Grade 1 · 1.OA.D.8

Finding the missing number in an equation

In \square + 6 = 13 the box comes first, and in 11 = \square + 4 it sits after the equal sign. How does a child find the missing number wherever the box is?

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Subtraction as an unknown addendGrade 1 The equal sign means the same amountGrade 1

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Finding the missing number in an equation

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When you reach for it

A child who has learned that the equal sign means the same amount can now meet equations with a hole in them. The hole is a box, \square, and it stands for one number that nobody has said yet. The box can sit anywhere:

\begin{gathered} \square + 6 = 13 \qquad 15 - \square = 9 \\ 11 = \square + 4 \qquad 8 + 5 = \square + 6 \end{gathered}

There is no story here, only the equation. (Stories that lead to equations like these have their own lesson, problems with the change or the start missing.) So the child cannot lean on “what happened first”. What they can lean on is the meaning of the sign. Every one of these equations asks the same question: which number in the box makes the two sides the same amount? A child who reads \square + 6 = 13 as “something and six is the same as thirteen” has the whole problem in hand, wherever the box sits.

Three ways find the number, and the child picks whichever is shortest for the numbers in front of them:

a related fact: an addition fact and a subtraction fact made of the same three numbers, as in fact families. If 13 - 6 = 7 is known, so is the box in \square + 6 = 13.
counting up (or back) on a number line, stopping at 10 on the way, as in counting on and making ten.
a known bond: a fact the child already owns, such as a double or a near double.

Then one more move finishes every problem: put the number back in the box and check that both sides are the same amount. The check is not an extra; it is what “the missing number” means.

All the numbers stay within 20. The lesson is taught over four days:

Part 1: a box in an addition equation, first or second: \square + 6 = 13.
Part 2: a box in a subtraction equation: 15 - \square = 9, and \square - 8 = 4.
Part 3: the box on the other side of the equal sign: 11 = \square + 4.
Part 4: a sum on each side: 8 + 5 = \square + 6.

The procedure

The picture for the whole lesson is the two-pan balance from the equal sign lesson, with counting bears instead of cubes, and one new thing: a mystery bag. Some bears are hidden in the bag. The bag is so light that the balance does not feel it; it feels only the bears inside. When the beam is level, the bears in the bag, together with any bears beside it, are the same amount as the other pan. The bag is the box.

Part 1: a box in an addition equation

Read the lead equation aloud: “Box plus six equals thirteen”, or better, “something and six is the same amount as thirteen.”

\square + 6 = 13

Sindre shows it on the balance. He hides some bears in the mystery bag and puts the bag in the left pan with 6 bears beside it. In the right pan he puts 13 bears. The beam is level.

Figure 1The beam is level, so the bag and the 6 bears are the same amount as 13 bears. That is \square + 6 = 13.

The level beam says the left pan holds 13 bears too. Six of them can be seen; the rest are in the bag. So the bag holds the number that goes with 6 to make 13. Here are the three ways to find it.

Counting up. The box comes first in \square + 6 = 13, but adding works in either order, so \square + 6 is the same amount as 6 + \square. That is the commutative property, and it lets the child start from the 6 they can see. Count up from 6 to 13, stopping at 10: four hops to 10, then three more to 13. Four and three make seven.

Figure 2Count up from 6 to 13, stopping at 10. Four and three make seven.

A related fact. The 13 is the whole and 6 is one part, so the missing part is 13 take away 6. A child who knows 13 - 6 = 7 is done. This is subtraction as finding a missing part, read the other way round.

A known bond. A child who knows the double 6 + 6 = 12 can say: 13 is one more than 12, so the other part must be one more than 6. That gives 6 + 7 = 13.

All three give 7. Check: put 7 in the box. The left side is 7 + 6 = 13 and the right side is 13. Same amount, so 7 is the missing number. Sindre opens the bag and counts 7 bears.

When the box is second, as in 6 + \square = 13, nothing changes: the same three ways apply and the same 7 comes out.

Part 2: a box in a subtraction equation

Read the lead equation: “Fifteen take away something is the same amount as nine.”

15 - \square = 9

Elof starts with 15 bears. He drops some into the mystery bag and ties it shut. 9 bears are left on the table. How many are in the bag?

The 15 bears are the whole. They are split into two parts: the bears in the bag and the 9 on the table. So the missing number is the part that goes with 9 to make 15, and the subtraction equation asks the same thing as

9 + \square = 15

Counting up from 9 is short: one hop to 10, then five more to 15. One and five make six.

