Kindergarten · 1.OA.C.6

Subtracting within ten

Ten take away seven. Can a child simply see that three are left, without counting back seven times?

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Adding within five and tenKindergarten Subtraction as taking awayKindergarten

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Counting on to add and counting back to subtractGrade 1 Making ten and bridging through tenGrade 1 Subtracting two-digit numbers without regroupingGrade 2

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Subtracting within ten

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When you reach for it

A child who understands taking away can find 7 - 3: put out seven counters, move three away, count what is left. That always works, and it should stay the thing the child falls back on. But counting every time is slow, it is easy to lose count, and it keeps the child’s attention on the counting instead of on the numbers.

This lesson is the next step: knowing each take-away with numbers up to 10 because the child can see it, in a ten frame, on their fingers, or in a number bond, without counting one at a time. It is the partner of adding within ten, and it is what the child reaches for every time a subtraction comes up later, inside bigger problems as much as on its own.

Quick here means quick because it is seen, not quick because it is rushed. It grows from many short, relaxed turns with objects and pictures, each one talked through, never from racing a clock. After enough of those turns, most children simply know that 10 - 7 is 3, and can still show why if asked.

The procedure

Underneath, every subtraction here is the same: the first number is the whole, the number taken away is one part, and the answer is the other part. That is Parts and the whole. What changes from fact to fact is how the child finds the other part quickly.

For the adult, it helps to know how much there is. Every whole from 0 to 10, with anything from 0 up to that whole taken away, gives 66 take-aways, from 0 - 0 to 10 - 10. That sounds like a lot, but none of them is new. Each is a split of the whole into two parts, one of the bonds to five or to ten read from the whole, and each is an adding fact read backwards: 7 - 3 = 4 because 4 + 3 = 7. A handful of pictures does nearly all the work, and each one works for a reason.

Ten is a full frame. When the whole is 10, fill a ten frame and cross out the part taken away. The counters left and the counters crossed out fill the frame between them, so what is left is the partner of the crossed-out number in ten. A child who knows that 7 goes with 3 knows 10 - 7 = 3 straight away.

Five is a full row. In a ten frame, any number from 5 to 10 is a full row of five with the rest in the bottom row. Taking away the bottom row leaves the full row, and taking away the top row leaves what was below: 8 - 3 = 5 and 8 - 5 = 3. A hand of five fingers works the same way.

A double splits into two equal parts. When the whole is a double the child knows, such as 4 + 4 = 8, taking away one of the equal parts leaves the other: 8 - 4 = 4. The doubles within ten give 2 - 1, 4 - 2, 6 - 3, 8 - 4 and 10 - 5.

Taking away a little. When only 1, 2 or 3 are taken away, the answer is just a few numbers before the whole: for 9 - 2, say “eight, seven”, so 9 - 2 = 7. That is counting back, which has its own lesson. Taking away 0 changes nothing.

Taking away nearly all. When the part taken away is close to the whole, only a few are left, and counting back all that way is long. Ask instead what goes with the part to make the whole: for 9 - 7, “7 and what make 9?” gives 9 - 7 = 2. That is subtraction as a missing part. Taking away everything leaves 0.

One fewer to start leaves one fewer. If the child knows 8 - 4 = 4, then 7 - 4 is one less, 7 - 4 = 3. Four are taken away both times, and the counter that is missing from the start was one of the counters that would have been left. So one fewer at the start, with the same number taken away, always leaves exactly one fewer, as long as the known fact leaves at least one: 4 - 4 = 0 has nothing left to lose, and 3 - 4 cannot be done. One more at the start leaves one more.

Any of these can be checked by putting the parts back together, because taking away only splits the whole: 3 + 7 = 10. And whenever a picture does not come to mind, taking away and counting what is left still gives the right answer.

Worked examples

Read each problem aloud. Let the child build it with counters in a ten frame, or on fingers, before anything is written down.

Example 1

Ten take away seven

Read aloud: “There are 10 cupcakes on a tray. We eat 7 at the party. How many cupcakes are left?”

Solution

Ten is a full ten frame, so put a counter in every box. Take away 7 by crossing out the last seven counters, as in Figure 1.

Figure 1Seven crossed out, three left. The bond says the same thing, 7 and 3 make 10.

Three counters are left at the start of the top row, and a child sees three without counting.

10 - 7 = 3

3 cupcakes are left. Now point out that the 7 crossed out and the 3 left fill the frame together, which is the bond 10, 7, 3. Once that bond is known, the frame is not needed: the partner of 7 in ten is 3. The same bond answers 10 - 3 = 7 as well.

Example 2

Eight, two ways

Read aloud: “There are 8 children on a bus. 5 sit in the front row and 3 sit in the back row. The 3 in the back get off. How many are still on the bus?” Then: “Start again with all 8. This time the 5 in the front get off. How many are still on the bus?”

