Curriculum guideMathsYear 8

Simplifying linear expressions: a Year 8 guide

Learn to simplify, expand and factorise linear expressions with checked step-by-step examples, common mistakes and Year 8 practice questions.

Prepared by PocketTutor Education Team

Curriculum and supporting sources last reviewed: August 2026

What you'll be able to do

  • Collect like terms, including terms with negative signs
  • Expand brackets using the distributive property, and handle a minus sign in front
  • Factorise a linear expression fully using its greatest common factor
  • Check that two expressions are equivalent

CurriculumAustralian Curriculum v9 - Year 8 Mathematics, Algebra strand (AC9M8A01) · view it on the curriculum site

To simplify a linear expression, rewrite it in an equivalent form with fewer terms or a clearer structure. The main tools are collecting like terms, using the distributive property to expand brackets, and factorising to put brackets back in. You are not finding the value of a variable unless the question gives one; an expression such as 3x + 5 has no equals sign to solve.

The goal is to make the structure easier to use while preserving exactly the same value.

Start with the parts of an expression

Consider

4x − 7 + 3x + 2

  • The variable is x.
  • The coefficient of the first variable term is 4.
  • The terms are 4x, −7, 3x and 2.
  • The constants are −7 and 2.
The parts of 4x − 7 + 3x + 2

The sign belongs to the term that follows it. Thinking of the expression as 4x + (−7) + 3x + 2 can make negative terms easier to track.

Method 1: collect like terms

Like terms have exactly the same variable part. 4x and 3x are like terms; −7 and 2 are like terms. Variable terms and constants are not like terms.

Worked example: collecting like terms

4x − 7 + 3x + 2

  1. 4x + 3x − 7 + 2

    Rearrange so like terms sit together. Addition is commutative, so terms may move - but their signs must move with them.

  2. (4 + 3)x + (−7 + 2) = 7x − 5

    Combine the coefficients, then the constants.
So 4x − 7 + 3x + 2 = 7x − 5.

Method 2: expand brackets

Expanding uses the distributive property: multiply the term outside a bracket by every term inside it.

Worked example: two brackets

3(x + 4) − 2(x − 1)

  1. 3x + 12 − 2x + 2

    Expand both brackets. The final term becomes +2 because (−2) × (−1) = +2.

  2. 3x − 2x + 12 + 2 = x + 14

    Now collect like terms.
Therefore 3(x + 4) − 2(x − 1) = x + 14.

Method 3: factorise a linear expression

Factorising reverses expanding. Look for the greatest common factor shared by every term.

Worked example: factorising

6x + 18

  1. 6x + 18 = 6(x + 3)

    Both terms share a factor of 6.

  2. 6(x + 3) = 6x + 18

    Check by expanding again - factorising and expanding are reverse processes.
For an expression such as 12y − 8, the greatest common factor is 4, so 12y − 8 = 4(3y − 2). Writing 2(6y − 4) is equivalent but is not fully factorised, because a further common factor remains inside the bracket.

How to check that two expressions are equivalent

The strongest check is to reverse the operation. If you factorised, expand again. If you expanded and simplified, you can also substitute a convenient value into the original and final expressions.

For example, test 3(x + 4) − 2(x − 1) = x + 14 with x = 2. The original expression gives 3(2 + 4) − 2(2 − 1) = 18 − 2 = 16, and the simplified expression gives 2 + 14 = 16.

Matching values provide a useful error check. One matching input is not a proof for every possible value, but it will often reveal an arithmetic or sign mistake.

From expressions to equations

Simplifying an expression makes later algebra easier, but it is not the same as solving. Once an equals sign appears, the task is to find a value that makes both sides equal. For example, 3x + 5 is an expression, while 3x + 5 = 20 is an equation.

When collecting like terms and expanding brackets feel reliable, continue to solving linear equations.

Common mistakes

  • Combining unlike terms
    3x + 4 is not 7x. The terms are not alike, so without a value for x, 3x + 4 cannot be combined further.
  • Adding coefficients and variables
    2x + 5x = 7x, not 7x². Adding like terms changes the coefficient; it does not multiply the variable by itself.
  • Multiplying only the first term in a bracket
    4(x − 3) is 4x − 12, not 4x − 3. The 4 multiplies both x and −3.
  • Losing a negative sign
    −3(x − 2) is −3x + 6, not −3x − 6, because a negative multiplied by a negative is positive.
  • Cancelling across addition
    In (2x + 6) ÷ 2, the whole numerator is divided by 2, giving x + 3. It is unsafe to “cancel the 2” from only one part without first treating each term correctly or factorising the numerator.

Try it yourself

Have a go on paper before revealing anything - the working is what makes it stick. The hint is there if you stall.

Simplify 5x + 3 + 2x − 8

Hint
Group the x terms, then the constants.
Show the answer
5x + 2x + 3 − 8 = 7x − 5.

Expand and simplify 4(2a − 3) − 3(a + 1)

Hint
Expand both brackets first. Watch the sign on the second one.
Show the answer
8a − 12 − 3a − 3 = 5a − 15.

Factorise 9y + 12 fully using integer factors

Hint
What is the largest number that divides both 9 and 12?
Show the answer
The greatest common factor is 3, so 9y + 12 = 3(3y + 4).

Explain why 2m + 5 cannot be simplified to 7m

Show the answer
2m is a variable term and 5 is a constant. They are not like terms, so they cannot be collected.

Sources

  1. Australian Curriculum v9: Year 8 Mathematics (ACARA)australiancurriculum.edu.au
  2. Mathematics scope and sequence, Years 7-10 (ACARA)australiancurriculum.edu.au

Related resources

Tutoring

Online Maths tutoring for Australian students

On-demand Maths help built around Australian curriculum content, from Foundation to Year 12.