AC9M5A02 • Year 5 Maths

AC9M5A02: Find unknown values in numerical equations involving multiplication and division using the properties of numbers and operations

find unknown values in numerical equations involving multiplication and division using the properties of numbers and operations

What students learn in AC9M5A02

Students identify the role of an unknown, use inverse operations, factor relationships and distributive properties, then substitute the solution to verify both sides of an equation.

Solving an equation means finding a value that makes the two expressions equal, not moving numbers by an unexplained rule.

The inverse operation finds the missing factor. Substitution confirms that the equality is true.

Identify whether the unknown is a factor, dividend or divisor. Multiplication can be regrouped; division order cannot be swapped casually.

Learning routine: Represent → Reason → Calculate → Interpret → Verify

Success looks like

  • Represent accurately
  • Explain the relationship
  • Choose a suitable method or unit
  • Apply in context
  • Check and justify
Key vocabulary
equation
statement that two expressions have equal value
unknown
value to be determined
substitution
replacing a symbol with a value to test the equation
Concept models and worked thinking

Solve 24 × □ = 1 008

stepreasonunknown is a factordivide product by known factor1 008 ÷ 2442substitute24 × 42 = 1 008solution□ = 42

The inverse operation finds the missing factor. Substitution confirms that the equality is true.

  1. Read every label and identify the quantities, parts or evidence.
  2. Explain the relationship shown—not just the final answer.
  3. Check the conclusion against the original question and units.

Use properties to simplify equations

□ × 25 = 1 2001 200 ÷ 25 = 48
6 × □ × 5 = 900group 6 × 5 = 30; □ = 30
1 440 ÷ □ = 24□ = 1 440 ÷ 24 = 60
□ ÷ 18 = 35□ = 35 × 18 = 630

Identify whether the unknown is a factor, dividend or divisor. Multiplication can be regrouped; division order cannot be swapped casually.

Now transfer the same relationship to a new situation and justify the result with precise vocabulary.

Curriculum coverage and elaborations

Content description: find unknown values in numerical equations involving multiplication and division using the properties of numbers and operations.

  • E1: using knowledge of equivalent number sentences to form and find unknown values in numerical equations; for example, given that 3\times5=15 and 30\div2=15 then 3\times5=30\div2 therefore the solution to 3\times5=30\div\square is 2
  • E2: using relational thinking, an understanding of equivalence and number properties to determine and reason about numerical equations; for example, explaining whether an equation involving equivalent multiplication number sentences is true, such as 15 ÷ 3 = 30 ÷ 6
  • E3: using materials, diagrams and arrays to demonstrate that multiplication is associative and commutative but division is not; for example, using arrays to demonstrate that 2 \times 3 = 3 \times 2 but 6 ÷ 3 does not equal 3 ÷ 6; demonstrating that 2 \times 2 \times 3 = 12 and 2 \times3 \times2 = 12 and 3 \times 2 \times 2 = 12; understanding that 8 ÷ 2 ÷ 2 = (8 ÷ 2) ÷ 2 = 2 but 8 ÷ (2 ÷ 2) = 8 ÷ 1 = 8
  • E4: using materials, diagrams or arrays to recognise and explain the distributive property; for example, where 4 \times 13 = 4 \times 10 + 4 \times 3
  • E5: constructing equivalent number sentences involving multiplication to form a numerical equation, and applying knowledge of factors, multiples and the associative property to find unknown values in numerical equations; for example, considering 3 \times 4 = 12 and knowing 2 \times 2 = 4 then 3 \times 4 can be written as 3\times (2 \times 2) and using the associative property (3 \times 2) \times 2 so 3 \times 4 = 6 \times 2 and so 6 is the solution to 3 \times 4 = \square\times 2
Guided learning activities

1. Model and annotate

Represent unknown values in multiplication and division equations and label the relationship or measurement being used.

stepreasonunknown is a factordivide product by known factor1 008 ÷ 2442substitute24 × 42 = 1 008solution□ = 42

2. Strategy or instrument comparison

Compare two valid methods or tools and justify which better suits the numbers, attribute or purpose.

□ × 25 = 1 2001 200 ÷ 25 = 48
6 × □ × 5 = 900group 6 × 5 = 30; □ = 30
1 440 ÷ □ = 24□ = 1 440 ÷ 24 = 60
□ ÷ 18 = 35□ = 35 × 18 = 630

3. Transfer and verify

Apply the concept to an unfamiliar example, state units or conditions and verify independently.

Solve 24 × □ = 1 008.Solve □ ÷ 18 = 35.Solve 1 440 ÷ □ = 24.Regroup 6 × □ × 5.
Revision Notes

Core idea: Solving an equation means finding a value that makes the two expressions equal, not moving numbers by an unexplained rule.

Remember

  • Represent accurately
  • Explain the relationship
  • Choose a suitable method or unit
  • Apply in context
  • Check and justify

Important questions

  • Solve 24 × □ = 1 008. Explain using the model or evidence above.
  • Solve □ ÷ 18 = 35. Explain using the model or evidence above.
  • Solve 1 440 ÷ □ = 24. Explain using the model or evidence above.
  • Regroup 6 × □ × 5. Explain using the model or evidence above.
  • Substitute and check. Explain using the model or evidence above.
How to use this unit

Learn from the Topic Guide and fixed Teacher Slides, complete the Practice Sheet, use Practice for supported feedback, then take the Test when ready.

AC9M5A02 Teacher Slides

Project the fixed branded slide deck one slide at a time.

Open Classroom View
Common misconceptions
  • Same inverse used for every unknown position — Identify factor, dividend or divisor first.
  • Equals sign read as an instruction only — It states balance between expressions.
  • Division reversed — a ÷ b is not generally b ÷ a.
  • Solution not checked — Substitute into the original equation.
🎥 Optional Video Lesson

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Recommended: Unknowns with Multiplication and Division

Khan Academy — Use related multiplication and division facts to find an unknown.

As you watch: Which inverse operation lets you find the missing value?

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Try it: Find the missing numbers in 9 × □ = 72 and 84 ÷ □ = 7, and check each by substitution.

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Curriculum equivalents: Victoria, NSW and international

Curriculum equivalents for Find unknown values in numerical equations involving multiplication and division...

Mapped skill: find unknown values in numerical equations involving multiplication and division using the properties of numbers and operations

These references identify matching or closely related learning. Curriculum sequence, terminology and depth vary, so teachers should use the mapped skill and lesson difficulty to confirm suitability.

RegionCurriculum frameworkClosest level or code
AustraliaAustralian Curriculum v9.0AC9M5A02 · Year 5
VictoriaVictorian Curriculum F–10 Version 2.0 — MathematicsVC2M5A02 · Level 5
New South WalesNSW Mathematics K–10 Syllabus (2022)MA3-MR-02 · Stage 3
United States (USA)Common Core State Standards for MathematicsGrade 5
Canada (Ontario)Ontario Curriculum — MathematicsGrade 5
United Kingdom (England)National Curriculum in England — MathematicsYear 6, Key Stage 2
IndiaNCERT / CBSE — MathematicsClass 5

Australian Curriculum v9.0 is the canonical source for this SkillrHub lesson. Victoria and NSW entries name the closest published state codes or outcomes; international entries are planning references rather than claims of identical curricula.

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Topic reference: AC9M5A02 — AC9M5A02: Find unknown values in numerical equations involving multiplication and division using the properties of numbers and operations

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Official curriculum reference
Related Year 5 Maths topics

Depth checkpoint

Choose the inverse operation that isolates the unknown: use division for an unknown factor and multiplication for an unknown dividend. Substitute the result back into the original equation to verify it.