Download MTH55_Lec-14_sec_3-3b_3Var_Sys_Apps

Chabot Mathematics
§3.3b 3-Var
System Apps
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
1
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Review § 3.3
MTH 55
 Any QUESTIONS About
• §3.3a → 3 Variable Linear Systems
 Any QUESTIONS About HomeWork
• §3.3a → HW-10
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Equivalent Systems of Eqns

Operations That Produce Equivalent
Systems of Equations
1. Interchange the position of any two eqns
2. Multiply (Scale) any eqn by a nonzero
constant; i.e.; multiply BOTH sides
3. Add a nonzero multiple of one eqn to
another to affect a Replacment
 A special type of Elimination called
Gaussian Elimination uses these steps
to solve multivariable systems
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Gaussian Elimination
 An algebraic method used to solve
systems in three (or more) variables.
 The original system is transformed to an
equivalent one of the form:
Ax + By + Cz = D
Ey + Fz = G
Hz = K
 The third eqn is solved for z and backsubstitution is used to find y and then x
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Gaussian Elimination
1. Rearrange, or InterChange, the
equations, if necessary, to obtain the
Largest (in absolute value) x-term
coefficient in the first equation. The
Coefficient of this large x-term is called
the leading-coefficient or pivot-value.
2. By adding appropriate multiples of the
other equations, eliminate any x-terms
from the second and third equations
Chabot College Mathematics
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Bruce Mayer, PE
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Gaussian Elimination
2. (cont.) Rearrange the resulting two
equations obtain an the Largest (in
absolute value) y-term coefficient in
the second equation.
3. If necessary by adding appropriate
multiple of the third equation from Step
2, eliminate any y-term from the third
equation. Solve the resulting equation
for z.
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Bruce Mayer, PE
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Gaussian Elimination
4. Back-substitute the values of z from
Steps 3 into one of the equations in
Step 3 that contain only y and z, and
solve for y.
5. Back-substitute the values of y and z
from Steps 3 and 4 in any equation
containing x, y, and z, and solve for x
6. Write the solution set (Soln Triple)
7. Check soln in the original equations
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Gaussian Elim
 Solve System by
Gaussian Elim
1
2
3
6 x  3 y  4 z  41
12 x  5 y  7 z  26
 5 x  2 y  6 z  14
 INTERCHANGE, or
Swap, positions of
Eqns (1) & (2) to get
largest x-coefficient
in the top equation
Chabot College Mathematics
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1
2
3
12 x  5 y  7 z  26
6 x  3 y  4 z  41
 5 x  2 y  6 z  14
 Next SCALE by
using Eqn (1) as the
PIVOT To Multiply
• (2) by 12/6
• (3) by 12/[−5]
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Gaussian Elim
 The Scaling
Operation
1
12 x  5 y  7 z  26
2
12
6 x  3 y  4 z  41
6
12
 5 x  2 y  6 z  14
5
3
1
2
3
12 x  5 y  7 z  26
12 x  6 y  8 z  82
12 x  4.8 y  14.4 z  33.6
Chabot College Mathematics
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 Note that the 1st
Coeffiecient in the
Pivot Eqn is Called
the Pivot Value
• The Pivot is used to
SCALE the Eqns
Below it
 Next Apply
REPLACEMENT by
Subtracting Eqs
• (2) – (1)
• (3) – (1)
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Gaussian Elim
 The Replacement
Operation Yields
1
2
3
12 x  5 y  7 z  26
0 x  11y  15 z  108
0 x  9.8 y  7.4 z  7.6
Or
1
2
3
12 x  5 y  7 z
 11y  15 z  108
 9.8 y  7.4 z  7.6
Chabot College Mathematics
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 26
 Note that the
x-variable has been
ELIMINATED below
the Pivot Row
• Next Eliminate in
the “y” Column
 We can use for the
y-Pivot either of −11
or −9.8
• For the best numerical
accuracy choose the
LARGEST pivot
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Gaussian Elim
 Our Reduced Sys
1
2
3
 11y  15 z  108
1
2
 9.8 y  7.4 z  7.6
3
12 x  5 y  7 z
 26
12 x  5 y  7 z
 26
 11y  15 z  108
 11
 9.8 y  7.4 z  7.6
 9.8
Or
 Since | −11| > | −9.8|
we do NOT need to
1 12 x  5 y  7 z
interchange (2)↔(3)
 Scale by Pivot
against Eqn-(3)
Chabot College Mathematics
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2
3
 11y  15 z
 26
 108
 11y  8.306 z  8.531
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Gaussian Elim
 Perform
Replacement by
Subtracting (3) – (2)
1
2
3
12 x  5 y  7 z
 11y  15 z
 26
 Find y & x by BACK
SUBSTITUTION
 108
 From Eqn (2)
 23.306 z  116.531
 Now Easily Find
the Value of z from
Eqn (3)
z  116.531 23.306  5
Chabot College Mathematics
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 The Hard Part is
DONE
108  15 z 108  75
y

