Year 4 Mathematics · AC9M4A01

Find unknown values in numerical equations involving addition and subtraction, using the properties of numbers and operations

An equation states that two expressions have the same value. Find a value that preserves equality

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Learning goalsGoals and vocabulary

An equation states that two expressions have the same value. Solving means finding a value that preserves equality.

Learning routine: Read both sides → Identify the unknown’s role → Use a number relationship → Replace the box and check.

Success looks like

  • I can find an unknown addend, starting amount, removed amount or result.
  • I can explain why both sides of an equation have equal value, including when both sides show calculations.
  • I can reorder and group addition, use a common difference, or use an inverse operation to calculate efficiently.
  • I can show my reasoning with equal counters, a number line or a part-part-whole bar, then check the original equation.
equation
a statement that two expressions have equal value
expression
a number or calculation, such as 170 or 80 + 90
unknown
a value to find; a box marks its place in an equation
addend and sum
an addend is a part being added; the sum is the total
minuend and subtrahend
the minuend is the starting amount in a subtraction; the subtrahend is the amount removed
difference
the result of subtraction, also shown as the gap between two numbers
inverse operations
operations that undo each other; addition and subtraction connect the same whole and parts
commutative property of addition
changing the order of the addends preserves their sum
grouping addition
combining the same addends in a different grouping preserves their sum
Clean visual examplesWorked models

Clean visual examples

AC9M4A01 · Equal values and unknowns
Example 1

1. Solve 348 + □ = 725 using a part-part-whole bar

A whole of 725 is split into a known part of 348 and an unknown part.
The box labels a part of the whole, not the whole itself.
  1. Identify 725 as the whole and 348 as one part.
  2. Find the other part: 725 − 348 = 377. One way is 725 − 300 − 48 = 425 − 48 = 377.
  3. Replace the box and check: 348 + 377 = 725.

Addition combines the parts; subtraction finds a part from the whole. These are inverse operations.

Example 2

2. Decide which role the unknown has

EquationMeaning and calculationCheck
□ + 46 = 100Missing addend: 100 − 46 = 54.54 + 46 = 100.
□ − 286 = 439Missing starting amount: put the parts together, 286 + 439 = 725.725 − 286 = 439.
900 − □ = 347Missing removed amount: 900 − 347 = 553.900 − 553 = 347.
□ = 70 − 26Missing result: 70 − 26 = 44.44 = 70 − 26.

The equals sign does not mean “the answer comes next”. A result may appear on either side. Decide whether the box is a whole, a part or a result before choosing a calculation.

Zero also has a role: 83 − □ = 83 needs 0 because nothing is removed; □ − 18 = 0 needs 18 because the whole starting amount is removed.

Example 3

3. Reorder and group addition to make useful pairs

Two number lines from 0 to 7. Jumps of 5 then 2 reach 7; jumps of 2 then an unknown length also reach 7.
The missing jump is 5: 5 + 2 = 2 + 5. The order changes; the total does not.

With three groups of counters, 2 + 2 + 3, 2 + 3 + 2 and 3 + 2 + 2 all total 7. Rearranging the groups does not add or remove counters.

For 38 + □ + 62 = 145, group the known addends first: 38 + 62 = 100. Then 100 + □ = 145, so □ = 45. Check: 38 + 45 + 62 = 145.

Subtraction is different: the starting amount and the amount removed have different roles. Reordering addition does not give permission to swap the numbers in a subtraction.

Example 4

4. Keep both sides balanced

A level scale has groups of 6 and 8 equal counters on the left, and an unknown group followed by 10 equal counters on the right.
Each counter is the same informal unit of mass. Equal masses correspond to equal total numbers of these counters.

The left pan has 6 + 8 = 14 counters. The right pan needs 4 more alongside its 10: □ = 14 − 10 = 4. Check: 6 + 8 = 4 + 10.

For 125 + □ = 80 + 90, first find the known side: 80 + 90 = 170. The missing addend is 170 − 125 = 45. Check: 125 + 45 and 80 + 90 both equal 170.

Equal changes to both sides preserve equality. Starting with 7 + 8 = 6 + 9, remove 3 equal counters from each pan. If they come from the group of 8 and the group of 9, one valid result is 7 + 5 = 6 + 6, with both sides equal to 12. Other group choices can also preserve equality.

Example 5

5. Compare differences without starting again

On number lines with the same scale, the interval from 14 to 27 and the interval from 4 to 17 are both labelled gap 13.
The second interval is the first shifted 10 places left. Its length is unchanged.

27 − 14 = 17 − 4 is true. Subtracting 10 from both numbers shifts both endpoints equally, so the common difference stays 13.

This can make calculations easier: 74 − 38 = 76 − 40 = 36. Adding 2 to both numbers preserves their difference. Changing just one number does not generally preserve it: 27 − 14 = 13, but 27 − 4 = 23.

For sums, a different balancing change is useful: 31 + 19 = 30 + 20 because one addend decreases by 1 while the other increases by 1. Both sums equal 50.

Example 6

6. Use counting on as an inverse strategy

For 27 + □ = 63, find the distance from 27 to 63.

27 → 30: +3→30 → 60: +30→60 → 63: +3

The total increase is 3 + 30 + 3 = 36. This is the same missing part found by 63 − 27 = 36. Check: 27 + 36 = 63.

Explain the connection: adding the missing part reaches the whole; subtracting the known part from the whole leaves the missing part.

Curriculum examplesExplore each relationship

The exact Australian Curriculum v9.0 content description and its four elaborations are listed below. The worked models and activities put each relationship into practice.

