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School District of

Palm Beach County

Summer Packet

Algebra EOC Review

Answers

Name Class Date

Prentice Hall Algebra 1 • Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

9

1-1 ReteachingVariables and Expressions

You can represent mathematical phrases and real-world relationships using symbols and operations. Th is is called an algebraic expression.

For example, the phrase 3 plus a number n can be expressed using symbols and operations as 3 1 n.

Problem

What is the phrase 5 minus a number d as an algebraic expression?

Th e phrase 5 minus a number d, rewritten as an algebraic expression, is 5 2 d .

Th e left side of the table below gives some common phrases used to express mathematical relationships, and the right side of the table gives the related symbol.

Exercises

Write an algebraic expression for each word phrase.

1. 5 plus a number d 2. the product of 5 and g

3. 11 fewer than a number f 4. 17 less than h

5. the quotient of 20 and t 6. the sum of 12 and 4

Write a word phrase for each algebraic expression.

7. h 1 6 8. m 2 5 9. q 3 10

10. 35r 11. h 1 m 12. 5n

SymbolPhrase

1

2

3

4

2

1

sum

difference

product

quotient

less than

more than

d

a number dminus

2

5

5

5 1 d 5 3 g

f 2 11 h 2 17

20 4 t 12 1 4

the sum of h and 6 5 less than a number m the product of q and 10

the quotient of 35 and r the sum of h and m the product of 5 and n

Name Class Date


10

Multiple operations can be combined into a single phrase.

Problem

What is the phrase 11 minus the product of 3 and d as an algebraic expression?

Th e phrase 11 minus the product of 3 and a number d, rewritten as an algebraic expression, is 11 – 3d.

Exercises

Write an algebraic expression for each phrase.

13. 12 less than the quotient of 12 and a number z

14. 5 greater than the product of 3 and a number q

15. the quotient of 5 1 h and n 1 3

16. the diff erence of 17 and 22t

Write an algebraic expression or equation to model the relationship expressed in each situation below.

17. Jane is building a model boat. Every inch on her model is equivalent to 3.5 feet on the real boat her model is based on. What would be the mathematical rule to express the relationship between the length of the model, m, and the length of the boat, b?

18. Lyn is putting away savings for his college education. Every time Lyn puts money in his fund, his parents put in $2. What is the expression for the amount going into Lyn’s fund if Lyn puts in L dollars?

1-1 Reteaching (continued)

Variables and Expressions

3 3 d

the product of 3 and a number dminus

2

11

11

12 4 z 2 12

5 1 3 3 q

5 1 hn 1 3

17 2 22t

3.5m 5 b

L 1 2


19

Name Class Date

Exponents are used to represent repeated multiplication of the same number. For example, 4 3 4 3 4 3 4 3 4 5 45. Th e number being multiplied by itself is called the base; in this case, the base is 4. Th e number that shows how many times the base appears in the product is called the exponent; in this case, the exponent is 5. 45 is read four to the fi fth power.

Problem

How is 6 3 6 3 6 3 6 3 6 3 6 3 6 written using an exponent?

Th e number 6 is multiplied by itself 7 times. Th is means that the base is 6 and the exponent is 7. 6 3 6 3 6 3 6 3 6 3 6 3 6 written using an exponent is 67.

Exercises

Write each repeated multiplication using an exponent.

1. 4 3 4 3 4 3 4 3 4 2. 2 3 2 3 2

3. 1.1 3 1.1 3 1.1 3 1.1 3 1.1 4. 3.4 3 3.4 3 3.4 3 3.4 3 3.4 3 3.4

5. (27) 3 (27) 3 (27) 3 (27) 6. 11 3 11 3 11

Write each expression as repeated multiplication.

7. 43 8. 54

9. 1.52 10. Q27R4

11. x7 12. (5n)5

13. Trisha wants to determine the volume of a cube with sides of length s. Write an expression that represents the volume of the cube.

1-2 ReteachingOrder of Operations and Evaluating Expressions

45 23

1.15 3.46

(27)4 113

4 3 4 3 4 5 3 5 3 5 3 5

1.5 3 1.5 Q27R 3 Q27R 3 Q

27R 3 Q

27R

x ? x ? x ? x ? x ? x ? x 5n 3 5n 3 5n 3 5n 3 5n

s3


20

Name Class Date

Th e order of operations is a set of guidelines that make it possible to be sure that two people will get the same result when evaluating an expression. Without this standard order of operations, two people might evaluate an expression diff erently and arrive at diff erent values. For example, without the order of operations, someone might evaluate all expressions from left to right, while another person performs all additions and subtractions before all multiplications and divisions.

You can use the acronym P.E.M.A. (Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction) to help you remember the order of operations.

Problem

How do you evaluate the expression 3 1 4 3 2 2 10 4 5?

3 1 8 2 10 4 5 There are no parentheses or exponents, so fi rst,5 3 1 8 2 2 do any multiplication or division from left to right.

5 11 2 2 Do any addition or subtraction from left to right. 5 9

Exercises

Simplify each expression.

14. (5 1 3)2 15. (8 2 5)(14 2 6)

16. (15 2 3) 4 4 17. Q22 1 35 R

18. 40 2 15 4 3 19. 20 1 12 4 2 2 5

20. (42 1 52)2 21. 4 3 5 2 32 3 2 4 6

Write and simplify an expression to model the relationship expressed in the situation below.

22. Manuela has two boxes. Th e larger of the two boxes has dimensions of 15 cm by 25 cm by 20 cm. Th e smaller of the two boxes is a cube with sides that are 10 cm long. If she were to put the smaller box inside the larger, what would be the remaining volume of the larger box?

1-2 Reteaching(continued)

Order of Operations and Evaluating Expressions

64 24

3 5

35 21

1681 17

15 3 25 3 20 2 103 5 6500 cm3

Name Class Date


29

1-3 ReteachingReal Numbers and the Number Line

A number that is the product of some other number with itself, or a number to the second power, such as 9 5 3 3 3 5 32, is called a perfect square. Th e number that is raised to the second power is called the square root of the product. In this case, 3 is the square root of 9. Th is is written in symbols as !9 5 3. Sometimes square roots are whole numbers, but in other cases, they can be estimated.

Problem

What is an estimate for the square root of 150?

Th ere is no whole number that can be multiplied by itself to give the product of 150.

10 3 10 5 100

11 3 11 5 121

12 3 12 5 144

13 3 13 5 169

You cannot fi nd the exact value of !150, but you can estimate it by comparing 150 to perfect squares that are close to 150.

150 is between 144 and 169, so !150 is between !144 and !169.

!144 , !150 , !169

12 , !150 , 13

Th e square root of 150 is between 12 and 13. Because 150 is closer to 144 than it is to 169, we can estimate that the square root of 150 is slightly greater than 12.

Exercises

Find the square root of each number. If the number is not a perfect square, estimate the square root to the nearest integer.

1. 100 2. 49 3. 9

4. 25 5. 81 6. 169

7. 15 8. 24 9. 40

10. A square mat has an area of 225 cm2. What is the length of each side of the mat?

10 7 3

5 9 13

4 5 6

15 cm

Name Class Date


30


Real Numbers and the Number Line

Th e real numbers can be separated into smaller, more specifi c groups, called subsets. Each of these subsets has certain characteristics. For example, a rational number can be expressed as a fraction of two integers, with the denominator of the fraction not equal to 0. Irrational numbers cannot be expressed as a fraction of two integers.

Every real number belongs to at least one subset of the real numbers. Some real numbers belong to multiple subsets.

Problem

To which subsets of the real numbers does 17 belong?

17 is a natural number, a whole number, and an integer.

But 17 is also a rational number because it can be written as 171 , a fraction of two

integers with the denominator not equal to 0.

A number cannot belong to both the subset of rational numbers and the subset of irrational numbers, so 17 is not an irrational number.

Exercises

List the subsets of the real numbers to which each of the given numbers belongs.

11. 5 12. 116 13. !3

14. 17.889 15. 225 16. 268

17. 21720 18. 0 19. !16

20. !20 21. !6.25 22. 7710

rational, whole, natural, integer

rational, whole, natural, integer

irrational

rational rational, integer rational, integer

rational rational, whole, integer

rational, natural, whole, integer

irrational rational rational


39

Name Class Date

Equivalent algebraic expressions are expressions that have the same value for all

values for the variable(s). For example x 1 x and 2x are equivalent expressions

since, regardless of what number is substituted in for x, simplifying each

expression will result in the same value. Certain properties of real numbers lead to

the creation of equivalent expressions.

Commutative Properties

Th e commutative properties of addition and multiplication state that changing the order of the addends does not change the sum and that changing the order of factors does not change the product.

Addition: a 1 b 5 b 1 a Multiplication: a ? b 5 b ? a

To help you remember the commutative properties, you can think about the root

word “commute.” To commute means to move. If you think about commuting or

moving when you think about the commutative properties, you will remember

that the addends or factors move or change order.

Problem

Do the following equations illustrate commutative properties?

a. 3 1 4 5 4 1 3 b. (5 3 3) 3 2 5 5 3 (3 3 2) c. 1 2 3 5 3 2 1

3 1 4 and 4 1 3 both simplify to 7, so the two sides of the equation in part (a) are

equal. Since both sides have the same two addends but in a diff erent order, this

equation illustrates the Commutative Property of Addition.

Th e expression on each side of the equation in part (b) simplifi es to 30. Both

sides contain the same 3 factors. However, this equation does not illustrate the

Commutative Property of Multiplication because the terms are in the same order

on each side of the equation.

1 2 3 and 3 2 1 do not have the same value, so the equation in part (c) is

not true. Th ere is not a commutative property for subtraction. Nor is there a

commutative property for division.

Associative Properties

Th e associative properties of addition and multiplication state that changing the grouping of addends does not change the sum and that changing the grouping of factors does not change the product.

Addition: (a 1 b) 1 c 5 a 1 (b 1 c) Multiplication: (a ? b) ? c 5 a ? (b ? c)

1-4 ReteachingProperties of Real Numbers


40

Name Class Date

Problem

Do the following equations illustrate associative properties? a. (1 1 5) 1 4 5 1 1 (5 1 4) b. 4 3 (2 3 7) 5 4 3 (7 3 2)

(1 1 5) 1 4 and 1 1 (5 1 4) both simplify to 10, so the two sides of the equation in part (a) are equal. Since both sides have the same addends in the same order but grouped diff erently, this equation illustrates the Associative Property of Addition.

Th e expression on each side of the equation in part (b) simplifi es to 56. Both sides contain the same 3 factors. However, the same factors that were grouped together on the left side have been grouped together on the right side; only the order has changed. Th is equation does not illustrate the Associative Property of Multiplication.

Other properties of real numbers include: a. Identity property of addition: a 1 0 5 0 12 1 0 5 12 b. Identity property of multiplication: a ? 1 5 a 32 ? 1 5 32 c. Zero property of multiplication: a ? 0 5 0 6 ? 0 5 0 d. Multiplicative property of negative one: 21 ? a 5 2a 21 ? 7 5 27

Exercises

What property is illustrated by each statement?

1. (m 1 7.3) 1 4.1 5 m 1 (7.3 1 4.1) 2. 5p ? 1 5 5p

3. 12x 1 4y 1 0 5 12x 1 4y 4. (3r)(2s) 5 (2s)(3r)

5. 17 1 (22) 5 (22) 1 17 6. 2(23) 5 3

Simplify each expression. Justify each step.

7. (12 1 8x) 1 13 8. (5 ? m) ? 7

9. (7 2 7) 1 12


Properties of Real Numbers

5 (8x 1 12) 1 13 Comm. Prop. of Add.5 8x 1 (12 1 13) Assoc. Prop. of Add.5 8x 1 25 Combine like terms.

5 (m ? 5) ? 7 Comm. Prop. of Mult.5 m ? (5 ? 7) Assoc. Prop. of Mult.5 35m Comm. Prop. of Mult.

5 0 1 12 Add. Ident.5 12 Simplify.

Associative Property of Addition Multiplicative Identity

Additive Identity Commutative Property of Multiplication

Commutative Property of Addition Multiplicative Property of 21


49

Name Class Date

You can add real numbers using a number line or using the following rules.

Rule 1: To add two numbers with the same sign, add their absolute values. Th e

sum has the same sign as the addends.

Problem

What is the sum of 27 and 24?

Use a number line.

Start at zero. Move 7 spaces to the left to represent 27. Move another 4 spaces to the left to represent 24.

Th e sum is –11.

Use the rule.

27 1 (24) The addends are both negative.

|27| 1 |24| Add the absolute values of the addends.

7 1 4 5 11 |27| 5 7 and |24| 5 4.

27 1 (24) 5 211 The sum has the same sign as the addends.

Rule 2: To add two numbers with diff erent signs, subtract their absolute values.

Th e sum has the same sign as the addend with the greater absolute value.

Problem

What is the sum of 26 and 9?

Use the rule.

9 1 (26) The addends have different signs.

|9| 2 |26| Subtract the absolute values of the addends.

9 2 6 5 3 |9| 5 9 and |26| 5 6.

9 1 (26) 5 3 The positive addend has the greater absolute value.

1-5 ReteachingAdding and Subtracting Real Numbers

21121029 28 27 26 25 24 23 22 21 0 1


50

Name Class Date

Exercises

Find each sum.

1. 24 1 212 2. 23 1 15 3. 29 1 1

4. 13 1 (27) 5. 8 1 (214) 6. 211 1 (25)

7. 4.5 1 (21.1) 8. 25.1 1 8.3 9. 6.4 1 9.8

Addition and subtraction are inverse operations. To subtract a real number,

add its opposite.

Problem

What is the diff erence 25 2 (28)?

25 2 (28) 5 25 1 8 The opposite of 28 is 8.

5 3 Use Rule 2.

Th e diff erence 25 2 (28) is 3.

Exercises

Find each diff erence.

10. 8 2 20 11. 6 2 (212) 12. 24 2 9

13. 28 2 (214) 14. 211 2 (24) 15. 17 2 25

16. 3.6 2 (22.4) 17. 21.5 2 (21.5) 18. 21.7 2 5.4

19. Th e temperature was 58C. Five hours later, the temperature had dropped

108C. What is the new temperature?

20. Reasoning Which is greater, 52 1 (277) or 52 2 (277)? Explain.


Adding and Subtracting Real Numbers

216 12 28

6 26 216

3.4

212 18 213

6 27 28

6 0 27.1

258C

52 2 (277) is greater. It is the same as 52 1 77 which is a positive number. The sum of 52 1 (277) is a negative number.

3.2 16.2

Name Class Date


59

You need to remember two simple rules when multiplying or dividing real

numbers.

1. Th e product or quotient of two numbers with the same sign is positive.

2. Th e product or quotient of two numbers with diff erent signs is negative.

Problem

What is the product –6(–30)?

26(230) 5 180 26 and 230 have the same sign so the product is positive.

Problem

What is the quotient 72 4 (26)?

72 4 (26) 5 212 72 and 26 have different signs so the quotient is negative.

Exercises

Find each product or quotient.

1. 25(26) 2. 7(220) 3. 23 3 22

4. 44 4 2 5. 81 4 (29) 6. 255 4 (211)

7. 262 4 2 8. 25 ? (24) 9. (26)2

10. 29.9 4 3 11. 27.7 4 (211) 12. 21.4(22)

13. 2

12 3

13 14. 2

23Q2

35R 15. 3

4 ? Q2

13R

16. Th e temperature dropped 2°F each hour for 6 hours. What was the total

change in temperature?

17. Reasoning Since 52 5 25 and (25)2 5 25, what are the two values for the

square root of 25?

1-6 ReteachingMultiplying and Dividing Real Numbers

30 2662140

22 29 5

231

23.3

2

16

2100

0.7

2128 F

5 and 25

25

36

2.8

2

14

Name Class Date


60


Multiplying and Dividing Real Numbers

Th e product of 7 and 17 is 1. Two numbers whose product is 1 are called

reciprocals. To divide a number by a fraction, multiply by its reciprocal.

Problem

What is the quotient 23 4 Q2

57R?

23 4 Q2

57R 5

23 3 Q2

75R To divide by a fraction, multiply by its reciprocal.

5 2

1415 The signs are different so the answer is negative.

Exercises

Find each quotient.

18. 12 4

13 19. 26 4

23 20. 2

25 4 Q2

23R

21. 12 4 Q2

14R 22. Q2

57R 4 Q2

12R 23. 2

23 4

14

24. Writing Another way of writing ab is a 4 b. Explain how you could evaluate

12

16

.

What is the value of this expression?

1

12

22

Change the problem to the equivalent division problem 12 416. To fi nd this

quotient, change this division problem to the multiplication problem 12 361. The

answer is 3.

29

1

37

35

22

23


69

Name Class Date

Th e Distributive Property states that the product of a sum and another factor can be rewritten as the sum of two products, each term in the sum multiplied by the other factor. For example, the Distributive Property can be used to rewrite the product 3(x 1 y) as the sum 3x 1 3y. Each term in the sum x 1 y is multiplied by 3; then the new products are added.

Problem

What is the simplifi ed form of each expression?

a. 4(x 1 5) b. (2x 2 3)(23) 5 4(x) 1 4(5) Distributive Property 5 2x(23) 2 3(23) Distributive Property 5 4x 1 20 Simplify. 5 26x 1 9 Simplify.

Th e Distributive Property can be used whether the factor being multiplied by a sum or diff erence is on the left or right.

Th e Distributive Property is sometimes referred to as the Distributive Property of Multiplication over Addition. It may be helpful to think of this longer name for the property, as it may remind you of the way in which the operations of multiplication and addition are related by the property.

Exercises

Use the Distributive Property to simplify each expression.

1. 6(z 1 4) 2. 2(22 2 k) 3. (5x 1 1)4 4. (7 2 11n)10

5. (3 2 8w)4.5 6. (4p 1 5)2.6 7. 4(y 1 4) 8. 6(q 2 2)

Write each fraction as a sum or diff erence.

9. 2m 2 59 10. 8 1 7z

11 11. 24f 1 15

9 12. 12d 2 166


13. 2(6 1 j) 14. 2(29h 2 4) 15. 2(2n 1 11) 16. 2(6 2 8 f)

1-7 ReteachingThe Distributive Property

6z 1 24

13.5 2 36w

26 2 j

2m9 2 5

9

24 2 2k

10.4p 1 13

9h 1 4

811 1

7z11

20x 1 4

4y 1 16

n 2 11

8f3 1

53

70 2 110n

6q 2 12

26 1 8f

2d 2 83


70

Name Class Date

Th e previous problem showed how to write a product as a sum using the Distributive Property. Th e property can also be used to go in the other order, to convert a sum into a product.

Problem

How can the sum of like terms 15x 1 6x be simplifi ed using the Distributive Property?

Each term of 15x 1 6x has a factor of x. Rewrite 15x 1 6x as 15(x) 1 6(x).Now use the Distributive Property in reverse to write 15(x) 1 6(x) as (15 1 6)x , which simplifi es to 21x.

Exercises

Simplify each expression by combining like terms.

17. 16x 1 12x 18. 25n 2 17n 19. 24p 1 6p

20. 215a 2 9a 21. 29k2 2 5k2 22. 12t2 2 20t2

By thinking of or rewriting numbers as sums or diff erences of other numbers that are easier to use in multiplication, the Distributive Property can be used to make calculations easier.

Problem

How can you multiply 78 by 101 using the Distributive Property and mental math?

78 3 101 Write the product.

78 3 (100 1 1) Rewrite 101 as sum of two numbers that are easy to use in multiplication.

78(100) 1 78(1) Use the Distributive Property to write the product as a sum.

7800 1 78 Multiply.

7878 Simplify.

Exercises

Use mental math to fi ntd each product.

23. 5.1 3 7 24. 24.95 3 4 25. 999 3 11 26. 12 3 95


The Distributive Property

35.7

28x

224a

8n

214k2

2p

28t2

99.8 10,989 1140


79

Name Class Date

An equation is a mathematical sentence with an equal sign. An equation can be true, false, or open. An equation is true if the expressions on both sides of the equal sign are equal, for example 2 1 5 5 4 1 3. An equation is false if the expressions on both sides of the equal sign are not equal, for example 2 1 5 5 4 1 2.

An equation is considered open if it contains one or more variables, for example x 1 2 5 8. When a value is substituted for the variable, you can then decide whether the equation is true or false for that particular value. If an open sentence is true for a value of the variable, that value is called a solution of the equation. For x 1 2 5 8, 6 is a solution because when 6 is substituted in the equation for x, the equation is true: 6 1 2 5 8.

Problem

Is the equation true, false, or open? Explain.

a. 15 1 21 5 30 1 6 The equation is true, because both expressions equal 36. b. 24 4 8 5 2 ? 2 The equation is false, because 24 4 8 5 3 and 2 ? 2 5 4; 3 2 4. c. 2n 1 4 5 12 The equation is open, because there is a variable in the expression

on the left side.

Tell whether each equation is true, false, or open. Explain.

1. 2(12) 2 3(6) 5 12 2. 3x 1 12 5 219 3. 14 2 19 5 25

4. 2(28) 1 4 5 12 5. 7 2 9 1 3 5 x 6. (28 1 12) 4 22 5 220

7. 14 2 (28) 2 14 5 8 8. (13 2 16) 4 3 5 1 9. 42 4 7 1 3 5 9

Problem

Is x 5 23 a solution of the equation 4x 1 5 5 27?

4x 1 5 5 27

4(23) 1 5 5 27 Substitute 23 for x.

27 5 27 Simplify.

Since 27 5 27, 23 is a solution of the equation 4x 1 5 5 27.

Tell whether the given number is a solution of each equation.

10. 4x 2 1 5 227; 27 11. 18 2 2n 5 14; 2 12. 21 5 3p 2 5; 9

13. k 5 (26)(28) 2 14; 262 14. 20v 1 36 5 2156; 26 15. 8y 1 13 5 21; 1

16. 224 2 17t 5 258; 2 17. 226 5 13 m 1 5; 27 18. 1

4 g 2 8 5 32 ; 38

1-8 ReteachingAn Introduction to Equations

false; 2(12) 2 3(6) 5 6

false; 2(28) 1 4 5 212

true

no

no

yes

open; it contains a variable

open; it contains a variable

false; (13 2 16) 4 3 5 21

yes

no

no

true

true

true

no

yes

yes


80

Name Class Date

A table can be used to fi nd or estimate a solution of an open equation. You will have to choose a value to begin your table. If you choose the value that makes the equation true, you have found the solution and are done. If your choice is not the solution, make another choice based on the values of both sides of the equation for your fi rst choice. If you choose one value that makes one side of the equation too high and then another value that makes that same side too low, you know that the solution must lie between the two values you chose. It may not be possible to determine an exact solution for each equation; estimating the solution to be between two integers may be all that is possible in some cases.

Problem

What is the solution of 6n 1 8 5 28?

If n 5 2, then the left side of the equation is 6(2) 1 8 or 20, which is too low.

If n 5 5, then the left side of the equation is 6(5) 1 8 or 38, which is too high.

Th e solution must lie between 2 and 5, so keep trying values between them.

If n 5 3, then the left side of the equation is 6(3) 1 8 or 26, which is too low.

If n 5 4, then the left side of the equation is 6(4) 1 8 or 32, which is too high.

Th e solution must lie between 3 and 4, but there are no other integers between 3 and 4.

You can give an estimate for the solution of 6n 1 8 5 28 as being between the integers 3 and 4.

Write an equation for each sentence.

19. 13 times the sum of a number and 5 is 91.

20. Negative 8 times a number minus 15 is equal to 30.

21. Jared receives $23 for each lawn he mows. What is an equation that relates the number of lawns w that Jared mows and his pay p?

22. Shariff has been working for a company 2 years longer than Patsy. What is an equation that relates the years of employment of Shariff S and the years of employment of Patsy P?

Use mental math to fi nd the solution of each equation.

23. h 1 6 5 13 24. 211 5 n 1 2 25. 6 2 k 5 14 26. 5 5 28 1 t

27. z5 5 22 28.

j26 5 12 29. 8c 5 248 30. 215a 5 245

Use a table to fi nd the solution of each equation.

31. 23b 2 12 5 15 32. 15y 1 6 5 21 33. 28 5 5y 1 22 34. 6t 2 1 5 249


An Introduction to Equations

13(n 1 5) 5 91

28n 2 15 5 30

p 5 23w

S 5 P 1 2

7

210

29

213 28 13

272

1

26

26

3

28


89

Name Class Date

Tables, equations, and graphs are some of the ways that a relationship between two quantities can be represented. You can use the information provided by one representation to produce one of the other representations; for example, you can use data from a table to produce a graph. You can also use any of the representations to draw conclusions about the relationship.

Problem

Are (2, 11) and (5, 3) solutions of the equation y 5 3x 1 5?

For each ordered pair, you can substitute the x- and y- coordinates into the equation for x and y and then simplify to see if the values satisfy the equation.

For (2, 11): For (5, 3):

11 5 3(2) 1 5 Substitute for x and y. 3 5 3(5) 1 5

11 5 11 Multiply and then add. 3 2 20

Since both sides of the equation have the same value, the ordered pair (2, 11) is a solution of the equation y 5 3x 1 5. Since the two sides of the equation have diff erent values, the ordered pair (5, 3) is not a solution of the equation y 5 3x 1 5.

Problem

Th e table shows the relationship between the number of hours Kaya works at her job and the amount of pay she receives. Extend the pattern. How much money would Kaya earn if she worked 40 hours?

Method 1: Write an equation.

y 5 12.50x Kaya earns $12.50 per hour.

5 12.50(40) Substitute 40 for x.

5 500 Simplify.

She would earn $500 in 40 hours.

Method 2: Draw a graph.

She would earn $500 in 40 hours.

1-9 ReteachingPatterns, Equations, and Graphs

HoursWorked

MoneyEarned ($)

37.50

75

112.50

150

3

6

9

12

x

y

4

50100150200250300350400450500

8 12 16 20 24 28 32 36 40Hours worked

Mon

ey E

arne

d ($

)

O


90

Name Class Date

Exercises

Tell whether the equation has the given ordered pair as a solution.

1. y 5 x 2 7; (2, 25) 2. y 5 x 1 6; (25, 11) 3. y 5 2x 1 1; (21, 0)

4. y 5 25x ; (23, 215) 5. y 5 x 2 8; (7, 21) 6. y 5 x 1 34; (21, 2

14 )

Use a table, an equation, and a graph to represent each relationship.

7. Tickets to the fair cost $17. 8. Brian is 5 years older than Sam.

Use the table to draw a graph and answer the question.

9. Th e table shows Jake’s earnings for the number of cakes he

baked. What are his earnings for baking 75 cakes?

Use the table to write an equation and answer the question.


Patterns, Equations, and Graphs

Cakes Earnings($)

5

10

120

240

36015

10. Th e table shows the number

of miles that Kate runs on a

weekly basis while training for a

race. How many total miles will

she have run after 15 weeks?

