Slide 5-1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Systems of Equationsand Matrices

Chapter 5


5.1Systems of Equations in

Two Variables Solve a system of two linear equations in two variables

by graphing. Solve a system of two linear equations in two variables

using the substitution and the elimination methods. Use systems of two linear equations to solve applied

problems.


Systems of Equations

A system of equations is composed of two or more equations considered simultaneously.

Example: 5x y = 5

4x y = 3

This is a system of two linear equations in two variables. The solution set of this system consists of all ordered pairs that make both equations true. The ordered pair (2, 5) is a solution of this system.


Solving Systems of Equations Graphically

When we graph a system of linear equations, each point at which the equations intersect is a solution of both equations and therefore a solution of the system of equations.

Let’s solve the previous system graphically.

5x y = 5 4x y = 3

Solution: We see that the graph

intersects at the single point (2, 5), so this is the solution of the system of equations.


Systems of Equations

If a system of equations has at least one solution, it is consistent. If the system has no solutions, it is inconsistent.

If a system of two linear equations in two variables has an infinite number of solutions, the equations are dependent. Otherwise, they are independent.


Illustration of Graphs

Graphs of linear equations may be related to each other in one of three ways.


Substitution Method

The substitution method is a technique that gives accurate results when solving systems of equations. It is most often used when a variable is alone on one side of an equation or when it is easy to solve for a variable. One equation is used to express one variable in terms of the other, then it is substituted in the other equation.

Example: Let’s use this method to solve the previous system. 5x y = 5 4x y = 3


Solution

Solve the first equation for y.

5x y = 5

y = 5x 5

Then we substitute 5x 5 for y in the second equation to give an equation in one variable.

4x (5x 5) = 3

4x 5x + 5 = 3

x = 2

Now we use back-substitution and substitute 2 for x in either original equation.

4x y = 3

4(2) y = 3

8 y = 3

y = 5

We find the solution to the system of equations to be (2, 5), once again.


Elimination Method

Using the elimination method, we eliminate one variable by adding the two equations. If the coefficients of a variable are opposites, that variable can be eliminated by simply adding the original equations. If the coefficients are not opposites, it is necessary to multiply one or both equations by suitable constants, before we add.


Example

Solve the system using the elimination method. 6x + 2y = 4 10x + 7y = 8 If we multiply the first equation by 5 and the second equation by 3,

we will be able to eliminate the x variable.

30x + 10y = 20 Substituting: 6x + 2y = 4 30x 21y = 24 6x + 2(4) = 4 11y = 44 6x 8 = 4 y = 4 6x = 12 The solution is (2, 4). x = 2


More Examples

Solve the system.

x 3y = 9 (1)

2x 6y = 3 (2)

Solution:

2x + 6y = 18 Mult. (1) by 2 2x 6y = 3

0 = 21

There are no values of x and y in which 0 = 21. So this system has no solution. The graphs of the equations are of parallel lines.

Solve the system. 9x + 6y = 48 (1) 3x + 2y = 16 (2)

Solution: 9x + 6y = 48 9x 6y = 48 Mult. (2) by 3 0 = 0

When we obtain the equation 0 = 0, we know the equations are dependent. There are infinitely many solutions. The graphs of the equations are identical.


Application

Ethan and Ian are twins. They have decided to save all of the money they earn, at their part-time jobs, to buy a car to share at college. One week, Ethan worked 8 hours and Ian worked 14 hours. Together they saved $256. The next week, Ethan worked 12 hours and Ian worked 16 hours and they earned $324. How much does each twin make per hour?


Solution

Letting E represent Ethan and I represent Ian, the following system can be obtained.

8E + 14I = 256 Mult by 12 96E + 168I = 3072 12E + 16I = 324 Mult by 8 96E 128I = 2592

40I = 480 I = 12 Solve for E. 8E + 14(12) = 256 8E = 88 E = 11

Ian makes $12 per hour while Ethan makes $11 per hour.


Graphical Solution

y1 = and y2 = 128 4

7 7

x

81 3

4 4

x


5.2Systems of Equations in Three

Variables Solve systems of linear equations in three variables. Use systems of three equations to solve applied

problems. Model a situation using a quadratic function.


