Level H Lesson 12
One-Step Equations – Addition and Subtraction with Integers
In lesson 12 the objective is, the student will solve one step equations with addition
and subtraction.
We have three essential questions that will be guiding our lesson. Number 1, why do
we use variables in solving equations? Number 2, what are the goals when solving
equations? Number 3, in an addition equation, how do we isolate the variable?
The SOLVE problem for this lesson is, Rick is saving money to buy a new computer.
He wants to buy the computer during tax-free week at the computer store. The total
cost of the computer is $579.89. Rick has saved $476.25 this summer working part
time. How much more money does he need?
We’re going to complete the S step, study the problem. We’re going to start by
underlining the question. How much more money does he need? Then we’re going
to complete the statement this problem is asking me to find, how much more money
Rick needs.
Today we’re going to be working with one-step equations, addition and subtraction.
We’re going to start by using our balance scale. And in order for our scale to stay
balanced both sides need to be equal. We’re going to be using yellow chips for our
positive, and red chips to represent the negative. We’re going to start by placing 2
yellow chips on the left side of the balance. In order to make this scale not tip this
way and to be balanced, what do we need to add to the right side? We would need to
add the same thing which is 2 yellow chips. Now my scale is balanced.
If I start with 3 red chips, on the left side of my balance scale, in order to make it
balance I would have to have the same number of chips on the right side, which
would be 3 red chips.
Before we move into modeling the equations, we’re going to talk about zero pairs. I
have a value of a yellow counter that is positive 1. What is the value of my red
counter? It is negative 1. If I combine 1 red and 1 yellow, that is called a zero pair,
because the value of adding 1, and negative 1 is 0.
I’m going to put 2 yellow chips on each side of my balance scale again. Now I’m
going to add a zero pair over here, 1 red, 1 yellow, zero pair. Did this change the
value of what is on the right side of my scale? And the answer is no, this still has a
value of positive 2 because this zero pair adding or taking it away does not change
the value.
Our first example we’re going to work with is c plus 2 equals 6. When we model an
equation there are always 2 goals to focus on. The first is to isolate the variable.
What does the word isolate mean? It means to be alone or by your self. The variable
in our equation is a c. It is representing a number. A variable is any letter that
represents a number. And a variable is usually written as a letter of the alphabet. So
our first goal is to isolate this variable, or to get it alone. The second goal in solving
equations is to keep the equation balanced. The equal sign in our equation means
that what ever is on this side, the left side is equivalent to what is on the other side.
So our c plus 2 must be equal in this case to 6. What ever you do to one side of your
equation, you must do to the other side in order to keep it balanced. We’re going to
model this equation, and we’re going to use a cup, as our representative of the
variable c. How are we going to model adding 2? We can do that by showing positive
2, or adding 2 chips to our cup. We now have, a cup plus 2 on the left side of my
equation. I can model the right side of the equation, a positive 6 using 6 yellow
chips. My equation as it exists now is balanced. What are our 2 goals when we’re
working with equations? The first is to isolate the variable, and the second is to keep
the equation balanced. How can we isolate our variable or our cup, or to get it alone?
In order to isolate our variable or our cup we’re going to use the inverse or the
opposite operation. Because this is an addition equation, we’re going to use the
inverse operation for addition which would be subtraction. If our equation was
subtraction we would use the inverse of addition. Because we’re going to subtract,
we’re going to take away 2 yellow chips. We have now isolated our variable, the
variable is alone. However, our equation is not balanced, because in order to keep
our equations balanced, we must do the same thing to both sides. Because we
subtracted a positive 2 here, we’re also going to need to take away or subtract a
positive 2 chips here. My equation is now balanced, because I have performed the
same operation on both sides. The value of c is 4 yellow chips or positive 4. We can
check our work by going back to original equation c plus 2 is equal to positive 6. We
know that when we solved our equation we found that c was equal to 4. So we can
substitute the cup or c for a value of 4 chips. If we have correctly solved the
equation, our equation should be balanced. 6 yellow chips on the left, 6 yellow chips
on the right. So our equation is balanced.
We’re now going to be modeling c plus negative 2 equals negative 6. We’re going to
represent the variable c using our cup. We’re going to represent the negative 2 with
2 chips. So we have cup or c plus negative 2. In order to keep the equation
balanced on the other side we have to have a negative 6. My equation is now
balanced, c plus negative 2 equals negative 6. In order to solve this addition
equation I’m going use the opposite operation which is subtraction. So I’m going to
take away 2 red chips. Remember when solving an equation we want to isolate the
variable, which we’ve done. But we also need to keep our equation balanced. Which
means what ever we do to one side of the equation we must do to the other. Since
we took away 2 red chips from the left side, we also need to take away 2 red chips
from the right side, and our equation is now balanced. C is equal to negative 4. We
can check that by going back to represent the original equation c plus negative 2
equals negative 6. We’re then going to substitute in the value of my cup which is
negative 4, add the negative 2 by pushing them together and I see that my balance
scale is in perfect balance, negative 6 is equal to negative 6.
