 1.6.18