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```							3.6 Solving Systems of Linear
Equations in Three Variables
Warm-Up

No Solution

Infinitely many
solutions
Here is a system of three linear
equations in three variables:
 x  2 y  3 z  3
The ordered triple (2,-1,1)
is a solution to this

 2 x  5 y  4 z  13
system since it is a
solution to all three
5 x  4 y  z  5
                        equations.

2  2(1)  3(1)  2  2  3  3

2(2)  5(1)  4(1)  4  5  4  13
5(2)  4(1)  1  10  4  1  5

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               The planes could
intersect in a                 intersect in a line.
single point.                  The system has
The system has                 infinitely many
exactly one                    solutions
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.
The linear combination method
in lesson 3.2 can be extended
to solve a system of linear
equations in three variables.
Solve this system
3 x  2 y  4 z  11   Our strategy will be to use two
                       of the equations to eliminate
2 x  y  3z  4       one of the variables.
5 x  3 y  5 z  1

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
Solve this system
3 x  2 y  4 z  11   Our strategy will be to use two
                       of the equations to eliminate
2 x  y  3z  4       one of the variables.
5 x  3 y  5 z  1

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
Solve this system
3 x  2 y  4 z  11     Equation 1

-3
2     2 x  y  3z  4         Equation 2

5 x  3 y  5 z  1
                         Equation 3

Multiply Eq. 2 by 2 and add it to Eq. 1.
Solve this new system of linear
Save this result
equation in two variables.
3x  2 y  4 z  11
                                                Multiply the bottom eq. by 7 and
4 x  2 y  6 z  8                             add it to the top eq.

7x         +10z = 19                            7 x  10 z  19

7 x  28z  91
Now multiply Eq. 2 by -3 and
add it to Eq. 3. Save this
result.
6 x  3 y  9 z  12                                       -18z=-72 or z = 4
                                 Substituting z=4 into either of the new equations will
...5 x  3 y  5 z  1          give x = -3……finally substituting these values into any
of the original equations give y = 2.
-x       -4z = -13                      Our final solution is (-3,2,4)
Here is a system with No
Solution. Watch what happens
when we try to solve it.
x  y  z  2          Equation 1

3x  3 y  3z  14     Equation 2
x  2 y  z  4
                       Equation 3

3x  3 y  3z  6       Add -3 times Eq 1 to Eq 2.

..3x  3 y  3z  14          Since this is a false equation, you can
conclude the original system of equations
0=8             has no solution.
Here is a system with MANY
solutions. Watch what happens
when we try to solve it.
x  y  z  2     Equation 1

x  y  z  2     Equation 2
2 x  2 y  z  4 Equation 3
                                     Solving this new system of two equation
by adding -3 times the first eq. to 2 times
x  y  z  2    Add Eq. 1 to Eq. 2   the second eq. produces the identity 0 =
                                      0. So, the system has infinitely many
x  y  z  2                         solution.
2x + 2y    =4    New EQ. 1            You could describe the solution this way:
divide New Eq 1 by 2 to get x+y=2, or
 x  y  z  2 Add Eq 2 to Eq 3         y=-x+2. Substituting this into the
                                        orignial Equation 1 produces z = 0. So
2 x  2 y  z  4                       any ordered triple of the form (x, -x+2,0) is
a solution. For example (0,2,0) and
3x +3y      =6                          (2,0,0) are solutions.
New EQ 2
Substitution Method
 x  y  z  24

5 x  3 y  z  56
x  y  z

Since x+y=z, substitute this for z in the
first two equations

 x  y  ( x  y)  24

5 x  3 y  ( x  y)  56
Simplify

Finally, solve this linear system of two equations
2 x  2 y  24                 and two variables to get x = 4 and y =8

6 x  4 y  56                Since z=x+y, z = 12. Our final solution is (4,8,12)
Assignment 3.6

Pg 157, 1-16 (1st column)
19 & 20
22-24

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