11. Th e table shows the amount of

money Kevin receives for

items that he sells. How much

will he earn if he sells 30 items?

TrainingWeeks

MilesRun

40

80

120

1

2

3

ItemsSold

Earnings($)

1125

1500

1875

15

20

25

Tickets

1

2

3

4

5

17

34

51

68

85

Cost ($)

yes no no

no

y 5 17x y 5 x 1 5

$1800

yes yes

20

2 4 6 8

40

60

80

x

y Sam (yrs)

0

3

6

9

12

5

8

11

14

17

Brain (yrs)

10

2 6 10 14 18

20

30

400

x

y

100

2 6 10 14

200

350

400

x

y

y 5 40x; 600 mi y 5 75x; $2250

Name Class Date


9

2-1 ReteachingSolving One-Step Equations

You can use the properties of equality to solve equations. Subtraction is the inverse of addition.

Problem

What is the solution of x 1 5 5 33?

In the equation, x 1 5 5 33, 5 is added to the variable. To solve the equation, you need to isolate the variable, or get it alone on one side of the equal sign. Undo adding 5 by subtracting 5 from each side of the equation.

Drawing a diagram can help you write an equation to solve the problem.

Solve x 1 5 5 33

x 1 5 2 5 5 33 2 5 Undo adding 5 by subtracting 5.

x 5 28 Simplify. This isolates x.

Check x 1 5 5 33 Check your solution in the original equation.

28 1 5 0 33 Substitute 28 for x.

33 5 33 3

Th e solution to x 1 5 5 33 is 28.

Division is the inverse of multiplication.

Problem

What is the solution of x5 5 12?

In the equation, x5 5 12, the variable is divided by 5. Undo

dividing by 5 by multiplying by 5 on each side of the equation.

Solve x5 5 12

x5 ? 5 5 12 ? 5 Undo dividing by 5 by multiplying by 5.

x 5 60 Simplify. This isolates x.

Th e solution to x5 5 12 is 60.

X

33

5PartPart

Whole

X

12 12 12 12 12

Name Class Date


10

Exercises

Solve each equation using addition or subtraction. Check your answer.

1. 23 5 n 1 9 2. f 1 6 5 26 3. m 1 12 5 22

4. r 1 2 5 7 5. b 1 1.1 5 211 6. t 1 9 5 4

Defi ne a variable and write an equation for each situation. Th en solve.

7. A student is taking a test. He has 37 questions left. If the test has 78 questions, how many questions has he fi nished?

8. A friend bought a bouquet of fl owers. Th e bouquet had nine daisies and some roses. Th ere were a total of 15 fl owers in the bouquet. How many roses were in the bouquet?

Solve each equation using multiplication or division. Check your answer.

9. z8 5 2 10. 226 5 c

13 11. q

11 5 26

12. 2a3 5 18 13. 225 5

g5 14. 20.4 5 s

2.5

15. A student has been typing for 22 minutes and has typed a total of 1496 words. Write and solve an equation to determine the average number of words she can type per minute.


Solving One-Step Equations

212

5

16

51

f 1 37 5 78; 41

9 1 r 5 15; 6

22a 5 1496; 68

212

212.1

254

2338 266

2125

25

10

Name Class Date


19

2-2 ReteachingSolving Two-Step Equations

Properties of equality and inverse operations can be used to solve equations that involve more than one step to solve. To solve a two-step equation, identify the operations and undo them using inverse operations. Undo the operations in the reverse order of the order of operations.

Problem

What is the solution of 5x 2 8 5 32?

5x 2 8 1 8 5 32 1 8 To get the variable term alone on the left side, add 8 to each side.

5x 5 40 Simplify.

5x5 5

405 Divide each side by 5 since x is being multiplied by 5 on the left

side. This isolates x.

x 5 8 Simplify.

Check 5x 2 8 5 32 Check your solution in the original equation.

5(8) 2 8 5 32 Substitute 8 for x.

32 5 32 3 Simplify.

To solve 216 5 x3 1 5, you can use subtraction fi rst to undo the addition, and

then use multiplication to undo the division.

Problem

What is the solution of 216 5 x3 1 5?

216 2 5 5 x3 1 5 2 5 To get the variable term alone on the right, subtract 5

from each side.

221 5 x3 Simplify.

3(221) 5 3ax3b Since x is being divided by 3, multiply each side by 3 to

undo the division. This isolates x.

263 5 x Simplify.

Name Class Date


20

Solve each equation. Check your answer.

1. 4f 2 8 5 20 2. 25 2 6b 5 55

3. 2z 1 7 5 28 4. w29 1 7 5 10

5. 25 5 8 1 n2 6.

y 2 83 5 27

Solve each equation. Justify each step.

7. 6d 2 5 5 31

8. p 2 722 5 5

Defi ne a variable and write an equation for each situation. Th en solve.

9. Ray’s birthday is 8 more than four times the number of days away from today

than Jane’s birthday. If Ray’s birthday is 24 days from today, how many days

until Jane’s birthday?

10. Jerud weighs 15 pounds less than twice Kate’s weight. How much does Kate

weigh if Jerud weighs 205 pounds?

11. A phone company charges a fl at fee of $17 per month, which includes free

local calling plus $0.08 per minute for long distance calls. Th e Taylor’s phone

bill for the month is $31.80. How many minutes of long distance calling did

they use during the month?

12. A delivery company charges a fl at rate of $3 for a large envelope plus an

additional $0.25 per ounce for every ounce over a pound the package weighs.

Th e postage for the package is $5.50. How much does the package weigh?

(Hint: remember the fi rst pound is included in the $3.)


Solving Two-Step Equations

22 ?p 2 722 5 5 ? 22 Multiply both sides by 22. (Mult. Prop. of Equal.)

p 2 7 5 210 Simplify.

p 2 7 1 7 5 210 1 7 Add 7 to both sides. (Add. Prop. of Equal.) p 5 23 Simplify.

24 5 8 1 4j; in 4 days

6d 2 5 1 5 5 31 1 5 Add 5 to each side. (Add. Prop. of Equal.)

6d 5 36 Simplify.

6d6 5 36

6 Divide both sides by 6. (Div. Prop. of Equal.)

d 5 6 Simplify.

7 25

227

213

15

34

205 5 2k 2 15; 110 lb

17 1 0.08m 5 31.80; 185 min

5.50 5 3 1 0.25e; 1 lb 10 oz;

Name Class Date


29

2-3 ReteachingSolving Multi-Step Equations

To solve multi-step equations, use properties of equality, inverse operations, the

Distributive Property, and properties of real numbers to isolate the variable. Like

terms on either side of the equation should be combined fi rst.

Problem

a) What is the solution of 23y 1 8 1 13y 5 252?

23y 1 13y 1 8 5 252 Group the terms with y together so that the like terms are grouped together.

10y 1 8 5 252 Add the coeffi cients to combine like terms.

10y 1 8 2 8 5 252 2 8 To get the variable term by itself on the left side, subtract 8 from each side.

10y 5 260 Simplify.

10y10 5

26010 Divide each side by 10 since y is being multiplied by 10 on

the left side. This isolates y.

y 5 26 Simplify.

b) What is the solution of 22(3n 2 4) 5 210?

26n 1 8 5 210 Distribute the 22 into the parentheses by multiplying each term inside by 22.

26n 1 8 2 8 5 210 2 8 To get the variable term by itself on the left side, subtract 8 from each side.

26n 5 218 Simplify.

26n26 5

21826 Divide each side by 26 since n is being multiplied by 26 on

the left side. This isolates n.

n 5 3 Simplify.


1. 4 2 6h 2 8h 5 60 2. 232 5 27n 2 12 1 3n 3. 14 1 12 5 215x 1 2x

4. 8(23d 1 2) 5 88 5. 222 5 2(x 2 4) 6. 35 5 25(2k 1 5)

7. 3m 1 6 2 2m 5 222 8. 4(3r 1 2) 2 3r 5 210 9. 218 5 15 2 3(6t 1 5)

10. 25 1 2(10b 2 2) 5 31 11. 7 5 5x 1 3(x 2 2) 1 5 12. 218 5 3(2z 1 6) 1 2z

13. Reasoning Solve the equation 14 5 7(2x 2 4) using two diff erent methods.

Show your work. Which method do you prefer? Explain.

24

23

228

2

3; Check students’ work.

5

26

22

1

22

26

1

36

Name Class Date


30


Solving Multi-Step Equations

Equations with fractions can be solved by using a common denominator or by eliminating the fractions altogether.

Problem

What is the solution of x4 223 5

712?

Method 1 Method 2

Get a common denominator fi rst. Multiply by the common denominator fi rst.

33Q

x4R 2

44Q

23R 5

712 12Qx4 2

23R 5 12Q 7

12R

3x12 2

812 5

712 123 Qx

4R 2 124 Q2

3R 5 12Q 7

12R

3x12 5

1512 3x 2 8 5 7

3x12 ?

123 5

1512 ?

123 3x 5 15

x 5 5 x 5 5

Decimals can be cleared from the equation by multiplying by a power of ten with the same number of zeros as the number of digits to the right of the decimal. For instance, if the greatest number of digits after the decimal is 3, like 4.586, you multiply by 1000.

Problem

What is the solution of 2.8x 2 4.25 5 5.55?

100(2.8x 2 4.25 5 5.55) Multiply by 100 because the most number of digits after the decimal is two.

280x 2 425 5 555 Simplify by moving the decimal point to the right 2 places in each term.

280x 5 980 Add 425 to each side to get the term with the variable by itself on the left side.

x 5 3.5 Divide each side by 280 to isolate the variable.


14. x16 2

12 5

38 15. 2a

3 189 5 4 16. 3n

7 2 1 5 18

17. 21.68j 1 1.24 5 13 18. 4.6 5 3.5w 2 6.6 19. 5.23y 1 3.02 5 22.21

14

27 3.2 21

143

218

Name Class Date


39

To solve equations with variables on both sides, you can use the properties of equality and inverse operations to write a series of simpler equivalent equations.

Problem

What is the solution of 2m 2 4 1 5m 5 13 2 6m 2 4?

7m 2 4 5 26m 1 9 Add the terms with variables together on the left side and the constants on the right side to combine like terms.

7m 2 4 1 6m 5 26m 1 9 1 6m To move the variables to the left side, add 6m to each side.

13m 2 4 5 9 Simplify.

13m 2 4 1 4 5 9 1 4 To get the variable term alone on the left, add 4 to each side.

13m 5 13 Simplify.

13m13 5

1313 Divide each side by 13 since x is being multiplied by 13 on the

left side. This isolates x.

m 5 1 Simplify.

Problem

What is the solution of 3(5x 2 2) 5 23(x 1 6)?

15x 2 6 5 23x 2 18 Distribute 3 on the left side and 23 on the right side into the parentheses by multiplying them by each term inside.

15x 2 6 1 6 5 23x 2 18 1 6 To move all of the terms without a variable to the right side, add 6 to each side.

15x 5 23x 2 12 Simplify.

15x 1 3x 5 23x 2 12 1 3x To get the variable terms to the left side, add 3x to each side.

18x 5 212 Simplify.

18x18 5 2

1218 Divide each side by 18 since x is being multiplied by 18 on the

left side. This isolates x.

x 5 2 23 Simplify and reduce the fraction.


1. 25x 1 9 5 23x 1 1 2. 14 1 7n 5 14n 1 28 3. 22(g 2 1) 5 2g 1 8

4. 2d 1 12 2 3d 5 5d 2 6 5. 4(m 2 2) 5 22(3m 1 3) 6. 2(4y 2 8) 5 2(y 1 4)

7. 5a 2 2(4a 1 5) 5 7a 8. 11w 1 2(3w 2 1) 5 15w 9. 4(3 2 5p) 5 25(3p 1 3)

2-4 ReteachingSolving Equations With Variables on Both Sides

4

2

21

22

15

1

1.5

0

275

Name Class Date


40


Solving Equations With Variables on Both Sides

An equation that is true for every value of the variable for which the equation is defi ned is an identity. For example, x 2 5 5 x 2 5 is an identity because the equation is true for any value of x. An equation has no solution if there is no value of the variable that makes the equation true. Th e equation x 1 6 5 x 1 3 has no solution.

Problem

What is the solution of each equation? a) 3(4x 2 2) 5 22(26x 1 3)

12x 2 6 5 12x 2 6 Distribute 3 on the left side and 22 on the right side into the parentheses by multiplying them by each term inside.

12x 2 6 2 12x 5 12x 2 6 2 12x To get the variable terms to the left side, subtract 12x from each side.

26 5 26 Simplify.

Because 26 5 26 is always true, there are infi nitely many solutions of the original equation. Th e equation is an identity.

b) 2n 1 4(n 2 2) 5 8 1 6n

2n 1 4n 2 8 5 8 1 6n Distribute 4 into the parentheses by multiplying it by each term inside.

6n 2 8 5 8 1 6n Add the variable terms on the left side to combine like terms.

6n 2 8 2 6n 5 8 1 6n 2 6n To get the variable terms to the left side, subtract 6n from each side.

28 5 8 Simplify.

Since 28 2 8 , the equation has no solution.

Determine whether each equation is an identity or whether it has no solution.

10. 23(2x 1 1) 5 2(23x 2 1) 11. 4(23x 1 4) 5 22(6x 2 8) 12. 3n 1 3(2n 1 3) 5 3

Solve each equation. If the equation is an identity, write identity. If it has no solution, write no solution.

13. 2(4n 1 2) 5 22(2n 2 1) 14. 2(2d 1 4) 5 2d 1 8 15. 2k 2 18 5 25 2 k 2 13

16. Open-Ended Write three equations with variables on both sides of the equal sign with one having no solution, one having exactly one solution, and one being an identity.

no solution

no solution

Answers may vary. Sample: no solution: 3y 1 7 5 3y 2 12; one solution:

3y 1 7 5 y 2 12; an identity: 2y 2 10 1 5y 5 7y 2 10;

an identity

0

no solution

an identity

Name Class Date


49

A literal equation is an equation that involves two or more variables. When you

work with literal equations, you can use the methods you have learned in this

chapter to isolate any particular variable. To solve for specifi c values of a variable,

simply substitute the values into your equation and simplify.

Problem

What is the solution of 4x 2 5y 5 3 for y? What is the value of y when x 5 10?

4x 2 5y 2 4x 5 3 2 4x To get the y-term by itself on the left side, subtract 4x from each side.

25y 5 24x 1 3 Simplify.

25y25 5

24x 1 325 Divide each side by 25 since y is being multiplied by 25 on

the left side. This isolates y.

y 5 45x 2 3

5 Simplify by dividing each term by 25. Notice, this changes

the sign of each term.

y 5 45(10) 2 3

5 To fi nd the value of y when x 5 10, substitute 10 in for x.

y 5 725 Simplify by multiplying fi rst, then subtracting.

When you rewrite literal equations, you may have to divide by a variable or

variable expression. When you do so in this lesson, assume that the variable or

variable expression is not equal to zero because division by zero is not defi ned.

Problem

Solve the equation ab 2 bc 5 cd for b.

b(a 2 c) 5 cd Since b is a factor of each term on the left side, it can be factored out using the Distributive Property.

b(a 2 c)

a 2 c 5cd

a 2 c To get b by itself, divide each side by a 2 c since b is being multiplied by a 2 c. Remember a 2 c 2 0.

b 5 cda 2 c Simplify.

Solve each equation for y. Th en fi nd the value of y for each value of x.

1. y 1 5x 5 2; 21, 0, 1 2. 6x 5 2y 2 4; 1, 2, 4 3. 6x 2 3y 5 29; 22, 0, 2

4. 4y 5 5x 2 8; 22, 21, 0 5. 3y 1 2x 5 25; 0, 2, 3 6. 5x 5 8y 2 6; 21, 0, 1

7. 3(y 2 2) 1 x 5 1; 21, 0, 1 8. x 1 2y 2 3 5 1; 21, 0, 1 9.

y 1 4x 2 5 5 23; 22, 2, 4

2-5 ReteachingLiteral Equations and Formulas

y 5 2 2 5x; 7; 2; 23;

y 5 5x 2 84 ; 24

12; 23

14; 22;

y 5 7 2 x3 ; 2

23; 2

13; 2;

y 5 3x 1 2; 5; 8; 14;

y 5 25 2 2x3 ; 212

3; 23; 23 23;

y 5 x 1 5; 4; 5; 6;

y 5 2x 1 3; 21; 3; 7;

y 5 5x 1 68 ; 18; 34; 1

38

y 5 23x 1 11; 17; 5; 21;

Name Class Date


50


Literal Equations and Formulas

A formula is an equation that states a relationship among quantities. Formulas

are special types of literal equations. Some common formulas are shown below.

Notice that some of the formulas use the same variables, but the defi nitions of the

variables are diff erent. For instance, r is the radius in the area and circumference

of a circle and the rate in the distance formula.

Formula Name Formula

Perimeter of a rectangle P 5 2l 1 2w

Circumference of a circle C 5 2πr

Area of a rectangle A 5 lw

Area of a triangle A 5 12bh

Area of a circle A 5 πr2

Distance traveled d 5 rt

Each of the formulas can be solved for any of the other unknowns in the equation

to produce a new formula. For example, r 5 C2π is a formula for the radius of a

circle in terms of its circumference.

Problem

What is the length of a rectangle with width 24 cm and area 624 cm2?

A 5 lw Formula for the area of a rectangle.

Aw 5

lww Since you are trying to get l by itself, divide each side by w.

l 5 Aw Simplify.

l 5 62424 Substitute 624 for A and 24 for w.

l 5 26 cm Simplify.

Solve each problem. Round to the nearest tenth, if necessary. Use 3.14 for π.

10. A triangle has base 6 cm and area 42 cm2. What is the height of the triangle?

11. What is the radius of a circle with circumference 56 in.?

12. A rectangle has perimeter 80 m and length 27 m. What is the width?

13. What is the length of a rectangle with area 402 ft2 and width 12 ft?

14. What is the radius of a circle with circumference 27 in.?

14 cm

about 8.9 in.

13 m

33.5 ft

about 4.3 in.

Name Class Date


59

A unit rate is a rate with denominator 1. For example, 12 in.1 ft is a unit rate. Unit rates

can be used to compare quantities and convert units.

Problem

Which is greater, 74 inches or 6 feet?

It is helpful to convert to the same units. Conversion factors, a ratio of two equivalent measures in diff erent units, are used to do conversions.

Multiply the original quantity by the conversion factor(s) so that units cancel out, leaving you with the desired units.

6 ft 3 12 in1 ft 5 72 in.

Since 72 in. is less than 74 in., 74 in. is greater than 6 ft.

Rates, which involve two diff erent units, can also be converted. Since rates involve two diff erent units, you must multiply by two conversion factors to change both of the units.

Problem

Jared’s car gets 26 mi per gal. What is his fuel effi ciency in kilometers per liter?You need to convert miles to kilometers and gallons to liters. Th is will involve multiplying by two conversion factors.

Th ere are 1.6 km in 1 mi. Th e conversion factor is either 1.6 km1 mi or 1 mi

1.6 km.

Since miles is in the numerator of the original quantity, use 1.6 km1 mi as the

conversion factor so that miles will cancel.

26migal 3

1.6 km1 mi

Th ere are 3.8 L in 1 gal. Th e conversion factor is either 3.8 L1 gal or

1 gal3.8 L.

Since gallons is in the denominator of the original quantity, use 1 gal3.8 L as the

conversion factor so that gallons will cancel.

26migal 3

1.6 km1 mi 3

1 gal3.8 L < 10.9

kmL

Jared’s vehicle gets 10.9 kilometers per liter.

2-6 ReteachingRatios, Rates, and Conversions

Name Class Date


60


Ratios, Rates, and Conversions

Exercises

Convert the given amount to the given unit.

1. 12 hours; minutes 2. 1000 cm; km 3. 45 ft; yd

4. 32 cups; gallons 5. 30 m; cm 6. 15 lbs; kilograms

7. 42 in.; cm 8. 10 miles; km 9. 25 ft; in.

10. Serra rode 15 mi in 1.5 hr. Phaelon rode 38 mi in 3.5 h. Justice rode 22 mi in 2.25 hr. Who had the fastest average speed?

11. Mr. Hintz purchased 12 gallons of drinking water for his family for $14.28. He knows that this should last for 2 weeks. What is the average cost per day for drinking water for the family?

12. Th e price for a particular herb is 49 cents for 6 ounces. What is the price of the herb in dollars per pound?

Copy and complete each statement.

13. 45 mi/h 5 ____ft/s 14. 7 g/s 5 ____kg/min 15. 50 cents/min 5 ____$/h

16. 22 m/h 5 ____cm/s 17. 15 km/min 5 ____mi/h 18. 6 gal/min 5 ____qt/h

19. Writing Describe the conversion factor you would use to convert feet to miles. How do you determine which units to place in the numerator and the denominator?

20. Writing Describe a unit rate. How do you determine the unit rate if the rate is not given as a unit rate. Illustrate using an example.

720 min

2 gal

106.68 cm

66 ft/s

0.6 cm/s

1 mile is the numerator and 5280 ft is the denominator; ft should cancel out so ft should be in the denominator

A unit rate is a rate with a denominator of 1; divide both the numerator and denominator by the denominator; 5 lbs

$3 5 5 4 3 lb$3 4 3 5

1.6 lb$1

Phaelon

$ 1.02/day

$ 1.31/lb

0.01 km

3000 cm

16.09 km

0.42 kg/min

558.9 mi/h

15 yd

6.81 kg

300 in.

$ 30/h

1440 qt/h

Name Class Date


69

A proportion is an equation that states that two ratios are equal. If a quantity in a proportion is unknown, you can solve a proportion to fi nd the unknown quantity as shown below.

Problem

What is the solution of 34 5x

14?

Th ere are two methods for solving proportions—using the Multiplication Property of Equality and the Cross Products Property.

1) Th e multiplication Property of Equality says that you can multiply both sides of an equation by the same number without changing the value.

34 5

x14

14Q34R 5 Q x

14R14 To isolate x, multiply each side by 14.

424 5 x Simplify.

10.5 5 x Divide 42 by 4.

2) Th e Cross Products Property says that you can multiply diagonally across the proportion and these products are equal.

34 5

x14

(4)(x) 5 (3)(14) Multiply diagonally across the proportion.

4x 5 42 Multiply.

4x4 5

424 To isolate x, divide each side by 4.

x 5 10.5 Simplify.

Real world situations can be modeled using proportions.

Problem

A bakery can make 6 dozen donuts every 21 minutes. How many donuts can the bakery make in 2 hours?

A proportion can be used to answer this question. It is key for you to set up the proportion with matching units in both numerators and both denominators.

For this problem, you know that 2 hours is 120 minutes and 6 dozen is 72 donuts.

Correct: Incorrect:

72 donuts21 min 5

x donuts120 min 72 donuts

21 min 5120 minx donuts

2-7 ReteachingSolving Proportions

Name Class Date


70


Solving Proportions

Th is proportion can be solved using the Multiplication Property of Equality or the Cross Products Property.

Problem

Solve this proportion using the cross products.

72 donuts21 min 5

x donuts120 min

21x 5 (72)(120) Cross Products Property

21x 5 8640 Multiply.

21x21 5

864021 Divide each side by 21.

x 5 411.43 Simplify.

Since you cannot make 0.43 donuts, the correct answer is 411 donuts.

Exercises

Solve each proportion using the Multiplication Property of Equality.

1. 34 5

n7 2. 1

3 5t

10 3. n5 5

820

4. z6 5

98 5. 15

5 5a

11 6. 72 5

d8

Solve each proportion using the Cross Products Property.

7. 35 5

b8 8. 12

m 583 9. z

2 596

10. 14v 5

73 11. 24

29 5f212 12. 13

h 5226

13. A cookie recipe calls for a half cup of chocolate chips per 3 dozen cookies. How many cups of chocolate chips should be used for 10 dozen cookies?

Solve each proportion using any method.

14. x 2 322 5

45 15. 12

10 5y 1 6

13 16. 5x 2 3 5

226

103

33

92

2163

9.6

214

274

245

6

75

1 23

2

28

3

239

212

Name Class Date


79

In similar fi gures, the measures of

corresponding angles are equal, and the

ratios of corresponding side lengths are

equal. It is important to be able to identify the

corresponding parts in similar fi gures.

Since /A > /D, /B > /E, and /C > /F ,

ABDE 5

BCEF ,

ABDE 5

ACDF . Th is fact can help you

to fi nd missing lengths.

Problem

What is the missing length in the similar fi gures?

First, determine which sides correspond. Th e side

with length 14 corresponds to the side with length

16. Th e side with length x corresponds to the side

with length 12. Th ese can be set into a proportion.

1416 5

x12 Write a proportion using corresponding lengths.

(16)(x) 5 (14)(12) Cross Products Property

16x 5 168 Multiply.

x 5 10.5 Divide each side by 16 and simplify.

Exercises

Th e fi gures in each pair are similar. Identify the corresponding sides and angles.

1. 2.

2-8 ReteachingProportions and Similar Figures

B

A C

E

D F

12

1614

x

A

B C

D E

F G

H

N

M O

Q

P R

40 m 41 m

48 m

60 m 61.5 m

72 m

AB and EF, BC and FG, CD and GH, HE and DA, lA and lE, lB and lF, lC and lG, lD and lH

MN and PQ, NO and QR, OM and RP, lM and lP, lN and lQ, lO and lR

Name Class Date


80

Exercises

Th e fi gures in each pair are similar. Find the missing length.

3. 4.

5. 6.

Problem

A map shows the distance between two towns is 3.5 inches where the scale on the

map is 0.25 in. : 5 mi. What is the actual distance between the towns?

Map scale: map distance

actual distance

If you let x be the actual distance between the towns, you can set up and solve a

the proportion to answer the question.

0.25 in.

5 mi 53.5 in.x mi

0.25x 5 17.5

x 5 70

Th e towns are 70 miles apart.

Exercises

Th e scale of a map is 1.5 in. : 50 mi. Find the actual distance corresponding to

each map distance.

7. 10 in. 8. 4.25 in. 9. 6.75 in.

10. Th e blueprints of an octagonal shaped hot tub are drawn with a 1 in. : 5 ft

scale. In the drawing the sides are 3.5 inches long. What is the perimeter of the

hot tub?


Proportions and Similar Figures

4 mx 8 m

5 m

11 cm14 cm

9 cm x

23 in.x

18 in.27 in.

4 ft 5 ft10 ft x

6.4 m

12.5 ft

33313 mi

140 ft

141 23 mi 225 mi

about 11.5 cm

15 13 in.

Name Class Date


89

Percents compare whole quantities, represented by 100%, and parts of the whole.

Problem

What percent of 90 is 27?

Th ere are two ways presented for fi nding percents.