 11
 11
y  33  11  3
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Gaussian Elim
 BackSub into (1)
12 x  5 y  7 z  26
1
2
3
6 x  3 y  4 z  41
12 x  5 y  7 z  26
7 z  5 y  26
x
 5 x  2 y  6 z  14
12
35  15  26 24
x

2
12
12
 x=2
 Thus the Solution
Set for Our Linear
System
Chabot College Mathematics
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 y = −3
 z=5
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Fuel Useage Rates
 A food service distributor conducted a
study to predict fuel usage for new delivery
routes, for a particular truck. Use the chart
to find the rates of fuel consumption in rush
hour traffic, city traffic, and on the highway.
Rush Hour
Hours
City Traffic
Hours
Highway
Hours
Total Fuel
Used (gal)
Week 1
2
9
3
15
Week 2
7
8
3
24
Week 3
6
18
6
34
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Fuel Usage Rates
 Familiarize: The Fuel Use Calc’d by
the RATE Eqn:
Quantity = (Rate)·(Time) = (Time)·(Rate)
 In this Case the Rate Eqn
(UseTime)·(UseRate) → (hr)·(Gal/hr)
• So LET:
– x ≡ Fuel Use Rate (Gal/hr) in Rush Hr Traffic
– y ≡ Fuel Use Rate (Gal/hr) in City Traffic
– z ≡ Fuel Use Rate (Gal/hr) in HiWay Traffic
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Fuel Usage Rates
 Translate: Use Data Table
Rush Hour
Gallons
Week 1
2x
Week 2
7x
Week 3
6x
City Traffic
Gallons



9y
8y
18y



Highway
Gallons
3z
3z
6z



Total Fuel
Used (gal)
15
24
34
2 x  9 y  3z  15 1
 Thus the
System of 7 x  8 y  3z  24 2
Equations 6 x  18 y  6 z  34 3
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Fuel Usage Rates
 Solve by Guassian Elimination:
Interchange to place largest
x-Coefficient on top
7x 
8y
 3z  24
2x 
9y
 3z  15
6 x  18 y  6 z  34
2
1
3
 Scale
• Multiply Eqn (1) by −7/2
• Multiply Eqn (2) by −7/6
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Fuel Usage Rates
 The new, equivalent system
8y

3z

24
63
21
105
 7x  
y   z  
2
2
2
119
 7 x   21 y   7 z  
3
7x

2
4
5
 Make Replacement by Adding Eqns
• {Eqn (2)} + {Eqn (4)}
• {Eqn (2)} + {Eqn (5)}
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Fuel Usage Rates
 The new, equivalent system
7x 
8y

3z
 24
47
15
57
 
y   z  
2
2
2
47
  13 y   4 z  
3
2
6
7 
 Notice how x has been Eliminated
below the top Eqn
 Clear Fractions by multiplying
Eqn (6) by −2
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Fuel Usage Rates
 The new, equivalent system
7x 
8y

3z


47 y

15 z

24
57
47
  13 y   4 z  
3
2
8
7 
 Now Scale Eqn (7) by the factor 47/13
47 
47  
188
2209 
 13 y  4 z      47 y 
z

13 
3 
13
39 
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Fuel Usage Rates
 The new, equivalent system
7x 
8y


47 y
3z

2
8
9
24

15 z

57
188
2209
  47 y  
z  
13
39
 Replace by Adding: {Eqn (8)}+{Eqn (9)}
7x 
8y

3z
 15
 47 y  15 z  57
7
14


z 
13
39
Chabot College Mathematics
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2
8
10
Bruce Mayer, PE
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Example  Fuel Usage Rates
 Solve Eqn (10) for z
14 
14 13 2 1
2
 13   7
  z    z      z 
39 
39 7 3 1
3
 7  13
 BackSub z = 2/3 into Eqn (8) to find y
2
47 y  15   57  47 y  10  57 
3
47
47 y  57  10  47  y 
 y 1
47
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Fuel Usage Rates
 BackSub z = 2/3 and y = 1 into Eqn (2)
to find x 7 x  81  3 2   24  7 x  8  2  24
3
7 x  24  10  7 x  14  x  2
 Chk x = 2, y = 1 & z = 2/3 in Original Eqns
72 
81
22 
91
62  181
Chabot College Mathematics
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2
 3   24
3
2
 3   15
3
2
 6   34
3
2
1
3
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Fuel Usage Rates
 Continue Chk of x = 2, y = 1 & z = 2/3
14 
8
 2  24
24  24

