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chapter 11 Equvalent Expressons and Equvalent EquatonsEquvalent Expressons and Equvalent Equatons | 549

Chapter

11

Equialent Expressions andEquialent Equations

Chapter Objecties

Use properties to generate equivalent expressions. Identify the terms of an expression and use the correct order of operations

when evaluating expressions.

Understand the concept of equivalent equations and know that equals added toequals are equal, and equals multiplied by equals are equal.

Solve simple equations by multiplying both sides by the same factor.

Lessons

Lesson 56

Dividing Using

Equivalent Expressions

Lesson 57

Generating Equivalent

Expressions

Lesson 58

Factors and Terms

Lesson 59

Simplifying

Expressions

Lesson 60

Equivalent Equations

Lesson 61

Inverse Property ofMultiplication

Are You reAdY?

1. What is 45% of 81?

2. Write9

__

5using percent notation.

3. Apply the distributive property then evaluate the expression. Write the

value in decimal notation.

7

__

6

11 +

7

__

6

4

Solve the following equation.

4. 7.72 + d = 9.01

5. If a vehicle travels at a constant rate of 6 meters per second, how long will

it take to drive 530.16 meters?

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550 | chapter 11 Equvalent Expressons and Equvalent Equatons

Lesson 56 Diiding Using EquialentExpressions

Objecties Divide by decimal divisors.

Form equivalent expressions using the multiplicative property of 1.

Concepts and Skills

RO.15 Divide whole numbers and decimals by decimal divisors by rst

rewriting the division as an equivalent expression with a whole number

divisor (e.g., rewriting7__0.4as 70__4 ).

EE.9 Determine if two expressions are equivalent.

Remember from Before

What are equivalent fractions?

What are complex fractions?

How do fractions represent division of whole numbers?

Get Your Brain in Gear

1. Find an equivalent fraction with a denominator of 8.

a.

9

__

4

b.

3

__

6

c.

70__802. Express the division as multiplication by the inverted fraction. Use mental math

to evaluate the expression.

a. 7 1

__

2

b.2

__

35

__

4

c. 8 1

__

3

Vocabulary

oplex raton

equvalent

expresson

equvalent raton

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Lesson 56 vdng sng Equvalent Expressonsvdng sng Equvalent Expressons | 551

Lesson 56 Diiding UsingEquialent Expressions

Conepts and Sklls: RO.15, EE.9 In the previous couple of lessons we discussed how long division works when

decimals are involved. We now can divide a decimal by a whole number and divide

whole numbers to get a decimal result. However, we still havent discussed what to

do when the divisor is a decimal. Well address that situation here.

Diidingby WholeNumbers

Lets use the number line to review the meaning of dividing by a whole number.

Here is a simple division expression:

3 5

The value of this expression tells us how big the jumps need to be in order to travel

to the number 3 using 5 equal jumps:

Here the size of each jump is 3 5. As weve seen in the past couple of lessons, we

can calculate the value of 3 5 using long division:

From this we see that 3 5 equals 0.6, and this means the size of each jump must

be 0.6 in order to travel to 3 using 5 equal jumps:

Diiding by

Decimals

That is what it means to divide by a whole number. Lets now examine what it

means to divide by a decimal. Well divide 3 by the decimal 0.5:

3 0.5

If we apply the same reasoning as above, then 3 0.5 should tell us how big the

jumps need to be in order to travel to number 3 in 0.5 equal jumps. But what does

0.5 equal jumps mean?

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552 | chapter 11 Equvalent Expressons and Equvalent EquatonsEquvalent Expressons and Equvalent Equatons

From earlier lessons we know that 0.5 is another name for1_2. As a result, 3 0.5

must tell us how to get to 3 using half of a jump. If we visualize this we see that if

we make a jump of +6, half of that jump will get us to the number 3:

From this we see that in order to get to number 3 using half of a jump, the jump

must be +6 . Therefore 3 0.5 equals 6:

3 0.5 = 6

Using this same reasoning, lets see how big the jump must be to get to number

4 in half a jump:

As we can see, in order to travel to the number 4 using 0.5 jumps, the jump must

be +8. This means:

4 0.5 = 8

Check forUnderstanding

1. Find the value of each expression:

a. 5 0.5 b. 7 0.5 c. 9 0.5

EquialentExpressions

Because its difcult to work with parts and pieces of jumps on the number line, lets

develop a better way to divide by a decimal. Heres the decimal division problem

we discussed earlier:

3 0.5

Lets write this division using fraction notation:

3

___0.5We should notice that this is a complex fraction because the denominator is not

a whole number. Later well learn that we can work with complex fractions in

the same ways that we work with regular fractions. For example, we can form

an equivalent expression by multiplying byn_n, just like we create equivalent

fractions. We can do this because multiplying byn_n is the same as multiplying by

1, which doesnt change the value of an expression.

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In this case, lets multiply by10__10 so that the denominator becomes a whole

number:

Now we have the equivalent fraction30__5 which we can express using division like

this:

30 5

We now have a new expression that is equivalent to our original one. In other

words, the following two expressions are equal:

3 0.5 = 30 5

The expression 30 5 involves division by a whole number, which we are very

good at doing. We know that 30 5 equals 6.

30 5 = 6

From this we conclude that 3 0.5 is also equal to 6:

3 0.5 = 6

This is the same result that we got earlier, but now we have a much better way of

carrying out the decimal division.

Lets review what we just did by looking at another example:

7 0.4

Here we are dividing by a decimal again. As we did before, we can create an

equivalent expression that has the same value but is easier to work with:

If we write70__4 as division we get:

70 4

Since this is an equivalent expression, it has the same value as 7 0.4, but its

easier to work with. Its easier because we can use our knowledge of dividing by

whole numbers to nd the value.

Lets nd the value of 70 4 using long division:

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This shows that 70 4 equals 17.5, which means that 7 0.4 is also equal to

17.5:

7 0.4 = 17.5

Check for

Understanding

2. Find the value of each expression.

a. 9 0.2 b. 3 0.8 c. 56 0.7 d. 19 0.4

EquialentComplexFractions

Earlier we made division problems easier to calculate by nding equivalent

expressions. For example, we found that the complex fraction7__0.4 is equivalent to

the fraction70__4 after multiplying by 10__10. Lets discuss this further.

Here is a complex fraction:

93.5____0.02

What equivalent expression do we get when we multiply this by10

__

10?

Lets nd the value of the numerator rst:

93.5 10

If we expand 93.5 using place value, the above expression becomes:

(90 + 3 +5

__10) 10Using the distributive property, we get the following:

90 10 + 3 10 +5

__10 10Now lets perform each multiplication:

This is the expanded form of the following whole number:

935

What this shows is that multiplying by 10 increased the power of ten for each digit.

Using this same reasoning, it follows that 0.02 10 becomes 0.2, which gives us our

equivalent expression:

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This is an equivalent expression, but we are still dividing by a decimal. In order to

make it so we are dividing by a whole number, lets multiply by 10__10one more time:

Now we have an equivalent expression where we can divide by a whole number.

In general, if we shift the power of ten of each digit in the numerator, and each digitin the denominator, we can create an equivalent expression. We can do this place

value shifting as many times as we want. For example, the following expressions

all have the same value:


3. Write an equivalent expression where the divisor is a whole number, then

divide.

a.7.2____0.06 b. 97_____0.002 c. 0.39 0.4 d. 8.35 0.1

Long DiisionNotation

We just discussed a powerful way to turn difcult division into equivalent expressions

that are easier to solve. Lets use this method to solve the following problem:

Here we are dividing 15 by 0.06, which we can write using fraction notation as:

15____0.06By multiplying by

10__10 then 10__10 again, we will get an equivalent expression with awhole number in the denominator:

In long division notation, this is:

61500All we did is create an equivalent expression by shifting each digit in the numerator

and denominator by 2 powers of 10. We can represent this in a shorthand notation

like this:

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Now that we are dividing by a whole number we can easily evaluate the

expression:

This means 15 0.06 is also equal to 250.


4. Write an equivalent expression where the divisor is a whole number, then

divide.

