Level H Lesson 13
One-Step Equations Multiplication and Division with Integers.
In Lesson 13 the objective is, the student will solve one-step equations with
multiplication and division.
We will have three essential questions that will be guiding the lesson. Number 1,
why do we use variables in solving equations? Number 2, what are the goals when
solving equations? Number 3, in a multiplication equation, how do we isolate the
variable?
The SOLVE problem for this lesson is, the art club is going on a field trip to the art
museum. A total of 138 students and teachers are going on the trip. They are going
by bus and each bus can hold a total of 46 people. How many buses will they need
for the trip?
We’re going to complete the S step at the beginning of the lesson. We’re going to
study the problem. We’re going to start by underlining the question. How many
buses will they need for the trip? Then we’re going to complete the statement, this
problem is asking me to find how many buses are needed for the trip.
Today we’re going to be working with multiplication and division equations. We have
a balance scale here. And in order for this scale to be balanced both sides need to
be equal. We’re going to be using our integer chips today. We have yellow chips to
represent positive and we have red chips that represent negative. I’m going to start
by placing 2 yellow chips on the left side of the scale. The question is what can we
place on the right side of the scale to balance the scale? Because we have 2 yellow
chips on this side, we have 2 yellow chips on the right side and now the scale is
balanced. Next I’m going to place 3 red chips on the left side. We want our scale to
be balanced, right now it is out of balance because we have a negative 3 on this side.
What can I place on the right side so that my scale is balanced? I can place 3 red
chips on the right side. Now my scale is balanced.
We now looking at the equation 3c equals 9. When we model an equation, what
equals 9? What is our first focus? Our first focus is to isolate the variable.
Remember that isolate means to be alone or by yourself. Here in our equation 3c
equals 9 the variable is the c. Remember in an equation a variable is a symbol that
represents an unknown value or number. Usually a variable is written as a letter of
the alphabet. The second focus in our equation is to keep it balanced. The equal
sign means that what ever is on the left side of the equation is equal or the same as
what is on the right side of the equation. If 2 things are equal then they have the
same value. Therefore in an equation what ever you do to the left side you must also
do to the right side. We’re now going to model this equation 3c equals 9. We’re
going to be using cups to represent our variable. One cup is going to represent c. If c
represents 1 cup. We need to find out how we can represent 3 c. We can do that by
using 3 cups. On this side of our balance scale we have 3c. Because each cup
represent c. We have 3c on the left. Now our question is, what are we going to place
on the right side of the scale? We have a positive 9, so we’re going to be using
yellow chips, and I’m going to place 9 yellow chips on the right side of the scale. We
now know that are equation is balanced. Because our equation that we were given
was 3c, 3 cups is going to be equivalent to 9, positive 9 we have 9 yellow chips.
Remember that our two goals when working with the equations are, isolate the
variable, and keep the equation balanced. In other words to keep the equation
balanced what ever we do to the left side equation we must also do to the right. The
first thing we need to do is isolate our variable. We only want to have 1 c here. So
we’re going to have to find a way to do that. Remember when we isolated the
variable in addition and subtraction we used the inverse operation. We now have
multiplication, so if the equation is a multiplication equation, we’re going to divide to
isolate the variable. If this was a division equation we would multiply to isolate the
variable. In this particular equation we’re going to use division, because it’s a
multiplication equation. We’re going to divide our 3 cups into 3 separate groups. So
I have 1 group, 2 groups, and I now have 3 separate groups on the left side of my
equation. Remember to keep our equation balanced we must do to the right side
what we’ve done to the left side. We’ve isolated our variable by dividing. We’re also
going to divide on the right side of the equation. We divided this into 3 groups, we’re
going to have to divide our 9 chips into 3 equal groups. I have 1 group, 2 groups, and
3 groups. Now my equation is balanced because I have divided both sides by 3.
Each of my groups on this side has an equal amount of chips. So the value of 1 cup,
is going to be 3. Each c represents 3. Now let’s go back to our original equation and
check this. We have 3c equals 9. We found that the value of each of our cups was
3. I’m going to substitute in 3 yellow chips for each cup. There’s the value of one
cup. There’s the 3 chips for the second cup, and then 3 chips for the third cup. If I
have correctly solved my equation I should have a balanced scale with 9 chips on
each side. 9 equals 9 so I know that I have correctly solved my equation.
The next equation we’re going to model is 3c is equal to negative 9. We’re going to
use the cup to represent the variable of c. Because we have 3 c we’re going to use 3
cups again. On the right side of the equation we have a value of negative 9.
Remember we represent negative with the red chips. I now have a value of negative
9, and my equation is balanced. When we’re solving equations, remember there are
2 goals. The first goal is to isolate the variable, and the second is to keep our
equation balanced. In order to isolate the variable, we use the opposite operation.
