Transcript 3.4: Solve a Linear System in Three Variables

Solve a Linear System in Three Variables
Objectives:
1. To geometrically interpret the solution to a
linear system in three variables
2. To solve a linear system in three variables
using substitution and elimination
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
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 yintercept 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 zintercept 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:
• Macs
• Click the Graphing
tab
• Choose 3D from the
drop down menu
• Type in the equation
and click Graph
Free Graph Paper
http://www1.itlnet.net/
users/smatheson/gr
aphpaper/
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.
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
In Step 1, you’ll have to eliminate the same
variable from 2 different sets of the
equations.
Elimination Method
The system has no solution if you obtain a
contradiction (ex. 0 = 1) while solving the
system.
Elimination Method
The system has infinitely many solutions if
you obtain an identity (ex. 0 = 0) while
solving the system.
Protip #1: Letter Equations
To help you through
the often
labyrinthine
process of solving
a 3-variable
system, letter each
of your equations.
2x – y + 6z = -4
6x + 4y – 5z = -7
-4x – 2y + 5z = 9
A
B
C
Protip #1: Letter Equations
In terms of these
A
2x
–
y
+
6z
=
-4
letters, write a
6x + 4y – 5z = -7 B
simple expression
-4x – 2y + 5z = 9 C
that tells you how
to add/subtract
6x + 4y – 5z = -7
multiples of each B + C
+ -4x – 2y + 5z = 9
equation.
Label the new
2x + 2y
=2
equation with a
1
D
new letter.
x+ y
=1
2
D
E
Protip #1: Letter Equations
Continue this
process until the
system is solved.
2x – y + 6z = -4
6x + 4y – 5z = -7
-4x – 2y + 5z = 9
A
B
C
5 A + 6 B 10x – 5y + 30z = -20
+ 36x + 24y – 30z = -42
46x + 19y
= -62
F
Protip #1: Letter Equations
Continue this
process until the
system is solved.
2x – y + 6z = -4
6x + 4y – 5z = -7
-4x – 2y + 5z = 9
-19 E + F
A
B
C
-19x – 19y = -19
+ 46x + 19y = -42
27x
= -81
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 variable in terms of a.
x+y+z=6
x
+z=6
Let x = a
Then by substitution
in the 2nd equation:
Then by substitution
in the 1st equation:
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 a
specific solution points.
x+y+z=6
x
+z=6
x=a
z=6–a
(a, 0, 6 – a)
y = 0 Let a = 0: (0, 0, 6)
Let a = 1: (1, 0, 5)
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
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?
Substitution Method
If it is convenient, you could use
substitution to help solve a linear system
in three variables.
1. Solve one of the equations for one of the
variables.
2. Substitute the expression from Step 1 into
both of the other equations.
3. Solve the remaining 2 variable system.
Exercise 7: SAT
If 5 sips + 4 gulps = 1 glass and 13 sips + 7
gulps = 2 glasses, how many sips equal a
gulp?