Objectives: 1. To identify properties of addition and multiplication.

Pre-AlgebraPre-Algebra

Objectives:1. To identify properties of addition and multiplication.

2. To use properties to solve problems.

Properties of NumbersProperties of Numbers

Lesson 2-1


New Terms: Commutative Properties – changing the order of the values you are adding or multiplying does not change the sum or product.

Associative Properties – Changing the grouping of the values you are adding or multiplying does not change the sum or product.

Additive Identity – when you add a number to 0, the sum equals the original number.

Multiplicative Identity – when you multiply a number to 1, the product equals the original number.


Lesson 2-1


Carlos spent $42 on his golf game. He then bought

a bottle of water for $2 and a chef’s salad for $8. What was

the total cost for his golf game and meal?


Lesson 2-1

42 + (2 + 8) = 42 + 10 = 52 Add 2 and 8 first.

Carlos’s total cost was $52.

You can use the Associative Property of Addition to find the total cost in two different ways.

(42 + 2) + 8 = 44 + 8 = 52 Add 42 and 2 first.

Additional Examples


Name each property shown.


Lesson 2-1

a. 17 + x + 3 = 17 + 3 + x

b. (36 2)10 = 36(2 10)

c. km = km • 1

d. (103 + 26) + 4 = 103 + (26 + 4)

Commutative Property of Addition

Associative Property of Multiplication

Identity Property of Multiplication

Associative Property of Addition

Additional Examples


Use mental math to simplify (48 + 7) + 2.


Lesson 2-1

(48 + 7) + 2

= (7 + 48) + 2 Use the Commutative Property of Addition.

= 7 + (48 + 2) Use the Associative Property of Addition.

= 7 + 50 Add within parentheses.

= 57 Add.

Additional Examples


Suppose you buy school supplies costing $.45,

$.65, and $1.55. Use mental math to find the cost of these

supplies.

0.45 + 0.65 + 1.55

= 0.65 + 0.45 + 1.55 Use the Commutative Property of Addition.

= 0.65 + (0.45 + 1.55) Use the Associative Property of Addition.

= 0.65 + 2.00 Add within parentheses.

= 2.65 Add.

The cost of the school supplies is $2.65.


Lesson 2-1

Additional Examples


Use mental math to simplify (20 • 13) • 5.


Lesson 2-1

(20 • 13) • 5 = (13 • 20) • 5 Use the Commutative Property of Multiplication.

= 13 • (20 • 5) Use the Associative Property of Multiplication.

= 13 • 100 Multiply within parentheses.

= 1,300 Multiply.

Additional Examples


Objectives:1. To use the Distributive Property with numerical expressions

2. To use the Distributive Property with algebraic expressions

The Distributive PropertyThe Distributive Property

Lesson 2-2


New Terms:Distributive Property – to multiply a sum or difference, multiply each number within the parentheses by the number outside the parantheses.

Tips: remember when multiplying by a negative number, the rules for integers still apply.


Lesson 2-2


Use the Distributive Property to find 15(110) mentally.


Lesson 2-2

15(110) = 15(100 + 10) Write 110 as (100 + 10).

= 15 • 100 + 15 • 10 Use the Distributive Property.

= 1,500 + 150 Multiply.

= 1,650 Add.

Additional Examples


Ms. Thomas gave 5 pencils to each of her 37

students. What is the total number of pencils she gave to

the students?


Lesson 2-2

(37)5 = (40 – 3)5 Write 37 as (40 – 3).

= 40 • 5 – 3 • 5 Use the Distributive Property.

= 200 – 15 Multiply.

= 185 Subtract.

Ms. Thomas gave the students 185 pencils.

Additional Examples


Simplify 11(23) + 11(7).


Lesson 2-2

11(23) + 11(7) = 11(23 + 7) Use the Distributive Property.

= 11(30) Add within parentheses.

= 330 Multiply.

Additional Examples


Multiply.


Lesson 2-2

a. –9(2 – 8y)

= –18 – (–72y) Multiply.

= –18 + 72y Simplify.

b. (5m + 6)11

= 55m + 66 Multiply.

–9(2 – 8y) = –9(2) – (–9)(8y) Use the Distributive Property.

