NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0

Review the basics

72 maths concepts for Years 7–10 in six areas. Pick an area, then open a concept for a worked example and quick check.

Formula sheets · Years 7–10 →

Browse 72 concepts in six areas.

Guides can relate to more than one year.

19 conceptsNumbersBuild confidence with quantities, calculations and number relationships.Y7 14Y8 8Y9 4Y10 1Open Numbers
15 conceptsAlgebraWork with variables, equations, graphs and mathematical models.Y7 5Y8 8Y9 6Y10 8Open Algebra
18 conceptsGeometryExplore shapes, angles, coordinates and spatial reasoning.Y7 9Y8 8Y9 4Y10 4Open Geometry
8 conceptsMeasurementUnderstand size, units, area, volume, scale and time.Y7 5Y8 4Y9 4Y10 1Open Measurement
7 conceptsStatisticsMake sense of data, displays, sampling and statistical claims.Y7 3Y8 2Y9 2Y10 2Open Statistics
5 conceptsProbabilityReason about chance, events and simulations.Y7 3Y8 3Y9 2Y10 2Open Probability

Formula sheets

Choose a year for formulas, symbols and worked substitutions. Use your browser’s Print or Save as PDF after choosing a year. Earlier-year sheets are useful for prerequisite revision.

Year grouping is a revision guide; topics may be taught in a different sequence. A = area, V = volume and SA = surface area. Use consistent units: lengths in cm, areas in cm² and volumes in cm³, for example. Round only at the final step.

Year 7 formula sheet · NavGanitMate

Percentage of an amount

part = (p ÷ 100) × whole

p is the percentage. Use the same units for part and whole.

Example: 15% of $80 = 0.15 × $80 = $12.

Rectangle

A = l × w; P = 2(l + w)

l = length, w = width. A = area; P = perimeter.

Example: l = 5 cm, w = 3 cm: A = 15 cm²; P = 16 cm.

Triangle and parallelogram

triangle: A = bh ÷ 2; parallelogram: A = bh

b = base; h = perpendicular height, not a sloping side.

Example: Triangle: b = 8 m, h = 3 m gives A = 12 m².

Rectangular prism

V = lwh

V = volume; l, w, h are perpendicular dimensions.

Example: 2 cm × 3 cm × 4 cm = 24 cm³.

Mean and range

mean = sum of values ÷ number of values; range = maximum − minimum

For a non-empty numerical data set.

Example: 2, 4, 9: mean = 15 ÷ 3 = 5; range = 9 − 2 = 7.

Equally likely probability

P(E) = favourable outcomes ÷ total outcomes

E is an event; use a finite sample space with equally likely outcomes.

Example: Fair die: P(even) = 3 ÷ 6 = 1/2.

Speed

v = d ÷ t

d = distance; t = positive elapsed time. Match units.

Example: 120 km ÷ 2 h = 60 km/h.

Year 8 formula sheet · NavGanitMate

Circle

C = 2πr = πd; A = πr²

r = radius, d = diameter = 2r, C = circumference; π ≈ 3.14159.

Example: r = 6 m: C = 12π m; A = 36π m². Keep π for exact answers.

Trapezium

A = (a + b)h ÷ 2

a and b are parallel side lengths; h is perpendicular distance between them.

Example: a = 4 cm, b = 8 cm, h = 3 cm: A = 18 cm².

Prism

V = A_base × h

A_base = cross-sectional base area; h = perpendicular distance between congruent parallel bases.

Example: A_base = 12 cm², h = 5 cm: V = 60 cm³.

Index laws

aᵐ × aⁿ = aᵐ⁺ⁿ; aᵐ ÷ aⁿ = aᵐ⁻ⁿ; (aᵐ)ⁿ = aᵐⁿ

Here m and n are integers; a ≠ 0 when dividing or using non-positive powers.

Example: 2³ × 2² = 2⁵ = 32; 5⁰ = 1.

Percentage change

change (%) = (new − original) ÷ original × 100

Original must be non-zero. A negative result denotes a decrease for positive starting amounts.

Example: From $80 to $100: 20 ÷ 80 × 100 = 25% increase.

Complementary events

P(not E) = 1 − P(E)

Probabilities lie between 0 and 1.

Example: P(rain) = 0.3 gives P(no rain) = 0.7.

Year 9 formula sheet · NavGanitMate

Pythagoras

a² + b² = c²

Right-angled triangles only; c is the hypotenuse opposite the right angle.

Example: a = 3 cm, b = 4 cm gives c = √25 = 5 cm.

Right-triangle trigonometry

sin θ = opposite/hypotenuse; cos θ = adjacent/hypotenuse; tan θ = opposite/adjacent

Sides are relative to acute angle θ. Set the calculator to degrees for degree angles.

Example: opposite = 3, hypotenuse = 5: sin θ = 3/5, so θ ≈ 36.87°.

Straight lines

m = (y₂ − y₁)/(x₂ − x₁); y = mx + b

m = gradient, b = y-intercept; gradient needs x₂ ≠ x₁. Vertical lines have equation x = constant.

Example: Through (1, 3), (3, 7): m = 2; y = 2x + 1.

Cylinder

V = πr²h; SA = 2πr² + 2πrh

V = volume; SA = surface area. Right circular cylinder, closed at both ends; r = radius, h = perpendicular height.

Example: r = 2 cm, h = 5 cm: V = 20π cm³; SA = 28π cm².

Simple interest

I = Prt; total = P + I

P = principal, r = annual rate as a decimal, t = years. No compounding.

Example: P = $500, r = 0.04, t = 2: I = $40; total = $540.

Independent events

P(A and B) = P(A) × P(B)

Use only when A and B are independent.

Example: Two independent fair coin tosses: P(two heads) = 1/2 × 1/2 = 1/4.

Year 10 formula sheet · NavGanitMate

Quadratic formula

x = (−b ± √(b² − 4ac))/(2a)

For ax² + bx + c = 0, a ≠ 0. Over real numbers: discriminant > 0 gives two roots, = 0 one repeated root, < 0 no real roots.

Example: x² − 5x + 6 = 0: x = (5 ± √1)/2 = 2 or 3.

Distance and midpoint

d = √((x₂ − x₁)² + (y₂ − y₁)²); M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Cartesian coordinates with equal unit scales on perpendicular axes.

Example: (0, 0) to (6, 8): d = 10; midpoint = (3, 4).

Compound growth and decay

A = P(1 + r)ⁿ; decay: A = P(1 − r)ⁿ

P = initial amount; r = decimal rate per period; n = number of periods. For decay, 0 ≤ r ≤ 1.

Example: P = $1,000, annual growth 5%, n = 2: A = $1,102.50.

Cone and sphere

cone: V = πr²h/3; sphere: V = 4πr³/3, SA = 4πr²

V = volume; SA = surface area; r = radius. Cone h is perpendicular height, not slant height.

Example: Sphere r = 3 cm: V = 36π cm³; SA = 36π cm².

Similar shapes

length factor k; area factor k²; volume factor k³

Corresponding lengths scale by k > 0; volume rule requires similar solids.

Example: k = 3: lengths ×3, areas ×9, volumes ×27.

Conditional probability

P(A given B) = P(A and B)/P(B)

P(B) > 0; restrict the sample space to B.

Example: P(A and B) = 0.2, P(B) = 0.5 gives P(A given B) = 0.4.

Further reference: AMSI measurement · AMSI quadratics · AMSI trigonometry.

How these learning connections work

Each guide explains why the idea matters and gives an example beyond maths. Year bands describe possible destinations, not a promise that a topic is taught next year. Connections may support current learning or later study; school sequences vary. Advanced uses may need additional teaching.

Examples draw on Australian Curriculum Science and Victorian Curriculum Geography. These are editorial applications, not official cross-subject mappings.

Multiplication

Year 7 · prerequisite review

Multiplication counts equal groups. Three groups of four means 3 × 4.

Four equal groups of three. 4 × 3 = 12 counters. Each box contains the same number.4 equal groups · 12 counters
4 × 3 = 12 counters. Each box contains the same number.

A useful approach

  1. Count the equal groups
  2. Identify how many are in each group
  3. Multiply groups by group size.

Worked example

  1. 4 groups of 3 counters contain 4 × 3 = 12 counters.

Common mistake

Equal groups are required; adding the two numbers gives a different operation.

Quick check

What is 5 × 3?

Check answer and reasoning

15. Five equal groups of three contain 3 + 3 + 3 + 3 + 3 = 15.

Year-level reference

Prerequisite review for the related secondary content.

Related Victorian content: VC2M7N05.

This is a starting concept.

Build on this: Division (Year 7 · prerequisite review) · Percentages (Year 7) · Ratios and rates (Years 7–8) · Squares and powers (Year 7) · Radius and diameter (Year 7) · Area and perimeter (Year 7 · prerequisite review) · Variables and like terms (Years 7–8) · Expanding brackets (Years 7–8) · Prime factors and common factors (Year 7) · Order of operations (Years 7–8 · prerequisite review) · Triangles and polygons (Year 7) · Volume, capacity and surface area (Years 7–9) · Units and conversions (Years 7–9 · prerequisite review)

Back to concept index

Division

Year 7 · prerequisite review

Division splits a total into equal groups or finds how many groups fit. It reverses multiplication.

Twenty counters shared equally. 20 ÷ 4 = 5 counters in each of four groups.4 equal groups · 20 counters
20 ÷ 4 = 5 counters in each of four groups.

A useful approach

  1. Identify the total
  2. Identify the number of equal groups
  3. Divide, then multiply to check.

Worked example

  1. 20 counters shared equally between 4 people gives 20 ÷ 4 = 5 each.

Common mistake

Sharing equally differs from subtracting the number of groups once.

Quick check

18 shared equally between 3 people gives how many each?

Check answer and reasoning

6. 3 × 6 = 18, so 18 ÷ 3 = 6.

Year-level reference

Prerequisite review for the related secondary content.

Related Victorian content: VC2M7N05.

Go one step back: Multiplication

Build on this: Fractions (Year 7 · prerequisite review) · Ratios and rates (Years 7–8) · Radius and diameter (Year 7) · Mean and data (Year 7 · prerequisite review) · Order of operations (Years 7–8 · prerequisite review) · Exponent laws (Years 8–9) · Gradient, midpoint and distance (Year 9) · Area of triangles and parallelograms (Year 7) · Elapsed time and time zones (Year 8) · Algorithms, testing and debugging (Year 8) · Median, mode, range and displays (Year 7)

Back to concept index

Fractions

Year 7 · prerequisite review

A fraction describes equal parts of a whole. The denominator counts the equal parts; the numerator counts the parts selected.

Three quarters of one whole. 3 selected parts out of 4 equal parts: 3/4.3 of 4 equal parts
3 selected parts out of 4 equal parts: 3/4.

A useful approach

  1. Check that the whole is split into equal parts
  2. Count selected parts
  3. Divide numerator and denominator by the same nonzero factor to find equivalents.

Worked example

  1. Three quarters means 3 of 4 equal parts. Half of 10 is 10 ÷ 2 = 5.

Common mistake

The same number of unequal pieces does not describe a fraction of equal parts.

Quick check

Which fraction is equivalent to 1/2?

Check answer and reasoning

2/4. 2/4 simplifies by dividing both numbers by 2, giving 1/2.

Year-level reference

Prerequisite review for the related secondary content.

Related Victorian content: VC2M7N03.

Go one step back: Division

Build on this: Percentages (Year 7) · Probability and sample spaces (Year 7) · Calculating with fractions (Years 7–8) · Fractions, decimals and recurring decimals (Years 7–8) · Rational, irrational and real numbers (Years 8–9) · Sample size and variability (Year 8) · Conditional probability and two-way tables (Year 10)

Back to concept index

Percentages

Year 7

Percent means out of 100. To find p% of a quantity, multiply the quantity by p/100.