SimplifyingWhole Number

Diision

We can also use this method to simplify division even though we are already

dividing by a whole number. For example, lets divide 135 by 20:

Here we are dividing by a larger whole number than we have to. We can make the

division easier by nding an equivalent expression like this:

Here we shifted each digit to smaller powers of ten. This gives us 13.5 divided by

2, which is easier to calculate:


5. Divide.

a.24__40 b. 30.0 500 c. 14,238 1,000

d. 18 90 e. 18 0.9 f. 0.18 0.09

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6. One teaspoon is about 0.005 liters. How many teaspoons are there in a

0.355 liter can of beverage?

7. There are 60 seconds in a minute. The average song on a music CD is

211.2 seconds. How many minutes is 211.2 seconds?

Problem Set Find the value of each expression.

1. 0.6 10 2. 0.007 10 3. 0.03 10

0.6 100 0.007 100 0.03 100

0.6 1,000 0.007 1,000 0.03 1,000

4. 3.6 10 5. 4.02 10 6. 621.03 10

3.6 100 4.02 100 621.03 100

3.6 1,000 4.02 1,000 621.03 1,000

Generate equivalent expressions.

7.2.1____0.07 = 21__? = 210___? = 2100____? 8. 82.15_____0.003 = ?____0.03 = 8215____? = ?__3 = ?__30

9.8.001_____0.18 = 800.1_____? = ?___180 = ?___1.8 10. 0.182_____0.0009 = ?__9 = 18.2____? = ?_____0.009

11.1.6_____0.008 = 160___? = ?__80 = ?__8

Generate an equivalent expression so that the divisor is a whole number. Then

divide. Use bar notation for repeating decimals.

12. 32.87 0.3 13. 0.287 0.004 14. 186 0.2

15. 7.003 0.06 16. 28.376 0.0007

Generate an equivalent expression so that the divisor is a whole number. Then

divide. Use bar notation for repeating decimals.

17. 0.28.64 18. 0.004127.3 19. 0.082320. 0.00987 21. 0.65.32Generate an equivalent expression where the divisor is a single digit. Then

divide.

22. 3072 23. 409.6 24. 80024.825. 600632 26. 8002.348

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Write an equation to describe the situation. Solve the equation to answer the

question.

27. When astronauts landed on the Moon, they placed mirrors on the Moons

surface. Scientists on the Earth can shine lasers at these mirrors to get some

accurate measurements of the Moons position. It takes about 0.04 minutes for

light to travel from the Earth to the Moon and back to the Earth again. There

are 1,440 minutes in a day. How many times in a day can light travel back andforth between the Earth and the Moon?

28. The weight of 2 pennies is 0.005 kilograms. How many pennies do you need

to weigh as much as a 7.2 kilogram bowling ball? How many dollars are all of

those pennies worth?

29. At an average speed of 60 miles per hour, how long will it take to drive 50.4

miles?

30. Going the same 50.4 miles as in the previous problem, how long will it take if

you go 70 miles per hour? What is the difference in the time it takes to get there

going the different speeds?

Challenge Problems

1. The race car engine averages 8,000 revolutions per minute. How many seconds

does each revolution of the engine take?

2. You have a two foot high stack of at plastic bags made from 2 mil, or 0.002

inches, thick plastic material. How many bags are in that pile? (Hint: each

complete bag is two layers of plastic thick.).

3. Sales tax consisted of $0.05 for the state and $0.02 for the county on every dollar

spent. If the sales tax totaled $1.12 on a purchase, how big was the purchase?

Multiple Choice Practice

1. Estimate the location of 2 0.5 on the number line.

2. Which expression is an equivalent expression to 0.7 56 ?7 0.56 7 5.67 56 7 560

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Math Journal Questions

1. Some people claim that when you divide you always get a smaller result. Give a

clear and logical argument that will show that this claim is wrong.

2. What is an equivalent expression? Explain how we can use equivalent expressions

to make it easier to divide by decimals.

Find the Errors A student made 3 mistakes below. Find and correct each mistake.

1. 2.

3. 4.

looking bAck

Vocabulary: oplex raton, equvalent expresson, equvalent

raton

Student Self Assessment: o I get t?

1. How does ultplyng by powers o 10 hange a deal nuber?

2. How do I reate equvalent expressons that are easer to alulate?

3. How do I dvde wth deal dvsors?

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Lesson 57 Generatng Equvalent Expressons | 551

Lesson 57 Generating EquialentExpressions

Objecties Know and understand the properties of rational numbers.

Understand the concept of equivalent expressions.

Apply properties to simplifying numerical expressions.

Concepts and Skills

EE.7 Use the correct order of operations to evaluate expressions.

EE.8 Write and evaluate expressions with parentheses.

EE.9 Determine if two expressions are equivalent.

PR.9 Simplify expressions, generate equivalent expressions and equations and

solve equations using the following properties of rational numbers: the

commutative and associative properties of addition and multiplication,

the distributive property, and the special properties of 0 and 1.

PR.10 Understand that multiplication and division are inverse operations.

Use the inverse relationship of multiplication and division to generate

equivalent expressions, evaluate expressions, verify the results of

computations, and solve equations.


What does it mean to simplify an expression?

What are some of the properties of rational numbers that we have learned?


1. Use mental math to nd the value of each expression.

a. 32 0.8 b. 32 80

c. 25 0.5 d. 25 500

2. Use mental math to solve each equation.

a. 12 a = 3 b. 12 b = 30

c. 15 c = 500

Vocabulary

addtve property

o 0

outatve

property o addton

outatve

dstrbutve property

o ultplaton

over addton

equvalentexpresson

ultplatve

property o 1

sply

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Lesson 57 Generatng Equvalent ExpressonsGeneratng Equvalent Expressons | 561

Lesson 57 Generating EquialentExpressions

Conepts and Sklls: EE.7, EE.8,

EE.9, PR.9, PR.10

The previous lesson demonstrated a powerful technique in mathematics. When an

expression is difcult, we can often nd an equivalent expression that is easier to

work with. We used this technique to turn division by decimals into division by

whole numbers. In this lesson well discuss the basic mathematical properties that

allow us to generate equivalent expressions.

MultiplicatieProperty of 1

Throughout this book weve learned about many different mathematical properties.

We usually represent such properties as an identity. For example, the property we

used in the previous lesson is called the multiplicative property of 1, and we

describe it with the following identity:

b = b 1

What makes this equation an identity is that b and b 1 are always equal

no matter what the value ofb is. Since b and b 1 are always equal to each

other, we call them equivalent expressions. The mathematical properties, such

as the multiplicative property of 1, provide the rules for generating equivalent

expressions.

The above property tells us that multiplying by 1 doesnt change the value of an

expression. When using this property, we often multiply byn_n, which is simply

another name for 1. For example, in the previous lesson, we multiplied by10__10 like

this:

This new expression has the same value as the old one, but its easier to solve. Its

easier because were dividing by a whole number instead of by a decimal.

In our everyday lives we use this property without even knowing it. Whenever we

cut a pizza into slices, we are using the multiplicative property of 1. For example,

here is a pizza:

If we multiply this pizza by6_6, well cut the pizza into 6 equal slices and keep all

6 slices:

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We still have the same amount of pizza as before, but now its easier for people

to eat. This is just one way in which equivalent expressions can make our lives

easier.

Whenever we turn 4 quarters into a dollar, we are using this property in the other

direction. For example, here we are turning 3 groups of 4 quarters into 3 dollars:

These two expressions are equivalent because they have the same value. We can

write this in symbols with the following equation (in units of dollars):

3 4_4 = 3This uses the multiplicative property of 1 in the other direction:

b1 = b

In general, all the properties we are going to talk about can be used in both directions

to generate useful equivalent expressions.


1. The following examples show the generation of equivalent expressions.

They all make use of the multiplicative property of 1. Explain how this

property is being used, and how the result might be useful.

2. Write equations for each of the above examples.

AdditieProperty of 0

We just discussed how multiplying by 1 doesnt change the value of an expression.Similarly, adding 0 doesnt change the value of an expression either. This is called

the additive property of 0:

b + 0 = b

It doesnt seem like we make use of this property very often, but in fact, we use it

all the time to simplify how we write numbers. Lets see how this works.

