Grade 1 · K.OA.A.4 · 1.OA.C.6

Making ten and bridging through ten

Eight and five make more than ten, and thirteen take away five drops below it. How can the ten hiding inside sums and differences like these be used to find the answer without counting one by one?

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Subtracting within tenKindergarten Parts, wholes and number bondsKindergarten

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Making ten and bridging through ten

Adding and subtracting within 20 from memory

When you reach for it

This lesson has two halves that are one idea. The first half, making ten, adds two numbers whose total goes past ten. The second half, bridging back through ten, takes away across ten. Both stop at 10 on the way, because from ten the rest can be said at once: 10 + 3 is thirteen, and 10 - 2 = 8 is a bond of ten.

Adding past ten

A child adding 3 + 4 can count on from 4 in three small steps, or simply know the answer. A sum like 8 + 5 is harder. Its total is past ten, so counting on takes five counts, “nine, ten, eleven, twelve, thirteen”, and each count is a chance to slip. Making ten is the move for these sums: the two numbers are each less than ten, but together they make more.

The move rests on three things the child already has. The first is the pairs that make ten, from the bonds to ten: 8 needs 2 more to be 10. The second is that a small number can be split into two parts, from the bonds to five and the bonds to ten: 5 is 2 and 3. The third is that ten and some more is a teen number: 10 + 3 is thirteen, with no counting at all (Ten more and ten less, Teen numbers as ten and some more).

Put together, they give this. To find 8 + 5, take 2 of the 5 to fill the 8 up to 10. That leaves 3. And 10 and 3 make 13.

Figure 1The 5 light counters split into 2 that finish the ten and 3 that start the next frame.

Two ten frames show it best (Figure 1). The 8 dark counters sit in the first frame, with two boxes empty. The 5 light counters go in next, filling the boxes in order, so 2 of them finish the first frame and the other 3 spill into the second. Now the first frame is a full ten and the second holds 3. Nobody needs to count: a full frame and 3 more is 13.

Taking away across ten: bridging back through ten

A child can find 13 - 5 by counting back five times: “twelve, eleven, ten, nine, eight”. It works, but five counts in a row is a lot to hold, and it is easy to count the 13 itself or to lose track of how many counts have been made. Bridging back through ten does the same job in two steps, each of which the child already knows:

13 - 3 = 10 \qquad\text{then}\qquad 10 - 2 = 8

The move is for one kind of subtraction: the whole is a number from 11 to 19, and the number taken away is less than ten but bigger than the ones of the whole. Every teen number is a ten and some ones (Teen numbers as ten and some more): 13 is ten and 3, a full ten frame and 3 in the second frame. 5 is more than 3, so the ones run out and some of the 5 has to come from the ten. When the number taken away is not bigger than the ones, as in 17 - 4, nothing crosses ten: the 4 comes straight out of the 7 and the ten is never touched.

Three things the child needs first. Teen numbers seen as ten and some more. The bonds of small numbers, so that 5 can be seen as 3 and 2 (the bonds to five and to ten). And subtracting from ten, which is the bonds of ten read backwards: 8 and 2 make 10, so 10 - 2 = 8 (Subtracting within ten).

The procedure

Making ten to add

Every sum in this half has the same shape: two numbers, each less than ten, that together make ten or more. The move has one idea in it, filling a ten, and three questions a careful adult will ask about it: why the split is allowed, which number to fill, and what is left at the end.

Filling the ten. Start with the number you will fill, 8 in Figure 1, and ask how many more it needs to be 10. That is a bond of ten, one the child already has from the bonds to ten, so it comes without counting: 8 needs 2. Take that many from the other number. Written out, 8 + 5 becomes

8 + 5 = 8 + 2 + 3 = 10 + 3 = 13

and a child can read the middle step aloud as “eight and two make ten, and three more makes thirteen”.

Why the split is allowed. It can look as if the 5 has been changed into something else. It has not. The five light counters in Figure 1 are the same five counters whether we think of them as one group or as a group of 2 and a group of 3, because 2 and 3 make 5. Moving two of them into the empty boxes does not add a counter or take one away. So 8 + 5 and 8 + 2 + 3 are the same pile of counters, and they must have the same total. The split has to be an exact bond of the number being split: 5 can become 2 and 3, never 2 and 4.