A quarter is 25 percent. One of four equal parts is 25/100. For a whole of 80, each part is 20.1 of 4 equal parts
One of four equal parts is 25/100. For a whole of 80, each part is 20.

A useful approach

  1. Write the percentage as a fraction over 100
  2. Multiply by the original quantity
  3. Label whether you found the part, change or final amount.

Worked example

  1. 25% of 80 = 25/100 × 80 = 20. A 25% discount leaves 80 − 20 = 60.

Common mistake

A percentage has to refer to a base quantity; use the one stated in the problem.

Quick check

What is 10% of 70?

Check answer and reasoning

7. 10/100 × 70 = 7.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7N07.

Go one step back: Fractions · Multiplication

Build on this: Percentage change and simple interest (Years 8–9) · Compound growth and decay (Years 8, 10) · Measurement accuracy and error (Year 9)

Back to concept index

Ratios and rates

Years 7–8

A ratio compares quantities in the stated order. A rate compares quantities with different units.

Two red counters for every three blue counters. Red:blue = 2:3. There are 5 counters in all, so the red fraction of the whole is 2/5.RedRedBlueBlueBlue2 red : 3 blue · 5 counters in all
Red:blue = 2:3. There are 5 counters in all, so the red fraction of the whole is 2/5.

A useful approach

  1. Write the quantities in the stated order
  2. Find a multiplier relating corresponding quantities
  3. Apply that multiplier to each part.

Worked example

  1. A ratio of red to blue of 2:3 means 2 red for every 3 blue. Travelling 120 km in 2 hours gives an average speed of 60 km/h.

Common mistake

A ratio compares parts; a fraction of the whole uses the sum of the parts.

Quick check

Red:blue is 2:3. If there are 6 red, how many blue?

Check answer and reasoning

9. 6 is 3 times 2, so the corresponding blue quantity is 3 × 3 = 9.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7N09, VC2M8M05.

Go one step back: Multiplication · Division

Build on this: What π means (Years 7–8) · Congruence, similarity and scale factors (Years 8–9) · Right-triangle trigonometry (Years 9–10) · Building and checking mathematical models (Years 7–10) · Scale drawings and proportional models (Years 7, 9, 10)

Back to concept index

Squares and powers

Year 7

An exponent counts repeated factors. For example, 3² means 3 × 3, not 3 × 2.

Four squared. 4² = 4 × 4 = 16 unit squares.4 units across4
4² = 4 × 4 = 16 unit squares.

A useful approach

  1. Identify the base and exponent
  2. Write the repeated factors
  3. Multiply the factors.

Worked example

  1. 4² = 4 × 4 = 16, while 2³ = 2 × 2 × 2 = 8.

Common mistake

An exponent is not a multiplier: 5² is not 5 × 2.

Quick check

What is 5²?

Check answer and reasoning

25. 5² = 5 × 5 = 25.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7N01, VC2M7N02.

Go one step back: Multiplication

Build on this: Square roots (Year 7) · Area of a circle (Year 8) · Prime factors and common factors (Year 7) · Exponent laws (Years 8–9) · Compound growth and decay (Years 8, 10) · Exponential relationships (Year 10) · Pythagorean theorem (Year 8) · Congruence, similarity and scale factors (Years 8–9)

Back to concept index

Square roots

Year 7

The principal square root is the non-negative number whose square gives the original non-negative number.

A square with area forty-nine. 49 unit squares form a 7 by 7 square. √49 = 7.7 units across7
49 unit squares form a 7 by 7 square. √49 = 7.

A useful approach

  1. Ask which non-negative number squares to the given number
  2. Use a known square or retain a radical
  3. Square the result to check.

Worked example

  1. √49 = 7 because 7² = 49. Although (−7)² is also 49, the symbol √49 means 7.

Common mistake

The principal square-root symbol gives the non-negative root.

Quick check

What is √81?

Check answer and reasoning

9. 9 × 9 = 81, so √81 = 9.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7N01.

Go one step back: Squares and powers

Build on this: Area of a circle (Year 8) · Rational, irrational and real numbers (Years 8–9) · Quadratic equations and graphs (Years 9–10) · Pythagorean theorem (Year 8)

Back to concept index

Radius and diameter

Year 7

The radius joins the centre of a circle to its edge. The diameter passes through the centre from edge to edge. Diameter = 2 × radius.

Radius and diameter. Radius = 4 cm; diameter = 8 cm. Schematic, not drawn to physical scale.r = 4 cmd = 8 cm
Radius = 4 cm; diameter = 8 cm. Schematic, not drawn to physical scale.

A useful approach

  1. Locate the centre and circumference
  2. Identify whether the given length is radius or diameter
  3. Use d = 2r or r = d/2.

Worked example

  1. A circle with radius 4 cm has diameter 8 cm. A diameter of 14 cm gives radius 7 cm.

Common mistake

The radius stops at the centre; a diameter spans the entire circle through its centre.

Quick check

A circle has diameter 10 cm. What is its radius?

Check answer and reasoning

5 cm. The radius is half the diameter: 10 ÷ 2 = 5 cm.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7M03.

Go one step back: Multiplication · Division

Build on this: Circumference (Years 7–8) · Area of a circle (Year 8)

Back to concept index

What π means

Years 7–8

For every circle, circumference divided by diameter is π. The symbol π is exact; 3.14 is only an approximation.

Circumference compared with diameter. A circle of diameter 1 unit has circumference π units. The straight line represents the unrolled circumference: a little more than three diameters. Schematic.d = 10123C = π × diameter
A circle of diameter 1 unit has circumference π units. The straight line represents the unrolled circumference: a little more than three diameters. Schematic.

A useful approach

  1. Identify the diameter
  2. Multiply by π for circumference
  3. Retain π if an exact result is requested.

Worked example

  1. A circle of diameter 5 cm has circumference 5π cm. Keep π when an exact answer is requested.

Common mistake

Neither 3.14 nor 22/7 is exactly π.

Quick check

Which is the exact value of π?

Check answer and reasoning

π. The symbol π represents the exact constant; both numerical alternatives are approximations.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7M03, VC2M8N01.

Go one step back: Ratios and rates

Build on this: Circumference (Years 7–8) · Area of a circle (Year 8)

Back to concept index

Circumference

Years 7–8

Circumference is the distance around a circle. Use C = πd or C = 2πr. It is a length, so use length units.

Around a circle. For radius 5 cm, diameter = 10 cm and circumference = 10π cm. Schematic.r = 5 cmd = 10 cm
For radius 5 cm, diameter = 10 cm and circumference = 10π cm. Schematic.

A useful approach

  1. Read whether the length is a radius or diameter
  2. Choose C = 2πr or C = πd
  3. Substitute and keep length units.

Worked example

  1. For radius 5 cm, C = 2 × π × 5 = 10π cm.

Common mistake

Squaring the radius gives an area ingredient, not a circumference.

Quick check

For diameter 8 cm, what is the exact circumference?

Check answer and reasoning

8π cm. C = π × 8 = 8π cm.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7M03, VC2M8M03.

Go one step back: Radius and diameter · What π means

Back to concept index

Area of a circle

Year 8

Area measures the surface inside a circle. A = πr² uses the radius, squared. To recover the radius from an area, first divide by π, then take the square root.

Inside a circle. For radius 3 cm, area = π × 3² = 9π cm². Schematic.r = 3 cmd = 6 cm
For radius 3 cm, area = π × 3² = 9π cm². Schematic.

A useful approach

  1. Identify the radius
  2. Square it
  3. Multiply by π and attach square units. Reverse these operations if the area is given.

Worked example

  1. If A = 16π cm², then r² = 16, so r = 4 cm. Its circumference would be 2π × 4 = 8π cm.

Common mistake

Do not use a diameter as the radius, or forget to square the radius.

Quick check

A circle has radius 3 cm. What is its area?

Check answer and reasoning

9π cm². A = π × 3² = 9π cm².

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8M03.

Go one step back: Radius and diameter · What π means · Squares and powers · Square roots

Back to concept index

Area and perimeter

Year 7 · prerequisite review

Area counts square units covering a surface. Perimeter measures distance around its boundary. For a rectangle, area = length × width.

Area and perimeter of a rectangle. 6 × 4 = 24 square units inside; 6 + 4 + 6 + 4 = 20 units around.6 units across4
6 × 4 = 24 square units inside; 6 + 4 + 6 + 4 = 20 units around.

A useful approach

  1. Identify the boundary or surface requested
  2. Use length × width for rectangle area
  3. Use the sum of exposed edge lengths for perimeter.

Worked example

  1. A 6 cm by 4 cm rectangle has area 24 cm² and perimeter 6 + 4 + 6 + 4 = 20 cm.

Common mistake

Area and perimeter have different meanings and units.

Quick check

What is the area of a 7 cm by 2 cm rectangle?

Check answer and reasoning

14 cm². Rectangle area = 7 × 2 = 14 cm².

Year-level reference

Prerequisite review for the related secondary content.

Related Victorian content: VC2M7M01.

Go one step back: Multiplication

Build on this: Area of triangles and parallelograms (Year 7) · Composite area and boundary length (Year 8) · Volume, capacity and surface area (Years 7–9) · Nets and two-dimensional views (Year 7)

Back to concept index

Negative numbers

Year 7

Negative numbers are less than zero. Moving right on a number line increases a value; moving left decreases it.

Adding five to negative three. Start at −3. Move five units right to reach 2.-5-4-3-2-1012345+5
Start at −3. Move five units right to reach 2.

A useful approach

  1. Locate the starting number
  2. Move right for a positive addition or left for a negative addition
  3. Check which side of zero the result lies on.

Worked example

  1. Starting at −3 and adding 5 moves five steps right to 2.

Common mistake

The negative sign belongs to a number; do not silently discard it.

Quick check

What is −4 + 7?

Check answer and reasoning

3. Moving 7 steps right from −4 lands at 3.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7N08.

This is a starting concept.

Build on this: Keeping equations balanced (Years 7–8) · Coordinates, gradient and linear graphs (Years 8–9) · Inequalities (Years 8, 10) · Coordinates and transformations (Year 7)

Back to concept index

Variables and like terms

Years 7–8

A variable stands for a number. Like terms have the same variable part; combine their coefficients.

Combining like terms as equal tiles. Each tile represents x. Three x-tiles and two x-tiles make five x-tiles. A constant tile represents a different quantity.xxxxx3x2x3x + 2x = 5x
Each tile represents x. Three x-tiles and two x-tiles make five x-tiles. A constant tile represents a different quantity.

A useful approach

  1. Identify the variable part of each term
  2. Group matching variable parts
  3. Combine only their numerical coefficients.

Worked example

  1. 3x + 2x = 5x. But 3x + 2 cannot be combined into 5x because 2 is a constant.

Common mistake

Terms such as a and a² are not like terms.

Quick check

Simplify 4a + 3a.

Check answer and reasoning

7a. 4a + 3a = (4 + 3)a = 7a.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7A02, VC2M8A01.

Go one step back: Multiplication

Build on this: Expanding brackets (Years 7–8) · Keeping equations balanced (Years 7–8) · Substitution and formulas (Year 7) · Factorising and binomial expansion (Years 8–10)

Back to concept index

Expanding brackets

Years 7–8

Multiply every term inside brackets by the factor outside. Keep each sign.

