6-1 study guide and intervention...chapter 6 6 glencoe precalculus 6-1 study guide and intervention...

Chapter 6 5 Glencoe Precalculus

6-1 Study Guide and Intervention Multivariable Linear Systems and Row Operations

Gaussian Elimination You can solve a system of linear equations using matrices. Solving a system by transforming it

into an equivalent system is called Gaussian elimination. First, create the augmented matrix. Then use elementary row

operations to transform the matrix so that it is in row-echelon form. Then write the corresponding system of equations

and use substitution to solve the system..

Example: Solve the system of equations using Gaussian elimination with matrices.

x – 2y + z = –1

2x + y – 3z = –7

3x – y + 2z = 0

Step 1 Write the augmented matrix

[1 −2 12 1 −33 −1 2

|−1−7

0]

Step 2 Apply elementary row operations to obtain a row–echelon form of the matrix.

a. b. c.

𝑅2 – 2𝑅1 → [1 −2 10 5 −53 −1 2

|−1−5

0]

𝑅3 – 3𝑅1 →[1 −2 10 5 −50 5 −1

|−1−5

3]

1

5𝑅2 → [

1 −2 10 1 −10 5 −1

|−1−1

3]

d. e.

𝑅3 – 5𝑅2 →[1 −2 10 1 −10 0 4

|−1−1

8]

1

4𝑅3 →

[1 −2 10 1 −10 0 1

|−1−1

2]

Step 3 Write the corresponding system of equations and use substitution to solve the system.

x – 2y + z = –1

y – z = –1

z = 2

The solution of the system is x = –1, y = 1, and z = 2 or (–1, 1, 2).

Exercises Solve each system of equations using Gaussian elimination with matrices.

1. –2x – y + z = 0 2. 5x – y = –13 3. –4x – y – z = –11

x + 2y – z = –3 –3x + 2y – z = 8 x – z = 2

3x + y – 2z = –1 x – 4y + z = –10 2y + 4z = 0

(2, –1, 3) (–2, 3, 4) (3, –2, 1)


6-1 Study Guide and Intervention (continued)

Multivariable Linear Systems and Row Operations

Gauss-Jordan Elimination If you continue to apply elementary row operations to the row-

echelon form of any augmented matrix, you can obtain a matrix in which every column has one

element equal to 1 and the remaining elements equal to 0. This is called the reduced row-

echelon form of the matrix. Solving a system by transforming an augmented matrix so that it is

in reduced row–echelon form is called Gauss–Jordan elimination..

[1 0 00 1 00 0 1

| 𝑎𝑏𝑐

]

Example : Solve the system of equations.

x – 2y + z = 15

–2x – y + 2z = –1

–x + y = –9

Write the augmented matrix. Apply elementary row operations to obtain a row–echelon form. Then apply elementary row operations to obtain zeros above the leading 1s in each row.

Augmented Matrix

[1 −2 1

−2 −1 2

−1 1 0

|15

−1−9

] 2𝑅1 + 𝑅2 → [1 −2 1

0 −5 4

−1 1 0

|15

29

−9]

𝑅1 + 𝑅3 →[1 −2 1

0 −5 4

0 −1 1

|15

29

6

]

Row–echelon form

𝑅2 – 6𝑅3 → [1 −2 1

0 1 −20 −1 1

|15

−76

]

𝑅2 + 𝑅3 →[1 −2 1

0 1 −20 0 −1

|15

−7−1

]

– 𝑅3 →[1 −2 1

0 1 −20 0 1

|15

−71

]

Reduced row-echelon form

𝑅2 + 2𝑅3 → [1 −2 1

0 1 0

0 0 1

|15

−51

]

𝑅1 + 2𝑅2 →[1 0 1

0 1 0

0 0 1

|5

−51

]

𝑅1 – 𝑅3 →[1 0 0

0 1 0

0 0 1

|4

−51

]

The solution of the system is x = 4, y = –5, and z = 1 or (4, –5, 1).

Exercises

Solve each system of equations using Gaussian or Gauss–Jordan elimination.

1. 3x – 2y + z = –22 2. x – 4z = 6 3. –2x – y – z = 1

–4x + z = 31 –2y + 3z = –2 –x + 3y – 2z = 24

2x – 5y = –24 2x – 5y = 6 4x + 2y + z = 2

(–7, 2, 3) (–2, –2, –2) (–1, 5, –4)

NAME _____________________________________________ DATE ____________________________ PERIOD _____________


6-1 Practice Multivariable Linear Systems and Row Operations

Write each system of equations in triangular form using Gaussian elimination. Then solve the system.