Because this is a multiplication equation we’re going to use division to isolate our
variable. We’re going to take our 3c which is represented by the 3 cups and divide
them, so that each group has a c, 1 cup. Our second goal in solving equations is to
keep it balanced. To keep it balanced we need to do the same thing to the right side
that we did to the left side. On the left side remember we divided by 3. So we’re
going to take our chips and divide them into 3 equal groups. We now see that each
cup is equivalent to negative 3. In order to check our equation we’re going to go
back to the original model and we found that when we solved the equation, that each
cup or c was equivalent to 3 chips. So what we’re going to do is substitute in c with 3
red chips. Our second c another 3 red chips, and the third c is also worth 3 chips. By
checking we see that our equation is balanced, negative 9 is equal to negative 9. So
our equation has been correctly solved.
We’re now going to move to the pictorial model for solving multiplication equations.
We’ve represented 3c equals 9 using our cup and our chips, and now we’re going to
use the pictorial model. 3c we’re going to represent as being equivalent to positive 9
which would be 9y’s. Now, what we need to do is to is isolate the variable. How
we’re going to do that is to divide the 3c, because we have a multiplication equation
we want to do the opposite which is division. So we have 3 c. If divide our 9y’s into
groups of 3, we’re going to have 3 y’s, which represents the yellow in each group. So
if 3c is equal to 9, c is equal to 3. We’re going to check our equation by substituting
our value for c back the original equation. So we have 3c, we have one group of c, 2
groups, 3 groups of c. We find that we have 9y’s on this side, and 9y’s on this side.
We have now modeled 3c equals 9. When we push those together we have 9y
equals 9y. Or 9 yellows equals 9 yellows. 9 equals 9, we correctly solved our
equation.
In our second pictorial model we have 3c is equal to negative 9. We’re going to
represent that by writing the equation using R to represent the negative 9. We now
have 3 c equals negative 9 with our 9 R’s. When solving the equation our pictorial
model, we’re going to model the number dividing, because this is a multiplication,
we’re going to divide this side into 3 groups. Each group has a c. We’re also going to
divide the right side into 3 groups. Because in order to keep our equation balanced,
what ever we do to the left side, we must do to the right side. We now have 3 groups
with 3R or negative 3 in each group. We’re going to check our equation by
substituting back in to the original equation. So we have each group of c’s, 1 group,
2 groups, 3 groups on the left side of our equation. On the right side of the equation
we have our 9 R’s. Our final step, is to push all these together. We now have
negative 9 equals negative 9, because we have 9 R’s on the left side, and 9 R’s on
the right side.
We’re now going to move to solving the equation. We have the equation 3c equals 9.
When we solve our equation the very first thing we need to do is to isolate the
variable. The way we isolate the variable is by using the inverse or the opposite
operation. Because this is a multiplication equation, we’re going to use division.
We’re going to start by dividing the left side by 3, so that we can isolate our variable.
3 divided by 3 is 1, so that will isolate our c, our variable. Is our equation balanced?
No, because we only divided the left side by 3. In order to keep our equation
balanced, we need to divide the right side by 3. We’ve now isolated our variable c,
because 3 divided by 3 is 1. and now we divide the right side by 3, 9 divided by 3 is
3. As we did in our pictorial and our concrete model we’re going to substitute back in
our original equation to check our answer. We start with the original equation 3c
equals 9. We found the value of c to be 3. 3 times 3 is 9. 9 is equal to 9. We have
correctly solved our equation.
Our second equation we’re going to solve 3c is equal to negative 9. We’re going to
use the same steps as with our model. We have two goals when solving the
equation. The first is to isolate the variable. Because this is a multiplication
equation we’re going to isolate the variable using division. We’re going to divide the
left side by 3. Our equation is not balanced at this point because we have only
divided the left side by 3. In order to balance our equation we need to use the same
operation on the right side of the equation. So we’ll divide the negative 9 by 3. 3
divided by 3 is 1. So we’ve isolated our variable and negative 9 divided by 3 is
negative 3. We want to check our equation to make sure we have the correct
solution. We start by writing the original equation; 3c is equal to negative 9. We
found the value of c to be negative 3. So we’re going to substitute that back into the
original equation. Our final step is to multiply 3 times negative 3 which is negative 9.
And we find that we correctly solved our equation. Negative 9 equals negative 9.
Let’s look at problem number 3. Negative 3c equals 9. It is not possible to model
this problem either with pictures or with cups. You cannot represent a negative value
with the cups or pictorially. What we’re going to do is use the same process we did in
problems 1 and 2. We’re going to start with our equation, negative 3c equals 9.
