EXPRESSIONS & EQUATIONS 20120803 www.njctl.org 1 Table of - - PDF document

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EXPRESSIONS & EQUATIONS 20120803 www.njctl.org 1 Table of - - PDF document

Advanced Unit 3: EXPRESSIONS & EQUATIONS Name: _____________________ Advanced Unit 3 EXPRESSIONS & EQUATIONS 20120803 www.njctl.org 1 Table of Contents Commutative and Associative Properties Combining Like Terms Click on a


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Advanced Unit 3: EXPRESSIONS & EQUATIONS

1

www.njctl.org

2012­08­03

Name: _____________________

Advanced Unit 3

EXPRESSIONS & EQUATIONS

2

Table of Contents

Inverse Operations One Step Equations Two Step Equations Multi­Step Equations Distributing Fractions in Equations Graphing & Writing Inequalities with One Variable

Click on a topic to go to that section.

The Distributive Property Combining Like Terms Simple Inequalities involving Addition & Subtraction Simple Inequalities involving Multiplication & Division Common Core Standards: 7.EE.1, 7.EE.3, 7.EE.4 Simplifying Algebraic Expressions Translating Between Words and Equations Commutative and Associative Properties Using Numerical and Algebraic Expressions and Equations

3

Commutative and Associative Properties

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

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Commutative Property of Addition: The order in which the terms of a sum are added does not change the sum. a + b = b + a 5 + 7 = 7 + 5 12= 12 Commutative Property of Multiplication: The order in which the terms of a product are added does not change the product. ab = ba 4(5) = 5(4)

Pull Pull

5

Associative Property of Addition: The order in which the terms of a sum are grouped does not change the sum. (a + b) + c = a + (b + c) (2 + 3) + 4 = 2 + (3 + 4) 5 + 4 = 2 + 7 9 = 9

Pull Pull

6

The Associative Property is particularly useful when you are combining integers. Example: ­15 + 9 + (­4)= ­15 + (­4) + 9= Changing it this way allows for the ­19 + 9 = negatives to be added together first. ­10

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

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Associative Property of Multiplication: The order in which the terms of a product are grouped does not change the product.

8

1

Identify the property of ­5 + 3 = 3 + (­5)

A

Commutative Property of Addition

B

Commutative Property of Multiplication

C

Associative Property of Addition

D

Associative Property of Multiplication

9

2

Identify the property of a + (b + c) = (a + c) + b

A

Commutative Property of Addition

B

Commutative Property of Multiplication

C

Associative Property of Addition

D

Associative Property of Multiplication

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

10

3

Identify the property of (3 x ­4) x 8 = 3 x (­4 x 8)

A

Commutative Property of Addition

B

Commutative Property of Multiplication

C

Associative Property of Addition

D

Asociative Property of Multiplication

11

Discuss why using the associative property would be useful with the following problems:

  • 1. 4 + 3 + (­4)
  • 2. ­9 x 3 x 0
  • 3. ­5 x 7 x ­2
  • 4. ­8 + 1 + (­6)

12

Combining Like Terms

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

13

An Expression ­ contains numbers, variables and at least one

  • peration.

14

Like terms: terms in an expression that have the same variable raised to the same power Examples: LIKE TERMS NOT LIKE TERMS 6x and 2x 6x2 and 2x 5y and 8y 5x and 8y 4x2 and 7x2 4x2y and 7xy2

15

4

Identify all of the terms like 2x A 5 B 3x2 C 5y D 12y E ­7y

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

16

5

Identify all of the terms like 8y A 9y B 4y2 C 7y D 8 E ­18x

17

6

Identify all of the terms like 8xy A 8x B 3x2y C 39xy D 4y E ­8xy

18

7

Identify all of the terms like 2y A 51w B 2x C

3y D 2w E ­10y

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

19

8

Identify all of the terms like 14x2 A ­5x B 8x2

C 13y2 D

x

E ­x2

20

If two or more like terms are being added or subtracted, they can be combined. To combine like terms add/subtract the coefficient but leave the variable alone. 7x +8x =15x 9v­2v = 7v

21

Sometimes there are constant terms that can be combined. 9 + 2f + 6= 9 + 2f + 6= 2f + 15 Sometimes there will be both coeffients and constants to be combined. 3g +7 + 8g ­2 11g + 5 Notice that the sign before a given term goes with the number.

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

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Try These: 8x + 9x 7y + 5y 2b +6g 3 + 4f + 9f 9j + 3 + 2 4h + 6 + 7h + 3 7a + 4 + 2a ­1 9 + 8c ­12 + 5c

23

9

8x + 3x = 11x

True False

24

10 7x + 7y = 14xy True

False

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

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11 2x + 3x = 5x

True False

26

12 9x + 5y = 14xy

True False

27

13 6x + 2x = 8x2

True False

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

28

14 ­15y + 7y = ­8y

True False

29

15 ­6 + y + 8 = 2y

True False

30

16 ­7y + 9y = 2y

True False

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

31

17 9x + 4 + 2x = A

15x

B

11x + 4

C

13x + 2x

D

9x + 6x

32

18 12x + 3x + 7 ­ 5 A

15x + 7 ­ 5

B

13x

C

17x

D

15x + 2

33

19 ­4x ­ 6 + 2x ­ 14 A

­22x

B

­2x ­ 20

C

­6x +20

D

22x

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

34

The Distributive Property

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35

An Area Model

Imagine that you have two rooms next to each

  • ther. Both are 4 feet long. One is 7 feet wide

and the other is 3 feet wide .

4 7 3

How could you express the area of those two rooms together?

36

4 7 + 3

Either way, the area is 40 feet2:

You could add 7 + 3 and then multiply by 4 4(7+3)= 4(10)= 40

OR

You could multiply 4 by 7, then 4 by 3 and add them 4(7) + 4(3) = 28 + 12 = 40

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

37

An Area Model

Imagine that you have two rooms next to each

  • ther. Both are 4 yards long. One is 3 yards

wide and you don't know how wide the other is.

4 x 3

How could you express the area of those two rooms together?