Distributive law as a rectangle. A rectangle of height 3 and total width x + 4 splits into areas 3x and 12. Therefore 3(x + 4) = 3x + 12. Algebraic lengths are schematic.x433x123(x + 4) = 3x + 12
A rectangle of height 3 and total width x + 4 splits into areas 3x and 12. Therefore 3(x + 4) = 3x + 12. Algebraic lengths are schematic.

A useful approach

  1. Identify the factor outside the brackets
  2. Multiply each term inside by that factor
  3. Collect like terms if possible.

Worked example

  1. 3(x + 4) = 3x + 12. Also 2(x − 5) = 2x − 10.

Common mistake

Multiplying only the first term misses part of the expression.

Quick check

Expand 2(x + 3).

Check answer and reasoning

2x + 6. 2 × x + 2 × 3 = 2x + 6.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7A02, VC2M8A01.

Go one step back: Variables and like terms · Multiplication

Build on this: Simultaneous equations (Year 10) · Factorising and binomial expansion (Years 8–10)

Back to concept index

Keeping equations balanced

Years 7–8

An equation states that two expressions are equal. Apply the same operation to both sides to keep them equal.

Keep both sides balanced. The level beam represents equality, not measured weights. Subtract 6 from both sides of x + 6 = 14 to leave x = 8.x + 614−6−6x = 8
The level beam represents equality, not measured weights. Subtract 6 from both sides of x + 6 = 14 to leave x = 8.

A useful approach

  1. Identify the operation attached to the unknown
  2. Undo it on both sides
  3. Substitute your solution back into the original equation.

Worked example

  1. For x + 6 = 14, subtract 6 from both sides: x = 8. Check: 8 + 6 = 14.

Common mistake

Changing only one side breaks the equality.

Quick check

Solve x + 4 = 11.

Check answer and reasoning

7. Subtracting 4 from both sides gives x = 11 − 4 = 7.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7A03, VC2M8A02.

Go one step back: Variables and like terms · Negative numbers

Build on this: Inequalities (Years 8, 10) · Simultaneous equations (Year 10) · Building and checking mathematical models (Years 7–10)

Back to concept index

Probability and sample spaces

Year 7

A sample space lists possible outcomes. For equally likely outcomes, probability = favourable outcomes ÷ total outcomes.

Six equally likely die outcomes. Outcomes 5 and 6 are greater than 4: 2 favourable outcomes out of 6, so probability = 1/3.12345✓6✓Each outcome has probability 1/6
Outcomes 5 and 6 are greater than 4: 2 favourable outcomes out of 6, so probability = 1/3.

A useful approach

  1. List possible outcomes
  2. Decide whether they are equally likely
  3. Count favourable outcomes and divide by the total when they are.

Worked example

  1. A fair six-sided die has 6 equally likely outcomes. Two are greater than 4, so P(greater than 4) = 2/6 = 1/3.

Common mistake

The simple counting formula requires equally likely elementary outcomes.

Quick check

What is the probability of heads on a fair coin toss?

Check answer and reasoning

1/2. One of the two equally likely coin outcomes is heads, giving 1/2.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7P01.

Go one step back: Fractions

Build on this: Complementary events and relative frequency (Years 7–8) · Combined events and probability trees (Years 8–10) · Probability experiments and simulations (Years 7–9)

Back to concept index

Mean and data

Year 7 · prerequisite review

The mean is the sum of all observations divided by the number of observations. Keep every observation, including repeats.

Balancing a data set. Values 2, 4 and 9 total 15. Their mean is 5, shown by the dashed line.05102A4B9C5
Values 2, 4 and 9 total 15. Their mean is 5, shown by the dashed line.

A useful approach

  1. Add every observation
  2. Count every observation
  3. Divide the sum by the count.

Worked example

  1. For 2, 4 and 9, the sum is 15 and the mean is 15 ÷ 3 = 5.

Common mistake

Do not divide by the largest observation or by the number of distinct values.

Quick check

What is the mean of 3, 6 and 9?

Check answer and reasoning

6. (3 + 6 + 9)/3 = 18/3 = 6.

Year-level reference

Prerequisite review for the related secondary content.

Related Victorian content: VC2M7ST01.

Go one step back: Division

Build on this: Median, mode, range and displays (Year 7)

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Place value and decimals

Year 7 · prerequisite review

Each place is ten times the place to its right. A decimal point separates whole units from fractional units.

Places on either side of the decimal point. In 34.27, each place is ten times the place to its right: 30 + 4 + 0.2 + 0.07.TensUnitsTenthsHundredths342730 + 4 + 0.2 + 0.07
In 34.27, each place is ten times the place to its right: 30 + 4 + 0.2 + 0.07.

A useful approach

  1. Locate the decimal point
  2. Name each place relative to it
  3. Write each digit multiplied by its place value.

Worked example

  1. In 34.27, 3 is tens and 4 is units
  2. 2 is tenths and 7 is hundredths
  3. 34.27 = 30 + 4 + 0.2 + 0.07.

Common mistake

Do not count decimal places from the last digit; start at the decimal point.

Quick check

What is the value of 6 in 2.64?

Check answer and reasoning

0.6. The first place to the right of the decimal point is tenths, so 6 means six tenths.

Year-level reference

Prerequisite review for the related secondary content.

Related Victorian content: VC2M7N02.

This is a starting concept.

Build on this: Fractions, decimals and recurring decimals (Years 7–8) · Scientific notation (Year 9) · Rounding and estimation (Year 7) · Units and conversions (Years 7–9 · prerequisite review)

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Calculating with fractions

Years 7–8

For addition and subtraction, use a common denominator. For multiplication, multiply numerators and denominators. Divide by a nonzero fraction by multiplying by its reciprocal.

Common denominators as equal strips. The first strip selects 4/12 = 1/3; the second selects 3/12 = 1/4. The result selects 7/12. Every strip represents the same whole.1/31/47/124/12 + 3/12 = 7/12
The first strip selects 4/12 = 1/3; the second selects 3/12 = 1/4. The result selects 7/12. Every strip represents the same whole.

A useful approach

  1. Identify the operation
  2. For addition use a common denominator; for division use the reciprocal
  3. Simplify and check the size.

Worked example

  1. Add 1/3 + 1/4
  2. Use denominator 12: 4/12 + 3/12
  3. The sum is 7/12.

Common mistake

Do not add denominators. They describe the size of each part.

Quick check

What is 1/2 + 1/3?

Check answer and reasoning

5/6. 1/2 = 3/6 and 1/3 = 2/6, so the sum is 5/6.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7N05, VC2M7N06, VC2M8N04.

Go one step back: Fractions

Build on this: Complementary events and relative frequency (Years 7–8)

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Fractions, decimals and recurring decimals

Years 7–8

A decimal can be written as a fraction using place value. A reduced fraction terminates in base ten exactly when its denominator has only factors 2 and 5.

Half in three forms. 1/2 = 0.5 = 50%. The whole is divided into two equal parts.1 of 2 equal parts
1/2 = 0.5 = 50%. The whole is divided into two equal parts.

A useful approach

  1. Write the decimal using place value
  2. Simplify the fraction
  3. Inspect the prime factors of the reduced denominator.

Worked example

  1. 0.75 = 75/100
  2. Divide top and bottom by 25
  3. 0.75 = 3/4
  4. For a recurring example, let x = 0.333…
  5. 10x = 3.333…
  6. Subtract x from 10x: 9x = 3
  7. X = 1/3 exactly.

Common mistake

A recurring decimal is exact as a fraction; cutting it off gives only an approximation.

Quick check

Which reduced fraction has a terminating decimal?

Check answer and reasoning

3/8. 8 = 2³ has only the prime factor 2, and 3/8 = 0.375.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7N03, VC2M8N03.

Go one step back: Place value and decimals · Fractions

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Prime factors and common factors

Year 7

A prime number is an integer greater than 1 with exactly two positive factors: 1 and itself. Prime factorisation expresses a positive integer greater than 1 as a product of primes.

Prime factor tree for twenty-four. 24 splits into 2 × 12; 12 into 2 × 6; 6 into 2 × 3. The prime leaves give 24 = 2³ × 3.24212262324 = 2³ × 3
24 splits into 2 × 12; 12 into 2 × 6; 6 into 2 × 3. The prime leaves give 24 = 2³ × 3.

A useful approach

  1. Try a small prime divisor
  2. Divide and continue with the quotient
  3. Stop when every factor is prime and multiply to check.

Worked example

  1. 24 = 2 × 12
  2. 12 = 2 × 6 and 6 = 2 × 3
  3. 24 = 2³ × 3
  4. Compare with 18 = 2 × 3²
  5. The shared prime factors with the smaller powers give HCF(24,18) = 2 × 3 = 6.

Common mistake

1 is not prime, and composite factors must be broken down further.

Quick check

Which is the prime factorisation of 18?

Check answer and reasoning

2 × 3². 18 = 2 × 9 = 2 × 3 × 3.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7N02.

Go one step back: Multiplication · Squares and powers

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Order of operations

Years 7–8 · prerequisite review

Evaluate brackets first, then powers, then multiplication/division left to right, then addition/subtraction left to right.

Multiplication before addition. In 2 + 3 × 4, multiply 3 × 4 before adding 2. The answer is 14, not 20.2 + 3 × 4Original→2 + 12Multiply first→14Then add
In 2 + 3 × 4, multiply 3 × 4 before adding 2. The answer is 14, not 20.

A useful approach

  1. Mark the brackets and powers
  2. Evaluate multiplication and division left to right
  3. Evaluate addition and subtraction left to right.

Worked example

  1. Evaluate 2 + 3 × 4
  2. Multiply 3 × 4 = 12 first
  3. 2 + 12 = 14.

Common mistake

Multiplication does not always come before division; they share priority and run left to right.

Quick check

What is 18 ÷ 3 × 2?

Check answer and reasoning

12. 18 ÷ 3 = 6, then 6 × 2 = 12.

Year-level reference

Prerequisite review for the related secondary content.

Related Victorian content: VC2M7N06, VC2M8N04.

Go one step back: Multiplication · Division

Build on this: Substitution and formulas (Year 7)

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Exponent laws

Years 8–9

For a nonzero base, multiplying like bases adds exponents, dividing subtracts exponents, and raising a power to a power multiplies exponents. Also a⁰ = 1 and a⁻ⁿ = 1/aⁿ.

Join repeated factors. x³ × x² has three x-factors followed by two more. That is five factors of x, so x³ × x² = x⁵.xxxxx3 factors2 factorsx³ × x² = x⁵
x³ × x² has three x-factors followed by two more. That is five factors of x, so x³ × x² = x⁵.

A useful approach

  1. Check that the bases match
  2. Select the exponent law for the operation
  3. Simplify without changing the base.

Worked example

  1. Simplify x³ × x²
  2. Keep base x and add 3 + 2
  3. X⁵.

Common mistake

The laws combine exponents only when the bases match. Do not use them across a sum.

Quick check

For x ≠ 0, simplify x⁵ ÷ x².

Check answer and reasoning

x³. Dividing like bases subtracts exponents: 5 − 2 = 3.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8N02, VC2M9A01.

Go one step back: Squares and powers · Division

Build on this: Scientific notation (Year 9) · Logarithms and logarithmic scales (Year 10)

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Scientific notation

Year 9

Scientific notation writes a positive number as a × 10ⁿ with 1 ≤ a < 10 and integer n. A negative power represents a small positive number.

Three place-value steps. Moving from 4.2 to 0.0042 divides by ten three times. Thus 0.0042 = 4.2 × 10⁻³.4.2÷ 10→0.42 → 0.042÷ 10 each→0.00424.2 × 10⁻³
Moving from 4.2 to 0.0042 divides by ten three times. Thus 0.0042 = 4.2 × 10⁻³.

