EXAMPLE 1 Use the elimination method Solve the system. 4x + 2y + 3z = 1 Equation 1 2x – 3y + 5z =...

EXAMPLE 1 Use the elimination method

Solve the system. 4x + 2y + 3z = 1 Equation 1

2x – 3y + 5z = –14 Equation 2

6x – y + 4z = –1 Equation 3

SOLUTION

STEP 1 Rewrite the system as a linear system in two variables.

4x + 2y + 3z = 1

12x – 2y + 8z = –2

Add 2 times Equation 3

to Equation 1.

16x + 11z = –1 New Equation 1

EXAMPLE 1

2x – 3y + 5z = –14

–18x + 3y –12z = 3

Add – 3 times Equation 3to Equation 2.

–16x – 7z = –11 New Equation 2

STEP 2 Solve the new linear system for both of its variables.

16x + 11z = –1 Add new Equation 1

and new Equation 2. –16x – 7z = –11

4z = –12

z = –3 Solve for z.

x = 2 Substitute into new Equation 1 or 2 to find x.

Use the elimination method

6x – y + 4z = –1

EXAMPLE 1 Use the elimination method

STEP 3 Substitute x = 2 and z = – 3 into an original equation and solve for y.

Write original Equation 3.

6(2) – y + 4(–3) = –1 Substitute 2 for x and –3 for z.

y = 1 Solve for y.

EXAMPLE 2 Solve a three-variable system with no solution

Solve the system. x + y + z = 3 Equation 1

4x + 4y + 4z = 7 Equation 2

3x – y + 2z = 5 Equation 3

SOLUTION

When you multiply Equation 1 by – 4 and add the result to Equation 2, you obtain a false equation.

Add – 4 times Equation 1

to Equation 2.

–4x – 4y – 4z = –12

4x + 4y + 4z = 7

0 = –5 New Equation 1

EXAMPLE 2 Solve a three-variable system with no solution

Because you obtain a false equation, you can conclude that the original system has no solution.

EXAMPLE 3 Solve a three-variable system with many solutions

Solve the system. x + y + z = 4 Equation 1

x + y – z = 4 Equation 2

3x + 3y + z = 12 Equation 3

SOLUTION

STEP 1 Rewrite the system as a linear system in two variables.

Add Equation 1

to Equation 2.

x + y + z = 4

x + y – z = 4

2x + 2y = 8 New Equation 1

EXAMPLE 3 Solve a three-variable system with many solutions

x + y – z = 4 Add Equation 2

3x + 3y + z = 12 to Equation 3.

4x + 4y = 16 New Equation 2

Solve the new linear system for both of its variables.

STEP 2

Add –2 times new Equation 1

to new Equation 2.

Because you obtain the identity 0 = 0, the system has infinitely many solutions.

–4x – 4y = –16

4x + 4y = 16

EXAMPLE 3 Solve a three-variable system with many solutions

STEP 3 Describe the solutions of the system. One way to do this is to divide new Equation 1 by 2 to get x + y = 4, or y = –x + 4. Substituting this into original Equation 1 produces z = 0. So, any ordered triple of the form (x, –x + 4, 0) is a solution of the system.

GUIDED PRACTICE for Examples 1, 2 and 3

Solve the system.

1. 3x + y – 2z = 10

6x – 2y + z = –2

x + 4y + 3z = 7

ANSWER (1, 3, –2)

GUIDED PRACTICE for Examples 1, 2 and 3

2. x + y – z = 22x + 2y – 2z = 65x + y – 3z = 8

ANSWER no solution

GUIDED PRACTICE for Examples 1, 2 and 3

3. x + y + z = 3x + y – z = 3

2x + 2y + z = 6

ANSWER Infinitely many solutions

EXAMPLE 4 Solve a system using substitution

Marketing The marketing department of a company has a budget of $30,000 for advertising. A television ad costs $1000, a radio ad costs $200, and a newspaper ad costs $500. The department wants to run 60 ads per month and have as many radio ads as television and newspaper ads combined. How many of each type of ad should the department run each month?

EXAMPLE 4 Solve a system using substitution

SOLUTION

STEP 1 Write verbal models for the situation.

EXAMPLE 4 Solve a system using substitution

STEP 2 Write a system of equations. Let x be the number of TV ads, y be the number of radio ads, and z be the number of newspaper ads.

x + y + z = 60 Equation 1

1000x + 200y + 500z = 30,000 Equation 2

y = x + z Equation 3

STEP 3 Rewrite the system in Step 2 as a linear system in two variables by substituting x + z for y in Equations 1 and 2.

EXAMPLE 4 Solve a system using substitution

x + y + z = 60 Write Equation 1.

x + (x + z) + z = 60 Substitute x + z for y.

2x + 2z = 60 New Equation 1

1000x + 200y + 500z = 30,000 Write Equation 2.

1000x + 200(x + z) + 500z = 30,000 Substitute x + z for y.

1200x + 700z = 30,000 New Equation 2

EXAMPLE 4 Solve a system using substitution

STEP 4 Solve the linear system in two variables from Step 3.

Add –600 times new Equation 1

to new Equation 2.

–1200x – 1200z = – 36,000

1200x +700z = 30,000

– 500z = – 6000

z = 12 Solve for z.

x = 18 Substitute into new Equation 1 or 2 to find x.

y = 30 Substitute into an original equation to find y.

EXAMPLE 4 Solve a system using substitution

ANSWER

The solution is x = 18, y = 30, and z = 12, or (18, 30, 12). So, the department should run 18 TV ads, 30 radio ads, and 12 newspaper ads each month.

GUIDED PRACTICE for Example 4

4. What IF? In Example 4, suppose the monthly budget is $25,000. How many of each type of ad should the marketing department run each month?

ANSWER

the department should run 8 TV ads, 30 radio ads, and 12 newspaper ads each month.