(5m + 6)11 = (5m)11 + (6)11 Use the Distributive Property.

Additional Examples


Objectives:1.To identify parts of a variable expression

2. To simplify expressions

Simplifying Variable ExpressionsSimplifying Variable Expressions

Lesson 2-3


New Terms:Term – a number or the product of a number and variable(s)

Constant – a term that has no variable

Like Terms – terms that have identical variables

Coefficient – a number that multiples a variable

Deductive Reasoning – the process of reasoning logically from given facts to a conclusion. As you use properties, rules, and definitions to justify the steps in a problem, you are using deductive reasoning.

Tips: some variable terms have an unwritten coefficient of 1, important to remember when adding like terms.


Lesson 2-3


Name the coefficients, the like terms, and the

constants in 7x + y – 2x – 7.


Lesson 2-3

Coefficients: 7, 1, –2 Like terms: 7x, –2x Constant: –7

Additional Examples


Simplify 2b + b – 4.


Lesson 2-3

2b + b – 4 = 2b + 1b – 4 Use the Identity Property of Multiplication.

= (2 + 1)b – 4 Use the Distributive Property.

= 3b – 4 Simplify.

Additional Examples


Simplify (7 – 3x)5 + 20x.

(7 – 3x)5 + 20x = 35 – 15x + 20x Use the Distributive Property.

= 35 + (–15x + 20x) Use the Associative Property of Addition.

= 35 + (–15 + 20)x Use the Distributive Propertyto combine like terms.

= 35 + 5x Simplify.


Lesson 2-3

Additional Examples


Objectives:1.To classify types of equations.

2. To check equations using substitution

Variables and EquationsVariables and Equations

Lesson 2-4


New Terms:Equation – is a mathematical sentence with an equal sign

Open Sentence – an equation with one or more variables

Solution to an Equation – a value to a variable that make the equation “true”

Tips:

≠ means not equal

The verb “is” indicates an equal sign


Lesson 2-4


State whether each equation is true, false, or an

open sentence. Explain.


Lesson 2-4

a. 3(b – 8) = 12

c. –9 + 5 = – 4

b. 7 – (–6) = 1

open sentence, because there is a variable

=/ false, because 13 1

true, because – 4 = – 4

Additional Examples


Write an equation for Six times a number added to the

number is the opposite of forty-two. State whether the equation is

true, false, or an open sentence. Explain.


Lesson 2-4

The equation is an open sentence, because there is a variable.

Equation 6x + =x –42

six times the numberWords

6x

added to

added to x is

isthe oppositeof forty-two

–42

thenumber

Additional Examples


Is 45 a solution of the equation 120 + x = 75?


Lesson 2-4

120 + x = 75

No, 45 is not a solution of the equation.

120 + 450 75 Substitute 45 for x.

165 75=/

Additional Examples


A gift pack must hold 20 lb of food. Apples weigh 9 lb and

cheese weighs 5 lb. Can the jar of jam that completes the package

weigh 7 lb?


Lesson 2-4

9 + 5 + j = 20

Equation 5 + =j 20

weight of cheese

Words plus

= weight of jam.

is 20 lbweight of

jam

9

j

weight of apples plus

Let

+

14 + j = 20 Add.

14 + 7 20 Substitute 7 for the variable.

No, the jar of jam cannot weigh 7 lb.

21 20 =/

Additional Examples


Objectives:1. To solve one-step equations using subtraction

2. To solve one-step equations using addition

Solving Equations by Adding or SubtractingSolving Equations by Adding or Subtracting

Lesson 2-5


New Terms:Inverse Operations – used to get the variable alone

Tips: The goal to solving any equation is to “isolate” the variable using inverse operations.

You should always simplify both sides of an equation before isolating the variable.

Remember to add or subtract BOTH sides by the same number


Lesson 2-5


Solve y + 5 = 13.


Lesson 2-5

Method 1: Method 2:

y + 5 = 13 Subtract 5

y + 5 – 5 = 13 – 5 from each side.

y = 8 Simplify.

y + 5 = 13

y = 8

Check: y + 5 = 13

8 + 5 13 Replace y with 8.

