Download Pre-Algebra

Properties of Numbers
Lesson 2-1
Pre-Algebra
Objectives:
1. To identify properties of
addition and multiplication.
2. To use properties to solve
problems.
Properties of Numbers
Lesson 2-1
Pre-Algebra
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.
Properties of Numbers
Lesson 2-1
Additional Examples
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?
You can use the Associative Property of Addition to find the total
cost in two different ways.
42 + (2 + 8) = 42 + 10 = 52 Add 2 and 8 first.
(42 + 2) + 8 = 44 + 8 = 52 Add 42 and 2 first.
Carlos’s total cost was $52.
Pre-Algebra
Properties of Numbers
Lesson 2-1
Additional Examples
Name each property shown.
a. 17 + x + 3 = 17 + 3 + x
Commutative Property of Addition
b. (36  2)10 = 36(2  10)
Associative Property of Multiplication
c. km = km • 1
Identity Property of Multiplication
d. (103 + 26) + 4 = 103 + (26 + 4)
Associative Property of Addition
Pre-Algebra
Properties of Numbers
Lesson 2-1
Pre-Algebra
Additional Examples
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.
Properties of Numbers
Lesson 2-1
Pre-Algebra
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.
Properties of Numbers
Lesson 2-1
Pre-Algebra
Additional Examples
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.
The Distributive Property
Lesson 2-2
Pre-Algebra
Objectives:
1. To use the Distributive Property with
numerical expressions
2. To use the Distributive Property with
algebraic expressions
The Distributive Property
Lesson 2-2
Pre-Algebra
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.
The Distributive Property
Lesson 2-2
Pre-Algebra
Additional Examples
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.
The Distributive Property
Lesson 2-2
Pre-Algebra
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?
(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.
The Distributive Property
Lesson 2-2
Pre-Algebra
Additional Examples
Simplify 11(23) + 11(7).
11(23) + 11(7) = 11(23 + 7)
Use the Distributive Property.
= 11(30)
Add within parentheses.
= 330
Multiply.
The Distributive Property
Lesson 2-2
Pre-Algebra
Additional Examples
Multiply.
a. –9(2 – 8y)
–9(2 – 8y) = –9(2) – (–9)(8y)
Use the Distributive Property.
= –18 – (–72y)
Multiply.
= –18 + 72y
Simplify.
b. (5m + 6)11
(5m + 6)11 = (5m)11 + (6)11 Use the Distributive Property.
= 55m + 66
Multiply.
Simplifying Variable Expressions
Lesson 2-3
Pre-Algebra
Objectives:
1. To identify parts of a variable expression
2. To simplify expressions
Simplifying Variable Expressions
Lesson 2-3
Pre-Algebra
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.
Simplifying Variable Expressions
Lesson 2-3
Pre-Algebra
Additional Examples
Name the coefficients, the like terms, and the
constants in 7x + y – 2x – 7.
Coefficients: 7, 1, –2
Like terms: 7x, –2x
Constant: –7
Simplifying Variable Expressions
Lesson 2-3
Pre-Algebra
Additional Examples
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.
Simplifying Variable Expressions
Lesson 2-3
Pre-Algebra
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 Property
to combine like terms.
= 35 + 5x
Simplify.
Variables and Equations
Lesson 2-4
Pre-Algebra
Objectives:
1. To classify types of equations.
2. To check equations using substitution
Variables and Equations
Lesson 2-4
Pre-Algebra
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
Variables and Equations
Lesson 2-4
Additional Examples
State whether each equation is true, false, or an
open sentence. Explain.
a. 3(b – 8) = 12
open sentence, because there is a variable
b. 7 – (–6) = 1
false, because 13 =/ 1
c. –9 + 5 = – 4
true, because – 4 = – 4
Pre-Algebra
Variables and Equations
Lesson 2-4
Pre-Algebra
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.
Words
Equation
six times the number
added to
the
number
6x
added to
x
is
–42
6x
+
x
=
–42
is
the opposite
of forty-two
The equation is an open sentence, because there is a variable.
Variables and Equations
Lesson 2-4
Pre-Algebra
Additional Examples
Is 45 a solution of the equation 120 + x = 75?
120 + x = 75
120 + 450
75
Substitute 45 for x.
165 =/ 75
No, 45 is not a solution of the equation.
Variables and Equations
Lesson 2-4
Pre-Algebra
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?
Words
weight of
apples
Let
Equation
j
plus
weight of
weight of
plus
cheese
jam
20 lb
=
20
= weight of jam.
