straight lines and other non-crooked objects
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Narrator: Charles Ebeling




Lesson 6 The Methods of Linear Systems
Lesson 7 Matrix Algebra
Lesson 8 Modeling Linear Systems
Just 3 easy lessons!
Did you know: The shortest distance between two points is not a straight-line when traveling on the surface of a sphere?
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|
-4
Cartesian Coordinates & Straight
Lines y
Quadrant II Quadrant I
3 __
B(x
2
,y
2
)
2 __
A(x
1
,y
1
)
1 __
| | | |
-1 __
| | | | x
-3 -2 -1 1 2 3 4
-2 __
Quadrant III
-3 __
-4 __
-5 __
Quadrant IV
3

General linear equation : Ax + By + C = 0
Solving for Y as the dependent variable: y = -C/B – Ax/B let b = -C/B and m = -A/B, then
Slope – Intercept form: y = mx + b where b is the y-intercept (i.e. when x = 0, y = b) and m is the slope.
4

Point-slope Formula : Given a point A:(x
1
,y
1
) and slope m, then y
1
 
1 or b
 
1 mx
1
Therefore y
 
1 mx
1
 mx or y
 
1
(
 x
1
)
5

Two-point Formula : Given to points A (x
1
,y
1
) and B (x
2
,y
2
), then
Slope y
2
 y
1 x
2
 x
1 since and y y
 y
1

(
 x
1
)
 y
1
 y
2
 y
1 x
2
 x
1
 x
 x
1

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|
-4
Intercept Form
Quadrant II x y a b y
1
2x – 4y = 8
3 __ x/4 + y/-2 = 1
2 __ y = 0; x = 4 x = 0; y = -2
1 __
| | | |
-1 __
| |
-3 -2 -1 1 2
-2 __
Quadrant III
-3 __
-4 __
-5 __
|
3
Quadrant I
|
4 x
Quadrant IV
7

UPS charges $54 to deliver a package of a specified weight 500 miles and charges $66 to deliver the same package 1000 miles.
Assuming that delivery costs are linear with distance, derive a model that will provide delivery costs (for the specified weight) as a function of distance and determine the delivery cost if the distance is $750 miles.
8

Let x = number of miles and y = delivery cost
Then (x
1
,y
1
) = (500, 54) and (x
2
,y
2
) = (1000, 66) using the 2-point formula: y

54
  x

500
 y
   x

500
 
.024
x

42 y

( ); (750)
   
42

60
9

n variables and m equations coefficients a x
11 1
 a x
12 2 a x
21 1
 a x
22 2 a x
1 n n
 b
1 a x
2 n n
 b
2 right hand side a x m 1 1
 a x m 2 2 a x mn n
 b m
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n variables and m equations
Case I. if m < n, there are an infinite number of solutions
(arbitrarily assign values to n-m variables and solve for the remaining)
Case II. If m > n, there may be no solutions (null set)
Case III. If m = n, there may be
1. a unique solution (consistent and independent)
2. an infinite number of solutions (consistent and dependent)
3. no solution (inconsistent)
11

Case I. (m < n)
One equation and two variables y x
Infinite number of solution
Any (x,y) satisfying
Ax + By = C
12

y
Case II. (m > n) Three equations and two variables y no solution x unique solution x
13

y
Case III. (m=n) Two equations and two variables y unique solution y x x no solution (parallel lines) y + x = 10
2y + 2x = 20 x infinite solutions (lines coincide)
14

2 Eqs, 2 variables – unique solution
2x + 3y = 10
4x - 2y = 12 method of substitution y = 10/3 – (2/3)x
4x - 2[10/3 – (2/3)x] = 12
4x –(20/3) + (4/3)x = 12
(16/3)x = 12 + 20/3 = 56/3 x = (3/16) (56/3) = 56/16 = 3.5
y = 10/3 – (2/3) (3.5) = 1.0
15

2x - 3y = 10
4x - 6y = 12 rewrite equations in slope-intercept form same slope parallel lines y = -10/3 + (2/3)x = -3.33 + (2/3)x y = -12/6 + (4/6)x = -2.0 + (2/3)x using substitution:
4x - 6 [-10/3 + (2/3)x ] = 12
4x + 20 – 4x 
12 an inconsistency, no solution exists
16

2 Eqs, 2 variables – infinite solution
2x – 3y = 12
5x – 7.5y = 30
Eq2 = 2.5 x Eq1 y = -12/3 + (2/3)x substituting:
5x – 7.5 [-12/3 + (2/3)x] = 30
5x - (15/2)(-12/3) - (15/2) (2/3)x = 30
5x + 30 - 5x = 30
30 = 30; an identity
17

2 Eqs, 2 variables – unique solution alternate method multiply the first equation by -2 then add to the second equation:
2x + 3y = 10
4x - 2y = 12
(-2)2x + (-2)3y = (-2)10
4x - 2y = 12
-8y = -8 y = 1 from the first equation: 2x +3(1) = 10 x = (10 – 3)/2 = 3.5
18

Solution by Elementary Operations
Given a n x n system of linear equations,
1.
2.
any 2 equations may be interchanged any equation may be multiplied by a constant
3.
a multiple of any equation may be added to another equation replacing the equation
Count them, there are only three!
Imagine that.
19

Solution by Elementary Operations
Given a n x n system of linear equations,
 any 2 equations may be interchanged
2 x
1

