ENM 503 Block 2Lesson 6 –Linear Systems - Methods

straight lines and other non-crooked

objects

Narrator: Charles Ebeling 1

Block 2 Linear Systems

Lesson 6 The Methods of Linear Systems

Lesson 7 Matrix Algebra Lesson 8 Modeling Linear Systems

Did you know: The shortest distance between two points is nota straight-line when traveling on the surface of a sphere?

Just 3 easy lessons!

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Quadrant II Quadrant I

Quadrant III Quadrant IV

Cartesian Coordinates & Straight Lines

x

y

| | | | | | | | |

-4 -3 -2 -1 1 2 3 4

3 __

2 __ 1 __

-1 __

-2 __

-3 __

-4 __

-5 __

A(x1,y1)

B(x2,y2)

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Equations of a straight line - 1

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.

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Equations of a straight line - 2

Point-slope Formula: Given a point A:(x1,y1) and slope m,then

1 1

1 1

1 1

1 1

or

Therefore

or ( )

y b mx

b y mx

y y mx mx

y y m x x

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Equations of a straight line - 3

Two-point Formula: Given to points A (x1,y1) and B (x2,y2),then

2 1

2 1

1 1

2 11 1

2 1

Slope

since ( )

and

y ym

x x

y y m x x

y yy y x x

x x

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Quadrant II Quadrant I

Quadrant III Quadrant IV

Intercept Form

x

y

| | | | | | | | |

-4 -3 -2 -1 1 2 3 4

3 __

2 __ 1 __

-1 __

-2 __

-3 __

-4 __

-5 __

2x – 4y = 8x/4 + y/-2 = 1y = 0; x = 4x = 0; y = -2

1x y

a b

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A Linear Example - 1

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.

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A Linear Model - 2

Let x = number of miles and y = delivery costThen (x1,y1) = (500, 54) and (x2,y2) = (1000,

66) using the 2-point formula:

66 5454 500

1000 50054 .024 500 .024 42

( ); (750) .024 750 42 60

y x

y x x

y f x f

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Systems of Linear Equations

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

...

...

...

n n

n n

m m mn n m

a x a x a x b

a x a x a x b

a x a x a x b

right hand sidecoefficients

n variables and m equations

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Systems of Linear Equations

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 be1. a unique solution (consistent and independent)2. an infinite number of solutions (consistent and

dependent)3. no solution (inconsistent)

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Case I. (m < n) One equation and two variables

x

y

Infinite number of solution

Any (x,y) satisfyingAx + By = C

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Case II. (m > n) Three equations and two variables

x

y

no solutionx

y

unique solution

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Case III. (m=n) Two equations and two variables

x

y

unique solution x

y

no solution (parallel lines)

x

y

infinite solutions (lines coincide)

y + x = 102y + 2x = 20

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2 Eqs, 2 variables – unique solution

2x + 3y = 104x - 2y = 12

y = 10/3 – (2/3)x

4x - 2[10/3 – (2/3)x] = 124x –(20/3) + (4/3)x = 12(16/3)x = 12 + 20/3 = 56/3

x = (3/16) (56/3) = 56/16 = 3.5y = 10/3 – (2/3) (3.5) = 1.0

method of substitution

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2 Eqs, 2 variables – no solution

2x - 3y = 104x - 6y = 12

y = -10/3 + (2/3)x = -3.33 + (2/3)xy = -12/6 + (4/6)x = -2.0 + (2/3)x

same slopeparallel lines

rewrite equations in slope-intercept form

an inconsistency, no solution exists

4x - 6 [-10/3 + (2/3)x ] = 124x + 20 – 4x 12

using substitution:

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2 Eqs, 2 variables – infinite solution

2x – 3y = 125x – 7.5y = 30

y = -12/3 + (2/3)xsubstituting:5x – 7.5 [-12/3 + (2/3)x] = 305x - (15/2)(-12/3) - (15/2) (2/3)x = 305x + 30 - 5x = 30

30 = 30; an identity

Eq2 = 2.5 x Eq1

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2 Eqs, 2 variables – unique solution alternate method

2x + 3y = 104x - 2y = 12

(-2)2x + (-2)3y = (-2)10 4x - 2y = 12 -8y = -8

y = 1from the first equation: 2x +3(1) = 10

x = (10 – 3)/2 = 3.5

multiply the first equationby -2 then add to the second equation:

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Solution by Elementary Operations

Given a n x n system of linear equations,1. any 2 equations may be interchanged2. any equation may be multiplied by a

constant3. a multiple of any equation may be added

to another equation replacing the equationCount them,

there are only three! Imagine

that.