13 = 13

– 5 = – 5

Additional Examples


Larissa wants to increase the number of books in her

collection to 327 books. She has 250 books now. Find the number

of books she needs to buy.


Lesson 2-5

target numberWords plusis 250 number to buy

= number to buy.xLet

Equation 327 + x250=

Additional Examples


(continued)


Lesson 2-5

327 = 250 + x

Larissa needs to buy 77 more books.

327 = x + 250 Use the Commutative Property of Addition.

327 – 250 = x + 250 – 250 Subtract 250 from each side.

77 = x Simplify.

Check: Is the answer reasonable?250 plus the number of books bought should be a total collection of 327.

250 + 77 = 327

Additional Examples


Solve c – 23 = – 40.


Lesson 2-5

c – 23 = – 40

c – 23 + 23 = – 40 + 23 Add 23 to each side.

c = –17 Simplify.

Additional Examples


Marcy’s CD player cost $115 less than her DVD player. Her

CD player cost $78. How much did her DVD player cost?


Lesson 2-5

78 = t – 115 Write an equation.

78 + 115 = t – 115 + 115 Add 115 to each side.

193 = t Simplify.

Marcy’s DVD player cost $193.

cost of CD player $115 cost of DVD playerWords less thanwas

t = cost of the DVD player.Let

t 115Equation 78 = –

Additional Examples


Objectives:1.To solve one-step equations using division

2. To solve one-step equations using multiplication

Solving Equations by Multiplying or DividingSolving Equations by Multiplying or Dividing

Lesson 2-6


Tips:The division property of equality suggests you can divide each side of an equation by the same nonzero number. Divisors are restricted to nonzero values because division by zero is undefined.

Remember to multiply or divide BOTH sides by the same number.


Lesson 2-6


288 pens are boxed by the dozen. How many boxes

are needed?


Lesson 2-6

Let

Equation 288

=

b

number of pensWords times

number of boxes.b

is 12

12

number of boxes

= •

Additional Examples


(continued)


Lesson 2-6

288 = 12b

24 = b Simplify.

24 boxes are needed.

Divide each side by 12.28812

12b12

=

Check: Is the answer reasonable? Twelve times the number of boxes is the number of pens. Since 12 24 = 288, the answer is reasonable.

Additional Examples


Solve –2v = –24.


Lesson 2-6

–2v = –24

v = 12 Simplify.

Divide each side by –2.–2v–2

–24–2

=

Check: –2v = –24–2 • (12) –24 Replace v with 12.

–24 = –24

Additional Examples


Solve = – 5.


Lesson 2-6

x = – 40 Simplify.

x8

= – 5x8x8

8 = 8(–5) Multiply each side by 8.

Additional Examples


Objectives:1.To solve a problem using the Guess, Check, Revise strategy

Problem Solving Strategy: Guess, Check, ReviseProblem Solving Strategy: Guess, Check, Revise

Lesson 2-7


During the intermission of the play, the Theater Club sold cups

of popcorn and soda. The club sold 79 cups of popcorn and 96 sodas for

a total of $271. What was the selling price of a cup of popcorn? Of a

soda?


Lesson 2-7

You can organize conjectures in a table. As a first conjecture, try both with a price of $1.

Additional Examples


(continued)


Lesson 2-7

Popcorn SodaPrice Price Total Price

$1 $1 79(1) + 96(1) = 79 + 96 The total is too low. Increase = 175 the price of the popcorn only.

$2 $1 79(2) + 96(1) = 158 + 96 The total is too low. = 254 Increase the price of the soda.

$2 $2 79(2) + 96(2) = 158 + 192 The total is too high. = 350 Decrease the price of the popcorn.

$1 $2 79(1) + 96(2) = 79 + 192 The total is correct. = 271

The popcorn price was $1, and the soda price was $2.

Additional Examples


Objectives:1.To graph inequalities

2. To write inequalities

Inequalities and Their GraphsInequalities and Their Graphs

Lesson 2-8


New Terms:Inequality – a mathematical sentence that contains <,>,≤,≥, or ≠.

Solution to an Inequality – any number that makes the inequality true.

Tips: Know what the different signs mean, and we do not have gators or crocodiles in this class

When reading the solution to an inequality, you should always start with the varible


Lesson 2-8


An open dot shows that –2 is not a solution.

