2-1 properties of numbers pre-algebra lesson 2-1 at the red valley sports camp, 15 kids went...

2-1

Properties of NumbersProperties of NumbersPRE-ALGEBRA LESSON 2-1PRE-ALGEBRA LESSON 2-1

At the Red Valley Sports Camp, 15 kids went horseback riding, 14played tennis, 23 went hiking, and the rest of the campers stayed intheir cabins. If 83 kids were in the camp, how many stayed indoors?

31


(For help, go to Lesson 1-5.)

Simplify.

1. –18 + (–7) 2. 32 – (–3)

3. (–13) + 6 4. 2 – 48

Check Skills You’ll Need

2-1


Solutions

1. –18 + (–7) = –25 2. 32 – (–3) = 32 + 3

= 35

32 – (–3) = 35

3. (–13) + 6 = –7 4. 2 – 48 = 2 + (–48)

= –46

2 – 48 = –46

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?


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.

Quick Check

2-1

Name each property shown.


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

2-1

Quick Check

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


(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.

2-1

Quick Check

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.


2-1

Quick Check

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


(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.

2-1

Quick Check


Name each property shown.

1. d 1 = d

2. 3 + (2 + 5) = (3 + 2) + 5

3. f g = g f

4. 32 + 45 + 102 = 45 + 32 + 102

Commutative Property of Multiplication

Identity Property of Multiplication

Associative Property of Addition

Commutative Property of Addition

2-1

2-2

The Distributive PropertyThe Distributive PropertyPRE-ALGEBRA LESSON 2-2PRE-ALGEBRA LESSON 2-2

Evaluate each expression.

a. 24 – 6 ÷ 2 • 4 + 36 b. 6 • (3 – 1) + 18 c. 10 + 6 • 5 – 4 ÷ 2

48 30 38



Simplify each expression.

1. 3 • 7 – 9 2. (9 – 5)6

3. 8 + 2 • 6 4. 2(6 – 3)

5. 4 • 5 – 4 • 3 6. 3 • 2 – 1 • 2


2-2


Solutions

1. 3 • 7 – 9 2. (9 – 5) • 6

21 – 9 4 • 6

12 24

3. 8 + 2 • 6 4. 2 • (6 – 3)

8 + 12 2 • 3

20 6

5. 4 • 5 – 4 • 3 6. 3 • 2 – 1 • 2

20 – 12 6 – 2

8 4

2-2

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


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.

Quick Check

2-2

Ms. Thomas gave 5 pencils to each of her 37

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

the students?


(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.

2-2

Quick Check

Simplify 11(23) + 11(7).


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

= 11(30) Add within parentheses.

= 330 Multiply.

2-2

Quick Check

Use algebra tiles to multiply 4(3x – 4).

So 4(3x – 4) = 12x – 16.


Model four groups of 3x – 4.

Group like tiles.

2-2

Quick Check

Multiply.


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.

2-2

Quick Check


Use the Distributive Property to simplify.

1. 15(203) 2. 7(180)

3. 6(26) – 6(16) 4. (8a – 4)9

5. 2(14 + y)

60

3,045 1,260

28 + 2y

72a – 36

2-2

2-3

Simplifying Variable ExpressionsSimplifying Variable ExpressionsPRE-ALGEBRA LESSON 2-3PRE-ALGEBRA LESSON 2-3

Use numbers to write the following word expression: eight andone-third minus five and two-sevenths.

8 – 513

27




1. 5(b + 4) 2. –3(2x + 5)

3. 4(–8 – 3q) 4. –6(2b – 7)


2-3


Solutions

1. 5(b + 4) = 5(b) + 5(4) 2. –3(2x + 5) = –3(2x) + (–3)(5) = 5b + 20 = –6x – 15

3. 4(–8 – 3q) = 4(–8) + 4(–3q) 4. –6(2b – 7) = –6(2b) + (–6)(–7)

= –32 – 12q = –12b + 42

2-3

Name the coefficients, the like terms, and the

constants in 7x + y – 2x – 7.


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

Quick Check

2-3

Simplify 9 + 4f + 3 + 2f.


9 + 4f + 3 + 2f 6f + 12

2-3

Quick Check

Simplify 2b + b – 4.


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

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

= 3b – 4 Simplify.

2-3

Quick Check

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.


2-3

Quick Check


Name coefficients, like terms, and constants.

1. 4f – 2f + 3 2. z + 2y – 14


3. 3(a + c – 1) – 2c 4. 4(4v) – 4(v – 9)

3a + c – 3

4, –2; 4f, –2f; 3 1, 2; none; –14

12v + 36

2-3

2-4

Variables and EquationsVariables and EquationsPRE-ALGEBRA LESSON 2-4PRE-ALGEBRA LESSON 2-4

Evaluate 10 – for c = 10.c5

8



Write a variable expression for each phrase.