The same fact explains why the 5 may be added in two steps. “Add 2, then add 3” is the 5 arriving in two handfuls instead of one, and the pile at the end is the same.

Which number to fill, and which to split. Either number can be filled. In 8 + 5 we could fill the 5 instead: 5 needs 5, so split the 8 into 5 and 3, and 5 + 5 + 3 = 13. The total agrees, as it must, because joining two groups gives the same pile whichever group comes first (see Addition as putting together). But the work is not the same. The number closer to ten needs the smaller piece, so it is usually easier to fill the bigger number and split the smaller one. With 9 in a sum, filling the 9 takes just 1; with 8 it takes 2.

That choice does not depend on which number is written first. In 4 + 9 the 9 is still the one to fill: take 1 from the 4, and 9 + 1 + 3 = 13. Starting from the bigger number is allowed only because the order of the two parts never changes the total, which is the same reason counting on may start from the bigger number.

What is left at the end. After the ten is full, what is left of the split number decides the answer, and two conditions keep it small. If the two numbers make more than ten, at least 1 is left. If both numbers are 9 or less, at most 8 is left, because the biggest such sum is 9 + 9 = 9 + 1 + 8 = 18. So the leftover is from 1 to 8, and the answer is a ten and some ones, a teen number from 11 to 18 that can be said at once. Drop either condition and this fails: 8 + 2 makes exactly ten with nothing left over, and a number bigger than nine, as in 8 + 15, leaves ten or more after the ten is full: 8 + 2 + 13 = 23.

An adult will notice a pattern: adding to 9, the answer is ten and one less than the other number, so 9 + 6 is 10 and 5; adding to 8 it is ten and two less. That pattern is only the first frame being filled, written in digits, and it is worth letting the child find it rather than telling it.

Bridging back through ten

Why the number taken away may be split. Taking away 5 counters in one handful, or 3 and then 2 more, moves the same five counters, so the same counters are left behind. The lesson on taking away showed this with birds leaving a fence in two groups. Since 3 and 2 make 5,

13 - 5 = 13 - 3 - 2

where the right-hand side is read from the left: take 3 away, then take 2 away from what is left. The only rule is that the two parts together make exactly the number taken away, so they are a bond of 5, and both of them are taken away.

Which split to choose. 5 splits as 1 and 4, 2 and 3, 3 and 2, or 4 and 1, and every one of these gives the right answer. One of them is special. 13 is ten and 3, so taking away 3 first removes exactly the ones and leaves one full ten, 13 - 3 = 10, which the child can see without working anything out. What is still owed, 2, then comes out of a ten, and 10 - 2 = 8 is a bond of ten. So the first part is always the ones of the whole, and the second part is whatever of the number taken away is left over. Any other first step misses ten. Taking 1 first gives 13 - 1 = 12, and then 12 - 4 still crosses ten, which is as hard as the question we started with.

Figure 2Take the 3 ones in the second frame first and exactly one full ten is left. The other 2 come out of the ten.

In two ten frames (Figure 2), 13 is a full frame and 3 more. Cross out the 3 in the second frame: the second frame is now empty and exactly one full frame is left, which is 10. The 2 still to take come off the end of the full frame. A frame with its last two crossed out shows 8 at a glance, a full top row and 3 more, so

13 - 5 = 8

On a number line the same two steps are two hops back: 3 from 13 lands on 10, and 2 more lands on 8. It is counting back, done in two big hops instead of five small steps, and the hop that stops on 10 is what makes the second hop easy.

Children often write the two steps as two short number sentences, 13 - 3 = 10 and 10 - 2 = 8, or draw a small number bond under the 5 showing its parts 3 and 2. Either records the thinking; neither is needed once the child can say it.

Another road: take from ten. Some classrooms teach taking all 5 from the ten instead. 13 is ten and 3; 10 - 5 = 5, and the 3 ones were never touched, so 5 + 3 = 8. In the frames that means crossing out 5 in the full frame and leaving the second frame alone. It gives the same answer, 10 - 5 + 3 = 8, because the same 5 counters leave and the same 8 stay; it just adds a step at the end. This lesson follows the first road, stopping at ten on the way down. A third road goes up instead of down, “5 and what make 13?”, which is subtraction as a missing part.