A closed dot shows that –5 is a solution.

Shade all the points to the right of –5.

Shade all the points to the right of –2.

Graph the solutions of each inequality on a number

line.


Lesson 2-8

a. x > –2

b. w –5 >–

Additional Examples


(continued)


Lesson 2-8

c. k 4

d. y < 6

<– A closed dot shows that 4 is a solution.

Shade all the points to the left of 4.

An open dot shows that 6 is not a solution.

Shade all the points to the left of 6.

Additional Examples


Write the inequality shown in each graph.


Lesson 2-8

a.

b.

x –3 >–

x < 3

Additional Examples


Food can be labeled very low sodium only if it meets the requirement established by the federal government. Use the table to write an inequality for this requirement.


Lesson 2-8

Label Definition

Sodium-free food Less than 5 mg per serving

Very low sodium food At most 35 mg per serving

Low-sodium food At most 140 mg per serving

Additional Examples


(continued)


Lesson 2-8

=

35 mg sodiumWords

number of milligrams of sodium in a serving ofvery low sodium food.

v

has at mosta serving of very low sodium food

Let

Inequality 35v <–

Additional Examples


Objectives:1. To solve one-step inequalities using subtraction

2. To solve one-step inequalities using addition

Lesson 2-9

Solving One-Step Inequalities by Adding or SubtractingSolving One-Step Inequalities by Adding or Subtracting


Tips:Just like the equality properties, you must add or subtract the same number from each side

When rewriting an inequality in reverse order, you must pay attention to the direction of the inequality symbol

Lesson 2-9



Solve each inequality. Graph the solutions.

Lesson 2-9

a.

b.

4 + s – 4 < 12 – 4 Subtract 4 from each side.

4 + s < 12

s < 8 Simplify.

4 + s < 12

–16 y – 14 >––16 y – 14 > –

–16 + 14 y – 14 + 14 Add 14 to each side.>– –2 y or y –2 Simplify.>– <–


Additional Examples


Suppose your computer’s hard drive has a capacity of 6

gigabytes (GB). The files you have stored on the hard drive occupy

at least 2 GB. How much storage space is left for other files?

Lesson 2-9

= storage space available.sLet

storage space for our filesWords

is less than or equal toplus

totalspace

storage space left

Inequality 2 s 6+ <–

2 + s 6<–

2 – 2 + s 6 – 2 Subtract 2 from each side.<–

s 4 Simplify.<–

At most 4 GB are left.


Additional Examples


Solve –10 < –13 + q.

Lesson 2-9

–10 < –13 + q

–10 + 13 < –13 + 13 + q Add 13 to each side.

3 < q Simplify.


Additional Examples


Objectives:1.To solve one-step inequalities using division

2. To solve one-step inequalities using multiplication

Lesson 2-10

Solving One-Step Inequalities by Multiplying or DividingSolving One-Step Inequalities by Multiplying or Dividing


Tips:When you multiply or divide each side of an inequality by a negative number, you must reverse the direction of the inequality symbol.

Lesson 2-10


Pre-AlgebraPre-AlgebraLesson 2-10

A 1-ton truck has the ability to haul 1 ton, or 2,000 lb.

At most, how many television sets can the truck carry if each

TV set weighs 225 lb?

225x 2,000<–

Let x = number of televisions.

Inequality

Wordsnumber oftelevisions

225 lb 2,000 lbtimesis less thanor equal to

x • 225 2,000<–


Additional Examples


(continued)

Divide each side by 225.<–2,000225

225x225

x 8.8 Simplify. Round the answer down to find a wholenumber of television sets.

<–

At most, the truck can carry 8 television sets.

Check: Is the answer reasonable? The total weight of 8 television sets is 8(225) = 1,800 lb, which is less than 2,000 lb but so close that another television set could not be carried. The answer is reasonable.


Additional Examples


Solve –2.z–8

<–

–2z

–8<–

Multiply each side by –8 and reverse the inequality symbol.

–8(–2)z

–8–8 >–

Simplify.z 16>–


Additional Examples

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Objectives: 1. To identify properties of addition and multiplication.

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