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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 two-digit 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.