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Algebraic Expressions and the Order of Operations
LESSON 1-1
Course 3
Problem of the Day
The only socks in Jerry’s drawer are 245 black socks and 246 green socks.
How many socks does he have to pull out in order to be sure that he has a
matched pair?
3
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Lesson
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Feature
Algebraic Expressions and the Order of Operations
LESSON 1-1
Course 3
Check Skills You’ll Need
(For help, go to the Skills Handbook page 632)
1. Vocabulary Review Which expression does
not use multiplication: 3 • 4, 3 , 3 x 4, or 3(4)?
4
Multiply.
2. 12 • 8
3. 2.5 • 4
4. 7.4 • 6
5. 0.6 • 5
Check Skills You’ll Need
Lesson
Main
Lesson
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Feature
Algebraic Expressions and the Order of Operations
LESSON 1-1
Course 3
Check Skills You’ll Need
Solutions
1. 3
2. 96
3. 10
4. 44.4
4
5. 3
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Algebraic Expressions and the Order of Operations
LESSON 1-1
Course 3
Additional Examples
A student earns $5 an hour babysitting. The hours vary from
week to week. Define a variable. Write an algebraic expression for how
much the student earns in a week.
Words
$5 an hour times number of hours spent babysitting
Let h = number of hours spent babysitting.
Equation
5
•
h
The algebraic expression 5h represents the student’s weekly earnings.
Quick Check
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Feature
Algebraic Expressions and the Order of Operations
LESSON 1-1
Course 3
Additional Examples
Evaluate p – 23 for p = 10.
p – 23 = 10 – 23
= – 13
Substitute 10 for p.
Simplify by subtracting 23 from 10.
Quick Check
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Feature
Algebraic Expressions and the Order of Operations
LESSON 1-1
Course 3
Additional Examples
Evaluate the expression t + (12 – t ) ÷ 2 for t = 6.
t + (12 – t ) ÷ 2
= 6 + (12 – 6) ÷ 2
Substitute 6 for t.
=6+6÷2
Work within parentheses.
=6+3
Divide.
=9
Add.
Quick Check
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Algebraic Expressions and the Order of Operations
LESSON 1-1
Course 3
Additional Examples
An Internet provider charges $25 for a connection fee and
then $16 per month. Write an expression to model the total cost and
then evaluate the expression for 1 to 5 months of Internet access.
Words
connection fee plus 16 times number of months
Let m = the number of months.
Equation
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25
+
16
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1-1
•
m
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Algebraic Expressions and the Order of Operations
LESSON 1-1
Course 3
Additional Examples
(continued)
m
25 + 16m
Total Cost
1
25 + 16(1)
$41
2
25 + 16(2)
$57
3
25 + 16(3)
$73
4
25 + 16(4)
$89
5
25 + 16(5)
$105
Evaluate the expression for
each value of the variable.
Use the values 1–5 for m.
Quick Check
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Feature
Algebraic Expressions and the Order of Operations
LESSON 1-1
Course 3
Lesson Quiz
1. Write an expression for the number of months in y years.
12y
2. Evaluate your expression in Exercise 1 for 7 years.
84 months
3. Evaluate 5n – n ÷ 2 for n = 10.
45
4. Evaluate 3c – 5 for c = 2, 4, and 6.
1, 7, 13
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Integers and Absolute Value
LESSON 1-2
Course 3
Problem of the Day
CDs take up half of Antancio’s music shelf. Cassette tapes fill up 1 as much
3
space as the CDs. If the shelf is 6 ft long, how much space remains?
2 ft
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Lesson
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Feature
Integers and Absolute Value
LESSON 1-2
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-1.)
1.
Vocabulary Review What is an algebraic expression?
Evaluate each expression for x = 4.
2. 2x + 3
3. 5(x – 1)
4. 7x – 4x
Check Skills You’ll Need
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Main
Lesson
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Feature
Integers and Absolute Value
LESSON 1-2
Course 3
Check Skills You’ll Need
Solutions
1. An algebraic expression is a mathematical phrase that uses numbers,
variables, and operation symbols.
2. 2(4) + 3 = 8 + 3 = 11
3. 5(4 – 1) = 5(3) = 15
4. 7(4) – 4(4) = 28 – 16 = 12
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Integers and Absolute Value
LESSON 1-2
Course 3
Additional Examples
Find each absolute value.
a. |3|
On the number line, 3 is 3 units from 0. This means |3| = 3.
b. |–2|
On the number line, –2 is 2 units from 0. This means |–2| = 2.
Quick Check
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Feature
Integers and Absolute Value
LESSON 1-2
Course 3
Additional Examples
Order –7, 5, and –4 from least to greatest.
Put the integers on the same number line.
