st math ch 11
TRANSCRIPT
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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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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 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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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:
Check forUnderstanding
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.
Check forUnderstanding
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:
Check forUnderstanding
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 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.
Remember from Before
What does it mean to simplify an expression?
What are some of the properties of rational numbers that we have learned?
Get Your Brain in Gear
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 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.
Check forUnderstanding
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.
Check forUnderstanding
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:
Check forUnderstanding
5. Explain how each of the following examples is an application of the inverse
relationship between multiplication and division:
6. Use equations to describe each of the above examples.
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.
8. Use equations to describe each of the above examples.
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
Check forUnderstanding
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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Check forUnderstanding
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)
Multiple Choice Practice
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.
The commutative property of multiplication.
The additive property of zero.
The multiplicative property of one.
3. Which property is illustrated by the equation on the following number line?
The commutative property of addition.
The commutative property of multiplication.
The additive property of zero.
The multiplicative property of one.
Math Journal Questions
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
Student Self Assessment: o I get t?
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 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
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.
Remember from Before
What do parentheses mean in an expression?
What are equivalent expressions?
Get Your Brain in Gear
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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Check forUnderstanding
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)
Check forUnderstanding
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
equivalent expression:
Check forUnderstanding
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.
Check forUnderstanding
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
equivalent expression:
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
equivalent expression:
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Check forUnderstanding
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:
Check forUnderstanding
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
Multiple Choice Practice
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)
Math Journal Questions
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
Student Self Assessment: o I get t?
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 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
commutative and associative properties of addition and multiplication,
the distributive property, and the special properties of 0 and 1.
Remember from Before
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?
Get Your Brain in Gear
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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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.
Check forUnderstanding
1. The following figure is made of two rectangles. Write an expression for the
total area. Simplify if possible.
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Check forUnderstanding
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.
Check forUnderstanding
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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Check forUnderstanding
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
Student Self Assessment: o I get t?
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 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
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.
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.
Remember from Before
What is an equation?
What is the denition of a solution to an equation?
Get Your Brain in Gear
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:
Check forUnderstanding
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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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.
Check forUnderstanding
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.
Check forUnderstanding
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.
Check forUnderstanding
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.
Check forUnderstanding
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 2: Evaluate the right side of the equation.
Step 3: Multiply both sides by 8.
Step 4: Evaluate the right side of the equation.
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
Step 1: Multiply both sides by 3.
Step 2: Evaluate the left side.
Step 3: Add 5 to both sides.
Step 4: Evaluate the left side.
25. 2 = b
Step 1: Add 7 to both sides.
Step 2: Evaluate the left side.
Step 3: Multiply both sides by 10.
Step 4: Apply the distributive property to the right side.
26. 99 = 9 (n +1)
Step 1: Apply the distributive property to the right side.
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.
Multiple Choice Practice
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.
Math Journal Questions
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
Student Self Assessment: o I get t?
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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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
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.
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.
Remember from Before
What is the multiplicative property of 1?
What operation is the inverse of multiplication?
Get Your Brain in Gear
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 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.
Check forUnderstanding
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.
Check forUnderstanding
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