Teaching unit · NZ Year 9 (Phase 4, ages 13 to 14)

Algebraic expressions and equations: expanding, solving and rearranging

Collecting like terms, expanding a bracket, factorising, forming and solving linear equations, and substituting into and rearranging formulae

About three lessons of 45 to 60 minutes

Student view

Start here · hook

Same rule, three different-looking jobs

3(2x + 4) and 6x + 12 look like different expressions, but they are exactly the same value for every possible x, because expanding the bracket and simplifying are just two ways of writing the same thing. Factorising runs that process backwards: starting from 6x + 12 and rebuilding 3(2x + 4).

Solving an equation like 3x + 5 = 26 asks a different question: for what ONE value of x are the two sides equal? And a formula like P = 2l + 2w can be substituted into (given l and w, find P) or rearranged (given P and l, find w instead) - the same three letters, used in two directions. This unit builds all four skills, expanding, solving, substituting and rearranging, on the same underlying idea: an equals sign means both sides always balance.

3(2x + 4) expandedis 6x + 12, the same expression written a different way
3x + 5 = 26 solvedgives the one value x = 7 that balances both sides
P = 2l + 2w rearrangedfor w gives w = (P - 2l) / 2
6x + 12 factorisedback to 3(2x + 4), by taking out the common factor 3

Learning objective

What students will be able to do

Students will simplify algebraic expressions by collecting like terms, expand a term over a bracket, factorise an expression by taking out the greatest common factor, form and solve linear equations (including ones with a bracket or with x on both sides), substitute given values into a real formula, and rearrange a formula to make a different letter the subject before solving.

Success criteria

I can simplify an expression by collecting like terms.
I can expand a single term over a bracket.
I can factorise an expression by taking out the greatest common factor.
I can form a linear equation from a word problem, and solve it.
I can solve a two-step linear equation, one with a bracket, and one with x on both sides.
I can substitute given values into a formula, and rearrange a formula to make a different letter the subject.

Curriculum anchor

Standards this unit teaches

Phase 4 (Years 9-10): Algebra, Equations and relationshipsNew Zealand Curriculum (NZC), Mathematics and Statistics
Equations and relationships
Paraphrased (see the licence note in lib/content_nzsecondarymath2.ts) from the Ministry of Education's Tāhūrangi curriculum site, "NZC - Mathematics and Statistics Phase 4 (Years 9-10)" (official policy for all English-medium state and state-integrated schools from 1 January 2026), Algebra strand, "Equations and relationships" section: during Year 9, students simplify and manipulate algebraic expressions by collecting like terms, factorising using a common factor, and expanding a single term over a bracket; form and solve linear equations with rational-number coefficients; substitute given values into an expression or formula; and rearrange a formula to make a different letter the subject (the source's own worked example rearranges P = 2l + 2w to make w the subject). Source: https://newzealandcurriculum.tahurangi.education.govt.nz/nzc---mathematics-and-statistics-phase-4-years-9-10/5637291579.p (verified live 2026-07-14).

Before you start

Prior knowledge

This unit builds on skills students should already have met. Revisit any that are shaky first.

Key vocabulary

Words to teach and display

Term: a single part of an expression, such as 5x or -3, separated from the rest by + or -
Like terms: terms with exactly the same variable part (e.g. 5x and 2x), which can be combined by adding or subtracting their coefficients
Expand: to multiply out a bracket, e.g. turning 3(2x + 4) into 6x + 12
Factorise: to write an expression as a product by taking out a common factor, the reverse of expanding
Formula: an equation that describes a relationship between quantities, such as P = 2l + 2w
Subject of a formula: the letter on its own on one side of a formula (e.g. P is the subject of P = 2l + 2w); rearranging changes which letter is the subject

Teaching sequence

Teach it: concrete, pictorial, abstract

The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.

1. Collecting like terms, expanding and factorising

Concrete

Collecting like terms means combining terms with the same variable part: 5x + 2x = 7x, because 5 groups of x plus 2 more groups of x is 7 groups of x. Expanding a bracket like 3(2x + 4) means multiplying EVERY term inside the bracket by the 3 outside it: 3 x 2x = 6x, and 3 x 4 = 12, so 3(2x + 4) = 6x + 12.

