Transcript 3.4

3.4
Systems of Equations in
Three Variables

Identifying Solutions

Solving Systems in Three Variables

Dependency, Inconsistency, and Geometric
Considerations
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Identifying Solutions
A linear equation in three variables is an
equation equivalent to one in the form
Ax + By + Cz = D, where A, B, C, and D
are real numbers. We refer to the form
Ax + By + Cz = D as standard form for a
linear equation in three variables.
A solution of a system of three equations in
three variables is an ordered triple (x, y, z)
that makes all three equations true.
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Example
Determine whether (2, –1, 3) is a solution of
the system
x  y  z  4,
2 x  2 y  z  3,
4 x  y  2 z  3.
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Solution We substitute (2, –1, 3) into the three
equations, using alphabetical order:
2x – 2y – z = 3
x+y+z=4
2 + (–1) + 3 4
4 = 4 TRUE
2(2) – 2(–1) – 3 3
3 = 3 TRUE
–4x + y + 2z = –3
–4(2) + (–1) + 2(3) –3
–3 = –3 TRUE
The triple makes all three equations true, so it is a
solution of the system.
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Solving Systems in Three
Variables
The elimination method allows us to
manipulate a system of three equations in
three variables so that a simpler system of
two equations in two variables is formed.
Once that simpler system has been solved,
we can substitute into one of the three
original equations and solve for the third
variable.
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Example
Solve the following system of equations:
x  y  z  6,
x  2 y  z  2,
x  y  3z  8.
(1)
(2)
(3)
Solution
We select any two of the three equations and
work to get one equation in two variables. Let’s
add equations (1) and (2):
x yz6
(1)
(2)
x  2y  z  2
Adding to
(4)
2x + 3y
=8
eliminate z
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Next, we select a different pair of equations and
eliminate the same variable. Let’s use (2) and (3) to
again eliminate z.
x  2y  z  2
x  y  3z  8
Multiplying
equation (2)
by 3
3x + 6y – 3z = 6
x – y + 3z = 8
4x + 5y
= 14.
(5)
Now we solve the resulting system of equations
(4) and (5). That will give us two of the numbers
in the solution of the original system,
(4)
2x + 3y = 8
(5)
4x + 5y = 14
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We multiply both sides of equation (4) by –2 and
then add to equation (5):
–4x – 6y = –16,
4x + 5y = 14
–y = –2
y=2
Substituting into either equation (4) or (5) we find
that x = 1.
Now we have x = 1 and y = 2. To find the value
for z, we use any of the three original equations
and substitute to find the third number z.
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Let’s use equation (1) and substitute our two
numbers in it:
x+y+z=6
1+2+z=6
z = 3.
We have obtained the ordered triple (1, 2, 3). It
should check in all three equations.
The solution is (1, 2, 3).
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Solving Systems of Three Linear
Equations
To use the elimination method to solve systems of
three linear equations:
1. Write all the equations in standard form
Ax + By+ Cz = D.
2. Clear any decimals or fractions.
3. Choose a variable to eliminate. Then select
two of the three equations and work to get
one equation in which the selected variable is
eliminated.
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Solving Systems of Three Linear
Equations (continued)
4. Next, use a different pair of equations and
eliminate the same variable that you did in
step (3).
5. Solve the system of equations that resulted
from steps (3) and (4).
6. Substitute the solution from step (5) into one
of the original three equations and solve for
the third variable. Then check.
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Dependency, Inconsistency, and
Geometric Considerations
The graph of a linear equation in three
variables is a plane. Solutions are points
common to the planes of each system.
Since three planes can have an infinite
number of points in common or no points at
all in common, we need to generalize the
concept of consistency in three dimensions.
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Consistency
A system of equations that has at least one
solution is said to be consistent.
A system of equations that has no solution
is said to be inconsistent.
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Example
Solve the following system of equations:
y  2 z  2,
 x  2 y  z  5,
x  y  z  1.
(1)
(2)
(3)
Solution
The variable x is missing in equation (1). Let’s
add equations (2) and (3) to get another
equation with x missing:
x  2 y  z  5
(2)
x  y  z  1 (3)
Adding
y + 2z = 6 (4)
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Equations (1) and (4) form a system in y and z.
Solve as before:
y + 2z = 2
y + 2z = 6
Multiplying
equation (1)
by –1
–y – 2z = –2
y + 2z = 6
0 = 4.
Since we ended up with a false equation, or
contradiction, we know that the system has no
solution. It is inconsistent.
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Recall that when dependent equations
appeared in Section 3.1, the solution sets
were always infinite in size and were written
in set-builder notation. There, all systems of
dependent equations were consistent. This is
not always the case for systems of three or
more equations. The following figures
illustrate some possibilities geometrically.
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