One stop at ten, both ways

The two halves are the same walk in opposite directions. Making ten takes 8 + 5 from 8 up to 10 and on to 13, using 5 as 2 and then 3. Bridging back takes 13 - 5 from 13 down to 10 and on to 8, using 5 as 3 and then 2. The walk passes the same three numbers, 8, 10 and 13; only the direction changes, so the two parts of 5 come in the other order. The same is true of every fact that crosses ten: 9 + 4 climbs 1 and then 3, and 13 - 4 comes down 3 and then 1. That is why each answer checks the other: 8 + 5 = 13 and 13 - 5 = 8 are one fact read two ways.

Worked examples

Read each problem aloud. Two ten frames drawn on paper, with counters, buttons or coins, let the child build every one, and a number line drawn on a strip of paper works as well. The first six add by making ten; the last six take away by bridging back through ten.

Example 1

Nine on the bus

There are 9 children on the bus. 4 more children get on. How many children are on the bus now?

Solution

Put 9 dark counters in the first frame for the children already on the bus. One box is empty. Now put in the 4 children who get on, as light counters, filling the boxes in order.

Figure 3The 4 light counters are the children who get on the bus.

The first light counter fills the empty box, so 1 of the 4 makes the ten. The other 3 go in the second frame. A full frame and 3 more is 13, so there are 13 children on the bus:

9 + 4 = 9 + 1 + 3 = 10 + 3 = 13

The 4 could be split as 1 and 3 because 1 and 3 make 4: the same four children, just thought of as one child and then three.

Example 2

A bond under the number

8 + 6 = \square

Solution

This time there is no picture, only the numbers, which is how the move looks once a child no longer needs the frames. 8 needs 2 to make ten, so split the 6 into 2 and 4, since 2 and 4 make 6. Children often draw the split as a small bond hanging under the 6.

Figure 4The same move written down. The bond under the 6 shows how it splits.

Now add in two steps. 8 and 2 make 10, and 10 and 4 make 14:

8 + 6 = 8 + 2 + 4 = 10 + 4 = 14

So \square = 14. Counting on from 8 also gives 14, “nine, ten, eleven, twelve, thirteen, fourteen”, and it is a fair check, but it takes six counts where the bond takes two small facts.

Example 3

The smaller number first

3 + 9 = \square

Solution

The 3 is written first, but it is the 9 that is close to ten. The order of the two numbers does not change the total, so the child may start with the 9. 9 needs 1, so split the 3 into 1 and 2:

3 + 9 = 9 + 1 + 2 = 10 + 2 = 12

So \square = 12. Filling the 3 also works, but it is more work: 3 needs 7, so the 9 must be split into 7 and 2, giving 3 + 7 + 2 = 12. The answer is the same either way. Choosing the easier ten is a choice, not a rule.

Example 4

Seven and six, two ways

7 + 6 = \square

Solution

Here the two numbers are close, so filling either one is reasonable. Figure 5 shows both.

Figure 5Fill the 7 or fill the 6. The counters differ, the total does not.

Fill the 7: it needs 3, and 6 is 3 and 3, so 7 + 6 = 7 + 3 + 3 = 10 + 3 = 13.

Fill the 6: it needs 4, and 7 is 4 and 3, so 7 + 6 = 6 + 4 + 3 = 10 + 3 = 13.

So \square = 13. In both pictures the first frame is full and 3 counters are left over, because the same 13 counters are being arranged. A child who knows the doubles has a third way, from Near doubles: 6 + 6 = 12, and 7 is one more than 6, so 7 + 6 = 13. Two different routes to one answer is a good check, and a good conversation.

Example 5

Apples in a basket

Lily picks 9 apples. Sam picks 8 apples. They put them all in one basket. How many apples are in the basket?

Solution

The basket holds both groups together, so this is 9 + 8. Fill the 9, which needs only 1. Split the 8 into 1 and 7, since 1 and 7 make 8:

9 + 8 = 9 + 1 + 7 = 10 + 7 = 17

There are 17 apples in the basket. Notice the pattern from adding to 9: the answer is ten and one less than 8. The same story could be done by filling the 8 instead: 8 needs 2, 9 is 2 and 7, and 8 + 2 + 7 = 17.