The place value notation that we use is an abbreviation of the expanded form. For

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example, 023 is an abbreviation for:

0 100 + 2 10 + 3 1

Since 0 100 equals 0, we can apply the additive property of 0 like this:

2 10 + 3 1

Now we can write the number in a simpler way:

We use this property in the other direction when dealing with money. For example,

the following means 3 and a half dollars in decimal notation:

$3.5

Here is the value in expanded form (in units of dollars):

3 1 +5

__

10Its customary to write money amounts with two places after the decimal point.

According to the additive property of 0, we can add on0___100 of a dollar without

changing the value of the expression:

3 1 +5

__10+ 0___100Now we can write the $3.5 in a more standard way:

Any time we hide or show the 0 digits of a number, we are making use of the

additive property of 0.


3. The following are some examples that use the additive property of 0 to

generate equivalent expressions. For each example, explain how the

property is being used, and how it could be helpful.

4. Use equations to describe each of the above examples.

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InerseRelationship of

Multiplicationand Diision

The following identity is actually the result of other more basic properties, but its

useful to discuss in this form. The following is true as long as b and c are not equal

to zero:

This shows that dividing by a fraction is equivalent to multiplying by the invertedfraction. This is a result of the inverse relationship between multiplication and

division.

Lets apply the above identity to the following expression:

a b

Sinceb is the same as the fractionb_1, we can turn the above division into multiplication

by the inverted fraction like this:


5. Explain how each of the following examples is an application of the inverse

relationship between multiplication and division:


CommutatieProperties

According to the commutative property of addition, we can change the order in

which things are added without changing the value of the expression. Here is an

identity that illustrates this property:

a + b = b + a

We see the effects of this property every day. A quarter plus a dollar is equal in valueto a dollar plus a quarter:

Both expressions have a value of $1.25 and are therefore equivalent due to the

commutative property of addition.

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Addition isnt the only operation that is commutative. We can also change the order

in which things are multiplied without changing the value of an expression:

ab = ba

Because of the commutative property of multiplication, 10 ves is equal to

5 tens:

In symbols, we can describe this as follows:

10 5 dollars = 5 10 dollars

Both of these expressions are equal to $50.

Check forUnderstanding. 7. The following examples show the generation of equivalent expressions.

Which of them are due to the commutative property of addition, and which

are due to the commutative property of multiplication? Explain your

reasoning.


DistributieProperty

The distributive property of multiplication over addition is extremely useful.

We can describe it with the following identity:

ab + ac = a (b + c)

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The distributive property has helped us out many times throughout this book. One

aspect of it that we havent pointed out explicitly is that the distributive property also

works with division over addition. For example, the following is also an identity (as

long as we dont divide by zero):

b a + c a = (b + c) a

To see why this is true, lets write each division in fraction notation:

Recognize this? This is simply the denition of fraction addition which we are very

familiar with. It is the distributive property that allows us to add fractions with a

common denominator.

Its important to realize that the distributive property of division is a direct result

of the distributive property of multiplication. To see why, take the identity we just

discussed, and rewrite each division as multiplication by the inverted fraction:

b1__a + c1__a = (b + c) 1__a


9. Explain how the distributive property is being used in each of these

situations:

10. Describe each of the above examples with an equation.

SimplifyingExpressionsOne of the reasons we generate equivalent expressions is to make problems simpler.However, to make a problem really simple, it often takes a sequence of steps and

applications of more than one property. For example, the following simplication

process takes 7 steps. See if you can identify the property used to generate the

equivalent expression at each step:

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The rst step uses the commutative property of multiplication to express

b 3_4 as 3_4 b.

The next step applies the commutative property of addition to swap the order of

4.0 and3_4 b.

The third step applies the distributive property to rewrite 1_2b + 3_4b as(1_2+ 3_4) b.The fourth step used the multiplicative property of 1 to form a common

denominator. We multiplied by 2_2 to turn 1_2 into the equivalent expression 2_4.The fth step used the distributive property again, this time to add the fractions.

The nal step applies the additive property of 0 to write 4.0 more simply as 4.

Here is a summary of the properties used at each step:

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11. Identify the properties used to generate each equivalent expression:

a. 2 m + 5 m + 13 (2 + 5) m + 13

b. g 7 + 9.4 + g g 7 + g+ 9.4

c. 7.14 + 32 h + 0 + 14.7 h 7.14 + 32 h + 14.7 h

12. All of the expressions below are equivalent. Identify the property used to

create the equivalent expression at each step:

7 d+ 16 + d

7 d+ d+ 16

7 d+ 1 d+ 16

(7 + 1) d+ 16

8 d+ 16

13. All of the expressions below are equivalent. Identify the property used to

create the equivalent expression at each step:

23 (z+ 1) + z 3 + 0 z

23 (z+ 1) + z 3 + 0

23 (z+ 1) + z 3

23 z+ 23 1 + z 3

23 z+ 23 + z 3

23 z+ z 3 + 23

23 z+ 3 z+ 23

(23 + 3) z+ 23

26 z+ 23

Problem Set Simplify each fraction.

1.9

__12 2. 15__54 3. 50___350Find the value of each expression.

4. 0.38.76 5. 8032.4

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6. Explain how you used the multiplicative property of 1 when answering questions

1 through 5 above.

Find the value of the following expressions. Look closely! Some are addition

problems and some are difference problems.

7. 8.

9. 10.

11. Explain where the additive property of 0 is used when calculating the above

expressions.

Find the value of the following expressions.

12. 6 7.94 13. 4.5 6.2

14. 0.23 8.7 15. 5 (200 + 60 + 1)

16. Explain how you used the distributive property of multiplication over addition

when performing the above calculations.

Use theinverse relationshipbetweenmultiplicationand divisionto ndthe

values of the following expressions.

17. 16 1

__

4 18. 24 0.3 19.4

__

57

__

2 20. 160

1

__

5

Solve for the variable in each equation. Name the property you used.

21. 2 53 g = 2 53 22. 9 4 + 9 2 = 9 k

23. 14 2

__

3 = 14m 24. 8 b = 4 8

25. 30 + 8 + m = 38 26. 32 w = 0

27. What property is demonstrated by the equation on the following number

line?

28. Apply the distributive property to the following equation, then evaluate

the expression.

7

__

8 9 +

7

__

8 7

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29. The following shows the steps in simplifying an expression. Identify the

property that generated the equivalent expression at each step.

2

__

3 5 +

2

__

3

2

__

3 5 +

2

__

3 1

2__

3 (5 + 1)

2

__

3 6

4

30. Here is another example of simplifying an expression. Determine the

property used at each step to generate the equivalent expression.

h + h + 3 (1 +h) + 0

h + h + 3

(1 +h)

h + h + 3 1 + 3 h

h + h + 3 + 3 h

h + h + 3 h + 3

1 h + 1 h + 3 h + 3

(1 + 1 + 3) h + 3

5 h + 3

Challenge Problems

As we did in problems 29 and 30 of the problem set, use properties to simplify the

following expressions. Show each step, and name what property you used.

1. 5 2

__

7+ 3

2

__

7+2

__

7 6 2. p + 0 p + 4 (2 + 1 p)


1. Which property is represented by the following equation?

4 1__

5 = 1__

5+ 4

The commutative property of addition.

The commutative property of multiplication.

The additive property of zero.

The multiplicative property of one.

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2. Which property is illustrated by the following equation?

5

__

4

3

__

3 =

15__12The commutative property of addition.




3. Which property is illustrated by the equation on the following number line?

The commutative property of addition.





1. Addition is commutative and multiplication is commutative. How would you

prove to someone that division is NOT commutative?

2. What is an equivalent expression? Explain how the mathematical properties

allow us to generate equivalent expressions.

3. Explain how the following equation demonstrates the distributive property of

multiplication over addition.

15 + 18 = 3 11

4. Explain what property is demonstrated by each situation.

a. I have 5 bags and I put 4 oranges in each bag. My friend has 4 bags and put 5

oranges in each bag. My friend and I have the same number of oranges.

b. I have 1 bag. I rst put 4 oranges in the bag, then I put 5 oranges in the bag.