Solving Systems of Equations in Three Variables

A linear equation in three variables is an equation equivalent to one of the form Ax + By + Cz = D. A, B, C, and D are real numbers and A, B, and C are not all 0.

A solution of a system of three equations in three variables is an ordered triple that makes all three equations true.

Example: The triple (4, 0, 3) is the solution of this system of equations. We can verify this by substituting 4 for x, 0 for y, and 3 for z in each equation.

x 2y + 4z = 8 2x + 2y z = 11

x + y 2z = 10


Gaussian Elimination

An algebraic method used to solve systems in three variables.

The original system is transformed to an equivalent one of the form:

Ax + By + Cz = D,

Ey + Fz = G,

Hz = K.

Then the third equation is solved for z and back-substitution is used to find y and then x.


Operations

The following operations can be used to transform the original system to an equivalent system in the desired form.

Interchange any two equations. Multiply both sides of one of the equations by a

nonzero constant. Add a nonzero multiple of one equation to another

equation.


Example

x + 3y + 2z = 9 x y + 3z = 163x 4y + 2z = 28

x 3y 2z = 9 Mult. (1) by 1 x y + 3z = 16 (2) 4y + z = 7 (4)

3x 9y 6z = 27 Mult. (1) by 3 3x 4y + 2z = 28 (3) 13y 4z = 1 (5)

Solution: Choose 1 variable to eliminate using 2 different pairs of equations. Let’s eliminate x from equations (2) and (3).


Example continued

Now we have…

x + 3y + 2z = 9 (1)

4y + z = 7 (4)

13y 4z = 1 (5)

Next, we multiply equation (4) by 4 to make the z coefficient a multiple of the z coefficient in the equation below it.

x + 3y + 2z = 9 (1)

16y + 4z = 28 (6)

13y 4z = 1 (5)


Example continued

Now, we add equations 5 and 6. 13y 4z = 1 (5) 16y + 4z = 28 (6) 29y = 29

Now, we have the system of equations: x + 3y + 2z = 9 (1) 13y 4z = 1 (5) 29y = 29 (7)

Next, we solve equation (7) for y: 29y = 29 y = 1


Example continued

Then, we back-substitute 1 in equation (5) and solve for z. 13(1) 4z = 1 13 4z = 1 4z = 12 z = 3

Finally, we substitute 1 for y and 3 for z in equation (1) and solve for x:

x + 3(1) + 2(3) = 9 x 3 + 6 = 9 x = 6

The triple (6, 1, 3) is the solution of this system.


Graphs

The graph of a linear equation in three variables is a plane. Thus the solution set of such a system is the intersection of three planes.


Application

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 in rush hour traffic, city traffic, and on the highway.

6

3

3

Highway Hours

34186Week 3

2487Week 2

1592Week 1

Total Fuel Used (gal)

City Traffic Hours

Rush Hour Hours


Solution

Familiarize. We let x, y, and z represent the hours in rush hour traffic, city traffic, and highway, respectively.

Translate. We have three equations: 2x + 9y + 3z = 15 (1) 7x + 8y + 3z = 24 (2) 6x + 18y + 6z = 34 (3) Carry Out. We will solve this equation by eliminating z from

equations (2) and (3). 2x 9y 3z = 15 Mult. (1) by 1 7x + 8y + 3z = 24 (2) 5x y = 9 (4)


Solution continued

Next, we can solve for x: 4x 18y 6z = 30 Mult. (1) by 2

6x + 18y + 6z = 34 (3) 2x = 4 x = 2 Next, we can solve for y by substituting 2 for x in equation (4): 5(2) y = 9 y = 1 Finally, we can substitute 2 for x and 1 for y in equation (1) to solve for z: 2(2) + 9(1) + 3z = 15 4 + 9 + 3z = 15 z = Solving the system we get (2, 1, ).

2

3 2

3


Solution continued

Check: Substituting 2 for x, 1 for y, and for z, we see that the solution makes each of the three equations true.

State: In rush hour traffic the distribution truck uses fuel at a rate of 2 gallons per hour. In city traffic, the same truck uses 1 gallon of fuel per hour. In highway traffic, the same truck used gallon of fuel per hour.

2

3

2

3


5.3Matrices and Systems of

Equations Solve systems of equations using matrices.