38

4 x + 3

You cannot add x and 3 because they aren't like terms, so you can

  • nly do it by multiplying 4 by x

and 4 by 3 and adding 4(x) + 4(3)= 4x + 12 The area of the two rooms is 4x + 12

(Note: 4x cannot be combined with 12)

39

The Distributive Property

Finding the area of the rectangles demonstrates the distributive

  • property. Use the distributive property when expressions are

written like so: a(b + c) 4(x + 2) 4(x) + 4(2) 4x + 8 The 4 is distributed to each term of the sum (x + 2)

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

40

Write an expression equivalent to: 5(y + 4) 5(y) + 5(4) 5y + 20 6(x + 2) 3(x + 4) 4(x ­ 5) 7(x ­ 1)

Remember to distribute the 5 to the x and the 3

41

The Distributive Property is often used to eliminate the parentheses in expressions like 4(x + 2). This makes it possible to combine like terms in more complicated expressions. EXAMPLE: ­2(x + 3) = ­2(x) + ­2(3) = ­2x + ­6 or ­2x ­ 6 3(4x ­ 6) = 3(4x) ­ 3(6) = 12x ­ 18 ­2 (x ­ 3) = ­2(x) ­ (­2)(­3) = ­2x ­ 6 TRY THESE: 3(4x + 2) = ­1(6m + 4) ­3(2x ­ 5) =

Be careful with your signs!

42

Keep in mind that when there is a negative sign on the

  • utside of the parenthesis it really is a ­1.

For example: ­(2x + 7) = ­1(2x + 7) = ­1(2x) + ­1(7) = ­2x ­ 7 What do you notice about the original problem and its answer? The numbers are turned to their opposites.

Remove to see answer.

Try these: ­(9x + 3) = ­(­5x + 1) = ­(2x ­ 4) = ­(­x ­ 6) =

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

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20 4(2 + 5) = 4(2) + 5

True False

44

21 8(x + 9) = 8(x) + 8(9)

True False

45

22 ­4(x + 6) = ­4 + 4(6)

True False

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

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23 3(x ­ 4) = 3(x) ­ 3(4)

True False

47

24

Use the distributive property to rewrite the expression without parentheses 3(x + 4) A 3x + 4 B 3x + 12 C x + 12 D 7x

48

25

Use the distributive property to rewrite the expression without parentheses 5(x + 7)

A

x + 35

B

5x + 7

C

5x + 35

D

40x

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

49

26

Use the distributive property to rewrite the expression without parentheses (x + 5)2 A 2x + 5 B 2x + 10 C x + 10 D 12x

50

27

Use the distributive property to rewrite the expression without parentheses 3(x ­ 4) A 3x ­ 4 B x ­ 12 C 3x ­ 12 D 9x

51

28 Use the distributive property to rewrite the

expression without parentheses 2(w ­ 6) A 2w ­ 6 B w ­ 12 C 2w ­ 12 D 10w

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

52

29 Use the distributive property to rewrite the

expression without parentheses ­4(x ­ 9) A ­4x ­ 36 B x ­ 36 C 4x ­ 36 D

­4x + 36

53

30 Use the distributive property to rewrite the

expression without parentheses 5.2(x ­ 9.3) A ­5.2x ­ 48.36 B 5.2x ­ 48.36 C ­5.2x + 48.36 D ­48.36x

54

31 Use the distributive property to rewrite the

expression without parentheses A B C D

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

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We can also use the Distributive Property in reverse. This is called Factoring. When we factor an expression, we find all numbers or variables that divide into all of the parts of an expression. Example: 7x + 35 Both the 7x and 35 are divisible by 7 7(x + 5) By removing the 7 we have factored the problem We can check our work by using the distributive property to see that the two expressions are equal.

56

We can factor with numbers, variables, or both. 2x + 4y = 2(x + 2y) 9b + 3 = 3(3b + 1) ­5j ­ 10k + 25m = ­5(j + 2k ­ 5m) *Careful of your signs 4a + 6a + 8ab = 2a(2 + 3 + 4b)

57

Try these: Factor the following expressions: 6b + 9c = ­2h ­ 10j = 4a + 20ab + 12abc =

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

58

32 Factor the following: 4p + 24q A

4 (p + 24q)

B

2 (2p + 12q)

C

4(p + 6q)

D

2 (2p + 24q)

59

33 Factor the following: 5g + 15h A

3(g + 5h)

B

5(g + 3h)

C

5(g + 15h)

D

5g (1 + 3h)

60

34 Factor the following: 3r + 9rt + 15rx A

3(r+ 3rt + 5rx)

B

3r(1 + 3t + 5x)

C

3r (3t + 5x)

D

3 (r + 9rt + 15rx)

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35 Factor the following: 2v+7v+14v A

7(2v + v + 2v)

B

7v(2 + 1 + 2)

C

7v (1 + 2)

D

v(2 + 7 + 14)

62

36 Factor the following: ­6a ­ 15ab ­ 18abc A

­3a(2 + 5b + 6c)

B

3a(2+ 5b + 6c)

C

­3(2a ­ 5b ­ 6c)

D

­3a (2 ­5b ­ 6c)

63

­

What divides into the expression: ­5n ­ 20mn ­ 10np

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

64

­

If a regular pentagon has a perimeter of 10x + 25, what does each side equal?

65

Simplifying Algebraic Expressions

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66

Now we will use what we know about combining like terms and the distributive property to simplify algebraic expressions. Remember, like terms have the same variable and same exponent.

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

67

To simplify: 4 + 5(x + 3) First Distribute 4 + 5(x) + 5(3) 4 + 5x + 15 Then combine Like Terms 5x + 19 Notice that when combining like terms, you add/subtract the coefficients but the variable remains the same. Remember that you can combine coefficient or constant terms.

68

37 7x +3(x ­ 4) = 10x ­ 4

True False

69

38 8 +(x + 3)5 = 5x + 11

True False

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

70

39 4 +(x ­ 3)6 = 6x ­14

True False

71

40 2x + 3y + 5x + 12 = 10xy + 12

True False

72

41 5x2 + 2x + 7(x + 1) + x2 = 6x2 + 9x + 7

True False

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

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42 2x3 + 4x2 + 6(x2 + 3x) + x = 2x3 + 10x2 + 4x

True False

74

43 The lengths of the sides of home plate in a baseball field are

represented by the expressions in the accompanying figure.

A

5xyz

B

x2 + y3z

C

2x + 3yz

D

2x + 2y + yz

yz y y x x

From the New York State Education Department. Office of Assessment Policy, Development and

  • Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June,

2011.

