Download C12 – LESSON 12

Level C Lesson 12
Multiplication and Division Equations
In lesson 12 the objective is, the student will determine the unknown whole number in a
multiplication or division equation using the related multiplication or division fact.
The skills students should have in order to help them in this lesson include, multiplication
and division facts 0 through 9 families, vocabulary for multiplication and division
including Product, Quotient, Factor, Divisor, and Dividend, and use of properties in
multiplication of whole numbers.
We will have three essential questions that will be guiding our lesson. Number 1, what is
the relationship between multiplication and division? Number 2, how do I use
multiplication to solve a division equation? And number 3, how do I use division to
solve a multiplication equation?
The SOLVE problem for this lesson is, Dianne collects cards. She has two copies of
some of the cards. She wants to give 4 cards each to 6 of her friends. How many cards
does Dianne have before giving them to her friends? Set up an equation with 6 as the
quotient.
We will begin by Studying the Problem. First we need to identify where the question is
located in the problem. And we will underline the question. How many cards does
Dianne have before giving them to her friends? Next, we will take this question and put
it in our own words in the form of a statement. This problem is asking me to find the
number of cards Dianne has before giving them to her friends.
During this lesson we will learn how to solve multiplication and division equations using
whole numbers. We will use this knowledge to complete this SOLVE problem at the end
of the lesson.
Throughout this lesson students will be working together in cooperative pairs. All
students should know their role as either Partner A or Partner B before beginning this
lesson.
We will start this lesson by investigating facts with arrays. On Grid A we will represent
the fact 2 times 3. And on Grid B we will represent the fact 3 times 2. We will use the
centimeter cubes that use see below the grid to represent these facts as an array. On Grid
A we are representing the fact 2 times 3, which means that we want to create an array that
shows 2 groups with 3 items in each group. Since each group will have 3 items we will
place our centimeter cubes on the grid in groups of 3; 1 group of 3, 2 groups of 3. This is
called an array. On Grid B we will create an array of the fact 3 times 2, which means that
we will have 3 groups with 2 items in each group, here’s 1 group of 2, 2 groups of 2, and
3 groups of 2. Now let’s take our concrete representation using the centimeter cubes and
turn it into a pictorial representation. We can do this by removing 1 cube from the grid at
a time and shading in the space where that cube was located using our colored pencils.
We will use 1 color to represent our array on Grid A, and a second color to represent our
array on Grid B. The product of each of our number facts is 6. We know this because
there are 6 squares that are shaded on each grid. 2 times 3 equals 6 and 3 times 2 equals
6.
On Grid C and D we will create a concrete representation of the division facts that are
given. On Grid C we want to represent the facts 6 divided by 3 using the centimeter
cubes. We will place 6 cubes on Grid C, with 3 cubes in each of the first 2 rows. On
Grid D, we want to represent the facts 6 divided by 2. So we will place 6 cubes on Grid
D, with 2 cubes in each of the first 3 rows. Now let’s change our concrete representation
of the facts into a pictorial representation, by shading in each square where a centimeter
cube currently sits. We will use 1 color to shade in Grid C, and a second color to shade
in Grid D. On Grid C the fact 6 divided by 3, means that we have a total of 6 items, with
3 items in each group. We will draw a box around each group of 3 items. On Grid D we
have fact 6 divided 2, which means that we have 6 total items, with 2 items in each
group. We will draw a box around each group of 2 items. Our quotient for Grid C is 2,
because there are 6 total items with 3 items in 2 groups. Our quotient for Grid D is 3,
because there are 6 total items with 2 items in 3 groups. Remember that our quotient
represents the number of groups that we were able to make.
Let’s go back to Grids A and B and answer a couple of questions about what we have
observe. What do we know about 2 times 3 and 3 times 2? They have the same product,
but the arrays differ. What three numbers do both facts have in common? Both facts
have the numbers 2, 3 and 6 in common. The numbers 2, 3 and 6 are known as a Fact
Family. This means, that the numbers can be used as factors and a product to create 2
multiplication number sentences.