Solution

Put 8 counters in a ten frame: a full top row of 5 and 3 in the bottom row, just like the bus (Figure 2).

Figure 2One frame of eight, two take-aways. The rows show each answer without counting.

In the first story the back row gets off, so cross out the bottom row. The full top row is left, and a full row is 5:

8 - 3 = 5

In the second story the front row gets off, so cross out the top row. The 3 in the bottom row are left:

8 - 5 = 3

Neither answer needed counting, because the rows of five show them. One frame gave two take-aways, since 8 is 5 and 3 and either part can be the one that leaves. On fingers it is the same: 8 fingers is one whole hand and 3 more, and folding the whole hand leaves 3.

Example 3

Half of a double

Read aloud: “8 ducks swim on a pond in two lines of 4. One line swims away. How many ducks are left?”

Solution

Put out two rows of 4 counters, one for each line of ducks, and cross out one row (Figure 3).

Figure 3Two equal rows. Take one away and the other is left.

The other row is left, 4 ducks:

8 - 4 = 4

A child who knows the double 4 + 4 = 8 can say this at once: the whole is two equal parts, so taking one part away leaves the other. The same reason gives 6 - 3 = 3 and 10 - 5 = 5.

Example 4

A little, or nearly all

Read aloud: “There are 9 pencils in a cup. You take out 2. How many are left?” Then: “There are 9 pencils again. This time you take out 7. How many are left?”

Solution

Only 2 are taken out the first time, so count back from 9 two numbers: “eight, seven”. Seven pencils are left.

9 - 2 = 7

The second time, counting back seven numbers from 9 is a long way, and it is easy to lose track of how many have been said. Instead notice that 7 is nearly all of the pencils, so only a few can be left. Ask “7 and what make 9?” and count on from 7: “eight, nine”. That is 2 more.

9 - 7 = 2

The two answers come from one bond: 9 is 7 and 2. Counting back is short when the part taken is small, and thinking of what goes with it is short when the part taken is large. Both find the same split of 9, so the child can use whichever is shorter.

Example 5

One fewer to start

Read aloud: “What is 7 take away 4?”

Solution

This one does not line up with a picture at a glance: the whole is not ten, 4 is not a full row, and 7 is not a double. So use a fact the child already knows. From the ducks, 8 - 4 = 4. Now 7 is one fewer than 8. Take away the same 4, and one fewer is left (Figure 4).

Figure 4Four are taken away both times. One fewer at the start means one fewer left.

7 - 4 = 3

There is a second way, using five. Seven is a full row and 2 more. Take away the 2 in the bottom row first. That is not yet 4, so 2 more still have to go, and they come from the row of five, which leaves 3. Both ways agree. Check. 3 + 4 = 7, which is also a near double, one more than 3 + 3. Taking away down to a full row and then a little more is the same move that, with ten in place of five, becomes bridging back through ten.

Example 6

Every way to leave three

Read aloud: “Some counters are on the table, never more than 10. Some are taken away, and 3 are left. How many counters could there have been, and how many were taken away? Find every way.”

Solution

Put out the 3 counters that will be left, and keep them. Now add counters that will be taken away, one at a time: none at first, then 1, then 2, and so on. Each time the whole is the 3 kept counters and the ones taken away. Stop when there are 10 counters on the table, which is when 7 are taken away (Figure 5).

Figure 5Keep three, and take away more each time, until there are ten counters.

3 - 0 = 3 \qquad 4 - 1 = 3

5 - 2 = 3 \qquad 6 - 3 = 3

7 - 4 = 3 \qquad 8 - 5 = 3

9 - 6 = 3 \qquad 10 - 7 = 3

There are 8 ways. There cannot be more: taking away 8 would need 3 + 8 = 11 counters, which is more than 10. The list also shows a pattern worth saying out loud: when the whole and the part taken away both go up by one, the number left stays the same. It fits the last example: one more at the start would leave one more, and taking one more away takes that one back. Much later it becomes a strategy of its own, shifting both numbers to keep the difference.

Boundaries

You cannot take away more than you have. With whole things, the number taken away is never bigger than the whole, so 3 - 5 has no answer here. Negative numbers give it one much later, but not by taking anything away.

The order matters. 9 - 2 starts with 9 and takes 2 away. Swapping the numbers gives 2 - 9, a different question, which has no answer with whole things. Addition can be swapped and subtraction cannot: subtraction has an order.

The pictures need the whole to fit in one frame. A ten frame holds a whole of at most 10, which is why every picture here works. For 13 - 5 the whole needs two frames, and taking 5 away uses up the 3 beyond ten and then dips into the full ten; that needs bridging back through ten or counting back, part of subtracting within twenty.