Which expression represents the perimeter of the figure?

75

44 A rectangle has a width of x and a length that is

double that. What is the perimeter of the rectangle?

A

4x

B

6x

C

8x

D

10x

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

76

Inverse Operations

Return to Table of Contents

77

What is an equation? An equation is a mathematical statement containing an equal sign to show that two expressions are equal. 2+3=5 9­2=7 5 + 3 = 1 + 7 An algebraic equation is just an equation that has algebraic symbols in one or both of the expressions. 4x = 24 9 + h = 15

78

Equations can also be used to state the equality of two expressions containing one or more variables. In real numbers we can say, for example, that for any given value of x it is true that 4x + 1 = 13 x = 3 because

4(3) + 1 = 13 12 + 1 = 13 13 = 13

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When defining your variables, remember... Letters from the beginning of the alphabet like a, b, c... often denote constants in the context of the discussion at hand. While letters from end of the alphabet, like x, y, z..., are usually reserved for the variables, a convention initiated by Descartes. Try It! Write an equation with a variable and have a classmate identify the variable and its value.

80

An equation can be compared to a balanced scale. Both sides need to contain the same quantity in order for it to be "balanced".

81

For example, 9+ 11 = 6 + 14 represents an equation because both sides simplify to 20. 9 + 11 = 6 + 14 20 = 20 Any of the numerical values in the equation can be represented by a variable. Examples: 15 + c = 25 x + 10 = 25 15 + 10 = y

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

82

When solving equations, the goal is to isolate the variable on

  • ne side of the equation in order to determine its value (the

value that makes the equation true).

83

In order to solve an equation containing a variable, you need to use inverse (opposite/undoing) operations on both sides

  • f the equation.

Let's review the inverses of each operation: Addition Subtraction MultiplicationDivision

84

There are two questions to ask when solving an equation: *What operation is in the equation? *What is the inverse of that operation (This will be the

  • peration you use to solve the equation.)?
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Advanced Unit 3: EXPRESSIONS & EQUATIONS

85

A good phrase to remember when doing equations is: Whatever you do to one side of the equation, you do to the other. For example, if you add three on one side of the equal sign you must add three to the other side as well.

86

To solve for "x" in the following equation... x + 7 = 32 Determine what operation is being shown (in this case, it is addition). Do the inverse to both sides (in this case, it is subtraction). x + 7 = 32 ­ 7 ­7 x = 25 In the original equation, replace x with 25 and see if it makes the equation true. x + 7 = 32 25 + 7 = 32 32 = 32

87

For each equation, write the inverse operation needed to solve for the variable. a.) y +7 = 14 subtract b.) a ­ 21 = 10 add 21 c.) 5s = 25 divide by 5 d.) x = 5 multiply by 12 12

move move

move

move

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

88

Think about this... To solve c ­ 3 = 12 Which method is better? Why? Kendra Added 3 to each side of the equation c ­ 3 = 12 +3 +3 c = 15 Ted Subtracted 12 from each side, then added 15. c ­ 3 = 12 ­12 ­12 c ­ 15 = 0 +15 +15 c = 15

89

45 What is the inverse operation needed to solve this

equation? 2x = 14

A

Addition B Subtraction C Multiplication D Division

90

46 What is the inverse operation needed to solve this

equation? x ­ 3 = ­12

A Addition B Subtraction C

Multiplication D Division

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

91

47 What is the inverse operation needed to solve this

problem? ­2 + x = 9

A

Addition

B

Subtraction

C

Multiplication

D

Division

92

One Step Equations

Return to Table of Contents

93

To solve equations, you must work backwards through the

  • rder of operations to find the value of the variable.

Remember to use inverse operations in order to isolate the variable on one side of the equation. Whatever you do to one side of an equation, you MUST do to the other side!

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

94

Examples: y + 3 = 13 ­ 3 ­3 The inverse of adding 3 is subtracting 3 y = 10 4m = 32 4 4 The inverse of multiplying by 6 is dividing by 6 m = 8 Remember ­ whatever you do to one side of an equation, you MUST do to the other!!!

95

x ­ 5 = 2 +5 +5 x = 7 x + 5 = ­14 ­5 ­5 x = ­19 2 = x ­ 4 +4 +4 6 = x 6 = x + 1 ­1 ­1 5 = x 12 = x + 17 ­17 ­17 ­5 = x x + 9 = 5 ­9 ­9 x = ­4

One Step Equations Solve each equation then click the box to see work & solution.

click to show inverse operation click to show inverse operation click to show inverse operation click to show inverse operation click to show inverse operation click to show inverse operation click to show inverse operation click to show inverse operation click to show inverse operation click to show inverse operation

96

One Step Equations 4x = 16 4 4 x = 4 ­2x = ­12 ­2 ­2 x = 6 ­20 = 5x 5 5 ­4 = x

click to show inverse operation click to show inverse operation click to show inverse operation

x 2 x = 18 = 9 (2) (2) x

­6 x = ­216 = 36

click to show inverse operation

(­6) (­6)

click to show inverse operation

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

97

48 Solve. x ­ 7 = 19

98

49 Solve. j + 15 = 17

99

50 Solve. 42 = 6y

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

100

51 Solve. ­115 = ­5x

101

52 Solve.

= 12 x 9

102

53 Solve. w ­ 17 = 37

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

103

54 Solve.

­3 = x 7

104

55 Solve. 23 + t = 11

105

56 Solve. 108 = 12r

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

106

Sometimes the operation can be confusing. For example: ­2 + x = 7 This looks like you should use subtraction to undo the problem. However, ­2 + x = 7 is the same as x ­ 2 = 7 so while it appears to be addition, it is really subtraction. In order to undo this we would have to add. ­2 + x = 7 x ­ 2 = 7 +2 +2 x = 9

107

­2 + x = 7 ­2 = ­2 ­4 + x = 5 This did not cancel out anything. ­2 + x = 7 +2 +2 x = 9 This did cancel out to find the answer.