Now let’s look at Grids C and D and answer a couple of questions about these 2 facts.
What do we know about 6 divided by 3 and 6 divided by 2? They have the same total
items, but the number of items circled differs. What three numbers do both facts have in
common? 2, 3 and 6. Again, 2, 3, and 6 are called a Fact Family. A dividend, divisor
and quotient create 2 division number sentences using this Facts Family.
In our next set of examples the grids have already been shaded in for us. We need to
figure out what the fact is that’s represented on each grid. On Grid A we are shown 3
groups with 4 items in each group. There are a total of 4 shaded squares. So the fact for
Grid A is 3 times 4 equals 12. On Grid B we are shown 4 groups with 3 items in each
group. There are a total of 12 shaded squares on Grid B. So the fact for Grid B is 4 times
3 equals 12. What do we know about Grids A and B? They have the same product, or
equal number of shaded squares, but the arrays differ.
Now let’s look at Grid C and D. Both of these grids show that the arrays have been
divided into groups. Since we see the circled groups on each of the arrays, so we know
that these are both going to represent division problems. Grid C shows a total of 12
items, with 4 items in each of 3 groups. The fact represented on Grid C is 12 divided by
4 equals 3. On Grid D we are shown a total of 12 items, with 3 items in each of 4 groups.
So the fact that is represented on Grid D is, 12 divided by 3 equals 4. So what do we
know about Grids C and D? They have the same number of total items, but the number
of items circled differs. So looking at the examples from Grids A, B, C and D, what
numbers are in this fact family? We have the numbers 3, 4 and 12. These numbers were
used as factors and a product to create 2 multiplication number sentences. These factors
were also used as a dividend, a divisor and a quotient to create 2 division number
sentences.
Now let’s determine the fact that’s represented on Grid A in this next example. There are
5 groups with 3 items in each group. There a total of 15 shaded squares. So the fact
that’s represented by this array is 5 times 3 equals 15. On Grid B we have an array that
shows 3 groups with 5 items in each group with a total of 15 shaded squares. So the fact
that is represented by this array is 3 times 5 equals 15. What do we know about Grids A
and B? They have the same product, or equal number of shaded squares, but the arrays
differ.
In our next example only Grid A has been shaded in with an array. We need to determine
what the fact is on Grid A. Since we have 5 groups with 3 items in each group our fact
is, 5 times 3 which equals 15 as we have a total of 15 shaded squares in the array.
Thinking about the fact families that we worked with in previous activities and the
pictures that they created, let’s determine another multiplication fact for the fact family of
5, 3 and 15. This fact would be 3 times 5 equals 15. Let’s shade Grid B to show the
array of 3 times 5 equals 15. We will have 3 groups with 5 items in each group. Here’s 1
group of 5 items, 2 groups of 5 items, and 3 groups of 5 items. What do we know about
Grids A and B? They have the same product, or equal number of shaded squares, but the
arrays differ.
Now let’s take our fact family of 5, 3 and 15 and create a division number fact using
these numbers. On Grid C, we will create the fact 15 divided by 3 equals 5. We need to
total number of 15 shaded squares. On Grid C these will be in groups of 3. Once we
have shaded in our groups of 3 for a total of 15 shaded squares, we want to place a box
around each individual group, 1, 2, 3, 4, 5. So our answer is 5 groups. Now on Grid D,
we want to create another division number fact, using the same numbers in our fact
family of 15, 3 and 5. Our fact for Grid D is, 15 divided by 5 equals 3. So we will shade
in a total of 15 squares in groups of 5. Now we want to place a box around each group, 1
group, 2 groups, 3 groups. What do we know about Grids C and D? They have the same
number of total items, but the number of items circled differs. Let’s look at the group of
facts that we just created. This fact family is, 3, 5, and 15.
Our next set of Grids, are already shaded in for us. Let’s determine the fact for Grid A.
