3-4 Solving Systems of Linear Equations in 3 Variables

3-4 Solving Systems of Linear Equations in 3 Variables

Algebra 2 Textbook

Terms to remember

Visualizing what this means

Using elimination to solve

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 1Rewrite 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 A

EXAMPLE 1

2x – 3y + 5z = – 14

– 18x + 3y – 12z = 3

Add – 3 times Equation 3to Equation 2.

– 16x – 7z = – 11 New Equation B

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

16x + 11z = –1 Add new Equation A

and new Equation B.– 16x – 7z = –11

4z = –12

z = – 3 Solve for z.

x = 2 Substitute into new Equation A or B to find x.

Use the elimination method

6x – y + 4z = – 1

EXAMPLE 1

Use the elimination method

STEP 3Substitute 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.

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)

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 3SOLUTION

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 A

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 A

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 B

Solve the new linear system for both of its variables.

STEP 2

Add –2 times new Equation A

to new Equation B.

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

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

ANSWERInfinitely many solutions

3.20

4.80

3.68

Step 1 Write the Equations

Step 2 Use substitution to solve the system

Step 3 Write the answers down with correct labels

Eq. 1 p + c = 100Eq. 2 3.2p + 4.8c = 368

Solve Eq. 1 for p p = 100-c Substitute 100-c in for p in Eq. 2 3.2(100-c) + 4.8c = 368Use distribute property to simplify 320 – 3.20c + 4.8c = 368Combine like terms 320 + 1.6c = 368Subtract 320 from each side 1.6c = 48Solve for c then p by substitution c = 30 p = 100 – 30 = 70

30 pounds of cashews & 70 pounds of peanuts