Matrices

A rectangular array of numbers is called a matrix (plural, matrices).

Example:

The matrix shown above is an augmented matrix because it contains not only the coefficients but also the constant terms.

The matrix is called the coefficient matrix.

4 2 3

1 5 4

4 2

1 5


Matrices continued

The rows of a matrix are horizontal. The columns of a matrix are vertical. The matrix shown has 2 rows and 3 columns.

A matrix with m rows and n columns is said to be of order m n.

When m = n the matrix is said to be square.

1 2 3

4 5 6


Gaussian Elimination with Matrices

Row-Equivalent Operations

1. Interchange any two rows.

2. Multiply each entry in a row by the same nonzero

constant.

3. Add a nonzero multiple of one row to another row.


Example

Solve the following system:

First, we write the augmented matrix, writing 0 for the missing y-term in the last equation.

Our goal is to find a row-equivalent matrix of the form

.

2 8

3 2 1

4 5 23

x y z

x y z

x z

2 1 1 8

1 3 2 1

4 0 5 23

1

10

0 0 1

a b c

d e

f


Example continued

3 2 1

2 1 1 8

4 0 5 23

1

2 1 1 8

1 3 2 1

4 0 5 23

New row 1 = row 2

New row 2 = row 1

1 Row 1 is unchanged

New row 2= 2(row 1) + r

3 2 1

5 5 1 ow 2

Ne

0

w row 3= 4(row 112 )+row31

0

270 3

We multiply the first row by 2 and add it to the second row. We also multiply the first row by 4 and add it to the third row.


Example continued

We multiply the second row by 1/5 to get a 1 in the second row, second column.

We multiply the second row by 12 and add it to the third row.

Now, we can write the system of equations that corresponds to the last matrix above:

15

1

1 New row 2= (row 2)

3 2 1

1 2

12 13 2

0

0 7

1

1

1 New row

3 2 1

1 2

3=

0

3 12(row 2) + row 30 0

3 2 1

2

3

x y z

y z

z


Example continued

We back-substitute 3 for z in equation (2) and solve for y.

Next, we back-substitute 1 for y and 3 for z in equation (1) and solve for x.

The triple (2, 1, 3) checks in the original system of equations, so it is the solution.

3

2

2

1

y z

y

y

3 2 1

3( ) 2( )1 1

3 6 1

3

3

1

2

x y z

x

x

x

x


Row-Echelon Form

1. If a row does not consist entirely of 0’s, then the first nonzero element in the row is a 1 (called a leading 1).

2. For any two successive nonzero rows, the leading 1 in the lower row is farther to the right than the leading 1 in the higher row.

3. All the rows consisting entirely of 0’s are at the bottom of the matrix.

If a fourth property is also satisfied, a matrix is said to be in reduced row-echelon form:

4. Each column that contains a leading 1 has 0’s everywhere else.


Example

Which of the following matrices are in row-echelon form?

a) b)

c) d)

Matrices (a) and (d) satisfy the row-echelon criteria. In (b) the first nonzero element is not 1. In (c), the row consisting entirely of 0’s is not at the bottom of the matrix.

1 6 7 5

0 1 3 4

0 0 1 8

0 2 4 1

0 0 0 0

1 2 7 6

0 0 0 0

0 1 4 2

1 0 0 3.5

0 1 0 0.7

0 0 1 4.5


Gauss-Jordan Elimination

We perform row-equivalent operations on a matrix to obtain a row-equivalent matrix in row-echelon form. We continue to apply these operations until we have a matrix in reduced row-echelon form.

Example: Use Gauss-Jordan elimination to solve the system of equations from the previous example; we had obtained the matrix

.0

0 0

3 2 1

1 2

3

1

1

1


Gauss-Jordan Elimination continued

We continue to perform row-equivalent operations until we have a matrix in reduced row-echelon form.

Next, we multiply the second row by 3 and add it to the first row.

1 New row 1 = 2(row 3) + row 1

1 New row

3 5

2 1 = 1(ro

0

0 0

0

w 3) + row 2

30 1

1 New row 1 = 3(row 2) + row 1

1

20 0

0 0

0 1

1

0 3


Gauss-Jordan Elimination continued

Writing the system of equations that corresponds to this matrix, we have

We can actually read the solution, (2, 1, 3), directly from the last column of the reduced row-echelon matrix.

