Transcript 1-9 - s3.amazonaws.com

1-9
1-9 Simplifying
SimplifyingAlgebraic
AlgebraicExpressions
Expressions
Warm Up
Problem of the Day
Lesson Presentation
Course
Course
22
1-9 Simplifying Algebraic Expressions
Warm Up
Evaluate each expression for y = 3.
1. 3y + y
12
2. 7y
21
3. 10y – 4y
18
4. 9y
27
5. y + 5y + 6y
36
6. 10y
30
Course 2
1-9 Simplifying Algebraic Expressions
Problem of the Day
Emilia saved nickels, dimes, and quarters in
a jar. She had as many quarters as dimes,
but twice as many nickels as dimes. If the
jar had 844 coins, how much money had
she saved?
$94.95
Course 2
1-9 Simplifying Algebraic Expressions
Learn to simplify algebraic expressions.
Course 2
1-9 Simplifying Algebraic Expressions
Vocabulary
term
coefficient
Course 2
1-9 Simplifying Algebraic Expressions
In the expression 7x + 9y + 15, 7x, 9y, and 15 are
called terms. A term can be a number, a variable,
or a product of numbers and variables. Terms in an
expression are separated by + and –.
7x + 5 – 3y2 + y + x
3
term
term
term
term term
In the term 7x, 7 is called the Coefficient
coefficient. A coefficient is a
number that is multiplied by a
variable in an algebraic
expression. A variable by itself,
like y, has a coefficient of 1.
So y = 1y.
Course 2
Variable
1-9 Simplifying Algebraic Expressions
Like terms are terms with the same variable
raised to the same power. The coefficients do
not have to be the same. Constants, like 5, 1
,
2
and 3.2, are also like terms.
Like Terms
Unlike
Terms
Course 2
w and w
5 and 1.8
7
5x2 and 2x
6a and 6b
3.2 and n
The exponents The variables Only one term
contains a
are different. are different
variable
3x and 2x
1-9 Simplifying Algebraic Expressions
Additional Example 1: Identifying Like Terms
Identify like terms in the list.
3t
5w2
7t
9v
4w2
8v
Look for like variables with like powers.
3t
5w2
7t
9v
Like terms: 3t and 7t
4w2
8v
5w2 and 4w2
9v and 8v
Helpful Hint
Use different shapes or colors to indicate sets of
like terms.
Course 2
1-9 Simplifying Algebraic Expressions
Check It Out: Example 1
Identify like terms in the list.
2x
4y3
8x
5z
5y3
8z
Look for like variables with like powers.
2x
4y3
8x
5z
Like terms: 2x and 8x
Course 2
5y3
8z
4y3 and 5y3
5z and 8z
1-9 Simplifying Algebraic Expressions
Combining like terms is like grouping similar objects.
x
x
x
x
4x
+
+
x
x
x
x
5x
x
=
=
x
x
x
x
x
x
x
x
x
9x
To combine like terms that have variables, add or
subtract the coefficients.
Course 2
1-9 Simplifying Algebraic Expressions
Additional Example 2: Simplifying Algebraic
Expressions
Simplify. Justify your steps using the
Commutative, Associative, and Distributive
Properties when necessary.
A. 6t – 4t
6t – 4t
6t and 4t are like terms.
2t
Subtract the coefficients.
B. 45x – 37y + 87
In this expression, there are no like terms
to combine.
Course 2
1-9 Simplifying Algebraic Expressions
Additional Example 2: Simplifying Algebraic
Expressions
Simplify. Justify your steps using the
Commutative, Associative, and Distributive
Properties when necessary.
C. 3a2 + 5b + 11b2 – 4b + 2a2 – 6
3a2 + 5b + 11b2 – 4b + 2a2 – 6
(3a2 + 2a2) + (5b – 4b) + 11b2 – 6
5a2 + b + 11b2 – 6
Course 2
Identify like
terms.
Group like
terms.
Add or subtract
the coefficients.
1-9 Simplifying Algebraic Expressions
Check It Out: Example 2
Simplify. Justify your steps using the
Commutative, Associative, and Distributive
Properties when necessary.
A. 5y + 3y
5y + 3y
8y
5y and 3y are like terms.
Add the coefficients.
B. 2(x2 – 13x) + 6
2x 2 – 26x + 6
Distributive Property.
There are no like terms to combine.
Course 2
1-9 Simplifying Algebraic Expressions
Check It Out: Example 2
Simplify. Justify your steps using the
Commutative, Associative, and Distributive
Properties when necessary.