There are 2 groups of 5 items. So our fact is 2 times 5. The fact for Grid B is, 5 groups,
with 2 items in each group. So our fact is 5 times 2. Both grids have a total of 10 shaded
squares so the third number in this fact family is 10. 2 times 5 equals 10. And 5 times 2
equals 10. What numbers are in this fact family? 2, 5 and 10. Now that we have found
the facts for multiplication in this fact family, let’s see if we can find the facts for division
in this fact family.
In this next example, this problem looks different from the problems (added the word
that) (19: we have been doing in the last few activities. There is a box to represent the
divisor or 1 of the numbers in the number sentence. Since we do not know what number
goes in the box right now we call this an Unknown Value. Mathematicians call a number
sentence with an unknown and an equal sign an Equation. This is a division equation that
has 2 of the numbers in the fact family of 2, 5, and 10. The divisor or the number to be
divided into 10 to find the quotient of 2 is what we are trying to find. How can we use
what we know about fact families to solve this equation? We can multiply to find the
product of 10. I know that 1 of the factors is 2. I can multiply 2 times 5 to find the
product of 10. Therefore, I can use the multiplication fact to determine the missing
divisor. The missing divisor is 5.
Let’s look at the second division equation that would be a part of this fact family. Again,
we have our Unknown Value and our equal sign which makes this an equation. 10
divided by what equals 5? The divisor or the number to be divided into the 10 will give
us a quotient of 5. I can multiply to find the product of 10. I know that 1 of the factors is
5. I can multiply 5 times 2 to find the product of 10. Therefore, I can use the
multiplication fact to determine the missing divisor. The missing divisor is 2.
Now let’s determine the facts that are given to us in Grids A and B seen here. Our first
fact is 12 divided by 4. As we have 12 total shaded squares and we can see that they
have been placed into groups of 4. On Grid B our fact is 12 divided 3, as we again have
12 shaded squares. But this time our shaded squares are put into groups of 3. 12 divided
by 4 equals 3 as I have a total of 3 groups. And 12 divided by 3 equals 4, as I have a total
of 4 groups. What numbers are in this fact family? 3, 4 and 12. Since we have found
the 2 division number facts in this fact family, let’s use an equation to find the number
fact that is in this fact family for multiplication. Again, we have an Unknown Value, and
a equal sign which tells us we are working with an equation. This is a multiplication
equation that has 2 of the numbers in the fact family of 3, 4, and 12. We are trying to
find 1 of the factors in this equation. We can use what we know about fact families to
solve this equation. We can divide 12 by 4 to find the missing factor. The missing factor
is 3. Our second multiplication fact for this fact family can also be found by solving an
equation. This time the 2 numbers that we know in this fact family are 3 and 12. We are
trying to find one of the factors to solve this equation. I can divide 12 by 3 to find the
missing factor. 12 divided by 3 equals 4. So our Unknown Value in this equation is 4.
Let’s look back at problems 2 and 3. Here we have division equations. To solve for the
unknown value in these equations, we multiply. How does multiplication relate to
division? It is the opposite operation. Looking at the equations from problems 5 and 6,
these were multiplication equations. How did we solve for the unknown value in these
multiplication equations? We used division. How does division relate to multiplication?
It is the opposite operation. So we can conclude that in order to solve for missing,
unknown in a multiplication equation we can break it apart and use division to solve.
And we can conclude that in order to solve for a missing unknown in a division equation,
you can break it apart and use multiplication to solve.
Let’s apply this to another problem. Here we have a division equation. Our known
numbers are 3 and 7. So we are looking for the third number in this fact family. The
operation in this equation is division. So the opposite operation is multiplication. If we
multiply our known numbers we can find our unknown number. 3 times 7 equals 21. So
the unknown value in our equation is 21. Let’s test the equation. 21 divided by 3 equals
7.
Let’s take a look at another example like this. Here our known numbers are 8 and 16.
Again, we are looking for the third number in this fact family. The operation in our
equation is multiplication, so the opposite operation is division. We will divide our
known numbers to find the unknown number. 16 divided by 8 equals 2. So 2 is our
unknown. Let’s test the equation, 8 times 2 equals 16.