2

1

3

x

y

z


Special Systems

When a row consists entirely of 0’s, the equations are dependent and the system is equivalent.

When we obtain a row whose only nonzero entry occurs in the last column, we have an inconsistent system of equations. For example, in the matrix

the last row corresponds to the false equation 0 = 9, so we know the original system has no solution.

1 0 4 6

0 1 4 8

0 0 0 9


5.4Matrix Operations

Add, subtract, and multiply matrices when possible. Write a matrix equation equivalent to a system of

equations.


Matrices

A capital letter is generally used to name a matrix, and lower-case letters with double subscripts generally denote its entries.

For example, a23 read “a sub two three,” indicates the entry in the second row and the third column.

Two matrices are equal if they have the same order and corresponding entries are equal.

11 1

1

[ ]n

ij

m mn

a a

A a

a a


Matrix Addition and Subtraction

To add or subtract matrices, we add or subtract their corresponding entries.

Addition and Subtraction of Matrices

Given two m n matrices A = [aij] and B = [bij], their sum is

A + B = [aij + bij]

and their difference is

A B = [aij bij].


Example

Find A + B for each of the following.

a)

b)

14

6 7A =

2

2 4B =

8 4

1 4

A 2 6

7 0

2 4

B 4 2

5 3


Example continued

We have a pair of 2 2 matrices in part (a) and a pair of 3 2 matrices in part (b). Since each pair has the same order we can add their corresponding entries.

14

1 344

A + B =

( ) 4 3

( ) ( ) 6

2 4

8 4

2

6 7

2

6 7

3

4

82 4

1 4 2 4

A + B 2 6 4 2

7 0 5 3

1 2 4 4 3 8

2 ( 4) 6 2 6 8

7 ( 5) 0 ( 3) 2 3


Example

Find C D for each of the following.

a)

Since the order of each matrix is 3 2, we can subtract corresponding entries.

b)

Since the matrices do not have the same order, we cannot subtract them.

1 2

C 3 0

4 2

1 3

D 4 7

0 3

5 6C

1 2

9D

2

1 3

4 7

0 3

1 3

C D

( ) (

1

) 2 5

2

3 0

4 2

4 7

1 2

3 0

4 42

7

0

7

53


Scalar Multiplication

When we find the product of a number and a matrix, we obtain a scalar product.

The scalar product of a number k and a matrix A is the matrix denoted kA, obtained by multiplying each entry of A by the number k. The number k is called a scalar.


Example

Find 4A and (2)A for .

Solution:

4 1A =

0 7

4 44

4

4 1 4 1

0 7 0 7

( ) ( ) 16 44A =

( ) ( ) 0 284

4 1 2( 4) 2(1) 8 22A = 2

0 7 2(0) 2(7) 0 14


Properties of Matrix Addition and Scalar Multiplication

For any m n matrices, A, B, and C and any scalars k and l:

Commutative Property of Addition

A + B = B + A Associative Property of Addition

A + (B + C) = (A + B) + C Associative Property of Scalar Multiplication

(kl)A = k(lA)


More Properties

Distributive Propertyk(A + B) = kA + kB(k + l)A = kA + lA

There exists a unique matrix 0 such that:A + 0 = 0 + A = A Additive Identity Property

There exists a unique matrix A such that:A + (A) = A + A = 0 Additive Inverse Property


Matrix Multiplication

For an m n matrix A = [aij] and an n p matrix B = [bij], the product AB = [cij] is an m p matrix, where

cij = ai1 • b1j + ai2 • b2j + ai3 • b3j + … + ain • bnj.

We can multiply two matrices only when the number of columns in the first matrix is equal to the number of rows in the second matrix.


Examples

For

,

find each of the following.a) AB

b) BA

c) AC

2 2 2A ,

1 0 4

0 4

B 2 7 ,

1 3

2 4and C

1 0


Solution AB

A is a 2 3 matrix and B is a 3 2 matrix, so AB will be a 2 2 matrix.