C. 4x2 + 4y + 3x2 – 4y + 2x2 + 5
4x2 + 4y + 3x2 – 4y + 2x2 + 5
Identify like
terms.
(4x2 + 3x2 + 2x2)+ (4y – 4y) + 5
Group like
terms.
9x2 + 5
Course 2
Add or subtract
the coefficients.
1-9 Simplifying Algebraic Expressions
Additional Example 3: Geometry Application
Write an expression for the perimeter of the
triangle. Then simplify the expression.
2x + 3
3x + 2
x
Write an expression using
the side lengths.
(x + 3x + 2x) + (2 + 3) Identify and group like
terms.
6x + 5
Add the coefficients.
2x + 3 + 3x + 2 + x
Course 2
1-9 Simplifying Algebraic Expressions
Check It Out: Example 3
Write an expression for the perimeter of the
triangle. Then simplify the expression.
2x + 1
2x + 1
x
Write an expression using
the side lengths.
(x + 2x + 2x) + (1 + 1) Identify and group like
terms.
5x + 2
Add the coefficients.
x + 2x + 1 + 2x + 1
Course 2
1-9 Simplifying Algebraic Expressions
Lesson Quiz: Part I
Identify like terms in the list.
1. 3n2 5n 2n3 8n
2. a5 2a2
a3
3a
5n, 8n
4a2
2a2, 4a2
Simplify. Justify your steps using the
Commutative, Associative, and Distributive
Properties when necessary.
3. 4a + 3b + 2a
6a + 3b
4. x2 + 2y + 8x2
9x2 + 2y
Course 2
1-9 Simplifying Algebraic Expressions
Lesson Quiz: Part II
5. Write an expression for the perimeter of
the given figure.
2x + 3y
x+y
x+y
2x + 3y
6x + 8y
Course 2
1-9
1-10Simplifying
Equations Algebraic
and Their Expressions
Solutions
Warm Up
Problem of the Day
Lesson Presentation
Course
Course
22
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Warm Up
Evaluate each expression for x = 12.
1. x + 2
2. x
4
3. x – 8
14
3
4
4. 10x – 4
116
5. 2x + 12
36
67
6. 5x + 7
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Problem of the Day
Alicia buys buttons at a cost of 8 for
$20. She resells them for $5 each.
How many buttons does Alicia need
to sell for a profit of $120?
48 buttons
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Learn to determine whether a
number is a solution of an equation.
Course 2
1-10
1-9 Equations
Insert Lesson
Title
Simplifying
and
Algebraic
TheirHere
Solutions
Expressions
Vocabulary
equation
solution
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Ella has 22 CDs. This is 9 more than her
friend Kay has.
This situation can be written as an equation.
An equation is a mathematical statement
that two expressions are equal in value.
An equation is like a balanced scale.
Number of is equal 9 more than
CDs Ella has
to
Kay has
22
=
j+9
Left expression
Course 2
Right expression
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Just as the weights on both sides of a
balanced scale are exactly the same, the
expressions on both sides of an equation
represent exactly the same value.
When an equation contains a variable, a
value of the variable that makes the
statement true is called a solution of the
equation.
22 = j + 9 j = 13 is a solution because 22 = 13
+ 9.
22 = j + 9 j = 15 is not a solution because 22  15
+ 9.
Reading Math
The symbol ≠ means “is not equal to.”
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Additional Example 1A: Determining Whether a
Number is a Solution of an Equation
Determine whether the given value of the
variable is a solution of t + 9 = 17.
26
t + 9 = 17
?
26 + 9 = 17
Substitute 26 for t.
?
35 = 17 
26 is not a solution of t + 9 = 17.
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Additional Example 1B: Determining Whether a
Number is a Solution of an Equation
Determine whether the given value of the
variable is a solution of t + 9 = 17.
8
t + 9 = 17
?
8 + 9 = 17
Substitute 8 for t.
?
17 = 17 
8 is a solution of t + 9 = 17.
Course 2
1-10
1-9 Insert
Lesson
Title
Simplifying
Equations
and
Algebraic
TheirHere
Solutions
Expressions
Check It Out: Example 1
Determine whether each number is a solution
of x – 5 = 12.
A. 22
x – 5 = 12
?
22 – 5 = 12
?
Substitute 22 for x.
17 = 12
22 is not a solution of x – 5 = 12.
B. 8
x – 5 = 12
?
8 – 5 = 12
?
Substitute 8 for x.