A useful approach

  1. Move the decimal to make a factor between 1 and 10
  2. Count places and choose the sign of the exponent
  3. Expand back to check.

Worked example

  1. Write 0.0042
  2. Move the decimal three places right to make 4.2
  3. 0.0042 = 4.2 × 10⁻³.

Common mistake

The leading factor must be at least 1 and less than 10.

Quick check

Which is scientific notation for 72,000?

Check answer and reasoning

7.2 × 10⁴. Moving the decimal four places left gives 7.2, so the compensating factor is 10⁴.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M9M02.

Go one step back: Place value and decimals · Exponent laws

Back to concept index

Rational, irrational and real numbers

Years 8–9

A rational number can be written as a/b with integers a and b, b ≠ 0. Irrational numbers cannot. Together they make up the real numbers.

Rational numbers inside the real numbers. Rational and irrational numbers are disjoint parts of the real numbers. The examples 4 and 1/2 are rational; √2 and π are irrational.Real numbersRationalIrrational4 1/2√2 π
Rational and irrational numbers are disjoint parts of the real numbers. The examples 4 and 1/2 are rational; √2 and π are irrational.

A useful approach

  1. Simplify the number if possible
  2. Ask whether it is a ratio of integers
  3. Distinguish a terminating or recurring decimal from a non-recurring infinite decimal.

Worked example

  1. √16 = 4, so it is rational
  2. √2 cannot be expressed as a fraction of integers
  3. √2 is irrational.

Common mistake

Having a square-root sign does not automatically make a number irrational.

Quick check

Which number is irrational?

Check answer and reasoning

√3. 3 is not a perfect square; √3 is irrational, while √9 = 3 and 0.25 = 1/4.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8N01, VC2M9N01.

Go one step back: Fractions · Square roots

Back to concept index

Rounding and estimation

Year 7

To round to a place, inspect the next digit. Digits 0–4 keep it; 5–9 increase it under the usual school rounding convention. An estimate is an approximation, not an exact answer.

Nearest hundredth on a number line. 6.347 lies between 6.34 and 6.35. It is beyond the midpoint 6.345, so rounding to the nearest hundredth gives 6.35.6.346.3456.356.3476.347 ≈ 6.35
6.347 lies between 6.34 and 6.35. It is beyond the midpoint 6.345, so rounding to the nearest hundredth gives 6.35.

A useful approach

  1. Identify the place requested
  2. Inspect the next digit
  3. Round once and retain the requested precision.

Worked example

  1. Round 6.347 to two decimal places
  2. Inspect the third decimal, 7
  3. 6.35.

Common mistake

Keep exact values through intermediate steps unless the task specifically asks for rounding.

Quick check

Round 4.26 to one decimal place.

Check answer and reasoning

4.3. The hundredths digit is 6, so the tenths digit increases from 2 to 3.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7N04.

Go one step back: Place value and decimals

Back to concept index

Percentage change and simple interest

Years 8–9

Percentage change compares the change to the original amount. Simple interest uses I = Prt with r as a decimal annual rate and t in years.

Compare the increase with the original. A price grows from 80 to 100. The added 20 is one quarter of the original 80: a 25% increase. Bar lengths share the same scale.BeforeAfter80802020 ÷ 80 × 100% = 25%
A price grows from 80 to 100. The added 20 is one quarter of the original 80: a 25% increase. Bar lengths share the same scale.

A useful approach

  1. Find the change or interest requested
  2. Identify the original base and convert percentages to decimals
  3. Calculate and distinguish the change from the final amount.

Worked example

  1. A price rises from 80 to 100
  2. The increase is 20
  3. Percentage increase = 20/80 × 100% = 25%.

Common mistake

Use the original amount as the denominator for percentage change. Interest is not the final balance.

Quick check

What is the simple interest on $200 at 5% per year for 2 years?

Check answer and reasoning

$20. I = 200 × 0.05 × 2 = 20; the final balance would be 220.

Year-level reference

Percentage change relates to Year 8; simple-interest modelling relates to Year 9.

Related Victorian content: VC2M8N05, VC2M9A06.

Go one step back: Percentages

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Compound growth and decay

Years 8, 10

Repeated percentage change multiplies by the same factor each period. A = P(1 + r)ⁿ for growth, and P(1 − r)ⁿ for decay, when r is constant.

Ten percent twice. Each period multiplies the current value by 1.1. The second increase is 11 rather than 10: 100 → 110 → 121.100Start→110× 1.1→121× 1.1 again
Each period multiplies the current value by 1.1. The second increase is 11 rather than 10: 100 → 110 → 121.

A useful approach

  1. Identify the initial amount and period
  2. Form 1 + r for growth or 1 − r for decay
  3. Raise the factor to the number of periods.

Worked example

  1. Start with 100 and increase 10% twice
  2. 100 × 1.1 = 110
  3. 110 × 1.1 = 121.

Common mistake

Two 10% changes are not a single 20% change because the second applies to a new amount.

Quick check

A value of 200 decreases 10% each period. What remains after two periods?

Check answer and reasoning

162. 200 × 0.9 × 0.9 = 162.

Year-level reference

Arc includes repeated percentage growth in Level 8. The compound-interest formula and growth/decay modelling extend to Level 10.

Related Victorian content: VC2M8N05, VC2M10A15.

Go one step back: Percentages · Squares and powers

Back to concept index

Substitution and formulas

Year 7

Substitution replaces a variable with its given value. Use brackets for negative values and follow the order of operations.

Replace x with its given value. For x = 4, the expression 2x + 3 becomes 2 × 4 + 3 = 11.2x + 3x = 4→2 × 4 + 3Substitute→11Evaluate
For x = 4, the expression 2x + 3 becomes 2 × 4 + 3 = 11.

A useful approach

  1. Write the formula
  2. Replace every relevant variable with its value using brackets
  3. Follow the order of operations.

Worked example

  1. Evaluate 2x + 3 for x = 4
  2. Replace x by 4: 2 × 4 + 3
  3. The result is 11.

Common mistake

A coefficient multiplies its variable; 3a is not 3 + a.

Quick check

Evaluate 3a − 2 for a = 5.

Check answer and reasoning

13. 3 × 5 − 2 = 15 − 2 = 13.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7A01, VC2M7A03.

Go one step back: Variables and like terms · Order of operations

Build on this: Coordinates, gradient and linear graphs (Years 8–9) · Functions and graph transformations (Years 9–10) · Exponential relationships (Year 10)

Back to concept index

Coordinates, gradient and linear graphs

Years 8–9

A point (x, y) gives horizontal position first and vertical position second. A straight-line equation y = mx + b has gradient m and y-intercept b.

A straight-line relationship. Illustration: y = x + 1. Each unit right adds one unit up; the y-intercept is 1.-2-2-1-111223344xy
Illustration: y = x + 1. Each unit right adds one unit up; the y-intercept is 1.

A useful approach

  1. Identify the coordinates or equation
  2. Calculate rise divided by run or read m in y = mx + b
  3. Check using another point.

Worked example

  1. For y = 2x + 1, choose x = 0, 1, 2
  2. Calculate y = 1, 3, 5
  3. Plot (0,1), (1,3), (2,5).

Common mistake

Gradient is change in y divided by change in x, not the other way around.

Quick check

What is the gradient of y = 3x − 4?

Check answer and reasoning

3. In y = mx + b, m = 3 and b = −4.

Year-level reference

Linear relations relate to Year 8; explicit gradient work extends to Year 9.

Related Victorian content: VC2M8A02, VC2M9A04.

Go one step back: Substitution and formulas · Negative numbers

Build on this: Gradient, midpoint and distance (Year 9) · Functions and graph transformations (Years 9–10) · Scatterplots, association and statistical claims (Year 10)

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Gradient, midpoint and distance

Year 9

For two points, gradient is Δy/Δx when Δx ≠ 0. Midpoint averages the coordinates. Distance follows Pythagoras. A vertical line has undefined gradient.

Rise, run and midpoint. From (1,2) to (5,6), run = 4 and rise = 4. Gradient = 1, midpoint = (3,4) and distance = √32 = 4√2. Equal coordinate scales.(1,2)(5,6)rise 4run 4(3,4)
From (1,2) to (5,6), run = 4 and rise = 4. Gradient = 1, midpoint = (3,4) and distance = √32 = 4√2. Equal coordinate scales.

A useful approach

  1. Label the two x-values and y-values
  2. Choose differences for gradient/distance or averages for midpoint
  3. Check order, sign and units.

Worked example

  1. Points (1,2) and (5,6) have changes Δx = 4 and Δy = 4
  2. Gradient = 4/4 = 1
  3. Midpoint = (3,4)
  4. Distance = √(4² + 4²) = √32 = 4√2 units.

Common mistake

Average x-values together and y-values together; do not subtract for a midpoint.

Quick check

What is the midpoint of (2,4) and (8,10)?

Check answer and reasoning

(5,7). ((2 + 8)/2, (4 + 10)/2) = (5,7).

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M9A04.

Go one step back: Coordinates, gradient and linear graphs · Division

Back to concept index

Inequalities

Years 8, 10

An inequality compares values using <, >, ≤ or ≥. Adding the same number preserves the direction. Multiplying or dividing by a negative reverses it.

An open endpoint and an increasing ray. Dividing −2x < 6 by −2 reverses the inequality: x > −3. The open circle excludes −3; the ray includes all greater real values.-4-3-2-1012x > −3
Dividing −2x < 6 by −2 reverses the inequality: x > −3. The open circle excludes −3; the ray includes all greater real values.

A useful approach

  1. Undo operations on both sides
  2. Reverse the inequality if multiplying or dividing by a negative
  3. Test a permitted value.

Worked example

  1. Solve −2x < 6
  2. Divide both sides by −2 and reverse the sign
  3. X > −3.

Common mistake

Reverse the inequality only when multiplying or dividing by a negative number.

Quick check

Solve x + 3 ≤ 8.

Check answer and reasoning

x ≤ 5. Subtract 3 from both sides: x ≤ 8 − 3 = 5.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8A02, VC2M10A08.

Go one step back: Keeping equations balanced · Negative numbers

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Simultaneous equations

Year 10

A solution must satisfy both equations at once. Substitution replaces one variable; elimination adds or subtracts equations to remove a variable.

The intersection satisfies both equations. The lines y = 7 − x and y = x − 1 meet at (4,3). Substituting x = 4 and y = 3 satisfies both x + y = 7 and x − y = 1.(4,3)y = 7 − xy = x − 1xy43
The lines y = 7 − x and y = x − 1 meet at (4,3). Substituting x = 4 and y = 3 satisfies both x + y = 7 and x − y = 1.

A useful approach

  1. Align the variables or isolate one variable
  2. Substitute or eliminate
  3. Check the pair in both original equations.

Worked example

  1. Given x + y = 7 and x − y = 1
  2. Add the equations: 2x = 8
  3. X = 4, then y = 3.

Common mistake

Check both equations; a value that works in only one is not a simultaneous solution.

Quick check

Given x + y = 5 and x − y = 1, what is x?

Check answer and reasoning

3. Adding gives 2x = 6, so x = 3; then y = 2.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M10A09.

Go one step back: Keeping equations balanced · Expanding brackets

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Factorising and binomial expansion

Years 8–10

Factorising reverses expansion. First look for a common factor. For binomials, multiply each term in one bracket by each term in the other.

An algebraic area model for two brackets. The four areas in (x + 2)(x + 3) are x², 3x, 2x and 6. Adding them gives x² + 5x + 6. Schematic algebraic lengths.x3x2x²3x2x6(x + 2)(x + 3)
The four areas in (x + 2)(x + 3) are x², 3x, 2x and 6. Adding them gives x² + 5x + 6. Schematic algebraic lengths.

