3.4: Solve a Linear System in Three Variables

3.4: Solve a Linear System in Three
Variables
1.
2.
Objectives:
To geometrically
interpret the solution
to a linear system in
three variables
To solve a linear
system in three
variables using
substitution and
elimination
•
•
•
•
•
Assignment:
P. 177: 1, 2
P. 182-185: 1, 2-32
even, 41, 42, 46, 48-50
Multiple Solutions
Worksheet: 1-3, 6, 9, 12
Extra Credit?
Print Review 2a
Objective 1
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 𝑥, 𝑦,
and 𝑧 can be written
𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑,
where 𝑎, 𝑏, 𝑐, and 𝑑 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 𝑥-, 𝑦-,
and a 𝑧-axis, dividing
space into eight
octants.
Points in space are
located with an
ordered triple 𝑥, 𝑦, 𝑧 .
Activity 1: Graphing in 3D
We graph this equation
in 3-D, on a coordinate
system with an 𝑥-, 𝑦-,
and a 𝑧-axis, dividing
space into eight
octants.
Points in space are
located with an
ordered triple 𝑥, 𝑦, 𝑧 .
Activity 1: Graphing in 3D
The solution to a linear
equation in three
variables is the set of
all points 𝑥, 𝑦, 𝑧 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
3𝑥 + 4𝑦 + 6𝑧 = 12.
Step 1: Start by
finding the 𝑥intercept.
Substitute 0 in for 𝑦
and 𝑧 and solve for
𝑥. Plot this point.
Activity 1: Graphing in 3D
Step 2: Next find the 𝑦-intercept by
substituting 0 in for 𝑥 and 𝑧 and solving for
𝑦. Plot this point.
Step 3: Finally find
the 𝑧-intercept by
substituting 0 in for
𝑥 and 𝑦 and solving
for 𝑧. Plot this point.
Activity 1: Graphing in 3D
Step 4: Connect your three points: 𝑥intercept to 𝑦-intercept, 𝑦-intercept to 𝑧intercept, and 𝑧-intercept to 𝑥-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.
3𝑥 + 9𝑦 − 3𝑧 = −18
Exercise 1
Sketch the graph of the
equation.
3𝑥 + 9𝑦 − 3𝑧 = −18
Linear System in 3 Variables
A linear equation in three variables 𝑥, 𝑦,
and 𝑧 can be written
𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑,
where 𝑎, 𝑏, 𝑐, and 𝑑 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 𝑥, 𝑦,
and 𝑧 can be written
𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑,
where 𝑎, 𝑏, 𝑐, and 𝑑 are real numbers, all of
which are not zero.
Linear System in 3 Variables
A linear equation in three variables 𝑥, 𝑦,
and 𝑧 can be written
𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑,
where 𝑎, 𝑏, 𝑐, and 𝑑 are real numbers, all of
which are not zero.
The solution to such a system is the ordered
triple 𝑥, 𝑦, 𝑧 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 2a
You will be
able to solve a
linear system
in 3 variables
by elimination
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 elimination method, you first
eliminate one of your variables so that you
have 2 equations with 2 variables. Easy.
Exercise 2
Solve the system.
2𝑥 − 𝑦 + 6𝑧 = −4
6𝑥 + 4𝑦 − 5𝑧 = −7
−4𝑥 − 2𝑦 + 5𝑧 = 9
Elimination Method
Use elimination
to write the 3
variable
Step 1
system as a 2
variable
system
Solve the
new 2
variable
Step 2
system for
both
variables
Substitute these
values into one of
the original
Step 3
equations and
solve for the
remaining variable
You’ll have to eliminate the same variable
from 2 different sets of the equations.
Elimination Method
Use elimination
to write the 3
variable
Step 1
system as a 2
variable
system
Solve the
new 2
variable
Step 2
system for
both
variables
Substitute these
values into one of
the original
Step 3
equations and
solve for the
remaining variable
The system has no solution if you obtain a
contradiction (ex. 0 = 1) while solving the system.
Elimination Method
Use elimination
to write the 3
variable
Step 1
system as a 2
variable
system
Solve the
new 2
variable
Step 2
system for
both
variables
The system has infinitely many solutions
if you obtain an identity (ex. 0 = 0) while
solving the system.
Substitute these
values into one of
the original
Step 3
equations and
solve for the
remaining variable
Protip #1: Letter Equations
To help you through
2𝑥 − 𝑦 + 6𝑧 = −4 A
the often
6𝑥 + 4𝑦 − 5𝑧 = −7 B
labyrinthine
−4𝑥 − 2𝑦 + 5𝑧 = 9 C
process of solving a
3-variable system,
letter each of your
equations.
