Transcript 3.6 Solving Systems of Linear Equations in Three Variables

7.1 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  3z  3

2 x  5 y  4 z  13
5 x  4 y  z  5

The ordered triple (2,-1,1)
is a solution to this
system since it is a
solution to all three
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 218).
The planes could
The planes could
intersect in a line.
intersect in a
The system has
single point.
infinitely many
The system has
solutions
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.
The linear combination method
in lesson 3.2 can be extended
to solve a system of linear
equations in three variables.
Solve this system
3x  2 y  4 z  11

2 x  y  3z  4
5 x  3 y  5 z  1

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
Solve this system
3x  2 y  4 z  11

2 x  y  3z  4
5 x  3 y  5 z  1

Equation 1
Equation 2
Equation 3
Multiply Eq. 2 by 2 and add it to Eq. 1.
Save this result
3x  2 y  4 z  11

4 x  2 y  6 z  8
7x
+10z = 19
Now multiply Eq. 2 by -3 and
add it to Eq. 3. Save this
result.
6 x  3 y  9 z  12

...5 x  3 y  5 z  1
-x
-4z = -13
Solve this new system of linear
equation in two variables.
Multiply the bottom eq. by 7 and
add it to the top eq.
7 x  10 z  19

7 x  28 z  91
-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.
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

3x  3 y  3z  14
x  2 y  z  4

3x  3 y  3z  6

..3x  3 y  3z  14
0=8
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.
Here is a system with MANY
solutions. Watch what happens
when we try to solve it.
Equation 1
x  y  z  2

Equation 2
x  y  z  2
2 x  2 y  z  4 Equation 3

x  y  z  2

x  y  z  2
2x + 2y
=4
Add Eq. 1 to Eq. 2
New EQ. 1
 x  y  z  2 Add Eq 2 to Eq 3

2 x  2 y  z  4
3x +3y
=6
New EQ 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.
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.
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
2 x  2 y  24

6 x  4 y  56
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)
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