3 = 12 
8 is not a solution of x – 5 = 12.
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Additional Example 2: Writing an Equation to
Determine Whether a Number is a Solution
Mrs. Jenkins had $32 when she returned home
from the supermarket. If she spent $17 at the
supermarket, did she have $52 or $49 before
she went shopping?
You can write an equation to find the amount
of money Mrs. Jenkins had before she went
shopping. If m represents the amount of
money she had before she went shopping,
then m - 17 = 32.
$52
m – 17 = 32
?
52 - 17 = 32 Substitute 52 for m.
?
35 = 32 
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Additional Example 2 Continued
Mrs. Jenkins had $32 when she returned home
from the supermarket. If she spent $17 at the
supermarket, did she have $52 or $49 before
she went shopping?
You can write an equation to find the amount
of money Mrs. Jenkins had before she went
shopping. If m represents the amount of
money she had before she went shopping,
$49 m - 17 = 32.
then
m – 17 = 32
?
49 - 17 = 32 Substitute 49 for m.
?
32 = 32 
Mrs. Jenkins had $49 before she went shopping.
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Check it Out: Additional Example 2
Mr. Rorke had $12 when he returned home
from buying a hat. If he spent $47 at the hat
store, did he have $61 or $59 before he
bought the hat?
You can write an equation to find the amount
of money Mr. Rorke had before he purchased a
hat. If m represents the amount of money he
had before he purchased a hat, then m – 47 =
12.
$61
m – 47 = 12
?
61 - 47 = 12 Substitute 61 for h.
?
14 = 12
Course 2

1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Check it Out: Additional Example 2 Continued
Mr. Rorke had $12 when he returned home
from buying a hat. If he spent $47 at the hat
store, did he have $59 or $61 before he
bought the hat?
You can write an equation to find the amount
of money Mr. Rorke had before he purchased a
hat. If m represents the amount of money he
had before he purchased a hat, then m – 47 =
12.
$59
m – 47 = 12
?
59 - 47 = 12 Substitute 59 for h.
?
12 = 12 
Mr. Rorke had $59 before he purchased a hat.
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Additional Example 3: Deriving a Real-World
Situation from an Equation
Which problem situation best matches the
equation 5 + 2x = 13?
Situation A:
Admission to the county fair costs $5 and
rides cost $2 each. Mike spent a total of
$13. How many rides did he go on?
$5 for admission
$2 per ride
5+
2x
Mike spent $13 in all, so 5 + 2x = 13.
Situation A matches the equation.
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Additional Example 3 Continued
Which problem situation best matches the
equation 5 + 2x = 13?
Situation B:
Admission to the county fair costs $2 and
rides cost $5 each. Mike spent a total of $13.
How many rides did he go on?
The variable x represents the number of
rides that Mike bought.
$5 per ride
5x
Since 5x is not a term in the given equation,
Situation B does not match the equation.
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Check It Out: Additional Example 3
Which problem situation best matches the
equation 13 + 4x = 25?
Situation A:
Admission to the baseball game costs $4 and
souvenir hats cost $13 each. Trina spent a total
of $25. How many souvenir hats did she buy?
The variable x represents the number of
souvenir hats Trina bought.
$13 per souvenir hat
13x
Since 13x is not a term in the given equation,
Situation A does not match the equation.
Course 2
1-10
1-9 Equations
Simplifyingand
Algebraic
Their Solutions
Expressions
Check It Out: Additional Example 3 Continued
Which problem situation best matches the
equation 13 + 4x = 25?
Situation B:
Admission to the baseball game costs $13 and
souvenir hats cost $4 each. Trina spent a total
of $25. How many souvenir hats did she buy?
$13 for admission
13 +
$4 per souvenir hat
4x
Trina spent $25 in all, so 13 + 4x = 25.
Situation B matches the equation.
Course 2
1-10
1-9 Equations
Simplifying
Insert Lesson
and
Algebraic
Title
TheirHere
Solutions
Expressions
Lesson Quiz
Determine whether the given value of the
variable is a solution of 5 + x = 47.
1. x = 42 yes
2. x = 52 no
Determine whether the given value of the
variable is a solution of 57 – y = 18.
3. y = 75 no
4. y = 39 yes
5. Kwan has 14 marbles. This is 7 more than
Drue has. Does Drue have 21 or 7 marbles?