The numbers from least to greatest are –7, –4, and 5.
Quick Check
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Integers and Absolute Value
LESSON 1-2
Course 3
Additional Examples
The lowest recorded temperature in Asia is –90ºF;
in North America it is –81ºF. Which continent has the lower
recorded temperature?
| –90| = 90
| –81| = 81
Find the negative integer with the greater absolute value.
The number with the greater absolute value is –90.
Asia has the lower recorded temperature.
Quick Check
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Integers and Absolute Value
LESSON 1-2
Course 3
Additional Examples
Evaluate 5 |z| for z = –2.4
5 |z| = 5 | –2.4|
Substitute –2.4 for z.
= 5(2.4)
Find the absolute value of –2.4.
= 12
Simplify.
Quick Check
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Feature
Integers and Absolute Value
LESSON 1-2
Course 3
Lesson Quiz
Compare. Write <, >, or =.
1. |7| < |–8|
2. |–1| < |–6|
3. |–5| = |5|
4. |–10| > –2
5. 14 = |14|
6. 9 = |–9|
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Feature
Adding and Subtracting Integers
LESSON 1-3
Course 3
Problem of the Day
How many weeks are in 363 days?
51 6 weeks
7
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Main
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Feature
Adding and Subtracting Integers
LESSON 1-3
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-1.)
1. Vocabulary Review How is simplifying an expression
different from evaluating an expression?
Evaluate each expression for x = 7.
2. x + 12
3. x – 5
4. 7x – 11
5. 12 + 9
x
Check Skills You’ll Need
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Feature
Adding and Subtracting Integers
LESSON 1-3
Course 3
Check Skills You’ll Need
Solutions
1. Answers may vary. Sample: When you simplify an expression, you write
its simplest name. When you evaluate an expression, you replace each
variable with a number and then simplify.
2. 7 + 12 = 19
3. 7 – 5 = 2
4. 7(7) – 11 = 49 – 11 = 38
5. 12 + 9 = 21 = 3
7
7
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Adding and Subtracting Integers
LESSON 1-3
Course 3
Additional Examples
Simplify each expression.
a. –7 + (–7)
–7 + (–7) = –14
Both 7s are negative, so the sum is
negative.
b. 8 + (–32)
|–32| – |8| = 24
32 – 8 = 24
8 + (–32) = –24
Find the absolute value of each integer.
Subtract.
|–32| > |8|, so the sum is negative.
Quick Check
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Feature
Adding and Subtracting Integers
LESSON 1-3
Course 3
Additional Examples
Simplify the expression 10 – 17.
10 – 17 = 10 + (–17)
= –7
Add the opposite of 17, which is –17.
Simplify.
Quick Check
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Adding and Subtracting Integers
LESSON 1-3
Course 3
Additional Examples
The elevation of the Dead Sea is 396 m below sea level. A group
of archaeologists leaves the shore of the Dead Sea. They stop after
ascending 500 m. What is their elevation?
Draw a diagram.
The diagram shows that the elevation of the archaeologists is given by
the expression –396 + 500.
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Adding and Subtracting Integers
LESSON 1-3
Course 3
Additional Examples
(continued)
–396 + 500 = 104
Simplify by using the rule for adding
integers with different signs.
The elevation of the archaeologists is 104 m above sea level.
Quick Check
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Feature
Adding and Subtracting Integers
LESSON 1-3
Course 3
Lesson Quiz
Simplify each expression.
1. 11 – 17
–6
4. 15 – 12
3
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2. 8 + (–6)
3. –1 – (–20)
19
2
5. –4 + (–6)
–10
6. 9 – (–2)
11
Lesson
1-3
Feature
Multiplying and Dividing Integers
LESSON 1-4
Course 3
Problem of the Day
How many numbers are in this sequence: 3, 6, 9, . . . 144?
48
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Lesson
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Feature
Multiplying and Dividing Integers
LESSON 1-4
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-3)
1. Vocabulary Review Integers are the set of whole
numbers, their
? , and 0.
Simplify each expression.
2. –3 + (–3)
3. –5 + (–5) + (–5)
4. –2 + (–2) + (–2)
Check Skills You’ll Need
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Lesson
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Feature
Multiplying and Dividing Integers
LESSON 1-4
Course 3
Check Skills You’ll Need
Solutions
1. opposites
2. –6
3. –10 + (–5) = –15
4. –6
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Feature
Multiplying and Dividing Integers
LESSON 1-4
Course 3
Additional Examples
Simplify –11 • 2 • (–1).
–11 and 2 have different signs,
so the product is negative.
–11 • 2 • (–1) = (–22)(–1)
–22 and –1 have the same sign,
so the product is positive.