1) You can use the percent proportion ab 5p

100. Th e percent is represented by p

100. Th e base, b, is the whole quantity and must be the denominator of the

other fraction in the proportion. Th e part of the quantity is represented by a.

2790 5

p100 Substitute given values into the percent proportion. Since you are

looking for percent, p is the unknown.

27(100) 5 (90)(p) Cross Products Property

2700 5 90p Multiply.

30 5 p Divide each side by 90 and simplify.

27 is 30% of 90.

2) Th e other way to fi nd percents is to use the percent equation. Th e percent equation is a 5 p% 3 b, where p is the percent, a is the part, and b is the base.

27 5 p% 3 90 Substitute 27 for a and 90 for b.

0.3 5 p% Divide each side by 90.

30% 5 p% Write the decimal as a percent.

27 is 30% of 90.

Exercises

Find each percent.

1. What percent of 125 is 50? 2. What percent of 14 is 35?

3. What percent of 24 is 18? 4. What percent of 50 is 75?

Problem

75% of 96 is what number?

In this problem you are given the percent p and the whole quantity (base) b.

a 5 p% 3 b Write the percent equation.

a 5 75% 3 96 5 72 Substitute 75 for p and 96 for b. Multiply.

2-9 ReteachingPercents

40%

75%

250%

150%

Name Class Date


90

Problem

28% of what number is 42?

You are given the percent p and the partial quantity a. You are looking for the base b.

a 5 p% 3 b Write the percent equation.

42 5 28% 3 b Substitute 28 for p and 42 for a.

42 5 0.28 3 b Write 28% as a decimal, 0.28.

150 5 b Divide each side by 0.28.

Exercises

Find each part.

5. What is 32% of 250? 6. What is 78% of 130?

Find each base.

7. 45% of what number is 90? 8. 70% of what number is 35?

Problems involving simple interest can be solved using the formula I 5 Prt , where I is the interest, P is the principal, r is the annual interest rate written as a decimal, and t is the time in years.

Problem

You deposited $2200 in a savings account that earns a simple interest rate of 2.8% per year. You want to keep the money in the account for 3 years. How much interest will you earn?

I 5 Prt Simple Interest Formula

I 5 (2200)(2.8%)(3) Substitute 2200 for P, 2.8% for r, and 3 for t.

I 5 184.8 Multiply.

You will earn $184.80 in interest.

Exercises

9. If you deposit $11,000 in a savings account that earns simple interest at a rate of 3.5% per year, how much interest will you have earned after 5 years?

10. If you deposit $500 in a savings account that earns simple interest at a rate of 4.25% per year, how much interest will you have earned after 10 years?


Percents

80

200

$ 1925

$ 212.50

101.4

50

Name Class Date


99

2-10 ReteachingChange Expressed as a Percent

A percent change occurs when the original amount changes and the change is expressed as a percent of the original amount. Th ere are two possibilities for percent change: percent increase or perent decrease. Th e following formula can be used to fi nd percents of increase/decrease.

percent change 5 amount of increase or decreaseoriginal amount

Problem

In its fi rst year, membership of the community involvement club was 32 members. Th e second and third years there were 28 members and 35 members respectively. Determine the percent change in membership each year.

From the fi rst to the second year, the membership went down from 32 to 28 members, representing a percent decrease. Th e amount of decrease can be found by subtracting the new amount from the original amount.

percent change 5original amount 2 new amount

original amount Percent Change Formula for percent decrease.

532 2 28

32 Substitute 32 for the original number and 28 for the new number.

54

32 5 0.125 Subtract. Then divide.

Membership decreased by 12.5% from the fi rst year to the second year.

From the second to the third year, the membership increased from 28 to 35 members, representing a percent increase. Th e amount of increase can be found by subtracting the original amount from the new amount.

percent change 5original amount 2 new amount

original amount Percent Change Formula for percent increase.

535 2 28

28 Substitute 28 for the original number and 35 for the new number.

57

32 < 0.22 Subtract. Then divide.

Membership increased by about 22% from the second year to the third year.

Exercises

Tell whether each percent change is an increase or decrease. Th en fi nd the percent change. Round to the nearest percent.

1. Original amount: 25 2. Original amount: 17 3. Original amount: 22 New amount: 45 New amount: 10 New amount: 21

increase; 80% decrease; 41% decrease; 5%

Name Class Date


100

Errors can occur when making measurements or estimations. Percents can be used to compare estimated or measured values to exact values. Th is is called relative error. Relative error can be determined with the following formula comparing the estimated value and the actual value.

Percent error 5umeasured or estimated value 2 actual value u

actual value

Problem

Mrs. Desoto estimated that her class would earn an average of $126 per person for the fundraiser. When the money was counted after the fundraiser ended, each student had raised an average of $138 per person. What is the percent error?

Th ere are two values given in this situation. Th e estimated value is $126 per person. Th e actual value that each person raised was $138.

Percent error 5umeasured or estimated value 2 actual value u

actual value Percent Error Formula

5u 126 2 138 u

138 Substitute 126 for the estimated

value and 138 for the actual value.

5u212 u

138 Subtract.

512

138 u212 u 5 12

< 0.09 Divide.

Th ere was a 9% error in her estimation.

Exercises

Find the percent error in each estimation. Round to the nearest percent.

4. You estimate that your baby sister weighs 22 lbs. She is actually 26 lbs.

5. You estimate that the bridge is 60 ft long. Th e bridge is actually 53 ft long.

6. You estimate the rope length to be 80 ft. Th e rope measures 72 ft long.

7. A carpenter estimates the roof to be 375 ft2. Th e rectangular roof measures 18 feet wide by 22 feet long. What is the percent error?


Change Expressed as a Percent

15%

13%

11%

5%


9

Name Class Date

3-1 ReteachingInequalities and Their Graphs

You use the following symbols for inequalities.

. is greater than $ is greater than or equal to

, is less than # is less than or equal to

Problem

What inequality represents “5 plus a number y is less than 210”?

5 plus a number y is less than 210

5 1 y , 210

Th e inequality 5 1 y , 210 represents the phrase.

Exercises

Write an inequality that represents each verbal expression.

1. p is greater than or equal to 5 2. a is less than or equal to 24

3. 2 times d is less than 10 4. r divided by 5 is greater than 0

Problem

Is 22 a solution of 3t 1 10 $ 5?

3t 1 10 $ 5 Original inequality

3(22) 1 10 $ 5 Substitute 22 for t.

26 1 10 $ 5 Simplify.

4 4 5 22 is not a solution.

Exercises

Determine whether each number is a solution of the given inequality.

5. 5b 2 7 . 13 a. 24 b. 4 c. 8

6. 2(m 1 1) , 26 a. 26 b. 24 c. 22

7. 8 1 h2 # 8 a. 6 b. 8 c. 10

?

?

p L 5 a K 24

2d R 10 r5 S 0

no no yes

yes no no

yes yes no


10

Name Class Date

When graphing an inequality on a number line, an open circle means the number is not included in the inequality. A closed circle means the number is included in the inequality.

Problem

What is the graph of w $ 21?

Since w is greater than or equal to 21, place a closed circle at 21.

Draw a dark line with an arrow to the right of the closed circle to show the numbers greater than 21.

Exercises

Graph each inequality.

8. y # 0 9. p . 24

10. a $ 22

Problem

What inequality represents the graph?

Th e circle is open so 4 is not included in the inequality.

Th e dark line and arrow are to the left indicating less than.

Th e graph represents “x is less than 4” or x , 4.

Exercises

Write an inequality for each graph.

11. 12.

13. 14.


Inequalities and Their Graphs

25 0 5

25 0 5

25 0 5

25 0 5

25 0 5

25 0 5

x R 21 x L 3

x K 2 x S 24

8 67 5 4 3 2 1 0 1 2 8 67 5 4 3 2 1 0 1 2

8 67 5 4 3 2 1 0 1 2

Name Class Date


19

3-2 ReteachingSolving Inequalities Using Addition or Subtraction

You can add the same number to each side of an equation. You can also add the same number to each side of an inequality.

Problem

What are the solutions of b 2 4 . 22? Graph and check the solutions.

b 2 4 . 22 Original inequality.

b 2 4 1 4 . 22 1 4 Add 4 to each side.

b . 2 Simplify.

To graph b . 2, place an open circle at 2 and shade to the right.

To check the endpoint of b . 2, make sure that 2 is the solution of the related equation b 2 4 5 22.

b 2 4 5 22

2 2 4 0 22

22 5 22 ✓

Th en check to see if a number greater than 2 is a solution of the inequality. 5 is greater than 2.

b 2 4 . 22

5 2 4 . 22

1 . 22 ✓

Exercises

Solve each inequality. Graph and check your solutions.

1. m 2 14 $ 210 2. t 2 2 , 4

3. y 2 3 # 4 4. d 2 9 $ 212

5. w 2 17 . 13 6. a 2 22 , 27

7. Writing Explain how you would solve t 2 15 # 5.

8. Anita is baking dinner rolls and pumpkin bread. She needs 4 cups of fl our for the rolls. She needs at least 7 cups of fl our left for the pumpkin bread. Write and solve an inequality to determine how much fl our Anita needs before she starts baking.

25 0 5

?

3 12 0 1 2 3 4 5 6 7 3 12 0 1 2 3 4 5 6 7

1 10 2 3 4 5 6 7 8 9 5 34 2 1 0 1 2 3 4 5

0 10 20 30 40 20 10 0 10 20 30

add 15 to both sides

c 2 4 L7; at least 11 cups

m L 4 t R 6

y K 7 d L 23

w S 30 a R 15

Name Class Date


20


Solving Inequalities Using Addition or Subtraction

You can subtract the same number from each side of an equation. You can also subtract the same number from each side of an inequality.

Problem

What are the solutions of h 1 7 # 4? Graph and check the solutions.

h 1 7 # 4 Original inequality.

h 1 7 2 7 # 4 2 7 Subtract 7 from each side.

h # 23 Simplify.

To graph h # 23, place a closed circle at 23 and shade to the left.

To check the endpoint of h # 23, make sure that 23 is the solution of the related equation h 1 7 5 4.

h 1 7 5 4

23 1 7 0 4

4 5 4 ✓

Th en check to see if a number less than 23 is a solution of the inequality. 24 is less than 23.

h 1 7 # 4

24 1 7 # 4

3 # 4 ✓

Exercises


9. s 1 7 $ 12 10. p 1 3 , 21

11. b 1 5 # 24 12. n 1 1 $ 8

13. v 1 18 . 212 14. k 1 26 , 6

15. A boat can hold up to 1000 pounds. Two friends get in the boat. Together they weigh 285 pounds. Write and solve an inequality to determine how much more weight can be added to the boat.

25 0 5

?

2 01 1 2 3 4 5 6 7 8 8 67 5 4 3 2 1 0 1 2

12 8 4 40 0 2 4 6 8

40 30 20 10 0 10 40 30 20 010

s L 5 p R 24

b K 29 n L 7

v S 230 k R 220

w 1 285 K1000; up to 715 pounds

Name Class Date


29

3-3 ReteachingSolving Inequalities Using Multiplication or Division

You can solve inequalities using multiplication or division using these two

important rules.

• You can multiply or divide each side of an inequality by a positive number.

• You can multiply or divide each side of an inequality by a negative number

only if you reverse the inequality sign.

Problem

What are the solutions of c5 # 22? Graph the solutions.

c5 # 22 Original inequality

5Qc5R # 5(22) Multiply each side by 5. Keep the inequality symbol the same.

c # 210 Simplify.

To graph c # 210, place a closed circle at 210 and shade to the left.

Problem

What are the solutions of 2 23 t . 4? Graph the solutions.

2 23 t . 4 Original inequality

2 32 Q2

23 tR , 2

32(4) Multiply each side by 2

32. Reverse the inequality symbol.

t , 26 Simplify.

To graph t , 26, place an open circle at 26 and shade to the left.

16 12 8 4 0 4

8 6 4 2 0 2

Name Class Date


30


Solving Inequalities Using Multiplication or Division

Problem

What are the solutions of 26h # 239? Graph the solutions.

26h # 239 Original inequality

26h26 $

23926 Divide each side by 26. Reverse the inequality symbol.

h $ 612 Simplify.

To graph h $ 612, place closed circle at 61

2 and shade to the right.

Exercises


1. x7 . 22 2. 8p # 32

3. 25 r $ 6 4. 2

k2 , 25

5. 23f # 12 6. 35t . 29

7. 22w . 28 8. 2 z5 $ 4

9. 2 34d , 2

38 10. 24n $ 14

11. A bus company charges $2 for each trip. It also sells monthly passes for $50.

Write and solve an inequality to fi nd how many trips you could make before

the monthly pass is cheaper.

2 0 2 4 6 8

x S 214 p K 4

r L 15 k S 10

f L 24 t S 215

w R 4 z K 220

d S 12 n K 27

2

2t S 50; For more than 25 trips, the monthly pass is cheaper.

16 12 8 4 0 4 3 12 0 1 2 3 4 5 6 7

6 03 3 6 9 12 15 18 21 24 4 02 2 4 6 8 10 12 14 16

7 56 4 3 2 1 0 1 2 3 24 18 12 6 0 6

3 12 0 1 2 3 4 5 6 7 40 30 20 010

2 1 0 1 2 5 4 3 12

Name Class Date


39

3-4 ReteachingSolving Multi-Step Inequalities

Solving inequalities is similar to solving equations. However, if you multiply or divide each side of an inequality by a negative number, the direction of the inequality sign is reversed.

Problem

What are the solutions of 6 2 3k . 45?

6 2 3k . 45 Original inequality

6 2 3k 2 6 . 45 2 6 Subtract 6 from each side.

23k . 39 Simplify.

23k23 ,

3923 Divide each side by 23 and reverse the sign.

k , 213 Simplify.

Problem

What are the solutions of 6(n 2 3) 1 4n # 42?

6(n 2 3) 1 4n # 42 Original inequality

6n 2 18 1 4n # 42 Distributive Property

10n 2 18 # 42 Combine like terms.

10n 2 18 1 18 # 42 1 18 Add 18 to each side.

10n # 60 Simplify.

10n10 #

6010 Divide each side by 10.

n # 6 Simplify.

Problem

What are the solutions of 7p 1 12 . 6p 2 15?

7p 1 12 . 6p 2 15 Original inequality

7p 1 12 2 6p . 6p 2 15 2 6p Subtract 6p from each side.

p 1 12 . 215 Simplify.

p 1 12 2 12 . 215 2 12 Subtract 12 from each side.

p . 227 Simplify.

Name Class Date


40


Solving Multi-Step Inequalities

Exercises

Solve each inequality.

1. 8w 1 9 , 231 2. 5h 2 6 $ 24

3. 17 2 2a # 29 4. 5 2 3t . 27

5. d7 1 4 . 22 6. 4 2 2x

3 # 28

7. 5(y 2 2) 2 2y $ 5 8. 8(2f 1 3) 1 4f # 216

9. 3(p 2 2) 2 7p , 6 10. 2(3b 1 5) 2 10b . 30

11. 7z 2 4 # 6z 1 18 12. 8m 1 7 $ 6m 2 9

13. 12c 1 6 . 9c 2 15 14. 7d 1 2 , 17 2 3d

15. A student had $45 when she went to the mall. She spent $9 on a pair of

earrings. Th en she wants to buy some CDs that cost $12 each. Write and solve

an inequality to determine how many CDs she can buy.

16. A friend needs at least $125 to go on the class trip. He has saved $45. He makes

$20 for each lawn he mows. Write and solve an inequality to determine how

many lawns he needs to mow to go on the trip.

17. You have earned 85, 92, 95, and 88 on tests this grading period. You have one

last test and want an average of at least 90. Write and solve an inequality to

determine what scores you can earn to achieve your goal.

w R 25 h L 6

a L 26 t R 4

d S 242 x L 18

y L 5 f K 22

p S 23 b R25

z K 22 m L 28

c S 27 d R 32

12c 1 9 K45; at most 3 CDs

20m 1 45 L125; at least 4 lawns

85 1 92 1 95 1 88 1 s5 L90; at least 90

Name Class Date


49

3-5 ReteachingWorking With Sets

Th ere are two ways to write a set.

• Roster form lists the elements of a set within braces, { }.

• Set-builder notation describes the properties an element must have to be

included in a set.

Problem

How do you write “R is the set of even whole numbers less than 10” in roster form

and in set-builder notation?

Roster Form

List the numbers 0, 2, 4, 6, and 8 in braces.

R 5 {0, 2, 4, 6, 8}

Set-Builder Notation

Describe the properties.

R 5 5x u x is an even whole number, x , 10}

Th is is read as “R is the set of all numbers x such that x is an even whole number

and x is less than 10.”

Exercises

Write each set in roster form and in set-builder notation.

1. D is the set of integers greater than 25 and less than 5.

2. N is the set of odd natural numbers less than 14.

3. P is the set of natural numbers less than or equal to 7.

4. T is the set of real numbers that are factors of 18.

5. A is the set of integers between 23 and 5.

Write the solutions of each inequality in set-builder notation.

6. 4b 1 8 . 212 7. 7n 2 14 $ 28

8. 5s 2 15 # 18 2 2s 9. 2(3p 2 5) 2 7p , 22

D 5 524, 23, 22, 21, 0, 1, 2, 3, 46; D 5 5x z x is an integer; 25 R x R 56

N 5 51, 3, 5, 7, 9, 11, 136; N 5 5x z x is an odd natural number; x R146

P 5 51, 2, 3, 4, 5, 6, 76; P 5 5x z x is a natural number, x K 76

T 5 51, 2, 3, 6, 9, 186; T 5 5x z x is a natural number, x is a factor of 186

A 5 522, 21, 0, 1, 2, 3, 46; A 5 5x z x is an integer, 23 R x R 56

5b z b is a real number; b S 256

5s z s is a real number; s K 337 6

5n z n is a real number; n L 66

5p z p is a real number; p S 286

Name Class Date


50


Working With Sets

Set A is a subset of set B if each element of A is also an element of B.

Problem

What are all the subsets of the set {r, s, t}?

; Start with the empty set because it is a subset of every set.

5r6, 5s6, 5t6 Each element of the given set is a subset of the given set, so list the subsets with one element.

5r, s6, 5r, t6, 5s, t6 List the subsets with two elements from the given set.

5r, s, t6 List the original set. It is always a subset of itself.

Th e subsets of 5r, s, t6 are ;, 5r6, 5s6, 5t6 , 5r, s6, 5r, t6, 5s, t6 and 5r, s, t6 .

Exercises

List all the subsets of each set.

10. {5, 10} 11. {mom, dad, child}

12. 522, 21, 06 13. {a, b, c, d}

Th e universal set U is the set containing all of the elements in a problem. Th e

complement of a set S is the set of all the elements in the universal set but not in

the set S. Th e complement of S is written Sr.

Problem

If U 5 {days of the week} and D 5 {Sunday, Friday}, what is Dr?

Th e days of the week that are not in D are Monday, Tuesday, Wednesday,

Th ursday, and Saturday.

Dr 5 {Monday, Tuesday, Wednesday, Th ursday, Saturday}

Exercises

14. If U 5 {natural numbers less than 20} and N 5 {factors of 18}, what is Nr?

15. If U 5 {months of the year} and M 5 {months starting with J}, what is Mr?

;, 5a6 , 5b6 , 5c6 , 5d6 , 5a, b6 , 5a, c6 , 5a, d6 , 5b, c6 , 5b, d6 , 5c, d6 , 5a, b, c6 , 5a, b, d6 , 5a, c, d6 , 5b, c, d6 , 5a, b, c, d6

;, 5mom6, 5dad6, 5child6, 5mom, dad6, 5mom, child6, 5dad, child6, 5mom, dad, child6

;, 556, 5106, 55, 106

Nr 5 54, 5, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 196

Mr 5 5Feb., Mar., April, May, Aug., Sept., Oct., Nov., Dec.6

;, 5226, 5216, 506, 522, 216, 522, 06 521, 06, 522, 21, 06

Name Class Date


59

3-6 ReteachingCompound Inequalities

A compound inequality with the word or means one or both inequalities must be true. Th e graph of the compound inequality a , 24 or a $ 3 is shown below.

A compound inequality with the word and means both inequalities must be true. Th e graph of the compound inequality b # 4 and b . 21 is shown below.

To solve a compound inequality, solve the simple inequalities from which it is made.

Problem

What are the solutions of 17 # 2x 1 7 # 29? Graph the solutions.

17 # 2x 1 7 # 29 is the same as 17 # 2x 1 7 and 2x 1 7 # 29. You can solve it as two inequalities.

17 # 2x 1 7 and 2x 1 7 # 29

17 2 7 # 2x 1 7 2 7 and 2x 1 7 2 7 # 29 2 7

10 # 2x and 2x # 22

102 #

2x2 and 2x

2 #222

5 # x and x # 11

To graph the compound inequality, place closed circles at 5 and 11. Shade between the two circles.

25 0 5

24 0 2 4 6

20 4 6 8 10 12

Name Class Date


60


Compound Inequalities

Problem

What are the solutions of 3t 2 5 , 28 or 2t 1 5 . 17? Graph the solutions.

Solve each inequality.

3t 2 5 , 28 or 2t 1 5 . 17

3t 2 5 1 5 , 28 1 5 or 2t 1 5 2 5 . 17 2 5

3t , 23 or 2t . 12

3t3 ,

233 or

2t2 .

122

t , 21 or t . 6

To graph the compound inequality, place open circles at 21 and at 6. Shade to the

left of 21 and to the right of 6.

Exercises

Solve each compound inequality. Graph the solutions.

1. h 2 7 $ 25 and h 1 4 , 10 2. r 2 2 # 21 or r 2 3 . 2

3. 27 , w 2 4 , 2 4. 22 #y2 # 1

5. 5p 1 3 # 22 or 3p 2 6 $ 3 6. 22n 2 5 $ 1 or 5n 1 7 . 2

7. 34a 2 6 , 0 and

23a 1 4 . 2 8. 24 # 4d 1 24 # 4

9. 5m 2 2 , 8 or 6m 2 2 . 6 1 5m 10. w2 1 1 $ 2 and w 2 5 # 1

23 70 3

2 K h R 6

23 R w R 6

p K 21 or p L 3

23 R a R 8

m R 2 or m S 8

24 K y K 2

n K 23 or n S 21

27 K d K 25

2 K w K 6

r K 1 or r S 5

1 10 2 3 4 5 6 7 8 9

3 12 0 1 2 3 4 5 6 7

5 34 2 1 0 1 2 3 4 5

8 4 0 4 8

01 1 2 3 4 5 6 7 8 9

0 21 3 4 5 6 7 8 9 10

5 34 2 1 0 1 2 3 4 5

5 34 2 1 0 1 2 3 4 5

8 67 5 4 3 2 1 0 1 2

1 10 2 3 4 5 6 7 8 9


79

Name Class Date

Th e union of two or more sets contains all elements of the sets. Th e symbol for union is < .

Problem

If A 5 {1, 2, 3, 4, 5, 6} and B 5 {2, 4, 6, 8, 10}, what is A < B?

Th e union of A and B is the set of elements that are in either A or B or both. List all elements that are in either or both of the sets, but do not repeat any elements. Even if an element is in both sets, it only appears once in the union.

A < B 5 {1, 2, 3, 4, 5, 6, 8, 10}

Th e intersection of two or more sets contains elements that are in every set. Th e symbol for intersection is >.

Problem

If M 5 {m k m is a positive factor of 24} and N 5 {n k n is a positive even number}, what is M > N ?

List the numbers in each set.

M 5 {1, 2, 3, 4, 6, 8, 12, 24} N 5 {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, …}

Th e intersection of M and N is the set of elements that are in both M and N. Because the largest number in M is 24, it is not necessary to list numbers greater than 24 for set N. Th e numbers that are in both sets are 2, 4, 6, 8, 12, and 24.

M > N 5 {2, 4, 6, 8, 12, 24}

Exercises

Find each union or intersection. Let P 5 {1, 2, 3, 4, 5}, R 5 {r z r is an odd natural number, r R 10}, S 5 {5, 10, 15}, and T 5 {t z t is a positive factor of 8}.

1. P < R 2. P > R 3. R > S

4. S < R 5. P < T 6. P > T

7. S < T 8. S > T 9. S > P

10. S < P 11. R > T 12. R < T

3-8 ReteachingUnions and Intersections of Sets

51, 2, 3, 4, 5, 7, 96

51, 3, 5, 7, 9, 10, 156

51, 2, 4, 5, 8, 10, 156

51, 2, 3, 4, 5, 10, 156

51, 3, 56

51, 2, 3, 4, 5, 86

;

516

556

51, 2, 46

556

51, 2, 3, 4, 5, 7, 8, 96


80

Name Class Date

Th e solutions of compound inequalities are unions or intersections of sets.

Problem

What are the solutions of x 1 5 . 28 and x 2 7 , 12? Write the solutions as either the union or the intersection of two sets.

First solve the two inequalities separately.

x 1 5 . 28 and x 2 7 , 12

x 1 5 2 5 . 28 2 5 and x 2 7 1 7 , 12 1 7

x . 213 and x , 19

Th e word and indicates the intersection of two sets. Th e solutions are numbers that satisfy both of the fi nal inequalities, numbers between 213 and 19. Th e solutions can be written as {x k x . 213} > {x k x , 19}.

Problem

What are the solutions of u 2y u $ 7? Write the solutions as either the union or the intersection of two sets.

First write the absolute value inequality as two inequalities and solve the two inequalities separately.

2y $ 7 or 2y # 27

2y2 $

72 or

2y2 #

272

y $ 3.5 or y # 23.5

Th e word or indicates the union of two sets. Th e solutions are numbers that satisfy either of the fi nal inequalities or both of them, numbers greater than or equal to 3.5 or numbers less than or equal to 23.5.Th e solutions can be written as {y k y $ 3.5} < {y k y # 23.5}.

Exercises

Solve. Write the solutions as either the union or intersection of two sets.

13. p 1 4 $ 3 or p 2 2 , 24 14. 5t $ 220 and 4t # 24

15. k k 2 4 k $ 15 16. ` n6 ` , 5

17. k 3t 2 6 k , 15 18. k 5d 1 10 k $ 5


Unions and Intersections of Sets

5p…p L 216 : 5p…p R 226 5t…t L 246 " 5t…t K 66

5k…k L 196 : 5k…k K 2116 5n…n S 2306 " 5n…n R 306

5t…t S 236 " 5t…t R 76 5d…d K 236 : 5d…d L 216


9

Name Class Date

4-1 ReteachingUsing Graphs to Relate Two Quantities

An important life skill is to be able to a read graph. When looking at a graph, you should check the title, the labels on the axes, and the general shape of the graph.

Problem

What information can you determine from the graph?

• Th e title tells you that the graph describes Trina’s trip.

• Th e axes tell you that the graph relates the variable of time to the variable of distance to the destination.