A useful approach

  1. Look for a common factor first
  2. Match the required product and sum when factorising a monic quadratic
  3. Expand to verify.

Worked example

  1. Expand (x + 2)(x + 3)
  2. X² + 3x + 2x + 6
  3. X² + 5x + 6.

Common mistake

For (x + a)(x + b), the constant is ab and the coefficient of x is a + b.

Quick check

Factorise x² + 5x + 6.

Check answer and reasoning

(x + 2)(x + 3). 2 × 3 = 6 and 2 + 3 = 5.

Year-level reference

Linear factorisation relates to Year 8; binomial expansion and quadratic factorisation extend through Years 9–10.

Related Victorian content: VC2M8A01, VC2M9A02, VC2M10A04.

Go one step back: Expanding brackets · Variables and like terms

Build on this: Quadratic equations and graphs (Years 9–10)

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Quadratic equations and graphs

Years 9–10

A quadratic has an x² term as its highest nonzero power. Its graph is a parabola. To solve an equation by factors, first put one side equal to zero.

A parabola with two real roots. The graph y = x² − 9 crosses the x-axis at −3 and 3. These are the solutions of x² − 9 = 0; the vertex is (0,−9).xy-33y = x² − 9(0,−9)
The graph y = x² − 9 crosses the x-axis at −3 and 3. These are the solutions of x² − 9 = 0; the vertex is (0,−9).

A useful approach

  1. Write one side as zero
  2. Factor if possible or use a suitable quadratic method
  3. Check all candidate solutions in the original equation.

Worked example

  1. Solve x² − 9 = 0
  2. Factor as (x − 3)(x + 3) = 0
  3. X = 3 or x = −3.

Common mistake

Solving x² = a differs from evaluating √a: an equation can have both positive and negative solutions.

Quick check

Solve x² = 16 over the real numbers.

Check answer and reasoning

x = 4 or x = −4. Both 4² and (−4)² equal 16.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M9A05, VC2M10A13.

Go one step back: Factorising and binomial expansion · Square roots

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Functions and graph transformations

Years 9–10

A function assigns one output to each allowed input. In y = f(x) + k, k shifts the graph vertically; y = f(x − h) shifts it right by h.

Moving a parabola up three units. The solid parabola is y = x² + 3; the dashed parabola is y = x². Every output increases by 3, moving the vertex from (0,0) to (0,3).xy(0,3)y = x² + 3
The solid parabola is y = x² + 3; the dashed parabola is y = x². Every output increases by 3, moving the vertex from (0,0) to (0,3).

A useful approach

  1. Identify the allowed inputs and rule
  2. Compare each transformed output or input with the original
  3. Test a known point such as a vertex or intercept.

Worked example

  1. Start with y = x²
  2. Y = x² + 3 adds 3 to every y-value
  3. Its vertex moves from (0,0) to (0,3).

Common mistake

Changing an input inside brackets is different from changing the output outside.

Quick check

Which graph is y = x² shifted up 2 units?

Check answer and reasoning

y = x² + 2. Adding 2 outside the square adds 2 to every output.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M9A07, VC2M10A11.

Go one step back: Substitution and formulas · Coordinates, gradient and linear graphs

Build on this: Building and checking mathematical models (Years 7–10)

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Exponential relationships

Year 10

In exponential change, the variable is in the exponent. Equal input steps multiply outputs by a constant factor rather than adding a constant amount.

Equal input steps double the output. For y = 3 × 2ˣ, outputs 3, 6 and 12 occur at x = 0, 1 and 2. Equal horizontal steps multiply the output by 2.xy3612x: 0 → 1 → 2 y: 3 → 6 → 12
For y = 3 × 2ˣ, outputs 3, 6 and 12 occur at x = 0, 1 and 2. Equal horizontal steps multiply the output by 2.

A useful approach

  1. Check for a constant multiplicative factor
  2. Identify the initial amount and base
  3. Substitute the exponent and calculate.

Worked example

  1. For y = 3 × 2ˣ, x = 0 gives 3
  2. X = 1 gives 6
  3. X = 2 gives 12.

Common mistake

Do not confuse 2ˣ with 2x; one repeats factors and the other multiplies by x.

Quick check

For y = 2ˣ, what is y when x = 3?

Check answer and reasoning

8. 2³ = 2 × 2 × 2 = 8.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M10A11, VC2M10A14, VC2M10A15.

Go one step back: Squares and powers · Substitution and formulas

Build on this: Logarithms and logarithmic scales (Year 10)

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Logarithms and logarithmic scales

Year 10

A logarithm answers an exponent question: log_b(a) = c means bᶜ = a, for a > 0, b > 0 and b ≠ 1. Equal steps on a logarithmic scale represent equal multiplicative factors.

Equal spacing for equal multiplicative factors. On this base-ten logarithmic scale, 10, 100 and 1000 are equally spaced because each step multiplies the value by 10. Their logarithms are 1, 2 and 3.101001000×10×10log = 1log = 2log = 3
On this base-ten logarithmic scale, 10, 100 and 1000 are equally spaced because each step multiplies the value by 10. Their logarithms are 1, 2 and 3.

A useful approach

  1. Identify the base and positive argument
  2. Rewrite the logarithm as an exponent equation
  3. Check the resulting power.

Worked example

  1. Since 10³ = 1000
  2. Log₁₀(1000) = 3
  3. Moving from 10 to 100 to 1000 adds one each time to the base-ten logarithm.

Common mistake

A logarithmic scale does not have equal differences in the original quantities.

Quick check

What is log₁₀(100)?

Check answer and reasoning

2. 10² = 100, so the required exponent is 2.

Year-level reference

Level 10 covers applied logarithmic scales. The guide’s general logarithm definition provides supporting background.

Related Victorian content: VC2M10M02.

Go one step back: Exponential relationships · Exponent laws

Back to concept index

Angles and parallel lines

Year 7

Angles on a straight line total 180°; angles around a point total 360°. Vertically opposite angles are equal. With parallel lines, corresponding and alternate angles are equal.

Adjacent angles on a straight line. The straight angle is 180°. The ray splits it into 65° and 115°, so the two adjacent angles sum to 180°. Drawing is schematic.65°115°65° + 115° = 180°
The straight angle is 180°. The ray splits it into 65° and 115°, so the two adjacent angles sum to 180°. Drawing is schematic.

A useful approach

  1. Mark stated parallel lines and angle facts
  2. Choose the straight-line, full-turn or parallel-line relationship
  3. Calculate and name the reason.

Worked example

  1. Two angles share a straight line and one is 65°
  2. Subtract from 180°
  3. The other is 115°.

Common mistake

Equal-looking angles in a sketch are not enough; use the stated line or shape properties.

Quick check

What is the missing angle on a straight line next to 72°?

Check answer and reasoning

108°. 180° − 72° = 108°.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7M04.

This is a starting concept.

Build on this: Triangles and polygons (Year 7) · Bearings and direction (Years 9–10 · prerequisite review) · Geometric constructions (Year 9)

Back to concept index

Triangles and polygons

Year 7

A triangle has interior-angle sum 180°. A simple n-sided polygon has interior-angle sum (n − 2) × 180°. A regular polygon has equal sides and equal angles.

Triangulating a pentagon. Two diagonals from one vertex split this pentagon into three triangles. Its interior-angle sum is 3 × 180° = 540°.(5 − 2) × 180° = 540°
Two diagonals from one vertex split this pentagon into three triangles. Its interior-angle sum is 3 × 180° = 540°.

A useful approach

  1. Count sides or identify triangles
  2. Use the correct total-angle relationship
  3. Divide equally only when regularity is stated.

Worked example

  1. A pentagon has 5 sides
  2. Its interior-angle sum is (5 − 2) × 180°
  3. The total is 540°.

Common mistake

The angle-sum formula gives the total, not one angle unless you also use regularity.

Quick check

What is the interior-angle sum of a quadrilateral?

Check answer and reasoning

360°. A quadrilateral splits into two triangles, giving 2 × 180° = 360°.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7M05, VC2M7SP02.

Go one step back: Angles and parallel lines · Multiplication

Build on this: Properties of quadrilaterals (Years 7–8) · Geometric constructions (Year 9)

Back to concept index

Properties of quadrilaterals

Years 7–8

Use stated properties to classify shapes. A parallelogram has two pairs of parallel opposite sides. A rectangle has four right angles; a rhombus has four equal sides. A square has both properties.

Squares belong to two broader families. A square is both a rectangle and a rhombus. The arrows mean “is a type of”, not a step in a calculation.RectangleRhombusSquare
A square is both a rectangle and a rhombus. The arrows mean “is a type of”, not a step in a calculation.

A useful approach

  1. List the properties actually given
  2. Compare side, angle and parallel-line conditions
  3. Use a definition or theorem to justify the classification.

Worked example

  1. A parallelogram has one right angle
  2. Adjacent angles sum to 180°
  3. All its angles are right angles, so it is a rectangle.

Common mistake

A diagram alone does not establish equal lengths, right angles or parallel lines.

Quick check

Which property is always true of a rectangle?

Check answer and reasoning

All four angles are 90°. A rectangle is defined by its four right angles; its adjacent side lengths need not match.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7SP02, VC2M8SP02.

Go one step back: Triangles and polygons

Build on this: Reasoning, counterexamples and geometric proof (Year 10) · Algorithms for classifying shapes (Years 7–8)

Back to concept index

Pythagorean theorem

Year 8

In a right triangle, a² + b² = c², where c is the hypotenuse opposite the right angle. The theorem needs a right angle.

A right triangle. The marked angle is 90°. 3² + 4² = 9 + 16 = 25 = 5². All sides use the same unit.345
The marked angle is 90°. 3² + 4² = 9 + 16 = 25 = 5². All sides use the same unit.

A useful approach

  1. Confirm a right angle
  2. Identify the hypotenuse
  3. Add or subtract squares, then take the positive square root for a length.

Worked example

  1. Legs 6 cm and 8 cm give c² = 36 + 64 = 100
  2. Take the positive square root
  3. C = 10 cm.

Common mistake

The hypotenuse must be the longest side; subtract when finding a missing leg.

Quick check

A right triangle has legs 3 cm and 4 cm. What is its hypotenuse?

Check answer and reasoning

5 cm. 3² + 4² = 9 + 16 = 25, so c = √25 = 5.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8M06.

Go one step back: Squares and powers · Square roots

Build on this: Right-triangle trigonometry (Years 9–10) · Three-dimensional coordinates (Year 8)

Back to concept index

Congruence, similarity and scale factors

Years 8–9

Congruent figures have the same shape and size. Similar figures have equal corresponding angles and proportional corresponding side lengths. A length scale factor k produces area factor k² and volume factor k³.

Corresponding lengths scale together. The right triangles have side lengths 3,4,5 and 9,12,15. The length scale factor is 3 and the area scale factor is 9. Both use the same drawing scale.34912155×3
The right triangles have side lengths 3,4,5 and 9,12,15. The length scale factor is 3 and the area scale factor is 9. Both use the same drawing scale.

A useful approach

  1. Match corresponding angles and sides
  2. Find the length scale factor
  3. Use its square for areas or cube for volumes.

Worked example

  1. A similar triangle has scale factor 3
  2. A 4 cm corresponding side becomes 12 cm
  3. Its area is multiplied by 9.

Common mistake

Match corresponding sides; area does not scale by the same factor as length.

Quick check

Lengths are enlarged by factor 2. By what factor does area change?

Check answer and reasoning

4. Area uses two length dimensions, so its factor is 2² = 4.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8SP01, VC2M9SP02.