My friend also has 1 bag, but put 5 oranges in rst then put 4 oranges in the

bag. My friend and I have the same number of oranges.

c. I have 5 bags and my friend has 5 bags. I have 4 oranges in each bag, but my

friend has 3 oranges in each bag. When we combined the oranges together

we had 5 bags, with 7 oranges in each bag.

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Find the Errors A student made 3 mistakes below. Find and correct each mistake.

1. 2.

3. 4.

looking bAck

Vocabulary: addtve property o 0, outatve property o addton,

outatve property o ultplaton, dstrbutve property,

equvalent expresson, ultplatve property o 1, sply


1. How do I generate equvalent equatons?

2. What s the ultplatve property o 1?

3. What s the addtve property o 0?

4. What operatons are outatve?

5. How does the dstrbutve property o ultplaton over addton

work?

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Lesson 58 ators and ers | 573

Lesson 58 Factors and Terms

Objecties

Know and understand the associative properties.

Use parentheses to indicate the part of an expression that gets evaluated rst.

Understand and apply the order of operations when evaluating expressions.

Apply properties of rational numbers to more sophisticated expressions.

Concepts and Skills

PR.1 Understand and identify the associative property of addition.

PR.2 Understand and identify the commutative property of addition.

PR.3 Understand and identify the associative property of multiplication.

PR.4 Understand and identify the commutative property of multiplication.PR.9 Simplify expressions, generate equivalent expressions and equations and





What do parentheses mean in an expression?

What are equivalent expressions?


1. Use mental math to nd the value of the expression.

a. 5 13 b. 37 + 18 c. 92 4

2. Use mental math to solve each equation.

a. 4 a = 12 b. 7 b + 3 = 59

c. (3 +c) 2 = 8 d. 5 d+ 4 d = 54

Vocabulary

assoatveproperty o

addton

assoatve

property o

ultplaton

sply

ter

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Lesson 58 Factors and Terms

Conepts and Sklls: PR.1, PR.2,

PR.3, PR.4, PR.9

As we begin to deal with more and more sophisticated expressions it becomes

important to have ways of describing these expressions. In this lesson well learn

how to describe the structure of an expression.

Adding Terms When we form an expression by adding things together, the things are called terms.For example, the following expression adds together four terms:

5 + (b + 3) + 7 k+ 2 (4 +n)

Lets discuss each of the above terms briey.

The rst term is simply the number5. Thats easy enough.

The second term is (b + 3). This term is actually a group of two terms. Well talk

more about groups of terms shortly.

The third term is 7 k. This term is a product of 7 and k. Due to the agreed-uponorder of operations, we multiply before we add. This is why 7 k is treated as a

single term.

The fourth term is 2 (4 + n). This term is the product of 2 and (4 + n).

This is what the word term means when describing an expression. Its important

to know that the word term has different meanings when used in other contexts.

For example, a term sometimes means a period of time. To visualize how this

meaning is related to what we just discussed, lets examine a timeline showing

when the rst three presidents of the United States held ofce:

1789 1793 1797 1801 1805 1809

Washington Adams Jefferson JeffersonWashington

When elected, a president holds the position for a term of 4 years. The rst

president, George Washington, held 2 terms starting in the year 1789. Next came

John Adams who only served as president for 1 term. After that, Thomas Jefferson

became president for 2 terms ending in year 1809. We can describe the rst 20 yearsof the U.S. presidency from 1789 to 1809 by adding up all of 5 terms:

Washington + Washington + Adams + Jefferson + Jefferson

From this we can see that the different meanings of the word term are related.

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Lesson 58 ators and ersators and ers | 575


1. Put parentheses around each term of the expression.

Example: 7 w + 19 + h12 3

Solution: (7 w) + (19) + (h12 3)

a. 13.7 + m 67 +3

__

4+ 128,200 9

b. n + 632 n + 100 d 2 45 + 22.8

c. 152,234 21 8.2 + 998 t+ t+ 12 + 4 y

d. 938.5 16 p 4 22.3 b

2. Put brackets [ ] around each term of the expression.

Example: 7 + 2 (15 + m) + 29 s

Solution: [7] + [2 (15 + m)] + [29 s]

a. 99.1 a + (12 + 3 + z) w + 3 w

b. 14 (a + b) + 186.70 + 5 c + 4 + 3 12 (15 + 0.1) g

c. 18.7 + 4 (1 + 2 + 3) w + (89.3 + 23 +q) (22.5 + 3.14) + 2.7 (8 + v)

GroupingTerms Together

Lets discuss how grouping terms works. Here is an expression with three terms:

3 + 4 + 2

We can visualize this expression using unit squares as follows:

To calculate how many unit squares this is, we nd the value of 3 + 4 + 2. Which

addition do we perform rst?

The standard convention is to perform addition from left to right. This means we

nd the value of 3 + 4 rst, and then add 2. To show that we perform 3 + 4 rst,

we put parentheses around it like this:

(3 + 4) + 2

What we are doing here is grouping the terms 3 and 4 together as a single term:

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Since 3 + 4 equals 7, we get:

7 + 2

Now we add 7 + 2, which equals 9. From this we conclude that there are 9 unit

squares in the above picture.

Even though it is conventional to add things from left to right, we dont have to do

it that way. For example, to calculate 3 + 4 + 2, we could add 4 + 2 together rst,and then add that result to 3:

3 + (4 + 2)

In this case, we are rst grouping the terms 4 and 2 together as a single term:

Its obvious that this is going to give us the same answer:

The value of the expression is 9, which is the same result that we got last time. This

means that (3+ 4) + 2 and 3 + (4+ 2) are equivalent expressions.

In general, we can group the terms of an expression together in any order that we

want and get the same value. This is called the associative property of addition.

The word associate simply means to group things together.

We can describe the associative property of addition with the following identity:

(a + b) + c = a + (b + c)


3. In lesson 5 we illustrated the associative property of addition on the number

line. Draw a picture of what that looks like.

Properties ofAddition

In the previous lesson we learned about several mathematical properties that allow

us to generate equivalent expressions. Its important to realize that the properties

of addition apply to the terms of an expression. For example, lets go back to the

expression we used at the beginning of this lesson:

5 + (b + 3) + 7 k+ 2 (4 +n)

Lets apply the commutative property of addition around the following addition

operation:

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The commutative property of addition says that we can change the order in which

we add two terms. Here are the two terms on either side of the addition operation:

After commuting these two terms, we get the following equivalent expression:

We can also apply the commutative property to terms within a group of terms. For

example, lets apply the commutative property around this addition operation:

We are now working inside a group of terms. Here are the two terms on either side

of this operation:

After applying the commutative property of addition, we get the following



4. Apply the commutative property of addition at the indicated operation.

GroupingFactors

Together

So far in this lesson weve concentrated on addition. Lets now shift our focus to

multiplication. When things are multiplied together, we can call the things factors.

For example, the following expression is a product of three factors:

3 2 4

To calculate the value of this expression, the convention is to multiply from left to

right. This means we multiply 3 2 rst, and then multiply the 4. We indicate that

we multiply 3 2 rst by putting parentheses around it:

(3 2) 4

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If we use this expression to describe a rectangle, it will be 3 2 units wide and 4

units tall:

We are treating the group of factors (3 2) as a single value. Since 3 2 equals 6,

we can describe the above rectangle as 6 4 unit squares:

This is a total of 24 unit squares.

As with addition, multiplication is also associative. This means we could have

calculated the expression 3 2 4 by multiplying 2 4 rst, and then multiplying by 3:

3 (2 4)

Using unit squares to visualize this, we get a rectangle that is 3 units wide and 2 4

units tall:

This time we are treating (2

4) as a single value. Since 2

4 equals 8, we candescribe the above rectangle as 3 8 unit squares:

Again, this is a total of 24 unit squares.

Its easy to visualize why these two different rectangles have the same area. By

rearranging the columns, we can turn the rst rectangle into the second rectangle:

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This means the following two expressions are equivalent:

(3 2) 4 = 3 (2 4)

In general, the way we group factors doesnt change the value of the expression. As

a result, the following is an identity:

(a

b)

c = a

(b

c)

This is called the associative property of multiplication.