0 4

AB 2 7

1 3

(0) (2) ( 1) ( 4) ( 7) ( )(3) 6 28

(0) (2)

2 2 2

1 0 4

2 2 2 2 2 2

1 0 4 1 0( 1) ( 4) ( 7) (3) 4 84


Solution BA

B is a 3 2 matrix and A is a 2 3 matrix, so BA will be a 3 3 matrix.

0 42 2 2

BA 2 71 0 4

1 3

0(2) ( 4)(1) 0(2) ( 4)(0) 0( 2) ( 4)(4) 4 0 16

2(2) ( 7)(1) 2(2) ( 7)(0) 2( 2) ( 7)(4) 3 4 32

1(2) 3(1) 1(2) 3(0) 1( 2) 3(4) 1 2 14


Solution AC

The product AC is not defined because the number of columns of A, 3, is not equal to the number of rows of C, 2.

Note that AB BA. Multiplication of matrices is generally not commutative.


Application

Dalton’s Dairy produces no-fat ice cream and frozen yogurt. The following table shows the number of gallons of each product that are sold at the dairy’s three retail outlets one week. On each gallon of no-fat ice cream, the dairy’s profit is $4, and on each gallon of frozen yogurt, it is $3. Use matrices to find the total profit on these items at each store for the given week.

120

80

Store 2

100160Frozen Yogurt (in gallons)

120100No-fat Ice Cream (in gallons)

Store 3Store 1


Application continued

We can write the table showing the distribution as a 2 3 matrix.

The profit per gallon can also be written as a matrix.

The total profit at each store is given by the matrix

product PD.

100 80 120D

160 120 100

P 4 3


Application continued

The total profit on no-fat ice cream and frozen yogurt for the given week was $880 at store 1, $680 at store 2, and $780 at store 3.

100 80 120PD

160 120 100

(100) (160) (80) (120) (120)

4 3

4 3 4 3 4 3(100)

880 680 780


Properties of Matrix Multiplication

For matrices A, B, and C, assuming that the indicated operations are possible:

Associative Property of MultiplicationA(BC) = (AB)C

Distributive PropertyA(B + C) = AB + AC(B + C)A = BA + CA


Matrix Equations

We can write a matrix equation equivalent to a system of equations.

Example:

Can be written as:

3 11

5 9

2 3 3

x y z

x z

x y z

3 1 1 11

5 0 1 9

1 2 3 3

x

y

z


5.5Inverses of Matrices

Find the inverse of a square matrix, if it exists. Use inverses of matrices to solve systems of equations.


The Identity Matrix


Example

For find each of the following.

a) AI b) IA

1 3 1 0A and I =

4 6 0 1

1 3AI

4 6

1( ) 3( ) 1( ) 3( ) 1 3A

4( ) 6( ) 4( )

1 0

0 1

1 0 0 1

1 0 0 16( ) 4 6

1 3IA

4 6

( )(1) ( )(4) ( )(3) ( )

1 0

0 1

1 0 1 0

0 1

(6) 1 3A

( )(1) ( )4 ( )(3) ( )(1 6) 4 60


Inverse of a Matrix

For an n n matrix A, if there is a matrix A1 for which A1 • A = I = A • A1, then A1 is the inverse of A.

Verify that is the inverse of .

We show that BA = I = AB.

3 4B

5 7

7 4A

5 3

3 4 7 4 1 0BA =

5 7 5 3 0 1

7 4 3 4 1 0AB

5 3 5 7 0 1


Finding an Inverse Matrix

To find an inverse, we first form an augmented matrix consisting of A on the left side and the identity matrix on the right side.

Then we attempt to transform the augmented matrix to one of the form

.

3 4 1 0

5 7 0 1

The 2 2

identity matrix

The 2 2

matrix A

1 0

0 1

a b

c d


Example

Find A1, where A = .

3 4

5 7

3 4 1 0

5 7 0 1

14 13 3 3new row 1 =1 0

5 7 0

(row 1)

1

4 13 3

513 3 new row 2 = 5(ro

1 0

w

1) + row 20 1


Example continued

Thus, A1 = .

4 13 3

new row

1 0

0 1 5 3 2 = 3(row 2)

-43new row 1 = 1 0 (7 4

0 1 5 3

row 3) + row 1

7 4

5 3


Notes

If a matrix has an inverse, we say that it is invertible, or nonsingular.