1. the sum of x and 46 2. four less than g

3. t decreased by five 4. the quotient of z and 26


2-4


Solutions

1. the sum of x and 46 2. four less than g x plus 46 g minus 4x + 46 g – 4

3. t decreased by five 4. the quotient of z and 26t minus 5 z divided by 26t – 5 z

26

2-4

State whether each equation is true, false, or an

open sentence. Explain.


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

Quick Check

2-4

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.


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

2-4

Quick Check

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


120 + x = 75

No, 45 is not a solution of the equation.

120 + 450 75 Substitute 45 for x.

165 75=/

2-4

Quick Check

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?


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 =/

2-4

Quick Check


1. Is b + 3b = 15 true, false, or an open sentence?

2. Write an equation for A number added to seven times the number is the opposite of thirty-two. Is the equation true, false, or an open sentence?

3. Is 25 a solution of the equation 175 – x = 150?

4. Can you stack eight books that are each 3 in. thick inside a box that is 18 in. tall?

yes

open sentence

7x + x = –32; open sentence

no

2-4

2-5

Solving Equations by Adding or SubtractingSolving Equations by Adding or SubtractingPRE-ALGEBRA LESSON 2-5PRE-ALGEBRA LESSON 2-5

Lauralee pitched 12 innings. Maureen pitched 3 fewer innings.How many innings did Maureen pitch?

23

8 13




1. 3 + 4 – 4 2. 7 + 9 – 7

3. 8 – 2 + 2 4. 6 + 2 – 2


2-5


Solutions

1. 3 + 4 – 4 2. 7 + 9 – 7 = 3 + (4 – 4) = 9 + (7 – 7)= 3 + 0 = 9 + 0= 3 = 9

3. 8 – 2 + 2 4. 6 + 2 – 2 = 8 + (2 – 2) = 6 + (2 – 2)= 8 + 0 = 6 + 0= 8 = 6

2-5

Solve y + 5 = 13.


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

Quick Check

2-5

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.


target numberWords plusis 250 number to buy

= number to buy.xLet

Equation 327 + x250=

2-5

(continued)


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

2-5

Quick Check

Solve c – 23 = – 40.


c – 23 = – 40

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

c = –17 Simplify.

2-5

Quick Check

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

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


78 = t – 115

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 = –

2-5

Quick Check


Solve each equation.

1. y + 8 = 12 2. 7 + f – 21 = –20

3. 67 = g – (–36)

4. Ricky rides his bike 12 miles every day. He stops after 7 miles to rest. How much farther does he have to ride?

5 mi

4 –6

31

2-5

2-6

Solving Equations by Multiplying or DividingSolving Equations by Multiplying or DividingPRE-ALGEBRA LESSON 2-6PRE-ALGEBRA LESSON 2-6

0.59521.6

Divide. Round each quotient to the nearest hundredth.

a. b. 10.51262.1

0.372, which rounds to 0.37 5.006, which rounds to 5.01



Simplify each quotient.

1. 2.

3. 4.

1818

21–21

–77

–13–13


2-6


Solutions

1. = 1 2. = –1

3. = –1 4. = 1

1818

21–21

–77

–13–13

2-6

288 pens are boxed by the dozen. How many boxes

are needed?


Let

Equation 288

=

b

number of pensWords times

number of boxes.b

is 12

12

number of boxes

= •

2-6

(continued)


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.

Quick Check

2-6

Solve –2v = –24.


–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

2-6

Quick Check

Solve = – 5.


x = – 40 Simplify.

x8

= – 5x8x8

8 = 8(–5) Multiply each side by 8.

2-6

Quick Check


–6


1. 8x = –48 2. –2x = 18

3. 108 = 9x 4. = 14

5. –6 =

v–3

n4

–9

12 –42

–24

2-6

2-7

Problem Solving Strategy: Guess, Check, ReviseProblem Solving Strategy: Guess, Check, RevisePRE-ALGEBRA LESSON 2-7PRE-ALGEBRA LESSON 2-7

Find three different odd numbers whose sum is 21. List all possibilities.

1, 3, 17; 1, 5, 15; 1, 7, 13; 1, 9, 11; 3, 5, 13; 3, 7, 11; 5, 7, 9



Simplify.