­2 + x = 7 x ­ 2 = 7 +2 +2 x = 9 This is the same as the middle problem

108

Try these: ­4 + b = 7 ­2 + r = 4 ­3 + w = 6 ­5 + c = 9

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

109

Think about this... In the expression To which does the "­" belong? Does it belong to the x? The 3? Both? The answer is that there is one negative so it is used once with either the variable or the 3. Generally, we assign it to the 3 to avoid creating a negative variable. So:

110

57

111

58 Solve.

­5 + q = 15

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

112

59 Solve.

113

60 Solve

114

61 Solve.

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

115

62 Solve.

116

63 Solve.

117

Sometimes you will have an equation where you are multiplying a variable by a fraction.

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

118

To undo the fraction you: Multiply by the Reciprocal of the Coefficent This means that you will flip the fraction and then multiply

119

1 times any number is itself so this is why it can cancel out.

120

64 Solve.

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

121

65 Solve

122

66 Solve.

123

Two­Step Equations

Return to Table of Contents

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

124

Sometimes it takes more than one step to solve an equation. Remember that to solve equations, you must work backwards through the order of operations to find the value of the variable. This means that you undo in the opposite order (PEMDAS): 1st: Addition & Subtraction 2nd: Multiplication & Division 3rd: Exponents 4th: Parentheses Whatever you do to one side of an equation, you MUST do to the other side!

125

Examples: 4x + 2 = 10 ­ 2 ­ 2 Undo addition first 4x = 8 4 4 Undo multiplication second x = 2 ­2y ­ 9 = ­13 + 9 + 9 Undo subtraction first ­2y = ­4 ­2 ­2 Undo multiplication second y = 2 Remember ­ whatever you do to one side

  • f an equation, you MUST do to the other!!!

126

5b + 3 = 18 ­3 ­3 5b = 15 5 5 b = 3 3j ­ 4 = 14 +4 +4 3j = 18 3 3 j = 6 ­2x + 3 = ­1 ­ 3 ­3 ­2x = ­4 ­2 ­2 x = 2

Two Step Equations

Solve each equation then click the box to see work & solution. ­6 = ­6 w = 14 ­2m ­ 4 = 22 +4 +4 ­2m = 26 ­2 ­2 m = ­13 +5 = +5 m = 15

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

127

67 Solve the equation.

5x ­ 6 = ­56

128

68 Solve the equation.

14 = 3c + 2

129

69 Solve the equation.

x 5 ­ 4 = 24

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

130

70 Solve the equation.

5r ­ 2 = ­12

131

71 Solve the equation.

14 = ­2n ­ 6

132

72 Solve the equation.

+ 7 = 13 x 5

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

133

73 Solve the equation.

+ 2 = ­10 x 3 ­

134

74 Solve the equation.

135

75 Solve the equation.

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

136

76 Solve the equation.

137

77 Solve the equation.

138

78 Solve

­3 5 1 2 x + = 1 10

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

139

79 Solve the equation.

140

80 Solve the equation.

141

Multi­Step Equations

Return to Table of Contents

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

142

Steps for Solving Multiple Step Equations

As equations become more complex, you should:

  • 1. Simplify each side of the equation.

(Combining like terms and the distributive property)

  • 2. Use inverse operations to solve the equation.

Remember, whatever you do to one side of an equation, you MUST do to the other side!

143

Examples: 5x + 7x + 4 = 28 12x + 4 = 28 Combine Like Terms ­4 ­ 4 Undo Addition 12x = 24 2 2 Undo Multiplication x = 12 ­1 = 2r ­ 7r +19 ­1 = ­5r + 19 Combine Like Terms ­19 = ­ 19 Undo Subtraction ­20 = ­5r ­5 ­5 Undo Multiplication 4 = r

144

Try these. 12h ­ 10h + 7 = 25 ­17q + 7q ­13 = 27 17 ­ 9f + 6 = 140 h = 9 q = 4 f = 117

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

145

Always check to see that both sides of the equation are simplified before you begin solving the equation. Sometimes, you need to use the distributive property in

  • rder to simplify part of the equation.

Remember: The distributive property is a(b + c) = ab + bc Examples 5(20 + 6) = 5(20) + 5(6) 9(30 ­ 2) = 9(30) ­ 9(2) 3(5 + 2x) = 3(5) + 3(2x) ­2(4x ­ 7) = ­2(4x) ­ (­2)(7)

146

Examples: 2(b ­ 8) = 28 2b ­ 16 = 28 Distribute the 2 through (b ­ 8) +16 +16 Undo subtraction 2b = 44 2 2 Undo multiplication b = 22 3r + 4(r ­ 2) = 13 3r + 4r ­ 8 = 13 Distribute the 4 through (r ­ 2) 7r ­ 8 = 13 Combine Like Terms +8 +8 Undo subtraction 7r = 21 7 7 Undo multiplication r = 3

147

Try these. 3(w ­ 2) = 9 4(2d + 5) = 92 6m + 2(2m + 7) = 54 w = 5 d = 9 m = 4

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

148

81 Solve.

9 + 3x + x = 25

149

82 ­8e + 7 +3e = ­13

150

83 Solve.

­27 = 8x ­ 4 ­ 2x ­ 11

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

151

84 Solve.

n ­ 2 + 4n ­ 5 = 13

152

85 Solve.

32 = f ­ 3f + 6f

153

86 Solve.

6g ­ 15g + 8 ­ 19 = ­38

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154

87 Solve.

3(a ­ 5) = ­21

155

88 Solve.

4(x + 3) = 20

156

89 Solve.

3 = 7(k ­ 2) + 17

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157

90 Solve.

2(p + 7) ­7 = 5

158

91 Solve.

159

92 Solve.

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160

93 Solve.

161

94 Solve.

162

Distributing Fractions in Equations

Return to Table of Contents

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163

Remember...

  • 1. Simplify each side of the equation.
  • 2. Solve the equation.

(Undo addition and subtraction first, multiplication and division second)

Remember, whatever you do to one side of an equation, you MUST do to the other side!

164

There is more than one way to solve an equation with a fraction

  • coefficient. While you can, you don't need to distribute.

Multiply by the reciprocal Multiply by the LCD

(­3 + 3x) = 3 5 72 5 (­3 + 3x) = 3 5 72 5 (­3 + 3x) = 3 5 72 5 (­3 + 3x) = 3 5 72 5 5 3 5 3 ­3 + 3x = 24 +3 +3 3x = 27 3 3 x = 9 (­3 + 3x) = 3 5 72 5 5 5 3(­3 + 3x) = 72 ­9 + 9x = 72 +9 +9 9x = 81 9 9 x = 9

165

Some problems work better when you multiply by the reciprocal and some work better multiplying by the LCM. Which strategy would you use for the following? Why?