The same distributive idea appears in ordinary number multiplication: 3 x (6 + 3) can be worked out as (3 x 6) + (3 x 3) = 18 + 9 = 27, splitting the second factor into parts and multiplying each part separately. Expanding 3(2x + 4) does exactly the same thing, just with an x term as one of the parts instead of another whole number.

Factorising is expanding in reverse. For 8x + 20, find the greatest common factor of 8 and 20 (which is 4), divide both terms by it, and write that factor outside a bracket: 8 / 4 = 2 and 20 / 4 = 5, so 8x + 20 = 4(2x + 5). Always check that the numbers left inside the bracket have no common factor of their own, otherwise the true greatest common factor was not fully taken out.

A 3 by 9 array split into a 3 by 6 part and a 3 by 3 part: 3 x 9 = (3 x 6) + (3 x 3) = 18 + 9 = 27. Expanding 3(2x + 4) into 6x + 12 uses exactly this same splitting idea.

Worked example

Simplify 5x + 3 + 2x - 7. Then expand and simplify 3(2x + 4) - x.

Collect like terms: 5x + 2x = 7x, and 3 - 7 = -4, so 5x + 3 + 2x - 7 = 7x - 4.
Expand: 3(2x + 4) = (3 x 2x) + (3 x 4) = 6x + 12.
Simplify: 6x + 12 - x = 5x + 12.

Answer: 5x + 3 + 2x - 7 = 7x - 4, and 3(2x + 4) - x = 5x + 12.

Check for understanding, ask

Why can 5x and 2x be combined into 7x, but 5x and 3 cannot be combined?
What do you multiply BOTH terms inside the bracket by, when expanding 4(3x + 5)?

2. Forming and solving linear equations

Pictorial

Solving an equation means finding the one value of x that makes both sides equal. For a two-step equation like 3x + 5 = 26, undo the operations in reverse order: subtract 5 from both sides first (3x = 21), then divide both sides by 3 (x = 7).

An equation with a bracket, like 4(x - 3) = 20, is usually easiest to solve by dividing both sides by the number outside the bracket FIRST: (x - 3) = 5, then adding 3 to both sides: x = 8. An equation with x on both sides, like 5x + 2 = 2x + 17, needs the x terms brought together first: subtract 2x from both sides (3x + 2 = 17), then solve as normal (x = 5).

Word problems ask you to form the equation yourself before solving it. 'I think of a number, multiply it by 3, then add 8, and the result is 29' becomes 3x + 8 = 29 (x standing for the unknown number), which solves the same way as any other two-step equation.

Solving 3x + 5 = 26: the bar shows the total 26 splits into 3x and 5, so 3x = 26 - 5 = 21, and x = 21 / 3 = 7.

Worked example

Solve 4x - 5 = 19. Then form and solve an equation for: I think of a number, multiply it by 3, then add 8, and the result is 29.

4x - 5 = 19: add 5 to both sides, 4x = 24; divide by 4, x = 6.
Word problem: 3x + 8 = 29 (x is the unknown number).
Subtract 8 from both sides: 3x = 21; divide by 3, x = 7.

Answer: 4x - 5 = 19 gives x = 6. The number in the word problem is x = 7 (check: 3 x 7 + 8 = 21 + 8 = 29).

Check for understanding, ask

Why do you divide both sides by 4 as the LAST step, not the first, when solving 4x - 5 = 19?
For 5x + 2 = 2x + 17, which term would you move first: an x term, or a constant?

3. Substituting into formulae and rearranging

Abstract

Substituting means replacing every letter in a formula with a given number, then working out the result. For P = 2l + 2w, substituting l = 9 and w = 5 gives P = 2(9) + 2(5) = 18 + 10 = 28.

Rearranging a formula changes which letter is the SUBJECT (the one on its own). Starting from P = 2l + 2w and wanting w instead of P as the subject: subtract 2l from both sides (P - 2l = 2w), then divide both sides by 2 (w = (P - 2l) / 2). The rearranged formula works for ANY values of P and l, not just one example.