Figure 3Count up from the 9 left to the 15 there were. One and five make six.

Counting back from 15 until reaching 9 also works, but it means keeping track of six steps backward. When the number left is close to the whole, counting up is the shorter road.

Check: 15 - 6 = 9. Both sides are 9, so the box holds 6.

The box can also be the whole. In

\square - 8 = 4

the box is the number of bears Elof started with. Taking 8 away left 4, so the whole was split into a part of 8 and a part of 4. Put the parts back together: 8 + 4 = 12. Check: 12 - 8 = 4. A minus sign in the equation does not mean the answer comes from subtracting. The child asks which number is the whole, and the answer here is “the box”, so the child adds.

Part 3: the box on the other side

Read the lead equation: “Eleven is the same amount as something and four.”

11 = \square + 4

Ylva sets it up on the balance with the pans the other way round from Sindre’s: 11 bears on the left, and on the right the mystery bag with 4 bears beside it. The beam is level.

Figure 4The bag is in the right pan this time. The question is the same, so the equation is 11 = \square + 4.

Nothing about the question has changed. The equal sign does not care which side is written first, because “the same amount as” works both ways. So 11 = \square + 4 asks exactly what \square + 4 = 11 asks: what goes with 4 to make 11?

This time the part that is known, 4, is small. Taking 4 away from 11 by counting back is quick: 10, 9, 8, 7. So the box holds 7. A child who prefers to count up gets there too: from 4, six hops reach 10 and one more reaches 11, and six and one make seven.

Check: 7 + 4 = 11, so the right side is 11, the same as the left. True.

Reading the equation aloud as “eleven is the same amount as something and four” keeps both sides in view. So does turning the balance around in the mind: the bag and its 4 bears weigh the same in either pan.

Part 4: a sum on each side

Read the lead equation: “Eight and five is the same amount as something and six.”

8 + 5 = \square + 6

Ylva puts 8 dark bears and 5 light bears in the left pan. In the right pan go the mystery bag and 6 light bears. The beam is level.

Figure 5A sum on each side. The 6 light bears on the right are one more than the 5 on the left, so the bag holds one fewer than 8.

There are two good ways in.

Find the side you can. The left side has no box, so work it out: 8 + 5 = 13 (eight and two make ten, and three more is thirteen). Now the equation says 13 = \square + 6, which is the Part 1 lead with its sides swapped. The box holds 7.

Compare the parts. Look at the light bears. The right pan has 6 and the left pan has 5: one more on the right. The pans balance, so the bag must make up for it by holding one fewer than the 8 dark bears. One fewer than 8 is 7. In bears: move one dark bear across to the light group on the left. The left pan becomes 7 + 6, the very same bears, and now it matches \square + 6 part for part.

Check: with 7 in the box the right side is 7 + 6 = 13, and the left side is 8 + 5 = 13. Same amount.

The second way is worth the time. It does not need the total at all, and it is the same reasoning the equal sign lesson used to see that a sum can be rearranged without changing its amount.

Worked examples

Read each equation aloud before solving it, and let the child build it with bears and a bag when that helps. Examples 1 and 2 go with Part 1, Examples 3 and 4 with Part 2, Examples 5 and 6 with Part 3, and Examples 7 and 8 with Part 4.

Example 1

Seven and what make sixteen

Find the number in the box: 7 + \square = 16.

Solution

The box is the part that goes with 7 to make 16. Count up from 7, stopping at 10: three hops to 10, then six more to 16. Three and six make nine, so the box holds 9.

A child who knows 9 + 7 = 16 can use it straight away: adding works in either order, so 7 + 9 = 16 as well.

Check: 7 + 9 = 16. Both sides are 16, so 9 is right.

Example 2

A double with the box in front

Find the number in the box: \square + 7 = 14.

Solution

What goes with 7 to make 14? This one is a known bond, the double 7 + 7 = 14. So the box holds 7.

The answer is the same number that is already written in the equation. That is allowed: the box asks for whichever number makes the sides match, and nothing says it must differ from the other numbers.

Check: 7 + 7 = 14. True.

Example 3

Seventeen take away what

Find the number in the box: 17 - \square = 8.

Solution

17 is the whole. The box and the 8 are its two parts, so the question is “8 and what make 17?” Count up from 8: two hops to 10, then seven more to 17. Two and seven make nine.

8 + 9 = 17 \qquad\text{so}\qquad 17 - 9 = 8

The box holds 9. Check: 17 - 9 = 8, and the right side is 8. Same amount.

Example 4

The box is the whole

Find the number in the box: \square - 6 = 8.