1. [1 −1

−3 2|−1232

] 2. [2 3 1

−3 −1 4−1 5 −1

|−23−5

−19] 3. [

−5 −3 00 −2 64 0 −7

|−2242]

Write the augmented matrix for each system of linear equations.

4. 5x – 2y = 14 5. 3x + 4y + 7z = –8 6. –4x – 2y – z = 5

–3x + y = –7 –2x – 3y + z = 6 2x – z = 8

5x – 2y + z = 4 y – 2z = –4

Solve each system of equations using Gauss–Jordan elimination.

7. –4x – 2y = –6 8. –2x – 5y + z = 6 9. 8x – y + 3z = –38

x + 3y = –11 3x + 2y – 4z = –1 –2x + 5y – 4z = 32

5x – y + 2z = –6 x – y + z = –9

10. FRUIT Three customers bought fruit at Michael’s Groceries. The table shows the amount of fruit bought by each

person. Write and solve a system of equations to determine the price of each type of fruit.

Name Apples Oranges Pears Total Cost ($)

Rosario 5 4 3 13.50

Lindsay 7 2 4 14.20

Edwin 3 8 2 15.30

(–8, 4) (–2, –5, –4) (4, –6, 2)

[𝟓 −𝟐

−𝟑 𝟏|𝟏𝟒−𝟕

] [𝟑 𝟒 𝟕

−𝟐 −𝟑 𝟏𝟓 −𝟐 𝟏

|−𝟖𝟔𝟒] [

−𝟒 −𝟐 −𝟏𝟐 𝟎 −𝟏𝟎 𝟏 −𝟐

|𝟓𝟖

−𝟒

]

(4, –5) (–1, –1, –1) (–3, 2, –4)

[𝟓 𝟒 𝟑𝟕 𝟐 𝟒𝟑 𝟖 𝟐

|𝟏𝟑. 𝟓𝟎𝟏𝟒. 𝟐𝟎𝟏𝟓. 𝟑𝟎

]; (1.1, 1.25, 1); apples: $1.10, oranges: $1.25, pears: $1

NAME _____________________________________________ DATE ____________________________ PERIOD _____________


6-2 Study Guide and Intervention Matrix Multiplication, Inverses, and Determinants

Multiply Matrices To multiply matrix A by matrix B, the number of columns in A must be equal to the number of rows

in B. If A has dimensions m × r and B has dimensions r × n, their product, AB, is an m × n matrix. If the number of columns in A does not equal the number of rows in B, the matrices cannot be multiplied.

[𝑎 𝑏𝑐 𝑑

]⋅[𝑒 𝑓𝑔 ℎ

] =[𝑎𝑒 + 𝑏𝑔 𝑎𝑓 + 𝑏ℎ𝑐𝑒 + 𝑑𝑔 𝑐𝑓 + 𝑑ℎ

]

Example: Use matrices A = [ 𝟒 −𝟐−𝟏 𝟑

] and B =[−𝟏 𝟐 𝟑−𝟐 𝟒 −𝟏

]to find AB, if possible.

AB = [4 −2

−1 3]⋅[

−1 2 3−2 4 −1

]

A is a 2 × 2 matrix and B is a 2 × 3 matrix. Because the number of columns for A is equal to the number of rows for B,

the product AB exists.

To find the first entry in AB, write the sum of the products of the entries in row 1 of A and in column 1 of B.

[4 −2

−1 3]⋅[

−1 2 3−2 4 −1

]= [𝟒(– 𝟏) + (– 𝟐)(– 𝟐)]

Follow this same procedure to find the entry for row 1, column 2 of AB.

[4 −2

−1 3]⋅[

−1 2 3−2 4 −1

]= [𝟒(– 𝟏) + (– 𝟐)(– 𝟐) 𝟒(𝟐) + (– 𝟐)(𝟒)]

Continue multiplying each row by each column to find the sum for each entry.

[4 −2

−1 3]⋅[

−1 2 3−2 4 −1

]= [4(−1) + (−2)(−2) 4(2) + (−2)(4) 4(3) + (−2)(−1)

(−1)(−1) + 3(−2) (−1)(2) + 3(4) (−1)(3) + 3(−1)]

Then simplify each sum.

[4 −2

−1 3]⋅[

−1 2 3−2 4 −1

]= [0 0 14

−5 10 −6]

Exercises

Find AB and BA, if possible.

1. A = [2 4

−3 −1], B = [

−1 50 −2

] 2. A = [−1 3−3 2

], B = [−2 4 0−3 −1 2

]

AB =[−𝟐 𝟐

𝟑 −𝟏𝟑]; BA = [

−𝟏𝟕 −𝟗𝟔 𝟐

] AB =[−𝟕 −𝟕 𝟔

𝟎 −𝟏𝟒 𝟒]; BA is undefined.