Example 6

How many more to make fifteen

Ben has 8 stickers. He wants to have 15. How many more stickers does he need?

Solution

This asks for a missing part: 8 + \square = 15. Make ten on the way. From 8, it takes 2 to reach 10. From 10, it takes 5 to reach 15, because fifteen is ten and five. So the whole jump is 2 and 5, which is 7:

8 + 7 = 15

Ben needs 7 more stickers. Check it by making ten forwards: 8 + 7 = 8 + 2 + 5 = 10 + 5 = 15. The same stop at ten, used to take away instead of to add, is bridging back through ten, which the next six examples work through: 15 - 8 is 15 - 5 - 3.

Example 7

Strawberries on a plate

Read aloud: “There are 12 strawberries on a plate. The children eat 4. How many strawberries are left?”

Solution

12 is ten and 2. The children eat 4, which is more than those 2 ones, so the strawberries eaten will reach into the ten. Split the 4 so the first part is the 2 ones: 4 is 2 and 2.

Eat 2 strawberries first, and exactly ten are left:

12 - 2 = 10

Then eat 2 more, from the ten:

10 - 2 = 8

8 strawberries are left, so 12 - 4 = 8. Check. Put the 4 back: 8 and 2 make ten, and 2 more make 12, the number we started with.

Example 8

Hops on a number line

Read aloud: “What is 15 take away 7?”

Solution

15 is ten and 5, so the first hop takes away the 5 ones: from 15 back to 10. The 7 is 5 and 2, so 2 are still to take: from 10 back to 8.

Figure 6Two hops back, stopping at 10 on the way. Five and two make the seven taken away.

The hops in Figure 6 land on 8:

15 - 5 = 10 \qquad\text{then}\qquad 10 - 2 = 8

So 15 - 7 = 8. Counting back seven single steps would pass through 10 anyway; the two big hops simply skip saying 14, 13, 12, 11 and 9 out loud.

Example 9

Choosing the split

Read aloud: “Mia wants to find 14 take away 6. How should she split the 6?”

Figure 714 is ten and 4, so the 6 is split as 4 and 2.

Solution

Look at the whole first, not at the 6. 14 is ten and 4, as the frames in Figure 7 show, so the first part of the 6 is 4. The bond says 6 is 4 and 2:

14 - 4 = 10 \qquad\text{then}\qquad 10 - 2 = 8

So 14 - 6 = 8. Splitting the 6 another way still gives 8, but it does not help. With 3 and 3: 14 - 3 = 11, and then 11 - 3 crosses ten again. The split is chosen by the whole. The same 6 splits as 3 and 3 for 13 - 6, since 13 is ten and 3, giving 10 - 3 = 7; and as 2 and 4 for 12 - 6, giving 10 - 4 = 6.

Example 10

Is there a ten to cross?

Read aloud: “Find 17 take away 4. Then find 17 take away 8.”

Figure 8In A the ones in the second frame are enough. In B they run out, and one counter comes from the ten.

Solution

17 is ten and 7. For 17 - 4, the 4 is less than the 7 ones, so all of it comes from them (frames A in Figure 8). 7 - 4 = 3, and the full ten is still there:

17 - 4 = 13

This is just 7 - 4 = 3 with a ten in front, and there is no stop at ten because nothing crosses it.

For 17 - 8, the 8 is more than the 7 ones, so they run out. Split 8 as 7 and 1 (frames B):

17 - 7 = 10 \qquad\text{then}\qquad 10 - 1 = 9

So 17 - 8 = 9. The question to ask first is always the same: is the number taken away bigger than the ones of the whole? Only then is there a ten to cross.

Example 11

Stickers, checked by adding

Read aloud: “Ava has 16 stickers. She gives 9 to her brother. How many stickers does Ava have now?”