Solution

The box is the number we start from. Taking 6 away from it left 8, so it was made of a part of 6 and a part of 8. Join the parts: 6 + 8. Make ten from the 8: it needs 2 more, taken from the 6, which leaves 4. So 6 + 8 = 10 + 4 = 14, and the box holds 14.

Check: 14 - 6 = 8. Both sides are 8.

Example 5

The box after the equal sign

Find the number in each box.

(a) \square = 9 + 4 (b) 12 = \square + 3 (c) 8 = 15 - \square

Solution

(a) The box stands alone on one side, so it must be the same amount as the whole of the other side. Work out 9 + 4: nine and one make ten, and three more is thirteen. The box holds 13. Check: 13 = 9 + 4. True.

(b) What goes with 3 to make 12? The known part is small, so count back three from 12: 11, 10, 9. The box holds 9. Check: 9 + 3 = 12, the same as the left side.

(c) Read it from the right: “15 take away something is the same as 8.” Then 15 is the whole and 8 is one part. Count up from 8: two hops to 10, five more to 15, seven in all. The box holds 7. Check: 15 - 7 = 8, the same as the left side.

The box sat in three different places, and every time the question was the same: what makes the two sides the same amount?

Example 6

Trying numbers and checking

Which of 6, 7 and 8 makes 10 = 18 - \square true?

Solution

Put each number in the box and work out the right side.

With 6: 18 - 6 = 12, and 12 is not 10. False.
With 7: 18 - 7 = 11, and 11 is not 10. False.
With 8: 18 - 8 = 10. True.

The box holds 8. Notice the pattern: each time one more is taken away, one fewer is left, so the right side drops by one. Only one number can land exactly on 10. A related fact gets there directly too: 10 + 8 = 18, so taking 8 from 18 leaves 10.

Putting a number back in is how every answer in this lesson is checked. Here it does the finding as well.

Example 7

Seven and five against something and four

Find the number in the box: 7 + 5 = \square + 4.

Solution

Find the side you can. The left side is 7 + 5: seven and three make ten, and two more is twelve. So the equation says 12 = \square + 4. What goes with 4 to make 12? Count back four from 12: 11, 10, 9, 8. The box holds 8.

Compare the parts. The right side has 4 where the left has 5: one fewer. To balance, the box must hold one more than 7, which is 8. Both ways agree.

Check: 8 + 4 = 12 and 7 + 5 = 12. Same amount.

Example 8

Comparing parts that differ by three

Find the number in each box.

(a) \square + 9 = 6 + 8 (b) 13 - 4 = 2 + \square

Solution

(a) Compare the parts. The 9 on the left is 3 more than the 6 on the right. For the sides to match, the box must be 3 less than the 8. Three less than 8 is 5, so the box holds 5. In bears: take 3 bears off the 9 and put them with the box. The 9 becomes 6, matching the 6 on the right, and the box with 3 more must match the 8, so the box is 5.

Check: 5 + 9 = 14 and 6 + 8 = 14. Same amount.

(b) Comparing parts does not help here, since one side subtracts and the other adds. Find the side you can: 13 - 4 is 9 (take 3 to reach 10, then 1 more). The equation says 9 = 2 + \square. What goes with 2 to make 9? Seven. The box holds 7.

Check: 2 + 7 = 9 and 13 - 4 = 9. Same amount.

Boundaries

Some boxes have no whole number. In \square + 9 = 5, the left side is at least 9 whatever goes in the box, and 9 is already more than 5. No number of bears can be put in the bag to make that balance. The same holds for 4 - \square = 7: taking away never leaves more than you started with. These equations are not wrong to write; they just have no answer among the numbers 0, 1, 2, 3 and so on.

One box, one number. In each equation here, exactly one number makes the sides the same amount. A bigger number in an addition box makes that side too big, and a smaller one makes it too small. Equations with two boxes, such as \square + \square = 10, are a different question.

The box can hold zero. In 9 = \square + 9 the box holds 0: an empty bag. Zero is a number like any other here.

Taking away cannot take more than the whole. In 15 - \square = 9 the box cannot be more than 15. When the box is the amount taken, it is always a part of the whole, never bigger.

A check uses the whole of each side. With the box filled in, work out everything on the left and everything on the right, and compare. In 8 + 5 = \square + 6 that means 13 against 7 + 6, not 13 against the 7 alone.

Where this goes next

Missing-number equations are what story problems turn into once they are written down. The stories themselves, and how to tell which number is the whole, are the work of problems with the change or the start missing and of addition and subtraction problems in words. The facts used to fill the boxes keep growing more automatic in adding and subtracting within 20.

