1. 3-4 Solving Systems of Linear Equations in 3 Variables
Algebra 2 Textbook

2. Terms to remember

3. Visualizing what this means

5. Using elimination to solve

6. Use the elimination method
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 z = – 1
New Equation A

7. Use the elimination method
EXAMPLE 1
Use the elimination method
2x – 3y + 5z = – 14
Add – 3 times Equation 3
to Equation 2.
– 18x + 3y – 12z = 3
– 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.

8. Use the elimination method
EXAMPLE 1
Use the elimination method
STEP 3
Substitute x = 2 and z = – 3 into an original
equation and solve for y.
6x – y + 4z = – 1
Write original Equation 3.
6(2) – y + 4(– 3) = – 1
Substitute 2 for x and –3
for z.
y = 1
Solve for y.

9. 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)

10. Solve a three-variable system with no solution
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
New Equation A
0 = – 5
Because you obtain a false equation, you can conclude that the original system has no solution.

11. Solve a three-variable system with many solutions
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.
x + y + z = 4
x + y – z = 4
Add Equation 1
to Equation 2.
2x + 2y = 8
New Equation A

12. Solve a three-variable system with many solutions
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
– 4x – 4y = – 16
4x + 4y = 16
Add –2 times new Equation A
to new Equation B.
Because you obtain the identity 0 = 0, the system has infinitely many solutions.

13. 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.

14. GUIDED PRACTICE for Examples 1, 2 and 3 2. x + y – z = 2
ANSWER
no solution

15. GUIDED PRACTICE
for Examples 1, 2 and 3
3. x + y + z = 3
x + y – z = 3
2x + 2y + z = 6
ANSWER
Infinitely many solutions

16. Step 1 Write the Equations
3.20
4.80
3.68
Step 1 Write the Equations
Eq. 1 p + c = 100
Eq p c = 368
Step 2 Use substitution to solve the system
Solve Eq. 1 for p p = 100-c
Substitute 100-c in for p in Eq (100-c) + 4.8c = 368
Use distribute property to simplify 320 – 3.20c + 4.8c = 368
Combine like terms c = 368
Subtract 320 from each side c = 48
Solve for c then p by substitution c = p = 100 – 30 = 70
Step Write the answers down with correct labels
30 pounds of cashews & 70 pounds of peanuts