We’re now going move to a pictorial model of addition equation. We’re going to look
back at problem 1, which was c plus 2 equals 6. Remember when we’re solving an
equation we want to isolate the variable, and we want to keep our equation
balanced. We’re going to write our equation as the c, we’re going to represent our
positive 2 with y (2 yellows) and that’s going to be equivalent to 6 (yyyyyy). Now we
wanted to isolate our variable, so in the first step, we’re going to model how to
subtract or take away our 2 yellow from the left side. We’re going to do that by simply
crossing them out. Remember what ever we do to one side of the equation, we must
do to the other. So we can cross out 2 yellow on the right side of the equation. We
then have a value for c of 4 yellow. In order to check our equation pictorially, we’re
going to go back to the original equation, c plus 2 y’s equals 6 y’s. and we’re going to
substitute in our value of 4 y’s for the c. When we combine or push those together
we end up with 6 y’s equals 6 y’s.
In problem 2, we had c plus negative 2 equals negative 6. We can model that
pictorially by using our c again, and this time we’re going to represent our negative
using the letter R. So we have C plus RR which represents negative 2 and that’s
going to be equivalent to (RRRRRR) to negative 6, which is represented by 6 R. In the
next step what we’re going to do is we’re going to represent how we’re going to
subtract those because we have an addition equation we’re going to use the
opposite operation. We’re going to simply mark then out modeling taking away. That
means my answer is c is equal to 4 R’s, which is negative 4. To check that I’m going
to substitute back in to my original equation the value of C and then I’m going to
model 4 R’s plus 2 R’s is equal to 6 R’s. If I add these, which means to push them
together, that means on the left side of the equations I have 6 R’s and on the right
side of the equation I have 6 R’s.
In the first 2 equations that I modeled we had both values as positive or both values
as negative. In this example I’m going to be modeling adding a positive and having
an answer that is negative on the right side. We have C plus 2 equals negative 6.
Remember C is our variable and we’re going to represent that with a cup. How can
we model adding 2 or positive 2? We can do that with 2 yellow chips. On the right
side of our equation we have 6 red chips or negative 6. Remember in order to solve
our equation we must isolate the variable and keep our equation balanced. This is
an addition equation so we’re going to subtract or take away 2 yellow. Remember we
need to isolate our variable which we’ve done but our equation is not balanced.
Because we have not taken away 2 yellow from this side. Is it possible right now to
take away 2 yellow chips from the right side? No it’s not possible because we have
all red chips. We’re going to create the possibility of taking away 2 yellow chips by
using 0 pairs. Remember at the beginning of the lesson, we learned when we
combine a red and a yellow chip, positive 1, negative 1, the value is 0. 1 red plus 1
yellow is a zero pair, and its value is 0. You can take away or add zero pairs to either
side without changing the value of the equation. I’m going to start by adding 1 red, 1
yellow, 1 zero pair. The equation still has the same value. This is still a negative 6.
Because adding 0 does not change the value. Can we take away 2 yellows from this
side now? The answer is no. I’m going to add another zero pair. Did I change the
value on the right side of the equation, no it is still a negative 6. Because the zero
pairs have a value of 0. However, do I have the ability to now take away 2 yellows?
And the answer is yes. I’m going to take those away and my equation is now
balanced. Is my variable isolated? Yes. Is my equation balanced? Yes. Because the
same operation has been performed on both sides. What is the value of C in this
equation? The value of C is negative 8 because all the chips are red. We can check
our problem by going back to my original equation. C plus positive 2 is equal to
negative 6. Remember when we solved the equation the value of C was negative 8.
We can substitute in for the value of C negative 8 which is 8 red chips. My equation
should now be balanced. We have chips of different colors however on the left side.
So we can create some zero pairs, and take those away without changing the value.
So I have 1 zero pair, and a second zero pair. I now have 6 red chips on the left side
and 6 red chips on the right side. The value on the left side is negative 6, the value
on the right side is negative 6, my equation is balanced.
The last equation I’m gong to model with the chips is C plus negative 2 equals
positive 6. Remember when I’m modeling equations my two goals are to isolate my
variable and to keep the equation balanced. I’m going to represent my C my variable
with a cup. I’m going to add to that negative 2, which is 2 red chips. On the right
side of the equation right now I have 6 yellow chips, which represent my positive 6.