3 x
2

4 x
3

20 x
1
6 x
2

3 x
3

71 x
1 x
2
2 x
3

55 x
1
6 x
2

3 x
3

71
2 x
1

3 x
2

4 x
3

20 x
1 x
2
2 x
3

55
20

Solution by Elementary Operations
Given a n x n system of linear equations,
2 x
1

 any equation may be multiplied by a constant
3 x
2

4 x
3

20
1
 
2 x
1

3 x
2 x
1

3
2 x
2

2 x
3

10

20
2 x
1

3 x
2

4 x
3

20 x
1
6 x
2

3 x
3

71 x
1 x
2
2 x
3

55
4 x
3

21

Solution by Elementary Operations
Given a n x n system of linear equations,
 a multiple of any equation may be added to another equation replacing the equation
2 x
1

3 x
2

4 x
3

20 x
1
6 x
2

3 x
3

71 x
1 x
2
2 x
3

55
+ x
1

3
2 x
2

2 x
3

10 x
1 x
2
2 x
3

55
2

5 x x
1
2
2

65
22

Equivalent Systems of Equations
2 x
1

3 x
2

4 x
3

20 x
1
6 x
2

3 x
3

71 x
1 x
2
2 x
3

55 x
1
6 x
2

3 x
3

71 x
1

3
2 x
2

2 x
3

10
2 x
1

5
2 x
2

65
Solution
21.065574
9.147541
12.393443
Solution
21.065574
9.147541
12.393443
23

Solving systems of eqs. by elementary operations



 a x
11 1
 a x
12 2 a x
21 1
 a x
22 2 a x m 1 1
 a x m 2 2 elementary operations a x
1 n n a x
2 n n

 b b
1
2
 


 x
1 a x mn n
 b m x
2
 b
1
'
 b
2
' x n
 b
' m
24

1.
Let’s solve a system of equations using elementary operations!

2 x

2 y

3 z

2
 3 x y 6 z 4
8 x

4 y

3 z

8
3.

 x y
3
2 z 1



 y
21
2 z

 y

9 z

0
1
2.




3
3 x y z
2 x
 
6 z

1
4
8 x

4 y

3 z

8
Work x variable multiply first row by ½
1/2 R
1
 R’
1 multiply first row by -3 and add to 2 nd
-3R
1
+ R
2
 R’
2 row multiply first row by -8 and add to 3 rd
-8R
1
+ R
3
 R’
3 row
25

keep solving…
Work y variable
1.
 x
 
3
2 z

1



3.
0 4 y

21
2 z

 y

9 z

0
1
2.





 x
 
3
2 z

0
 y y

21
8
9 z z

1

0

1
4

 x

 y
 0 0

9 z

5
8 4
21

1 z

8 4
2 z 1 multiply 2 nd
-1/4 R
2
 R’
2 row by -1/4 multiply 2 nd row by -1 and add to 1 st row
- R
2
+ R
1
 R’
1 multiply 2 nd
4 R
2
+ R
3 row by 4 and add to 3 rd
 R’
3 row
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not much longer now…
1.

 x

9 z

5
8 4




0 y
21
8 z


4
0 0
3
2 z
 
1
1
1
Work z variable
2.

 x

9 z

5
8 4




0 y
21
0 0 z
8 z

2
3

1
4 multiply 3
2/3 R
3 rd
 R’ row by 2/3
3 multiply 3 rd row by 9/8 and add to1 st row
9/8 R
3
+ R
1
 R’
1

 x
3.

 y 0
2
3
2
2 multiply 3
-21/8 R
3 rd row by -21/8 and add to 2 nd
+ R
2
 R’
2 row
 0 0 z
3 x = 1/2, y = 3/2, z = -2/3
27

Excel with solver
There must be an easier way to do this!

2 x

2 y

3 z

2
 3 x y 6 z 4
8 x 4 y 3 z 8
28


The XYZ Company makes three products: A, B, and C. Each product requires processing on three machines: M1, M2, and
M3. Based upon the data in the following table, how many units of each product should manufactured in one month if all
3 machines are to be fully utilized?
Machine hours per unit
Product
A
B
C
Available hrs / month
M1
1
2
2
200
M2
2
8
1
525
M3
2
3
4
350
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Product
A
B
C
Available hrs / month
M1
1
2
2
200
M2
2
8
1
525
M3
2
3
4
350
A + 2B + 2C = 200
2A + 8B + C = 525
2A + 3B + 4C = 350 solution:
A = 50, B = 50, C = 25
30

Find the intersection of the following sets:
A = {x| 3x-5 < 40}, B = {x| -2x+10 < 66},
C = {x| 6x+50 > 20}, D = {x| -4x – 20 > -24} solution:
3x-5 < 40  3x < 45 or x < 15
-2x+10 < 66  -2x < 56 or x > 56/-2 = -28
6x+50 > 20  6x > -30 or x > -5
-4x – 20 > -24  -4x > -44 or x < 11
Therefore : -5 < x < 11
|
-15
|
-10 -5
| |
0 5
| |
10
|
15 31

The Makit Manufacturing Company
The Makit Manufacturing Company produces two products A and B. Each unit of product A requires 3 hours of machine time and Each unit of product B requires 2 hours of machine time. There are 300 hours of machine time available this month for production.
Let x = the number of units of A produced and y = number of units of B produced this month.
Then 3x + 2y = 300 provides the allowable combinations of X and Y if the machine is at full capacity.
and 3x + 2y  300 represents all feasible production levels.
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y
200
3x + 2y = 300 boundary equation
100
3x + 2y

300; x

0; y

0 feasible region
100 200 300 x
33



In addition to machine time, each unit of product A requires 2 hours of manual assembly time and each unit of product B requires 5 hours of assembly time.
There are 500 labor hours of assembly time available.
Therefore: 2x + 5y  500 hours
Product A assembly
34

y
200
100 2x + 5y = 500 hours boundary equation feasible region
100 200 300 x
35







Life is good
It doesn’t get any better than this
I can solve systems of linear equations
The professor is our friend
Being an ENM student is the (pinnacle)
(nadir) of my career
Solving equations is the best thing in the whole world
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