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166

95 Solve.

167

96 Solve.

168

(8 ­ 3c) = 2 3 16 3

97 Solve.

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169

98 Solve.

170

99 Solve.

171

Translating Between Words and Expressions

Return to Table

  • f Contents
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172

PULL List words that indicate addition

173

PULL List words that indicate subtraction

174

PULL List words that indicate multiplication

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175

PULL List words that indicate division

176

PULL List words that indicate equals

177

Be aware of the difference between "less" and "less than". For example: "Eight less three" and "Three less than Eight" are equivalent

  • expressions. So what is the difference in wording?

Eight less three: 8 ­ 3 Three less than eight: 8 ­ 3 When you see "less than", you need to switch the order of the numbers.

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178

As a rule of thumb, if you see the words "than" or "from" it means you have to reverse the order

  • f the two items on either side of the word.

Examples:

  • 8 less than b means b ­ 8
  • 3 more than x means x + 3
  • x less than 2 means 2 ­ x

click to reveal

179

The many ways to represent multiplication...

How do you represent "three times a"? (3)(a) 3(a) 3 a 3a The preferred representation is 3a When a variable is being multiplied by a number, the number (coefficient) is always written in front of the variable. The following are not allowed: 3xa ... The multiplication sign looks like another variable a3 ... The number is always written in front of the variable

180

Representation of division...

How do you represent "b divided by 12"? b ÷ 12 b ∕ 12 b 12

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181

When choosing a variable, there are some that are

  • ften avoided:

l, i, t, o, O, s, S Why might these be avoided? It is best to avoid using letters that might be confused for numbers or operations. In the case above (1, +, 0, 5)

182

Three times j Eight divided by j j less than 7 5 more than j 4 less than j

1 2 3 4 5 6 7 8 9 + ­

.

÷ TRANSLATE THE WORDS INTO AN ALGEBRAIC EXPRESSION j

183

23 + m The sum of twenty­three and m Write the expression for each statement. Then check your answer.

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184

d - 24 Twenty­four less than d Write the expression for each statement. Then check your answer.

185

4(8­j) Write the expression for each statement. Remember, sometimes you need to use parentheses for a quantity. Four times the difference of eight and j

186

7w 12 The product of seven and w, divided by 12 Write the expression for each statement. Then check your answer.

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187

(6+p)2 Write the expression for each statement. Then check your answer. The square of the sum of six and p

188

100 The quotient of 200 and the quantity of p times 7

A 200 7p B 200 ­ (7p) C 200 ÷ 7p D 7p 200

189

101 35 multiplied by the quantity r less 45

A 35r ­ 45 B 35(45) ­ r C 35(45 ­ r) D 35(r ­ 45)

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190

102 Mary had 5 jellybeans for each of 4 friends.

A 5+4 B 5 ­ 4 C 5 x 4 D 5 ÷ 4

191

103 If n + 4 represents an odd integer, the next

larger odd integer is represented by A n + 2 B n + 3 C n + 5 D n + 6

From the New York State Education Department. Office of Assessment Policy, Development and

  • Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

192

104 a less than 27

A 27 ­ a B a 27 C a ­ 27 D 27 + a

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193

105 If h represents a number, which equation is a

correct translation of: “Sixty more than 9 times a number is 375”? A 9h = 375 B 9h + 60 = 375 C 9h ­ 60 = 375 D 60h + 9 = 375

From the New York State Education Department. Office of Assessment Policy, Development and

  • Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

194

Using Numerical and Algebraic Expressions and Equations

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195

There are some problems that do not need algebra to solve them. Here is an example: Yessica, Zanaya, and Paulo were all reading the same book. Yessica said that she was 42% done, Zanaya said she read of the book and Paulo said he read 0.56 of the book. Who read the most?

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196

Yessica, Zanaya, and Paulo were all reading the same book. Yessica said that she was 42% done, Zanaya said she read of the book and Paulo said he read 0.56 of the book. Who read the most? In this case, you do not need to use algebra. You simply need to convert all of the numbers to the same form and then order them. 42% = 0.42 = 0.6 0.56 = 0.56 Zanaya read the most.

197

Teshawn buys a pair of sneakers for 35% off. If the

  • riginal price was $125, how much does Teshawn

actually pay? There are two ways to solve this problem. 125 x 0.35 = 43.75 125 x 0.65 = 81.25 125 ­ 43.75 = 81.25 Where did the 0.65 come from? 100% ­ 35% = 65% Because we are taking 35% off, we can subtract it from 100%. Teshawn pays $81.25 for sneakers.

198

Let's look at a little different example. Demi goes out for lunch and the bill comes to $29.52. She wants to leave an 18% tip. What is the total she spends on her lunch? Again, there are two ways to solve this problem. 29.52 x 0.18 5.31 29.52 X 1.18 = 34.83 29.52 + 5.31 = 34.83 100% + 18% = 118% She paid 100% of the bill and the 18% tip so she paid 118%. 29.52 x 1.18 = 34.83 Demi pays a total of $34.83.

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199

106 Peter went to the zoo and saw 15 penguins, and 8

  • bears. He saw 7 times as many monkeys as bears.

How many animals did he see altogether?

A

30

B

46

C

79

D

128

200

107 Daniela is making pizza for 30 people. Half of the

people want plain, 30% want sausage and the rest want mushroom. How many people want mushroom pizza?

A

5

B

10

C

15

D

20

201

108 Alicia buys a sweater for $46.25 and a scarf for

$19.84. The sales tax is 6%. If Alicia pays with $100, how much change will she get?

(Hint: Round to the nearest cent.)

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202

109 Alexander is shopping for a new phone. At Phones R

Us the phone cost $196 and there is a 20% off sale. At Mobile Mart, the phone is $206 with a off sale. Which is the better deal?

A

Phones R Us

B

Mobile Mart

C

They are the same

1 5

203

110 A school surveyed 234 students about school

  • lunches. liked hamburgers the most, 0.34 liked

fried chicken best, 42% preferred pizza, and the rest liked beef and macaroni. From greatest to least, what were the preferences?