5. Verify that the associative property of multiplication holds by evaluating

each side of this equation:

(5 10) 1

__

2 = 5 (10

1

__

2)

Properties of

Multiplication

We can apply the properties of multiplication that we discussed in the previous lesson

to the factors of an expression. For example, consider the following expression:

(k+ 8) m

Lets apply the commutative property of multiplication to this expression.

Here are the two factors on either side of the multiplication operation:

After applying the commutative property of multiplication, we get the following


m (k+ 8)

We can also apply the properties of multiplication to the factors within a term. For

example, the following expression has 3 terms:

a + 7 b + 4

Lets apply the commutative property of multiplication to the following operation:

This multiplication operation is inside of the second term. Here are the factors oneither side of the multiplication sign:

After applying the commutative property of multiplication, we get the following


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6. Apply the commutative property of multiplication at the indicated

operation.

EaluatingExpressions

Lets discuss how to evaluate relatively complicated expressions that have addition,

multiplication and parentheses. The way we do this is we rst evaluate what is

grouped by parentheses. Then we multiply together the factors inside each term.

Finally, we add the terms together.

Lets practice this by evaluating the following expression:

Oops. Notice that there is a division operation:

We havent discussed what to do with division, but we know how to turn division

into multiplication by the inverse. Lets turn the division by 2 into multiplication

by 1_2 to get the following equivalent expression:

Now we just have multiplication and addition operations, so we can evaluate the

expression as we just discussed.

First we evaluate what is inside the parentheses. Here (1 + 6) equals 7, so we get:

Now we multiply the factors within each term. Here we have 3 terms. Lets separate

the terms out so we can see them better:

The second term has 3 factors. We know that 3 6 equals 18, so we get:

Now 18 1_2 equals 9, which gives us:

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The last term is 4 7 which is 28:

The nal step is to add all the terms together. The rst two terms are 2 + 9 which

equals 11:

Finally we add 11+ 28:

From this we conclude that the value of our original expression is 39:


7. Evaluate each expression.

a. (1 + 2) (3 + 4)

b. 6.1 + 7 (8 + 2) + 3 10 + 9 100

c. 4 (1 + 2 2) + 4 15 3 + 20

Problem Set Describe the properties represented by each situation.

1. To get to school I walk 6 blocks on B Street, then 3 blocks on Holt. To get home

I walk 3 blocks on Holt, then 6 blocks on B street. I walk the same distancegoing to school as I do going home.

2. The following change was on the table.

I added it up this way: (10 + 10 + 5) + (10 + 10 + 5)

My friend added it up like this: (10 + 10 + 10 + 10) + (5 + 5)

We both concluded that it equaled 50 cents.

3. We wanted to nd the area of the following rectangle.

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I calculated it this way: (3 5) 4

My friend calculated it this way: 3 (5 4)

We both calculated the area to be 60.

Putparentheses( ) around each term of the expression.

4. 8.3 + g 13 +7

__

8

5. w + 42.8 w + 30 b 2 9 + 22.8

6. 538.2 21 9.3 + 100 + h + h + 15 d

7. 19 16 99.9 p

Put brackets [ ] around each term of the expression.

8. 64 n + (8 + k) 14 + 7 4

9. 2 (5 + 3 + 15) + 7 (2 + p) + 3.4 (h + 8) 4

Match the equivalent expressions. Name what properties relate the equivalent

expressions.

10. 3 k+ 5 k a. 3 5 + k 5 + k

11. 3 (k+ 5) k b. 3 + (k+ k) + 5

12. (3 +k) 5 + k c. 3 k (k+ 5)

13. (3 +k) + (k+ 5) d. k (3 + 5)

Apply the commutative property of addition to the operation indicated by the

arrow .

14. 15.

16. 17.

Apply the commutative property of multiplication to the operation indicated

by the arrow .

18. 19.

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20. 21.

Find the area of the rectangle.

22. 23.

Evaluate the expression. First rewrite any division as multiplication by the

inverted fraction. Next calculate what is in the parentheses. Then multiply the

factors within each term. Finally, add the terms together.

24. (4 + 6) (7 + 8) 25. 8 2 + (2 + 5) (3 + 4)

26. 5 + 8 2 + 3 (5 + 2)

Evaluate each expression when b = 5, h = 7, k= 10, and m = 2.

27. k+ bh 28. h + m (3 + b)

29. (b + m) (h + 3 + k) 30. b + h + (k+ 1) m + 100.8

Challenge Problems

1. The following expressions are all equivalent. Identify what property was used to

generate each equivalent expression.

3 (m + 5) + 7 (8 +m)

3 (m + 5) + 7 (m + 8)

3 m + 3 5 + 7 (m + 8)

3 m + 3 5 + 7 m + 7 8

3 m + 7 m + 3 5 + 7 8

3 m + 7 m + (15 + 56)

3 m + 7 m + 71

(3 + 7) m + 71

10 m + 71


1. Find the equivalent expression to: k+ m 9

m + k 9 9 k+ m 9 k+ 9 m (k+ m) 9

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2. Find the expression equivalent to: k 8 + k

kk+ 8 k (8 +k) (k+ 1) 8 k (8 + 1)


1. Write an expression that has four terms. Make one of the terms a product of two

factors.

2. The word term means has a boundary. In this lesson, we discussed terms

where the boundary is an addition operation. Explain what this means.

3. Is division associative? In other words, are the following two expressions

equivalent?

(a b) c a (b c)

If you say they are equivalent, show why they are equivalent for ALL values

ofa, b and c. If you say they are NOT equivalent, you only need to give one

example of values a, b and c that shows that the two equations are not equal.

4. Describe the steps used when evaluating complicated expressions like those in

problems 24 - 30 of the problem set.

Find the Errors

A student made 3 mistakes below. Find and correct each mistake.

1. 2.

3. 4.

looking bAck

Vocabulary: assoatve property o addton, assoatve property o

ultplaton, sply, ter


1. What s a ter?

2. What does the assoatve property o addton ean? How do Iake an equvalent expresson usng ths property?

3. What does the assoatve property o ultplaton ean? How do

I ake an equvalent expresson usng ths property?

4. How do I use the order o operatons to evaluate expressons?

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Lesson 59 Splyng Expressons | 585

Lesson 59 Simplifying Expressions

Objecties

Use symbols to express verbal information.

Apply properties of rational numbers to simplifying expressions.

Use symbols to express geometric relationships.

Solve problems involving percentages.

Concepts and Skills

WO.7 Represent the area of a rectangle using multiplication.

RO.16 Calculate percentages of numbers.

SN.2 Translate verbal descriptions into mathematical expressions.

PR.9 Simplify expressions, generate equivalent expressions and equations andsolve equations using the following properties of rational numbers: the




How do you nd the area of a rectangle?

What is the distributive property?

What is a percent?

How do you nd a percent of a value?


1. Use mental math to nd the value of the expression.

a. 0.1 5 b. 5 + 0.1 5

c. 20 + 20 0.1 d. 20 + 20 0.2

2. Express each value as a fraction.

a. 7% b. 70%

c. 25% d. 50%

Vocabulary

splyngexpressons

perent nrease

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Lesson 59 Simplifying Expressions

Conepts and Sklls: WO.7,

RO.16, S.2, PR.9

We spent the past couple of lessons discussing the rules for generating equivalent

expressions. These rules are very useful because they give us ways to make difcult

problems easier. This is called simplifying expressions, and its the topic of the

current lesson.

ComplicatedAreas

Well start the lesson by nding the area of the following shape made up of 3

identical pieces:

This is a more complicated shape than what we have worked with in the past,

but we can use our knowledge of equivalent expressions to describe the area in a

relatively simple way.

Since this shape is built from three pieces of equal area, lets start with the top

piece:

This shape is actually two rectangles stuck together. Lets break them apart:

The rectangle on the left is k wide and s tall, so we describe the area as k s:

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Lesson 59 Splyng ExpressonsSplyng Expressons | 587

The rectangle on the right is s wide and b tall, so it has an area of s b:

If we add these together, we get the area of the top piece of our shape:

Here we have an expression with two terms. Notice that each term has s as a factor.

