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