Download presentation

Presentation is loading. Please wait.

Published byMichelle Kerr Modified over 3 years ago

1
**3.6 Solving Systems of Linear Equations in Three Variables**

2
Warm-Up No Solution Infinitely many solutions

3
**Here is a system of three linear equations in three variables:**

The ordered triple (2,-1,1) is a solution to this system since it is a solution to all three equations.

4
**The graph of a linear equation in three variables is a plane**

The graph of a linear equation in three variables is a plane. Three planes in space can intersect in different ways (pg 152). The planes could intersect in a line. The system has infinitely many solutions The planes could intersect in a single point. The system has exactly one solution The planes could have NO point of intersection. The left figure shows planes that intersect pairwise, but all 3 do not have a common point of intersection. The right figure shows parallel planes. Each system has NO solution.

5
**The linear combination method in lesson 3**

The linear combination method in lesson 3.2 can be extended to solve a system of linear equations in three variables.

6
Solve this system Our strategy will be to use two of the equations to eliminate one of the variables. We will then use two other equations to eliminate the same variable. Once we have two equations with two variables, we can use the technique we learned in lesson 3.2

7
Solve this system Our strategy will be to use two of the equations to eliminate one of the variables. We will then use two other equations to eliminate the same variable. Once we have two equations with two variables, we can use the technique we learned in lesson 3.2

8
**Solve this system 7x +10z = 19 -x -4z = -13 2 -18z=-72 or z = 4**

Equation 1 Equation 2 2 -3 Equation 3 Multiply Eq. 2 by 2 and add it to Eq. 1. Save this result Solve this new system of linear equation in two variables. Multiply the bottom eq. by 7 and add it to the top eq. 7x z = 19 Now multiply Eq. 2 by -3 and add it to Eq. 3. Save this result. -18z=-72 or z = 4 Substituting z=4 into either of the new equations will give x = -3……finally substituting these values into any of the original equations give y = 2. -x z = -13 Our final solution is (-3,2,4)

9
**Here is a system with No Solution**

Here is a system with No Solution. Watch what happens when we try to solve it. Equation 1 Equation 2 Equation 3 Add -3 times Eq 1 to Eq 2. Since this is a false equation, you can conclude the original system of equations has no solution. 0=8

10
**Here is a system with MANY solutions**

Here is a system with MANY solutions. Watch what happens when we try to solve it. Equation 1 Equation 2 Solving this new system of two equation by adding -3 times the first eq. to 2 times the second eq. produces the identity 0 = 0. So, the system has infinitely many solution. Equation 3 Add Eq. 1 to Eq. 2 You could describe the solution this way: divide New Eq 1 by 2 to get x+y=2, or y=-x+2. Substituting this into the orignial Equation 1 produces z = 0. So any ordered triple of the form (x, -x+2,0) is a solution. For example (0,2,0) and (2,0,0) are solutions. 2x + 2y = 4 New EQ. 1 Add Eq 2 to Eq 3 3x +3y = 6 New EQ 2

11
Substitution Method Since x+y=z, substitute this for z in the first two equations Simplify Finally, solve this linear system of two equations and two variables to get x = 4 and y =8 Since z=x+y, z = 12. Our final solution is (4,8,12)

12
Assignment 3.6

Similar presentations

Presentation is loading. Please wait....

OK

Elimination Using Multiplication.

Elimination Using Multiplication.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on search engine optimization Jit ppt on manufacturing business Ppt on provident fund act 1952 Ppt on induced abortion Slideshare ppt on internet Ppt on credit policy in banks Ppt on air powered engine Ppt on stock exchange Ppt on data handling for class 3 Ppt on object oriented programming concepts