• In general, the more time that has elapsed, the closer Trina gets to her destination. In the middle of the trip, the distance does not change, showing she stops for a while.

Exercises

What are the variables in each graph? Describe how the variables are related at

various points on the graph.

1. 2. 3.

Time

Dis

tanc

eto

Des

tina

tion

Trina’s Trip

Time

Tile

s In

stal

led

Tiling Job

Age

Hei

ght

Dion’s GrowthChart

Time

Hei

ght

KickedFootball

time and total tiles installed; The number of tiles installed increases as time increases, and then there is a rest during which no tiles are installed, then more tiles are installed, another rest, and then more tiles are installed.

age and height; Up until Dior reaches a certain age, his height increases with age at various rates. Then he stops growing.

time and height; When a football is kicked, its height increases with time and then its height decreases with time.


10

Name Class Date


Using Graphs to Relate Two Quantities

A graph can show the relationship described in a table.

Problem

Which graph shown below represents the information in the table at the right?

Notice that for each additional CD purchased, the total cost increases by $15. Th e points on the graph should be in a straight line that goes up from left to right. Th e graph that shows this trend is Graph B.

A. B. C.

Exercises

Match each graph with its related table. Explain your answers.

4. 5. 6.

A. B. C.

11 115

115

115

115

11

11

11

1

2

3

4

5

$15

$30

$45

$60

$75

CDsPurchased

TotalCost

CDs Purchased

Tota

l Cos

t

Day

Tick

ets

Sold

1

2

3

4

60

45

40

75

Day Tickets Sold

Day

Tick

ets

Sold

1

2

3

4

70

65

50

45

Day Tickets Sold

Day

Tick

ets

Sold

1

2

3

4

35

45

55

65

Day Tickets Sold

CDs Purchased

Tota

l Cos

t

CDs PurchasedTo

tal C

ost

C; as days increase, tickets sold increases

A; tickets sold decreases until day 4

B; as days increase, tickets sold decreases


19

Name Class Date

4-2 ReteachingPatterns and Linear Functions

A relationship can be represented in a table, as ordered pairs, in a graph, in words, or in an equation.

Problem

Consider the relationship between the number of squares in the pattern and the perimeter of the fi gure. How can you represent this relationship in a table, as ordered pairs, in a graph, in words, and in an equation?

Table

For each number of squares determine the perimeter of the fi gure. Write the values in the table. Remember to focus on the perimeter of the fi gure, not the squares.

Ordered Pairs

Let x represent the number of squares and y represent the perimeter. Use the numbers in the table to write the ordered pairs.

(1, 20), (2, 30), (3, 40), (4, 50), (5, 60)

Graph

Use the ordered pairs to draw the graph.

Words

Th e pattern shows the perimeter is the number of squares times 10 plus 10.

Equation

Write an equation for the words. y 5 10x 1 10

555

55

5

55

5

55

5

5

5

55

5

55

5

5

5

5

5

55

5

55

5

5

5

5

5

5

5

55

5

5

1

20

Number of squares

Perimeter

2

30

3

40

4

50

5

60

x

y

20

40

60

4 62O

Number of Squares

Peri

met

er


20

Name Class Date


Patterns and Linear Functions

Exercises

Consider each pattern.

1. 2.

a. Make a table to show the relationship between the number of trapezoids and the perimeter.

a. Make a table to show the relationship between the number of cubes and the surface area.

b. Write the ordered pairs for the relationship.

b. Write the ordered pairs for the relationship.

c. Make a graph for the relationship.

c. Make a graph for the relationship.

d. Use words to describe the relationship.

d. Use words to describe the relationship.

e. Write an equation for the relationship.

e. Write an equation for the relationship.

6

33 3

6

33

6 3

3 63 3

6 3

3 63

11

1

1

1 1

11

1 1

1

1

1

1

11

1 1

1

1

1

1

1

1

(1, 15), (2, 24), (3, 33) (1, 6), (2, 10), (3, 14)

The surface area is 2 more than 4 times the number of cubes.

The perimeter is 6 more than 9 times the number of trapezoids.

p 5 9n 1 6 s 5 4n 1 2

1

15

Number of trapezoids

Perimeter

2

24

3

33

1

10

20

30

2 3 4Number of trapezoids

Peri

met

er

n

p

O 1

5

10

15

2 3 4Number of cubes

Surf

ace

area

n

s

O

1

6

Number of cubes

Surface area

2

10

3

14


29

Name Class Date

4-3 ReteachingPatterns and Nonlinear Functions

If the points of the graph of a function are in a straight line, the function is a linear function. If the points of the graph of a function are not in a straight line, the function is a nonlinear function.

Problem

Is the function given by the table at the rightlinear or nonlinear?

Graph the function.

Th e points are not in a straight line, so the function is nonlinear.

Do you like to solve puzzles? When you are given a list of function values and you are asked to fi nd the rule for the function, you are solving a puzzle. You are looking for a rule that works for all pairs of numbers.

Problem

What is a rule that represents the function given by the table below?

Try a rule. Is there an operation or sequence of operations that relates the values in the fi rst column of the table to the values in the second column?

Try division: 6 4 2 5 3, but 8 4 2 2 5.

Try another rule. 6 2 3 5 3 and 8 2 3 5 5.

Check to make sure this works for all pairs of numbers. 9 2 3 5 6 and 12 2 3 5 9.

Th e function can be represented by the rule y 5 x 2 3.

1

2

3

6

6

3

2

1

x y

6

8

9

12

3

5

6

9

x y

O

y

x

2

4

6

4 62


30

Name Class Date


Patterns and Nonlinear Functions

Graph the function shown by each table. Tell whether the function is linear or nonlinear.

1. 2. 3.

4. 5. 6.

Each set of ordered pairs represents a function. Write a rule that represents the function.

7. (2, 10), (4, 20), (5, 25), (7, 35), (9, 45) 8. (2, 5), (4, 9), (5, 11), (7, 15), (10, 21)

9. (0, 0), (1, 1), (2, 8), (3, 27), (4, 64) 10. (2, 5), (3, 10), (4, 17), (5, 26), (6, 37)

0

2

3

6

1

3

4

7

x y

2

3

4

6

6

4

3

2

x y

2

3

4

5

1

3

5

7

x y

4

3

0

2

4

3

0

2

x y

1

2

3

4

4

1

0

1

x y

0

2

4

6

1

2

3

4

x y

linear; linear; nonlinear;

nonlinear;

y 5 5x

y 5 x3

y 5 2x 1 1

y 5 x2 1 1

nonlinear; linear;

2 3 51

2

4

67

1

3

5

4 6 7x

y

O

2 3 51

2

4

67

1

3

5

4 6 7x

y

O

2 3 51

2

4

67

1

3

5

4 6 7x

y

O

x

O

y

2 2

2

4

6

4

2 3 51

2

4

67

1

3

5

4 6 7x

y

O

2 3 51

2

4

67

1

3

5

4 6 7x

y

O


39

Name Class Date

4-4 ReteachingGraphing a Function Rule

By fi nding values that satisfy a function rule, you can graph points and discover

the shape of its graph.

Problem

What is the graph of the function rule y 5 3x 1 5?

First, choose any values for x and fi nd the

corresponding values of y. Make a table of

your values.

Th en, graph the points from your table. In

this case, the points are in a line. Draw the

line.

Problem

What is the graph of the function rule y 5 ux 2 2 u ?

First, choose any values for x and fi nd the

corresponding values of y. Make a table of

your values.

Th en, graph the points from your table. In

this case, the points make a V shape. Draw

the V.

x (x, y)y 3x 5

y 3( 2) 5 12 ( 2, 1)

y 3( 1) 5 21 ( 1, 2)

y 3(0) 5 50 (0, 5)

y 3(1) 5 81 (1, 8)

y 3(2) 5 112 (2, 11)

x (x, y)

0

1

2

3

4

(0, 2)

(1, 1)

(2, 0)

(3, 1)

(4, 2)

y x 2y 0 2 2

y 1 2 1

y 2 2 0

y 3 2 1

y 4 2 2

O

y

x

4

8

4 4 8

O

y

x42

2

4


40

Name Class Date


Graphing a Function Rule

Exercises

Graph each function rule.

1. y 5 x2 1 3 2. y 5 2x 2 3

3. y 5 x2 2 3 4. y 5 u x u 1 1

x (x, y)y x2

3

x (x, y)y x2 3

x (x, y)y x 3

x (x, y)y x 1

xO

y

4

4

8

4

4

8

88x

O

y

2

2

4

2

2

4

44

xO

y

2

2

4

2

2

4

44x

O

y

2

2

4

2

2

4

44


49

Name Class Date

4-5 ReteachingWriting a Function Rule

When writing function rules for verbal descriptions, you should look for key words.

Problem

Twice a number n increased by 4 equals m. What is a function rule that represents the sentence?

Th e function rule is 2n 1 4 5 m.

Exercises

Write a function rule that represents each sentence.

1. t is 4 more than the product of 7 and s. 2. Th e ratio of a to 5 equals b.

3. 8 fewer than p times 3 equals x. 4. y is half of x plus 10.

5. k equals the sum of h and 23. 6. 15 minus twice a equals b.

7. m equals 5 times n increased by 6. 8. 17 decreased by three times d equals c.

9. 5 more than the product of 6 and n is 17. 10. d is 8 less than the quotient of b and 4.

Words that SuggestAddition

Words that SuggestSubtraction

Words that SuggestMultiplication

Words that SuggestDivision

plussummore thanincreased bytotalin all

differenceless thandecreased byfewer thansubtracted by

minus timesproductofeachfactorstwice

divided byquotientrateratiohalfa third of

mtwice a number n increased by 4 equals

2n 4 m

t 5 7s 1 4 a5 5 b

3p 2 8 5 x y 5 12 x 1 10

k 5 h 1 23 15 2 2a 5 b

m 5 5n 1 6 17 2 3d 5 c

6n 1 5 5 17 d 5 b4 2 8


50

Name Class Date


Writing a Function Rule

You can write functions to represent situations and then evaluate the function to determine a particular value.

Problem

A sales associate earns $500 per week plus 4% of his sales. Write a function rule for the amount he makes in a week if he sells s dollars of merchandise. How much will he make if he sells $4000 worth of merchandise?

First write the function rule.

Use this function rule to calculate how much he will make.

e 5 500 1 0.04s

5 500 1 0.04(4000)

5 700

He will make $700.

Exercises

11. Twelve cans of peaches are placed into each box. Write a function rule for the number of boxes needed for c cans. How many boxes are needed for 1440 cans?

12. Tara plans to rent a car for the weekend. Th e cost to rent the car is $45 plus $0.15 for each mile she drives. Write a function rule for the total cost of the rental. How much is the rental if she travels 500 miles?

13. A plumber charges $60 for a service call plus $55 for each hour she works. Write a function rule for the total bill for a plumbing job. What is the total bill for a job that takes the plumber 3 hours of work?

14. Tickets to a concert cost $45 per ticket plus a $10 processing fee for each order. Write a function rule for the total cost of ordering tickets. What is the total cost to order 6 tickets?

500 4% of salesearnings equals plus

e 500 0.04s

b 5 c12; 120 boxes

c 5 0.15m 1 45; 120

B 5 55h 1 60; $225

c 5 45t 1 10; $280


59

Name Class Date

When a relation is represented as a set of ordered pairs, the domain of the relation is the set of x-values. Th e range is the set of y-values.

A relation where each value in the domain is paired with just one value in the range is called a function.

Problem

Identify the domain and range of the relation 5(22, 3), (0, 2), (1, 3), (3, 4)6 . Represent the relation with a mapping diagram. Is the relation a function?

Th e domain (or x-values) is {–2, 0, 1, 3}.

Th e range (or y-values) is {2, 3, 4}.

Notice that each number in the domain is mapped to only one number in the range. Th is relation is a function.

Exercises

Identify the domain and range of each relation. Use a mapping diagram to determine whether the relation is a function.

1. {(2, 3), (4, 6), (1, 5), (2, 5), (0, 5)} 2. {(3, 4), (5, 4), (7, 4), (8, 4), (10, 4)}

4-6 ReteachingFormalizing Relations and Functions

Domain Range

234

22013

D: 50, 1, 2, 46 ; R: 53, 5, 66 D: 53, 5, 7, 8, 106 ; R: 546Domain Range

356

0124

Domain Range

4

35

87

10

The relation is a function.The relation is not a function.


60

Name Class Date


Formalizing Relations and Functions

You can determine whether or not a relation is a function by looking at the graph of the relation. If a vertical line is drawn anywhere on the graph and passes through two points of the relation, the relation is not a function. Th is is called the vertical line test.

Problem

Is the relation shown below a function? Use a vertical line test.

Notice that two of the dashed vertical lines pass through just one point on the graph.

However, one of the dashed vertical lines passes through three points.

Th e relation is not a function.

Exercises

Use the vertical line test to determine whether the relation is a function.

3. 4. 5.

6. 7. 8.

x

y

O

x

y

O

x

y

Ox

y

Ox

y

O

x

y

O

x

y

O

a function a functionnot a function

not a function not a function a function


69

Name Class Date

An orderly list of numbers is called a sequence. Each number in a sequence is called a term. Many sequences follow a pattern. To fi nd the pattern, you will be solving a puzzle.

Problem

Describe the pattern of the sequence 8, 4, 0,24, 28, … . What are the next two terms of the sequence?

You can divide 8 by 2 to get 4, but 4 divided by 2 is not 0. Th e pattern cannot be “divide by 2.” Look at the pattern again. You can subtract 4 from each number to get the next number.

Th e pattern is “subtract 4 from the previous term.” Th e next two terms are 28 2 4 or 212 and 212 2 4 or 216.

Exercises

Describe the pattern in each sequence. Th en fi nd the next two terms of the sequence.

1. 1, 5, 25, 125, … 2. 3, 9, 15, 21, … 3. 64, 32, 16, 8, …

4. 25, 23, 21, 1, … 5. 1, 23, 9, 227, … 6. 10, 3,24,211, …

7. 1000, 2100, 10,21, … 8. 23, 31, 39, 47, … 9. 24, 212, 236, 2108, …

10. 25, 29, 213, 217, … 11. 3.6, 4.1, 4.6, 5.1, … 12. 281, 227, 29, 23, …

4-7 ReteachingSequences and Functions

28,...24,

24 242424

0,4,8,

multiply the previous term by 5; 625, 3125

add 2 to the previous term; 3, 5

divide the previous term by 210; 0.1, –0.01

subtract 4 from the previous term; 221, 225




add 0.5 to the previous term; 5.6, 6.1

divide the previous term by 2; 4, 2

subtract 7 from the previous term; 218, 225


divide the previous

term by 3; 21, 213


70

Name Class Date


Sequences and Functions

An arithmetic sequence is a sequence in which the diff erence between consecutive terms is constant. Th is diff erence is called the common diff erence. You can fi nd the nth term of an arithmetic sequence by using the following formula.

A(n) 5 A(1) 1 (n 2 1)d

In this formula,

• A(n) represents the nth term

• A(1) represents the fi rst term

• n represents the term number

• d represents the common diff erence

Problem

Write a formula for the arithmetic sequence 15, 10, 5, 0, 25, … . What is the tenth term of the sequence?

Th e pattern is to “add –5 to the previous term.”

A(n) 5 A(1) 1 (n 2 1)d

A(n) 5 15 1 (n 2 1)(25) A(1) 5 15 and d 5 25

Use the formula to fi nd the tenth term.

A(n) 5 15 1 (n 2 1)(25)

A(10) 5 15 1 (10 2 1)(25) n 5 10

A(10) 5 230 Simplify.

Exercises

Write a formula for each arithmetic sequence. Th en, fi nd the tenth term.

13. 3, 10, 17, 24, … 14. 24, 1, 6, 11, … 15. 44, 40, 36, 32, …

16. 8, 2,24,210, … 17. 22, 32, 42, 52, … 18. 55, 44, 33, 22, …

25,...0,

1(25) 1(25)1(25)1(25)

5,10,15,

A(n) 5 3 1 (n 2 1)(7); 66

A(n) 5 8 1 (n 2 1)(26); 246

A(n) 5 24 1 (n 2 1)(5); 41

A(n) 5 22 1 (n 2 1)(10); 112

A(n) 5 44 1 (n 2 1)(24); 8

A(n) 5 55 1 (n 2 1)(211); 244


9

Name Class Date

5-1 ReteachingRate of Change and Slope

Th e rate of the vertical change to the horizontal change between two points on a line is called the slope of the line.

slope 5vertical change

horizontal change 5riserun

Th ere are two special cases for slopes.

• A horizontal line has a slope of 0.

• A vertical line has an undefi ned slope.

Problem

What is the slope of the line?



513

Th e slope of the line is 13.

In general, a line that slants upward from left to right has a positive slope.

Problem

What is the slope of the line?



5221

5 22

Th e slope of the line is 22.

In general, a line that slants downward from left to right has a negative slope.

xO

y4

42

2

4

2

2 6

(1, 1)(4, 2)

rise 5 1run 5 3

xO

y4

2

4

2

4 4

(1, 1)

(0, 3)

2run 5 1

rise 5 22


10

Name Class Date


Rate of Change and Slope

Exercises

Find the slope of each line.

1. 2. 3.

Suppose one point on a line has the coordinates (x1, y1) and another point on the same line has the coordinates (x2, y2). You can use the following formula to fi nd the slope of the line.

slope 5 riserun 5

y2 2 y1x2 2 x1

, where x2 2 x1 2 0

Problem

What is the slope of the line through R(2, 5) and S(21, 7)?

slope 5y2 2 y1x2 2 x1

57 2 521 2 2

Let y2 5 7 and y1 5 5.

Let x2 5 21 and x1 5 2.

5223 5 2

23

Exercises

Find the slope of the line that passes through each pair of points.

4. (0, 0), (4, 5) 5. (2, 4), (7, 8) 6. (22, 0), (23, 2)

7. (22, 23), (1, 1) 8. (1, 4), (2,23) 9. (3, 2), (25, 3)

xO

y4

42

2

4

2

2 6

(1, 0)

(5, 21) xO

y6

42

4

2

2

2 6

(1, 2)

(6, 5)

xO

y6

42

4

2

2

2 6

(1, 4) (5, 4)

214

35

0

54

43

45

27

22

218


19

Name Class Date

5-2 ReteachingDirect Variation

A direct variation is a relationship that can be represented by a function in the

form y 5 kx where k 2 0. Th e constant of variation for a direct variation k is the

coeffi cient of x. Th e equation y 5 kx can also be written as yx 5 k.

Problem

Does the equation 6x 1 3y 5 9 represent a direct variation? If so, fi nd the constant of variation.

If the equation represents a direct variation, the equation can be rewritten in the

form y 5 kx . So, solve the equation for y to determine whether the equation can

be written in this form.

6x 1 3y 5 9

3y 5 9 2 6x Subtract 6x from each side.

y 5 3 2 2x Divide each side by 3.

You cannot write the equation in the form y 5 kx . So 6x 1 3y 5 9 does not

represent a direct variation.

Problem

Does the equation 5y 5 3x represent a direct variation? If so, fi nd the constant of variation.

Again, if the equation represents a direct variation, the equation can be rewritten

in the form y 5 kx . So, solve the equation for y to determine whether the equation

can be written in this form.

5y 5 3x

y 5 35 x Divide each side by 5.

Th e equation has the form y 5 kx , so the equation represents a direct variation.

Th e coeffi cient of x is 35, so the constant of variation is

35.

Exercises

Determine whether each equation represents a direct variation. If it does, fi nd the constant of variation.

1. 2y 5 x 2. 3x 1 2y 5 1 3. 24y 5 8x

4. 2x 5 y 2 5 5. 4x 2 3y 5 0 6. 5x 5 2y

yes; 12

yes; 43 yes; 52

no

no

yes; 22


20

Name Class Date


Direct Variation

To write an equation for direct variation, fi nd the constant of variation k using an

ordered pair. Th en use the value of k to write an equation.

Problem

Suppose y varies directly with x, and y 5 24 when x 5 8. What direct variation equation relates x and y? What is the value of y when x 5 10?

You are given that x and y vary directly. Th is means that the relationship between x

and y can be written in the form y 5 kx , where k is a constant.

y 5 kx Start with the direct variation equation.

24 5 k(8) Substitute the given values: 8 for x and 24 for y.

3 5 k Divide each side by 8 to solve for k.

y 5 3x Write the direct variation equation that relates x and y by substituting 3 for k in y 5 kx .

Th e equation y 5 3x relates x and y. When x 5 10, y 5 3(10)or 30.

Exercises

Suppose y varies directly with x. Write a direct variation equation that relates x and y. Th en fi nd the value of y when x 5 6.

7. y 5 14 when x 5 2. 8. y 5 3 when x 5 9.

9. y 5 12 whenx 5 224. 10. y 5 281 when x 5 9.

11. y 5 216 when x 5 24. 12. y 5 5 when x 5 20.

13. Consider the direct variation y 5 3x .

a. List three ordered pairs that satisfy the equation.

b. Plot your three ordered pairs from

part (a) on a coordinate grid.

c. Complete the graph of y 5 3x on the grid.

y 5 7x; 42

y 5 29x; 254

y 5 4x; 24

Answers may vary. Sample: (0,0), (1, 3), (2. 6)

Sample:

Sample:

y 5 13 x; 2

y 5 2

12 x; 23

y 5 14 x; 32

2

4

6

2 4 6Ox

y

(0, 0)

(1, 3)

(2, 6)


29

Name Class Date

5-3 ReteachingSlope-Intercept Form

Th e slope-intercept form of a linear equation is y 5 mx 1 b. In this equation, m is the slope and b is the y-intercept.

Problem

What are the slope and y-intercept of the graph of y 5 22x 2 3?

Th e equation is solved for y, but it is easier to determine the y-intercept if the right side is written as a sum instead of a diff erence.

y 5 22x 2 3

y 5 22x 1 (23) Write the subtraction as addition.

Th e slope is 22 and the y-intercept is 23.

Problem

What is an equation for the line with slope 23 and y-intercept 9?

When the slope and y-intercept are given, substitute the values into the slope-intercept form of a linear equation.

y 5 mx 1 b

y 5 23 x 1 9 Substitute 2

3 for m and 9 for b.

Problem

What is an equation in slope-intercept form for the line that passes through the points (1, 23) and (3, 1)?

Substitute the two given points into the slope formula to fi nd the slope of the line.

m 51 2 (23)

3 2 1 542 5 2

Th en substitute the slope and the coordinates of one of the points into the slope-intercept form to fi nd b.

y 5 mx 1 b Use slope-intercept form.

23 5 2(1) 1 b Substitute 2 for m, 1 for x, and 23 for y.

25 5 b Solve for b.

Substitute the slope and y-intercept into the slope-intercept form.

y 5 mx 1 b Use slope-intercept form.

y 5 2x 1 (25) Substitute 2 for m and 25 for b.


30

Name Class Date


Slope-Intercept Form

Exercises

Find the slope and y-intercept of the graph of each equation.

1. y 5 12 x 1 7 2. y 5 25x 1 1 3. y 5 2

25 x 2 3

4. y 5 x 1 5 5. y 5 16 x 2 2 6. y 5 4x

Write an equation for the line with the given slope m and y-intercept b.

7. m 5 23, b 5 7 8. m 523, b 5 8 9. m 5 4, b 5 23

10. m 5 2 15, b 5 21 11. m 5 2

56, b 5 0 12. m 5 7, b 5 22

Write an equation in slope-intercept form for the line that passes through the given points.

13. (1, 3) and (2, 5) 14. (2,21) and (4, 0) 15. (1, 2) and (2,21)

16. (1,25) and (3,23) 17. (3, 3) and (6, 5) 18. (4,23) and (8,24)

19. Consider the equation y 5 22x 1 4. a. What is the y-intercept of the graph of the equation?

b. Graph the y-intercept.

c. What is the slope of the graph of the equation?

d. Use the point you graphed in part (b) and the slope to fi nd another point on the graph of the equation.

e. Graph the equation.

m 5 12 ; b 5 7

m 5 16 ; b 5 22

y 5 23 x 1 8

y 5 215x 2 1 y 5 25

6x

m 5 225 ; b 5 23m 5 25 ; b 5 1

y 5 23x 1 7

y 5 2x 1 1

4

22

Answers may vary. Sample: (1, 2)

See graph in part (b).

y 5 23x 1 5

y 5 x 2 6

y 5 12x 2 2

y 5 23x 1 1 y 5 21

4x 2 2

y 5 4x 2 3

y 5 7x 2 2

m 5 4 ; b 5 0m 5 1 ; b 5 5

x

y

4

(0, 4)

848

8

4

8

4

O


39

Name Class Date

5-4 ReteachingPoint-Slope Form

Th e point-slope form of a nonvertical linear equation is y 2 y1 5 m(x 2 x1). In this equation, m is the slope and (x1, y1) is a point on the graph of the equation.

Problem

A line passes through (5, 22) and has a slope 23. What is an equation for this line in point-slope form?

y 2 y1 5 m(x 2 x1) Use point-slope form.

y 2 (22) 5 23(x 2 5) Substitute (5, 22) for (x1, y1) and 23 for m.

y 1 2 5 23(x 2 5) Simplify.

Problem

A line passes through (1, 4) and (2, 9). What is an equation for this line in point-slope form? What is an equation for this line in slope-intercept form?First use the two given points to fi nd the slope.

m 59 2 42 2 1 5

51 5 5

Use the slope and one point to write an equation in point-slope form.

y 2 y1 5 m(x 2 x1) Use point-slope form

y 2 4 5 5(x 2 1) Substitute (1, 4) for (x1, y1) and 5 for m.

y 2 4 5 5x 2 5 Distributive Property

y 5 5x 2 1 Add 4 to each side.An equation in point-slope form is y 2 4 5 5(x 2 1). An equation in slope-intercept form is y 5 5x 2 1.

Exercises

Write an equation for the line through the given point and with the given slope m.

1. (21, 3); m 5 214 2. (7, 25); m 5 4 3. (22, 25); m 5

23

Write an equation in point-slope form of the line through the given points. Th en write the equation in slope-intercept form.

4. (1, 4) and (2, 7) 5. (2, 0) and (3, 22) 6. (4, 25) and (22, 22)

y 2 3 5 2 14 (x 1 1) y 1 5 5 4(x 2 7) y 1 5 5 2

3 (x 1 2)

y 2 4 5 3(x 2 1) or y 2 7 5 3(x 2 2);y 5 3x 1 1

y 5 22(x 2 2) or y 1 2 5 22(x 2 3); y 5 22x 1 4

y 1 5 5 2 12 (x 2 4) or

y 1 2 5 2 12 (x 1 2);

y 5 2 12 x 2 3


40

Name Class Date


Point-Slope Form

You can use the point-slope form of an equation to help graph the equation. Th e point given by the point-slope form provides a place to start on the graph. Plot a point there. Th en use the slope from the point-slope form to locate another point in either direction. Th en draw a line through the points you have plotted.