NAME _____________________________________________ DATE ____________________________ PERIOD _____________


6-2 Study Guide and Intervention(continued)

Matrix Multiplication, Inverses, and Determinants

Inverses and Determinants The identity matrix is an n × n matrix consisting of all 1s on its main diagonal, from

upper left to lower right, and 0s for all other elements. Let𝐼𝑛be the identity matrix of order n and let A be an n × n matrix.

If there exists a matrix B such that AB = BA = 𝐼𝑛 , then B is called the inverse of A and is written as 𝐴−1. If a matrixhas an

inverse, it is invertible. The determinant of a 2 × 2 matrix can be used to determine whether or not a matrix is invertible.

If A =[𝑎 𝑏𝑐 𝑑

], det(A) = ad –cb. If ad –cb ≠ 0, then 𝐴−1=1

𝑎𝑑 −𝑐𝑏[

𝑑 −𝑏−𝑐 𝑎

].

Determine whether A = [𝟕 −𝟒

−𝟓 𝟑] and B = [

𝟑 𝟒𝟓 𝟕

]are inverse matrices.

If A and B are inverse matrices, then AB = BA = I.

AB = [7 −4

−5 3]⋅[

3 45 7

] = [7(3) + (−4)(5) 7(4) + (−4)(7)

−5(3) + 3(5) −5(4) + 3(7)] or [

1 00 1

]

BA = [3 45 7

]⋅[7 −4

−5 3] = [

3(7) + 4(−5) 3(−4) + 4(3)

5(7) + 7(−5) 5(−4) + 7(3)] or [

1 00 1

]

Because AB = BA = I, B = 𝐴−1and A = 𝐵−1.

Find the determinant of A = [𝟐 −𝟐𝟑 −𝟔

].Then find 𝑨−𝟏, if it exists.

det(A) = |2 −23 −6

| A−1 = −1

6[−6 2−3 2

]

= 2(–6) –3(–2) or –6 = [1 −

1

31

2−

1

3

]

Since det(A) ≠ 0, A is invertible.

Exercises

Determine whether A and B are inverse matrices. Explain your reasoning.

1. A=[11 5

2 1], B = [

1 −5

−2 11] 2.A = [

3 2

4 1], B = [

1 5

4 3]

3. Find the determinant of A = [5 −1

−10 2]. Then find 𝐴−1, if it exists.

4. Find the determinant of A = [3 2

1 −1]. Then find 𝐴−1, if it exists.

yes; AB = BA = I2 no; AB ≠ BA ≠ I2

det(A) = 0; A–1 does not exist.

det(A) = –5, A–1 =[𝟎. 𝟐 𝟎. 𝟒𝟎. 𝟐 −𝟎. 𝟔

]

NAME _____________________________________________ DATE ____________________________ PERIOD _____________


6-2 Practice Matrix Multiplication, Inverses, and Determinants

Find AB and BA, if possible.

1. A = [−1 6 0

3 −2 1] , B = [

2 −4−1 3

] 2. A = [3 0

−1 2], B = [

3 5−2 0

]

3. GOLF The number of golf clubs manufactured daily by two different companies is shown, as well as the selling price of each type of club. Use this information to determine which company’s daily production has the highest retail value.

How much greater is the value?

Company Club Type and Quantity

1-Wood 3-Wood 5-Wood Putter

A 600 520 310 300

B 210 400 450 400

Club Club Value ($)

1-Wood 210

3-Wood 170

5-Wood 150

Putter 120

Write each system of equations as a matrix equation, AX = B. Then use Gauss-Jordan elimination on the

augmented matrix to solve for X.

4. 𝑥1 – 2𝑥2 + 3𝑥3 = 4 5. 2𝑥1 + 𝑥2 + 2𝑥3 = 11

5𝑥1 + 3𝑥2 – 𝑥3 = 13 –5𝑥1 – 𝑥2 + 4𝑥3 = 1

4𝑥1 – 𝑥2 + 4𝑥3 = 11 3𝑥1 – 2𝑥2 + 8𝑥3 = 28

Determine whether A and B are inverse matrices.

6. A = [1 21 3

], B = [3 −2

−1 1] 7. A = [

5 −24 −3

], B = [−1 0

2 −8]

Find the determinant of each matrix. Then find its inverse, if it exists.

8. [6 52 2

] 9. [−2 4

3 −6]

Evaluate.