Lets apply the commutative property of multiplication to the rst term so thatthe s is the rst factor of both terms:

sk+ sb

In the context of area, this means that we rotated the left rectangle like this:

Lets now apply the distributive property to generate the following equivalent

expression:

s (k+ b)

In the context of area, this simply means that since both rectangles have the same

width, we can stack them on top of each other and create a single rectangle:

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We now have a simple expression to describe the area of the top piece of our

shape:

Since there are three of these pieces with equal areas, we need to multiply our

expression by 3 to get the area of the overall shape:

3 (s (k+ b))

Here we have a factor of 3 multiplying a group of two factors s and (k+ b).

According to the associative property of multiplication, we can change the way

we group these factors and generate an equivalent expression:

(3 s) (k+ b)

This shows us that our overall shape has the same area as a rectangle that is (3 s)

wide and (k+ b) tall:

We just used equivalent expressions to turn a complicated shape into a simple

rectangle.


1. The following figure is made of two rectangles. Write an expression for the

total area. Simplify if possible.

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2. The figure below is made of a stack of 4 identical small rectangles on the

left and one tall rectangle on the right. The tall rectangle is as tall as the

stack. Write a simplified expression for the total area of all 5 rectangles.

Sales Tax The same expressions that describe the area of shapes are used in other situations aswell. A good example of this is sales tax.

If the price of a book is $6, what is the total cost including an 8% sales tax?

An 8% sales tax means that we have to pay the government 8% of the $6 when we

buy the book. Since 8% means8___100, we need to pay the following amount of tax:

8___100

6

Lets write the fraction 8___100 in decimal notation:0.08 6

This is just the tax part. The tax gets added to the $6 that we pay the store. This

means we end up paying the following total amount:

6 + 0.08 6

Lets simplify this. According to the multiplicative property of 1, we can rewrite

6 as 1 6 and generate the following equivalent expression:

1 6 + 0.08 6

In the context of area, this is two rectangles added together. The rst rectangle is 1

unit wide and 6 units tall. The second is 0.08 units wide and 6 units tall:

The left rectangle is how much the store gets. The skinny rectangle on the right is

how much tax the government gets.

Since both rectangles have the same height, we can use the distributive property

to form a single rectangle:

(1 + 0.08) 6

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Since 1 + 0.08 equals 1.08, we can simplify this as:

1.08 6

Lets multiply this to nd how much money we need to give the cashier when we

buy the book:

From this we see that we need to give the cashier $6.48, which means the government

gets $0.48 in taxes when we buy a $6 book.


3. A ticket to the magic show was originally only $3.50, but once the show

became popular the ticket price got marked up by 40%. What is the new

price? Use the figure below to help you visualize this:

4. The bridge was originally 17 feet tall, but this was too short. To fix it they

made the bridge 40% taller. How tall is it now?

5. The regular granola bar weighs 59.5 grams. The super sized one is 20%more! How much does the super size granola bar weigh?

Color Blindness Expressions like the ones weve been dealing with in this lesson come up often inreal world situations. Here is an example:

People who are colorblind usually have a hard time seeing the difference between

the color green and the color red. Colorblindness is much more common in men

than in women. About 7% of men have this problem, but only 0.4% of women

are colorblind. If a school has 520 male students and 530 female students, how

many of the students are likely to be colorblind?

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Since 7% is 0.07 in decimal notation, we can express the number of colorblind

male students as:

Colorblind male students: 0.07 520

Similarly, 0.4% is 0.004 in decimal notation. This gives us the following number

of colorblind female students:

Colorblind female students: 0.004 530

By adding these expressions together, we get the total number of students that are

likely to be colorblind:

0.07 520 + 0.004 530

This time, the terms dont share a common factor, so we cant simplify using the

distributive property. Real situations sometimes produce expressions that are hard

to simplify. To calculate this, we rst nd the value of each term.

Here is the number of male students that are likely to be colorblind:

And here is the number of female students that we might expect to be colorblind:

Now that we know the value of each term, we get the following equivalentexpression:

36.40 + 2.120

Finally, we add these two terms together to get the total number of colorblind

students:

From this we conclude that the value of our expression is 38.52:

0.07 520 + 0.004 530 = 38.52

This means we can expect the school to have somewhere around 38 or 39 students

that are colorblind.

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Write an expression for each situation. Simplify the expression when

possible.

6. The turf company needed to buy artificial grass to cover the entire new

football field. The main field is 65 yards wide and 100 yards long, and in

addition there are two end zones the same width but 10 yards long. What

is the total area to be covered?

7. About 3% of all mothers giving birth deliver more than one newborn

baby: twins, triplets, or more. Last year at the local hospital, 613 mothers

gave birth. How many of those mothers likely gave birth to more than

one baby?

8. Using the information from problem 7, how many of those 613 mothers

likely gave birth to just one baby?

Problem Set Write an expression to represent the area of each shape. Simplify the expressionwhen possible.

1. A rectangle: 2. Four identical rectangles:

3. 4.

Write each percent in decimal notation.

5. 37% 6. 840% 7. 6% 8. 3.2%

3.7% 8.4% 0.6% 0.87%

0.37% 84% 60% 8.12%

370% 0.84% 600% 38.5%

0.06% 0.05%

Find the value of each expression.

9. 7 8% 10. 13 9% 11. 15.3 10%

12. 275 25% 13. 50 0.2%

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Here is an area of 2 unit squares:

Answereachquestionbelowbyndingthearea.

14. What is 30% of 2? 15. What is 2 increased by 30%?

16. What is 30% of 3.2? 17. What is 3.2 increased by 30%?

18. What is 50% of 2? 19. What is 2 increased by 50%?

20. What is 600% of 2?

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21. What is 2 increased by 600%?

22. What is 2 increased by 100%? 23. What is 2 increased by 9%?

Write an expression showing the value of each purchase. Find the value of the

expression.

SALE!

Batteries: pack of 8 AAA for $4.79

Cell phone charger for $18.95

CD-R data discs 50pk for $10.00

24. How much is a pack of batteries plus 7% tax?

25. How much for 3 packs of batteries plus 7% tax?

26. How much for a cell phone charger and a pack of CD-R discs plus 7% tax?

Represent each situation with an equation. Solve the equation to answer the

question.

27. Only 0.8% of the students won a trip in the limousine. If there are 625 students,

how many students got to go on the trip?

28. The restaurant increased all of their prices by 5%. If the salmon steak was

originally $18.40, how much is it now?

29. At the sale you buy 3 sweatshirts for $15 each, and 2 pairs of jeans for $24

each. If the tax rate is 7%, how much will the total charge equal?

30. The cell phone company charges $29.95 for 300 minutes. If you go over your

300 minutes, then they charge you $0.45 for each additional minute. How much

will they charge you if you use 412 minutes?

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Find the Errors A student made 3 mistakes below. Identify and correct each mistake.

1.

2.

3.

4.

looking bAck

Vocabulary: perent nrease, sply expressons


1. How do I fnd the area o oposte shapes?

2. How an I sply expressons?

3. How do I fnd the aount o tax on an te?

4. How do I fnd a perent o a value?

5. How do I nrease a value by a perent?

6. How do I fnd the total pre o an te nludng tax?

7. How do I translate stuatons nto expressons and equatons?

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Lesson 60 Equvalent Equatons | 597

Lesson 60 Equialent Equations

Objecties

Understand the concept of equivalent equations.

Generate equivalent equations by adding equal amounts to both sides.

Generate equivalent equations by multiplying both sides by equal factors.

Solve simple equations using symbolic manipulation.

Concepts and Skills





EE.13 Determine if two equations are equivalent, i.e., have the same solutionset.

EE.14 Know that adding the same number to both sides of an equation results

in an equivalent equation.

EE.15 Know that multiplying or dividing both sides of an equation by the same

non-zero number results in an equivalent equation.

EE.16 Solve one-step and multi-step linear equations in one variable.


What is an equation?

What is the denition of a solution to an equation?


1. Use mental math and the correct order of operations to evaluate each

expression.

a. 3 + 5 7

b. 7 3 + 8

c. 4 (2 + 3)

d. (7 + 8) 1

__

5

Vocabulary

equvalentequaton

soluton

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Lesson 60 Equialent Equations

Conepts and Sklls: PR.9, EE.13,

EE.14, EE.15, EE.16

We spent the past several lessons discussing equivalent expressions. Here we will

move on to the concept of an equivalent equation, which is one of the most powerful

ideas in algebra.