When we cannot obtain the identity matrix on the left using the Gauss-Jordan method, then no inverse exists.


Solving Systems of Equations

Matrix Solutions of Systems of Equations

For a system of n linear equations in n variables, AX = B, if A is an invertible matrix, then the unique solution of the system is given by X = A1B.

Since matrix multiplication is not commutative in general, care must be taken to multiply on the left by A1.


Example

Use an inverse matrix to solve the following system of equations:3x + 4y = 55x + 7y = 9We write an equivalent matrix, AX = B:

In the previous example we found A1 = .

A X =

3 4 5

5 7 9

B

x

y

7 4

5 3


Example continued

We now have X = A1B.

The solution of the system of equations is (1, 2).

7 4 5 7(5) ( 4)(9) 1

5 3 9 5(5) 3(9) 2

x

y


5.6Determinants and

Cramer’s Rule Evaluate determinants of square matrices. Use Cramer’s rule to solve systems of equations.


Determinants of Square Matrices

The determinant of the matrix is denoted

and is defined as

a c

b d

a c

b d

.a c

ad bcb d


Example

Evaluate: .

Solution:

3 2

6 3

3 23 3 ( 6)( 2)

6 3

3 12

15


Evaluating Determinants Using Cofactors

Minor For a square matrix A = [aij], the minor Mij of an

element aij is the determinant of the matrix formed by deleting the ith row and the jth column of A.

Example: For the matrix

find each of the following.

a) M11 b) M22

2 0 1

A [ ] 4 5 3

1 2 4ija


Solution

For M11, we delete the first row and the first column and find the determinant of the 2 2 matrix formed by the remaining elements.

2 0 1

4 5 3

1 2 4

11

5 3

2 4

5( 4) (2)( 3)

20 ( 6)

20 6

14

M


Solution

For M22, we delete the second row and the second column and find the determinant of the 2 2 matrix formed by the remaining elements.

2 0 1

4 5 3

1 2 4

22

2 1

1 4

( 2)( 4) ( 1)( 1)

8 1

7

M


Cofactor

For a square matrix A = [aij], the cofactor Aij of an element aij is given by

Aij = (1)i + jMij,

where Mij is the minor of aij.

Example:

Find each of the following.

a) A11 b) A22

2 0 1

A= 4 5 3

1 2 4


Solution

a) We found M11 = 14, then

A11 = (1)1+1(14) = 14.

b) We found M22 = 7, then

A22 = (1)2+2(7) = 7.


Determinant of Any Square Matrix

For any square matrix A of order n n (n > 1), we define the determinant of A, denoted |A|, as follows. Choose any row or column. Multiply each element in that row or column by its cofactor and add the results. The determinant of a 1 1 matrix is simply the element of the matrix. The value of a determinant will be the same no matter which row or column is chosen.


Cramer’s Rule for 2 2 Systems


Example

Solve using Cramer’s Rule:

3 4 1

7 5 31

x y

x y

4

5 (1)( 5) (4)(31) 1293

3 4 (3)

1

( 5) (7)(4) 43

1

7 5

3x

3

7 (3)(31) (7)(1) 862

3

1

31

4 43 43

7 5

y

The solution is (3, 2).


Cramer’s Rule for 3 3 Systems


Example

Solve using Cramer’s rule:

Solution: We have

2 3 14

0

2 2

x y z

x y z

x y z

1 2 3

1 1 1 4

1 2 1

D

2 314

0

2

1 1 4

2 1xD


Example continued

Then

14

0

1 3

1 1

1 2

8

1yD

141 2

1 120

2

1

1 2zD

41

4

82

412

34

x

y

z

Dx

DD

yDD

zD

The solution is (1, 2, 3).


5.7Systems of Inequalities and

Linear Programming Graph linear inequalities. Graph systems of linear inequalities. Solve linear programming problems.


Linear Inequalities

A linear inequality in two variables is an inequality that can be written in the form

Ax + By < C, where A, B, and C are real numbers and A and B are not both zero. The symbol < may be replaced with , >, or .

The solution set of an inequality is the set of all ordered pairs that make it true. The graph of an inequality represents its solution set.


