This is Common Core State Standards Support Video for Mathematics. The standard is 2.NBT.B.9.
This standard reads: Explain why addition and subtraction strategies work using place value and the properties of operations. It's a fairly simplistic standard. So let's look at the related and connected standards to this. If we look at 2.NBT.5 that one states fluently add or subtract within a hundred. This is also connected to 2.NBT.6, which involves adding up to 4 two-digit numbers using strategies based on place value and properties of operations. Also 2NBT.7 states add and subtract within a thousand using concrete models or drawings and strategies based on place value and properties of operations. Also related to standard 2.NBT.9 is standard 2.NBT.8, which states mentally add or subtract 10 or 100 to a given number 100 to 900.
At this level, strategies for adding and subtracting whole numbers will be based primarily on definitions and conventions regarding place value in the base ten system, so that's very important. It's the foundation, the fundamental idea that only like items can be added or subtracted. This is critical. Also we will use the commutative property of addition and the associative property of addition. Now it's important to go ahead and use the formal terms; in this way, students will already know the proper terminology. That way they won't have to use a simpler term that's been taught to them and throw it out and re-learn the proper terms later. The concept of equality will be used extensively. A lot of basic arithmetic is about composing, decomposing and rearranging items. Conservation of numbers is important, as is the substitution principle.
Just a little bit of review, the conservation of number refers to the idea that if a group of objects is rearranged the number of objects still remains the same. So if I started out with two sets of three like I have here, I could rearrange them to say a set of four and a set of two. But I still have the same amount. Nothing changed, other than the positions. Related to this is the substitution principle that states that one expression can be replaced with another as long as they are of equal value. That deals more with written symbolism. For example, the two groups of three I could express as three plus three. I can replace that with four plus two because they are equivalent. In turn I could substitute six for the four plus two because again they are of equal value.
With that out of the way, we need to understand that at this stage any strategy for addition or subtraction will pretty much typically involve a combination of the substitution principle, place value, the commutative property, and the associative property. What we'll do here is we'll go ahead and use primarily three-digit numbers. Depending on where your students are, they might be more at the two-digit number stage. For the purposes of this video, we'll be concentrating in using three-digit numbers. Your second graders may not be ready for three-digit adding and subtracting. You might work with them with twodigit numbers instead, but it's still the same strategy. We'll still be using the same properties but use them with two-digit numbers instead of three-digit numbers, until your students transition from using concrete manipulatives to adding and subtracting with three-digit numbers like we'll be doing here.
So let's start off with this example here. Let's use the basic idea of place value and break this problem into three smaller addition problems. Again we took the initial addition problem, and we're breaking it up based on place value where we're adding the 100s, 10s and 1s separately. If we do that we've got 500 plus 150 plus 11. We can combine the hundreds—500 and one 50 to be 650. At the same time, we can break the 11 down to 10 plus 1, which makes it a little bit simpler. Now we can just combine the 650 and the 10 to be 660 add our 1 to get our final sum of 661.
Let's try combining three numbers. Let's use the idea of place value again where we break this up into again their appropriate 100s 10s and 1s. Here's what we did. Now we're using the commutative and associative properties where we're changing the order so we have all of our 100s together, we have all of our 10s together, and all of our 1s together. But notice that we still have the numbers that we started off with. For example, we still have 243, we still have 178, and we still have 361, so in that sense again we haven't changed anything we just used our commutative and associate properties to rearrange our place values. Now it's a matter of actually doing the computing. We can add our hundreds to be 600. Notice on the tens, if students can fluently add within a hundred they should recognize that the 40 plus a 60 is 100. If you rearrange those, to make it a little bit simpler, now we can combine that to be 100. Now we continue with the computation. Combine 600 and 100 that's 780. Again notice three plus seven is 10 so if you convert that very quickly, we have 780 plus 10 would be 790, plus one more is 791.
At a later stage, depending on where you students are, probably at a subsequent grade level, they can almost take this strategy and do this mentally and if nothing else even estimate for reasonableness as far as what kind of answer they should get. So if you look and take your 100s just like you did before, but were doing it mentally we would get two and one that's three, that's 600, and then our 10s, that is 640, add eight more that's 720, with the six that's 780, and then 783, 790, and 791. Again it's possible that students can do this mentally although at this stage they're probably aren't ready.