Equations andSolutions

At the very beginning of this book we discussed equations and solutions to equations.

Lets quickly review this material.

An equation is a statement claiming that two expressions are equal. In other words,

an equation tells us that two expressions describe the same point on the number

line. For example, here is an equation:

This equation says that 2 jumps of + j are equal to one jump of +6. We can write

this equation using symbols like this:

2 j = 6

The values of j that make these two expressions equal are called solutions to the

equation. In this case, there is only one solution. Its clear that j must equal 3

because 2 3 equals 6:


1. Which two number lines below show equations that have n=4 as the

solution?

EquialentEquations

Lets continue with our example equation:

2 j = 6

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Lesson 60 Equvalent EquatonsEquvalent Equatons | 599

We said that j= 3 is the solution to this equation. However, the above isnt the only

equation that has j= 3 as the solution. For example, here is another equation with

the same solution:

A jump of + j in this equation is exactly the same size jump as it was in the earlier

equation. In this case 4 jumps of + j equals one jump of +12. We write this in

symbols as:

4 j = 12

This equation has the same solution of j= 3 as the earlier equation. In this case

j= 3 is the solution because 4 3 equals 12. This means the following two equations

have the same solution:

2 j = 6 4 j = 12

Because they share the same solution, we call these equivalent equations.

Equivalent equations are equations that share the exact same set of solutions.

Next we will discuss some basic rules for generating equivalent equations.


2. Which number line below shows an equation that is equivalent to 4 + b = 9?

In other words, which equation has the same solution forb?

Adding theSame Term to

Both Sides

Lets continue with our earlier example:

2 j = 6

This equation tells us that 2 j and 6 describe the same point on the number line.

Both expressions describe the point 6:

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As we discussed before, j= 3 is the solution to this equation.

If we add a jump of +1 to each expression, well make two new expressions:

Now both expressions describe the point 7 on the number line. In symbols, we write

the above equation as:

2 j + 1 = 6 + 1

This is an equivalent equation because j=3 is still the only solution.

As long as we add equal amounts to both sides, well generate an equivalent

equation. Instead of adding 1, lets add 4 to both sides:

Now both sides of the equation describe the point 10. In symbols we write the

above equation as:

2 j + 4 = 6 + 4

Again, this is still an equivalent equation because it has the same solution as

before.

We now have our rst rule for generating equivalent equations:

Adding the same term to both sides of an equation generates an equivalent

equation.


3. Which equation below is equivalent to a+ 4 =b?

a. a + 4 + 5 = 7 + 5 b. a + 4 + 20 = b + 4

c. a + 4 + 9 = b + 9 d. a = b + 4

4. Which equation below is equivalent to k= 7 p ?

a.k+ 7 = p + 7 b. 5 + k= 5 + 7 p

c. 1 + k= 8 + 7 p d.k 9 = 7 p + 9

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UsingEquialent

Expressions

Lets examine another rule for generating equivalent equations. Here is the equation

we just nished discussing:

2 j + 4 = 6 + 4

We can generate an equivalent equation by replacing one of the sides with an

equivalent expression. For example, lets apply the commutative property of

addition to the expression on the right:

Changing an expression for an equivalent one doesnt change the solution, so we

end up with an equivalent equation. Here is what it looks like on the number

line:

Since 4 +6 equals 10, we can replace the right side expression with the number 10:

This produces another equivalent equation:

2 j + 4 = 10

We now have another rule for generating equivalent equations:

If we apply the rules for generating equivalent expressions to either side of

an equation, we will end up with an equivalent equation.


5. Consider the following equation:

18 = 5 2 + 2 a

Use each of the following properties to generate an equivalent equation.

a. Commutative property of multiplication

b. Commutative property of addition

c. Distributive property of multiplication over addition

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Multiplyingby the Same

Factor

Another way to generate an equivalent equation is to multiply both sides by the

same factor. To see how this works, lets return to our original example:

2 j = 6

If we multiply each side by 2, we get the following:

In symbols this is:

2 (2 j) = 2 6

Since we havent changed the value of j, we still have j=3 as the only solution.

We can also multiply both sides by a fraction. Instead of multiplying by 2, lets

multiply both sides by 1_2. This will give us:1_2 (2 j ) = 1_2 6

Since 1_2 of 6 equals 3, we can replace the right side of the equation with thenumber 3:

1

__

2 (2 j ) = 3

By applying the associative property of multiplication to the left expression we

get the following equivalent equation:

( 1_2 2) j = 3Since 1_2 2 equals 1, we can replace the left side of the equation with 1 j:

1j = 3

By applying the multiplicative property of 1 to the left side, we get the following

equivalent equation:

j = 3

Here is this equation on the number line:

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This equation is great because it shows us instantly that j= 3 is the solution. It also

conrms that j= 3 is the solution to all of these equivalent equations:

2 j = 6

4 j = 12

2 j + 1 = 6 + 1

2 j + 4 = 10

2 (2 j) = 2 6

In the next lesson well learn a powerful technique that involves turning complicated

equations like the ones above into equivalent equations that are much simpler. The

goal is to nd a super simple equivalent equation like j= 3 that basically tells us the

solution to the more complicated ones.

Oops, we almost forgot to state the third rule for generating equivalent equations:

Multiplying both sides of an equation by the same factor generates an

equivalent equation.

The only exception to the above rule is that we cant multiply both sides by 0. To

see why this is not allowed, lets multiply both sides of the equation j= 3 by 0:

0 j = 0 3

Since multiplying by 0 always equals 0, this equation has all values as solutions.

However, there is only one solution to the equation j= 3:

In order for equations to be equivalent, they must have the exact same set of solutions.

In the equation on the left j can be any number and it will form a solution. As for

the equation on the right, the only solution is when j is 3. Since the above two

equations have different solution sets, they are not equivalent.


6. Which equation below is equivalent to a + 4 = b?

a. 2 a + 4 = 2 b b. a + 4 7 = b + 7

c. 0 (a + 4) = 0 b d. 3 a + 12 = 3 b

7. Generate an equation that is equivalent to w = 12 by applying the following

sequence of steps:

Step 1: Add 8 to both sides of the equation.

Step 2: Multiply both sides of the equation by 3.

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Step 3: Apply the distributive property on the left side of the equation.

Step 4: Evaluate the right side of the equation.

You just made an equation more complicated. In the next several lessons

we are going to learn how to undo this and make complicated equations

into equivalent simple equations.

8. What sequence of steps turns the equation on the top into the equivalent

equation on the bottom?

a. c = 9 b. d = 5

2 (c + 1) = 20 3 d+ 2 = 17

c. 8 t = 56 d. m 8 + m 2 = 50

t = 7 m = 5

Problem Set Each equation below has 1 solution only. Find the solution.

1. 7 j = 56 2. 47 = 7 + g 3. 36 =n__2

4. 46 + d = 85 5.1

__

3+ k =5

__

6 6. 32 = 3p

Find four different solutions to each of the following equations. Organize the

solutions in a table.

7. a + b = 100 8. ab = 100 9.1

__

2p = k

Generate an equivalent equation as directed.

10. c = 5 Add 7 to both sides.

11. 7 = 2 k+ 5 Multiply both sides by 3.

12. 2 k = 12 Multiply both sides by1

__

2.

13. 3 (d+ 8) = 30 Use the commutative property of multiplication.

14. 3 (w 2) = 12 2 Use the associative property of multiplication.

15. 18 = d

7 + d

2 Use the distributive property.

What rule was used to turn the equation on the top into the equivalent equation

on the bottom?

16. 5 = k 17. 8 = 3 + k 18. 16 = 2 (3 +k)

3 + 5 = 3 + k 2 8 = 2 (3 +k) 16 = 2 3 + 2 k

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19. 16 = 6 + 2 k 20. 2 m = 18 21. 1m = 9

4 + 16 = 4 + 6 + 2 k1

__

2 2 m =

1

__

2 18 m = 9

Generate an equivalent equation as directed.

22. p = 10

Step 1: Add 5 to both sides.