A

hamburgers, chicken, pizza, macaroni

B

chicken, macaroni, hamburgers, pizza

C

pizza, chicken, hamburgers, macaroni

D

macaroni, hamburgers, chicken, pizza

1 6

204

111 Patty and 3 friends went out to dinner. The bill came

to $124. They want to leave a 15% tip and then split it evenly among all four of them. How much will each person spend?

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205

We can use our algebraic translating skills to solve other problems. We can use a variable to show an unknown. A constant will be any fixed amount. If there are two separate unknowns, relate one to the

  • ther.

206

The school cafeteria sold 225 chicken meals today. They sold twice the number of grilled chicken sandwiches than chicken tenders. How many of each were sold? 2c + c = 225

chicken sandwiches chicken tenders total meals

c + 2c = 225 3c = 225 3 3 c = 75 The cafeteria sold 125 grilled chicken sandwiches and 75 tenders.

207

Julie is matting a picture in a frame. Her frame is 9 inches wide and her picture is 7 inches wide. How much matting should she put on either side? 2m + 7 = 9 2m + 7 = ­7 ­7 2m = 2 2 2 m = 1 Julie needs 1 inches on each side.

1 4 1 2 1 2 1 4

9

both sides

  • f the mat

size of picture size of frame

1 2 1 2

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208

Many times with equations there will be one number that will be the same no matter what (constant) and one that can be changed based on the problem (variable and coefficient). Example: George is buying video games online. The cost of the video is $30.00 per game and shipping is a flat fee of $7.00. He spent a total of $127.00. How many games did he buy in all?

209

George is buying video games online. The cost of the video is $30.00 per game and shipping is a flat fee of $7.00. He spent a total of $127.00. How many games did he buy in all? Notice that the video games are "per game" so that means there could be many different amounts of games and therefore many different prices. This is shown by writing the amount for one game next to a variable to indicate any number of games.

30g

cost of

  • ne video

game number

  • f games

210

George is buying video games online. The cost of the video is

$30.00 per game and shipping is a flat fee of $7.00. He spent a total of $127.00. How many games did he buy in all? Notice also that there is a specific amount that is charged no matter what, the flat fee. This will not change so it is the constant and it will be added (or subtracted) from the other part of the problem.

30g + 7

cost of

  • ne video

game number

  • f games

the cost

  • f the

shipping

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211

George is buying video games online. The cost of the video is $30.00 per game and shipping is a flat fee of $7.00. He spent a total of $127.00. How many games did he buy in all? "Total" means equal so here is how to write the rest of the equation.

30g + 7 = 127

cost of

  • ne video

game number

  • f games

the total amount the cost of the shipping

212

George is buying video games online. The cost of the video is $30.00 per game and shipping is a flat fee of $7.00. He spent a total of $127.00. How many games did he buy in all?

Now we can solve it. 30g + 7 = 127 ­7 ­7 30g = 120 30 30 g = 4 George bought 4 video games.

213

112 Lorena has a garden and wants to put a gate to her

fence directly in the middle of one side. The whole length of the fence is 24 feet. If the gate is 4 feet, how many feet should be on either side of the fence?

3 4 1 2

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214

113 Lewis wants to go to the amusement park with his family. The cost is $12.00 for parking plus $27.00 per person to enter the

  • park. Lewis and his family spent $147. Which equation shows

this problem? A 12p + 27 = 147 B 12p + 27p = 147 C 27p + 12 = 147 D 39p = 147

215

114 Lewis wants to go to the amusement park with his family. The cost is $12.00 for parking plus $27.00 per person to enter the park. Lewis and his family spent $147. How many people went to the amusement park with Lewis?

Pull Pull

216

115 Mary is saving up for a new bicycle that is $239. She has

$68.00 already saved. If she wants to put away $9.00 per week, how many weeks will it take to save enough for her bicycle? Which equation represents the situation? A 9 + 68 = 239 B 9d + 68 = 239 C 68d + 9 = 239 D 77d = 239

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217

116 Mary is saving up for a new bicycle that is $239. She has

$68.00 already saved. If she wants to put away $9.00 per week, how many weeks will it take to save enough for her bicycle?

218

117

You are selling t­shirts for $15 each as a fundraiser. You sold 17 less today then you did yesterday. Altogether you have raised $675. Write and solve an equation to determine the number

  • f t­shirts you sold today.

Be prepared to show your equation!

219

118 Rachel bought $12.53 worth of school supplies. She still

needs to buy pens which are $2.49 per pack. She has $20.00. How many packs of pens can she buy? Write and solve an equation to determine the number of packs of pens Rachel can buy. Be prepared to show your equation!

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220

119 The length of a rectangle is 9 cm greater than its width

and its perimeter is 82 cm. Write and solve an equation to determine the width of the rectangle. Be prepared to show your equation!

221

120 The product of ­4 and the sum of 7 more than a number is ­96.

Write and solve an equation to determine the number. Be prepared to show your equation!

222

121 A magazine company has 2,100 more subscribers this

year than last year. Their magazine sells for $182 per

  • year. Their combined income from last year and this year

is $2,566,200. Write and solve an equation to determine the number of subscribers they had each year. Be prepared to show your equation!

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223

122 The perimeter of a hexagon is 13.2 cm.

Write and solve an equation to determine the length of a side of the hexagon. Be prepared to show your equation!

224

Graphing and Writing Inequalities with One Variable

Return to Table of Contents

225

When you need to use an inequality to solve a word problem, you may encounter one of the phrases below.

Important Words Sample Sentence Equivalent

Translation

is more than is greater than must exceed

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226

When you need to use an inequality to solve a word problem, you may encounter one of the phrases below. Important Words

Sample Sentence Equivalent Translation

cannot exceed is at most is at least

227

How are these inequalities read? 2 + 2 > 3 Two plus two is greater than 3 2 + 2 ≥ 4 Two plus two is greater than or equal to 4 2 + 2 < 5 Two plus two is less than 5 2 + 2 ≤ 5 Two plus two is less than or equal to 5 2 + 2 ≤ 4 Two plus two is less than or equal to 4 2 + 2 > 3 Two plus two is greater than or equal to 3

228

Writing inequalities

Let's translate each statement into an inequality. x is less than 10 20 is greater than or equal to y x < 10 words inequality statement translate to 20 > y

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229

You try a few:

  • 1. 14 is greater than a
  • 2. b is less than or equal to 8
  • 3. 6 is less than the product of f and 20
  • 4. The sum of t and 9 is greater than or equal to 36
  • 5. 7 more than w is less than or equal to 10
  • 6. 19 decreased by p is greater than or equal to 2
  • 7. Fewer than 12 items
  • 8. No more than 50 students
  • 9. At least 275 people attended the play

Answers

230

Do you speak math?