A = [−1 5

3 0] B = [

−4 2 −10 −5 3

] C = [−1 0 −4

3 −2 1]

10. AB + C 11. A(B – C)

AB is undefined; BA = [−𝟏𝟒 𝟐𝟎 −𝟒

𝟏𝟎 −𝟏𝟐 𝟑] AB = [

𝟗 𝟏𝟓−𝟕 −𝟓

]; BA = [𝟒 𝟏𝟎

−𝟔 𝟎]

Company A; $69,300

[𝟏 −𝟐 𝟑𝟓 𝟑 −𝟏𝟒 −𝟏 𝟒

] ⋅ [

𝒙𝟏

𝒙𝟐

𝒙𝟑

] = [𝟒

𝟏𝟑𝟏𝟏

] (4, –3, –2) [𝟐 𝟏 𝟐

−𝟓 −𝟏 𝟒𝟑 −𝟐 𝟖

] ⋅ [

𝒙𝟏

𝒙𝟐

𝒙𝟑

] = [𝟏𝟏𝟏

𝟐𝟖] (2, 1, 3)

yes no

2; [𝟏 −𝟐. 𝟓

−𝟏 𝟑] 0; singular

[𝟑 −𝟐𝟕 𝟏𝟐

−𝟗 𝟒 −𝟐] [

−𝟏𝟐 −𝟏𝟕 𝟕−𝟗 𝟔 𝟗

]

NAME _____________________________________________ DATE ____________________________ PERIOD _____________


6-3 Study Guide and Intervention Solving Linear Systems Using Inverses and Cramer’s Rule

Use Inverse Matrices A square system has the same number of equations as variables. If a square matrix has an

inverse, the system has one unique solution.

Example : Use an inverse matrix to solve each system of equations, if possible.

a. 3x – 7y = –16

–x + 2y = 8

Write the system in matrix form.

A ⋅ X = B

[3 −7

−1 2] ⋅ [

𝑥𝑦] = [

−168

]

Use the formula for an inverse of a 2 × 2 matrix to find

the inverse 𝐴−1.

𝐴−1 = 1

𝑎𝑑 − 𝑐𝑏 [

𝑑 −𝑏−𝑐 𝑎

]

= 1

(2)(3) − (−7)(−1) [

2 71 3

]

Multiply 𝐴−1 by B to solve the system.

X = 𝐴−1 ⋅ B

= [−2 −7−1 −3

] ⋅ [−16

8] = [

−24−8

]

So, the solution of the system is (–24, –8).

b. –2x + y + z = 0

x + 2z = 9

x – 2y – 9z = –31

Write the system in matrix form.

A ⋅ X = B

[−2 1 1

1 0 21 −2 −9

] ⋅ [𝑥𝑦𝑧

] = [09

−31]

Use a graphing calculator to find 𝐴−1.

Multiply 𝐴−1 by B to solve the system.

X = 𝐴−1 ⋅ B

= [4 7 2

11 17 5−2 −3 −1

] ⋅ [09

−3] = [

1−2

4]

So, the solution of the system is (1, –2, 4).

Exercises

Use an inverse matrix to solve each system of equations, if possible.

1. –2x + 5y = 24 2. x – y + 2z = 5

3x – y = –10 x – z = –4

3x + 2y + z = 0

3. 3x + y = 7 4. x + y – z = –5

–2x – 5y = 43 2x – 3y + 2z = 20

y + 4z = 18

(–2, 4) (–1, 0, 3) (6, –11) (2, –2, 5)

NAME _____________________________________________ DATE ____________________________ PERIOD _____________



Solving Linear Systems Using Inverses and Cramer’s Rule

Use Cramer’s Rule Another method, known as Cramer’s Rule, can be used to solve a square system of equations.

Let A be the coefficient matrix of a system of n linear equations in n variables given by AX = B. If det(A) ≠ 0, then the

unique solution of the system is given by

𝑥1 = |𝐴1|

|𝐴|, 𝑥2 =

|𝐴2|

|𝐴|, 𝑥3 =

|𝐴3|

|𝐴|…, 𝑥𝑛 =

|𝐴𝑛|

|𝐴|,

where Ai is the matrix obtained by replacing the ith column of A with the column of constants B. If det(A) = 0, then AX = B has either no solution or infinitely many solutions.

Example: Use Cramer’s Rule to find the solution of the system of linear equations, if a unique solution exists.

–2𝒙𝟏 + 𝒙𝟐 = –7

5𝒙𝟏 – 2𝒙𝟐 = 17

The coefficient matrix is A = [−2 1

5 −2]. Calculate the determinant of A.

|𝐴| = |−2 1

5 −2| = (–2) (–2) – 5(1) or –1

Because the determinant of A does not equal zero, you can apply Cramer’s Rule.