Problem

What is the graph of the equation y 2 2 5 13(x 2 1)?

Th e equation is in point-slope form, so the line passes through (1, 2) and has a

slope of 13.

Plot the point (1, 2).

Use the slope, 13. From (1, 2), go up 1 unit

and then right 3 units. Draw a point.

Draw a line through the two points.

Because 13 52123, you can start at (1, 2) and go down 1 unit and to the left 3 units to

locate a third point on the line.

Exercises

Graph each equation.

7. y 2 3 5 2(x 1 1) 8. y 1 2 523(x 2 2) 9. y 2 4 5 2

12(x 1 1)

xO

y4

42

2

4

2

2 6

(1, 2)(4, 3)

rise 5 1run 5 3

x

y

4 848

8

4

8

4

Ox

y

4 848

8

4

8

4

Ox

y

4 848

8

4

8

4

O


59

Name Class Date

Nonvertical lines are parallel if they have the same slope and diff erent y-intercepts. Th e

graphs of y 5 2x 2 6 and y 5 2x 1 3 are parallel because they have the same slope, 2,

but diff erent y-intercepts, 26 and 3.

Problem

What is an equation in slope-intercept form of the line that passes through

(8, 7) and is parallel to the graph of y 5 34 x 1 2?

Th e slope of y 5 34 x 1 2 is

34. Because the desired equation is for a line parallel to

a line with slope 34, the slope of the parallel line must also be

34. Use the slope and

the given point in the point-slope form of a linear equation and then solve for y to

write the equation in slope-intercept form.

y 2 y1 5 m(x 2 x1) Start with the point-slope form.

y 2 7 534 (x 2 8) Substitute (8, 7) for (x1, y1) and 34 for m.

y 2 7 534 x 2 6 Distributive Property

y 5 34 x 1 1 Add 7 to each side.

Th e graph of y 5 34 x 1 1 passes through (8, 7) and is parallel to the

graph of y 5 34 x 1 2.

Exercises

1. Writing Are the graphs of y 5 25 x 1 3 and y 5 3

5 x 2 4 parallel? Explain

how you know.

Write an equation in slope-intercept form of the line that passes through the given point and is parallel to the graph of the given equation.

2. (3, 1); y 5 2x 1 4 3. (1, 3); y 5 7x 1 5 4. (1, 6); y 5 9x 2 5

5. (0, 0); y 5 2

12 y 2 4 6. (25, 7); y 5 2

25 x 2 3 7. (6, 6); y 5 1

3 x 2 1

5-6 ReteachingParallel and Perpendicular Lines

No, because the slopes 25 and 35 are not equal.

y 5 7x 2 4 y 5 9x 2 3y 5 2x 2 5

y 5 2

12 x y 5 2

25 x 1 5 y 5 1

3 x 1 4


60

Name Class Date

Two lines that are neither horizontal nor vertical are perpendicular if the

product of their slopes is 21. Th e graphs of y 5 2

45 x 2 5 and y 5 5

4 x 1 4 are

perpendicular because 2

45 Q

54R 5 21.

Problem

What is an equation in slope-intercept form of the line that passes through

(2, 11) and is perpendicular to the graph of y 5 14 x 2 5?

Th e slope of y 5 14 x 2 5 is

14 . Since

14 (24) 5 21, the slope of the line

perpendicular to the given line is 24.

Use this slope and the given point to write an equation in point-slope form. Th en

solve for y to write the equation in slope-intercept form.

y 2 y1 5 m(x 2 x1) Start with the point-slope form.

y 2 11 5 24(x 2 2) Substitute (2, 11) for (x1, y1) and 24 for m.

y 2 11 5 24x 1 8 Distributive Property

y 5 24x 1 19 Add 11 to each side.

Th e graph of y 5 24x 1 19 passes through (2, 11) and is perpendicular to the

graph of y 5 14 x 2 5.

Exercises

8. Writing Are the graphs of y 5 23 x 1 6 and y 5 2

32 x 2 4 parallel,

perpendicular, or neither? Explain how you know.

Write an equation in slope-intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation.

9. (5, 23); y 5 5x 1 3 10. (4, 8); y 5 22x 2 4 11. (22, 25); y 5 x 1 3

12. (6, 0); y 5 32 x 2 6 13. (5, 3); y 5 5x 1 2 14. (7, 1); y 5 2

72 x 1 6


Parallel and Perpendicular Lines

perpendicular; the slopes 23 and 2

23 have a product of 21

y 5 2

15 x 2 2

y 5 2

23 x 1 4 y 5 2

15 x 1 4 y 5 2

7 x 2 1

y 5 2x 2 7y 5 12 x 1 6


69

Name Class Date

5-7 ReteachingScatter Plots and Trend Lines

A scatter plot is a graph that relates two diff erent sets of data by displaying them

as ordered pairs. A scatter plot can show a trend or correlation, which may be

either positive or negative. Or the scatter plot may show no trend or correlation.

It is often easier to determine whether there is a correlation by looking at a scatter

plot than it is to determine by looking at the numerical data.

If the points on a scatter plot generally slope up to the right, the two sets of data

have a positive correlation. If the points on a scatter plot generally slope down to

the right, the two sets of data have a negative correlation. If the points on a scatter

plot do not seem to generally rise or fall in the same direction, the two sets of data

have no correlation.

Problem

Th e table below compares the average height of girls at diff erent ages. Make a scatter plot of the data. What type of correlation does the scatter plot indicate?

Treat the data as ordered pairs. Th e average height of a

2-year-old girl is 34 inches, so one ordered pair is (2, 34).

Plot this point. Th en plot (3, 37), (4, 40), (5, 42),

(6, 45), (7, 48), (8, 50), (9, 52), and (10, 54).

Notice that the height increases as the age increases.

Th ere is a positive correlation for this data.

A trend line is a line on a scatter plot that is drawn near

the points. You can use a trend line to estimate other

values.

Age in years

Height in Inches

2

34

3

37

4

40

5

42

6

45

7

48

8

50

9

52

10

54

0 42 6 8 10 12 14

40

50

60

30

10

20

0

Age in years

Hei

ght

(in.)

Girls’ Growth chart

0062_hsm11a1_te_0507tr.indd 69 1/19/11 6:52:51 PM


70

Name Class Date


Scatter Plots and Trend Lines

Problem

Draw a trend line for the scatter plot in the previous problem. What is the equation for your trend line? What would you estimate to be the average height of a girl who is 12 years old?Draw a line that seems to fi t the data. Th e line drawn for this data goes through

(4, 40) and (8, 50). Use these points to write an equation.

m 550 2 40

8 2 4 5 2.5

Use the point-slope form of the line.

y 2 y1 5 m(x 2 x1)

y 2 40 5 2.5(x 2 4)

y 2 40 5 2.5x 2 10

y 5 2.5x 1 30

Use this equation to estimate the average height of

12-year-old girls.

y 5 2.5(12) 1 30

y 5 60

Exercises

Ryan practices throwing darts. From each distance listed below, he throws 10 darts and records how many times he hits the center.

1. Use the space at the right to make a scatter plot of the data.

2. Describe the type of correlation that is shown in the scatter plot.

3. Draw a trend line.

4. What equation represents your trend line?

5. How many hits do you estimate Ryan would make from 6 feet?

Distance (in feet)

Number of Center Hits

2 5 7 8 10 12 15

10 9 8 6 5 1 2

0 42 6 8 10 12 14

40

50

60

30

10

20

0

Age in years

Girls’ Growth chart

negative correlation; as the distance from the target increases, the number of center hits out of 10 decreases

See points on graph in Ex. 3.

4

8

12

4 8 12Ox

y

Distance (in feet)

Num

ber

of C

ente

r H

its

y 5 234x 1 12

about 7 hits

0062_hsm11a1_te_0507tr.indd 70 1/19/11 6:53:53 PM


79

Name Class Date

Th e graph of y 5 |x| is shown at the right.

Th e graph of y 5 |x| 1 k has the same shape as the graph of y 5 |x| but is a translation of y 5 |x| up or down by k units. If k is positive the translation is up. If k is negative, the translation is down.

Problem

What is the graph of y 5 |x| 2 4?

y 5 |x| 2 4 can be rewritten as y 5 |x| 1 (24). Th e equation is now in the form y 5 |x| 1 k. In this case, k 5 24, so translate the graph of y 5 |x| four units down.

Exercises

Graph each function by translating y 5 |x| .

1. y 5 |x| 1 1 2. y 5 |x| 2 2 3. y 5 |x| 1 4

4. Writing Compare and contrast the graphs of y 5 |x| and y 5 |x| 2 10.

5-8 ReteachingGraphing Absolute Value Functions

xO

y4

2

2

4

2

4 42

(3, 3)

(0, 0)

(23, 3)

xO

y4

2

2

4

2

4 42

translate 4units down

y 5 zx z

y 5 zx z24

xO

y

2

2

2

2

4

6

44

xO

y

2

2

4

2

2

4

44x

O

y

4

2

4

4

8

12

88

If you translate the fi rst graph down 10 units, you will make the second graph. They have the same shape.


80

Name Class Date

Th e graph of y 5 |x 1 h| has the same shape as the graph of y 5 |x| but is a translation of y 5 |x| right or left by h units. If h is positive, the translation is to the left. If h is negative, the translation is to the right.

Problem

What is the graph of y 5 |x 2 2|?

y 5 |x 2 2| can be rewritten as y 5 |x 1 (22)|. Th e equation is now in the form y 5 |x 1 h|. In this case, h 5 22, so translate the graph of y 5 |x| two units to the right.

Exercises

Graph each function by translating y 5 |x| .

5. y 5 |x 1 1| 6. y 5 |x 2 4| 7. y 5 |x 1 3|

8. Which equation is the translation 8 units up of y 5 |x|? A. y 5 |x| 1 8 B. y 5 |x| 2 8 C. y 5 |x 1 8| D. y 5 |x 2 8|

9. Which equation is the translation 7 units to the right of y 5 |x| ? F. y 5 |x| 1 7 G. y 5 |x| 2 7 H. y 5 |x 1 7| I. y 5 |x 2 7|


Graphing Absolute Value Functions

xO

y

2

2

4

2

4 42translate 2units right

y 5 zx22z

y 5 zx z

xO

y

2

2

2

2

4

6

44

xO

y

48 4 8 12 16

4

8

4

8

12

xO

y

2 2

2

4

2

4

6

468

A

I


9

Name Class Date

6-1 ReteachingSolving Systems by Graphing

Graphing is useful for solving a system of equations. Graph both equations and look for a point of intersection, which is the solution of that system. If there is no point of intersection, there is no solution.

Problem

What is the solution to the system? Solve by graphing. Check.

x 1 y 5 4 2x 2 y 5 2

Solution

y 5 2x 1 4 Put both equations into y-intercept form, y 5 mx 1 b. y 5 2x 2 2

y 5 2x 1 4 The fi rst equation has a y-intercept of (0, 4).

0 5 2x 1 4 Find a second point by substituting in 0 for y and solve for x.

x 5 4 You have a second point (4, 0), which is the x-intercept.

y 5 2x 2 2 The second equation has a y-intercept of (0, 22).

0 5 2(x) 2 2 Find a second point by substituting in 0 for y and solve for x.

2 5 2x, x 5 1 You have a second point for the second line, (1, 0).

Plot both sets of points and draw both lines. Th e lines appear to intersect (2, 2), so (2, 2) is the solution.

Check

If you substitute in the point (2, 2), for x and y in your original equations, you can double-check your answer.

x 1 y 5 4 2 1 2 0 4, 4 5 4 3

2x 2 y 5 2 2(2) 2 2 0 2, 2 5 2 3

x

y

4

8

4

4

8

8 84

(2, 2)

O


10

Name Class Date

If the equations represent the same line, there is an infi nite number of solutions, the coordinates of any of the points on the line.

Problem

What is the solution to the system? Solve by graphing. Check.2x 2 3y 5 6

4x 2 6y 5 18

Solution

What do you notice about these equations? Using the y-intercepts and solving for the x-intercepts, graph both lines using both sets of points.

Graph equation 1 by fi nding two points: (0, 22) and (3, 0). Graph

equation 2 by fi nding two points (0, 23) and (4.5, 0).

Is there a solution? Do the lines ever intersect? Lines with the same slope are parallel. Th erefore, there is no solution to this system of equations.

Exercises

Solve each system of equations by graphing. Check.

1. 2x 5 2 2 9y

21y 5 4 2 6x

2. 2x 5 3 2 y

y 5 4x 2 12

3. y 5 1.5x 1 4

0.5x 1 y 5 22

4. 6y 5 2x 2 14

x 2 7 5 3y

5. 3y 5 26x 2 3

y 5 2x 2 1

6. 2x 5 3y 2 1213x 5 4y 1 5

7. 2x 1 3y 5 11

x 2 y 5 27

8. 3y 5 3x 2 6

y 5 x 2 2

9. y 5 12x 1 9

2y 2 x 5 1


Solving Systems by Graphing

y 5 23x 2 2

y 5 23x 2 3

x

y

4

8

4

4

8

8 84O

a212, 13b a5

2, 22b a23, 212b

infi nitely many solutions

infi nitely many solutions no solution(22, 5)

(0, 21) (29, 22)


19

Name Class Date

6-2 ReteachingSolving Systems Using Substitution

You can solve a system of equations by substituting an equivalent expression for one variable.

Problem

Solve and check the following system:

x 1 2y 5 4

2x 2 y 5 3

Solution x 1 2y 5 4 The fi rst equation is easiest to solve in terms of one variable.

x 5 4 2 2y Get x to one side by subtracting 2y.

2(4 2 2y) 2 y 5 3 Substitute 4 2 2y for x in the second equation.

8 2 4y 2 y 5 3 Distribute.

8 2 5y 5 3 Simplify.

8 2 8 2 5y 5 3 2 8 Subtract 8 from both sides.

25y 5 25 Divide both sides by 25.

y 5 1 You have the solution for y. Solve for x.

x 1 2(1) 5 4 Substitute in 1 for y in the fi rst equation.

x 1 2 2 2 5 4 2 2 Subtract 2 from both sides.

x 5 2 The solution is (2, 1).

Check Substitute your solution into either of the given linear equations.

x 1 2y 5 4

2 1 2(1) 0 4 Substitute (2, 1) into the fi rst equation.

4 5 4 3 You check the second equation.

Exercises

Solve each system using substitution. Check your answer.

1. x 1 y 5 3

2x 2 y 5 0 2. x 2 3y 5 214

x 2 y 5 22

3. 2x 2 2y 5 10

x 2 y 5 5 4. 4x 1 y 5 8

x 1 2y 5 5

(1, 2) (4, 6)

infi nitely many solutions Q117 , 12

7 R


20

Name Class Date


Solving Systems Using Substitution

Problem

Solve and check the following system:

x2 2 3y 5 10

3x 1 4y 5 26

Solve x2 2 3y 5 10 First, isolate x in the fi rst equation.

x2 5 10 1 3y Add 3y to both sides and simplify.

x 5 20 1 6y Multiply by 2 on both sides.

3x 1 4y 5 26 Substitute 20 1 6y for x in second equation.

3(20 1 6y) 1 4y 5 26 Simplify.

60 1 22y 5 26 Subtract 60 from both sides.

22y 5 266, y 5 23 Divide by 22 to solve for y.

x2 2 3(23) 5 10 Substitute 23 in the fi rst equation.

x2 1 9 5 10 Simplify.

x 5 2 Solve for x.

Th e solution is (2, 23)..

Check 3(2) 1 4(23) 0 26

26 5 26 3

Now you check the fi rst equation.

Exercises

Solve each system using substitution. Check your answer.

5. 22x 1 y 5 8

3x 1 y 5 22 6. 3x 2 4y 5 8

2x 1 y 5 9

7. 3x 1 2y 5 25

2x 1 3y 5 26

8. 6x 2 5y 5 3

x 2 9y 5 25

(22, 4) (4, 1)

(22, 23)(1725, 2133

5)


29

Name Class Date

Elimination is one way to solve a system of equations. Th ink about what the word “eliminate” means. You can eliminate either variable, whichever is easiest.

Problem

Solve and check the following system of linear equations. 4x 2 3y 5 24

22x 1 3y 5 34Solution Th e equations are already arranged so that like terms are in columns.

Notice how the coeffi cients of the y-variables have the opposite sign and the same value.

4x 2 3y 5 24

2x 1 3y 5 34 Add the equations to eliminate y.

6x 5 30 Divide both sides by 6 to solve for x.

x 5 5

4(5) 2 3y 5 24 Substitute 5 for x in one of the original equations and solve for y.

20 2 3y 5 24

23y 5 224

y 5 8

Th e solution is (5, 8).

Check Substitute your solution into both of 4x 2 3y 5 24 the original equations to check.

4(5) 2 3(8) 0 24

20 2 24 0 24

24 5 24

✓

You can check the other equaton.

Exercises

Solve and check each system.

1. 3x 1 y 5 3

23x 1 y 5 3

2. 6x 2 3y 5 214

6x 2 y 5 22

3. 3x 2 2y 5 10

x 2 2y 5 6

4. 4x 1 y 5 8

x 1 y 5 5

6-3 ReteachingSolving Systems Using Elimination

(0, 3)

(2, 22) (1, 4)

Q23, 6R

Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

30

Name Class Date

If none of the variables has the same coefficient, you have to multiply before you eliminate.

Problem

Solve the following system of linear equations. 2x 3y 1

5x 4y 6

Solution

5( 2x 3y) ( 1)5

2(5x 4y) (6)2

Multiply the first equation by 5 (all terms, both sides) and the second equation by 2. You can eliminate the x variable when you add the equations together.

10x 15y 5

10x 8y 12 Distribute, simplify and add.

7y 7

y 1 Divide both sides by 7.

5x 4( 1) 6 Substitute –1 in for y in the second equation to find the value of x.

5x 4 6 Simplify.

5x 10 Add 4 to both sides.

x 2 Divide by 5 to solve for x.

The solution is (2, 1).

Check 2x 3y 1 Substitute your solution into both original equations.

2(2) 3( 1) 1

1 1 ✓ You can check the other equation.

Exercises

Solve and check each system.

5. x 3y 3

2x 7y 10

6. 2x 6y 0

3x 11y 4

7. 3x 10y 5

7x 20y 11

8. 4x y 8

x y 5


Solving Systems Using Elimination

(9, 4)

(1, 4)

( 6, 2)

1, 15


39

Name Class Date

You can solve systems of linear equations by graphing, substitution, or elimination. Deciding which method to use depends on the exactness needed and the form of the equations.

Problem

You just bought a coff ee shop for $153,600. Th e prior owner had an average monthly revenue of $8600 and an average monthly cost of $5400. If your monthly costs and revenues remain the same, how long will it take you to break even?

Write equations for revenue and costs, including the price you paid for the shop, after t months. Th en solve the system by graphing.

y 5 8600t Equation for revenue

y 5 5400t 1 153,600 Equation for cost

It appears that the point of intersection is where t is equal to 48 months. Substitute t 5 48 into either equation to fi nd the other coordinate (y), which is 412.8. Th erefore, your breakeven point is after you have run the shop for 48 months, at which point your revenue and cost are the same: $412,800.

Problem

A perfume is made from t ounces of 15% scented Th alia and b ounces of 40% Th alia. You want to make 60 oz of a perfume that has a 25% blend of the Th alia. How many ounces of each concentration of Th alia are needed to get 60 oz of perfume that is 25% strength of Th alia?

Write your systems of equations: 60(0.25) 5 0.15t 1 0.4b60 5 t 1 b

Solve the system by using substitution:

60(0.25) 5 0.15t 1 0.4b Solve the second equation for t and substitute in the fi rst equation.

15 5 0.15(60 2 b) 1 0.4b Substitute 60 2 b for t in the fi rst equation.

15 5 9 2 0.15b 1 0.4b Distributive property

24 5 b Solve for b.

Substitute 24 for b in second equation to fi nd that t = 36. Th e answer is (36, 24). Th e blend requires 36 oz of the 15% perfume and 24 oz of the 25% perfume.

6-4 ReteachingApplications of Linear Systems

t

y

128

256

384

512

32 48 6416O


40

Name Class Date

Exercises

1. You have a coin bank that has 275 dimes and quarters that total $51.50. How many of each type of coin do you have in the bank?

2. Open-Ended Write a break-even problem and use a system of linear equations to solve it.

3. You earn a fi xed salary working as a sales clerk making $11 per hour. You get a weekly bonus of $100. Your expenses are $60 per week for groceries and $200 per week for rent and utilities. How many hours do you have to work in order to break even?

4. Reasoning Find A and B so that the system below has the solution (1, 21). Ax 1 2By 5 0 2Ax 2 4By 5 16

5. You own an ice cream shop. Your total cost for 12 double cones is $24 and you sell them for $2.50 each. How many cones do you have to sell to break even?

6. Multi-Step A skin care cream is made with vitamin C. How many ounces of a 30% vitamin C solution should be mixed with a 10% vitamin C solution to make 50 ounces of a 25% vitamin C solution?

• Defi ne the variables.

• Make a table or drawing to help organize the information.

7. Your hot-air balloon is rising at the rate of 4 feet per second. Another aircraft nearby is at 7452 feet and is losing altitude at the rate of 30 feet per second. In how many seconds will your hot-air balloon be at the same altitude as the other aircraft?


Applications of Linear Systems

115 dimes; 160 quarters

Check students’ work.

10 ice cream cones

37.5 oz of 30% solution; 12.5 oz of 10% solution

about 219 s

about 14.5 h

A = 4; B = 2

0032_hsm11a1_te_0604tr.indd 40 1/19/11 2:07:00 AM


59

Name Class Date

6-6 ReteachingSystems of Linear Inequalities

A system of linear inequalities is a set of linear inequalities in the same plane. Th e solution of the system is the region where the solution regions of the inequalities of the system overlap.

Problem

What is the graph of the system of linear inequalities: x 2 y . 21

y # 2x 1 3?

Put the fi rst inequality into slope-intercept form, y , x 1 1. Use a dashed line since , does not include the points on the boundary line in the solution. Using the point (0, 0), decide where to shade the fi rst inequality. Th e point (0, 0) makes the inequality true, so shade the region including (0, 0).

Th en graph the boundary line of the second inequality, y # 2x 1 3. It is a solid line because of the # sign. Use the point (0, 0) to decide where to shade the second inequality. Th e point (0, 0) makes the second inequality true, so shade the region including (0, 0).

Th e overlapping region of the 2 inequalities is the solution to the system. It includes the points (0, 0), (1, 1), (3, 1). You can test any point in the region in both equations to see if it makes both equations true. In word problems, the solutions often cannot be negative (cars, tickets sold, etc.). Two requirements are that x $ 0 and y $ 0. Keep this in mind when graphing word problems.

Problem

A cash register has fewer than 200 dimes and quarters worth more than $39.95. How many of each coin are in the register?

Th e system of inequalities that you get from the table is: q 1 d , 200

25q 1 10d . 3995

x

y

4

8

4

4

8

8 84O

quarters

dimes

TOTAL

$0.25

$0.10

25q

10d

3995200

type of coin value in centsvalue of coinquantity

q

d


60

Name Class Date


Systems of Linear Inequalities

Using elimination, solve for q by multiplying all terms in the fi rst equation by 210and eliminating d: (q 1 d , 200)(210).

210q 2 10d . 22000 Now add the 2 systems together to solve for q.

25q 1 10d . 3995

15q . 1995

q . 133

q 1 d , 200 Write fi rst inequality.

133 1 d , 200, d , 67 Substitute in 133 for q, subtract 133 from both sides and solve for d.

Th e register contains at least 133 quarters and no more than 67 dimes.

Exercises

Graph the following systems of inequalities.

1. x 2 2y , 3y2 . 3x 1 6

2. y $ 2x 1 5

2x # 22y 2 3

3. x 1 3y $ 24

3x 2 2y , 5

4. 3y $ x4

2y # x 1 2 5. 2x 2 y , 1

x 1 2y , 24 6. 5x 2 4y $ 3

2x 1 3y # 22

xO

y

2

2

4

2

2

4

44

xO

y

2

2

4

2

2

4

44

xO

y

2

2

4

2

2

4

4 6 8

xO

y

2

4

2

2

468

2

xO

y

2

2

4

2

2

4

44

xO

y

2

2

4

6

8

2

2

44


9

Name Class Date

For every nonzero number a, a0 5 1.

For every nonzero number a and integer n, a2n 51

an . In other words, when the

exponent is negative, raise the reciprocal of the base to the opposite of the exponent.

Problem

What is the simplifi ed form of each expression?

a. 3.90 5 1 Since the exponent is 0 but the base of the expression is 3.9, which is not 0, the expression has a value of 1.

b. 922 51

92 The exponent is negative, so raise the reciprocal of 9, or 19 , to the exponent 2(22), or 2.

51

81 Simplify.

Problem

What is the simplifi ed form of 7b23

a2 using only positive exponents?

7b23

a2 57

a2 ? b23 Rewrite the expression as a product of factors with positive exponents and factors with negative exponents.

57

a2 ?1

b3 Rewrite the factor with the negative exponent by raising the reciprocal of the base to a positive exponent.

57

a2 b3 Simplify by multiplying.

7-1 ReteachingZero and Negative Exponents


10

Name Class Date

Exercises

Write each expression as an integer, a simple fraction, or an expression that contains only positive exponents. Simplify.

1. 2.30 2. 1024

3. 2a25 4. 113.70

5. 1921 6. 323

p

7. (7q)21 8. a27

8b22

9. 1.8c0 10. (29.7)0

Write each expression so that it contains only positive exponents. Simplify.

11. 2623 12. 22rs25

13. 7x28 y0 14. a5a

3bb22

15. (28v)22 w3 16. 223

m0n21

17. (3xy)0 z 18. 2323

uv22


Zero and Negative Exponents

1

1.8

2 1216

2 v2

27u

22rs5

9b2

25a27x8

n8

w3

64v2

z

1

12a5

110,000

119

6449

127p

17q


19

Name Class Date

Scientifi c notation is used to write very large numbers and very small numbers in a more compact form.

Scientifi c notation makes use of the fact that our number system is a base 10 system.

Th e number 10,000 can be expanded as 10 3 10 3 10 3 10. It can be written as a power of 10 as 104. Th e exponent 4 is the number of places the decimal point moves to the left when the number is written in scientifi c notation. In scientifi c notation, 10,000 is written as 1 3 104. Th e number 0.0001 can be written as a power of 10 as 1024.

Problem What is 0.0034 written in scientifi c notation?

In scientifi c notation, numbers are written in the form a 3 10n, where a is at least 1, but less than 10. Move the decimal point 3 places to the right. Th e exponent of 10 in scientifi c notation will be 23. Since 0.0034 is smaller than 3.4, the exponent must be negative.