9
+
5
+
9 + 5 + j = 20
14 + j = 20
Add.
14 + 7
Substitute 7 for the variable.
20
is
21 =/ 20
No, the jar of jam cannot weigh 7 lb.
j
Solving Equations by Adding or Subtracting
Lesson 2-5
Pre-Algebra
Objectives:
1. To solve one-step equations using subtraction
2. To solve one-step equations using addition
Solving Equations by Adding or Subtracting
Lesson 2-5
Pre-Algebra
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
Solving Equations by Adding or Subtracting
Lesson 2-5
Pre-Algebra
Additional Examples
Solve y + 5 = 13.
Method 1:
y + 5 = 13
y + 5 – 5 = 13 – 5
y=8
Method 2:
Subtract 5
from each side.
Simplify.
Check: y + 5 = 13
8 + 5 13
13 = 13
Replace y with 8.
y + 5 = 13
–5=–5
y=
8
Solving Equations by Adding or Subtracting
Lesson 2-5
Pre-Algebra
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.
Words
target number
Let
Equation
x
327
is
250
plus
number to buy
= number to buy.
=
250
+
x
Solving Equations by Adding or Subtracting
Lesson 2-5
Pre-Algebra
Additional Examples
(continued)
327 = 250 + x
327 = x + 250
327 – 250 = x + 250 – 250
77 = x
Use the Commutative Property of Addition.
Subtract 250 from each side.
Simplify.
Larissa needs to buy 77 more books.
Check: Is the answer reasonable?
250 plus the number of books bought should
be a total collection of 327.
250 + 77 = 327
Solving Equations by Adding or Subtracting
Lesson 2-5
Pre-Algebra
Additional Examples
Solve c – 23 = – 40.
c – 23 = – 40
c – 23 + 23 = – 40 + 23
c = –17
Add 23 to each side.
Simplify.
Solving Equations by Adding or Subtracting
Lesson 2-5
Pre-Algebra
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?
Words
cost of CD player
Let
t
Equation
was
$115
less than
cost of DVD player
= cost of the DVD player.
78
78 = t – 115
78 + 115 = t – 115 + 115
193 = t
Marcy’s DVD player cost $193.
=
t
–
Write an equation.
Add 115 to each side.
Simplify.
115
Solving Equations by Multiplying or Dividing
Lesson 2-6
Pre-Algebra
Objectives:
1. To solve one-step equations using division
2. To solve one-step equations using multiplication
Solving Equations by Multiplying or Dividing
Lesson 2-6
Pre-Algebra
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.
Solving Equations by Multiplying or Dividing
Lesson 2-6
Pre-Algebra
Additional Examples
288 pens are boxed by the dozen. How many boxes
are needed?
Words
number of pens
Let
Equation
b
288
is
12
times
number of boxes
•
b
= number of boxes.
=
12
Solving Equations by Multiplying or Dividing
Lesson 2-6
Pre-Algebra
Additional Examples
(continued)
288 = 12b
288
12b
=
12
12
24 = b
Divide each side by 12.
Simplify.
24 boxes are needed.
Check: Is the answer reasonable?
Twelve times the number of boxes is the number of pens.
Since 12  24 = 288, the answer is reasonable.
Solving Equations by Multiplying or Dividing
Lesson 2-6
Pre-Algebra
Additional Examples
Solve –2v = –24.
–2v = –24
–2v
–24
=
–2
–2
v = 12
Check:
Divide each side by –2.
Simplify.
–2v = –24
–2 • (12) –24
–24 = –24
Replace v with 12.
Solving Equations by Multiplying or Dividing
Lesson 2-6
Pre-Algebra
Additional Examples
Solve x = – 5.
8
x
8 =–5
8 x = 8(–5) Multiply each side by 8.
8
x = – 40
Simplify.
Problem Solving Strategy: Guess, Check, Revise
Lesson 2-7
Pre-Algebra
Objectives:
1.To solve a problem using the Guess, Check, Revise strategy
Problem Solving Strategy: Guess, Check, Revise
Lesson 2-7
Additional Examples
Pre-Algebra
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.
Problem Solving Strategy: Guess, Check, Revise
Lesson 2-7
Pre-Algebra
Additional Examples
(continued)
Popcorn Soda
Price
Price Total Price
$1
$1
79(1) + 96(1) = 79 + 96
= 175
The total is too low. Increase
the price of the popcorn only.