𝑥1 = |𝐴1|

|𝐴| =

|−7 117 −2

|

|−1| =

−7(−2) − 17(1)

−1 =

−3

−1 or 3

𝑥2 = |𝐴2|

|𝐴| =

|−2 −7

5 17|

−1 =

−2(17) − 5(−7)

−1 =

1

−1 or –1

Therefore, the solution is 𝑥1 = 3 and 𝑥2 = –1 or (3, –1).

Exercises

Use Cramer’s Rule to find the solution of each system of linear equations, if a unique solution exists.

1. x – 2y = –5 2. 3x – 3y = –18

–2x – 5y = –8 –x + 4y = 9

3. 3x + y = 21 4. –2x – 4y = 2

–x + 2y = 14 x + 3y = –3

(–1, 2) (–5, 1) (4, 9) (3, –2)

NAME _____________________________________________ DATE ____________________________ PERIOD _____________


6-3 Practice Solving Linear Systems Using Inverses and Cramer’s Rule

Use an inverse matrix to solve each system of equations, if possible.

1. 4x – 7y = 30 2. –2x – 8y = –36

–6x + 2y = –11 4x + 3y = 7

3. x – 2y + 7z = –33 4. x + y – 2z = 5

–4x + 5y – z = 18 x + 2y + z = 8

5x – 3y = –11 2x + 3y – z = 1

5. TELEVISION During the summer, Manuel watches television M hours per day, Monday through Friday. Harry

watches television H hours per day, Friday and Saturday. Ellen watches television E hours per day, Friday through

Sunday. Altogether, they watch television 37 hours each week. On Fridays, they watch a total of 11 hours of television. If the number of hours Ellen spends watching television on any given day is twice the number of hours that Manuel

spends watching television on any given day, how many hours of television does each of them watch each day?

Use Cramer’s Rule to find the solution of each system of linear equations, if a unique solution exists.

6. –4x – 5y = 1 7. x + y + z = 8

–2x – 3y = –1 3x – z = –22

y + 2z = 20

8. PAPER ROUTE Payton, Santiago, and Queisha each have a paper route. Payton delivers 5 times as many papers as

Santiago. Santiago delivers twice as many papers as Queisha. If 20 papers were added to Payton’s route, he would then deliver four times the total number of papers that Santiago and Queisha deliver. How many papers does each person

deliver?

(𝟏

𝟐, −𝟒) (–2, 5)

(–1, 2, –4) no solution

Manuel: 3 hours, Harry: 2 hours, Ellen: 6 hours

(– 4, 3)

(–5, 6, 7) Payton delivers 100; Santiago delivers 20; Queisha delivers 10

NAME _____________________________________________ DATE ____________________________ PERIOD _____________


6-4 Study Guide and Intervention Partial Fractions

Linear Factors The function g(x) shown below can be written as the sum of two fractions with denominators that are

linear factors of the original denominator.

g(x) = 3𝑥 − 1

2𝑥2 − 3𝑥 +1 =

2

𝑥 − 1 +

−1

2𝑥 − 1

Each fraction in the sum is a partial fraction. The sum of these partial fractions make up the partial fraction

decomposition of the original rational function. If the denominator of a rational expression contains a repeated linear

factor, the partial fraction decomposition must include a partial fraction with its own constant numerator for each power of this factor.

Example : Find the partial fraction decomposition of 𝒙 + 𝟏𝟏

𝒙𝟐 − 𝟑𝒙 − 𝟒.

Rewrite the equation as partial fractions with constant numerators, A and B, and denominators that are the linear factors of

the original denominator.

𝑥 +11

𝑥2− 3𝑥−4 =

𝐴

𝑥 − 4 +

𝐵

𝑥 +1 Form a partial fraction decomposition.

x + 11 = A(x + 1) + B(x – 4) Multiply each side by the LCD, 𝑥2 – 3x – 4.

x + 11 = Ax + A + Bx – 4B Distributive Property

1x + 11 = (A + B)x + (A + (–4B)) Group like terms.

Equate the coefficients on the left and right side of the equation to obtain a system of two equations. To solve the system,

write it in matrix form CX = D and solve for X.

A + B = 1

A + (–4B) = 11 ⟶

C ⋅ X = D

[1 11 −4

] ⋅ [𝐴𝐵

] = [1

11]

Solving for X yields A = 3 and B = –2. Therefore 𝑥 +11

𝑥2− 3𝑥−4 =

3

x − 4 +

−2

x +1.

Exercises

Find the partial fraction decomposition of each rational expression.

1. 5𝑥 − 34

𝑥2− 𝑥 −12 2.