Go one step back: Ratios and rates · Squares and powers

Build on this: Scale drawings and proportional models (Years 7, 9, 10)

Back to concept index

Coordinates and transformations

Year 7

Coordinates state positions in order. In 2D use (x,y); in 3D use (x,y,z). A translation adds the same displacement to every point; rotations and reflections preserve lengths and angles.

Coordinates on a plane. Illustration: the point (2, 2) lies on y = x. Read x across, then y up.-2-2-1-111223344xy
Illustration: the point (2, 2) lies on y = x. Read x across, then y up.

A useful approach

  1. Label the axes and original point
  2. Apply the stated displacement, reflection or rotation
  3. Check distances and coordinate order.

Worked example

  1. Translate (2,3) by 4 right and 1 down
  2. Add 4 to x and subtract 1 from y
  3. The image is (6,2).

Common mistake

Reflection in the y-axis changes x, not y. Always identify the axis or rotation centre.

Quick check

Reflect (3,2) in the y-axis. What is the image?

Check answer and reasoning

(−3,2). The horizontal coordinate changes sign while the vertical coordinate stays 2.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7A05, VC2M7SP03.

Go one step back: Negative numbers

Build on this: Three-dimensional coordinates (Year 8)

Back to concept index

Bearings and direction

Years 9–10 · prerequisite review

A three-figure bearing is measured clockwise from north. North is 000°, east 090°, south 180° and west 270°. Reverse bearings differ by 180°, reduced to the interval from 000° to less than 360°.

Bearings are clockwise from north. The route makes a 060° clockwise turn from north. The opposite route has bearing 240°. Compass directions are shown; the diagram is schematic.NESW060°240°
The route makes a 060° clockwise turn from north. The opposite route has bearing 240°. Compass directions are shown; the diagram is schematic.

A useful approach

  1. Draw north at the starting point
  2. Turn clockwise through the bearing
  3. Check three-figure notation and reverse direction if needed.

Worked example

  1. A route has bearing 060°
  2. The reverse direction adds 180°
  3. Its reverse bearing is 240°.

Common mistake

Bearings start at north and turn clockwise, rather than starting at east.

Quick check

What is the bearing of due west?

Check answer and reasoning

270°. A clockwise turn from north through east and south to west is 270°.

Year-level reference

Bearing notation supports direction problems in Years 9–10; these codes do not specify a first teaching year for the notation.

Related Victorian content: VC2M9M03, VC2M10M03.

Go one step back: Angles and parallel lines

Back to concept index

Right-triangle trigonometry

Years 9–10

Relative to a chosen acute angle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse and tan = opposite/adjacent. These ratios apply to right triangles; identify the angle first.

Choose the angle before naming sides. For angle θ at the lower-right vertex, opposite = 3, adjacent = 4 and hypotenuse = 5. Therefore sin θ = 3/5. The triangle is right-angled.opp. 3adj. 4hyp. 5θsin θ = 3/5
For angle θ at the lower-right vertex, opposite = 3, adjacent = 4 and hypotenuse = 5. Therefore sin θ = 3/5. The triangle is right-angled.

A useful approach

  1. Choose the reference angle
  2. Label opposite, adjacent and hypotenuse
  3. Select the ratio containing the known and required quantities.

Worked example

  1. An angle has opposite side 3 and hypotenuse 5
  2. Use sin θ = 3/5
  3. Sin θ = 0.6.

Common mistake

Opposite and adjacent depend on the chosen angle. Keep your calculator in the required angle mode.

Quick check

A right triangle has opposite side 4 and adjacent side 3. What is tan θ?

Check answer and reasoning

4/3. Tangent compares opposite to adjacent, so tan θ = 4/3.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M9SP01, VC2M9M03, VC2M10M03.

Go one step back: Pythagorean theorem · Ratios and rates

Back to concept index

Area of triangles and parallelograms

Year 7

A parallelogram has area base × perpendicular height. A triangle is half a matching parallelogram, so its area is base × perpendicular height ÷ 2.

Base and perpendicular height. Using base 4 and perpendicular height 3, area = ½ × 4 × 3 = 6 square units.345
Using base 4 and perpendicular height 3, area = ½ × 4 × 3 = 6 square units.

A useful approach

  1. Choose a base
  2. Find its perpendicular height
  3. Multiply base and height, then halve for a triangle.

Worked example

  1. A triangle has base 8 cm and perpendicular height 5 cm
  2. 8 × 5 ÷ 2
  3. Its area is 20 cm².

Common mistake

The height is perpendicular to the base, not necessarily a sloping side.

Quick check

A triangle has base 6 cm and height 4 cm. What is its area?

Check answer and reasoning

12 cm². 6 × 4 ÷ 2 = 12.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7M01.

Go one step back: Area and perimeter · Division

Build on this: Composite area and boundary length (Year 8)

Back to concept index

Composite area and boundary length

Year 8

Split an area into shapes with known formulas, or subtract cut-outs. For perimeter, trace the exposed boundary and include each exposed edge once.

Subtract a rectangular corner cut-out. The outer rectangle is 10 cm by 6 cm. Removing a 2 cm by 3 cm top-right corner leaves 60 − 6 = 54 cm². All dimensions are supplied; drawing is to one consistent scale.10 cm6 cm2 cm3 cm54 cm²
The outer rectangle is 10 cm by 6 cm. Removing a 2 cm by 3 cm top-right corner leaves 60 − 6 = 54 cm². All dimensions are supplied; drawing is to one consistent scale.

A useful approach

  1. Sketch the stated parts
  2. Add non-overlapping areas or subtract cut-outs
  3. Trace exposed edges separately for perimeter.

Worked example

  1. An outer rectangle is 10 cm by 6 cm
  2. Remove a 2 cm by 3 cm corner
  3. Remaining area = 60 − 6 = 54 cm².

Common mistake

Adding perimeters of touching pieces counts internal joins that are not exposed boundary.

Quick check

A 6 cm by 5 cm rectangle has a 2 cm by 2 cm square removed. What area remains?

Check answer and reasoning

26 cm². The outer area is 30 cm² and the removed area is 4 cm², giving 26 cm².

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8M01.

Go one step back: Area and perimeter · Area of triangles and parallelograms

Back to concept index

Volume, capacity and surface area

Years 7–9

A prism has volume cross-sectional area × perpendicular length. Surface area is the sum of exposed face areas. 1 cm³ = 1 mL and 1000 cm³ = 1 L.

A rectangular prism. A prism of length 4 cm, width 3 cm and height 2 cm has volume 24 cm³ = 24 mL. The perspective drawing is schematic.4 cm3 cm2 cm
A prism of length 4 cm, width 3 cm and height 2 cm has volume 24 cm³ = 24 mL. The perspective drawing is schematic.

A useful approach

  1. Identify the solid or prism cross-section
  2. Calculate cross-sectional area or individual exposed faces
  3. Attach cubic units for volume and square units for surface area.

Worked example

  1. A rectangular prism is 4 cm by 3 cm by 2 cm
  2. Volume = 4 × 3 × 2
  3. Volume = 24 cm³ = 24 mL.

Common mistake

Surface area uses square units; volume uses cubic units. Convert every length to consistent units first.

Quick check

A prism has cross-sectional area 12 cm² and length 5 cm. What is its volume?

Check answer and reasoning

60 cm³. 12 × 5 = 60 cubic centimetres.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7M02, VC2M8M02, VC2M9M01.

Go one step back: Area and perimeter · Multiplication

Build on this: Nets and two-dimensional views (Year 7)

Back to concept index

Units and conversions

Years 7–9 · prerequisite review

A measurement combines a number and a unit. Convert related units using their scale factor. Area and volume conversions require squared or cubed length factors.

Square the length conversion factor. A 1 m by 1 m square is 100 cm by 100 cm. Its area is 10,000 cm². The 10-by-10 grid shows 100 squares, each 10 cm by 10 cm.1 m = 100 cm1 m100 cm
A 1 m by 1 m square is 100 cm by 100 cm. Its area is 10,000 cm². The 10-by-10 grid shows 100 squares, each 10 cm by 10 cm.

A useful approach

  1. Write the starting and required units
  2. Apply the conversion factor with the correct length/area/volume power
  3. Check whether the numerical value should grow or shrink.

Worked example

  1. 1 m = 100 cm
  2. 1 m² = 100 × 100 cm²
  3. 1 m² = 10,000 cm².

Common mistake

Do not use a linear conversion factor directly for square or cubic units.

Quick check

How many centimetres are in 2.5 metres?

Check answer and reasoning

250. Each metre contains 100 cm, so 2.5 × 100 = 250.

Year-level reference

Unit conversion is supporting knowledge for the listed measurement content; this is not its first teaching year.

Related Victorian content: VC2M7M02, VC2M8M02, VC2M9M01.

Go one step back: Place value and decimals · Multiplication

Build on this: Measurement accuracy and error (Year 9) · Scale drawings and proportional models (Years 7, 9, 10)

Back to concept index

Measurement accuracy and error

Year 9

Recorded measurements depend on instrument precision. Absolute error is the magnitude of the difference from a reference. Percentage error compares that error with the magnitude of a nonzero reference.

Measurement compared with a reference. Measured length 49 cm differs from the reference 50 cm by 1 cm. Absolute error is 1 cm and percentage error is 2%. The axis is a close-up from 48 to 51 cm.48 cm49 cm50 cm51 cmMeasured: 49Reference: 50|49 − 50| = 1 cm
Measured length 49 cm differs from the reference 50 cm by 1 cm. Absolute error is 1 cm and percentage error is 2%. The axis is a close-up from 48 to 51 cm.

A useful approach

  1. Identify the reference and observed measurement
  2. Calculate the magnitude of their difference
  3. Divide by the reference magnitude only if percentage error is requested.

Worked example

  1. A reference length is 50 cm and the measured value is 49 cm
  2. Absolute error = 1 cm
  3. Percentage error = 1/50 × 100% = 2%.

Common mistake

Greater displayed precision does not prove greater accuracy. Absolute error is non-negative.

Quick check

A reference value is 20 and a measured value is 21. What is the absolute error?

Check answer and reasoning

1. The magnitude of 21 − 20 is 1.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M9M04.

Go one step back: Units and conversions · Percentages

Back to concept index

Elapsed time and time zones

Year 8

Use the same time unit before calculating. For time-zone conversions, apply the stated UTC offset and check whether the date changes. Offsets may change with daylight saving.

Break elapsed time at a whole hour. 09:40 to 10:00 is 20 minutes; 10:00 to 11:15 is 75 minutes. Total elapsed time is 95 minutes. Timeline spacing is proportional to elapsed minutes.09:4010:0011:1520 min75 min20 + 75 = 95 minutes
09:40 to 10:00 is 20 minutes; 10:00 to 11:15 is 75 minutes. Total elapsed time is 95 minutes. Timeline spacing is proportional to elapsed minutes.

A useful approach

  1. Put times on the same date and offset
  2. Count whole hours and remaining minutes or convert to one unit
  3. Check any midnight crossing.

Worked example

  1. From 09:40 to 11:15
  2. 20 minutes to 10:00, then 75 minutes
  3. Elapsed time is 95 minutes.

Common mistake

60 minutes make an hour; clock time is not decimal hours. Use the offset given for the date.

Quick check

How long is it from 14:50 to 15:20 on the same day?

Check answer and reasoning

30 minutes. There are 10 minutes to 15:00 and another 20 minutes to 15:20.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8M04.

Go one step back: Division

Back to concept index

Algorithms, testing and debugging

Year 8

An algorithm is a precise sequence of steps. Define inputs, operations, conditions and outputs. Test typical and boundary cases; a counterexample can disprove a claimed general rule.