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Solution by Elementary Operations

Given a n x n system of linear equations, any 2 equations may be interchanged

1 2 3

1 2 3

1 2 3

2 3 4 20

6 3 71

2 55

x x x

x x x

x x x

1 2 3

1 2 3

1 2 3

6 3 71

2 3 4 20

2 55

x x x

x x x

x x x

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Solution by Elementary Operations

Given a n x n system of linear equations, any equation may be multiplied by a

constant1 2 32 3 4 20x x x

1 2 3

1 2 3

1 1 1 12 3 4 20

2 2 2 2

32 10

2

x x x

x x x

1 2 3

1 2 3

1 2 3

2 3 4 20

6 3 71

2 55

x x x

x x x

x x x

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Solution by Elementary OperationsGiven a n x n system of linear equations,

a multiple of any equation may be added to another equation replacing the equation

1 2 3

1 2 3

32 10

22 55

x x x

x x x

1 2

52 65

2x x

+

1 2 3

1 2 3

1 2 3

2 3 4 20

6 3 71

2 55

x x x

x x x

x x x

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Equivalent Systems of Equations

1 2 3

1 2 3

1 2

6 3 71

32 10

25

2 652

x x x

x x x

x x

1 2 3

1 2 3

1 2 3

2 3 4 20

6 3 71

2 55

x x x

x x x

x x x

Solution21.0655749.147541

12.393443

Solution21.0655749.147541

12.393443

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Solving systems of eqs. by elementary operations

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

'1 1

'2 2

'

...

...

...

n n

n n

m m mn n m

n m

a x a x a x b

a x a x a x b

a x a x a x b

x b

x b

x b

elementaryoperations

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Let’s solve a system of equations using elementary operations!

312 2 3 2 2

3 6 4 3 6 4

8 4 3 8 8 4 3 8

31

221

0 4 12

0 4 9 0

x y zx y z

x y z x y z

x y z x y z

x y z

y z

y z

1. 2.multiply first row by ½1/2 R1 R’1

multiply first row by -3 and add to 2nd row-3R1 + R2 R’2

multiply first row by -8 and add to 3rd row-8R1 + R3 R’3

3.

Work x variable

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keep solving…3 3

1 12 221 21 1

0 4 1 02 8 4

0 4 9 0 0 4 9 0

9 50

8 421 1

08 43

0 0 12

x y z x y z

y z y z

y z y z

x z

y z

z

multiply 2nd row by -1/4-1/4 R2 R’2

multiply 2nd row by -1 and add to 1st row- R2 + R1 R’1

multiply 2nd row by 4 and add to 3rd row4 R2 + R3 R’3

Work y variable

1. 2.

3.

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not much longer now…9 5 9 5

0 08 4 8 421 1 21 1

0 08 4 8 43 2

0 0 1 0 02 3

10 0

23

0 022

0 03

x z x z

y z y z

z z

x

y

z

multiply 3rd row by 2/32/3 R3 R’3

multiply 3rd row by 9/8 and add to1st row9/8 R3 + R1 R’1

multiply 3rd row by -21/8 and add to 2nd row-21/8 R3 + R2 R’2

Work z variable

x = 1/2, y = 3/2, z = -2/3

1. 2.

3.

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Excel with solver

2 2 3 2

3 6 4

8 4 3 8

x y z

x y z

x y z

There must be an easier way to do

this!

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XYZ makes A, B, and C’s 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 M1 M2 M3A 1 2 2B 2 8 3C 2 1 4

Available hrs / month

200 525 350

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How many A, B, and C’s?

A + 2B + 2C = 2002A + 8B + C = 5252A + 3B + 4C = 350solution:A = 50, B = 50, C = 25

Product M1 M2 M3A 1 2 2B 2 8 3C 2 1 4

Available hrs / month

200 525 350

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Systems of Linear Inequalities

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 = -286x+50 > 20 6x > -30 or x > -5-4x – 20 > -24 -4x > -44 or x < 11Therefore: -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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The feasible region

x

y200

100

100 200 300

3x + 2y = 300

boundary equation

feasible region

3x + 2y 300; x 0; y 0

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More of Makit 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

The feasible region

x

y200

100

100 200 300

2x + 5y = 500 hours

boundary equation

feasible region

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Conclusion – pick one…

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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