7
Course 2
1-9
Algebraic
Expressions
Rates, Ratios,
and Proportions
2-6 Simplifying
Warm Up
Lesson Presentation
Lesson Quiz
Course
2
Holt
Algebra
1
1-9 Simplifying Algebraic Expressions
Warm Up
Solve each equation. Check your answer.
1. 6x = 36 6
2.
48
3. 5m = 18 3.6
4.
–63
5. 8y =18.4 2.3
Multiply.
6.
7
Course 2
7.
10
1-9 Simplifying Algebraic Expressions
Objectives
Write and use ratios, rates, and unit rates.
Write and solve proportions.
Course 2
1-9 Simplifying Algebraic Expressions
Vocabulary
ratio
rate
scale
unit rate
conversion factor
Course 2
proportion
cross products
scale drawing
scale model
1-9 Simplifying Algebraic Expressions
A ratio is a comparison of two quantities by
division. The ratio of a to b can be written a:b
or , where b ≠ 0. Ratios that name the same
comparison are said to be equivalent.
A statement that two ratios are equivalent, such
as
, is called a proportion.
Course 2
1-9 Simplifying Algebraic Expressions
Example 2: Finding Unit Rates
Raulf Laue of Germany flipped a pancake 416
times in 120 seconds to set the world record.
Find the unit rate. Round your answer to the
nearest hundredth.
Write a proportion to find an equivalent
ratio with a second quantity of 1.
Divide on the left side to find x.
The unit rate is about 3.47 flips/s.
Course 2
1-9 Simplifying Algebraic Expressions
Check It Out! Example 2
Cory earns $52.50 in 7 hours. Find the unit
rate.
Write a proportion to find an equivalent
ratio with a second quantity of 1.
Divide on the left side to find x.
The unit rate is $7.50.
Course 2
1-9 Simplifying Algebraic Expressions
A rate such as
in which the two quantities
are equal but use different units, is called a
conversion factor. To convert a rate from one
set of units to another, multiply by a conversion
factor.
Course 2
1-9 Simplifying Algebraic Expressions
Example 3A: Converting Rates
Serena ran a race at a rate of 10 kilometers
per hour. What was her speed in kilometers
per minute? Round your answer to the
nearest hundredth.
To convert the second quantity in a
rate, multiply by a conversion
factor with that unit in the first
quantity.
The rate is about 0.17 kilometer per minute.
Course 2
1-9 Simplifying Algebraic Expressions
Helpful Hint
In example 3A , “1 hr” appears to divide out,
leaving “kilometers per minute,” which are
the units asked for. Use this strategy of
“dividing out” units when converting rates.
Course 2
1-9 Simplifying Algebraic Expressions
Example 3B: Converting Rates
A cheetah can run at a rate of 60 miles per
hour in short bursts. What is this speed in
feet per minute?
Step 1
2 Convert the speed to feet per hour.
minute.
To convert the first quantity in a
rate, multiply by a conversion
factor with that unit in the second
first
quantity.
316,800
hour.
The speed is 5280
feetfeet
perper
minute.
Course 2
1-9 Simplifying Algebraic Expressions
Example 3B: Converting Rates
The speed is 5280 feet per minute.
Check that the answer is reasonable.
• There are 60 min in 1 h, so 5280 ft/min is
60(5280) = 316,800 ft/h.
• There are 5280 ft in 1 mi, so 316,800 ft/h
is
rate in the problem.
Course 2
This is the given
1-9 Simplifying Algebraic Expressions
Check It Out! Example 3
A cyclist travels 56 miles in 4 hours. What is the
cyclist’s speed in feet per second? Round your
answer to the nearest tenth, and show that your
answer is reasonable.
Step 1 Convert the speed to feet per hour.
Change to miles in 1 hour.
To convert the first quantity in a
rate, multiply by a conversion
factor with that unit in the
second quantity.
The speed is 73,920 feet per hour.
Course 2
1-9 Simplifying Algebraic Expressions
Check It Out! Example 3
Step 2 Convert the speed to feet per minute.
To convert the second quantity in a
rate, multiply by a conversion
factor with that unit in the first
quantity.
The speed is 1232 feet per minute.
Step 3 Convert the speed to feet per second.
To convert the second quantity in a
rate, multiply by a conversion
factor with that unit in the first
quantity.
The speed is approximately 20.5 feet per second.
Course 2
1-9 Simplifying Algebraic Expressions
Check It Out! Example 3
Check that the answer is reasonable. The answer
is about 20 feet per second.
• There are 60 seconds in a minute so 60(20)
= 1200 feet in a minute.