In later years the box becomes a letter, as in using a letter for the unknown number, and the balance comes back unchanged: finding the letter rests on the properties of equality, which say that doing the same thing to both pans keeps the beam level. The question asked of the letter is the one asked of the box here: what makes both sides the same amount?

Exercises

Part 1: a box in an addition equation

What number goes in the box? Check by putting it back.

\square + 5 = 11
8 + \square = 13
\square + 9 = 16
6 + \square = 12
Sindre’s balance below is level. How many bears are in the bag? Write the equation.

Figure 6The balance for Exercise 5.

Part 2: a box in a subtraction equation

12 - \square = 8
16 - \square = 7
\square - 4 = 7
\square - 9 = 8
14 - \square = 14

Part 3: the box on the other side

13 = \square + 8
9 = 11 - \square
\square = 8 + 6
18 = 9 + \square
6 = \square - 6
Which of 7, 8 and 9 makes 15 = 6 + \square true?

Part 4: a sum on each side

6 + 7 = \square + 8
9 + 4 = 10 + \square
\square + 5 = 7 + 7
5 + 8 = 4 + \square
16 - 6 = \square + 3

Explain

The box in \square + 8 = 15 holds 7. Without working anything out, say what goes in the box in 8 + \square = 15, and why.
Can two different numbers both make \square + 4 = 12 true? Explain.
Find the number in the box in 7 + 9 = \square + 10 without adding either side. Explain how.
Find the number in the box in \square + 2 = 2 + 9 without adding. Explain how.

Answers

6. Five and five make ten, and one more is eleven, so 6 + 5 = 11. Check: both sides are 11.
5. Count up from 8: two hops to 10, three more to 13. Two and three make five, and 8 + 5 = 13.
7. Count up from 9: one hop to 10, six more to 16. One and six make seven, and 7 + 9 = 16.
6. The double 6 + 6 = 12.
5 bears. The equation is \square + 7 = 12. Count up from 7: three hops to 10, two more to 12, five in all. Check: 5 + 7 = 12.
4. 12 is the whole and 8 is a part: 8 + 4 = 12. Check: 12 - 4 = 8.
9. “7 and what make 16?” Three hops to 10, six more to 16: 7 + 9 = 16. Check: 16 - 9 = 7.
11. The box is the whole, made of 4 and 7: 4 + 7 = 11. Check: 11 - 4 = 7.
17. The box is the whole, made of 9 and 8: nine and one make ten, seven more is seventeen, so 9 + 8 = 17. Check: 17 - 9 = 8.
0. Taking nothing away leaves all 14: 14 - 0 = 14.
5. What goes with 8 to make 13? Two hops to 10, three more to 13: 5 + 8 = 13.
2. Read it as “11 take away something is 9”. Count up from 9: 10, 11, two hops. Check: 11 - 2 = 9.
14. The box is the same amount as 8 + 6. Eight and two make ten, and four more is fourteen, so 8 + 6 = 14.
9. The double 9 + 9 = 18.
12. The box is the whole, made of 6 and 6: 6 + 6 = 12. Check: 12 - 6 = 6.
9. With 7: 6 + 7 = 13, false. With 8: 6 + 8 = 14, false. With 9: 6 + 9 = 15, true.
5. The left side is 6 + 7 = 13, and 5 + 8 = 13. Or compare: 8 is one more than 7, so the box is one less than 6.
3. The left side is 9 + 4 = 13, and 10 + 3 = 13. Or compare: 10 is one more than 9, so the box is one less than 4.
9. The right side is the double 7 + 7 = 14, and 9 + 5 = 14.
9. Compare: 4 is one less than 5, so the box is one more than 8. Check: 5 + 8 = 13 and 4 + 9 = 13.
7. The left side is 16 - 6 = 10, and 7 + 3 = 10.
7. 8 + \square is the same amount as \square + 8, because adding works in either order (the commutative property). So the same number makes it 15: 8 + 7 = 15.
No. Only 8 works: 8 + 4 = 12. A number bigger than 8 makes the left side more than 12, and a number smaller than 8 makes it less.
6. The 10 on the right is one more than the 9 on the left, so the box must be one less than the 7. Check: 7 + 9 = 16 and 6 + 10 = 16.
9. The right side is 2 + 9 and the left side already has the 2. The two sides balance when they have the same parts, so the box holds 9: 9 + 2 is the same amount as 2 + 9, because adding works in either order.