Solution

16 is ten and 6, and 9 is more than 6, so the stickers given away reach into the ten. 9 is 6 and 3:

16 - 6 = 10 \qquad\text{then}\qquad 10 - 3 = 7

Ava has 7 stickers. Check. Give the 9 back by making ten, as in the first half of this lesson: 7 and 3 make ten, and the other 6 make 16, so 7 + 9 = 16.

Example 12

A look past twenty

This one goes beyond twenty on purpose, to show where the move leads. Skip it with a child who is not yet at home with numbers above twenty.

Read aloud: “There are 23 children in the class. 5 go home early. How many children are still in the class?”

Solution

23 is two tens and 3 ones, so the ten to stop at is 20. The 5 is more than the 3 ones, so split it as 3 and 2:

23 - 3 = 20 \qquad\text{then}\qquad 20 - 2 = 18

18 children are still in the class. The move stops at whichever ten is just below the whole.

Boundaries

Making ten

The total has to reach ten. Making ten needs the other number to be at least as big as the piece the filled number needs. In 6 + 3, the 6 needs 4 and there are only 3, so the total, 9, never gets to ten and there is no ten to make. Such sums belong to adding within ten. When the piece uses up the whole second number, as in 8 + 2, the total is exactly 10 with nothing left over.
The split must be an exact bond. The two pieces of the split number have to make that number, no more and no less. That is what keeps the total the same.
Either number may be filled. Filling the one closer to ten is easier, not more correct. It is allowed to start with the second number only because two parts give the same total in either order.
Two numbers under ten. For two numbers from 1 to 9 that make more than ten, what is left after the ten is from 1 to 8, so the answer is a teen number read off at once. With a bigger first number, such as 28 + 5, the same move fills the next ten instead, 28 + 2 + 3 = 30 + 3, which is place value work for later.
Three numbers are a different question. In 8 + 2 + 3 the three numbers come from splitting one. When a problem starts with three numbers, choosing which two to add first is taken up in Adding three numbers.

Bridging back through ten

Both parts are taken away. The two parts of the split together make the number taken away, and each is subtracted: 15 - 7 is 15 - 5 and then take away 2 more, giving 8. Adding the 2 instead would give 12, and a check shows it: 12 + 7 = 19, not 15.
There has to be a ten to cross. The move is needed only when the number taken away is bigger than the ones of the whole. When it equals them, as in 13 - 3, there is only one step, and it lands on 10. When it is smaller, as in 17 - 4, the answer stays above ten.
The number taken away is no bigger than the whole. With whole things, 13 - 15 cannot be done: after all 13 are gone there are none left to take.
Here the number taken away is less than ten. Taking away 10 or more removes the whole ten as well, which is a different move, ten less. And a whole of exactly 20 has no ones beyond its two tens, so 20 - 7 is read from a bond to twenty, 13 and 7, with no stop on the way.
The order of the two steps does not change the answer, only the ease. 13 - 2 - 3 is also 8, because the same five leave. Taking the ones first is simply the order that stops on ten.
It is not always the shortest way. For 12 - 9 the numbers are close together, and counting up from 9, one to reach 10 and two more to 12, is quicker than taking away 9 (subtraction as a missing part). For 11 - 2, counting back two is as quick as two hops of 1. Every road gives the same answer, since a subtraction has only one.

Where this goes next

Making ten is the first strategy that uses ten as a landmark, and bridging back through ten is the same landmark used on the way down. Together they turn the facts within twenty into things a child can work out rather than recite, which is the heart of adding and subtracting within twenty. Next to them sit the other routes to the same facts, doubles and Near doubles, and a child who can choose among them is thinking about numbers rather than counting. Each subtraction found by bridging comes with an addition, as the sticker check showed: 16 - 9 = 7 and 7 + 9 = 16 belong to one fact family.

In adding three numbers the same eye looks for two numbers that make ten, as in 7 + 4 + 3. Later, both moves stop at the next ten in bigger numbers: 38 + 5 = 38 + 2 + 3 = 43, and 43 - 5 is 43 - 3 = 40, then 40 - 2 = 38. On the open number line they become two drawn hops, and they lead into regrouping in addition, subtracting bigger numbers and quick mental strategies. The step where the 2 come out of a full ten is the seed of regrouping in subtraction: when the ones run out, a ten is opened up. The fact that the 5 may be split and added in two steps is named much later, as part of the properties of addition.