= 22
Quick Check
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Feature
Multiplying and Dividing Integers
LESSON 1-4
Course 3
Additional Examples
A diver descended 80 ft in 4 min. What was the
rate of this descent?
Represent a descent of 80 feet with –80.
Then divide the descent by the number of minutes to find
the change in elevation per minute.
–80
= –20
4
The quotient of two integers with different
signs is negative.
The diver descended an average of 20 feet per minute.
Quick Check
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Feature
Multiplying and Dividing Integers
LESSON 1-4
Course 3
Additional Examples
Evaluate y ÷ (x – y) – zy for x = 6, y = 9, and z = –2.
y ÷ (x – y) – zy = 9(6 – 9) – (–2)(9)
Substitute 6 for x, 9 for y, and –2 for z.
= 9 ÷ (–3) – (–2)(9)
Simplify 6 – 9 first.
= –3 – (–18)
Multiply and divide.
= 15
Subtract.
Quick Check
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Feature
Multiplying and Dividing Integers
LESSON 1-4
Course 3
Lesson Quiz
Simplify each expression.
1. 7(–4)
2. 10(2)
3. (–3)(6)
–28
20
–18
4. (–5)(–9)(–2)
–90
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5. –21
–3
6. 18
–2
–9
7
Lesson
1-4
Feature
Properties of Numbers
LESSON 1-5
Course 3
Problem of the Day
Divide. Round answers to the nearest tenth.
a. 75 ÷ 8
b. 740 ÷ 7
9.4
105.7
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Feature
Properties of Numbers
LESSON 1-5
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-4.)
1. Vocabulary Review What does it mean to simplify an
expression?
Simplify each expression.
2. 23 + 15 + 73 – 12
3. 5 • 7 + 5 • 13
4. 6 • 7 + (–9) • 7
5. 8(–1) – 8(–5)
Check Skills You’ll Need
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Feature
Properties of Numbers
LESSON 1-5
Course 3
Check Skills You’ll Need
Solutions
1. To simplify an expression means to replace it with its simplest name.
2. 38 + 73 – 12 = 111 – 12 = 99
3. 35 + 65 = 100
4. 42 + (–63) = –21
5. (–8) – (–40) = 32
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Properties of Numbers
LESSON 1-5
Course 3
Additional Examples
Use mental math to simplify the expression.
3.8 + 17 + 6.2
What you think
Look for numbers that are easy to add. The sum of 3.8 and 6.2 is 10.
The sum of 10 and 17 is 27. So, 3.8 + 17 + 6.2 = 27.
Why it works
3.8 + 17 + 6.2 = 3.8 + 6.2 + 17
Commutative Property
= 10 + 17
Order of operations
= 27
Simplify.
Quick Check
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Properties of Numbers
LESSON 1-5
Course 3
Additional Examples
Use mental math to simplify the expression.
89 – 67
What you think
Make the problem easier. Split 89 into parts. First, do 87 – 67 which is 20.
Then add the rest of 89: 2 + 20 = 22.
Why it works
89 – 67 = (2 + 87) – 67
Write 89 as 2 + 87.
= 2 + (87 – 67)
Associative Property
= 2 + 20
Order of operations
= 22
Simplify.
Quick Check
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Properties of Numbers
LESSON 1-5
Course 3
Additional Examples
Use mental math to simplify –2 • 13 • 5.
What you think
It is easy to multiply with multiples of 10. –2 times 5 is –10, a multiple of
10. –10 times 13 is –130. So, –2 • 13 • 5 = –130.
Why it works
–2 • 13 • 5 = –2 • 5 • 13
Commutative Property
= –10 • 13
Order of operations
= –130
Simplify.
Quick Check
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Properties of Numbers
LESSON 1-5
Course 3
Additional Examples
Find each product.
a. 7(t – 5)
7(t – 5) = 7t – 7 • 5
Distributive Property
= 7t – 35
Simplify.
b. (d + 23)(–4)
Distributive Property
(d + 23)(–4) = d(–4) + 23(–4)
= –4d + (–92)
Simplify.
= –4d – 92
Rewrite as a subtraction expression.
Quick Check
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Properties of Numbers
LESSON 1-5
Course 3
Additional Examples
A student buys 11 CDs that cost $6.10 each. How
much will the CDs cost?
11(6.1) = 11 (6 + 0.1)
Replace 6.1 with 6 + 0.1.
= 11(6) + 11(0.1)
Distributive Property
= 66 + 1.1
Multiply.
= 67.1
Add.
The CDs will cost $67.10
Quick Check
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Properties of Numbers
LESSON 1-5
Course 3
Lesson Quiz
Use the Distributive Property to rewrite each expression.