−7𝑥 + 13

𝑥2 − 5𝑥 −14

3. 𝑥2 + 1

2𝑥(𝑥 − 1)2 4. 5𝑥2 − 𝑥 − 1

𝑥2(𝑥 − 1)

𝟕

𝒙 + 𝟑 +

−𝟐

𝒙 − 𝟒

−𝟒

𝒙 − 𝟕 +

−𝟑

𝒙 + 𝟐

𝟏

𝟐𝒙 +

𝟏

(𝒙−𝟏)𝟐 𝟐

𝒙 +

𝟏

𝒙𝟐 + 𝟑

𝒙−𝟏

NAME _____________________________________________ DATE ____________________________ PERIOD _____________



Partial Fractions

Irreducible Quadratic Factors Not all rational expressions can be written as the sum of partial fractions using only

linear factors in the denominator. If the denominator of a rational expression contains an irreducible quadratic factor, the partial fraction decomposition must include a partial fraction with a linear numerator of the form Bx + C

for each power of this factor.

Example : Find the partial fraction decomposition of 𝟒𝒙𝟒 − 𝟐𝒙𝟑 − 𝟏𝟑𝒙𝟐 + 𝟕𝒙 + 𝟗

𝒙(𝒙𝟐 − 𝟑)𝟐

This expression is proper. The denominator has one linear factor and one irreducible factor of multiplicity 2.

𝟒𝒙𝟒 − 𝟐𝒙𝟑 − 𝟏𝟑𝒙𝟐 + 𝟕𝒙 + 𝟗

𝒙(𝒙𝟐− 𝟑)𝟐 = 𝑨

𝒙 +

𝑩𝒙 + 𝑪

𝒙𝟐 − 𝟑 +

𝑫𝒙 + 𝑬

(𝒙𝟐 − 𝟑)𝟐

4𝑥4 – 2𝑥3 –13𝑥2 + 7x + 9 = A(𝑥2 − 3)2 + (Bx + C)x(𝑥2 – 3) + (Dx + E)x

4𝑥4 – 2𝑥3 –13𝑥2 + 7x + 9 = A𝑥4 + B𝑥4 + C𝑥3– 6A𝑥2 – 3B𝑥2 + D𝑥2 – 3Cx + Ex + 9A

4𝑥4 – 2𝑥3 –13𝑥2 + 7x + 9 = (A + B) 𝑥4 + C𝑥3 + (–6A – 3B + D) 𝑥2 + (–3C + E)x + 9A

Write and solve the system of equations obtained by equating coefficients.

A + B = 4 A = 1

C = –2 B = 3

–6A – 3B + D = –13 → C = –2

–3C + E = 7 D = 2

9A = 9 E = 1

Therefore, 4𝑥4 − 2𝑥3 − 13𝑥2 + 7𝑥 + 9

𝑥(𝑥2 − 3)2 = 1

𝑥 +

3𝑥 + 2

𝑥2 − 3 +

2𝑥 + 1

(𝑥2 − 3)2

Exercises


1. 5

𝑥3+ 5𝑥 2.

3𝑥3 − 2𝑥2− 8𝑥 + 5

(𝑥2 − 3)2

3. 2𝑥3 + 𝑥 + 3

(𝑥2 + 1)2 4. 𝑥3 + 2𝑥2 + 2

(𝑥2 + 1)2

𝟏

𝒙 –

𝒙

𝒙𝟐+𝟓

𝟑𝒙 − 𝟐

𝒙𝟐−𝟑 +

𝒙 − 𝟏

(𝒙𝟐−𝟑)𝟐

𝟐𝒙

𝒙𝟐+𝟏 +

𝟑−𝒙

(𝒙𝟐+𝟏)𝟐

𝒙 + 𝟐

𝒙𝟐+ 𝟏 +

−𝒙

(𝒙𝟐+𝟏)𝟐

NAME _____________________________________________ DATE ____________________________ PERIOD _____________


6-4 Practice Partial Fractions


1. 3𝑥 − 7

𝑥2 − 7𝑥 + 12 2.

6𝑥2 − 10𝑥 – 2

𝑥3 + 𝑥2 − 2𝑥

3. 9𝑥 + 15

𝑥2 + 3𝑥 + 2 4.

𝑥

2𝑥2 − 9𝑥 + 9

Find the partial fraction decomposition of each improper rational expression.

5. 3𝑥2 + 5𝑥 + 2

𝑥2 + 2𝑥 6.

−5𝑥2 − 11𝑥 + 54

𝑥2 + 2𝑥 − 8

7. 6𝑥2 + 17𝑥 + 2

𝑥2 + 𝑥 8.

−8𝑥2 + 22𝑥 − 10

(2𝑥 − 3)𝟐

Find the partial fraction decomposition of each rational expression with repeated factors.