9
 2  15 
15  15
4
12  18  4  34
34  34

 State: The Delivery Truck Uses
• 2 Gallons per Hour in Rush Hour traffic
• 1 Gallons per Hour in City traffic
• 2/3 Gallons per Hour in HighWay traffic
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Theater Concessions
 At a movie theatre, Kara buys one
popcorn, two drinks and 2 candy bars,
all for $12. Jaypearl buys two popcorns,
three drinks, and one candy bar for $17.
Nyusha buys one popcorn, one drink
and three candy bars for $11. Find the
individual cost of one popcorn, one
drink and one candy bar
Chabot College Mathematics
25
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Theater Concessions
 Familiarize: Allow UNITS to guide us to
the Total Cost Equation:


$
 $ 
 $ 
  No. CBars 
$Cost  No. Drinks 
  No. PopCorn 

 Drink 
 CBar 
 PopCorn 
 This Eqn does yield the Total Cost as
required. Thus LET
• c ≡ The UnitCost of Candy Bars
• d ≡ The UnitCost of Soft Drinks
• p ≡ The UnitCost of PopCorn Buckets
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Theater Concessions
 Translate: Translate the Problem
Description, Cost Eqn, and Variable
Definitions into a 3 Equation System


$
 $ 
 $ 
  No. Drinks 
No. PopCorn 
  No. CBars 
  $Cost
 Drink 
 CBar 
 PopCorn 
p
 2d
2 p  3d
p

Chabot College Mathematics
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d
 2c  $12

Kara
 $17
Jaypearl
 3c  $11
Nyusha
c
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Theater Concessions
 Solve by Guassian Elim: Interchange to
place largest x-Coefficient on top
2 p  3d

c
 17
p
 2d
 2c  12
p

 3c  11
d
1
2
3
 Scale
• Multiply Eqn (2) by −2
• Multiply Eqn (3) by −2
Chabot College Mathematics
28
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Theater Concessions
 The new, equivalent system
2p

3d

c

17
 2 p   4d
  4c   24
 2 p   2d
  6c   22
1
4
5
 Make Replacement by Adding Eqns
• {Eqn (1)} + {Eqn (4)}
• {Eqn (1)} + {Eqn (5)}
Chabot College Mathematics
29
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Theater Concessions
 The new, equivalent system
2p 
3d

 17
c
 d
  3c   7

  5c   5
d
1
6
7
 p Eliminated below the top Eqn
 Elim d by Adding {Eqn (6)} + {Eqn (7)
2p 

17
 d
  3c 
7

  8c   12
Chabot College Mathematics
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3d

c
1
6
8
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Theater Concessions
 Solve Eqn (8) for c
 8c  12
8
3
 c   c  $1.50
2
 BackSub c = 3/2 into Eqn (6) to find d
 3  14
d  3c  7  d  3   
2 2
14 9 5
d     d  $2.50
2 2 2
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Theater Concessions
 BackSub c = 3/2 & d = 5/2 into (1) find p
15 3
5 3
2 p  3      17  2 p    17
2 2
2 2
15 3 34
34 18
 2p   
 2p 

2 2 2
2 2
16
1
 2p 
 8  2 p  8 
2
2
8
p   p  $4.00
2
 The Chk is left for you to do Later
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Theater Concessions
 A Quick Summary
3
c
2
5
d
2
p4
 State: The Cost for the Movie Theater
Concessions:
• $4.00 for a Tub of PopCorn
• $2.50 for a Soft Drink
• $1.50 for a Candy Bar
Chabot College Mathematics
33
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Missing Term
 In triangle ABC, the measure of angle B
is three times the measure of angle A.
The measure of angle C is 60° greater
than twice the measure of angle A.
Find the measure of each angle.
 Familiarize: Make a
C
sketch and label the
A
B
angles A, B, and C.
Recall that the measures of the angles
in any triangle add to 180°.
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Bruce Mayer, PE
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Example  Missing Term
 Translate: This geometric fact about
triangles provides
A + B + C = 180.
one equation:
 Translate Relationship Statements
Angle B is three times the measure of angle A.
B
=
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35
3A
Bruce Mayer, PE
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Example  Missing Term
 Translate Relationship Statements
Angle C is 60o greater than twice the measure of A
C
=
 Translation
Produces the
3-Equation
System
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36
60 + 2A
A  B  C  180
B  3A
C  60  2 A
Bruce Mayer, PE
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Example  Missing Term
 Since this System has
A

