Brackets and variables on both sides | Year 8

01 / LEARN

Understand the idea.

Distribute before collecting

Distributing means multiplying every term in a bracket by its outside factor. For example, $3 (x - 2) = 3 x - 6$ : the factor multiplies both the unknown and the negative constant.

After expanding, gather variable terms on one side by subtracting the same variable expression from both sides. Choosing the smaller coefficient often keeps the remaining coefficient positive, but another balanced route is also valid.

Choose a useful route

Choose a route that simplifies the whole equation. If $4 (x + 1) = 20$ , dividing both sides by $4$ first is efficient. When different variable terms occur on both sides, expansion often makes comparison easier.

A term does not simply jump across an equals sign. Its apparent sign change comes from adding or subtracting it on both sides. Verify the result using the original brackets, especially when substituting a negative number.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

One bracket

Solve $3 (x + 2) = 2 x + 11$ .

Worked solution

1
Expand the left bracket.
$3 x + 6 = 2 x + 11$
The factor $3$ multiplies both terms inside the bracket.
2
Subtract the variable term on the right.
$x + 6 = 11$
Subtracting $2 x$ from both sides gathers the unknown on the left.
3
Subtract 6 from both sides.
$x = 11 - 6 = 5$
This isolates the unknown while preserving equality.
4
Check both original sides.
$\begin{aligned} 3 (5 + 2) & = 21 \\ 2 (5) + 11 & = 21 \end{aligned}$
The original equation is true at $x = 5$ .

Answer

$x = 5$ .

EXAMPLE 02

Two brackets

Solve $2 (x - 3) = 3 (x - 5)$ .

Worked solution

1
Expand both brackets.
$2 x - 6 = 3 x - 15$
Each outside factor multiplies the variable and the negative constant.
2
Subtract 2x from both sides.
$- 6 = x - 15$
The variable terms are now together on the right.
3
Add 15 to both sides.
$9 = x$
Adding $15$ removes the negative constant, giving $x = 9$ .
4
Check using the original brackets.
$\begin{aligned} 2 (9 - 3) & = 12 \\ 3 (9 - 5) & = 12 \end{aligned}$
The two original expressions agree.

Answer

$x = 9$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve $2 (x + 4) = x + 13$ .

Show answer & reasoning

$x = 5$ .

Expand to $2 x + 8 = x + 13$ . Subtract $x$ and then $8$ to get $x = 5$ ; both original sides are $18$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Solve $2 (x + 3) = 18$ .
2 marks
Solve $3 (x - 1) = 2 x + 5$ .
2 marks
Solve $5 (x + 2) = 3 x + 18$ .
2 marks
Solve $4 (x - 2) = 2 (x + 3)$ .
2 marks
Explain and correct the expansion $2 (x + 5) = 2 x + 5$ .
2 marks
Plan A costs AUD 8 plus AUD 3 per hour; B costs AUD 2 plus AUD 5 per hour. When are they equal?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $2 (x + 3) = 18$ .
2 marks
Divide by 2 to get $x + 3 = 9$ .
Subtract 3.
$x = 6$ .
Solve $3 (x - 1) = 2 x + 5$ .
2 marks
Expand: $3 x - 3 = 2 x + 5$ .
Subtract $2 x$ and add 3.
$x = 8$ .
Solve $5 (x + 2) = 3 x + 18$ .
2 marks
Expand, then subtract $3 x$ and 10: $2 x = 8$ .
Divide by 2.
$x = 4$ .
Solve $4 (x - 2) = 2 (x + 3)$ .
2 marks
Expand to $4 x - 8 = 2 x + 6$ .
Subtract $2 x$ , add 8 and divide by 2.
$x = 7$ .
Explain and correct the expansion $2 (x + 5) = 2 x + 5$ .
2 marks
The outside factor acts on every term.
Both $x$ and 5 must be doubled.
The correct expansion is $2 x + 10$ .
Plan A costs AUD 8 plus AUD 3 per hour; B costs AUD 2 plus AUD 5 per hour. When are they equal?
3 marks
Set $8 + 3 h = 2 + 5 h$ .
Subtract $3 h$ and 2: $6 = 2 h$ , so $h = 3$ .
At 3 hours, both cost AUD 17.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics (2026)

Multi-step equations; brackets and variables on both sides; equations with rational solutions; substitution checks; one-variable inequalities and number-line solutions; tables and straight-line graphs; horizontal/vertical lines; how coefficients change a graph; graphical solutions and intersection checks; real-world linear models with digital exploration.

Official curriculum source ↗

In this lesson

Understand the idea.

Distribute before collecting

Choose a useful route

See the method in action.

One bracket

Expand the left bracket.

Subtract the variable term on the right.

Subtract 6 from both sides.

Check both original sides.

Two brackets

Expand both brackets.

Subtract 2x from both sides.

Add 15 to both sides.

Check using the original brackets.

Try it yourself.

Check your understanding

Keep practising