Curriculum guideMathsYear 8

Solving linear equations: a Year 8 guide

Learn how to solve one-step and multi-step linear equations, keep both sides balanced and check answers by substitution, with Year 8 practice.

Prepared by PocketTutor Education Team

Curriculum and supporting sources last reviewed: August 2026

What you'll be able to do

  • Solve one-step, two-step and multi-step linear equations
  • Handle brackets and equations with the variable on both sides
  • Check a solution by substituting it into the original equation

CurriculumAustralian Curriculum v9 - Year 8 Mathematics, Algebra strand (AC9M8A02) · view it on the curriculum site

To solve a linear equation, find the value of the variable that makes both sides equal. Keep the equation balanced by applying the same operation to both sides, and undo operations in a sensible order until the variable is isolated. Then substitute the value into the original equation to check that it works.

An expression such as 2x + 7 represents a value. An equation such as 2x + 7 = 15 states that two expressions are equal, and a solution is a value of x that makes that statement true. Think of an equation as a balanced scale: changing only one side destroys the equality, while applying the same valid operation to both sides produces an equivalent equation with the same solution.

Why both sides must change together
  1. 2x + 7 = 15 — then subtract 7 from both sides
  2. 2x = 8 — then divide both sides by 2
  3. x = 4

The inverse-operations method

Inverse operations undo one another: addition and subtraction are inverses, and multiplication and division are inverses. When solving, identify what has been done to the variable and undo those operations in reverse order.

Example 1: a two-step equation

2x + 7 = 15

  1. 2x + 7 − 7 = 15 − 7, so 2x = 8

    Subtract 7 from both sides. The + 7 was the last thing done to x, so it is the first to undo.

  2. 2x ÷ 2 = 8 ÷ 2, so x = 4

    Divide both sides by 2 to isolate x.
Check in the original equation: 2(4) + 7 = 8 + 7 = 15. The left side equals the right side, so x = 4 is verified.

Example 2: an equation with a bracket

5(x − 2) = 20

  1. x − 2 = 4

    There are two valid approaches. The shortest here is to divide both sides by 5 first.

  2. x = 6

    Add 2 to both sides.
Check: 5(6 − 2) = 5 × 4 = 20. Expanding first would also work - 5x − 10 = 20, then 5x = 30 and x = 6. Choosing an efficient path is part of mathematical fluency.

Example 3: the variable appears on both sides

4x + 3 = 2x + 15

  1. 2x + 3 = 15

    Subtract 2x from both sides, so the unknown lives on one side only.

  2. 2x = 12

    Subtract 3 from both sides.

  3. x = 6

    Divide both sides by 2.
Check both sides of the original equation: 4(6) + 3 = 27 and 2(6) + 15 = 27. Both sides equal 27, so the solution is correct.

A reliable solving routine

For a multi-step equation:

  1. Simplify each side if needed. Expand brackets and collect like terms.
  2. Move variable terms to one side. Apply the same addition or subtraction to both sides.
  3. Move constants to the other side. Again, do the same to both sides.
  4. Isolate the variable. Divide or multiply both sides as required.
  5. Check by substitution. Use the original equation, not a later line that may already contain the mistake.

Writing one algebraic change per line makes errors much easier to find.

Equations with rational solutions

Not every answer is a whole number. Solving 4x + 1 = 8 gives 4x = 7, then x = 7/4 = 1.75. A fraction or decimal can be a perfectly valid solution - do not round unless the question or context requires it.

What about solving equations on a graph?

An equation can also be solved graphically. For 4x + 3 = 2x + 15, graph y = 4x + 3 and y = 2x + 15. Their intersection has x = 6, matching the algebraic solution. A graph can make the meaning visible, while algebra often gives a more exact result.

To build that connection, continue to graphing linear relationships.

Common mistakes

  • Changing sides and signs without explaining why
    The shortcut “move it across and change the sign” can hide the real operation. Writing “subtract 7 from both sides” makes a sign error much harder to make.
  • Dividing only one term
    From 3x + 6 = 18, dividing both sides by 3 gives x + 2 = 6, because every term on the left is divided by 3. Dividing only 3x would not preserve equality.
  • Expanding a bracket incorrectly
    2(x + 5) = 2x + 10, not 2x + 5. If the expansion is wrong, the later equation may be solved neatly but still produce the wrong answer.
  • Stopping without a check
    A substitution check is quick and can catch arithmetic, sign and copying errors. The solution must make the original left and right sides equal.
  • Ignoring the context
    An algebraic answer may need a unit or a practical interpretation. A negative number of tickets or 3.6 people is not sensible, even if the algebra was performed correctly.

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.

Solve 3x + 8 = 20

Hint
Undo the + 8 first, then the × 3.
Show the answer
3x = 12, so x = 4. Check: 3(4) + 8 = 20.

Solve 7(x − 2) = 35

Hint
Dividing both sides by 7 first keeps the numbers small.
Show the answer
x − 2 = 5, so x = 7. Check: 7(7 − 2) = 35.

Solve 5x + 4 = 3x + 18

Hint
Collect the x terms on one side first: what removes 3x from the right?
Show the answer
Subtract 3x: 2x + 4 = 18. Subtract 4: 2x = 14. Therefore x = 7.

Check your solution to question 3 by substitution

Show the answer
Left side: 5(7) + 4 = 39. Right side: 3(7) + 18 = 39. The solution is verified.

Sources

  1. Australian Curriculum v9: Year 8 Mathematics (ACARA)australiancurriculum.edu.au
  2. Year 8 work sample: linear relationships in the real world (ACARA)australiancurriculum.edu.au

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