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3-2: Solving Systems of 3 Equations in 3 Variables

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3-2: Solving Systems of 3 Equations in 3 Variables
3-2: Solving Systems of 3 Equations in 3
Variables
1.
2.
Objectives:
To solve a linear
system in three
variables using
substitution and
elimination
To formulate
systems of 3 linear
equations to model
real-world situations
•
•
•
•
Assignment:
P. 41: 10-14
P. 53: 8-12
Multiple Solutions
Worksheet: 1-3, 6, 9, 12
Print Review 1a
Objective 0
You will be able to
geometrically interpret the
solution to a linear system in
three variables
Activity 1: Graphing in 3D
A linear equation in three variables , ,
and  can be written
 +  +  = ,
where , , , and  are real numbers, all of
which are not zero.
Activity 1: Graphing in 3D
We graph this equation
in 3-D, on a coordinate
system with an -, -,
and a -axis, dividing
space into eight
octants.
Points in space are
located with an
ordered triple , ,  .
Activity 1: Graphing in 3D
We graph this equation
in 3-D, on a coordinate
system with an -, -,
and a -axis, dividing
space into eight
octants.
Points in space are
located with an
ordered triple , ,  .
Activity 1: Graphing in 3D
The solution to a linear
equation in three
variables is the set of
all points , ,  that
satisfy the equation.
In this activity, we will
discover the shape of
the graph of a linear
equation in 3 variables.
Activity 1: Graphing in 3D
We are going to use a three-dimensional
coordinate system to graph the equation
3 + 4 + 6 = 12.
Step 1: Start by
finding the intercept.
Substitute 0 in for 
and  and solve for
. Plot this point.
Activity 1: Graphing in 3D
Step 2: Next find the -intercept by
substituting 0 in for  and  and solving for
. Plot this point.
Step 3: Finally find
the -intercept by
substituting 0 in for
 and  and solving
for . Plot this point.
Activity 1: Graphing in 3D
Step 4: Connect your three points: intercept to -intercept, -intercept to intercept, and -intercept to -intercept.
What
shape is the
graph of a
linear
equation in 3 variables?
Activity 1: Graphing in 3D
Recall a postulate from geometry which
states:
Through any 3 noncollinear points, there exists
exactly one plane.
Thus, we can conclude that the graph of a
linear equation in 3 variables is a plane.
Activity I: Graphing in 3D
Microsoft Mathematics
4.0:
• Free download (PC)
• Click the Graphing tab
• Choose 3D from the
drop down menu
• Type in the equation
and click Graph
Exercise 1
Sketch the graph of the
equation.
3 + 9 − 3 = −18
Exercise 1
Sketch the graph of the
equation.
3 + 9 − 3 = −18
Linear System in 3 Variables
A linear equation in three variables , ,
and  can be written
 +  +  = ,
where , , , and  are real numbers, all of
which are not zero.
A linear system of equations in three
variables has 3 such equations.
Linear System in 3 Variables
A linear equation in three variables , ,
and  can be written
 +  +  = ,
where , , , and  are real numbers, all of
which are not zero.
Linear System in 3 Variables
A linear equation in three variables , ,
and  can be written
 +  +  = ,
where , , , and  are real numbers, all of
which are not zero.
The solution to such a system is the ordered
triple , ,  that satisfies all the equations.
Graphs of 3D Systems
Recall that a system of linear equations in
two variables can be either consistent or
inconsistent, and that consistent systems
can be either independent or dependent.
Possible Solutions
Geometrically, the solution to any system of
equations is the point or points of intersection.
Possible Solutions
Geometrically, the solution to any system of
equations is the point or points of intersection.
Objective 1a
You will be able
to solve a linear
system in three
variables using
substitution
Objective 2b
You will be able
to solve a linear
system in three
variables using
substitution
(Insert Substitute Here)
Solving Algebraically
We’d probably not want to solve a linear
system in 3 variables by graphing.
Instead, there would probably be far less
bloodshed if we solved such a system
algebraically, using either elimination or
substitution.
For the substitution method, you basically
turn your 3-variable system into a 2variable system. Easy.
Substitution Method
If it is convenient, you could use substitution to
help solve a linear system in three variables.
Solve one of the equations
Step for
1 one of the variables.
Substitute this expressionStep
into both
2 of the other equations.
Step23variable system.
Solve the remaining
Example A
Solve using substitution.
2 + 7 +  = −53
−2 + 3 +  = −13
6 + 3 +  = −45
You will be able to solve a linear
system by Gaussian
elimination
Objective 1b
Exercise 2
Solve the system of equations.
x  2 y  2z  9
y  2z  5
z  3
Equivalent Systems
Two systems are equivalent if they have
the same solution.
x
 2y
x
 3y
2x
 5y
 2z

