Solve Systems of Linear Equations in 3 Variables 1.7 (M3)

Solve Systems of Linear Solve Systems of Linear Equations in 3 VariablesEquations in 3 Variables

1.7 (M3)1.7 (M3)

General Steps for Solving Systems General Steps for Solving Systems with 3 variableswith 3 variables

1.1. Combine 2 equations to make a new equation with 2 Combine 2 equations to make a new equation with 2 unknowns (eliminate 1 of the variables)unknowns (eliminate 1 of the variables)

2.2. Do the same with 2 different equations (make sure you Do the same with 2 different equations (make sure you eliminate the same variable)eliminate the same variable)

3.3. Solve the system of the 2 new equations from steps #1 Solve the system of the 2 new equations from steps #1 and #2and #2

4.4. Solve for 1 variable.Solve for 1 variable.5.5. Substitute back into 1 of the new equations to find a 2Substitute back into 1 of the new equations to find a 2ndnd

variable.variable.6.6. Substitute both back into one of the original equations to Substitute both back into one of the original equations to

find the 3find the 3rdrd variable. variable.

Special SituationsSpecial Situations

If you get a false statement (like 0 = -1) when If you get a false statement (like 0 = -1) when you are trying to solve, the original system has you are trying to solve, the original system has no solution.no solution.

If you get 0 = 0 when solving, the system has If you get 0 = 0 when solving, the system has infinitely many solutions.infinitely many solutions.

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

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


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


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)

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

ANSWER no solution

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?


SOLUTION

STEP 1 Write verbal models for the situation.


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.


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


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.

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.

Do #’s 1-3 on p. 35 with a partner.Do #’s 1-3 on p. 35 with a partner.

Documents

Solve Systems of Linear Equations in 3 Variables 1.7 (M3)