Exercises

Pictures for adding

Use the frames in Figure 9. The dark counters are filled up to ten with some of the light counters.

Figure 9Frames for Exercises 1 to 3.

Frames A show 9 + 5. How many light counters fill up the first frame? How many counters are there in all?
Frames B show 8 + 3. How many light counters fill up the first frame? How many counters are there in all?
Frames C show 7 + 5. How many light counters fill up the first frame? How many counters are there in all?

Make ten to add

9 + 3 = \square
8 + 4 = \square
7 + 4 = \square
9 + 7 = \square
6 + 8 = \square
4 + 9 = \square
8 + 7 = 10 + \square
7 + 7 = 10 + \square

Adding stories

There are 9 cars parked on the street. 6 more cars park there. How many cars are parked on the street now?
Kai has 6 red beads and 5 blue beads. How many beads does Kai have in all?
Ella has 9 stickers. She wants to have 13. How many more stickers does she need?

Explain making ten

To add 9 + 5, Zoe moves 1 counter across. Why 1? Show it with ten frames.
Omar says 8 + 5 is the same as 10 + 3. Is he right? How do you know?
Can you make ten to add 6 + 3? How do you know?
Two children add 7 + 8. One fills the 7 first and the other fills the 8 first. Do they get the same answer? Show both ways.

Stop at ten to take away

First take away just enough to land on 10. Then take away the rest. Say both steps.

12 - 5
14 - 5
11 - 6
16 - 8
13 - 7
15 - 9

Pictures for taking away

Figure 10Frames for Exercises 25 and 26.

Look at frames A. Which take-away do they show? Say it as two steps.
Look at frames B. Which take-away do they show? Say it as two steps.

Figure 11The number line for Exercise 27.

Look at the number line. Which take-away does it show, and what is the answer?

Take-away stories

“There are 14 ducks on a pond. 8 swim away. How many ducks are still on the pond?”
“Rosa has 11 crayons. She gives 3 to her friend. How many crayons does Rosa have now?”
“There are 17 children on a bus. At the stop, 9 get off. How many children are still on the bus?”
Find 16 - 4 and 16 - 7. For which one do you stop at 10 on the way?

Explain bridging back

Show with counters why you can take away 5 from 13 by taking away 3 and then 2.
Maya finds 14 - 6 like this: “14 - 3 = 11, then 11 - 3 = 8.” Is she right? Can you split the 6 so that the first step lands on 10?
Leo says: “15 - 7: first 15 - 5 = 10, then 10 + 2 = 12.” Is he right? How do you know?
Why does 17 - 7 need only one step? What is the answer?