  • Content description: find unknown values in numerical equations involving addition and subtraction, using the properties of numbers and operations
  • E1: demonstrating the commutative properties of addition using materials, diagrams and number lines; for example, using number lines to demonstrate that 5 + 2 = 2 + 5, demonstrating that 2 + 2 + 3 = 7 and 2 + 3 + 2 = 7 and 3 + 2 + 2 = 7
  • E2: using balance scales and informal uniform units to create addition or subtraction number sentences showing equivalence, such as 7 + 8 = 6 + 9, and to find unknowns in equivalent number sentences, such as 6 + 8 = □ + 10
  • E3: using relational thinking and knowledge of equivalent number sentences to explain whether equations involving addition or subtraction are true; for example, explaining that 27 − 14 = 17 − 4 is true and using a number line to show the common difference is 13
  • E4: using part-part-whole diagrams or bar models to recognise and explain the inverse relationship between addition and subtraction, using this to make calculations easier; for example, solving 27 + □ = 63 using subtraction, □ = 63 − 27

1. Build and rearrange

Make groups of 2, 2 and 3 equal counters. Rearrange them into three orders and write an equation for each. Draw jumps of 5 then 2 and 2 then 5 from 0 on matching number lines.

Discuss: 2 + 2 + 3 = 7, 2 + 3 + 2 = 7 and 3 + 2 + 2 = 7. The two number-line journeys both end at 7. The groups or jumps are unchanged; only their order changes.

2. Create and preserve a balance

Use a balance scale and counters of equal mass to represent 7 + 8 = 6 + 9. Remove 3 counters from each side. Write the new equation and explain why it stays balanced. Then model 6 + 8 = □ + 10.

Discuss: The first totals are both 15. One valid choice removes 3 from the group of 8 and 3 from the group of 9, giving 7 + 5 = 6 + 6, with both totals 12. Other group choices are valid if exactly 3 counters are removed from each side. In the unknown equation, the missing group is 4 because 6 + 8 = 4 + 10.

3. Compare matching gaps

On number lines using the same scale, mark the intervals from 14 to 27 and from 4 to 17. Explain why 27 − 14 = 17 − 4 is true. Predict the difference between 24 and 37 without recounting each step.

Discuss: both original gaps are 13. Moving 14 and 27 up by 10 gives 24 and 37, so that gap is also 13.

4. Choose and check a strategy

Solve 27 + □ + 73 = 160 by grouping. Then test the proposed values 72 and 82 in 48 + □ = 120. Draw a bar to explain the successful value.

Discuss: 27 + 73 = 100, so the first box is 60. For the second equation, 48 + 72 = 120 is true but 48 + 82 = 130 is not. A whole of 120 has parts 48 and 72.

Questions and answersWorked answers

Core idea: both sides of an equation must have the same value. Find the role of the unknown, choose a useful number relationship, then replace the box and check.

Important questions

  • What does the equals sign mean? The expressions on its two sides have equal value. In 25 = 18 + 7, both sides are 25.
  • Solve 348 + □ = 725. The missing part is 725 − 348 = 377. Check: 348 + 377 = 725.
  • Solve □ − 286 = 439. The starting amount is 286 + 439 = 725. Check: 725 − 286 = 439.
  • Solve 900 − □ = 347. The removed part is 900 − 347 = 553. Check: 900 − 553 = 347.
  • Use grouping in 38 + □ + 62 = 145. 38 + 62 = 100, leaving 45 for the missing addend.
  • Solve 125 + □ = 80 + 90. The known total is 170, so the missing addend is 45.
  • Why is 27 − 14 = 17 − 4 true? Both endpoints decrease by 10, preserving the difference of 13.
  • Why is 83 − □ = 83 solved by 0? The amount has not changed, so nothing was removed.
Practice and reviewCheck understanding
  • “The equals sign tells me an answer comes next.” It states equal value. Read and compare the complete expressions on both sides.
  • “Always subtract to find the box.” In □ − 35 = 62, the unknown is the starting whole, so add 35 + 62 = 97. In 97 − □ = 62, the unknown is the removed part, so calculate 97 − 62 = 35.
  • “Subtraction can be reordered like addition.” The starting amount and removed amount have different roles. The commutative property of addition does not apply to subtraction.
  • “Any change preserves an equation.” Check the relationship. For a constant difference, change both subtraction numbers by the same amount; for an unchanged sum, an increase in one addend must be balanced by an equal decrease in the other.
  • “Finding a number finishes the work.” Put it into the original box and evaluate both sides. A check must keep the original operations and order.

Quick exit check

  1. Find the box in □ − 24 = 58. Explain its role.
  2. Complete 47 + 26 = 50 + □ without first finding the entire sum.
  3. Explain why 43 − 18 = 48 − 23 is true.

Expected responses: 82 is the starting whole because 24 + 58 = 82; the missing addend is 23 because increasing 47 by 3 requires decreasing 26 by 3; the last equation is true because both subtraction numbers increase by 5, preserving the difference of 25.

Support and extension

Support: start with whole numbers below 20 and real equal counters. Label whole and parts before recording an equation. Move the result to the other side and ask whether its meaning changes.

Extension within this code: give two different whole-number values that make □ + 18 = 18 + □ true when the same value fills both boxes. For example, 5 and 42 both work. Explain why any shared whole-number value works using the commutative property of addition.

Teach with the Topic Guide and Classroom View, then use Homework, Practice and Test. The worksheet asks for written working and models; Practice gives feedback, and Test checks independent application.

Curriculum alignmentAustralian Curriculum v9.0
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