Adding IntegersLesson 2
Our warm up involve adding numbers we will start by adding whole numbers because our lesson is on adding integers. We will be begin our lesson by S-ing our SOLVE problem. The Math Club held a bake sale to raise money for new calculators. Each calculator cost 45.00. Club members are able to raise a total of 320.00. Their expenses for supplies were 35.00. What was their total profit or loss?
Our first step in S is to underline the questions. What was their total profit or loss? The second step in S is to answer the questions what is this problem asking me to find, this problem is asking me to find the Math Club total profit or loss from the bake sale, remind your students that by the end if the lesson they will be able to find the correct answer to this SOLVE problem.
Positive 6 plus positive 2 to represent addition we will represent both groups and push then together. To represent positive six we will use six yellow chip, yellow represent positive numbers. To represent the positive two we will use two yellow chips, addition is a process of pushing together, so when we push the two groups together we have a total of 1,2,3,4,5,6,7,8 yellow chips. Our answer will be positive eight. To represent this problem pictorially, instead of using yellow chips we will use Y’s, to represent a yellow. We will write six Y’s to represent 6 and 2 Y’s to represent 2. We have a total of 1,2,3,4,5,6,7,8 Y’s that gives us an answer of positive 8.
Problem 2 is negative 6 plus negative 2, to represent a negative we will use red. To represent negative 6 we will use 6 red integer chips, to represent our negative 2 we will use 2 red integer chips. The process of addition is to push together, so we are going to push these two groups together and they have a total of 1,2,3,4,5,6,7,8, but they are red, so our answer is negative 8. To represent problem 2 pictorially we will use an R to represent the red. Our negative 6 will be represented with 6 R’s, our negative 2 will be represented with 2 R’s, which gives us an answer in all one color red, and the total of 1,2,3,4,5,6,7,8 so 8 reds is an answer of negative 8
In problem 5 we will add a positive 6 plus a negative 3. To represent our positive 6 we will use six yellow integer chips, to represent our negative three we will use three red integer chips, when we push the two groups together we see that our answer is not all one color. We cannot represent our answer in two different colors, one yellow and one red create a zero pair. Taking one step forward, and taking one step back puts you in the same place you started. It has a value of zero. One yellow one red ahs a value of zero. SO we will remove our zero pairs. One yellow one red zero pair, one yellow one red zero pair, one yellow one red zero pair and our final answer is positive three. To represent this pictorially we will use Y’s and R’s, we will represent our positive 6 with 6 Y’s, we will represent our negative 3 with 3 R’s. When representing pictorially you can draw a line through zero pairs and our final answer is positive three
Problem 6 is negative 6 plus 3. We will represent negative 6 with six red integer chips, we will represent our positive 3 with 3 yellow, once again our answer is in two different colors, so we are going to take away our zero pairs. One yellow one red zero pair, one yellow one red zero pair, one yellow one red zero pair. Our final answer is three reds or negative three. To represent this problem pictorially we will use R’s and Y’s. We will represent our negative 6 with six R’s for red, we will represent our positive 3 with 3 Y’s. We will cross out our zero pairs one yellow one red, one yellow one red, one yellow one red and be left with three reds which is an answer of negative 3.
Five plus three we will represent 5 with 5 yellow integer chips, to represent our three we will use 3 yellow integer chips. Addition is the process of pushing together so our two groups together are 1,2,3,4,5,6,7,8 yellows which is an answer of positive 8. to represent this problem pictorially we will use Y’s to represent the 5 we will represent our 5 with 5 yellows we will represent our 3 with 3 yellows which gives us 1,2,3,4,5,6,7,8 yellows and an answer of positive 8.
To represent negative 7 plus four, we will represent the negative 7 with 7 red integer chips and we will represent the positive 4 with 4 yellow integer chips, when we push the two groups together we see that they are two different colors, we have to remove zero pairs. One yellow one red zero pair, one yellow one red zero pair, one yellow one red zero pair, leaves us with a final answer of three reds or negative 3. To represent this problem pictorially we will use R’s and Y’s, we will represent our negative 7 with 7 R’s, we will represent our positive 4 with 4 Y’s, we will draw a line through our zero pairs, one yellow one red zero pair, zero pair, zero pair and our final answer is 3 reds which is an answer of negative 3.
To represent negative 3 plus negative 3 we will represent our first number negative 3 with 3 reds, we will represent our second number with negative 3 with three reds, when we push the two together their answer is all the same color and there are 1,2,3,4,5,6 reds which is a final answer of negative 6. To represent this problem pictorially we will use R’s negative 3 is 3 R’s plus another negative 3, our answer is all the same color we have all reds so our final answer is 1,2,3,4,5,6 and their reds so our final answer is negative six.
To represent four plus negative one we will represent the four with four yellows and the negative one with one red, our answer is in two different colors so we must remove the zero pairs, which is one yellow one red, one zero pair and we have final answer with 3 yellows which is positive 3. To represent this problem pictorially we will represent the 4 with 4 Y’s and the negative 1 with one R, one yellow one red zero pair that leaves us with three yellow or 3 Y’s which is a positive three.
After developing our rules for addition we are going to move to our graphic organizer. We will just complete the sections on addition. When we add two integers with the same signs we add them together and keep the signs. When we add two integers with different signs we subtract the value and take the sign of the larger value.
We are going to move back to our SOLVE problem from the beginning of the lesson, now that we know how to add integers. We had first underlined our question what was their total profit or loss, and we had S the problem saying that this problem was asking me to find the Math Club’s total profit or loss from the bake sale. In O we will first identify our facts the Math Club held a bake sale to raise money for new calculators, that’s one fact. Each calculator cost 45.00, Club members were able to raise a total of 320.00, fact and their expenses for supplies were 35.00,fact We have to decide which facts are necessary and unnecessary, eliminate the unnecessary facts and list the necessary facts, keeping in mind our question is what was their total profit or loss. The Math Club held a bake sale to raise money for new calculators, this is probably unnecessary. Each calculator cost 45.00, we are trying to find their total profit or loss so the fact that the calculators cost 45.00 is unnecessary, we will draw a line through it. Club members were able to raise a total of 320.00, this fact is necessary, the bake sale earned 320.00 their expenses for supplies were 35.00, this fact is also necessary. Their expenses were 35.00.
In L we line up our plan, the first step in L is to choose the operation or operations, because we are trying to find their total profit or loss we will use addition. Our second step in L is to write in word what your plan of action will be. We want to add the amount earned at the bake sale to the amount of the expenses, but we must represent out expenses as a negative number.
In V you are going to verify your plan with actions estimate your answer, because we know they earned a total of 320.00, but that total is going to go down because of expenses. We can estimate that our answer is going to be less than 320.00. We will also carry out our plan. Our plan said that we were going to add the amount earned by the bake sale which was 320.00 to the amount of expenses, which is negative 35.00. When we add to integers with different signs we subtract the values and take the sign of the larger value. In this case 320 minus 35 is 285.00, the larger value is 320.00, so it will stay as a positive number.
In E examine your results we will answer the first three questions, does your answer make sense, we go back to the question, what was their total profit or loss. 285.00 does make sense. Is you answer reasonable, we will go back to our estimate, less than 300.00. 285.00 is less than 300.00, and is you answer accurate. Your students should turn their paper over and rework the problem or use a calculator to check. Our last step in E is to write you answer in a complete sentence. The Math Club made a profit of 285.00.
To close the lesson we will review the essential questions. The first essential question is, how do you add integers that have the same sign? We will add them together and keep the sign
The second question is how do you add integers that have different signs? We will subtract the values and take the sign of the large value.