Then/NowNew VocabularyExample 1: A System with One SolutionExample 2:No Solution and Infinite SolutionsExample 3: Write and Solve a System of Equations

You solved linear equations with two variables. (Lesson 3–2)

• Solve systems of linear equations in three variables.

• Solve real-world problems using systems of linear equations in three variables.

• ordered triple

A System with One Solution

Step 1 Use elimination to make a system of two equations in two variables.

Solve the system of equations.5x + 3y + 2z = 22x + y – z = 5x + 4y + 2z = 16

5x + 3y + 2z = 2 5x + 3y + 2z = 2 First equation

9x + 5y = 12 Add to eliminate z.

Multiply by 2.

2x + y – z = 5 (+)4x + 2y – 2z = 10 Second equation

A System with One Solution

5x + 3y + 2z = 2 First equation

4x – y = –14 Subtract to eliminate z.

(–) x + 4y + 2z = 16 Third equation

Notice that the z terms in each equation have been eliminated. The result is two equations with the same two variables x and y.

A System with One Solution

Step 2 Solve the system of two equations.

9x + 5y = 12 9x + 5y = 12

29x = –58 Add to eliminate y.

Multiply by 5

4x – y = –14 (+) 20x – 5y = –70

x = –2 Divide by 29.

A System with One Solution

Substitute –2 for x in one of the two equations with two variables and solve for y.

4x – y = –14 Equation with two variables

4(–2) – y = –14 Replace x with –2.

–8 – y = –14 Multiply.

y = 6 Simplify.

The result is x = –2 and y = 6.

A System with One Solution

Step 3 Solve for z using one of the original equations with three variables.

2x + y – z = 5 Original equation with three variables

2(–2) + 6 – z = 5 Replace x with –2 and y with 6.

–4 + 6 – z = 5 Multiply.

z = –3 Simplify.

Answer: The solution is (–2, 6, –3). You can check this solution in the other two original equations.

A. A

B. B

C. C

D. D

What is the solution to the system of equations shown below? 2x + 3y – 3z = 16x + y + z = –3x – 2y – z = –1

A.

B. (–3, –2, 2)

C. (1, 2, –6)

D. (–1, 2, –4)

No Solution and Infinite Solutions

6x + 3y – 9z = 15 (–)6x + 3y – 9z = 15

0 = 0

Eliminate y in the first and third equations.

A. Solve the system of equations.2x + y – 3z = 5x + 2y – 4z = 76x + 3y – 9z = 15

2x + y – 3z = 5 6x + 3y – 9z = 15

Multiply by 3.

No Solution and Infinite Solutions

6x + 3y – 9z = 15 (–)6x + 3y – 9z = 15

9y – 15z = 27

The equation 0 = 0 is always true. This indicates that the first and third equations represent the same plane. Check to see if this plane intersects the second plane.

x + 2y – 4z = 7 6x + 12y – 24z = 42

Multiply by 6.

Divide by the GCF, 3. 3y – 5z = 9

Answer: The planes intersect in a line. So, there is an infinite number of solutions.

Write and Solve a System of Equations

0 = 39

Eliminate x in the last two equations.

B. Solve the system of equations.3x – y – 2z = 46x + 4y + 8z = 119x + 6y + 12z = –3

6x + 4y + 8z = 11 18x + 12y + 24z = 33

Multiply by 3.

9x + 6y + 12z = –3(–) 18x + 12y + 24z = –6

Multiply by 2.

Answer: The equation 0 = 39is never true. So, there is no solution of this system.

A. A

B. B

C. C

D. D

A. (1, 2, 0)

B. (2, 2, 0)

C. infinite number of solutions

D. no solution

A. What is the solution to the system of equations shown below? x + y – 2z = 3–3x – 3y + 6z = –92x + y – z = 6

A. A

B. B

C. C

D. D

A. (0, 6, 1)

B. (1, 0, –2)

C. infinite number of solutions

D. no solution

B. What is the solution to the system of equations shown below? 3x + y – z = 5–15x – 5y + 5z = 11x + y + z = 2

Write and Solve a System of Equations

SPORTS There are 49,000 seats in a sports stadium. Tickets for the seats in the upper level sell for $25, the ones in the middle level cost $30, and the ones in the bottom level are $35 each. The number of seats in the middle and bottom levels together equals the number of seats in the upper level. When all of the seats are sold for an event, the total revenue is $1,419,500. How many seats are there in each level?Explore Read the problem and define the variables.

u = number of seats in the upper levelm = number of seats in the middle levelb = number of seats in the bottom level

Write and Solve a System of Equations

Plan There are 49,000 seats.

u + m + b = 49,000

When all the seats are sold, the revenue is 1,419,500. Seats cost $25, $30, and $35.

25u + 30m + 35b = 1,419,500

The number of seats in the middle and bottom levels together equal the number of seats in the upper level.

m + b = u

Write and Solve a System of Equations

Solve Substitute u = m + b in each of the first two equations.

(m + b) + m + b = 49,000 Replace u with m + b.2m + 2b = 49,000 Simplify.

m + b = 24,500 Divide by 2.

25(m + b) + 30m + 35b = 1,419,500 Replace u with m + b.

25m + 25b + 30m + 35b = 1,419,500Distributive

Property55m + 60b = 1,419,500 Simplify.

Write and Solve a System of Equations

Now, solve the system of two equations in two variables.

–5b = –72,000

m + b = 24,500 55m + 55b = 1,347,500

Multiply by 55.

55m + 60b = 1,419,500 (–) 55m + 60b = 1,419,500

b = 14,400

Write and Solve a System of Equations

Substitute 14,400 for b in one of the equations with two variables and solve for m.

m + b = 24,500 Equation with two variablesm + 14,400 = 24,500 b = 14,400

m = 10,100 Subtract 14,400 from each side.

Write and Solve a System of Equations

Substitute 14,400 for b and 10,100 for m in one of the original equations with three variables.

m + b = u Equation with three variables

10,100 + 14,400 = u m = 10,100, b = 14,400

24,500 = u Add.

Answer: There are 24,500 upper level, 10,100 middle level, and 14,400 bottom level seats.

Write and Solve a System of Equations

Check Check to see if all the criteria are met.

24,500 + 10,100 + 14,400 = 49,000

The number of seats in the middle and bottom levels equals the number of seats in the upper level.

10,100 + 14,400 = 24,500

When all of the seats are sold, the revenue is $1,419,500.

24,500($25) + 10,100($30) + 14,400($35) = $1,419,500

1. A2. B3. C4. D

BUSINESS The school store sells pens, pencils, and paper. The pens are $1.25 each, the pencils are $0.50 each, and the paper is $2 per pack. Yesterday the store sold 25 items and earned $32. The number of pens sold equaled the number of pencils sold plus the number of packs of paper sold minus 5. How many of each item did the store sell?A. pens: 5; pencils: 10; paper: 10B. pens: 8; pencils: 7; paper: 10C. pens: 10; pencils: 7; paper: 8D. pens: 11; pencils: 2; paper: 12