1 / 11

# 3.5 Solving Linear Systems in Three Variables - PowerPoint PPT Presentation

3.5 Solving Linear Systems in Three Variables. 10/4/13. Intersection of 3 planes. We’ve been solving system of equations in 2 variables. The solution is a point where the lines intersect. For systems of equations with 3 variables, the solution is a point where all 3 planes intersect. Solve:.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' 3.5 Solving Linear Systems in Three Variables' - locke

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### 3.5 Solving Linear Systems in Three Variables

10/4/13

We’ve been solving system of equations in 2 variables. The solution is a point where the lines intersect.

For systems of equations with 3 variables, the solution is a point where all 3 planes intersect.

Example 1

Equation 1

Equation 2

Equation 3

Notice Eqn1 has only 2 variables. Solve for one variable (x).

Substitute -3z +5 for x in the other 2 equations.

Combine Like terms

New Eqn 2

New Eqn 3

-2( )

New Eqn 2

New Eqn 3

Solve by Elimination

Substitute z = 3 in

+

+

Solution (x, y, z)

(-4, -1, 3)

Substitute z= 3 in any of the new Eqns.

Step 1: Pick any 2 original equations and eliminate a variable. Eliminate the same variable from a second pair of original equations.

Step 2: With the 2 new equations from Step 1 eliminate one of the 2 variables and solve for the remaining variable. Substitute the value you obtained for the variable into one of the 2 new equations and solve for the other variable.

Step 3: Substitute the values of the 2 variables obtained in Step 2 into one of the 3 original equations and solve for the last variable (the one you eliminated in step 1).

Step 4: Check the solution in each of the original equations.

Solve.

Step 1

New Eqn 1

Step 1

Step 2

New Eqn 2

Step 3

-1( )

Solve the system.

Equation 1

3x

+

2y

+

4z

11

=

( 3, 2, 4).

Equation 2

2x

y

4

+

3z

=

5x

3y

1

Equation 3

+

5z

=

-

y

-

z

3

=

+

(2, -2, 1)

-x

+

2y

-1

5z

=

Example 4

Solve the system. Then check your solution.

+

x

y

+

4z

=

4

3.5 p.156 #7, 16-19