1. 158 + 20 2. 158 + 30

3. 158 + 25 4. 158 + 22

5. In Exercises 1–4, which result came closest to 181?


2-7


Solutions

1. 158 + 20 = 178 2. 158 + 30 = 188

3. 158 + 25 = 183 4. 158 + 22 = 180

5. |181 – 178| = |3| = 3 |181 – 188| = |–7| = 7

|181 – 183| = |–2| = 2 |181 – 180| = |1| = 1

The result of Exercise 4, 180, is the closest to 181.

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?


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

2-7

(continued)


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. Quick Check

2-7


Solve using any strategy.

1. A sporting goods store manager ordered twice as many pairs of basketball shoes as tennis shoes. He ordered 96 pairs in all. How many pairs of each did he order?

2. Manuel wrote a 512-word paper over the weekend. He wrote 180 more words on Sunday than he did on Saturday. How many words

did he write on Saturday?166

32 pairs of tennis shoes; 64 pairs of basketball shoes

2-7

2-8

Inequalities and Their GraphsInequalities and Their GraphsPRE-ALGEBRA LESSON 2-8PRE-ALGEBRA LESSON 2-8

Order from least to greatest: , 0.25, , – 0.5, 0.3,– 212

16

–2, – 0.5, , 0.25, 0.3, 12

16



Graph each set of numbers on a number line. Order the numbers from least to greatest.

1. –3, 7, –9 2. –2, –10, –8

3. 0, 3, –5 4. 3, –6, 10


2-8


Solutions

1. –9, –3, 7

2. –10, –8, –2

3. –5, 0, 3

4. –6, 3, 10

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.


a. x > –2

b. w –5 >–

2-8

(continued)


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. Quick Check

2-8

Write the inequality shown in each graph.


a.

b.

x –3 >–

x < 3

2-8

Quick Check

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.


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

2-8

(continued)


=

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

Equation 35v <–

2-8

Quick Check


x > –30

1. Graph –7 w.

2. Write an inequality for the graph.

3. A child must be at least 50 in. tall to ride on the roller coaster. Write an inequality for this situation.

>–

h 50>–

2-8

Solving One-Step Inequalities by Adding or SubtractingSolving One-Step Inequalities by Adding or SubtractingPRE-ALGEBRA LESSON 2-9PRE-ALGEBRA LESSON 2-9

2-9


a. n + 4 = 3 b. 4 – p = 8

n = –1 p = – 4

PRE-ALGEBRA LESSON 2-9PRE-ALGEBRA LESSON 2-9



1. m + 7 = 5 2. k – 8 = 11

3. 12 + h = 21 4. 6 = n – 23


2-9

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


Solutions

1. 2. m + 7 – 7 = 5 – 7 k – 8 + 8 = 11 + 8

3. 4.12 – 12 + h = 21 – 12

m + 7 = 5

m = – 2

12 + h = 21

h = 9

6 = n – 23

29 = n6 + 23 = n – 23 + 23

k – 8 = 11

k = 19

2-9


Solve each inequality. Graph the solutions.


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.>– <–

Quick Check

2-9


You have 6 GB on your computer. The files you have

stored on the hard drive occupy at least 2 GB. How much storage

space is left for other files?


= 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.<–

No more than 4 GB are left.

2-9


Quick Check

Solve –10 < –13 + q.


–10 < –13 + q

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

3 < q Simplify.

2-9


Quick Check


Solve each inequality.

1. e + 4 14 2. –22 g – 6

3. A number q plus the opposite of 5 is less than or equal to 0.

>–<––16 g>–e 10<–

q 5<–

2-9


2-10

Solving One-Step Inequalities by Multiplying or DividingSolving One-Step Inequalities by Multiplying or DividingPRE-ALGEBRA LESSON 2-10PRE-ALGEBRA LESSON 2-10


a. q ÷ 3 = 6 b. 7r = 21

q = 18 r = 3




1. 6x = 24 2. 63 = –7v

3. = 10 4. = 48 t6

x–2


2-10

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


Solutions

1. 2.

3. 4.

–2 = –2(10)

6x = 24

x = 4

= 10

x = –20

= 48

t = 288

6 = 48(6)

63 = –7v

–9 = v

x–2

6x6

246

63–7

–7v–7

x–2

==

t6

t6

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<–

2-10



(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.

Quick Check

2-10



Solve –2.z–8

<–

–2z

–8<–

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

–8(–2)z

–8–8 >–

Simplify.z 16>–

2-10


Quick Check

>–

Solve each inequality.

1. 3x –27 2. –5w > 15

3. y 4. 0

5. 4f > –12

–x5

<–14

12

<–


w < –3

x 0>–

x –9>–

f > –3

y 2<–

2-10


2-1 properties of numbers pre-algebra lesson 2-1 at the red valley sports camp, 15 kids went...

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