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.