1. 3.4 Solving Systems of Linear Equations in 3 Variables
Algebra II

2. A system of lin. eqns. in 3 variables
Looks something like:
x+3y-z=-11
2x+y+z=1
5x-2y+3z=21
A solution is an ordered triple (x,y,z) that makes all 3 equations true.

3. Steps for solving in 3 variables
Using the 1st 2 equations, cancel one of the variables.
Using the last 2 equations, cancel the same variable from step 1.
Use the results of steps 1 & 2 to solve for the 2 remaining variables.
Plug the results from step 3 into one of the original 3 equations and solve for the 3rd remaining variable.
Write the solution as an ordered triple (x,y,z).

4. 1)Solve the system. x+3y-z=-11 2x+y+z=1 5x-2y+3z=21 x+3y-z=-11
z’s are easy to cancel!
3x+4y=-10
2. 2x+y+z=1
5x-2y+3z=21
Must cancel z’s again!
-6x-3y-3z=-3
-x-5y=18
2(2)+(-4)+z=1
4-4+z=1
3. 3x+4y=-10
-x-5y=18
Solve for x & y.
3x+4y=-10
-3x-15y=54
-11y=44
y=- 4
3x+4(-4)=-10
x=2
(2, - 4, 1)
z=1

5. 2)Solve the system. -x+2y+z=3 2x+2y+z=5 4x+4y+2z=6 -2(2x+2y+z=5)
Cancel z’s again.
-4x-4y-2z=-10
0=- 4
Doesn’t make sense!
No solution
-x+2y+z=3
-(2x+2y+z=5)
z’s are easy to cancel!
-2x-2y-z=-5
-3x=-2
x=2/3

6. 3)Solve the system. -2x+4y+z=1 3x-3y-z=2 5x-y-z=8 -2x+4y+z=1 3x-3y-z=2
z’s are easy to cancel!
x+y=3
5x-y-z=8
Cancel z’s again.
-3x+3y+z=-2
2x+2y=6
3. x+y=3
2x+2y=6
Cancel the x’s.
-2x-2y=-6
0=0
This is true.
IMS

7. Assignment