Change the following expressions from English into math. Double a number is at most four. Three plus a number is at least six. 2x ≤ 4 2 + x ≥ 6 Answer Answer

231

Five less than a number is less than twice that number. The sum of two consecutive numbers is at least thirteen. Three times a number plus seven is at least nine. x ­ 5 < 2x x + (x + 1) ≥ 13 3x + 7 > 9 Answer Answer Answer

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232

0 1 2 3 4 5 6 7 8 9 10 7.5

$7.50 7.5 at least > An employee earns e A store's employees earn at least $7.50 per hour. Define a variable and write an inequality for the amount the employees may earn per hour. Let e represent an employee's wages.

233

Try this: The speed limit on a road is 55 miles per hour. Define a variable and write an inequality.

Answer

234

123 You have $200 to spend on clothes. You already

spent $140 and shirts cost $12. Which equation shows this scenario?

A

200 < 12x + 140

B

200 12x + 140

C

200 > 12x + 140

D

200 12x + 140 ≤ ≥

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235

124 A sea turtle can live up to 125 years. If one is already

37 years old, which scenario shows how many more years could it live? 125 < 37 + x 125 37 + x ≤

A B C

125 > 37 + x

D

125 37 + x ≥

236

125 The width of a rectangle is 3 in longer than the

  • length. The perimeter is no less than 25 inches.

A 4a + 6 < 25 B 4a + 6  25 C 4a + 6 > 25 D 4a + 6 ≥ 25

237

126 The absolute value of the sum of two numbers is

less than the sum of the absolute values of the same two numbers.

A B C D

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238

A solution to an inequality is NOT a single number. It will have more than one value.

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

This would be read as the solution set is all numbers greater than or equal to negative 5.

Solution Sets

239

Let's name the numbers that are solutions of the given inequality. r > 10 Which of the following are solutions? {5, 10, 15, 20} 5 > 10 is not true So, not a solution 10 > 10 is not true So, not a solution 15 > 10 is true So, 15 is a solution 20 > 10 is true So, 20 is a solution Answer: {15, 20} are solutions of the inequality r > 10

240

Let's try another one. 30 ≥ 4d; {3, 4, 5, 6, 7, 8} 30 ≥ 4d 30 ≥ (4)3 30 ≥ 12 30 ≥ 4d 30 ≥ (4)4 30 ≥ 16 30 ≥ 4d 30 ≥ (4)5 30 ≥ 20 30 ≥ 4d 30 ≥ (4) 6 30 ≥ 24 30 ≥ 4d 30 ≥ (4)7 30 ≥ 28 30 ≥ 4d 30 ≥ (4)8 30 ≥ 32

click to reveal click to reveal click to reveal click to reveal click to reveal click to reveal

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241 Graphing Inequalities ­ The Circle

An open circle on a number shows that the number is not part of the solution. It is used with "greater than" and "less than". The word equal is not included.< > A closed circle on a number shows that the number is part of the solution. It is used with "greater than

  • r equal to" and "less than or equal to". < >

242

Graphing Inequalities ­ The Arrow The arrow should always point in the direction of those numbers who satisfy the inequality. *If the variable is on the left side of the inequality, then < and ≤ will show an arrow pointing left. *If the variable is on the left side of the inequality, then > and ≥ will show an arrow pointing right.

243

Notice that < and ≤ look like an arrow pointing left and that > and ≥ look like an arrow pointing right. But what if the variable isn't on the left? Do the opposite of where the inequality symbol points.

­1 ­2 ­3 ­4 ­5 1 2 3 4 5

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244

What is the number in the inequality? What kind of circle should be used? In what direction does the line go?

Graphing Inequalities 245

Step 1: Rewrite this as x < 5. Step 2: What kind of circle? Because it is less than, it does not include the number 5 and so it is an open circle.

­1 ­2 ­3 ­4 ­5 1 2 3 4 5

Graphing Inequalities x is less than 5 246

Step 4: Draw a line, thicker than the horizontal line, from the dot to the arrow. This represents all of the numbers that fulfill the inequality. Step 3: Draw an arrow on the number line showing all possible solutions. Numbers greater than the variable, go to the right. Numbers less than the variable, go to the left.

­1 ­2 ­3 ­4 ­5 1 2 3 4 5

x < 5

­1 ­2 ­3 ­4 ­5 1 2 3 4 5

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247

Step 1: Rewrite this as x ≤ 5. Step 2: What kind of circle? Because it is less than or equal to, it does include the number 5 and so it is a closed circle.

­1 ­2 ­3 ­4 ­5 1 2 3 4 5

Graphing Inequalities x is less than or equal to 5 248

Step 4: Draw a line, thicker than the horizontal line, from the dot to the arrow. This represents all of the numbers that fulfill the inequality. Step 3: Draw an arrow on the number line showing all possible solutions. Numbers greater than the variable, go to the right. Numbers less than the variable, go to the left.

­1 ­2 ­3 ­4 ­5 1 2 3 4 5 ­1 ­2 ­3 ­4 ­5 1 2 3 4 5

x ≤ 5 249

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

You try Graph the inequality x > 2 Graph the inequality ­3 > x

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

click 2 on the number line for answer click ­3 on the number line for answer

Pull Pull

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250

Try these. Graph the inequalities.

  • 1. x > ­3

­1 ­2 ­3 ­4 ­5 1 2 3 4 5

  • 2. x < 4

­1 ­2 ­3 ­4 ­5 1 2 3 4 5

251

Try these. State the inequality shown. 1.

­1 ­2 ­3 ­4 ­5 1 2 3 4 5 ­1 ­2 ­3 ­4 ­5 1 2 3 4 5

2.