1. 8(a + 2)
8a + 16
2. –2(3x + 1)
–6x – 2
3. 4 • 3 + 6 • 3
(4 + 6)3
4. (6 – r)(–4)
–24 + 4r
5. 5 • 1 + 5 • 9
5(1 + 9)
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Solving Equations by Adding and Subtracting
LESSON 1-6
Course 3
Problem of the Day
How many different ways can the first five letters of the alphabet
be arranged?
120
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Feature
Solving Equations by Adding and Subtracting
LESSON 1-6
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-3.)
1. Vocabulary Review 4 and –4 are additive
?
.
Simplify each expression.
2. –4 + (–7)
3. 12 + (–12)
4. 3 – 10
5. –5 – 1
Check Skills You’ll Need
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Feature
Solving Equations by Adding and Subtracting
LESSON 1-6
Course 3
Check Skills You’ll Need
Solutions
1. inverses
2. –11
4. 3 + (–10) = –7
5. –5 + (–1) = –6
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3. | 12 | – | –12 | = 0; 0
Feature
Solving Equations by Adding and Subtracting
LESSON 1-6
Course 3
Additional Examples
Solve –3 = m – 16.
–3 = m – 16
–3 + 16 = m – 16 + 16
13 = m
Check –3 = m – 16
–3
13 – 16
–3 = –3
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Since 16 is subtracted from m, you
must add 16 to each side to isolate m.
Add 16 to each side.
Simplify.
Check the solution in the original equation.
Substitute 13 for m.
Subtract.
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Quick Check
Feature
Solving Equations by Adding and Subtracting
LESSON 1-6
Course 3
Additional Examples
After Vida adds 17 new shells to her collection, she has a
total of 52 shells. Write and solve an equation to find how many
shells she had before adding the new ones.
Words original number of shells plus new shells = total
Let s = the original number of shells.
Equation
s + 17 = 52
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s
+
17
= 52
Since 17 is added to s, you must
subtract 17 from each side to isolate s.
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Feature
Solving Equations by Adding and Subtracting
LESSON 1-6
Course 3
Additional Examples
(continued)
s + 17 – 17 = 52 – 17
s = 35
Subtract 17 from each side.
Simplify.
Vida had 35 shells before adding the new ones.
Check s + 17 = 52
35 + 17
52
52 = 52
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Check the solution in the original equation.
Substitute 35 for s.
Add.
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Quick Check
Feature
Solving Equations by Adding and Subtracting
LESSON 1-6
Course 3
Lesson Quiz
Solve each equation.
1. 8 = –7 + m
15
2. – 61 + t = 23
84
3. n – 15 = – 12
4. – 82 = r – 36
3
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–46
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Feature
Solving Equations by Multiplying and Dividing
LESSON 1-7
Course 3
Problem of the Day
Find the GCF for each pair of numbers.
a. 12 and 18
6
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b. 15 and 60
15
c. 54 and 60
6
Lesson
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Feature
Solving Equations by Multiplying and Dividing
LESSON 1-7
Course 3
Check Skills You’ll Need
(For help, go to Lesson 1-4.)
1. Vocabulary Review Inverse operations are operations
that ? each other.
Simplify each expression.
2. 6 • 4
3. –7 • 3
4. 10
5. –27
–5
–9
Check Skills You’ll Need
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Lesson
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Feature
Solving Equations by Multiplying and Dividing
LESSON 1-7
Course 3
Check Skills You’ll Need
Solutions
1. undo
2. 24
3. –21
4. –2
5. 3
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Feature
Solving Equations by Multiplying and Dividing
LESSON 1-7
Course 3
Additional Examples
Solve y = 14.
–3
y
= 14
–3
Since y is divided by –3, you must
multiply each side by –3 to isolate y.
y
Multiply each side by –3.
–3 • –3 = –3 • 14
Check
y = –42
Simplify.
y
= 14
–3
Check the solution in the
original equation.
–42
–3
14
14 = 14
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Substitute –42 for y.
Quick Check
Divide.
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Feature
Solving Equations by Multiplying and Dividing
LESSON 1-7
Course 3
Additional Examples
Solve 265 = –5x.
265 = –5x
Since x is multiplied by –5, you must
divide each side by –5 to isolate x.
265
–5x
=
–5
–5
Divide each side by –5.
–53 = x
Simplify.
Check 265 = –5x
265
–5(–53)
265 = 265
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Check the solution in the
original equation.
Substitute –53 for x.
Multiply.
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1-7
Quick Check
Feature
Solving Equations by Multiplying and Dividing
LESSON 1-7
Course 3
Lesson Quiz
Solve each equation.
1. –12w = –60
5
3. 5x = –20
–4
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n
=9
–3
2.
–27
4. –2 =
r
–16
32
Lesson
1-7
Feature