0.0034 5 3.4 3 1023

Problem What is 35,100,000 written in scientifi c notation?

To write 35,100,000 in scientifi c notation fi rst identify a. Th e value of a must be greater than or equal to 1 and less than 10. So a 5 3.51. Now determine what power of 10 you need to multiply a by to get 35,100,000.

3.51 3 10,000,000 5 35,100,000

Because 10,000,000 5 107, 35,100,000 5 3.51 3 107.

To change a number from scientifi c notation to standard notation, start with the value of a. Th en move the decimal point to the left or right depending on the exponent of 10. For example, 4.72 3 104 5 47,200. Th e decimal point is moved 4 places to the right, and zeros are added as placeholders.

7-2 ReteachingScientifi c Notation


20

Name Class Date

For each number in standard notation, fi rst identify a and then write the number in scientifi c notation.

1. 4300 2. 0.0029 3. 87,000,000

4. 0.0000402 5. 834,000 6. 0.0000090

For each number in scientifi c notation, rewrite in standard notation.

7. 8.3 3 105 8. 4.01 3 1028 9. 5.11 3 1011

10. 6.1 3 1022 11. 5.83 3 100 12. 3.6 3 10212

13. Error Analysis A student knew that the number 478.2 3 105 was not written in scientifi c notation, but was having trouble explaining why. Using the conditions placed on the value of a, explain why 478.2 3 105 is not written in scientifi c notation. Th en rewrite the expression so that it is in scientifi c notation.

14. Reasoning A classmate suggests that instead of using scientifi c notation to write very large and very small numbers, it would be easier to use octagonal

notation with numbers written in the form a ? 8b. Do you agree? Explain.

15. Open-Ended Describe three situations in which it is easier to use scientifi c notation than standard notation. Give examples for each situation.

16. Error Analysis Two students came up with the answers 3.8 ? 105 km and

3.8 ? 108 m for the same problem. Could they both be right? Explain. What if

the answers were 2.5 ? 104 mi and 2.5 ? 107 ft?


Scientifi c Notation

4.3; 4.3 3 103 2.9; 2.9 3 1023

9.0; 9.0 3 10268.34; 8.34 3 105

8.7; 8.7 3 107

4.02; 4.02 3 1025

830,000

0.061

a must be greater than or equal to 1 and less than 10; 4.782 3 107

no; Because our number system is a decimal system, it is easier to use place value and move the decimal point.

Answers may vary. Sample: very long distances, such as a light year, the distance light travels in one year, which is about 9.461 3 1015 m; very small distances, such as in molecules where the angstrom (1.0 3 10210 m) is used; and the mass of planets and stars, such as the sun, which has a mas of 1.98892 3 1030 kg.

Both stueents are right, because 3.8 3 105 km is equal to 3.8 3 108 m. This is true because 1 km 5 103. This does not work for mi and ft: 2.5 3 104 mi or 25,000 mi is not equal to 2.5 3 107 ft, or 25,000,000 ft.

5.83

0.0000000401 511,000,000,000

0.0000000000036

Name Class Date


29

7-3 ReteachingMultiplying Powers With the Same Base

When multiplying powers with the same base, you add the exponents. Th is is true for numerical and algebraic expressions.

Problem

What is each expression written as a single power? a. 34 ? 32 ? 33

All three powers have the same base, so this expression can be written as a single power by adding the exponents.

34 ? 32 ? 33 5 34 1 2 1 3 All powers have the same base. Add the exponents.

5 39 Simplify the exponent.

34 represents 4 factors of 3, 32 represents 2 factors of 3, and 33 represents 3 factors

of 3. Th is is a total of 9 factors of 3, so the answer is reasonable.

Even when some of the exponents are negative, exponents can be added when the bases are the same in a product of powers.

b. 1123 ? 114 ? 1125

1123 ? 114 ? 1125 5 1123 1 4 1 (25) All powers have the same base. Add the exponents.

5 1124 Simplify the exponent.

Problem

What is the simplifi ed form of (1.8 3 1011)(2.7 3 108)? Write the answer in scientifi c notation.

Use the Associative and Commutative Properties of Multiplication to regroup and reorder the factors so that the powers of 10 are grouped together and numbers that are not powers of 10 are grouped separately from the powers of 10.

(1.8 3 1011)(2.7 3 108) 5 (1.8 ? 2.7)(1011 ? 108) Associative and Commutative Prop. of Mult.

5 (4.86)(1011 1 8) Multiply the numbers in the fi rst set of parentheses. Add the exponents for the powers of 10.

5 4.86 3 1019 Simplify the exponent.

Name Class Date


30


Multiplying Powers With the Same Base

Exercises


1. a2a3 2. 3n3n5 3. 8k3 ? 3k6

4. (8p5)(6p4) 5. 21d7 ? 2d3 6. (26.1m4)(3m2)

7. h5 ? h2 ? h10 8. (29q28)(6q11) 9. (16r27)(22r)

10. (y3z13)(y2z26) 11. (23x2)(5w 8)(4x3) 12. (15fg

2)(f 3g23)(28f21g 6)

13. m26 ? m3 ? n22 14. 26j23k ? 7jk5 15. 22uvw21 ? 3u2v22w

Simplify each expression. Write each answer in scientifi c notation.

16. (4 3 103)(2 3 105) 17. (1 3 104)(6 3 103) 18. (7 3 102) ? 105

19. (8 3 109)(3 3 1025) 20. (2 3 105)(5 3 106) 21. (7 3 1028)(3 3 1026)

Write each answer in scientifi c notation.

22. Th e distance light travels in one year (one light-year) is about 5.87 3 1012 mi. A star called Proxima Centauri is 4.2 light-years away from Earth. About how many miles from Earth is Proxima Centauri?

23. After the Revolutionary War, the U.S. national debt was approximately 7.5 3 107 dollars. In 2008, the debt was approximately 1.33 3 105 times the original amount. What was the national debt in 2008?

Complete each equation.

24. 4u ? 43 5 413 25. 86 ? 85 5 8u 26. 34 ? 3u 5 310

27. k11 ? ku 5 k2 28. wu ? w 5 w4 29. x2 ? xu ? x 5 x9

30. p25 ? pu 5 p3 ? p2 31. n5 ? n217nu 5 n13 32. t5u2 ? tuu 5 t24u3

a5

48p9

h17

y5z7

8.0 3 108

2.4 3 105

2.4654 3 1013 mi

$9.975 3 1012

10

10

29

29

11

25

3 6

6

1.0 3 1012 2.1 3 10213

6.0 3 107 7.0 3 107

1m3n2 242k6

j226u3

v

42d10 218.3m6

254q3

260x5w8

2120f 3g5

232r6

3n8 24k9

0022_hsm11a1_te_0703tr.indd 30 1/19/11 7:43:19 AM

Name Class Date


39

7-4 ReteachingMore Multiplication Properties of Exponents

When a power is raised to another power, like (xy)z, multiply the exponents.

Problem

What is the simplifi ed form of (d3)4?

(d3)4 5 d3?4 The expression is a power, d3, raised to another power, 4. Multiply the exponents.

5 d12 Simplify.

Simplifying powers may require you to use multiple properties of exponents. You should follow the order of operations when simplifying exponential expressions.

Problem

What is the simplifi ed form of (n23)6n4?

Using the order of operations, fi rst simplify the power (n23)6.

(n23)6n4 5 (n23?6)n4 5 n218n4

Next, multiply. Th e two powers have the same base, so simplify by adding the exponents.

n218n4 5 n21814 5 n214

Finally, write the expression using positive exponents. Rewrite the expression using the reciprocal of the base and the opposite of the exponent.

n214 51

n14

You should follow the same rules when simplifying numbers written in scientifi c notation raised to a power.

Problem

What is the simplifi ed form of (4.2 3 1027)2 written in scientifi c notation?

(4.2 3 1027)2 5 (4.2)2(1027)2 This is a product raised to the exponent 2, so each factor of the product must be raised to the exponent 2.

5 17.64 3 10214 Multiply 4.2 by itself. Multiply the exponents on the expression with base 10.

5 1.764 3 10213 Move the decimal point one place left and adjust the exponent on 10 to write in scientifi c notation.

Name Class Date


40


More Multiplication Properties of Exponents

Exercises


1. (y2)3 2. (v9)6 3. (h4)5 4. (n4)11

5. (p21)5 6. (z3)26 7. (x24)25x 8. (f 5)21f 8

9. (3a)4 10. (6c)23 11. (7k)0 12. (10s23)2

13. (2y25)3(x11y210)2 14. u29(u21v)4u25 15. (x13y6)22(y25x10)6 16. 4m0n0(6m5)2

Simplify. Write each answer in scientifi c notation.

17. (2 3 1028)3 18. (3 3 105)3 19. (9 3 10215)3 20. (6 3 105)2

21. (6.7 3 1011)2 22. (9.5 3 107)3 23. (4.7 3 10211)22 24. (5.14 3 106)2

25. Th e radius of a cylinder is 6.8 3 105 m. Th e height of the cylinder is

2.2 3 103 m. What is the volume of the cylinder? (Hint: V 5 3.14r 2h)

Complete each equation.

26. (y3)u 5 y6 27. (6p3qu)2 5 36p6 28. (4au)3 5 64a26

29. (k11)u 5 1 30. (t28)u 5 t16 31. 15(c21)u 5 15c10

y6

1p5

1z18

1216c3

100s6

8x22

y35

8 3 10224

4.489 3 1023

3.19426 3 1015 m3

2

0

0 22

22 210

8.57375 3 1023 4.53 3 1020 2.64196 3 1013

2.7 3 1016 7.29 3 10243 3.6 3 1011

v4

u18x34

y42144m10

1

v54 h20

x21

81a4

f3

n44


49

Name Class Date

Understanding division properties of exponents allows you to simplify quotients involving exponents.

Problem

What is 66 divided by 64 ?

Method 1: Evaluate 66 and 64, and then divide the results.

66 5 46,656

64 5 1296

46,656 ÷ 1296 = 36

Method 2: Expand the numerator and denominator.

66

64 56 ? 6 ? 6 ? 6 ? 6 ? 6

6 ? 6 ? 6 ? 6

After dividing out the common factors, you are left with 6 3 6 5 36.

When you divide powers with the same base, subtract the exponents. In the

example above, 66 and 64 are powers with the same base and when you divided

them, the result was 36 5 62. Th is is the same result you get by subtracting the

exponents: 66

64 5 6624 5 62.

Th e division property of exponents also allows you to simplify quotients that contain variables.

Problem

How can you use the division property of exponents to show that x2 5x5

x3 when x u 0?

Expand the numerator and denominator.

x5

x3 5x ? x ? x ? x ? x

x ? x ? x

After dividing out the common factors you are left with x ? x 5 x2 .

Division properties of exponents work whether the bases in the problem are constants or variables. When you divide powers with the same base, subtract the

exponents. In this example, x5 and x3 are powers with the same base and when

you divided them, the result was x2 5 x523.

7-5 ReteachingDivision Properties of Exponents


50

Name Class Date


1. 75

72 2. 39

32

3. 52

5 4.

4z

44

5. m4

m2 6. p6

p5

7. r3

r 8.

x5y4

x3y

9. a3

a5 10. 10x5

15x2

11. Use properties of exponents to show that a0 5 1. (Hint: Write the quotient of two powers that have a as their base and have the same exponent.)

12. Compare multiplying and dividing powers with the same base.


Division Properties of Exponents

73 37

5

m2

r2

1a2

23 x3

Because a5

a5 5 1, a5

a5 5 a525 5 a0 5 1

To multiply powers of the same base, you add the exponents. To divide powers of the same base, you subtract the exponents.

x2y3

p

4z24


59

Name Class Date

Functions that can be modeled by an equation of the form y 5 a ? bx are exponential functions. Th ese functions have properties that are diff erent from the properties of linear and quadratic functions.

Consider the three function tables below.

Notice that in each table, the x-value increases by a constant amount. If these functions were linear, the y-values would also increase by a constant amount. You used these values to fi nd the slope of linear equations.

Th is does not hold true for exponential functions. See if you can determine the property that holds true for all exponential functions by: a. Finding the sum of some random pairs of consecutive y-values in each

table. b. Finding the diff erence between some random pairs of consecutive y-values

in each table. c. Finding the product of some random pairs of consecutive y-values in each

table. d. Finding the quotient of some random pairs of consecutive y-values in each

table.

You should have noticed that, for each table, the quotients remain the same.

Exponential functions model an initial amount, a, that is repeatedly multiplied by the same positive number, b. Th e number of times the multiplication occurs is determined by the independent variable, x, which is the exponent in the power bx.

012345

3612244896

x y012345

2105025012506250

x y012345

392781243729

x y

7-6 ReteachingExponential Functions

0052_hsm11a1_te_0706tr.indd 59 1/19/11 7:19:15 AM


60

Name Class Date

1. For each of the tables on the previous page, extend them two units in each direction. Use the common diff erence in the x-values and the common ratio in the y-values to do the extension. Th e fi rst table is done for you.

2. Plot the points in each of your extended tables on separate coordinate grids. Connect the points with a smooth curve. Th e domain of each function is all real numbers and that the range is all positive real numbers. Explain why there are negative values for x but not for y.

3. For each of the tables, identify the starting value a and the common ratio b. For the fi rst table, a is 1 and b is 3. Next, write the exponential function that describes each table. Th e function for the fi rst table is f (x) 5 1 ? 3x. Check if your function is correct by substituting in x-values and seeing if the function produces values for y that match the values in the table.

01234567

39278124372921876561

x y

131

2

1


Exponential Functions

012345

3

2 0.751 1.5

612244896

6 1927 384

x y

0

21

0.080.4

12345

2105025012506250

6 31,2507 156,250

x y

xO

y

4000

2000

6000

8000

2 4 62x

O

y

200

100

300

400

2 4 62x

O

y

80,000

40,000

120,000

160,000

2 4 62

The domain is all real numbers because x can have any value, but the range is all positive real numbers because a S 0 and bx

S 0, so a ? bx will always be S 0.

Table 1: 1; 3; 1 ? 3x; Table 2: 3; 2; 3 ? 2x ; Table 3: 2; 5; 2 ? 5x ;

0052_hsm11a1_te_0706tr.indd 60 1/19/11 8:28:38 AM

Name Class Date


69

7-7 ReteachingExponential Growth and Decay

Exponential functions can model the growth or decay of an initial amount.

Th e basic exponential function is y 5 a ? b x where

a represents the initial amount

b represents the growth (or decay) factor. Th e growth factor equals 100% plus the percent rate of change. Th e decay factor equals 100% minus the percent rate of decay.

x represents the number of times the growth or decay factor is applied.

y represents the result of applying the growth or decay factor x times.

Problem

A gym currently has 2000 members. It expects to grow 12% per year. How many members will it have in 6 years?

Th ere are 2000 members to start, so a 5 2000. Th e growth per year is 12%, or 0.12, so b 5 1 1 0.12 or 1.12. Th e desired time period is 6 years, so x 5 6. Th e function is y 5 2000 ? 1.12x.

When x 5 6, y 5 2000 ? 1.126 < 3948. So, the gym will have about 3948 members

in 6 years.

In Exercises 1–3, identify a, b, and x. Th en use them to write the exponential function that models each situation. Finally use the function to answer the question.

1. When a new baby is born to the Johnsons, the family decides to invest $5000 in an account that earns 7% interest as a way to start the baby’s college fund. If they do not touch that investment for 18 years, how much will there be in the college fund?

2. Th e local animal rescue league is trying to reduce the number of stray dogs in the county. Th ey estimate that there are currently 400 stray dogs and that through their eff orts they can place about 8% of the animals each month. How many stray dogs will remain in the county 12 months after the animal control eff ort has started?

3. A basket of groceries costs $96.50. Assuming an infl ation rate of 1.8% per year, how much will that same basket of groceries cost in 20 years?

a 5 5000, b 5 1.07, x 5 18; y 5 5000(1.07)18; $16,899.66

a 5 400, b 5 0.92, x 5 12; y 5 400(0.92)12; 147 stray dogs

a 5 96.5, b 5 1.018, x 5 20; y 5 96.5(1.018)20; $137.87

Name Class Date


70

While it is usually fairly straightforward to determine a, the initial value, you often need to read the problem carefully to make sure that you are correctly identifying b and x. Th is is especially true when considering situations where the given growth rate is applied in intervals that are not the same as the given value of x.

Problem

You invest $1000 in an account that pays 8% interest. If nothing changes, a 5 1000, b 5 1.08, and x 5 3. Th e function is y 5 1000 ? 1.083. How much money will you have left after 3 years?

In this case, the interest is compounded annually, meaning it is added to your account at the end of each year. But what happens if the interest is compounded quarterly, meaning it is added to your account at the end of each quarter of a year?

Th ere will be 12 compounding periods in the 3-year period.

In each compounding period you add the appropriate fraction of the total annual interest, in this case one-fourth of the interest.

Th e new function is y 5 1000 ? a1 1 0.084 b

12, given that a is the initial

investment, b is the amount of growth for each compounding period, and x is the number of compounding periods.

Compounded annually, the value of the investment will be $1259.71 after 3 years. Compounded quarterly, the value will be $1268.24.

In Exercises 4 and 5, identify a, b, and x. Th en write the exponential function that models the situation. Finally, use the function to answer the question.

4. You invest $2000 in an investment that earns 6% interest, compounded quarterly. How much will the investment be worth after 5 years?

5. You invest $3000 in an investment that earns 5% interest, compounded monthly. How much will the investment be worth after 8 years?

6. Th e formula that fi nancial managers and accountants use to determine the value of investments that are subject to compounding interest is

A 5 P (1 1rn

)nt where A is the fi nal balance, P is the initial deposit, r is the

annual interest rate, n is the number of times the interest is compounded per year and t is the number of years. Redo Exercises 4 and 5 using this formula.


Exponential Growth and Decay

a 5 2000, b 5 1.015, x 5 20; y 5 2000(1.015)20; $2693.71

a 5 3000, b 5 1.004167, x 5 96; y 5 3000(1.004167)96; $4471.89

Redo 4: A 5 PQ1 1 rnR

nt5 2000Q1 1 0.06

4 R4?5

5 2000(1.015)20 5 $2693.71

Redo 5: A 5 PQ1 1 rnR

nt5 3000Q1 1 0.05

12 R12?8

5 3000(1.004167)96 5 $4471.89

Name Class Date


9

You can add and subtract polynomials by lining up like terms and then adding or subtracting each part separately.

Problem

What is the simplifi ed form of (3x2 2 4x 1 5) 1 (5x2 1 2x 2 8)?

Write the problem vertically, lining up the like terms. Th en add each pair of like terms.

Solve Add the x2 terms. Add the x terms. Add the constant terms.

3x2 1 5x2 5 8x2 24x 1 2x 5 22x 5 1 (28) 5 23

3x2 2 4x 1 5

1 5x2 1 2x 2 8

8x2 2 2x 2 3

Add the sums.

Check Check your solution using subtraction.

8x2 2 5x2 5 3x2 22x 2 2x 5 24x 23 2 (28) 5 5

Solution: (3x2 2 4x 1 5) 1 (5x2 1 2x 2 8) 5 8x2 2 2x 2 3

Exercises

Simplify.

1. 5b2 1 3b

1 2b2 2 5b 2.

3c2 1 3c1 4c2 1 2c

3. 4d2 2 3d 1 6

1 2d2 1 5d 2 3

4. 23e2 2 5e 1 2

1 e2 1 2e 2 7 5.

4f 31 2f 2 1 5f1 2f 32 4f 22 3f

6. 5g3 2 2g2 1 3g

1 2g3 1 5g2 2 2g

7. (3h2 1 5) 1 (25h2 2 3) 8. (2j2 1 4j 2 6) 1 (4j2 2 3j 2 3)

8-1 ReteachingAdding and Subtracting Polynomials

3x2 2 4x 1 5

1 5x2 1 2x 2 8

7b2 2 2b

22e2 2 3e 2 5

22h2 1 2 6j2 1 j 2 9

6f 32 2f 21 2f 7g3 1 3g2 1 g

7c2 1 5c 6d2 1 2d 1 3

Name Class Date


10

To subtract polynomials, follow the same steps as in addition.

Problem

What is the simplifi ed form of (6x3 1 4x2 2 3x) 2 (2x3 1 3x2 2 5x)?

Write the problem vertically, lining up the like terms. Th en subtract each pair of like terms.

Solve

Subtract the x3 terms. Subtract the x2 terms. Subtract the x terms.

6x3 2 2x3 5 4x3 4x2 2 3x2 5 x2 23x 2 (25x) 5 2x

6x3 1 4x2 2 3x2 (2x3 1 3x2 2 5x)

4x3 1 x2 1 2x

Add the differences.

Check Check your solution using subtraction.

4x3 1 2x3 5 6x3 x2 1 3x2 5 4x2 2x 1 (25x) 5 23x

Solution: (6x3 1 4x2 2 3x) 2 (2x3 1 3x2 2 5x) 5 4x3 1 x2 1 2x

Exercises

Simplify.

9. 4k2 1 5k

2 (3k2 1 2k) 10.

5m2 2 4m2 (2m2 1 3m)

11. 7n2 1 4n 1 9

2 (4n2 1 3n 1 5)

12. 5p2 1 6p 1 4

2 (7p2 1 4p 1 8) 13. 3q3 1 2q2 1 7q

2 (6q3 2 4q2 2 5q) 14.

2r3 2 2r2 1 5r2 (4r3 1 5r2 1 3r)

15. (6s2 2 5s) 2 (22s2 1 3s) 16. (3w2 1 6w 2 5) 2 (5w2 2 4w 1 2)


Adding and Subtracting Polynomials

6x3 1 4x2 2 3x2 (2x3 1 3x2 2 5x)

k2 1 3k

22p2 1 2p 2 4

8s2 2 8s 22w2 1 10w 2 7

3m2 2 7m

23q3 1 6q2 1 12q 22r3 2 7r2 1 2r

3n2 1 n 1 4


19

Name Class Date

You can multiply a monomial and a trinomial by solving simpler problems. You can use the Distributive Property to make three simpler multiplication problems.

Problem

What is the simplifi ed form of 3x(2x2 1 4x 2 1)?

Use the Distributive Property to rewrite the problem as three separate multiplication problems.

3x(2x2 1 4x 2 1) 5 (3x ? 2x2) 1 (3x ? 4x) 1 (3x ? (21))

Remember that when you multiply same-base terms containing exponents, you

add the exponents.

Solve 3x ? 2x2 5 6x3 Multiply inside the fi rst pair of parentheses.

3x ? 4x 5 12x2 Multiply inside the second pair of parentheses.

3x ? (21) 5 23x Multiply inside the third pair of parentheses.

6x3 1 12x2 2 3x Add the products.

Check 6x3 4 2x2 5 3x Check your solution using division.

12x2 4 4x 5 3x

23x 4 (21) 5 3x

Solution: 3x(2x2 1 4x 2 1) 5 6x3 1 12x2 2 3x

Exercises

Simplify each product.

1. 4x(2x 2 7) 2. 3y(3y 1 4) 3. 2z2(2z 2 3)

4. 3a(24a 2 6) 5. 6b(3b2 1 2b 2 4) 6. 3c2(2c2 2 4c 1 3)

7. 22d(4d2 1 3d 2 2) 8. 5e2(23e2 2 2e 2 3) 9. 4f(23f3 1 2f2 1 6)

8-2 ReteachingMultiplying and Factoring

8x2 2 28x

212a2 2 18a

28d3 2 6d2 1 4d

9y2 1 12y

18b3 1 12b2 2 24b

215e4 2 10e3 2 15e2

4z3 2 6z2

6c4 2 12c3 1 9c2

212f 4 1 8f 31 24f


20

Name Class Date

To factor a polynomial, fi nd the greatest common factor (GCF) of the coeffi cients and

constants and also the GCF of the variables.

Problem

What is the factored form of 8x4 1 12x2 2 16x?

Solve Find the GCF of the coeffi cients. Use prime factorization.

8 5 2 ? 2 ? 2

12 5 2 ? 2 ? 3

16 5 2 ? 2 ? 2 ? 2

Th e GCF of the numbers is 4.

Each term has a variable. Remember, x 5 x1.

Th e GCF is the least exponent.

Th e GCF of the variables is x.

Th e GCF is 4x. Combine the GCFs.

Factor out the GCF of each term.

4(2 1 3 2 4) Factor the coeffi cients.

4x(2x3 1 3x 2 4) Insert the variables.

Check 4x(2x3 1 3x 2 4) 5 8x4 1 12x2 2 16x Check by multiplying.

Solution: Th e factored form of 8x4 1 12x2 2 16x is 4x(2x3 1 3x 2 4).

Exercises

Find the GCF of the terms of each polynomial.

10. 12x2 2 6x 11. 4y2 1 12y 1 8 12. 6z3 1 15z2 2 9z

Factor each polynomial.

13. 8a 1 10 14. 12b2 2 18b 15. 9c3 1 12c2

16. 5d3 2 10d2 1 20d 17. 6e2 1 10e 2 8 18. 8g3 2 24g2 1 16g


Multiplying and Factoring

6x

2(4a 1 5)

5d(d2 2 2d 1 4)

4

6b(2b 2 3)

2(3e2 1 5e 2 4)

3z

3c2(3c 1 4)

8g(g2 2 3g 1 2)


29

Name Class Date

You can multiply binomials by using the FOIL method. FOIL stands for First, Outer, Inner, and Last.

Problem

What is the simplifi ed form of (4x 1 3)(2x 1 6)?

Use the FOIL method to simplify the binomial.

Solve 4x ? 2x 5 8x2 Multiply the First terms.

4x ? 6 5 24x Multiply the Outer terms.

3 ? 2x 5 6x Multiply the Inner terms.

3 ? 6 5 18 Multiply the Last terms.

8x2 1 24x 1 6x 1 18 Add the products.

8x2 1 30x 1 18 Add the like terms.

Check Substitute any number for x. Try x 5 2. If the two sides of the equation are equal the simplifi cation may be correct.

(4x 1 3)(2x 1 6) 0 8x2 1 30x 1 18

(4 ? 2 1 3)(2 ? 2 1 6) 0 (8 ? 22) 1 (30 ? 2) 1 18

(11)(10) 0 32 1 60 1 18

110 5 110 3

Solution: Th e simplifi ed form of (4x 1 3)(2x 1 6) is 8x2 1 30x 1 18.