Protip #1: Letter Equations
In terms of these
2𝑥 − 𝑦 + 6𝑧 = −4 A
letters, write a simple
6𝑥 + 4𝑦 − 5𝑧 = −7 B
expression that tells −4𝑥 − 2𝑦 + 5𝑧 = 9 C
you how to
add/subtract
6𝑥 + 4𝑦 − 5𝑧 = −7
B+ C
multiples of each
+ −4𝑥 − 2𝑦 + 5𝑧 = 9
equation.
=2
D 2𝑥 + 2𝑦
Label the new
1
2 D 𝑥+𝑦 =1 E
equation with a new
letter.
Protip #1: Letter Equations
Continue this process
until the system is
solved.
5A +6 B
2𝑥 − 𝑦 + 6𝑧 = −4 A
6𝑥 + 4𝑦 − 5𝑧 = −7 B
−4𝑥 − 2𝑦 + 5𝑧 = 9 C
10𝑥 − 5𝑦 + 30𝑧 = −20
+ 36𝑥 − 24𝑦 − 30𝑧 = −42
F 46𝑥 − 29𝑦
= −62
Protip #1: Letter Equations
Continue this process
until the system is
solved.
2𝑥 − 𝑦 + 6𝑧 = −4 A
6𝑥 + 4𝑦 − 5𝑧 = −7 B
−4𝑥 − 2𝑦 + 5𝑧 = 9 C
−19 E + F
−19𝑥 − 19𝑦 = −19
+ 46𝑥 + 19𝑦 = −62
27𝑥
Substitute into D
= −81
𝑥 = −3
Protip #1: Letter Equations
Continue this process
until the system is
solved.
2𝑥 − 𝑦 + 6𝑧 = −4 A
6𝑥 + 4𝑦 − 5𝑧 = −7 B
−4𝑥 − 2𝑦 + 5𝑧 = 9 C
−3 + 𝑦 = 1
𝑦=4
Substitute into A
𝑥 = −3
Protip #1: Letter Equations
Continue this process
until the system is
solved.
2𝑥 − 𝑦 + 6𝑧 = −4 A
6𝑥 + 4𝑦 − 5𝑧 = −7 B
−4𝑥 − 2𝑦 + 5𝑧 = 9 C
2 −3 − 4 + 6𝑧 = −4
−3 + 𝑦 = 1
−10 + 6𝑧 = −4
𝑦=4
6𝑧 = 6
𝑧=1
𝑥 = −3
Exercise 3
Solve the system.
𝑥+𝑦−𝑧=2
3𝑥 + 3𝑦 − 3𝑧 = 8
2𝑥 − 𝑦 + 4𝑧 = 7
Exercise 4
Solve the system.
𝑥+𝑦+𝑧=6
𝑥−𝑦+𝑧=6
4𝑥 + 𝑦 + 4𝑧 = 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 𝑎.
Now re-write the other variables in terms of 𝑎.
𝑥+𝑦+𝑧=6
𝑥
+𝑧=6
Let 𝑥 = 𝑎
𝑎+𝑧=6
𝑧 =6−𝑎
𝑎+𝑦+ 6−𝑎 =6
𝑦=0
Protip #2: Multiple Solutions
Finally, use your new expressions to write an
ordered triple. Substitute values in for 𝑎 to get
specific solution points.
𝑥+𝑦+𝑧=6
𝑥
Let 𝑎 = 0:
0, 0, 6
+𝑧=6
𝑥=𝑎
𝑧 =6−𝑎
𝑦=0
Let 𝑎 = 1:
1, 0, 5
𝑎, 0, 6 − 𝑎
Let 𝑎 = −1:
−1, 0, 7
Exercise 5
Solve each system.
1. 3x + y – 2z = 10
6x – 2y + z = -2
x + 4y + 3z = 7
2.
x+y–z=2
2x + 2y – 2z = 6
5x + y – 3z = 8
3.
x+y+z=3
x+y–z=3
2x + 2y + z = 6
Objective 2b
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)
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
Exercise 6
At a carry-out pizza restaurant, an order of 3
slices of pizza, 4 breadsticks, and 2 soft drinks
cost $13.35. A second order of 5 slices of
pizza, 2 breadsticks, and 3 soft drinks cost
$19.50. If four bread sticks and a can of soda
cost $.30 more than a slice of pizza, what is
the cost of each item?
Exercise 7: SAT
If 5 sips + 4 gulps = 1 glass and 13 sips + 7
gulps = 2 glasses, how many sips equal a
gulp?
3.4: Solve a Linear System in Three
Variables
1.
2.
Objectives:
To geometrically
interpret the solution
to a linear system in
three variables
To solve a linear
system in three
variables using
substitution and
elimination
Assignment
• P. 177: 1, 2
• P. 182-185: 1, 2-32
even, 41, 42, 46, 4850
• Multiple Solutions
Worksheet: 1-3, 6, 9,
12
• Extra Credit?
• Print Review 2a
“I’m in 3-D!”