Because this is an addition equation, I’m going to take away the 2 red chips, so that I
can isolate my variable on the left side. Remember to keep my equations balanced I
need to be able to take away 2 red chip on the right side. However, I can’t do that
right now because I have all yellow chips. In the previous equation that we modeled
we learned that we could create the possibility using 1 red, 1 yellow, a zero pair.
Remember if I add this to this side of the scale it doesn’t change the value of this
because the zero pair has a value of 0. This still has a total value of positive 6. I still
have not been able to take away a negative 2 from this side of the equation so I’m
going to add another zero pair, 1 red, 1 yellow, 0 pair. Did I change the value of the
right side of the equation? No, the value of the right side is still positive 6. However,
I have now with my zero pairs, created the possibility of taking away 2 red chips on
the right. When I take those away, I now have a value of C my variable of C is being
equivalent to positive 8. I can use my original equation and substitute in that value
to model the check for this equation. Remember we just found out that the value of
C is going to be positive 8. So what I’m going to do is use 8 yellow chips to model the
value of C or the cup and I’m going to substitute in that value into my original
equation. This side is now equivalent to this side, however, we do have some zero
pairs that we can take off without changing the value. 1 red, 1 yellow, zero pair. 1
red, 1 yellow, zero pair. We have 6 yellow chips on the left and 6 yellow chips on the
right. My equation is balanced.
We’re now going to model the 2 equations that we just modeled with the balance
scales pictorially. I have C plus 2 equals negative 6. My pictorial model looks like
this: I have C plus the 2 Y representing the 2 yellow and that’s equivalent to negative
6(RRRRRR). Remember I want to take away my 2 yellows, because I want to isolate
my variable. But I can’t take away 2 yellows right now. There are no yellows on this
side. I can create the possibility however, by adding a red and yellow which is a zero
pair. It doesn’t change the value of the right side because the value is 0. Can I take
away 2 Y yet? No, I’m going to have to add another zero pair. 1 red, 1 yellow, zero
pair. I’ve now created the possibility of taking away 2 yellows. And my final answer is
C is equal to 8 reds. I can check that by substituting in for C in my original equation
the 8 R’s which represents 8 reds. I’m going to add to that 2 yellows or 2 Y’s. And
then I’m going to have on this side of my equation, I have 6 R’s. On the left side of
my equation R and a Y which makes a zero pair, R and a y which makes a zero pair.
I’m left with 6 R’s on the left side, and 6 R’s on the right side. My equation is
balanced.
The last addition equation I’m going to model pictorially is C plus negative 2 equals 6.
I’m going to model that by using my C and I’m going to add my negative 2 which is
represented with 2 R’s. That’s going to be equal to positive 6 which I can model with
6 Y’s. I want to take away my 2 R’s on this side of the equation so I can isolate my
variable. However, I can not take 2 R’s away on the right side of the equation
because there are no R’s at this point. We did learn however, with our chips that we
could create the possibility by adding a zero pair. This does not change the value on
the right side but it does help us create the possibility to take away 2 R’s. My answer
then is that C is equal to 8 which is represented by 8 Y’s. In order to check that I’m
going to substitute back in to the original equation the value I discovered for C which
was 8 Y’s. When I add the 2 R’s to that I should have an answer of 6 Y’s. Because I
have I have some zero pairs here, I can take away the 2 zero pairs and my final
answer on the left side of the equation I have 6 Y’s and on the right side of the
equation I have 6 Y’s.
We’re now moving on subtraction equations. We will not use concrete models to
show equations but we can use pictorial models which are similar to the ones we
used for addition.
We want to start by talking about the rules for integers. If we have C minus 2, when
we’re subtracting integers, remember our rule is that subtraction is as adding the
opposite. So C minus 2 is the same as saying C plus a negative 2. So what we’re
going to do is we’re going to represent C minus 2 equals 6 by rewriting that using our
integer rules as C plus negative 2 equals 6. Now to represent that pictorially we’re
going to use our model of C plus negative 2 equals 6. We have our cup, we’re going
to add the 2 R’s, we’re modeling this equation, and our answer is going to be positive
6, which is 6 yellow. We want to isolate our variable and in order to do that we’re
going to take away the 2 R’s from the left side, crossing them out. Can we take away
2 R’s from the right side? No we can not. However, we can do it by creating the
possibility with zero pairs. I’m going to add one zero pair to this side. Does that
change the value? No, but can we take away 2 R’s yet? And the answer is no. So
what we’re going to do is add a second zero pair. Now we’ve created the possibility
of subtracting or taking away 2 R’s. We now have a value for C which is 8. The check
for this equation is to go back to our original equation, C plus negative 2 equals 6.
I’m going to substitute in the value of C which is 8 yellows or 8 Y’s. I’m going to add
my 2 R’s, and I’m going to represent the right side of the equation with my 6 Y’s.
