Download Section 9.4

9.4
Solving Systems of Linear Equations in
Three Variables; Applications
1. Determine whether an ordered triple is a solution for a
system of equations.
2. Understand the types of solution sets for systems of
three equations.
3. Solve a system of three linear equations using the
elimination method.
4. Solve application problems that translate to a system
of three linear equations.
Is (2, –1, 3) a solution of the system?
Ordered
Triple
(x, y, z)
x+y+z=4
2 + (–1) + 3 = 4
4=4
TRUE
x yz 4


 2x  2 y  z  3
 4 x  y  2 z  3

2x – 2y – z = 3
– 4x + y + 2z = –3
2(2) – 2(–1) – 3 = 3 – 4(2) + (–1) + 2(3) = –3
3=3
–3 = –3
TRUE
TRUE
Because (2, 1, 3) satisfies all three equations in
the system, it is a solution for the system.
Determine if (2, –5, 3) is a solution to
the given system.
x  5 y  2z  7

 3 y  5 z  16

z  5

a) Yes
b) No
9.4
Copyright © 2011 Pearson Education, Inc.
Slide 4- 3
Determine if (2, –5, 3) is a solution to
the given system.
x  5 y  2z  7

 3 y  5 z  16

z  5

a) Yes
b) No
9.4
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Slide 4- 4
Types of Solution Sets
A Single Solution:
If the planes intersect at a
single point, that ordered
triple is the solution to the
system.
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Types of Solution Sets
Infinite Number of Solutions:
If the three planes intersect
along a line, the system has an
infinite number of solutions,
which are the coordinates of
any point along that line.
Infinite Number of Solutions:
If all three graphs are the same
plane, the system has an
infinite number of solutions.
They are the coordinates of all
points in the plane.
Copyright © 2011 Pearson Education, Inc.
Types of Solution Sets
No Solution:
If all of the planes are
parallel, the system has
no solution.
No Solution:
Pairs of planes also can
intersect, as shown.
However, because all three
planes do not have a
common intersection, the
system has no solution.
Copyright © 2011 Pearson Education, Inc.
Solve the system using elimination.
 x yz 6

x  2 y  z  2
 x  y  3z  8

(1)
(2)
(3)
We select any two of the three equations and
work to get one equation in two variables.
Add equations (1) and (2):
x yz6
x  2y  z  2
2x + 3y
=8
(1)
(2)
(4) Add to eliminate z
2x + 3y = 8
 x yz 6

x  2 y  z  2
 x  y  3z  8

(4)
*(1)
*(2)
(3)
Next, we select a different pair of equations and
eliminate the same variable.
Use (2) and (3) to again eliminate z.
x  2y  z  2
x  y  3z  8
Multiply by 3
3x + 6y – 3z = 6
x – y + 3z = 8
4x + 5y
= 14 (5)
Now solve the resulting system of equations (4) and (5).
2x + 3y = 8
4x + 5y = 14
(4)
(5)
-2
2x + 3y = 8
(4)
4x + 5y = 14
(5)
 x yz 6

x  2 y  z  2
 x  y  3z  8

(1)
(2)
(3)
Multiply equation (4) by –2 and then add to equation (5):
Multiply by -2
2x + 3y = 8 (4)
4x + 5y = 14 (5)
Substitute into either
equation (4) or (5) to find x.
2x  3y  8
–4x – 6y = –16
4x + 5y = 14
–y = –2
y=2
To find z, substitute into any
2 x  32   8
of the original equations.
2x  6  8
x+y+z=6
2x  2
1+2+z=6
(1,
2,
3)
x 1
z = 3.
Ordered triple
Solve the system using elimination.
3 x  9 y  6 z  3

 2x  y  z  2
 x yz 2

(1)
(2)
(3)
Eliminate z from equations (2) and (3).
2x  y  z  2
x yz2
3x + 2y = 4
(2)
(3)
(4)
continued
3x + 2y = 4
(4)
Eliminate z from equations (1) and (2).
3x  9 y  6 z  3
2x  y  z  2
Multiply by 6
3 x  9 y  6 z  3 (1)

 2 x  y  z  2 *(2)
 x  y  z  2 *(3)

3x  9 y  6 z  3
12 x  6 y  6 z  12
15x + 15y = 15
Eliminate x from equations (4) and (5).
15x – 10y = 20
15x + 15y = 15
5y = 5
Use y = 1, to find x in equation 4.
y = 1
3x + 2y = 4
3x + 2(1) = 4
x=2
3x + 2y = 4
15x + 15y = 15
Multiply by 5
(5)
continued
x=2
y = 1
3 x  9 y  6 z  3 (1)

 2 x  y  z  2 (2)
 x  y  z  2 (3)

Substitute x = 2 and y = 1 to find z.
x+y+z=2
2–1+z=2
1+z=2
z=1
The solution is the ordered triple (2, 1, 1).
Solving Systems of Three Linear Equations
Using Elimination
1. Write each equation in the form Ax + By+ Cz = D.
2. Eliminate one variable from one pair of equations
using the elimination method.
3. If necessary, eliminate the same variable from
another pair of equations.
continued
4. Steps 2 and 3 result in two equations with the same
two variables. Solve these equations using the
elimination method.
5. To find the third variable, substitute the values of
the variables found in step 4 into any of the three
original equations that contain the third variable.
6. Check the ordered triple in all three original
equations.
Solve the system using elimination.
(1)
 x  3y  z  1

(2)
2 x  y  2 z  2
 x  2 y  3 z  1 (3)

Eliminate x from equations (1) and (2).
x  3y  z  1
2x  y  2z  2
Multiply by -2
2 x  6 y  2 z  2 (1)
2 x  y  2 z  2 (2)
(4)
5y  4z = 0
continued
5y 4z = 0
(4)
Eliminate x from equations (1) and (3).
x  3y  z  1
x  2 y  3z  1
Multiply by -1
 x  3 y  z  1 *(1)

2 x  y  2 z  2 *(2)
 x  2 y  3 z  1 (3)

x  3y  z  1
 x  2 y  3z  1
5y + 4z = 2
(5)
Eliminate y from equations (4) and (5).
5y  4z = 0
5y + 4z = 2
0=2
All variables are eliminated and the resulting equation
is false. This system has no solution. It is inconsistent.
Solve the system.
2 x  4 y  z  1

5 x  2 y  z  9
3x  y  2 z  14

a) (–2, 2, –5)
b) (–5, 2, –2)
c) infinite number of solutions
d) no solution
9.4
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Slide 4- 18
Solve the system.
2 x  4 y  z  1

5 x  2 y  z  9
3x  y  2 z  14

a) (–2, 2, –5)
b) (–5, 2, –2)
c) infinite number of solutions
d) no solution
9.4
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Slide 4- 19