An even-number decision. Divide integer n by 2. If the remainder is 0, n is even; otherwise it is odd. The diagram is an algorithm, not a statement about non-integer inputs.Integer nRemainder of n ÷ 2= 0?YesNoEvenOdd
Divide integer n by 2. If the remainder is 0, n is even; otherwise it is odd. The diagram is an algorithm, not a statement about non-integer inputs.

A useful approach

  1. State the inputs and desired output
  2. Write ordered steps and conditions
  3. Test ordinary, boundary and counterexample inputs.

Worked example

  1. To test whether a number is even
  2. Divide by 2 and inspect the remainder
  3. Remainder 0 means even.

Common mistake

A few successful examples do not prove that an algorithm works for every input.

Quick check

Which test best exposes a rule that incorrectly calls every positive number even?

Check answer and reasoning

Test an odd number such as 3. The odd input 3 contradicts the incorrect rule.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8A04.

Go one step back: Division

Build on this: Algorithms for classifying shapes (Years 7–8)

Back to concept index

Networks and connectedness

Year 10

A network consists of vertices joined by edges. A path follows edges. A network is connected when every pair of vertices can be joined by a path. The degree of a vertex counts incident edges; a loop counts twice.

Degree counts incident edges. This connected path has edges AB and BC. The degrees are 1 at A, 2 at B and 1 at C. Every pair of vertices is connected by a path.ABCdegree 1degree 2degree 1
This connected path has edges AB and BC. The degrees are 1 at A, 2 at B and 1 at C. Every pair of vertices is connected by a path.

A useful approach

  1. Identify vertices and edges
  2. Trace the required path or count incident edges
  3. Check whether every required vertex is reachable.

Worked example

  1. A path A—B—C has edges AB and BC
  2. B touches two edges
  3. B has degree 2.

Common mistake

Crossing lines are not necessarily a vertex unless a connection is marked.

Quick check

In the path A—B—C, what is the degree of A?

Check answer and reasoning

1. Only edge AB meets A, so its degree is 1.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M10SP02.

This is a starting concept.

Back to concept index

Median, mode, range and displays

Year 7

Sort values to find the median. The mode is the most frequent value and need not be unique. Range is maximum minus minimum. Choose a display suited to the type of variable and comparison.

Median, mode and range in sorted data. The ordered values are 2,3,3,8,9. The middle value is 3, the most frequent value is 3, and the range is 9 − 2 = 7. The row uses equal spacing, not a number-line scale.23389↑ medianmode 3 range 7
The ordered values are 2,3,3,8,9. The middle value is 3, the most frequent value is 3, and the range is 9 − 2 = 7. The row uses equal spacing, not a number-line scale.

A useful approach

  1. Sort observations where required
  2. Select the measure or display suited to the question
  3. Report the statistic with context and units.

Worked example

  1. For 2, 3, 3, 8, 9
  2. The median and mode are 3
  3. The range is 9 − 2 = 7.

Common mistake

With an even number of observations, average the two middle values after sorting.

Quick check

What is the median of 1, 4, 7, 10?

Check answer and reasoning

5.5. The middle pair is 4 and 7, whose mean is (4 + 7)/2 = 5.5.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7ST01, VC2M7ST02.

Go one step back: Mean and data · Division

Build on this: Quartiles, boxplots and distributions (Year 10) · Populations, sampling and bias (Years 8–9) · Scatterplots, association and statistical claims (Year 10) · Choosing data displays (Years 7, 9)

Back to concept index

Quartiles, boxplots and distributions

Year 10

A boxplot summarises median, quartiles and endpoints under a stated convention. The interquartile range is Q3 − Q1. Compare centre, spread, shape and unusual values rather than one statistic alone.

A five-number summary. Illustrative five-number summary: minimum 1, Q1 4, median 7, Q3 10, maximum 13. IQR = 10 − 4 = 6. Whiskers use the minimum/maximum convention.1471013IQR = Q3 − Q1 = 6
Illustrative five-number summary: minimum 1, Q1 4, median 7, Q3 10, maximum 13. IQR = 10 − 4 = 6. Whiskers use the minimum/maximum convention.

A useful approach

  1. Identify the five-number summary and convention
  2. Compare medians and IQRs
  3. Describe spread and shape without inferring unsupported frequencies.

Worked example

  1. Given Q1 = 4 and Q3 = 10
  2. IQR = 10 − 4
  3. IQR = 6.

Common mistake

Quartile conventions can differ for small samples. Use the convention supplied; a box width does not measure frequency.

Quick check

If Q1 = 7 and Q3 = 15, what is the IQR?

Check answer and reasoning

8. The interquartile range is 15 − 7 = 8.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M10ST01.

Go one step back: Median, mode, range and displays

Back to concept index

Populations, sampling and bias

Years 8–9

The population is the entire group of interest. A sample is the part observed. Random selection can reduce selection bias; a larger sample alone cannot correct a biased method.

A sample is part of a population. The large outline represents a population of 24 members. Six filled members illustrate a sample. This picture does not establish that a selection process was random.6 selected from 24
The large outline represents a population of 24 members. Six filled members illustrate a sample. This picture does not establish that a selection process was random.

A useful approach

  1. Define the target population
  2. Examine how cases were selected and who could be missed
  3. Distinguish sampling variation from systematic bias.

Worked example

  1. To estimate a school opinion
  2. Randomly sample students across the school
  3. Surveying only one club may miss other views.

Common mistake

Random sampling and random assignment are different ideas; a sample can still have non-response bias.

Quick check

Which method best reduces selection bias in a school survey?

Check answer and reasoning

Randomly select students across the school. Selecting across the target population gives students a chance to be represented beyond one friendship or club.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8ST01, VC2M8ST02, VC2M9ST02.

Go one step back: Median, mode, range and displays

Build on this: Sample size and variability (Year 8)

Back to concept index

Sample size and variability

Year 8

Different random samples can give different results. Increasing a well-designed sample size generally reduces sampling variability, but does not remove bias or guarantee an exact result.

Larger samples can vary less. Illustrative proportions from repeated samples: n=10 gives 0.2,0.4,0.5,0.6,0.8; n=100 gives 0.45,0.48,0.50,0.52,0.55. These are authored illustrations, not measured experiment results or a guarantee.n=10n=10000.51Observed sample proportion
Illustrative proportions from repeated samples: n=10 gives 0.2,0.4,0.5,0.6,0.8; n=100 gives 0.45,0.48,0.50,0.52,0.55. These are authored illustrations, not measured experiment results or a guarantee.

A useful approach

  1. Compare the sampling methods first
  2. Distinguish random variation from bias
  3. Judge how the number of independent observations affects stability.

Worked example

  1. Two random samples of 10 may give visibly different proportions
  2. Samples of 100 usually vary less
  3. The sampling method still matters.

Common mistake

A larger sample cannot fix a systematically biased question or selection process.

Quick check

What does a larger random sample generally reduce?

Check answer and reasoning

Sampling variability. More independent observations generally make sample summaries more stable.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8ST03.

Go one step back: Populations, sampling and bias · Fractions

Back to concept index

Scatterplots, association and statistical claims

Year 10

Bivariate data record two variables for each case. A scatterplot can show direction, strength and form of association. An association does not by itself establish causation.

Association between paired values. Authored example pairs show a positive association: larger x tends to accompany larger y. A scatterplot alone cannot establish causation. Axes show arbitrary numerical units.xyPaired observations
Authored example pairs show a positive association: larger x tends to accompany larger y. A scatterplot alone cannot establish causation. Axes show arbitrary numerical units.

A useful approach

  1. Identify both variables and units
  2. Examine direction, strength and form
  3. Separate association from a causal conclusion.

Worked example

  1. As study time rises, a sample may show higher marks
  2. The scatterplot has positive association
  3. Other variables or study design still matter for a causal claim.

Common mistake

A trend is not proof of cause, and extrapolating far beyond the data is risky.

Quick check

Points tend downward as x increases. What association is suggested?

Check answer and reasoning

Negative. A decreasing y as x increases is a negative association.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M10ST02, VC2M10ST05.

Go one step back: Coordinates, gradient and linear graphs · Median, mode, range and displays

Back to concept index

Complementary events and relative frequency

Years 7–8

An event and its complement cover all outcomes without overlap. Their probabilities add to 1. Relative frequency is observed event count divided by trial count.

An event and its complement fill the whole. The first 3 of 10 equal parts represent rain, probability 0.3. The other 7 represent no rain, probability 0.7. Together they sum to 1.Rain: 0.3No rain: 0.70.3 + 0.7 = 1
The first 3 of 10 equal parts represent rain, probability 0.3. The other 7 represent no rain, probability 0.7. Together they sum to 1.

A useful approach

  1. Identify the event and all outcomes outside it
  2. Use P(not A) = 1 − P(A)
  3. Check that the two probabilities sum to 1.

Worked example

  1. If P(rain) = 0.3
  2. P(no rain) = 1 − 0.3
  3. P(no rain) = 0.7.

Common mistake

A relative frequency is an observation; it need not equal the theoretical probability in a short experiment.

Quick check

If P(A) = 2/5, what is P(not A)?

Check answer and reasoning

3/5. 1 − 2/5 = 5/5 − 2/5 = 3/5.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7P01, VC2M8P01.

Go one step back: Probability and sample spaces · Calculating with fractions

Build on this: Combined events and probability trees (Years 8–10) · Probability experiments and simulations (Years 7–9)

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Combined events and probability trees

Years 8–10

List ordered outcomes systematically using a table or tree. Multiply probabilities along a path. Add probabilities of disjoint paths. For dependent events, later branch probabilities must reflect earlier outcomes.

Two independent fair coin tosses. HH, HT, TH and TT are equally likely. Two outcomes have exactly one head: probability = 2/4 = 1/2.HHHT✓TH✓TTExactly one head: HT or TH
HH, HT, TH and TT are equally likely. Two outcomes have exactly one head: probability = 2/4 = 1/2.

A useful approach

  1. List ordered outcomes or draw branches
  2. Update probabilities after each event
  3. Multiply along paths and add only disjoint paths.

Worked example

  1. Two independent fair coin tosses have outcomes HH, HT, TH, TT
  2. One outcome is HH
  3. P(HH) = 1/2 × 1/2 = 1/4.

Common mistake

Do not assume every named event is equally likely. Independence is a condition, not a consequence of drawing a tree.

Quick check

For two independent fair coin tosses, what is P(exactly one head)?

Check answer and reasoning

1/2. HT and TH are two of four equally likely ordered outcomes.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8P02, VC2M9P01, VC2M10P02.

Go one step back: Probability and sample spaces · Complementary events and relative frequency

Build on this: Conditional probability and two-way tables (Year 10)

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Conditional probability and two-way tables

Year 10

Conditional probability restricts the sample space to cases meeting a condition. P(A given B) = P(A and B)/P(B), when P(B) > 0. In a table, use the conditioned group as the denominator.

Restrict the sample space before counting. All 12 dots are cyclists. Three filled dots also represent red clothing. Within the cyclist group, P(red given cycles) = 3/12 = 1/4.12 cyclists · 3 wearing red
All 12 dots are cyclists. Three filled dots also represent red clothing. Within the cyclist group, P(red given cycles) = 3/12 = 1/4.

A useful approach

  1. Identify the condition
  2. Restrict the denominator to that group
  3. Count cases meeting both the event and the condition.

Worked example

  1. Of 12 students who cycle, 3 wear red
  2. Restrict attention to those 12 cyclists
  3. P(red given cycles) = 3/12 = 1/4.