Answers

1 fills up the first frame; 14 in all. 9 needs 1, and 4 light counters are left in the second frame, so 9 + 5 = 10 + 4 = 14.
2 fill up the first frame; 11 in all. 8 needs 2, and 1 light counter is left, so 8 + 3 = 10 + 1 = 11.
3 fill up the first frame; 12 in all. 7 needs 3, and 2 light counters are left, so 7 + 5 = 10 + 2 = 12.
12. 9 needs 1, and 3 is 1 and 2, so 9 + 3 = 9 + 1 + 2 = 10 + 2 = 12.
12. 8 needs 2, and 4 is 2 and 2, so 8 + 4 = 8 + 2 + 2 = 10 + 2 = 12.
11. 7 needs 3, and 4 is 3 and 1, so 7 + 4 = 7 + 3 + 1 = 10 + 1 = 11.
16. 9 needs 1, and 7 is 1 and 6, so 9 + 7 = 9 + 1 + 6 = 10 + 6 = 16.
14. Fill the 8: it needs 2, and 6 is 2 and 4, so 6 + 8 = 8 + 2 + 4 = 10 + 4 = 14.
13. Fill the 9: it needs 1, and 4 is 1 and 3, so 4 + 9 = 9 + 1 + 3 = 10 + 3 = 13.
5. 8 needs 2, and 7 is 2 and 5, so 8 + 7 = 8 + 2 + 5 = 10 + 5, which is 15.
4. 7 needs 3, and 7 is 3 and 4, so 7 + 7 = 7 + 3 + 4 = 10 + 4, which is 14.
15 cars. 9 needs 1, and 6 is 1 and 5, so 9 + 6 = 9 + 1 + 5 = 10 + 5 = 15.
11 beads. 6 needs 4, and 5 is 4 and 1, so 6 + 5 = 6 + 4 + 1 = 10 + 1 = 11.
4 more. From 9 it takes 1 to reach 10, and from 10 it takes 3 to reach 13. So 1 and 3, which is 4: 9 + 4 = 13.
9 counters leave one box of the first frame empty, and 9 and 1 make 10. Moving 1 of the 5 fills the frame and leaves 4 in the second, so 9 + 5 = 10 + 4 = 14.
Yes. Take 2 of the 5 to fill the 8 up to ten: 8 + 5 = 8 + 2 + 3 = 10 + 3. No counter is added or taken away, only moved, and both are 13.
No. 6 needs 4 to make ten, but there are only 3. 6 + 3 = 9, which does not reach ten, so there is no ten to make.
Yes, both get 15. Filling the 7: 7 + 3 + 5 = 10 + 5 = 15. Filling the 8: 8 + 2 + 5 = 10 + 5 = 15. The same counters are joined either way, so the total cannot change.
7. 12 is ten and 2, and 5 is 2 and 3: 12 - 2 = 10, then 10 - 3 = 7.
9. 14 is ten and 4, and 5 is 4 and 1: 14 - 4 = 10, then 10 - 1 = 9.
5. 11 is ten and 1, and 6 is 1 and 5: 11 - 1 = 10, then 10 - 5 = 5.
8. 16 is ten and 6, and 8 is 6 and 2: 16 - 6 = 10, then 10 - 2 = 8.
6. 13 is ten and 3, and 7 is 3 and 4: 13 - 3 = 10, then 10 - 4 = 6.
6. 15 is ten and 5, and 9 is 5 and 4: 15 - 5 = 10, then 10 - 4 = 6.
12 - 3 = 9. The frames hold 12 and 3 are crossed out: first the 2 in the second frame, 12 - 2 = 10, then 1 from the full frame, 10 - 1 = 9.
16 - 9 = 7. The frames hold 16 and 9 are crossed out: first the 6 in the second frame, 16 - 6 = 10, then 3 from the full frame, 10 - 3 = 7.
13 - 8 = 5. The line starts at 13 and hops back 3 to 10, then 5 more to 5. The hops make 3 + 5 = 8 taken away: 13 - 3 = 10, then 10 - 5 = 5.
6 ducks. 8 is 4 and 4: 14 - 4 = 10, then 10 - 4 = 6, so 14 - 8 = 6.
8 crayons. 3 is 1 and 2: 11 - 1 = 10, then 10 - 2 = 8, so 11 - 3 = 8.
8 children. 9 is 7 and 2: 17 - 7 = 10, then 10 - 2 = 8, so 17 - 9 = 8.
16 - 4 = 12 and 16 - 7 = 9. Stop at 10 only for 16 - 7. In 16 - 4 the 4 comes out of the 6 ones, 6 - 4 = 2, and the ten stays, so 12. In 16 - 7 the 7 is more than 6, so 16 - 6 = 10, then 10 - 1 = 9.
Put out 13 counters, a full ten frame and 3 more. Push away the 3 in the second frame, then 2 from the ten: 8 are left. Put all 13 back and push away 5 in one handful: again 8 are left. The same five counters went away both times, because 3 and 2 make 5, so 13 - 5 = 13 - 3 - 2 = 8.
Yes, 14 - 6 = 8, since 3 and 3 make 6 and both were taken away. But her second step, 11 - 3, crosses ten again. Splitting the 6 as 4 and 2 lands on ten first: 14 - 4 = 10, then 10 - 2 = 8.
No. The 7 is 5 and 2, and both parts have to be taken away: 10 - 2 = 8, so 15 - 7 = 8. Adding back checks it: 8 + 7 = 15, while 12 + 7 = 19, which is not 15.
17 is ten and 7, and taking away 7 removes exactly the ones, so the first step already lands on ten and nothing is left to take: 17 - 7 = 10.