Now let’s take these Steps to Solve an Equation and write them down together. First,
identify what the question is asking me to find. Second, identify the numbers, I already
know. Third, identify the operation. Fourth, identify the opposite operation. Fifth,
perform the operation. And sixth, rewrite the equation filling in the unknown value.
Now we are going to go back to the SOLVE problem from the beginning of the lesson.
Dianne collects cards. She has two copies of some of the cards. She wants to give 4
cards each to 6 of her friends. How many cards does Dianne have before giving them to
her friends? Set up an equation with 6 as the quotient.
In Step S, at the beginning of the lesson, we Studied the Problem. First we underlined the
questions. How many cards does Dianne have before giving them to her friends? Then
we put this question in our own words in the form of a statement. This problem is asking
me to find the number of cards Dianne has before giving them to her friends.
In Step O, we will Organize the Facts. First we will identify the facts. Dianne collects
cards, fact. She has two copies of some of the cards, fact. She wants to give 4 cards
each, fact, to 6 of her friends, fact. How many cards does Dianne have before giving
them ot her friends? Set up an equation with 6 as the quotient, fact. Next, we will
eliminate the unnecessary facts. These are the facts that will not help us find the number
of cards Dianne has before giving them to her friends. Dianne collects cards. Knowing
this is not going to help us to find the number of cards she has before giving them to her
friends. So we will cross out this fact. She has two copies of some of the cards.
Knowing how many copies she has of some of the cards also will not help us to know
how many cards she has before giving them to her friends, so we cross out this fact. She
wants to give 4 cards each. This is going to be important to knowing the total number of
cards she starts with. To 6 of her friends, this is also important in knowing the number of
cards that she has, before giving them to her friends. Set up an equation with 6 as the
quotient. This fact is going to help us when we line up our plan. So we will keep this
fact as well. Next we will list the necessary facts. 4 cards each, 6 friends, 6 is the
quotient.
In Step L, we will Line up a Plan. First we will choose an operation or operations. Since
we were told that one of our facts is, that we want to have 6 as the quotient in our
equation. Quotient tells us that we’re going to use division. We also know from today’s
lesson that we will use the opposite operation, so we will be using both division and
multiplication. Write in words what your plan of action will be. Set up a division
equation showing the number of cards Dianne has divided by the number of cards she
will give to each friend, with the quotient given. Solve by using the opposite operation of
division.
In Step V, we Verify Your Plan with Action. First we will estimate your answer. We
can say that there are about 25 cards before Dianne gives them her friends. Now we will
carry out your plan. Our plan said that we would set up a division equation showing the
number of cards Dianne has. This is our unknown. Divide it by the number of cards she
will give to each friend, which is 4. With the quotient given, so our unknown divided by
4 equals 6. We will solve by using the opposite operation of division. 4 times 6 equals
24, so our unknown value is 24. 24 divided by 4 equals 6. So Dianne starts with 24
cards.
In Step E, we will Examine Your Results. The first question we ask ourselves is, does
your answer make sense? Here compare your answer to the question. Yes, because I am
looking for the total number of cards. Our second question is, is your answer reasonable?
Here we compare your answer to the estimate. Yes, because it is close to my estimate of
about 25 cards. And our third question is, is your answer accurate? Here you want to
check your work. Verify that 24 divided by 4 equals 6. You can also verify this by using
the opposite operation. 4 times 6 equals 24. So yes, our answer is accurate. Then we
write your answer in a complete sentence. Dianne has a total of 24 cards to give to her
friends.
Now, let’s go back and discuss the essential questions from this lesson.
Our first question was, what is the relationship between multiplication and division?
Multiplication and division are opposite operations.
Number 2, how do I use multiplication to solve a division equation? Multiply the divisor
by the quotient to find an unknown number.
And number 3, how do I use division to solve a multiplication equation? Divide the
product by the factor to find the unknown factor.