We’re going to still follow the same guidelines for all equations. We’re first going to
isolate the variable. We want to have the c by itself, so we have a negative 3, we can
divide by negative 3. We now need to balance our equation, by doing to same thing
to the right side as we have to the left. Because we had a multiplication equation we
divided by negative 3, we’re also going to divide the right side by negative 3.
Negative 3 divided by negative 3 is a 1. So we’ve isolated our variable, and 9 divided
by negative 3, is negative 3. We can check our equation to see if we have the correct
solution. We start again just as we have with all of our models by writing the
equation. We discovered when solving this equation that the value of c was negative
3. So we come back to our original equation and we substitute in the value of
negative 3 for our variable c. Our last step is to multiply on the left side, negative 3
times negative 3 is a positive 9. Our equation is balanced. We have two equal
amounts on either side of the equal sign. So we know we have the correct solution
for our equation. 9 equals 9, c has a value of negative 3.
We’re now going to be working with solving a division equation. When we have a
division equation, the division problems are written as fractions. The fraction bar
represents the division symbols in each problem. We’re not going to be modeling
with chips or using the pictorial model. However, we are going to follow the same
steps as we do with multiplication equations.
We have the first problem here, c divided by 3 is equal to 3. The first thing we want
to do with that equation is to isolate our variable. Because this is a division equation
we can isolate our variable using the inverse operation of multiplication. We need to
find our what number we need to multiply, by in order to isolate our variable. We’re
going to multiply by 3. Which gives us 3c over 3. 3 divided by 3 is 1. Because any
number divided by itself is 1. So now we’ve isolated our variable. The equation is not
balanced however, because we’ve only multiplied the left side by 3. We need to do
the same thing to the right side. We now have 3c over 3 which gives a value of c.
Remember 3 divided by 3 was 1. We’re now going to multiply 3 times 3 which is 9.
The value of c in this equation is 9. We can also check our equation just as we did
with our multiplication equation by writing the original equation, and substituting in
the value we found for c. We found c to be equal to 9. So now we have 9 divided by
3, because remember our fraction bar is a division sign here, 9 divided by 3 is equal
to 3. 9 divided by 3 is 3, and it is equal to the right side of the equation so we have
correctly solved our division equation. c is equal to 9.
We’re now going to back to the SOLVE problem from the beginning of the lesson. The
art club is going on a field trip to the art museum. A total of 138 students and
teachers are going on the trip. They are going by bus and each bus can hold a total
of 46 people. How many buses will they need for the trip?
At the beginning of the lesson we complete the S step by underlining the question
and completing the statement this problem is asking me to find, the number of buses
that are needed for the trip.
In the O step we’re going to organize our facts. We’re going to begin by marking each
fact in the word problem. The art club is going on a field trip to the art museum, that
is a fact. A total of 138 students and teachers are going on the trip, that is a second
fact. They are going by bus, that’s another fact. And each bus can hold a total of 46
people, that’s my final fact. Then we’re going to eliminate any unnecessary facts.
Remember a fact is not necessary if it will not help us determine how many buses
they will need for the trip. We’re going to cross out the first fact because we don’t
need to know that they are going to the art museum. We do need to know the total
number of people on the trip, so that is a necessary fact. We don’t need to know that
they are going by bus, but we do need to know how many student each bus can hold.
We complete the O step by listing the necessary facts, 138 total people, 46 people
on each bus.
In the L step we’re going to line up a plan. We’re going to start by choosing an
operation. We know we have a total number of people and we know how much each
bus can hold, so we’re going to have to figure out how many buses we’ll need. Our
plan is to write an equation that I can use to solve the problem and then solve the
equation.
In the V Step we’re going to estimate our answer. Our estimate is about 4 buses.
Then we’re going to carry out our plan. We write our equation, we know that each
bus can hold 46 people, and we’re using the variable x to represent how many buses
we’ll need. Now there’s a total of 138 people on the trip. We go ahead and carry out
our steps for solving a multiplication equation by using division. We divide by 46, and
we find the value of x to be 3 buses.
In the E step we’re going to examine the results. Does your answer make sense?
Compare your answer to the question. Our answer was 3 buses. And the question
was, how many buses will they need for the trip? Yes, because we were looking for
how many buses were needed. Is your answer reasonable? Compare your answer
to the estimate. Yes, because it is close to the estimate of about 4 buses. Is your
answer accurate? Go back and check your work. We complete the E step by writing
your answer in a complete sentence. There are 3 buses needed for the trip.
We’re going to go back and answer our essential question from the beginning of the
lesson. Why do we use variables in solving equations? To take the place of
unknown values. What are the goals when solving equations? Isolate the variable
and keep the equation balanced. In a multiplication equation, how do we isolate the
variable? With division.