Each shortcut has its own condition. Halving works only when the whole is a double, so only for 2, 4, 6, 8 and 10. Seven is not a double, so 7 - 3 has to be found another way, for example from the near double 3 + 4 = 7. The row of five gives an answer at a glance only when the part taken away is the whole top row or the whole bottom row; otherwise it takes two steps, as in the fifth example. One fewer to start needs a known fact that leaves at least one. Counting back is quick only when the part taken away is small. Taking away and counting what is left works every time.

Where this goes next

These facts are used everywhere subtraction goes. Bridging back through ten works out 13 - 5 as 13 - 3 - 2, which needs 5 split into 3 and 2, a fact within ten. Counting back turns the “taking away a little” picture into a number line strategy. Each bond, read every way, gives a fact family of two adding and two taking-away sentences.

The same facts answer questions where nothing is taken away at all. “Ana has 8 shells and Ben has 5; how many more does Ana have?” is 8 - 5 = 3, the comparing meaning of minus, and story problems of every kind come back to these facts. Later still, subtracting two-digit numbers finds 47 - 23 by working out 7 - 3 and 4 - 2, two facts from this lesson.

Exercises

Read the pictures

Use the pictures for Exercises 1 to 3.

Figure 6Pictures for Exercises 1 to 3.

Picture A. Write the take-away sentence it shows.
Picture B. Write the take-away sentence it shows.
Picture C. What number goes where the question mark is? Say the take-away sentence.

Take away from ten

Use a full ten frame or ten fingers.

10 - 6
10 - 1
10 - 9
10 - 4

Rows, doubles and nearly all

6 - 5
9 - 4
6 - 3
8 - 7

Stories and puzzles

Read each one aloud and show it with counters or fingers.

“Rosa has 7 balloons. 1 floats away. How many balloons does she have now?”
“There are 6 pears in a bowl. We eat 4. How many pears are left?”
“Omar has 10 toy cars. 8 are in the box and the rest are under the bed. How many cars are under the bed?”
“Some counters are on the table, never more than 10. Some are taken away, and 6 are left. Find every way it could happen.”

Explain

Jin says 8 - 5 = 2. Put 8 counters in a ten frame and take 5 away. Is Jin right? How do you know?
Ella knows that 10 - 3 = 7. How can that help her find 9 - 3?
To find 9 - 8, would you count back 8, or think “8 and what make 9”? Why? What is the answer?

Answers

10 - 2 = 8. A full frame is 10 and the last 2 are crossed out, so the top row and 3 in the bottom row are left: 5 + 3 = 8.
7 - 2 = 5. Seven is a full row and 2 more; the 2 are crossed out and the row of 5 is left.
3, and the sentence is 9 - 6 = 3. The whole is 9 and one part is 6; count on from 6, “seven, eight, nine”, so 6 + 3 = 9.
4. 6 and 4 make 10, so 10 - 6 = 4.
9. One finger folded down leaves one less than ten: 10 - 1 = 9.
1. 9 and 1 make 10, so only one counter is not crossed out: 10 - 9 = 1.
6. 4 and 6 make 10, so 10 - 4 = 6.
1. Six is a full row and 1 more; take away the row of five and 1 is left: 6 - 5 = 1.
5. Nine is a full row and 4 more; take away the 4 in the bottom row and the full row is left: 9 - 4 = 5.
3. 3 + 3 = 6 is a double, so taking one 3 away leaves the other: 6 - 3 = 3.
1. 7 is nearly all of 8, and 7 and 1 make 8, so 8 - 7 = 1.
6 balloons. Take away one: one less than 7 is 6, so 7 - 1 = 6.
2 pears. 4 and 2 make 6, so 6 - 4 = 2.
2 cars. The whole is 10 and the box holds 8; 8 and 2 make 10, so 10 - 8 = 2.
5 ways: 6 - 0 = 6, 7 - 1 = 6, 8 - 2 = 6, 9 - 3 = 6 and 10 - 4 = 6. Keep the 6 that are left and add counters to take away, from none up to 4. One more would make 6 + 5 = 11 counters, more than 10.
No. With 8 in a ten frame, taking away 5 crosses out the whole top row and leaves the 3 in the bottom row, so 8 - 5 = 3. Check by putting back: 3 + 5 = 8, but 2 + 5 = 7.
9 - 3 = 6. Nine is one fewer than ten. Taking away the same 3 leaves one fewer than 7, which is 6. Check: 6 + 3 = 9.
Think “8 and what make 9”: just one more, so 9 - 8 = 1. Counting back 8 numbers is long and easy to lose count, but 8 is nearly all of 9, so only a few can be left. Check: 8 + 1 = 9.