9
 4
 3z
Linear System

16
x  2 y  2z  9
y  2z  5
z  3
Equivalent Linear
System
Triangular Form
A linear system of equations is in
triangular form if:
The equations have
a stair-step pattern
x  2 y  2z  9
y  2z  5
z  3
The leading
coefficient of each
equation is 1
Triangular Form
Solving a system of equations
in triangular form is almost
too easy. The question is,
how do we rewrite a linear
system as an equivalent
system in triangular form?
The answer involves
elementary row
operations, a.k.a.
Gaussian elimination.
Money Man: Carl Gauss
(~1800s)
Protip #1: Letter Equations
To help you through
2 −  + 6 = −4 A
the often
6 + 4 − 5 = −7 B
labyrinthine
−4 − 2 + 5 = 9 C
process of solving a
3-variable system,
letter each of your
equations.
Elementary Row Operations
↔
1
2
2 + 
↔
1
2
2 + 
Used to create equivalent
linear systems
Interchange two
equations
Multiply one
equation by a
non-zero constant
Add a multiple of
one equation to
another equation
to replace the
latter equation
Exercise 3
Solve the system by rewriting it in triangular form.
A : x  2 y  2z 
B : x
 3y
C : 2x
 5y
9
 4
 3z 
16
2 xx 42yy 42zz
y 2 z
A+B:
2 x 5 y 3z
1. Eliminate  in all but the first equation.
−2 A + C :
 918
 5
 16
x 2 y 2 z 
9
y 2 z 
5
y
 z  2
x 2 y 2 z  9
2. Eliminate  in the last equation.
y 2 z  5
B+C:
z  3
Triangular Form
To rewrite a linear system in three variables
in triangular form, follow these helpful
hints:
Eliminate  in
all but the first
equation
Step 1
Eliminate 
in the last
equation
Step 2
Use
multiplication to
make the
leading
coefficients
equal
Step 13
Example B
Solve the system by rewriting it in triangular form.
2 +  −  = 4
−2 +  + 2 = 6
 + 2 +  = 11
Exercise 4
Solve the system.
+−=2
3 + 3 − 3 = 8
2 −  + 4 = 7
Exercise 5
Solve the system.
++=6
−+=6
4 +  + 4 = 24
Protip #2: Multiple Solutions
When you discover that you have a
consistent, dependent system of
equations, how do you write your answer?
Graphically, the equations in this system
intersect in a line, so you could just write
the equation of that line.
But what if you want specific solutions, in the
form of ordered triples?
Protip #2: Multiple Solutions
To write your answers as a set of ordered pairs, set
one of the variables in your equation equal to .
Now re-write the other variables in terms of .
++=6

+=6
Let  = 
+=6
 =6−
++ 6− =6
=0
Protip #2: Multiple Solutions
Finally, use your new expressions to write an
ordered triple. Substitute values in for  to get
specific solution points.
++=6

Let  = 0:
0, 0, 6
+=6
=
 =6−
=0
Let  = 1:
1, 0, 5
, 0, 6 − 
Let  = −1:
−1, 0, 7
Exercise 6: SAT
If 5 sips + 4 gulps = 1 glass and 13 sips + 7
gulps = 2 glasses, how many sips equal a
gulp?
3-2: Solving Systems of 3 Equations in 3
Variables
1.
2.
Objectives:
To solve a linear
system in three
variables using
substitution and
elimination
To formulate
systems of 3 linear
equations to model
real-world situations
•
•
•
•
Assignment:
P. 41: 10-14
P. 53: 8-12
Multiple Solutions
Worksheet: 1-3, 6, 9, 12
Print Review 1a
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