The same two skills, substituting and rearranging, work for every formula, not just perimeter: y = mx + c (a straight line's equation), A = (1/2)bh (a triangle's area), and d = st (distance = speed x time) all substitute and rearrange the same way.

Substituting x = -2 through x = 3 into y = 2x + 1 and plotting each result: the same substitution skill as any other formula, here forming a straight line.

Worked example

A rectangle has perimeter P = 2l + 2w. If l = 9 and w = 5, find P. Then, a different rectangle has P = 38 and l = 12: rearrange the formula to make w the subject, then find w.

Substitute: P = 2(9) + 2(5) = 18 + 10 = 28.
Rearrange: w = (P - 2l) / 2.
Substitute the second rectangle's values: w = (38 - 2 x 12) / 2 = (38 - 24) / 2 = 14 / 2 = 7.

Answer: The first rectangle has P = 28. The second rectangle (P = 38, l = 12) has w = 7.

Check for understanding, ask

In P = 2l + 2w, why is 2l subtracted from both sides BEFORE dividing by 2, when rearranging for w?
Once a formula is rearranged, does it only work for the one example used to rearrange it, or for any values?

Watch for

Common misconceptions and how to address them

MisconceptionWhen expanding a(bx + c), only the x term gets multiplied by a, not the constant.

Why it happens: Students multiply the first term inside the bracket but forget the distributive law applies to EVERY term, treating the constant as if it were outside the bracket's influence.

How to address it: Draw an arrow from the outside number to BOTH terms inside the bracket before multiplying either. 3(2x + 4) needs two multiplications: 3 x 2x = 6x, AND 3 x 4 = 12, giving 6x + 12, never just 6x + 4.

MisconceptionTo solve an equation with x on both sides, like 5x + 2 = 2x + 17, add the x terms together instead of subtracting one from both sides.

Why it happens: Students see two x terms and default to combining them by addition, rather than recognising that isolating x requires REMOVING one x term from both sides (subtraction), the same 'do the same thing to both sides' rule as any other step.

How to address it: Treat the smaller x term as something to subtract from both sides, exactly like a constant: subtracting 2x from both sides of 5x + 2 = 2x + 17 gives 3x + 2 = 17, which solves normally from there.

MisconceptionRearranging a formula only works for the specific numbers used in that one example, so it has to be redone from scratch for different values.

Why it happens: Students conflate substituting (using specific numbers) with rearranging (which changes the formula's structure for ANY numbers), because both skills are practised together.

How to address it: Once P = 2l + 2w is rearranged to w = (P - 2l) / 2, that new formula is permanent and reusable: substitute any P and l values directly into it, with no need to re-derive it each time.

MisconceptionFactorising 8x + 20 as 2(4x + 10) is fully factorised, because 2 is a common factor.

Why it happens: Students stop at the first common factor they spot (2), without checking whether the numbers left inside the bracket (4 and 10) still share a further common factor (2), which means the GREATEST common factor was not fully taken out.

How to address it: After factorising, check the two numbers left inside the bracket for a further common factor. 4 and 10 still share a factor of 2, so 2(4x + 10) is not fully factorised; taking out the full greatest common factor of 4 gives 4(2x + 5), and 2 and 5 share no further common factor, so that IS fully factorised.

4 x 8 splits into 4 x 5 and 4 x 3 (20 + 12 = 32): checking that the numbers left after factorising (5 and 3 here) share no further common factor confirms the greatest common factor was fully taken out.

Do it together

Guided practice (with answers)

1. Simplify: 6x + 4 - 2x + 9
Answer: 4x + 13, because 6x - 2x = 4x and 4 + 9 = 13.
2. Expand: 5(x - 3)
Answer: 5x - 15, because 5 x x = 5x and 5 x (-3) = -15.
3. Factorise by taking out the greatest common factor: 8x + 20
Answer: 4(2x + 5), because the greatest common factor of 8 and 20 is 4: 8 / 4 = 2 and 20 / 4 = 5.
4. Solve for x: 7x + 3 = 24
Answer: x = 3, because 7x = 21 and 21 / 7 = 3.
5. Solve for x: 2(x + 6) = 20
Answer: x = 4, because x + 6 = 10 (dividing both sides by 2), so x = 10 - 6 = 4.
6. The formula for the area of a triangle is A = (1/2)bh. Find A when b = 10 and h = 6.
Answer: A = 30, because (1/2) x 10 x 6 = 60 / 2 = 30.