B

C

180
Missing Terms in two
B  3A
of the Equations,
Substitution is faster
C  60  2 A
than Elimination
 Sub into Top Eqn A  3A  60  2 A  180
• B = 3A
• C = 60+2A
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60  6 A  180
6 A  120
A  20
Bruce Mayer, PE
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Example  Missing Term
 BackSub A = 20° into the other eqns
B  3A
C  60  2 A
 3(20)
 60  2(20)
 60
 100
 Check → 20° + 60° + 100° = 180° 
 State: The angles in the triangle
measure 20°, 60°, and 100°
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  Missing Term
 In triangle ABC, the measure of angle B
is three times the measure of angle A.
The measure of angle C is 60° greater
than twice the measure of angle A.
Chabot College Mathematics
39
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  CAT Scan
 Let A, B, and C
be three grid
cells as shown
 A CAT scanner
reports the
data on the
following slide
for a patient
named Satveer
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  CAT Scan

Linear Attenuation Units For the Scan
i.
Beam 1 is weakened by 0.80 units as it
passes through grid cells A and B.
ii. Beam 2 is weakened by 0.55 units as it
passes through grid cells A and C.
iii. Beam 3 is weakened by 0.65 units as it
passes through grid cells B and C

Using the following table, determine
which grid cells contain each of the
type of tissue listed
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  CAT Scan
 CAT Scan Tissue-Type Ranges
LAU  Linear Attenuation Units
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  CAT Scan
 Familiarize: Suppose grid cell A
weakens the beam by x units, grid cell B
weakens the beam by y units, and grid
cell C weakens the beam by z units.
 Thus LET:
• x ≡ The Cell-A Attenuation
• y ≡ The Cell-B Attenuation
• z ≡ The Cell-C Attenuation
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  CAT Scan

Translate: the Attenuation Data
i.
Beam 1 is weakened by 0.80 units as it
passes through grid cells A and B.
x + y = 0.80
ii. Beam 2 is weakened by 0.55 units as it
passes through grid cells A and C
x + z = 0.55
iii. Beam 3 is weakened by 0.65 units as it
passes through grid cells B and C
+ z = 0.65
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  CAT Scan
 Thus the
Equation
System
 0.80
x  y

 z  0.55
x

y  z  0.65

(1)
(2)
(3)
 Even with Missing Terms Elimination is
sometimes a good solution method
x  y
 0.80
 Add −1 times
Equation (1)
x
 z  0.55
(2)
to Equation (2)
y  z  0.25
(4)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  CAT Scan
 0.80 (1)
 The Replacement  x  y

Operation Produces  y  z  0.25 (4)
the Equivalent System

y  z  0.65 (3)

 Add Equation (4) to Equation (3) to get
 0.80 (1)
x  y

 y  z  0.25 (4)

2z  0.40 (5)

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46
0.40
 z
2
Or z  0.20
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  CAT Scan
 Back-substitute
z = 0.20 into
Eqn (4) to Obtain
 Back-substitute
y = 0.45 into
Eqn (1) and
solve for x
Chabot College Mathematics
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y  0.20  0.25
y  0.45
y  0.45
x  y  0.80
x  0.45  0.80
x  0.35
Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Example  CAT Scan
 Summarizing
Results
 Recall
Tissue-Type
Table
x  0.35 y  0.45 z  0.20
 Thus Conclude
• Cell A contains tumorous tissue (x = 0.35)
• Cell B contains a bone (y = 0.45)
• Cell C contains healthy tissue (z = 0.20)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
WhiteBoard Work
 Problems From §3.3 Exercise Set
• 46

An
Inconsistent
System
WHY?
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Bruce Mayer, PE
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All Done for Today
Carl
Friedrich
Gauss
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-14_sec_3-3a_3Var_Sys_Apps.ppt