Step 3: Multiply both sides by 8.


23. 32 = h 5 + h 3

Step 1: Apply the distributive property to the right side.

Step 2: Evaluate what is in the parentheses on the right side.

Step 3: Multiply both sides by1

__

8

.

Step 4: Evaluate the left side.

24. 15 = m





25. 2 = b





26. 99 = 9 (n +1)


Step 2: Apply the multiplicative property of 1 to the right side.

Step 3: Add 1 to both sides of the equation.Step 4: Evaluate the left side of the equation.

What sequence of steps turns the equation on the top into the equivalent

equation on the bottom?

27. w = 4 28. w = 4

5 w = 20 5 w + 3 = 23

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29. w = 4 30. w = 4

w + 3 = 7 5 (w + 3) = 35

Challenge Problems

1. Find the solutions to the equations in problems 12 through 15.


1. When two equations are equivalent...

they equal the same value.

they share some of the same solutions.

they have the exact same set of solutions.

they look the same.

2. Which of the following is NOT a way to generate an equivalent equation?

Add the same term to both sides of the equation.

Multiply both sides by the same non-zero factor.

Replace one of the sides with an equivalent expression.

Move a number from one side to the other side.


1. Dene the following concepts in your own words: equation, solution to an

equation, and equivalent equation.

2. What is the most complicated equation you can make (that will t on one line ofyour piece of paper) that is equivalent to the following:

m = 9

Show all of the steps that you used to create the complicated equation. How do

you know that this complicated equation is equivalent to m = 9?

3. Explain why we cant multiply both sides of an equation by zero to create an

equivalent equation.

4. Are the following two equations equivalent? Explain your reasoning.

5 +k = g g = 5 +k

Find the Errors A student made two mistakes generating equivalent equations. Identify andcorrect both mistakes.

1. 2.

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looking bAck

Vocabulary: equvalent equaton, soluton


1. What does t ean when two equatons are equvalent?

2. How an I generate equvalent equatons? Lst the varous ways andgve exaples.

3. How do I solve sple equatons.

4. Why ant I ultply both sdes o an equaton by 0 and get an

equvalent equaton?

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608 | chapter 11 Equvalent Expressons and Equvalent Equatons

Lesson 61 Inerse Property of Multiplication

Objecties

Understand and apply the inverse property of multiplication to solve equations.

Generate equivalent equations by multiplying both sides by equal factors, and

use this to solve equations.

Write and solve simple equations involving percentages.

Concepts and Skills





EE.15 Know that multiplying or dividing both sides of an equation by the same

non-zero number results in an equivalent equation.

EE.16 Solve one-step and multi-step linear equations in one variable.

SN.9 Solve word problems involving percents and word problems involving a

percentage increase or decrease.


What is the multiplicative property of 1?

What operation is the inverse of multiplication?


1. Use mental math to evaluate each expression.

a.2

__

3

4

__

5

b.3

__

5

5

__

3

c.7

__

5 20

d. 45 3

__

9

Vocabulary

nverse property oultplaton

ultplatve

nverse

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Lesson 61 Inverse Property o MultplatonInverse Property o Multplaton | 609

Lesson 61 Inerse Property of Multiplication

Conepts and Sklls: PR.9, EE.15,

EE.16, S.9

In the previous lesson we discussed some rules for generating equivalent equations.

These rules are useful for simplifying problems so that the solutions are easier to

nd. We will see how this works here by focusing on the rule involving multiplying

both sides of an equation by the same factor.

Rectangleswith Unknown

Sides

The following is a oor plan for a small rectangular apartment:

The plan says that the area of the apartment is 750 square feet. One of the sides is

shown as 20 feet, but the length of the longer side is not indicated. Instead of havingto call up the architect or go measure the apartment ourselves, we can simply use

mathematics to nd the length of the other side. Lets see how.

The following shows the rectangular shape of the apartment:

We have represented the unknown length with the variable u. We can represent the

area of the above rectangle with the following expression:

u 20

The above expression describes the area of the apartment, but we already know that

the area of the apartment is 750. Since these two expressions equal the same area,

we have the following equation:

u 20 = 750

We can nd the value ofu by using the techniques we learned in the previouslesson. The goal is to create an equivalent equation that tells us the value of u. Lets

start by multiplying both sides of the equation by 1__20:(u 20) 1__20 = 750 1__20

You will see why this is useful in a minute.

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According to the associative property of multiplication, we can change how we

group the factors on the left side expression:

u (20 1__20) = 750 1__20We know that 20 1__20equals 20__20:

u (20__20) = 750 1__20

Since20__20 equals 1, we can use the multiplicative property of 1 to simplify the left

side of this equation:

u = 750 1__20Lets now express the right side as a fraction:

u =750___20

We can simplify the fraction750___20 like this:

This tells us that u equals75__2:

u =75__2

Lets write75__2 in decimal notation using long division:

From this we conclude that u = 37.5 is the solution to our equation:

u = 37.5

This means the length of the apartment is 37.5 feet. We can now ll in the missing

value on the oor plan:

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Verifying ourSolution

Just to be sure that we didnt make a mistake, lets verify our result. We expressed

the area of the apartment as:

u 20

Now that we know the value of u, we can write this expression as:

37.5 20

Lets multiply to verify that this expression equals 750 as it should:

We have just veried our solution. Since it is very easy to make a mistake, it is

important to check that the solution works in the original equation.


1. Solve for the missing side of each rectangle.

InerseProperty of

Multiplication

Earlier we simplied an equation by multiplying both sides by the same factor. The

way we simplify with multiplication is by nding a product that equals 1.

Earlier we multiplied 20 by 1__20 to produce 1:20 1__20 = 1

In general, we can multiply a fraction by the inverted fraction to equal 1. For

example, lets nd the value of3_2 times 2_3:

2

__

3

3

__

2

By simplifying the expression, we get:

This shows us that 2_33_2 equals 1.

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This is called the inverse property of multiplication, and we can describe how it

works in general with the following identity:

Its easy to visualize this property in the context of area. For example, here is a unit

square:

If we break the width into 3 equal parts we get:

If we now break the height into 2 equal parts, we get:

This is still 1 unit square, but weve sliced it up into 6 equal parts. Now we can

rearrange the parts like this:

We still have an area of 1 unit square, but now the shape is a rectangle that is 2_3wideand 3_2 tall. From this we can see that

2_3 3_2 equals 1.


2. Solve for the value that gives a product of 1.

a. g5

__

7 = 1 b.18___124m = 1 c. 0.5 h = 1

d. k 15% = 1 e. 42 r = 1 f. 19.2 s = 1

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Finding thePrice from

the Tax

We can use the inverse property of multiplication to help us solve some interesting

problems. For example consider the following situation:

My friend got me a video game for my birthday. He gave me the receipt so I

could return it if I wanted. My friend didnt want me to know how much he

paid for the gift, so he crossed out the price on the receipt like this:

The only problem is that he didnt cross out the sales tax. Using our knowledge

of equivalent equations and the inverse property of multiplication, we

cangureoutthepriceofthevideogamefromthetax.Letsseehowthis

works.

We want to gure out the price of the video game, so lets create a variable p to

represent this unknown price:

Price of video game: p

The receipt says the sales tax is 7%. This means the amount of sales tax paid is 7%

of the price of the video game. Since the price of the video game is p, we have the

following expression for the sales tax:

Sales tax: 7% p

We know that 7% is just another way of writing7___100, so lets rewrite the

expression as:

Sales tax:7___100p

This expression tells us how much sales tax my friend paid. Since the receipt tells

us that the amount of sales tax is $0.91, we have the following equation:

7___100p = 0.91

Both sides of this equation describe the amount of sales tax that was paid. We can

pose the above equation as a question: 7___100 of what value equals 0.91?

Lets use the inverse property of multiplication to solve for the unknown value

p. To do this, we multiply both sides by100___7 :

100___7 7___100p = 100___7 0.91

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According to the inverse property of multiplication, we know that100___7 7___100

equals 1, which gives us:

1 p =100___7 0.91

The multiplicative property of 1 lets us simplify the left expression:

p =100___7 0.91

Since 100 0.91 equals 9

st math ch 11

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