On their own

Independent practice worksheets

Practise collecting like terms, expanding and factorising, forming and solving linear equations, and substituting into and rearranging formulae, with computed, never-wrong answer keys.

Year 9 expanding and simplifying expressions worksheetcollect like terms, expand a bracket, and factorise, the core skill from section 1 Year 9 forming and solving linear equations worksheettwo-step equations, bracketed equations, x on both sides, and word problems, the core skill from section 2 Year 9 substituting and rearranging formulae worksheetsubstitute into and rearrange real formulae, the core skill from section 3

Reach every student

Differentiation

Support

For collecting like terms, underline or colour-code matching terms before combining them, so only genuinely like terms get added or subtracted.
For expanding, draw two separate arrows from the outside number to each term inside the bracket, and complete one multiplication at a time before combining.
For solving equations, write out every 'do this to both sides' step on its own line, rather than trying to combine steps mentally.
For rearranging, work through the SAME steps as solving a normal equation, just with letters instead of one specific number as the 'answer'.

Extension

Investigate: does expanding a(bx + c) and then substituting a value for x always give the same result as substituting first and then working through with numbers alone? Try it with a = 3, b = 2, c = 4, x = 5.
Pose a mixed challenge: form an equation from a word problem that needs a bracket (e.g. 'twice the sum of a number and 5 gives 24'), then solve it.
Rearrange a formula for a DIFFERENT letter than the obvious one, e.g. rearrange A = (1/2)bh for h, then again for b, and compare the two rearrangements.
Preview quadratic expressions by asking what happens when a term is multiplied by a BRACKET instead of a single number, e.g. (x + 2)(x + 3) - this is next year's topic.

Check it stuck

Assessment: exit ticket

A three-question exit ticket sampling collecting like terms, an equation with x on both sides, and a rearranged formula.

1. Simplify: 3x + 8 - x - 2
Answer: 2x + 6, because 3x - x = 2x and 8 - 2 = 6.
2. Solve for x: 5x - 4 = 3x + 10
Answer: x = 7, because subtracting 3x from both sides gives 2x - 4 = 10, so 2x = 14 and x = 7.
3. A journey covers a distance of d = 180 km at a speed of s = 60 km/h, using d = st. Rearrange to make t the subject, then find t.
Answer: t = d / s = 180 / 60 = 3 hours.

For the teacher

Teacher notes and timings

Rough timing across three lessons: Lesson 1 collecting like terms, expanding and factorising (section 1), Lesson 2 forming and solving linear equations, including brackets and x on both sides (section 2), Lesson 3 substituting into and rearranging formulae, plus the exit ticket (section 3).
This is the site's second teaching unit anchored to the New Zealand Curriculum (NZC), following nzYear9FinancialMathematics. See the licence note at the top of lib/content_nzsecondarymath2.ts: the NZC's Tāhūrangi content is licensed CC BY-NC 4.0 (NonCommercial), so this unit paraphrases the curriculum rather than quoting it, with the source URL cited for attribution.
Scope note: this unit deliberately covers Year 9's algebra practices (collecting like terms, factorising, expanding, forming/solving linear equations, substitution, rearranging formulae) and leaves Year 10's continuation (factorising and solving quadratic expressions and equations) for a later unit, matching how the source page itself splits Year 9 from Year 10.
Language to keep repeating: expanding and factorising are opposite processes on the SAME two expressions, and rearranging a formula produces a permanent, reusable new formula, not a one-off answer.
Use Student view to project this lesson. Print saves the full teacher unit, including answers and teacher notes; use the linked independent-practice worksheets for student handouts.

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