Now, my equation is balanced, but notice that I have 1 red, 1 yellow, 1 red, 1 yellow.
I have 0 pairs there. I now have 6 Y’s on the left, and my answer on the right is also
6 Y’s. So I have balanced my equation.
The second subtraction equation we’re going to model is C minus 2 is equal is equal
to negative 6. We can model this pictorially, C plus our 2 R’s, is going to be
equivalent to 6 R’s. We can take away 2 R’s from the left side. And we can also take
away 2 R’s from the right side. Which leaves us a value of C being equal to 4 R’s.
The last step is to check that. We’re going to start by writing our original equation
and substituting in, the 4 R’s into that equation. When we add or push those
together, we have 6 R’s on the left side and that’s going to be equal to the 6 R’s on
my right side. And my equation is done.
We’re now going to be moving to Solving the Equations. Remember the 2 steps when
solving the equation are to isolate the variable and balancing the equation. How can
we isolate the variable? We’re going perform the opposite or inverse operation. This
is an addition equation so the inverse or opposite is to subtract. Subtract 2 from the
left side. The question is does my equation balanced? No, what do we need to do to
balance the equation? We now need to subtract 2 from the right side of the
equation. Because what ever operation you do to one side you must do to the other.
We now subtract 2 from this side of the equation, 6 minus 2 is 4. We bring down our
equal sign and we have solved the equation C equals 4. In order to check our
equation we going to write the original equation. We’re going to substitute in the
value of C and we’re going to and we’re gong to determine if those 2 sides are
balanced, 4 plus 2 equals 6. On this side we have a value of 6, and on the right side
we have a value of 6. So the equation is now balanced.
I’m going to model solving a subtraction equation. We have the subtraction equation
C minus 2 equals 6. Before solving the subtraction equation we’re going to rewrite
the equation as an addition equation using the rules of integers. Subtraction is the
same as adding to opposite. When we’re solving our equation we want to start by
isolating our variable. Because this is an addition equation we can subtract. We’re
going to subtract a negative 2. When we subtract a negative 2, we’re left with C on
the left side of the equation. Is the equation now balanced? No, because we’ve only
subtracted a negative 2 from one side of the equation. In order to balance it we
need to subtract a negative 2 from the right side of the equation. Because what ever
you do to one side you must do to the other. We’re now going to subtract negative 2
from the right side of the equation. When we subtract a negative it’s the same as
adding the opposite so the value of C is 8. We’re now going to check our equation by
writing the original equation, C plus negative 2 equals 6. We substitute in the value
of C, 8 plus a negative 2 equals 6. 8 plus negative 2 does equal 6. The value on
both sides of the equation is 6. My equation is balanced.
We’re now gong back to our SOLVE problem from the beginning of the lesson. Rick is
saving money to buy a new computer. He wants to buy the computer during tax free
week at the computer store. The total cost of the computer is $579.89. Rick has
saved $476.25 this summer working part time. How much more money does he
need?
At the beginning of the lesson we completed the S step, we underlined the question
and then we competed the statement this problem is asking me to find how much
more money Rick needs.
In the O step we’re going to organize our facts. We’re gong to first identify them.
Rick is saving money to buy a new computer, that’s a fact. He wants to buy the
computer during tax free week at the computer store, that’s another fact. The total
cost of the computer is $579.89, that’s another fact. Rick has saved $476.25 this
summer working part time, that’s my final fact. Then we’re going to eliminate
unnecessary facts. We really don’t need to know he’s saving the money. We also
don’t need to know that he’s buying it during tax free week, in order to figure out how
much more money he needs. But we can go ahead and list the necessary facts. The
computer costs $579.89, Rick has $476.25.
In the L step we’re going to line up a plan. We’re going to start by choosing an
operation Since we know how much he needs and we know how much he has we’re
going to subtract. We’re going to write our plan of action. Write an equation I can
use to solve the problem and then solve the equation.
In the V step we’re going to estimate your answer. We have an estimate here of
about $100.00. Then we’re going to carry out our plan. We’ve written our equation
and then we completed the equation by using the rules for equations. We’ve isolated
our variable, and we’ve balanced our equation by doing the same thing to both sides.
We have a value of X of $103.64.
The last part of SOLVE is to examine your results. Does your answer make sense?
Compare our answer to the question. Yes, because we were looking for how much
money Rick needs. Is your answer reasonable? Compare your answer to the
estimate. Yes, because it is close to the estimate of about $100.00. Is your answer
accurate? Go ahead and check your work. Yes it is. Write your answer in a complete
sentence. Rick still needs $103.64 to buy his computer.
We now going to go back and answer the essential questions 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 in solving equations? To isolate the variable
and keep the equation balanced. In an addition equation, how do we isolate the
variable? By using subtraction.