Example

Graph y > x 4. We begin by graphing the

related equation y = x 4. We use a dashed line because the inequality symbol is >. This indicates that the line itself is not in the solution set.

Determine which half-plane satisfies the inequality.

y > x 40 ? 0 40 > 4 True


To Graph a Linear Inequality:

Replace the inequality symbol with an equals sign and graph this related equation. If the inequality symbol is < or >, draw the line dashed. If the inequality symbol is or , draw the line solid.

The graph consists of a half-plane on one side of the line and, if the line is solid, the line as well. To determine which half-plane to shade, test a point not on the line in the original inequality. If that point is a solution, shade the half-plane containing that point. If not, shade the opposite half-plane.


Example

Graph: 4x + 2y 8 1. Graph the related equation,

using a solid line. 2. Determine which half-plane

to shade.

4x + 2y 8

4(0) + 2(0) ? 8

0 8

We shade the region containing (0, 0).


Example

Graph x > 2 on a plane.

1. Graph the related equation.

2. Pick a test point (0, 0).x > 20 > 2 FalseBecause (0, 0) is not a solution, we shade the half-plane that does not contain that point.


Example

Graph y 2 on a plane.

1. Graph the related equation.

2. Select a test point (0, 0).

y 2

0 2 True

Because (0, 0) is a solution, we shade the region containing that point.


Systems of Linear Inequalities

Graph the solution set of the system.

First, we graph x + y 3 using a solid line. Choose a test point (0, 0) and shade the correct plane.

Next, we graph x y > 1 using a dashed line. Choose a test point and shade the correct plane.

3

1

x y

x y

The solution set of the system of equations is the region shaded both red and green, including part of the line x + y 3.


Example

Graph the following system of inequalities and find the coordinates of any vertices formed:

We graph the related equations using solid lines. We shade the region common to all three solution sets.

2 0

2

0

y

x y

x y

To find the vertices, we solve three systems of equations. The system of equations from inequalities (1) and (2):

y + 2 = 0

x + y = 2The vertex is (4, 2).


Example continued

The system of equations from inequalities (1) and (3):

y + 2 = 0

x + y = 0

The vertex is (2, 2). The system of equations from

inequalities (2) and (3):

x + y = 2

x + y = 0

The vertex is (1, 1).


Linear Programming

In many applications, we want to find a maximum or minimum value. Linear programming can tell us how to do this.

Constraints are expressed as inequalities. The solution set of the system of inequalities made up of the constraints contains all the feasible solutions of a linear programming problem.

The function that we want to maximize or minimize is called the objective function.


Linear Programming Procedure

To find the maximum or minimum value of a linear objective function subject to a set of constraints:

1. Graph the region of feasible solutions.

2. Determine the coordinates of the vertices of the region.

3. Evaluate the objective function at each vertex. The largest and smallest of those values are the maximum and minimum values of the function, respectively.


Example

A tray of corn muffins requires 4 cups of milk and 3 cups of wheat flour. A tray of pumpkin muffins requires 2 cups of milk and 3 cups of wheat flour. There are 16 cups of milk and 15 cups of wheat flour available, and the baker makes $3 per tray profit on corn muffins and $2 per tray profit on pumpkin muffins. How many trays of each should the baker make in order to maximize profits?

Solution: We let x = the number of corn muffins and

y = the number of pumpkin muffins. Then the profit P is given by the function P = 3x + 2y.


Example continued

We know that x muffins require 4 cups of milk and y muffins require 2 cups of milk. Since there are no more than 16 cups of milk, we have one constraint. 4x + 2y 16

Similarly, the muffins require 3 and 3 cups of wheat flour. There are no more than 15 cups of flour available, so we have a second constraint. 3x + 3y 15

We also know x 0 and y 0 because the baker cannot make a negative number of either muffin.


Example continued

Thus we want to maximize the objective function P = 3x + 2y subject to the constraints

4x + 2y 16,

3x + 3y 15,

x 0,

y 0.

We graph the system of inequalities and determine the vertices. Next, we evaluate the objective function P at each vertex.