Let's take this addition problem. Students should notice that 87 is pretty close to 100; so 287 is pretty close to 300. So why don't we convert the 287 to 300 minus 13? Now the addition becomes just 374 plus 300 that's 674, but now we have to subtract 13. That shouldn't be too difficult. We just do our basic subtraction and we get 661 just like we did previously but we did it using a totally different strategy.
Let's take this problem again. Notice that 74 is pretty close to 100; in fact it's 26 away. It sure would be nice if we could combine 74 with 26. We can't take 26 out of the 287, because if we do that, we'll get 261. In essence, your 287 will break down to 26 plus 261. Now we have a nice combination here. We can combine 374 and 26 to be 400. Now we just add in our 261, which would give us 661 just like before.
Let's try a subtraction problem. Notice what's going to be difficult here is that we're going to have to do some decomposing because of the ones place and also the tens place. But first let's review a little bit, it's really important that students understand that operations just like numbers can be separated into different chunks or steps. In other words, something like subtracting nine would be the same thing as subtracting seven and then subtracting two. Numerically it would look like this. No we're not jumping the gun. We're not dealing with negatives here. What we have here and again you have to think of it this way, I am simply saying, what we said here is that subtracting nine is the same thing as a subtracting seven and then subtracting two. So taking that basic idea here subtracting 287 will be the same thing as subtracting 200 then subtracting 80 and subtracting seven. In fact that's what we do in the standard algorithm but we do it in reverse order. First we subtract seven in the ones place, then we subtract 80, and then we subtract 200. Now we're not limited to place value. For example, subtracting 287 would be the same thing as subtracting 250 and then subtracting 37. In fact, we could break this down to whatever we needed to, depending on the context. So if we look at our problem again and start thinking in this manner, let's break it up into place value. Again what I've done is taken this one subtraction problem and actually split the subtraction up into three different subtractions, where we subtracted 200, subtracting 80, and subtracting seven. But this still leaves us with the problem of how to compose and decompose because of these two situations with the 1s and he 10s places. Why don't we do this? Isn't subtracting 80 the same thing as subtracting 70 and subtracting 10? Then over here subtracting seven will be the same thing as subtracting four and then subtracting three. So see the advantage here? What's happened here is that this here will give us a 0, and this will give us a 0 when we do the subtracting. Now we're left with a fairly simple problem. Now we have 100, and then we have to subtract 10 and subtract 3. So 100 minus 10 is 90, and then 90 minus the 3 is 87.
Let's take that same problem again, and this time it sure would be nice if this was a 74 here. If it were 274 it would be again a nice simple subtraction. So what's the difference between the 87 and the 74 here? Well it's 13. So we should be able to take a 13 out of the 287, and if we do, again it is a difference of 13. So the subtracting of 287 would be the same thing as subtracting 274 and then subtracting 13. So with that in my list I rework it and break the subtracting of 287 to subtracting 274 then subtracting 13. Now we're left with 100 minus 13. If subtracting 13 is still a little bit too difficult, we can break that down to subtracting 10 and subtracting 3. Now we have 90 minus three, which is again our 87.
Oh, remember these problems! These were difficult because way back when it was called borrowing. So, now let's go with decomposing. So even with the new terminology this is still difficult. Now isn't 600 the same thing as 599 plus one? If we did this it would make the subtraction a lot simpler because there is no decomposing to be done because we have nines for our ones digit and our tens digit. So they'll be no decomposing that we have to do because a nine is as big as you get as far as your place value. Now we simply do our subtraction, of 131 and of course don't forget the plus one to get our answer of 132. Let's look at this problem. It's also difficult because of that same situation where we would have to decompose both our ones and tens places. If we do something similar to what we did a while ago, we could take the 416 and break it down to 400 plus 16. Then take the 400 and break it down to 399 plus one. Then we can combine the one and the 16 to be 17 and so now we're set. Now we can convert the 416 to 399 plus 17. We can do our subtraction where we don't have to do any decomposing. That's 151 plus 17 and we can break that 17 down to 10 plus seven. Combine that and we get 161 and we can get our final solution of 168.
These are some basic examples of what you can do using some of your basic properties at this level, again focusing on place value, the substitution principal, and then when needed use commutative and associative properties to change the order or to regroup. If your students are not ready for three-digit computation, focus on the two-digit computation and then slowly build up to the three digits that we were doing here as examples.