252

127 This solution set would be x > ­4.

True

False

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

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253

­1 ­2 ­3 ­4 ­5 1 2 3 4 5

128

A

x > 3

B

x < 3

C

x < 3

D

x > 3

State the inequality shown.

254

5 6 7 8 9 10 11 12 13 14 15

129 A

11 < x B 11 > x C 11 > x D 11 < x

State the inequality shown.

255

­1 ­2 ­3 ­4 ­5 1 2 3 4 5

130

A x > ­1 B x < ­1 C x < ­1 D x > ­1

State the inequality shown.

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256

­1 ­5 1 5 ­2 ­3 ­4 2 3 4

131

A ­4 < x B ­4 > x C

­4 < x D ­4 > x

State the inequality shown.

257

­1 ­2 ­3 ­4 ­5 1 2 3 4 5

132

A

x > 0

B

x < 0

C

x < 0

D

x > 0

State the inequality shown.

258 Simple Inequalities Involving Addition and Subtraction

Return to Table of Contents

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259

x + 3 = 13 ­ 3 ­ 3 x = 10 Remembers how to solve an algebraic equation? Does 10 + 3 = 13 13 = 13 Be sure to check your answer! Use the inverse of addition

260

  • Solving one­step inequalities is much like

solving one­step equations.

  • To solve an inequality, you need to isolate

the variable using the properties of inequalities and inverse operations.

  • Remember, whatever you do to one side, you

do to the other.

261

12 > x + 5 ­5 ­5 Subtract to undo addition 7 > x To find the solution, isolate the variable x. Remember, it is isolated when it appears by itself

  • n one side of the equation.
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262

7 > x The symbol is > so it is an open circle and it is numbers less than 7 so it goes to the left.

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

263

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

  • A. j + 7 > ­2

Solve and graph. ­5 is not included in solution set; therefore we graph with an open circle.

  • A. j + 7 > ­2

­7 ­7 j > ­9

264

  • B. r ­ 2 > 4

Solve and graph.

11 10 12 13 14 9 8 7 6 5 4 3 2 1

r ­ 2 > 4 +2 +2 r > 6

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265

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

5 > w ­ 4 ­ 4 9 > w + 4 w < 5

  • C. 9 > w + 4

Solve and graph.

266

133 Solve the inequality.

3 < s + 4 ____ < s

267

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

A

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

B

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

C

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

D 134 Solve the inequality and graph the solution.

­4 + b < ­2

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268

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

A

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

B

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

C

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

D 135 Solve the inequality and graph the solution.

­8 > b ­ 5

269

136 Solve the inequality.

m + 6.4 < 9.6 m < ______

270 Simple Inequalities Involving Multiplication and Division

Return to Table of Contents

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271

Since x is multiplied by 3, divide both sides by 3 for the inverse

  • peration.

Multiplying or Dividing by a Positive Number 3x > ­36 3x > ­36 3 3 x > ­12

272

Solve the inequality. 2 3 r < 4 3 2

( )

r < 6 Since r is multiplied by 2/3, multiply both sides by the reciprocal of 2/3. 2 3 r < 4 3 2

( )

273

137 3k > 18

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

A B C

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

D

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

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

274

138

A B C

­30 > 3q 10 > q ­10 < q ­10 > q

D

10 < q

275

139 X

2 A

B C D

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

< ­3

276

140

A B C

g > 27 g > 36 g > 108 g > 36

D

g > 108

3 4

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

277

141

A B C

­21 > 3d d > ­7 d > ­7 d < ­7

D

d < ­7

278

  • Sometimes you must multiply or divide to

isolate the variable.

  • Multiplying or dividing both sides of an

inequality by a negative number gives a surprising result. Now let's see what happens when we multiply

  • r divide by negative numbers.

279

  • 1. Write down two numbers and put the

appropriate inequality (< or >) between them.

  • 2. Apply each rule to your original two numbers

from step 1 and simplify. Write the correct inequality(< or >) between the answers.

  • A. Add 4
  • B. Subtract 4
  • C. Multiply by 4
  • D. Multiply by ­5
  • E. Divide by 4
  • F. Divide by ­4
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Advanced Unit 3: EXPRESSIONS & EQUATIONS

280

  • 3. What happened with the inequality symbol in

your results?

  • 4. Compare your results with the rest of the

class.

  • 5. What pattern(s) do you notice in the

inequalities? How do different operations affect inequalities? Write a rule for inequalities.

281

Let's see what happens when we multiply this inequality by ­1. 5 > ­1 ­1 • 5 ? ­1 • ­1 ­5 ? 1 ­5 < 1 We know 5 is greater than ­1 Multiply both sides by ­1 Is ­5 less than or greater than 1? You know ­5 is less than 1, so you should use < What happened to the inequality symbol to keep the inequality statement true?

282

The direction of the inequality changes only if the number you are using to multiply or divide by is negative.

Helpful Hint

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

283

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

Dividing each side by ­3 changes the > to <. ­3y > 18 ­3y < 18 ­3 ­3 y < ­6 Solve and graph. A.

284

Divide each side by ­7 Flip the sign because you divided by a negative.

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

­7m < 28 ­7m < 28 ­7 ­7 m > 4 Solve and graph. B.

285

Divide each side by 5. The sign does not change because you did not divide by a negative. 5m > ­25 5m > ­25 5 5 m > ­5 Solve and graph. C.

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

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

286

  • D. ­8y > 32

Solve and graph.

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

  • E. ­9f > ­54

287

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

You multiplied by a negative. ­r 2 < 5 ­2

( )

r > ­10 Multiply both sides by the reciprocal of ­1/2. ­r 2 > 5 ­2

( )

Why did the inequality change?

288

  • 1. ­6h < 42

Try these. Solve and graph each inequality.

  • 2. 4x > ­20

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

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SLIDE 97

Advanced Unit 3: EXPRESSIONS & EQUATIONS

289

  • 3. 5m < 30

Try these. Solve and graph each inequality.

  • 4. > ­3

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

a ­2

290

142

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

Solve and graph. 3y < ­6

291

143

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

Solve and graph.

x ­4 < ­2

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Advanced Unit 3: EXPRESSIONS & EQUATIONS

292

144

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

Solve and graph. ­5y ≤ ­25

293

145

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

Solve and graph.

n ­2 > 2