Exercises


1. (a 1 6)(a 2 3) 2. (b 2 4)(b 1 5) 3. (c 1 3)(c 1 7)

4. (2d 1 4)(3d 2 2) 5. (4e 2 5)(3e 1 3) 6. (3f 2 2)(2f 2 4)

7. (5g 1 3)(g 2 3) 8. (4h 1 4)(2h 1 5) 9. (3j 2 5)(4j 2 3)

8-3 ReteachingMultiplying Binomials

a2 1 3a 2 18

6d2 1 8d 2 8

5g2 2 12g 2 9

b2 1 b 2 20

12e2 2 3e 2 15

8h2 1 28h 1 20

c2 1 10c 1 21

6f 22 16f 1 8

12j2 2 29j 1 15


30

Name Class Date

To multiply a trinomial by a binomial, use the same steps as you would to multiply a 3-digit number by a 2-digit number. Find the partial products for each term of the binomial and then add the like terms of the partial products.

Problem

What is the simplifi ed form of (2x2 1 3x 2 4)(3x 1 2)?

Solve Start by arranging the polynomials vertically.

Multiply each part of the trinomial by 2.

2x2 1 3x 2 42x2 1 3x 1 24x2 1 6x 2 8

Multiply each part of the trinomial by 3x.

6x3 1 2x2 1 3x 2 46x2 1 4x2 1 3x 1 2

6x3 1 4x2 1 6x 2 8

6x3 1 9x2 2 12x 2 8

2x2 ? 3x 5 6x3

3x ? 3x 5 9x2

24 ? 3x 5 212x

Add the partial products.

6x3 1 4x2 1 6x 2 8

6x3 1 9x2 2 12x 2 8

6x3 1 13x2 2 6x 2 8

Check Substitute any number for x. Try x 5 2. If the two sides of the equation are equal, the simplifi cation may be correct.

(2x2 1 3x 2 4)(3x 1 2) 0 6x3 1 13x2 2 6x 2 8

(8 1 6 2 4)(6 1 2) 0 48 1 52 2 12 2 8

80 5 80 3

Solution: Th e simplifi ed form of (2x2 1 3x 2 4)(3x 1 2) is 6x3 1 13x2 2 6x 2 8.

ExercisesSimplify each product.

10. (w2 1 3w 2 4)(2w 1 3) 11. (x2 2 8x 1 6)(3x 2 4)

12. (2y2 1 4y 2 5)(4y 1 2) 13. (3z2 2 6z 1 4)(4z 1 1)


Multiplying Binomials

2x2 ? 2 5 4x2

3x ? 2 5 6x24 ? 2 5 28

2w3 1 9w2 1 w 2 12

8y3 1 20y2 2 12y 2 10

3x3 2 28x2 1 50x 2 24

12z3 2 21z2 1 10z 1 4


39

Name Class Date

A binomial is squared when it is multiplied by itself. Th e square of a binomial is the square of the fi rst term plus the twice the product of the two terms plus the

square of the last term. Th is can be expressed as (a 1 b)2 5 a2 1 2ab 1 b2.

Problem

What is the simplifi ed form of (x 1 5)2?

Use the rules for squaring a binomial.

Solve x ? x 5 x2 Square the fi rst term.

2(5 ? x) 5 10x Multiply the product of the two terms by 2.

5 ? 5 5 25 Square the last term.

So, (x 1 5)2 5 x2 1 10x 1 25.

Check (x 1 5)2 5 (x 1 5)(x 1 5) Rewrite the binomials.

x ? x 5 x2 Multiply the First addends.

x ? 5 5 5x Multiply the Outer addends.

5 ? x 5 5x Multiply the Inner addends.

5 ? 5 5 25 Multiply the Last addends.

x2 1 5x 1 5x 1 25 Add the products.

x2 1 10x 1 25 Combine the like terms.

Solution: Th e simplifi ed form of (x 1 5)2 is x2 1 10x 1 25.

Exercises


1. (a 1 7)2 2. (b 2 4)2 3. (2c 1 3)2 4. (3d 2 5)2

5. (4e 1 1)2 6. (2f 2 6)2 7. (g 2 10)2 8. (5h 1 8)2

9. (3j 2 3)2 10. (2k 1 4)2 11. (4m 2 2)2 12. (3n 1 6)2

8-4 ReteachingMultiplying Special Cases

a2 1 14a 1 49

16e2 1 8e 1 1

9j2 2 18j 1 9

b2 2 8b 1 16

4f 2 2 24f 1 36

4k2 1 16k 1 16

4c2 1 12c 1 9

g2 2 20g 1 100

16m2 2 16m 1 4

9d2 2 30d 1 25

25h2 1 80h 1 64

9n2 1 36n 1 36


40

Name Class Date

Th e product of the sum and the diff erence of the same two terms produces a pattern

that can be expanded algebraically as (a 1 b)(a 2 b) 5 a2 2 ab 1 ab 2 b2. Th e

sum of the two ab- terms is 0. Th erefore, (a 1 b)(a 2 b) 5 a2 2 b2. Th e product is the square of the fi rst term minus the square of the last term.

Problem

What is the simplifi ed form of (2x 2 3)(2x 1 3)?

Use the rules for fi nding the product of the sum and the diff erence of the same two terms.

Solve 2x ? 2x 5 4x2 Square the fi rst term.

3 ? 3 5 9 Square the last term.

Remember, the product is the diff erence of the two squares.

Th e product is 4x2 2 9.

Check Multiply the binomials using the FOIL Method.

2x ? 2x 5 4x2 Multiply the First addends.

2x ? 3 5 6x Multiply the Outer addends.

23 ? 2x 5 26x Multiply the Inner addends.

23 ? 3 5 29 Multiply the Last addends.

4x2 1 6x 2 6x 2 9 Add the products.

4x2 2 9 Combine the like terms.

Solution: Th e simplifi ed form of (2x 2 3)(2x 1 3) is 4x2 2 9.

Exercises


13. (p 2 4)(p 1 4) 14. (q 1 5)(q 2 5) 15. (3r 1 2)(3r 2 2)

16. (4s 2 6)(4s 1 6) 17. (2t 2 1)(2t 1 1) 18. (5u 2 3)(5u 1 3)

19. (6v 2 4)(6v 1 4) 20. (3w 2 8)(3w 1 8) 21. (7x 2 9)(7x 1 9)


Multiplying Special Cases

p2 2 16

16s2 2 36

36v2 2 16

q2 2 25

4t2 2 1

9w2 2 64

9r2 2 4

25u2 2 9

49x2 2 81


49

Name Class Date

8-5 ReteachingFactoring x2 1 bx 1 c

If a trinomial of the form x2 1 bx 1 c can be written as the product of two

binomials, then:

• Th e coeffi cient of the x-term in the trinomial is the sum of the constants in the

binomials.

• Th e trinomial’s constant term is the product of the constants in the binomials.

Problem

What is the factored form of x2 1 12x 1 32?

To write the factored form, you are looking for two factors of 32 that have a sum of 12.

Solve Make a table showing the factors of 32.

x2 1 12x 1 32 5 (x 1 4)(x 1 8)

Check (x 1 4)(x 1 8)

x2 1 8x 1 4x 1 32 Use FOIL Method.

x2 1 12x 1 32 Combine the like terms.

Solution: Th e factored form of x2 1 12x 1 32 is (x 1 4)(x 1 8).

Exercises

Factor each expression.

1. x2 1 9x 1 20 2. y2 1 12y 1 35 3. z2 1 8z 1 15

4. a2 1 11a 1 28 5. b2 1 10b 1 16 6. c2 1 12c 1 27

7. d2 1 6d 1 5 8. e2 1 15e 1 54 9. f 2 1 11f 1 24

Factors of 32 Sum of Factors

33

18

12

1 and 32

2 and 16

4 and 8

(x 1 5)(x 1 4) (y 1 7)(y 1 5) (z 1 5)(z 1 3)

(c 1 9)(c 1 3)

(f 1 8)(f 1 3)

(b 1 8)(b 1 2)

(e 1 9)(e 1 6)

(a 1 4)(a 1 7)

(d 1 5)(d 1 1)


50

Name Class Date


Factoring x2 1 bx 1 c

Some factorable trinomials in the form of x2 1 bx 1 c will have negative

coeffi cients. Th e rules for factoring are the same as when the x-term and the

constant are positive.

• Th e coeffi cient of the x-term of the trinomial is the sum of the constants in the

binomials.

• Th e trinomial’s constant term is the product of the constants in the binomials.

However, one or both constants in the binomial factors will be negative.

Problem

What is the factored form of x2 2 3x 2 40?

To write the factored form, you are looking for two factors of 240 that have a sum

of 23. Th e negative constant will have a greater absolute value than the positive

constant.

Solve Make a table showing the factors of 240.

x2 2 3x 2 40 5 (x 2 8)(x 1 5)

Check (x 2 8)(x 1 5)

x2 1 5x 2 8x 2 40 Use FOIL Method.

x2 1 (23x) 2 40 Combine the like terms.

Solution: Th e factored form of x2 2 3x 2 40 is (x 2 8)(x 1 5).

Exercises


10. s2 1 2s 2 35 11. t2 2 4t 2 32 12. u2 1 6u 2 27

13. v2 2 2v 1 48 14. w2 2 8w 2 9 15. x2 1 3x 2 18

3

Factors of ]40 Sum of Factors39186

1 and 402 and 204 and 105 and 8

(s 1 7)(s 2 5)

(v 2 8)(v 1 6)

(t 2 8)(t 1 4)

(w 2 9)(w 1 1)

(u 1 9)(u 2 3)

(x 1 6)(x 2 3)


59

Name Class Date

You can use your knowledge of prime numbers to help you factor some trinomials as two binomials. A prime number has only 1 and itself as factors. For trinomials of the form

ax2 1 bx 1 c, if a is a prime number then you already know the fi rst term of each binomial: ax and 1x. Th en list the factors that will multiply to produce c. Use guess and check to fi nd the factor pair that will add to b.

Problem

What is the factored form of 7x2 1 31x 1 12?

7x2 1 31x 1 12 5 (7x )(1x ) a is 7, which is prime, so the factors are 7 and 1.

5 (7x )(x ) You don’t need the 1 in front of the variable, so drop it.

7x2 1 31x 1 12 5 (7x 1 )(x 1 ) The trinomial has two plus signs, so the

binomials also have plus signs.

Because c is 12, fi nd factor pairs that multiply to 12: (1 and 12), (2 and 6), (3 and 4).

Try each pair in the expression to see if the INNER and OUTER products add to b, or 31.

(7x 1 1)(x 1 12) 5 7x2 1 x 1 84x 5 7x2 1 85x 1 12 (NO)

(7x 1 2)(x 1 6) 5 7x2 1 2x 1 42x 5 7x2 1 44x 1 12 (NO)

(7x 1 3)(x 1 4) 5 7x2 1 3x 1 28x 5 7x2 1 31x 1 12 (YES)

Th e factored form of 7x2 1 31x 1 12 is (7x 1 3)(x 1 4).

Exercises


1. 3x2 1 14x 1 8 2. 5y2 1 43y 1 24 3. 2z2 1 19z 1 42

4. 11a2 1 39a 1 18 5. 13b2 1 58b 1 24 6. 23c2 1 56c 1 20

7. 7d2 1 d 2 8 8. 3e2 1 20e 2 32 9. 19f 21 10f 2 9

10. 5s2 2 18s 1 16 11. 17t2 2 12t 2 5 12. 29u2 1 48u 2 20

8-6 ReteachingFactoring ax2 1 bx 1 c

(3x 1 2)(x 1 4)

(11a 1 6)(a 1 3)

(7d 1 8)(d 2 1)

(5s 2 8)(s 2 2)

(5y 1 3)(y 1 8)

(13b 1 6)(b 1 4)

(3e 2 4)(e 1 8)

(17t 1 5)(t 2 1)

(2z 1 7)(z 1 6)

(23c 1 10)(c 1 2)

(19f 2 9)(f 1 1)

(29u 2 10)(u 1 2)


60

Name Class Date

If you are given the area and one side of a rectangle, you can fi nd the second side by factoring the trinomial. One binomial is the width and the other binomial is the length.

Problem

Th e area of a rectangular swimming pool is 6x2 1 11x 1 3. Th e width of the pool is 2x 1 3. What is the length of the pool?

You are given the area and length of the pool. Set up an equation with what you are given and solve or length.

6x2 1 11x 1 3 5 (2x 1 3)(uuu ) Area = length × width.

6x2 1 11x 1 3 5 (2x 1 3)(3x uu ) 6x2 5 (2x)(3x), so the fi rst term of the second

binomial is 3x.

6x2 1 11x 1 3 5 (2x 1 3)(3x 1 u ) The trinomial has two plus signs, so the sign for

the second binomial must also be plus.

6x2 1 11x 1 3 5 (2x 1 3)(3x 1 1) The value of c is 3. Since 3 5 3 3 1, the second

term must be 1.

Multiply to check your answer. Use FOIL.

(2x 1 3)(3x 1 1) 5 6x2 1 2x 1 9x 1 3 5 6x2 1 11x 1 3 3

Th e length of the swimming pool is 3x 1 1.

Exercises

13. Th e area of a rectangular cookie sheet is 8x2 1 26x 1 15. Th e width of the cookie sheet is 2x 1 5. What is the length of the cookie sheet?

14. Th e area of a rectangular lobby fl oor in the new offi ce building is

15x2 1 47x 1 28. Th e length of one side of the lobby is 5x 1 4. What is the width?

15. Th e area of a rectangular school banner is 12x2 1 13x 2 90. Th e width of the banner is 3x 1 10. What is the length of the banner?

16. Th e distance a train has traveled is 6x2 2 23x 1 20. Th e train’s average speed is 3x 2 4. How long has the train been traveling?


Factoring ax2 1 bx 1 c

4x 1 3

3x 1 7

4x 2 9

2x 2 5

Name Class Date


69

8-7 ReteachingFactoring Special Cases

Th e area of a square is given by A 5 s2, where s is a side length. When

the side length is a binomial, the area can be written as a perfect-square

trinomial. If you are given the area of such a square, you can use

factoring to write an expression for a side length.

Problem

A mosaic is made of small square tiles called tesserae. Suppose the area of one

tessera is 9x2 1 12x 1 4. What is the length of one side of a tessera?

Because the tile is a square, you know the side lengths must be equal. Th erefore,

the binomial factors of the trinomial must be equal.

9x2 1 12x 1 4 5 ( u u u )2 This is a perfect square trinomial and can be factored as the square of a binomial.

9x2 5 (3x)2 9x2 and 4 are perfect squares. Write them as squares.

4 5 22

2(3x)(2) 5 12x Check that 12x is twice the product of the fi rst and last terms. It is, so you are sure that you have a perfect-square trinomial.

9x2 1 12x 1 4 5 (3x 1 2)2 Rewrite the equation as the square of a binomial.

Multiply to check your answer.

(3x 1 2)(3x 1 2) 5 9x2 1 6x 1 6x 1 4 5 9x2 1 12x 1 43

Th e length of one side of the square is 3x 1 2.

ExercisesFactor each expression to fi nd the side length.

1. Th e area of a square oil painting is 4x2 1 28x 1 49. What is the length of one

side of the painting?

2. You are installing linoleum squares in your kitchen. Th e area of each linoleum

square is 16x2 2 24x 1 9. What is the length of one side of a linoleum square?

3. You are building a table with a circular top. Th e area of the tabletop is

(25x2 2 40x 1 16)π. What is the radius of the tabletop?

4. A fabric designer is making a checked pattern. Each square in the pattern has

an area of x2 2 16x 1 64. What is the length of one side of a check?

s

s A 5 s2

2x 1 7

4x 2 3

5x 2 4

x 2 8

Name Class Date


70


Factoring Special Cases

Some binomials are a diff erence of two squares. To factor these expressions, write

the factors so the x-terms cancel and you are left with two perfect squares.

Problem

What is the factored form of 4x2 2 9?

4x2 2 9 5 ( u 1 u )( u 2 u ) Both 4x2 and 9 are perfect squares. You know the signs of the factors will be opposite, so the x-terms will cancel out.

"4x2 5 2x Find the square root of each term.

!9 5 3

(2x 1 3)(2x 2 3) Write each term as a binomial with opposite signs, so the x-terms will cancel out.


(2x 1 3)(2x 2 3) 5 4x2 1 6x 2 6x 2 9

5 4x2 2 93

Th e factored form of 4x2 2 9 is (2x 1 3)(2x 2 3).

ExercisesFactor each expression.

5. 9x2 2 4 6. 25x2 2 49 7. 144x2 2 1

8. 64x2 2 25 9. 49x2 2 16 10. 36x2 2 49

11. 81x2 2 16 12. 16x2 2 121 13. 25x2 2 144

14. 16x2 2 9 15. x2 2 81 16. 4x2 2 49

(3x 1 2)(3x 2 2)

(8x 1 5)(8x 2 5)

(9x 1 4)(9x 2 4)

(4x 1 3)(4x 2 3)

(5x 1 7)(5x 2 7)

(7x 1 4)(7x 2 4)

(4x 1 11)(4x 2 11)

(x 1 9)(x 2 9)

(12x 1 1)(12x 2 1)

(6x 1 7)(6x 2 7)

(5x 1 12)(5x 2 12)

(2x 1 7)(2x 2 7)


79

Name Class Date

8-8 ReteachingFactoring by Grouping

You can factor some higher-degree polynomials by grouping terms and factoring out the GCF to fi nd the common binomial factor. Make sure to factor out a common GCF from all terms fi rst before grouping.

Problem

What is the factored form of 2b4 2 8b3 1 10b2 2 40b?

2b4 2 8b3 1 10b2 2 40b 5 2b(b3 2 4b2 1 5b 2 20) 2b is the GCF of all four terms. Factor out 2b from each term.

5 2bfb2(b 2 4) 1 5(b 2 4)g Group terms into pairs and look for the GCF of each pair. b2 is the GCF of the fi rst pair, and 5 is the GCF of the second pair.

5 2b(b2 1 5)(b 2 4) b 2 4 is the common binomial factor. Use the Distributive Property to rewrite the expression.


2b(b2 1 5)(b 2 4) 5 2b(b3 1 5b 2 4b2 2 20) Multiply b2 1 5 and b 2 4.

5 2b4 1 10b2 2 8b3 2 40b Multiply by 2b.

5 2b4 2 8b3 1 10b2 2 40b 3 Reorder the terms by degree.

Th e factored form of 2b4 2 8b3 1 10b2 2 40b is 2b(b2 1 5)(b 2 4).

Exercises

Factor completely. Show your steps.

1. 4x4 1 8x3 1 12x2 1 24x 2. 24y4 1 6y3 1 36y2 1 9y

3. 72z4 1 48z3 1 126z2 1 84z 4. 2e4 2 8e3 1 18e2 2 72e

5. 12f 32 36f 2 1 60f 2 180 6. 16g4 2 56g3 1 64g2 2 224g

7. 56m3 2 28m2 2 42m 1 21 8. 40n4 2 60n3 2 50n2 1 75n

9. 60x3 2 90x2 2 30x 1 45 10. 12p5 1 8p4 1 18p3 1 12p2

11. 6r3 1 9r2 2 60r 12. 20s6 2 50s5 2 30s4

4x(x2 1 3)(x 1 2)

6z(4z2 1 7)(3z 1 2)

12(f 21 5)(f 2 3)

7(4m2 2 3)(2m 2 1)

15(2x2 2 1)(2x 2 3)

3r(2r 2 5)(r 1 4)

3y(2y2 1 3)(4y 1 1)

2e(e2 1 9)(e 2 4)

8g(g2 1 4)(2g 2 7)

5n(4n2 2 5)(2n 2 3)

2p2(2p2 1 3)(3p 1 2)

10s4(2s 1 1)(s 2 3)


80

Name Class Date


Factoring by Grouping

Polynomials can be used to express the volume of a rectangular prism. Th ey can sometimes be factored into 3 expressions to represent possible dimensions of the prism. Th e three factors are the length, width, and height.

Problem

Th e plastic storage container to the right has a volume of 12x3 1 8x2 2 15x . What linear expressions could represent possible dimensions of the storage container?

12x3 1 8x2 2 15x 5 x(12x2 1 8x 2 15) Factor out x, the GCF for all three terms.

5 x(12x2 1 18x 2 10x 2 15) ac is –180 and b is 8. Break 8x into two terms that have a sum of 8x and a product of 2180x2.

5 xf6x(2x 1 3) 2 5(2x 1 3)g Group the terms into pairs and factor out the GCF from each pair. The GCF of the fi rst pair is 6x. The GCF of the second pair is 25.

5 x(6x 2 5)(2x 1 3) 2x 1 3 is the common binomial term. Use the Distributive Property to reorganize the factors.


x(6x 2 5)(2x 1 3) 5 x(12x2 1 18x 2 10x 2 15) Multiply 6x 2 5 and 2x 1 3.

5 x(12x2 1 8x 2 15) Combine like terms.

5 12x3 1 8x2 2 15x 3 Multiply by x.

Possible dimensions of the storage container are x, 6x 2 5, and 2x 1 3.

Exercises

Find linear expressions for the possible dimensions of each rectangular prism.

13. 14.

15. 16.

V = 12x3 + 8x2 -15x

V = 60x3 - 68x2 -16x V = 12x3 - 15x2 -18x

V = 10x3 + 65x2 +105xV = 12x3 + 34x2 +14x

2x, 3x 1 7, 2x 1 1

4x, 5x 1 1, 3x 2 4

5x, 2x 1 7, x 1 3

3x, 4x 1 3, x 2 2

Name Class Date


9

A U-shaped graph such as the one at the right is called a parabola.

• A parabola can open upward or downward.

• A parabola that opens upward has a minimum or lowest point.

• A parabola that opens downward has a maximum or highest point.

• Th e vertex of a parabola is its minimum or maximum point.

All parabolas have a line or axis of symmetry.

Problem

What is the vertex of the graph below? Is it a minimum or maximum?

Th e graph opens downward, so you are looking for the highest point. Th e vertex is

(23, 2) and it is a maximum.

Exercises

Identify the vertex of each graph. Tell whether it is a minimum or a maximum.

1. 2. 3.

9-1 ReteachingQuadratic Graphs and Their Properties

x

y

2

4

6

8

2 2O

x

y

2

2

24O

x

y

2

4

2O 4

xy

2O

4

2

4

6

x

y O246

2

4

6

(3, 1); minimum (1, 26); minimum (23, 21); maximum

Name Class Date


10

Any function in the form y 5 ax2 1 bx 1 c where a 2 0 is called a quadratic

function. Th e graph of a quadratic function is a parabola.

Problem

What is the graph of y 5 12x

2 2 4?

Th is is a quadratic function where a 5 12, b 5 0 and c 5 24. Th e graph will be a

parabola. Use a table to fi nd some points on the graph. Th en use what you know

about parabolas to complete the graph.

Exercises

Graph each function.

4. y 5 2x2 1 5 5. y 5 x2 2 4 6. y 5 2x2 2 1


Quadratic Graphs and Their Properties

4

2

0

2

4

x (x, y)

( 4, 4)

( 2, 2)

(0, 4)

(2, 2)

(4, 4)

x2 4y 12

( 4)2 4 4y 12

( 2)2 4 2y 12

(0)2 4 4y 12

(2)2 4 2y 12

(4)2 4 4y 12

x

y

2 4

2

4

4

2

x

y

2 44 2

4

2

4

2

O x

y

2 44 2

4

2

4

2

O

x

y

2 44 2

2

2

4

6

8

10

O

Name Class Date


19

Recall that the general equation for a quadratic function is y 5 ax2 1 bx 1 c.

Using this general equation, the equation for the axis of symmetry is x 5 2b2a .

Since the vertex lies on the axis of symmetry, the x-coordinate of the vertex is 2b2a .

Problem

What are the equation of the axis of symmetry and the coordinates of the vertex of the graph of y 5 3x2 1 6x 2 4?

x 5 2b2a Equation for axis of symmetry

x 5 262(3)

a 5 3 and b 5 6

x 5 21 Simplify.

Now, fi nd the value of y when x 5 21.

y 5 3x2 1 6x 2 4

y 5 3(21)2 1 6(21) 2 4

y 5 27

Th e equation of the axis of symmetry is x 5 21 and the coordinates of the vertex of the graph are (21, 27).

Exercises

Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of each function.

1. y 5 x2 1 8x 2. y 5 2x2 1 12x 1 10 3. y 5 2x2 1 4x 2 8

4. y 5 2x2 2 4x 2 5 5. y 5 23x2 1 18x 2 25 6. y 5 22x2 1 2x 2 6

7. f (x) 5 6x2 2 7 8. f (x) 5 25x2 2 10x 1 1 9. f (x) 5 4x2 2 16x 2 2

9-2 ReteachingQuadratic Functions

(24, 216); x 5 24 (23, 28); x 5 23 (2, 24); x = 2

(1, 27); x 5 1 (3, 2); x 5 3 Q12, 2112 R ; x 5

12

(0, 27), x 5 0 (21, 6); x 5 21 (2, 218); x 5 2

Name Class Date


20

You can use the axis of symmetry and the vertex to help graph a quadratic equation. Use the equation x 5 2

b2a to fi nd the equation of the axis of symmetry.

Because the vertex lies on the axis of symmetry, this value is also the x-coordinate of the vertex.

Problem

What is the graph of y 5 2x2 2 4x 1 1?

1. Find the equation of the axis of symmetry.

x 5 2b2a

x 52(24)

2(2) a 5 2 and b 5 24

x 5 1 Simplify.

2. Find the vertex.

y 5 2x2 2 4x 1 1

y 5 2(12) 2 4(1) 1 1 x 5 1

y 5 21 Simplify.

Th e vertex is (1, 21)

3. Graph the axis of symmetry x 5 1 and the vertex (1, 21).

4. Find a couple points on the graph.

For x 5 0, y 5 2(02) 2 4(0) 1 1 or 1.

Plot (0, 1).

For x 5 21, y 5 2(21)2 2 4(21) 1 1 or 7.

Plot (21, 7) .

5. Use the axis of symmetry to complete the graph.

Exercises

Graph each function. Label the axis of symmetry and the vertex.

10. y 5 x2 2 3 11. y 5 2x2 2 4x 1 1 12. y 5 2x2 1 8x 1 6


Quadratic Functions

x

y

6

2

2

2O(1, 1)

(0, 1)

( 1, 7)

x 1

x

y

2 424

4

2

4

2

O

(0, 3)

x = 0 x

y

2 424

4

2

4

2

O

( 2, 5)

x = 2

x

y

2 424

4

2

4

2

O

( 2, 2)x = 2

Name Class Date


29

An equation in the form ax2 1 bx 1 c 5 0 where a 2 0 is called a

quadratic equation. Its related quadratic function is y 5 ax2 1 bx 1 c. If you

graph the related quadratic function, the solutions of the quadratic equation are

x-values where the graph crosses the x-axis.

A linear equation can have only one solution. However, a quadratic equation can

have 2, 1, or 0 real-number solutions.

Th e related function of

2x2 1 4 5 0 is

y 5 2x2 1 4. Th e graph of

y 5 2x2 1 4 is shown

below.


x2 2 2x 1 1 5 0 is

y 5 x2 2 2x 1 1.