Common mistake

P(A given B) is generally different from P(B given A). Do not divide by the whole population when the condition specifies a subgroup.

Quick check

Of 10 swimmers, 4 wear glasses. What is P(glasses given swimmer)?

Check answer and reasoning

2/5. The condition selects 10 swimmers, and 4 of them wear glasses: 4/10 = 2/5.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M10P01.

Go one step back: Combined events and probability trees · Fractions

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Probability experiments and simulations

Years 7–9

A simulation uses a model to imitate a chance process. Match the model to the stated probabilities, repeat trials and compare relative frequencies with theory.

A fair-coin simulation model. Choose 0 or 1 with equal probability: 0 means tails and 1 means heads. Repeat independent trials and count outcomes. No experiment results are asserted here.Choose 0 or 1Equal chance→0=T 1=HMap outcome→RepeatCount heads
Choose 0 or 1 with equal probability: 0 means tails and 1 means heads. Repeat independent trials and count outcomes. No experiment results are asserted here.

A useful approach

  1. Specify the theoretical chance model
  2. Map random outcomes to the process with matching probabilities
  3. Repeat trials and compare relative frequencies.

Worked example

  1. To model a fair coin
  2. Assign 0 to tails and 1 to heads with equal chance
  3. Repeat many independent trials and count the heads.

Common mistake

A simulation approximates behaviour under its assumptions; more trials do not repair an incorrect model.

Quick check

Which is suitable for modelling a fair six-sided die?

Check answer and reasoning

Choose uniformly from 1 to 6. Each of the six outcomes needs probability 1/6.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7P02, VC2M8P03, VC2M9P03.

Go one step back: Probability and sample spaces · Complementary events and relative frequency

Back to concept index

Reasoning, counterexamples and geometric proof

Year 10

A proof links stated facts to a conclusion using valid reasons. A diagram or a collection of examples can suggest a result but does not establish it for every case. One valid counterexample disproves a universal claim.

A counterexample to a universal claim. A 3 cm by 2 cm rectangle has four right angles but unequal adjacent sides. It disproves the claim that every rectangle is a square. One counterexample is enough.3 cm2 cmRectangle ≠ necessarily square
A 3 cm by 2 cm rectangle has four right angles but unequal adjacent sides. It disproves the claim that every rectangle is a square. One counterexample is enough.

A useful approach

  1. State the assumptions and claim
  2. Justify each inference with a definition or theorem
  3. Look for a counterexample when testing a universal claim.

Worked example

  1. Claim: all rectangles are squares
  2. A 2 cm by 3 cm rectangle has four right angles but unequal adjacent sides
  3. The claim is false.

Common mistake

Do not assume the feature you are trying to prove. State why each equality or angle fact holds.

Quick check

What disproves a claim about every member of a group?

Check answer and reasoning

One valid counterexample. A universal claim is false if even one allowed case contradicts it.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M10SP01.

Go one step back: Properties of quadrilaterals

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Nets and two-dimensional views

Year 7

A net unfolds the faces of a solid into a plane. Plans and elevations show views from chosen directions. A valid net must fold without overlap into the intended solid.

A cube net with six square faces. Four squares in a row, with one above and one below the second square, form a valid cube net. Each face has the same side length.1234566 faces
Four squares in a row, with one above and one below the second square, form a valid cube net. Each face has the same side length.

A useful approach

  1. Identify the solid and viewing direction
  2. Match adjacent faces or projected features
  3. Check how the representation folds or aligns.

Worked example

  1. A cube has 6 square faces
  2. Its net must contain 6 matching squares arranged to fold correctly
  3. Counting squares alone is not sufficient.

Common mistake

A view can hide edges or faces; a flat sketch need not show every face separately.

Quick check

How many faces does a cube have?

Check answer and reasoning

6. A cube has a top, bottom, front, back, left and right face.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7SP01.

Go one step back: Area and perimeter · Volume, capacity and surface area

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Three-dimensional coordinates

Year 8

A point (x,y,z) describes position along three perpendicular axes. For distances in this Cartesian system, add the squares of the three coordinate changes and take the square root.

Three perpendicular coordinate directions. A schematic projection of perpendicular x,y,z axes. The point (2,3,6) is reached by changes 2 in x, 3 in y and 6 in z. Perspective spacing is illustrative, not an angle or distance measurement.xyz(2,3,6)236
A schematic projection of perpendicular x,y,z axes. The point (2,3,6) is reached by changes 2 in x, 3 in y and 6 in z. Perspective spacing is illustrative, not an angle or distance measurement.

A useful approach

  1. Identify all three axis changes
  2. Square and add them when calculating Cartesian distance
  3. Take the non-negative square root and attach length units.

Worked example

  1. From (0,0,0) to (2,3,6), the changes are 2, 3 and 6
  2. Distance² = 4 + 9 + 36 = 49
  3. Distance = 7 units.

Common mistake

Keep the axis order consistent; a 2D projection may hide depth.

Quick check

Which point has x = 2, y = −1 and z = 4?

Check answer and reasoning

(2,−1,4). The ordered triple lists the stated x, y and z coordinates in that order.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M8SP03.

Go one step back: Coordinates and transformations · Pythagorean theorem

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Geometric constructions

Year 9

A geometric construction is a sequence of justified steps producing a shape or locus. With straightedge and compass, equal-radius arcs can locate points equidistant from two endpoints.

Equal-radius circles locate a perpendicular bisector. Equal-radius circles centred at A and B meet above and below AB. The line through those intersections is the perpendicular bisector. The radius exceeds half the distance AB.ABEqual radii
Equal-radius circles centred at A and B meet above and below AB. The line through those intersections is the perpendicular bisector. The radius exceeds half the distance AB.

A useful approach

  1. State the target property
  2. Follow the permitted construction steps
  3. Justify the result using equal distances or angle facts.

Worked example

  1. To bisect segment AB, draw equal-radius arcs from A and B with radius greater than half AB
  2. Join the two arc intersections
  3. That line is the perpendicular bisector of AB.

Common mistake

A sketch that looks correct is not a construction argument. Keep compass radii equal when the method requires it.

Quick check

What is true of every point on the perpendicular bisector of AB?

Check answer and reasoning

It is equally distant from A and B. The perpendicular bisector is the locus of points equidistant from the two endpoints.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M9SP03.

Go one step back: Angles and parallel lines · Triangles and polygons

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Algorithms for classifying shapes

Years 7–8

Translate shape definitions into explicit tests. Ask only questions supported by the supplied measurements or properties. Classification categories can overlap: every square is also a rectangle and a rhombus.

Classifying from known properties. For a quadrilateral already known to have four right angles, check whether all four sides are equal. Yes gives a square; no gives a rectangle that is not a square. A square remains a rectangle too.4 right anglesRectangle→4 equal sides?Check lengths→Yes: squareNo: non-square
For a quadrilateral already known to have four right angles, check whether all four sides are equal. Yes gives a square; no gives a rectangle that is not a square. A square remains a rectangle too.

A useful approach

  1. List precise shape definitions
  2. Turn them into ordered property tests
  3. Check overlapping categories and boundary cases.

Worked example

  1. For a quadrilateral, test whether all angles are right angles
  2. If yes, it is a rectangle
  3. If its four sides are also equal, it is a square.

Common mistake

Do not treat overlapping categories as mutually exclusive. A square meets more than one broader definition.

Quick check

A quadrilateral has four right angles and four equal sides. Which is its most specific name?

Check answer and reasoning

Square. Four right angles and four equal sides meet the definition of a square.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7SP04, VC2M8SP04.

Go one step back: Algorithms, testing and debugging · Properties of quadrilaterals

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Choosing data displays

Years 7, 9

Choose a graph based on the variables and purpose. Bar charts compare categories, histograms display a grouped numerical distribution, line graphs show ordered change and scatterplots show paired numerical variables.

Comparing three category counts. Illustrative category counts: A = 2, B = 4, C = 9. Separate bars compare categories; the vertical scale starts at zero.05102A4B9C5
Illustrative category counts: A = 2, B = 4, C = 9. Separate bars compare categories; the vertical scale starts at zero.

A useful approach

  1. Identify categorical or numerical variables
  2. Decide whether you need comparison, distribution, change or association
  3. Choose a display and label its scale honestly.

Worked example

  1. To compare counts for favourite colours
  2. The variable is categorical
  3. Use separated bars, one per category.

Common mistake

A histogram uses intervals on a numerical scale; a categorical bar chart does not imply continuous intervals.

Quick check

Which display best shows association between two numerical variables?

Check answer and reasoning

Scatterplot. A scatterplot places each paired observation at its two numerical coordinates.

Year-level reference

Numerical displays relate to Year 7; choosing and justifying displays is revisited in Year 9.

Related Victorian content: VC2M7ST02, VC2M9ST04.

Go one step back: Median, mode, range and displays

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Building and checking mathematical models

Years 7–10

A model translates a situation into variables, relationships and assumptions. Define units and the domain, solve the mathematical problem, then check whether the answer makes sense in context.

Translate a situation into a model. With a fixed charge of $4 and $2 per kilometre, cost C = 4 + 2d. At d = 3 km the cost is $10. Assume this tariff applies for the trip.$4 + $2/kmDefine tariff→C = 4 + 2dBuild rule→d=3 → C=$10Interpret
With a fixed charge of $4 and $2 per kilometre, cost C = 4 + 2d. At d = 3 km the cost is $10. Assume this tariff applies for the trip.

A useful approach

  1. Define variables, units and assumptions
  2. Form a relationship and solve within its domain
  3. Interpret and check the result against the original situation.

Worked example

  1. A taxi charges a fixed $4 plus $2 per kilometre
  2. For d kilometres, C = 4 + 2d
  3. At d = 3, C = $10.

Common mistake

Do not extrapolate a model beyond its stated assumptions or domain without justification.

Quick check

A service costs $5 plus $3 per hour. Which model gives cost C for h hours?

Check answer and reasoning

C = 5 + 3h. The fixed $5 is paid once, while $3 is multiplied by the number of hours.

Year-level reference

Related Victorian Curriculum 2.0 content; this short guide covers selected ideas.

Related Victorian content: VC2M7N10, VC2M8A03, VC2M9A06, VC2M10A15.

Go one step back: Keeping equations balanced · Functions and graph transformations · Ratios and rates

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Scale drawings and proportional models

Years 7, 9, 10

A scale drawing compares a represented length to a real length. Put both lengths in the same units before using the ratio. For a 1:n scale, real length is n times drawing length.

A one-to-one-hundred length scale. At scale 1:100, a 3 cm drawing length represents 300 cm = 3 m. The two bars are symbolic and not shown at a 1:100 size ratio.Drawing: 3 cmReal: 300 cm = 3 m×100
At scale 1:100, a 3 cm drawing length represents 300 cm = 3 m. The two bars are symbolic and not shown at a 1:100 size ratio.

A useful approach

  1. Write the scale ratio in a consistent direction
  2. Convert corresponding lengths to matching units
  3. Calculate and convert the final unit if necessary.

Worked example

  1. At scale 1:100, a 3 cm drawing length represents 300 cm
  2. Convert 300 cm to metres
  3. The real length is 3 m.

Common mistake

Scale factors apply to corresponding lengths. Convert units after applying the ratio, or consistently before it.

Quick check

At scale 1:50, what real length does 4 cm represent?

Check answer and reasoning

2 m. 4 × 50 = 200 cm, which is 2 m.

Year-level reference

Proportional measurement relates to Year 7, with scale modelling extended in Years 9–10.

Related Victorian content: VC2M7M06, VC2M9M05, VC2M10M04.

Go one step back: Ratios and rates · Units and conversions · Congruence, similarity and scale factors

Back to concept index

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