Example continued

P = 3(3) + 2(2) = 13(3, 2)

P = 3(0) + 2(5) = 10(0, 5)

P = 3(4) + 2(0) = 12(4, 0)

P = 3(0) + 2(0) = 0(0, 0)

Profit P = 3x+ 2yVertices

Maximum

The baker will make a maximum profit when 3 trays of corn muffins and 2 trays of pumpkin muffins are produced.


5.8Partial Fractions

Decompose rational expressions into partial fractions.


Partial Fractions


Example

Decompose into partial fractions: .

Solution: The degree of the numerator is less than the degree of the denominator. We begin by factoring the denominator: (x + 2)(2x 3). We know that there are constants A and B such that

.

To determine A and B, we add the expressions on the right:

2

4 13

2 6

x

x x

4 13

( 2)(2 3) 2 2 3

x A B

x x x x


Example continued

.

Next, we equate the numerators:4x 13 = A(2x 3) + B(x + 2).Since the last equation containing A and B is true for all x, we can substitute any value of x and still have a true equation. If we choose x = 3/2, then 2x 3 = 0 and A will be eliminated when we make the substitution. This gives us 4(3/2) 13 = A[2(3/2) 3] + B(3/2 + 2) 7 = 0 + (7/2)B.Solving we obtain B = 2.

4 13 (2 3) ( 2)

( 2)(2 3) ( 2)(2 3)

x A x B x

x x x x


Example continued

If we choose x = 2, then x + 2 = 0 and B will be eliminated when we make the substitution. This gives us 4(2) 13 = A[2(2) 3] + B(2 + 2)

21 = 7A.

Solving, we obtain A = 3. The decomposition is as follows:

.2

4 13 3 2 3 2 or

2 6 2 2 3 2 2 3

x

x x x x x x


Another Example

Decompose into partial fractions:

.

Solution: The degree of the numerator is 2 and the degree of the denominator is 3, so the degree of the numerator is less than the degree of the denominator. The denominator is given in factored form. The decomposition has the following form:

.

2

2

7 29 24

(2 1)( 2)

x x

x x

2

2 2

7 29 24

(2 1)( 2) 2 1 2 ( 2)

x x A B C

x x x x x


Another Example continued

Next, we add the expression on the right:

Then, we equate the numerators. This gives us

Since the equation containing A, B, and C is true for all of x, we can substitute any value of x and still have a true equation. In order to have 2x – 1 = 0, we let x = . This gives us

1

2

2 2

2 2

7 29 24 ( 2) (2 1)( 2) (2 1)

(2 1)( 2) (2 1)( 2)

x x A x B x x C x

x x x x

2 27 29 24 ( 2) (2 1)( 2) (2 1)x x A x B x x C x

2 21 1 17( ) 29 24 ( 2) 0.

2 2 2A



Solving, we obtain A = 5. In order to have x 2 = 0, we let x = 2. Substituting

gives us Solving, we obtain C = 2 .

To find B, we choose any value for x except or 2 and replace A with 5 and C with 2 . We let x = 1:

27(2) 29(2) 24 0 (2 2 1).C

1

2

2 27 1 29 1 24 5(1 2) (2 1 1)(1 2) ( 2)(2 1 1)

2 5 2

1

B

B

B



The decomposition is as follows:

2

2 2

7 29 24 5 1 2.

(2 1)( 2) 2 1 2 ( 2)

x x

x x x x x


Another Example

Decompose into partial fractions: .

Solution: The decomposition has the following form.

Adding and equating numerators, we get

or

2

2

11 8 7

(2 1)( 3)

x x

x x

2

2 2

11 8 7

(2 1)( 3) 2 1 3

x x Ax B C

x x x x

2 2

2 2

2 2

11 8 7 ( )( 3) (2 1)

3 3 2 ,

11 8 7 ( 2 ) ( 3 ) ( 3 ).

x x Ax B x C x

Ax Ax Bx B Cx C

x x A C x A B x B C



We then equate corresponding coefficients:

11 = A + 2C, The coefficients of the x2-terms

8 = 3A + B, The coefficients of the x-terms

7 = 3B C. The constant terms

We solve this system of three equations and obtain

A = 3, B = 1, and C = 4.

The decomposition is as follows:

2

2 2

11 8 7 3 1 4.

(2 1)( 3) 2 1 3

x x x

x x x x

Slide 5-1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

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Transcript of Slide 5-1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.