Th e graph of

y 5 x2 2 2x 1 1 is

shown below.


x2 2 x 1 2 5 0 is

y 5 x2 2 x 1 2. Th e

graph of y 5 x2 2 x 1 2

is shown below.

Th e graph crosses the x-axis

where x 5 22 and x 5 2.

Th e equation 2x2 1 4 5 0

has two solutions, 22 and 2.

Th e graph touches the

x-axis where x 5 1.

Th e equation

x2 2 2x 1 1 5 0 has

one solution, 1.

Th e graph does not

touch the x-axis.

Th e equation

x2 2 x 1 2 5 0 has no

real-number solutions.

Exercises

Solve each equation by graphing the related function. If the equation has no real-number solution, write no solution.

1. x2 1 3 5 0 2. x2 1 4x 1 4 5 0 3. x2 1 x 2 2 5 0

4. How many times does the graph of y 5 x2 2 4 cross the x-axis? Explain.

9-3 ReteachingSolving Quadratic Equations

x

y

2

3

2

3Ox

y4

2 2Ox

y4

2

2 2O

no solution

twice; at x 5 2 and x 5 22

22 22; 1

Name Class Date


30


Solving Quadratic Equations

You can solve a quadratic equation by taking the square root of each side of the equation.

Problem

What are the solutions of 81x2 5 49?

81x2 5 49

81x2

81 54981 Divide each side by 81.

x2 54981 Simplify.

"x2 5 4Å4981 Take the square root of each side.

x 5 479 Simplify.

Problem

What are the solutions of x2 1 9 5 0?

x2 1 9 5 0

x2 1 9 2 9 5 0 2 9 Subtract 9 from each side.

x2 5 29 Simplify.

Since x2 cannot equal 29 in the real numbers, x2 1 9 5 0 has no real-number

solutions.

Exercises

Solve each equation by fi nding square roots. If the equation has no real-number solution, write no solution. If a solution is irrational, round to the nearest tenth.

5. x2 5 100 6. x2 2 144 5 0 7. 5x2 2 125 5 0

8. 9x2 2 16 5 0 9. 3x2 1 27 5 0 10. 7x2 2 49 5 0

11. 64x2 2 25 5 0 12. 3x2 2 30 5 0 13. x2 1 7 5 0

210; 10 12; 212 5; 25

43; 2

43

no solution 2.6; 22.6

58; 2

58

3.2; 23.2 no solution

Name Class Date


39

9-4 ReteachingFactoring to Solve Quadratic Equations

If the product of two or more numbers is 0, then one of the factors must be 0. You

can use this fact to solve quadratic equations.

Problem

What are the solutions of the equation (4a 1 12)(5a 2 20) 5 0?

Since the product is 0, either (4a 1 12) or (5a 2 20) must equal 0.

4a 1 12 5 0 or 5a 2 20 5 0

4a 1 12 2 12 5 0 2 12 or 5a 2 20 1 20 5 0 1 20

4a 5 212 or 5a 5 20

4a4 5

2124 or

5a5 5

205

a 5 23 or a 5 4

Th e solutions are 23 and 4.

Exercises

Solve each equation.

1. b(b 1 7) 5 0 2. 8y(2y 2 12) 5 0 3. (d 2 8)(d 2 2) 5 0

4. (m 1 1)(m 2 4) 5 0 5. (2a 1 14)(3a 1 12) 5 0 6. (5p 2 10)(2p 1 20) 5 0

7. (8t 1 4)(3t 1 6) 5 0 8. (4h 2 1)(2h 1 1) 5 0 9. (8n 2 16)(5n 2 12) 5 0

10. (s 1 6)(4s 2 6) 5 0 11. (5w 2 30)(2w 2 1) 5 0 12. (3g 1 1)(2g 2 5) 5 0

0; 27 0; 6 8; 2

21; 4 27; 24 2; 210

2

12; 22 21

2; 14 2; 125

26; 32 6; 12 213; 52

Name Class Date


40


Factoring to Solve Quadratic Equations

If you can rewrite a quadratic equation as a product of factors that equals zero, you can solve the equation. To solve equations in this manner, you must use all your factoring skills.

Problem

What are the solutions of the equation x2 2 x 5 20?

First rewrite the equation so that one side equals zero.

x2 2 x 5 20

x2 2 x 2 20 5 20 2 20 Subtract 20 from each side.

x2 2 x 2 20 5 0 Simplify.

Now, factor to rewrite the equation as a product of factors equal to zero. Find two integers whose product is 220 and whose sum is 21. Th e product of 4 and 25 is 220, and the sum of 4 and 25 is 21.

x2 2 x 2 20 5 0

(x 1 4)(x 2 5) 5 0

x 1 4 5 0 or x 2 5 5 0

x 1 4 2 4 5 0 2 4 or x 2 5 1 5 5 0 1 5

x 5 24 or x 5 5


Exercises

Solve each equation by factoring.

13. y2 1 3y 1 2 5 0 14. a2 2 a 2 20 5 0 15. m2 2 7m 1 6 5 0

16. 2d2 1 7d 2 4 5 0 17. 6t2 1 13t 1 6 5 0 18. 5p2 1 29p 2 6 5 0

19. s2 1 9s 5 220 20. x2 2 5x 5 14 21. b2 1 7b 5 8

22. 2h2 2 9h 5 5 23. 3s2 2 13s 5 212 24. 6v2 1 13v 5 5

21; 22 24; 5 1; 6

12; 24 22

3; 232

15; 26

24; 25 22; 7 1; 28

212; 5 4

3; 3 13; 2

52

Name Class Date


49

You have learned to square binomials. Notice how the coeffi cient of the a term is related to the constant value in every perfect-square trinomial.

(a 1 1)2 5 (a 1 1)(a 1 1) 5 a2 1 2a 1 1 S Q22R25 1

(a 2 1)2 5 (a 2 1)(a 2 1) 5 a2 2 2a 1 1 S Q222 R

25 1

(a 2 2)2 5 (a 2 2)(a 2 2) 5 a2 2 4a 1 4 S Q242 R

25 4

(a 1 3)2 5 (a 1 3)(a 1 3) 5 a2 1 6a 1 9 S Q62R25 9

In each case, half the coeffi cient of the a term squared equals the constant term. You can use this pattern to fi nd the value that makes a trinomial a perfect square.

Problem

What is the value of c such that x2 2 14x 1 c is a perfect-square trinomial?

Th e coeffi cient of the x term is 214. Using the pattern, c 5 Q2142 R

2 or 49.

So, x2 2 14x 1 49 is a perfect-square trinomial.

Exercises

Find the value of c such that each expression is a perfect-square trinomial.

1. a2 1 8a 1 c 2. x2 2 16x 1 c 3. m2 1 20m 1 c

4. p2 2 14p 1 c 5. y2 2 10y 1 c 6. b2 1 18b 1 c

7. d2 1 12d 1 c 8. n2 2 n 1 c 9. w2 1 3w 1 c

9-5 ReteachingCompleting the Square

16 64 100

49 25 81

36 14

94

Name Class Date


50


Completing the Square

You can use completing the square to solve quadratic equations.

Problem

What are the solutions of the equation x2 1 2x 2 48 5 0?

First rewrite the equation so that the constant is on one side of the equation and the other terms are on the other side.

x2 1 2x 2 48 5 0

x2 1 2x 2 48 1 48 5 0 1 48 Add 48 to each side.

x2 1 2x 5 48 Simplify.

Since Q22R25 1, add 1 to each side.

x2 1 2x 1 1 5 48 1 1 Add 1 to each side.

(x 1 1)2 5 49 Simplify.

x 1 1 5 4!49 Take the square root of each side.

x 1 1 5 47 Simplify.

x 1 1 5 27 or x 1 1 5 7

x 1 1 2 1 5 27 2 1 or x 1 1 2 1 5 7 2 1

x 5 28 or x 5 6


Exercises

Solve each equation by completing the square. If necessary, round to the nearest hundredth.

10. b2 1 10b 5 75 11. y2 2 18y 5 63 12. n2 2 20n 5 275

13. a2 1 16a 5 215 14. t2 1 8t 2 9 5 0 15. h2 2 12h 2 9 5 0

16. m2 2 2m 2 8 5 0 17. s2 1 6s 1 1 5 0 18. v2 1 4v 2 2 5 0

5; 215 21; 23 15; 5

215; 21 29; 1 12.71; 20.71

4; 22 25.83; 20.17 24.45; 0.45


59

Name Class Date

9-6 ReteachingThe Quadratic Formula and the Discriminant

If a quadratic equation is written in the form ax2 1 bx 1 c 5 0, the solutions can be found using the following formula.

x 5 2b 4 "b2 2 4ac2a

Th is formula is called the quadratic formula.

Problem

What are the solutions of x2 1 7x 5 60? Use the quadratic formula.

First rewrite the equation in the form ax2 1 bx 1 c 5 0.

x2 1 7x 5 60

x2 1 7x 2 60 5 60 2 60 Subtract 60 from each side.

x2 1 7x 2 60 5 0 Simplify.

Th erefore, a 5 1, b 5 7, and c 5 260.

x 5 2b 4 "b2 2 4ac2a

x 527 4 "72 2 4(1)(260)

2(1)

x 5 27 4 "2892

x 5 27 4 172

Th e two solutions are 27 2 172 or 212 and 27 1 17

2 or 5.

Exercises

Use the quadratic formula to solve each equation.

1. x2 2 19x 1 70 5 0 2. x2 1 32x 1 175 5 0 3. 2x2 1 37x 2 19 5 0

4. x2 2 10x 5 75 5. x2 1 x 5 132 6. 6x2 1 13x 5 28

7. 20x2 1 11x 5 3 8. 4x2 1 24x 5 235 9. 15x2 1 20 5 40x

14; 5 225; 27 219; 0.5

15; 25 212; 11 23.5; 1.3

234; 15

23.5; 22.5 2; 23


60

Name Class Date


The Quadratic Formula and the Discriminant

In the quadratic equation, the expression under the radical sign, b2 2 4ac, is

called the discriminant. Consider the quadratic formula.

x 5 2b 4 "b2 2 4ac2a

• If b2 2 4ac is a negative number, the square root cannot be found in the real

numbers. Th ere are no real-number solutions of the equation. Th e graph of

the quadratic does not cross the x-axis.

• If b2 2 4ac equals 0, x 5 2b 4 !02a or

2b2a . Th ere is only one solution of the

equation. Th e vertex of the quadratic is on the x-axis.

• If b2 2 4ac is a positive number, there are two solutions of the equation,

2b 2 "b2 2 4ac2a and

2b 1 "b2 2 4ac2a . Th e graph of the quadratic intersects

the x-axis twice.

Problem

What is the number of solutions of x2 1 13 5 25x?

First rewrite the equation in the form ax2 1 bx 1 c 5 0.

x2 1 13 5 25x

x2 1 5x 1 13 5 0 Add 5x to each side.

Th erefore, a 5 1, b 5 5, and c 5 13.

b2 2 4ac 5 52 2 4(1)(13)

5 227

Since b2 2 4ac is a negative number, there are no real-number solutions of the

equation.

Exercises

Find the number of solutions of each equation.

10. 4x2 1 12x 1 9 5 0 11. x2 2 12x 1 32 5 0 12. x2 2 10x 1 1 5 0

13. 3x2 1 6x 1 8 5 0 14. 3x2 2 5x 5 26 15. x2 1 100 5 20x

16. 5x2 2 7x 5 2 17. 9x2 1 4 5 12x 18. 3x2 1 5x 5 2

one

one

one

two two

twotwo

no real solutions no real solutions


69

Name Class Date

9-7 ReteachingLinear, Quadratic, and Exponential Models

Data can resemble a linear function, a quadratic function, or an exponential

function. Recall the general shapes of these functions.

Linear Function Quadratic Function Exponential Function

Problem

Which model is most appropriate for the data points(0.5, 1.75), (1, 1), (1.5, 1.75), (2, 4) and (2.5, 7.75)?

Graph the data points.

Notice that the points are not in a straight line. Th e points do

not have an exponential shape. A quadratic function would best

represent the data because the graph appears to be U-shaped.

Exercises

Graph each set of points. Which model is most appropriate for each set?

1. (0, 0.25), (1, 0.75), 2. (0.5, 0.5), (1, 3), 3. (1, 1.5), (1.5, 1.75),

(1.5, 1.3), (2, 2.25), (1.5, 4.5), (2, 5), (2, 2), (2.5, 2.25),

(2.5, 3.9), (3, 6.75) (2.5, 4.5), (3.5, 0.5) (3, 2.5), (4, 3)

xO

y

2

2

2

2x

O

y

2

2

2

4

xO

y

2

2

2

4

xO

y

2

2

6

2

4

exponential quadratic linear

x

y

2

4

6

8

O 2 4 6 8x

y

2

4

6

8

O 2 4 6 8x

y

2

4

6

8

O 2 4 6 8


70

Name Class Date


Linear, Quadratic, and Exponential Models

You can fi nd the best model for a function using a table of values.

Linear Function

y 5 22x 1 5

Quadratic Function

y 5 x2 1 7

Exponential Function

y 5 3x

For each increase of 1 for

the x values, the y values

have a common diff erence.



change at diff erent rates.

But, the diff erences have

a common diff erence.



have a common ratio.

Exercises

Which kind of function best models the data in each table? Use diff erences or ratios.

4. 5. 6.

7. 8. 9.

22

2

2

2

1

1

1

1

9

1 7

0 5

1 3

2 1

x y

2

0 1

1 5

2 25

x y

125

1 15

2 11

1 1

0 5

1 7

2 5

x y

2 13

1 6

0 1

1 2

2 3

x y

2 9

1 7

0 5

1 3

2 1

x y

2 5

1 2

0 1

1 4

2 7

x y

2

x y

1501101

120

521

2522

23

1

1

3

2

2

2

1

1

1

1

11

1 8

0 7

1 8

2 11

x y

21

1

1

1

3

3

3

3

0 1

1 3

2 9

x y19

1 13

exponential

quadratic

linear

quadratic

exponential

linear


19

Name Class Date

You can remove perfect-square factors from a radicand.

Problem

What is the simplifi ed form of "80n5?

In the radicand, factor the coeffi cient and the variable separately into perfect square factors, and then simplify. Factor 80 and n5 completely and then fi nd paired factors.

Solve 80 5 8 ? 10 5 2 ? 2 ? 2 ? 2 ? 5 Factor 80 completely.

5 (2 ? 2)(2 ? 2) ? 5 5 (2 ? 2)2 ? 5 Find pairs of factors.

!80 5 "42 ? 5 5 "42 ? !5 Use the rule !ab 5 !a ? !b.

5 4 ? !5 5 4!5 The square root of a number squared

is the number: "a2 5 a.

n5 5 n ? n ? n ? n ? n Factor n5 completely.

5 (n ? n) ? (n ? n) ? n 5 (n ? n)2 ? n Find pairs of factors.

"n5 5 "(n ? n)2 ? !n Separate the factors.

5 n2 ? !n 5 n2!n Remove the perfect square.

"80n5 5 4 ? n2!5 ? n 5 4n2!5n Combine your answers.

Check "80n5 0 4n2!5n Check your solution.

"80n5

!5n0

4n2!5n

!5n Divide both sides by !5n .

"16n4 0 4n2 Simplify.

4n2 5 4n2 3

Solution: Th e simplifi ed form of "80n5 is 4n2!5n .

Exercises

Simplify each radical expression.

1. "100n3 2. "120b4 3. "66t5

4. !32x 5. "525c7 6. "86t2

7. "50g3 8. "54h6 9. !35y

10-2 ReteachingSimplifying Radicals

10n!n 2b2!30 t2!66t

4!2x 5c3!21c t!86

5g!2g 3h3!6 !35y


20

Name Class Date


Simplifying Radicals

Problem

What is the simplifi ed form of Å27t4

48t2 ?

Begin by cancelling the common factors in the numerator and denominator. Simplify the numerator and denominator separately when the denominator is a perfect square. Remember that the radical is not simplifi ed if there is a radical in the denominator. Multiply to remove the radical from the denominator.

Solve Å27t3

48t4 5% 3 ? 3 ? 3 ? t ? t ? t3 ? 4 ? 4 ? t ? t ? t ? t Factor the numerator and denominator

completely.

5%3 ? 3 ? 3 ? t ? t ? t

3 ? 4 ? 4 ? t ? t ? t ? t Cancel the common factors.

5"(3 ? 3)

"(4 ? 4)t5"32

"42t Find pairs of factors. These are the perfect-

square factors.

53

4"t Simplify the numerator and denominator

separately to remove the perfect-square

factors. "32 5 3 and "42t 5 4!t

53

4!t (!t)

(!t) Multiply the numerator and denominator by

!t to remove !t from the denominator.

53!t

4!t ? t5

3!t

4"t25

3!t

4t Remove the perfect-square factor from the

denominator.

Solution: Th e simplifi ed form of Å27t3

48t4 is 3!t

4t.

Exercises


10. Å4981

11. Å18x4

200 12. Å

28s

s3

13. Å25a5

9a7 14. Å40b4

12b3 15. Å48

6t6

16. Å50z3

4x2 17. Åt5

64 18. Å

32tt

79

3x2

102!7

s

53a

!30b3

2!2t3

5z!2z2x

t2!t8 4!2


29

Name Class Date

10-3 ReteachingOperations with Radical Expressions

You can use the Distributive Property with radical expressions.

Problem

What is the simplifi ed form of 4!2 2 !18?

You need to simplify the radical expressions before you know if there are any like radicals that can be subtracted.

Solve 4!2 2 !18 Look for a common radical in 4!2 and !18. 4!2 is factored completely, but !18 can be factored further.

!18 5 !3 ? 3 ? 2 Factor !18 completely.

5 !(3 ? 3) ? 2 5 "32 ? 2 Find pairs of factors that you can factor out. These are perfect-square factors.

5 3!2 Remove the perfect-square factor.

4!2 2 !18 5 4!2 2 3!2 Now you can see that each term in the expression shares the common radical !2.

5 (4 2 3)!2 Use the Distributive Property to combine like radicals. a!b 2 c!b 5 (a 2 c)!b

5 1 ? !2 Subtract.

5 !2 Simplify.

Check 4!2 2 !18 5 !2 Check your solution.

3!2 2 !18 5 0 Subtract !2 from both sides.

3!2 5 !18 Add !18 to both sides.

3!2 5 3!2 3 Simplify !18.

Solution: Th e simplifi ed form of 4!2 2 !18 is !2.

Exercises

Simplify each sum or diff erence.

1. 2!5 2 4!5 2. !7 1 !7 3. 3!6 1 2!6

4. 5!2 1 !32 5. 3!3 2 !75 6. 4!54 1 2!24

7. 10!5 2 5!20 8. 2!8 1 !200 9. 3!12 1 !108

22!5 2!7 5!6

9!2 22!3 16!6

0 14!2 12!3


30

Name Class Date


Operations with Radical Expressions

When you have two binomial factors that include radical expressions, treat them like any other binomials and multiply using FOIL (First, Outer, Inner, Last).

Problem

What is the simplifi ed form of (2!3 1 5!7)(3!7 2 !3)?

Solve Use FOIL to fi nd the product of each pair of terms. Multiply the coeffi cients and then multiply the radicals. Remove all perfect-square factors.

First: 2!3 ? 3!7 5 6!21

Outer: 2!3 ? (2!3) 5 22!9 5 22 ? 3 5 26

Inner: 5!7 ? 3!7 5 15!49 5 15 ? 7 5 105

Last: 5!7 ? (2!3) 5 25!21

5 6!21 2 6 1 105 2 5!21

5 (6!21 2 5!21) 1 (26 1 105) Group like terms.

5 (6 2 5)!21 1 99 Distributive Property

5 1!21 1 99 5 !21 1 99 Simplify.

Solution: Th e simplifi ed form of (2!3 1 5!7)(3!7 2 !3) is !21 1 99.

Exercises


10. !4(!3 1 !5) 11. !10(!8 2 9) 12. 2!3(2 2 !3)

13. 2!8(5 2 3!5) 14. 4!6(!2 1 4!3) 15. 2!6(!11 1 7)

16. (3!7 1 !3)2 17. (1 1 !3)(1 2 !3) 18. (3!6 1 2!2)(!2 2 4!6)

FIRST LAST

OUTER

INNER

( 2 3 5 7 ) ( 3 7 3 )

2!3 1 2!5 4!5 2 9!10 4!3 2 6

210!2 1 6!10 8!3 1 48!2 2!66 1 14!6

66 1 6!21 22 210!3 2 68

Name Class Date


9

11-1 ReteachingSimplifying Rational Expressions

Rational expressions may be in the form of monomials or polynomials.

Simplifying rational expressions is similar to simplifying numerical fractions where common factors are taken out.

Example: x 2 23x 2 6 5

x 2 23(x 2 2)

513

Excluded values are those that make the denominator 0. A denominator can not equal 0, so these values are not part of the solution. Consider not only the solution, but also the original expression to fi gure out the excluded values.

Problem

What is the simplifi ed form of 2a3

4a2? State any excluded values.

Solve Monomials: reduce numbers; cancel out like variables

2a3

4a2 52 ? a ? a ? a2 ? 2 ? a ? a 5

a2

Th e simplifi ed form is a2 when a 2 0.

What is the simplifi ed form of x2 1 4x 1 4

x 1 2 ? State any excluded values.

Solve Polynomials: cancel out factors or groups of factors

x2 1 4x 1 4

x 1 2 5(x 1 2)(x 1 2)

x 1 2 5 x 1 2

Th e simplifi ed form is x 1 2 when x 2 22.

Recognizing Opposite Factors

You can fi nd the opposite of a number by multiplying by 21. For example, the opposite of 3 is (21)(3) 5 23.

Similarly, multiplying a polynomial by 21 results in its opposite. For example, the opposite of x 2 2 is (21)(x 2 2) 5 2x 1 2. It can also be written as 2 2 x .

Problem

Write the opposite of (20 2 x) two ways.

Solve Multiply by (−1) to fi nd the opposite.

(21)(20 2 x) 5 220 1 x or x 2 20

Name Class Date


10

Exercises

Simplify each expression. State any excluded values.

1. a4

a 2. 4x3

16x2 3. cd2

3c2d

4. lml2m2n

5. 64y

16y2x 6. 2x2 2 4x

x

7. 5x3 2 15x2

x 2 3 8. x2 1 5x 1 6x 1 3

9. 2b 1 44 10. 3a 1 15

15

11. 3p 2 21

18 12. 44y 2 8

13. 7z 2 2814z 14. 9

18 2 81a

15. 535 2 5c 16.

2q 1 2

q2 1 4q 1 3

17. a 1 2a2 1 4a 1 4

18. 2x 2 22 2 2x

19. 9 2 x2

x 2 3 20. 2a 1 42

Write the opposite expression and simplify the opposite expression.

21. 10b5

40b4 22. 36 2 z2

4z 2 24

23. x2 2 16x 2 4 24. 30 1 2z

14 1 4z


Simplifying Rational Expressions

a3, a u 0

5x2; x u 3

b 1 22

a 1 55

p 2 76

z 2 42z ; z u 0

17 2 c

; c u 7

1a 1 2

; a u 22

2x 2 3, x u 3

210b5

40b4; 2b4, b u 0 z2 2 36

4z 2 24; z 1 64 , z u 6

2x2 1 16x 2 4 ; 2x 2 4, x u 4

2(30 1 2z)14 1 4z ; 215 2 z

7 1 2z , z u 3.5

a 1 2

21; x u 1

12 2 9a

; a u 29

2q 1 3

; q u 23 or 21

1y 2 2

; y u 2

x 1 2; x u 23

1lmn; l u 0, m u 0, n u 0

x4; x u 0

4xy; x u 0, y u 0

d3c; d u 0, c u 0

2x 2 4; x u 0


29

Name Class Date


29

A few important rules are needed for successful division of a polynomial. When

dividing by a monomial, a single term, remember to divide each polynomial term

by the monomial and reduce the fraction.

Problem

What is (8x3 2 3x2 1 16x) 4 2x2 ?

Solve

(8x3 2 3x2 1 16x) 4 2x2 Division equals multiplication by the reciprocal.

5 (8x3 2 3x2 1 16x) ?1

2x2 Multiply by 12x2, the reciprocal of 2x2.

58x3

2x2 23x2

2x2 116x2x2 Use the Distributive Property.

5 4x1 232x0 1

8x Subtract exponents when dividing powers

with the same base.

5 4x 2 32 1

8x Simplify.

Exercises

Divide.

1. (2x2 2 9x 1 18) 4 2x 2. (16x4 2 64) 4 4x3

3. (x5 2 3x4 1 10x3 234x2 2 6) 4 3x2 4. (5x3 2 25x2 2 1) 4 5x

When a polynomial (many terms) is divided by a binomial (2 terms), the

polynomial terms should be in order from highest to lowest exponent.

To make the polynomial 24 1 2x 2 16x2 1 3x3 division ready, put it in the

correct order for division, from greatest exponent to lowest. Th e correct order for

24 1 2x 2 16x2 1 3x3 to be division ready is 3x3 2 16x2 1 2x 2 4.

For any gaps or missing exponents, the place is held with 0. For example, x2 1 1

becomes x2 1 0x 1 1, 0x being the placeholder for the x term.

11-3 Reteaching Dividing Polynomials

x 2 92 1

9x 4x 2 16

x3

x3

3 2 x2 1 10x3 2 1

4 22x2 x2 2 5x 2 1

5x


30

Name Class Date


30

Dividing a polynomial by a binomial is similar to long division.

Problem

What is (p2 2 3p 1 2) 4 (p 2 2)?

Solve

p 2 2qp2 2 3p 1 2 Write the problem as long division.

p

p 2 2qp2 2 3p 1 2 Ask how many times p goes into p2 (p times). The variables must

align by exponent, so the p goes above 23p since both match.

pp 2 2qp2 2 3p 1 2

p2 2 2p0000

21p 1 2

Multiply p times (p 2 2). Subtract the product p2 2 2p. Bring down 2.

p 2 1

p 2 2qp2 2 3p 1 2

p2 2 2p0000

21p 1 2

21p 1 2

Determine how many times p goes into 21p ( 21 times). Multiply (21) times (p 2 2) to get 21p 1 2. Subtract the product 21p 1 2. There is no remainder.

So p 2 2 goes into p2 2 3p 1 2 exactly p 2 1 times with no remainder.

Exercises

Divide.

5. (d2 1 4d 2 12) 4 (d 2 2) 6. (y2 2 4y 1 4) 4 (y 2 2)

7. (x2 2 2x 1 1) 4 (x 2 1) 8. (b2 2 b 2 20) 4 (b 2 5)

11-3 Reteaching(continued) Dividing Polynomials

d 1 6 y 2 2

x 2 1 b 1 4

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