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Chapter 1 Systems of Equations. 1.6. Solving Systems of Linear Equations in Three Variables. 1.6. 1. MATHPOWER TM 11, WESTERN EDITION. Solving Systems of Three Equations in Three Variables- Principles. To solve a linear system of three equations - PowerPoint PPT Presentation
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MATHPOWERTM 11, WESTERN EDITION 1.6.1
1.6
Chapter 1 Systems of Equations
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: Multiply any equation by a constant
without changing the equation. Add or subtract any two equations
without changing the solution.
1.6.2
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
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
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 bemore convenient than others.
1.6.5
Solving a System of Three Equations [cont’d]
Group Equations 1 and 2 and use elimination:x + 4y + 3z = 5x + 3y + 2z = 4 y + z = 1
Group Equations 2 and 3 and use elimination:x + 3y + 2z = 4x + 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
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 = 22y + 3z = 5
-z = -3 z = 3Solve for y:
y + z = 1y + 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
1.6.8
Note: It doesn’t matter how the equations are paired as long as the same variable iseliminated from each pair. You could combine
– Equations 1 and 3– Then 1 and 2
Or, you could combine– Equations 1 and 2– Then 2 and 3
Or, you could combine– Equations 1 and 3– Then 2 and 3
1.6.9
Solving Systems of Three Equations-Principles
Note: It may be easier to solve a systemby 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
x + y + z = 25 1y = x + 2 2z = y - 3 3
Rewrite Equation 2 as x = y – 2.Now both y and z can bereplaced in Equation 1 leaving y as the only variable.
x + y + z = 25y - 2 + y + y - 3 = 25 3 y = 30 y = 10
Solve for z:z = y - 3z = 10 - 3z = 7Solve for x:x = y - 2x = 10 - 2x = 8
The solution is (8, 10, 7).
VerifyYourSolution!
Solving Another Systems of Three Equations [cont’d]
1.6.11
x + y + z = 25 (1)8 + 10 + 7 = 25 25 = 25 y = x + 2 (2)10 = 8 + 210 = 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]
Solve: x + 5y + 3z = 4 (1) 2x + y + 4z = 1 (2) 2x - y + 2z = 1 (3)
2x - 1y + 2z = 1 (3)
2y + 2z = 02x - 1y + 2z = 1 (3)2x + 10y + 6z = 8 (1)
11y + 4z = 7
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)
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
-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
The operator of a ski resort sells threetypes 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
x + y + z = 517 1 x + y = 425 240x + 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 50540x + 10 625 - 25x + 1380 = 16 505 15x +12 005 = 16 505 15x = 4 500 x = 300y = 125
A Problem Involving a System [cont’d]
1.6.16
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
Homework Questions:Pages 44 #1(odds), 3, 4(odds), 6, 9, 12
1.6.18