$2
$1
79(2) + 96(1) = 158 + 96
= 254
The total is too low.
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
= 271
The total is correct.
The popcorn price was $1, and the soda price was $2.
Inequalities and Their Graphs
Lesson 2-8
Pre-Algebra
Objectives:
1. To graph inequalities
2. To write inequalities
Inequalities and Their Graphs
Lesson 2-8
Pre-Algebra
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
Inequalities and Their Graphs
Lesson 2-8
Additional Examples
Graph the solutions of each inequality on a
number line.
a. x > –2
An open dot shows that –2 is not a solution.
Shade all the points to the right of –2.
b. w >
– –5
A closed dot shows that –5 is a solution.
Shade all the points to the right of –5.
Pre-Algebra
Inequalities and Their Graphs
Lesson 2-8
Additional Examples
(continued)
c. k <
–4
A closed dot shows that 4 is a solution.
Shade all the points to the left of 4.
d. y < 6
An open dot shows that 6 is not a solution.
Shade all the points to the left of 6.
Pre-Algebra
Inequalities and Their Graphs
Lesson 2-8
Additional Examples
Write the inequality shown in each graph.
a.
x >
– –3
b.
x<3
Pre-Algebra
Inequalities and Their Graphs
Lesson 2-8
Pre-Algebra
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.
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
Inequalities and Their Graphs
Lesson 2-8
Pre-Algebra
Additional Examples
(continued)
Words
a serving of very
low sodium food
Let
Inequality
v
=
v
has at most
35 mg sodium
number of milligrams of sodium in a serving of
very low sodium food.
<
–
35
Solving One-Step Inequalities by Adding or Subtracting
Lesson 2-9
Pre-Algebra
Objectives:
1. To solve one-step inequalities using subtraction
2. To solve one-step inequalities using addition
Solving One-Step Inequalities by Adding or Subtracting
Lesson 2-9
Pre-Algebra
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
Solving One-Step Inequalities by Adding or Subtracting
Lesson 2-9
Pre-Algebra
Additional Examples
Solve each inequality. Graph the solutions.
a. 4 + s < 12
4 + s < 12
4 + s – 4 < 12 – 4
s<8
b.
Subtract 4 from each side.
Simplify.
–16 >
– y – 14
–16 >
– y – 14
–16 + 14 >
– y – 14 + 14
–2 >
– y or y <
– –2
Add 14 to each side.
Simplify.
Solving One-Step Inequalities by Adding or Subtracting
Lesson 2-9
Pre-Algebra
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?
storage space
for our files plus
Words
Let
s
Inequality
2
storage
space left
total
space
<
–
6
= storage space available.
+
s
2+s<
–6
2–2+s<
–6–2
is less than
or equal to
Subtract 2 from each side.
s<
Simplify.
– 4
At most 4 GB are left.
Solving One-Step Inequalities by Adding or Subtracting
Lesson 2-9
Pre-Algebra
Additional Examples
Solve –10 < –13 + q.
–10 < –13 + q
–10 + 13 < –13 + 13 + q
3<q
Add 13 to each side.
Simplify.
Solving One-Step Inequalities by Multiplying or Dividing
Lesson 2-10
Pre-Algebra
Objectives:
1. To solve one-step inequalities
using division
2. To solve one-step inequalities
using multiplication
Solving One-Step Inequalities by Multiplying or Dividing
Lesson 2-10
Pre-Algebra
Tips:
When you multiply or divide each side of an inequality by a negative
number, you must reverse the direction of the inequality symbol.
Solving One-Step Inequalities by Multiplying or Dividing
Lesson 2-10
Pre-Algebra
Additional Examples
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?
Words
is less than
number of
times 225 lb
2,000 lb
or equal to
televisions
Let x = number of televisions.
Inequality
225x <
– 2,000
x
•
225
<
–
2,000
Solving One-Step Inequalities by Multiplying or Dividing
Lesson 2-10
Pre-Algebra
Additional Examples
(continued)
225x < 2,000
225 – 225
x<
– 8.8
Divide each side by 225.
Simplify. Round the answer down to find a whole
number 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.
Solving One-Step Inequalities by Multiplying or Dividing
Lesson 2-10
Pre-Algebra
Additional Examples
Solve z <
–2.
–8 –
z
< –2
–8 –
z
–8 –8 >
– –8(–2)
z>
– 16
Multiply each side by –8 and reverse
the inequality symbol.
Simplify.