• There are 60 minutes in an hour so 60(1200)
= 72,000 feet in an hour.
• Since there are 5,280 feet in a mile 72,000 ÷
5,280 = about 14 miles in an hour.
• The cyclist rode for 4 hours so 4(14) = about
56 miles which is the original distance traveled.
Course 2
1-9 Simplifying Algebraic Expressions
In the proportion
, the products a • d and
b • c are called cross products. You can solve
a proportion for a missing value by using the
Cross Products property.
Cross Products Property
WORDS
In a proportion, cross
products are equal.
Course 2
ALGEBRA
NUMBERS
If
2•6=3•4
and b ≠ 0
and d ≠ 0
then ad = bc.
1-9 Simplifying Algebraic Expressions
Example 4: Solving Proportions
Solve each proportion.
A.
B.
Use cross
products.
Use cross
products.
3(m) = 5(9)
3m = 45
Divide both
sides by 3.
m = 15
Course 2
6(7) = 2(y – 3)
42 = 2y – 6
+6
+6 Add 6 to
both sides.
48 = 2y
24 = y
Divide both
sides by 2.
1-9 Simplifying Algebraic Expressions
Check It Out! Example 4
Solve each proportion.
A.
B.
Use cross
products.
2(y) = –5(8)
2y = –40
Divide both
sides by 2.
4(g +3) = 5(7)
4g +12 = 35
–12 –12
4g
= 23
y = −20
g = 5.75
Course 2
Use cross
products.
Subtract 12
from both
sides.
Divide both
sides by 4.
1-9 Simplifying Algebraic Expressions
A scale is a ratio between two sets of measurements,
such as 1 in:5 mi. A scale drawing or scale model
uses a scale to represent an object as smaller or
larger than the actual object. A map is an example of
a scale drawing.
Course 2
1-9 Simplifying Algebraic Expressions
Example 5A: Scale Drawings and Scale Models
A contractor has a blueprint for a house
drawn to the scale 1 in: 3 ft.
A wall on the blueprint is 6.5 inches long.
How long is the actual wall?
blueprint
actual
1 in.
3 ft.
Write the scale as a fraction.
Let x be the actual length.
x • 1 = 3(6.5)
Use the cross products to solve.
x = 19.5
The actual length of the wall is 19.5 feet.
Course 2
1-9 Simplifying Algebraic Expressions
Example 5B: Scale Drawings and Scale Models
A contractor has a blueprint for a house
drawn to the scale 1 in: 3 ft.
One wall of the house will be 12 feet long when
it is built. How long is the wall on the blueprint?
blueprint
actual
1 in.
3 ft.
Write the scale as a fraction.
Let x be the actual length.
12 = 3x
Use the cross products to solve.
Since x is multiplied by 3, divide
both sides by 3 to undo the
multiplication.
4=x
The wall on the blueprint is 4 inches long.
Course 2
1-9 Simplifying Algebraic Expressions
Check It Out! Example 5
A scale model of a human heart is 16 ft. long.
The scale is 32:1. How many inches long is
the actual heart it represents?
model
actual
32x = 192
32 in.
1 in.
Write the scale as a fraction.
Convert 16 ft to inches.
Let x be the actual length.
Use the cross products to solve.
Since x is multiplied by 32, divide
both sides by 32 to undo the
multiplication.
x=6
The actual heart is 6 inches long.
Course 2
1-9 Simplifying Algebraic Expressions
Lesson Quiz: Part 1
1. In a school, the ratio of boys to girls is 4:3.
There are 216 boys. How many girls are there?
162
Find each unit rate. Round to the nearest
hundredth if necessary.
2. Nuts cost $10.75 for 3 pounds. $3.58/lb
3. Sue washes 25 cars in 5 hours. 5 cars/h
4. A car travels 180 miles in 4 hours. What is the
car’s speed in feet per minute? 3960 ft/min
Course 2
1-9 Simplifying Algebraic Expressions
Lesson Quiz: Part 2
Solve each proportion.
5.
6.
6
16
7. A scale model of a car is 9 inches long. The
scale is 1:18. How many inches long is the car
it represents? 162 in.
Course 2
1-9 Simplifying Algebraic Expressions
SMUE DIRECTIONS
For Smue question 18,19,20,and21,
• 18,write the response and tell how you got your
answer
• 19,write the answer in expand if you has a
answer in numberic form
• 20,21 compare and contrast the number
Course 2
1-9 Simplifying Algebraic Expressions
Question 18
Question 18
Course 2