9. −2𝑥2 + 29𝑥 − 100

𝑥3 − 10𝑥2 + 25𝑥 10.

5𝑥4 − 7𝑥3 − 12𝑥2 + 6𝑥 + 21

(𝑥 – 3)(𝑥2 – 2)2

11. 2𝑥2 + 5

𝑥3 + 6𝑥2 + 9𝑥 12.

4𝑥4 + 8𝑥3 + 6𝑥2 + 6𝑥 + 5

(3𝑥 + 2)(𝑥2 + 1)2

13. GROWTH When working with exponential growth in calculus, it is often necessary to work with functions of the

form f(x) = 1

𝑥(50 − 𝑥) and to decompose these functions into the sum of its partial fractions. Find the partial

decomposition of f(x).

𝟓

𝒙 − 𝟒 +

−𝟐

𝒙 − 𝟑

𝟏

𝒙 +

−𝟐

𝒙 − 𝟏 +

𝟕

𝒙 + 𝟐

𝟔

𝒙 + 𝟏 +

𝟑

𝒙 + 𝟐

𝟏

𝒙 − 𝟑 +

−𝟏

𝟐𝒙 − 𝟑

3 + 𝟏

𝒙 +

−𝟐

𝒙 + 𝟐 –5 +

𝟐

𝒙 − 𝟐 +

−𝟑

𝒙 + 𝟒

6 + 𝟐

𝒙 +

𝟗

𝒙 + 𝟏 –2 –

𝟏

𝟐𝒙 − 𝟑 +

𝟓

(𝟐𝒙 − 𝟑)𝟐

−𝟒

𝒙 +

𝟐

𝒙 − 𝟓 +

−𝟏

(𝒙 − 𝟓)𝟐 𝟑

𝒙 − 𝟑 +

𝟐𝒙 − 𝟏

𝒙𝟐−𝟐 +

𝒙 − 𝟓

(𝒙𝟐−𝟐)𝟐

𝟓

𝟗𝒙 +

𝟏𝟑

𝟗(𝒙 + 𝟑) +

–𝟐𝟑

𝟑(𝒙 + 𝟑)𝟐 𝟏

𝟑𝒙 + 𝟐 +

𝒙 + 𝟐

𝒙𝟐 + 𝟏 +

−𝒙

(𝒙𝟐+𝟏)𝟐

𝟏

𝟓𝟎

𝒙 +

𝟏

𝟓𝟎

𝟓𝟎−𝒙 or

𝟏

𝟓𝟎𝒙 +

𝟏

𝟓𝟎(𝟓𝟎−𝒙)

NAME _____________________________________________ DATE ____________________________ PERIOD _____________


6-5 Study Guide and Intervention Linear Optimization

Linear Programming Linear programming is a process for finding a minimum or maximum value for a specific

quantity. The following steps can be used to solve a linear programming problem.

Step 1 Write an objective function and a list of constraints to model the situation.

Step 2 Graph the region corresponding to the solution of the system of constraints.

Step 3 Find the coordinates of the vertices of the region formed.

Step 4 Evaluate the objective function at each vertex to find the minimum or maximum.

Example: A leather company wants to add belts and wallets to its product line. Belts require 2 hours of cutting

time and 6 hours of sewing time. Wallets require 3 hours of cutting time and 3 hours of sewing time. The cutting

machine is available 12 hours a week and the sewing machine is available 18 hours per week. Belts will net $18 in

profit and wallets will net $12. How much of each product should be produced to achieve maximum profit?

Let x represent the number of belts and y represent the number of wallets.

The objective function is then given by f(x, y) = 18x + 12y.

Write the constraints.

x ≥ 0; y ≥ 0 Numbers of items cannot be negative.

2x + 3y ≤ 12 Cutting time

6x + 3y ≤ 18 Sewing time

Graph the system. The solution is the shaded region, including its boundary segments.

Find the coordinates of the four vertices by solving the system of boundary equations

for each point of intersection. The coordinates are (0, 0), (0, 4), (1.5, 3), and (3, 0).

Evaluate the objective function for each ordered pair.

Point f(x, y) = 18x + 12y Result

(0, 0) f(0, 0) = 18(0) + 12(0) 0

(0, 4) f(0, 4) = 18(0) + 12(4) 48

(1.5, 3) f(1.5, 3) = 18(1.5) + 12(3) 63 ← Maximum

(3, 0) f(3, 0) = 18(3) + 12(0) 54

Since f is greatest at (1.5, 3), the company will maximize profit if it makes and sells 1.5 belts for every 3 wallets.

Exercises

Find the maximum and minimum values of the objective function f(x, y) and for what values of x and y they occur,

subject to the given constraints.

1. f(x, y) = 3x – 2y 2. f(x, y) = x + 2y

2x + y ≤ 10 x + y ≤ 4

x + 2y ≤ 8 x + 3y ≤ 6

x ≥ 0 x ≥ 0

y ≥ 0 y ≥ 0

max at (5, 0) = 15, max at (3, 1) = 5, min at (0, 4) = –8 min at (0, 0) = 0

NAME _____________________________________________ DATE ____________________________ PERIOD _____________



Linear Optimization

No or Multiple Optimal Solutions Linear programming models can have one, multiple, or no optimal solutions.

If the graph of the objective function f to be optimized is coincident with one side of the region of feasible solutions, f has multiple optimal solutions. If the region does not form a polygon, but instead is unbounded, f may have no

minimum value or maximum value.

Example: Find the maximum value of the objective function f(x, y) = 6x + 3y and for what values of x and y it

occurs, subject to the following constraints.

2x + y ≤ 8 y ≤ 4 x ≤ 3

x ≥ 0 y ≥ 0

Graph the region bounded by the given constraints. Find the value of the objective function f(x, y) = 6x + 3y at each vertex.

f(0, 0) = 6(0) + 3(0) or 0

f(0, 4) = 6(0) + 3(4) or 12

f(2, 4) = 6(2) + 3(4) or 24

f(3, 2) = 6(3) + 3(2) or 24

f(3, 0) = 6(3) + 3(0) or 18

Because f(x, y) = 24 at (2, 4) and (3, 2), the problem has multiple optimal solutions. An equation of the line through these

two vertices is y = –2x + 8. Therefore, f has a maximum value of 24 at every point on y = –2x + 8 for 2 ≤ x ≤ 3.

Exercises

Find the maximum and minimum values of the objective function f(x, y) and for what values of x and y they occur,

subject to the given constraints.

1. f(x, y) = 2x + y 2. f(x, y) = 3x + 6y

2x + y ≤ 11 x - y ≥ –4

y ≤ 5 x + 2y ≤ 20

y ≥ 0 x ≥ 0

x ≤ 5 x ≤ 8

x ≥ 0 y ≥ 0

max. of 11 at every max. of 60 at every

point on y = –2x + 11 point on y = – 𝟏

𝟐x + 10

for 3 ≤ x ≤ 5; for 4 ≤ x ≤ 8; min. at (0, 0) = 0 min. at (0, 0) = 0

NAME _____________________________________________ DATE ____________________________ PERIOD _____________


6-5 Practice Linear Optimization

Find the maximum and minimum values of the objective function f(x, y) and for what values of x and y they

occur, subject to the given constraints.

1. f(x, y) = 2x + 5y 2. f(x, y) = 4x + 3y

x ≥ 0 x ≥ 0

y ≥ 0 y ≥ 0

x + y ≤ 7 2x + 3y ≥ 6

2x + 3y ≤ 18 x + y ≤ 8

3. f(x, y) = 2x – 3y 4. f(x, y) = 3x + 3y

x ≥ 0 x ≥ 0

x ≤ 7 y ≥ 0

y ≥ 0 y ≤ 8

y ≤ 5 x + y ≤ 10

x + 2y ≥ 14 3x + 2y ≤ 24

5. SKATES A manufacturer produces roller skates and ice skates.

Manufacturer Information

Roller Skates Ice Skates Maximum Time Available

Assembling 5 minutes 4 minutes 200 minutes

Checking and Packaging 1 minute 4 minutes 120 minutes

Profit per Skate $40 $30

a. Write an objective function and list the constraints that model the given situation.

b. Sketch a graph of the region determined by the constraints from part a to find the

set of feasible solutions for the objective function.

c. How many roller skates and ice skates should be manufactured to maximize profit?

What is the maximum profit?

d. Describe why the company would choose a number of roller skates and ice skates

different from the answer in part c.

max. at (0, 6) = 30, max. at (8, 0) = 32, min. at (0, 0) = 0 min. at (0, 2) = 6

max. at (7, 3.5) = 3.5, max. of 30 at every point on min. at (4, 5) = –7 y = –x + 10 for 2 ≤ x ≤4,

min. at (0, 0) = 0 f(x, y) = 40x + 30y; x ≥ 0; y ≥ 0; 5x + 4y ≤ 200; x + 4y ≤ 120

40 roller skates and no ice skates; $1600

Sample answer: If customers cannot get ice skates, they might go somewhere else. They should combine the math model with customer needs.

6-1 study guide and intervention...chapter 6 6 glencoe precalculus 6-1 study guide and intervention...

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