Equation-solving consolidation | Year 9

01 / LEARN

The idea, step by step

Keep an equation balanced

An equation states that two expressions have the same value. An inverse operation undoes an operation: addition undoes subtraction, and division undoes multiplication. Perform the same operation on both sides, so the equality remains true.

When the unknown appears on both sides, collect its terms on one side and number terms on the other. Do not move a term and guess its sign. Write the operation you apply, such as subtracting $2 x$ from both sides.

5 x + 4 = 2 x + 19 \Rightarrow 3 x + 4 = 19 \Rightarrow x = 5

Clear brackets and check

Expand a bracket before collecting terms, unless dividing the entire equation gives a simpler route. A minus sign outside a bracket affects every term inside it. If a fraction contains a whole expression, multiply both sides by its denominator before simplifying.

Substitute your solution into the original equation. Both sides must give the same number. This check can reveal a lost negative sign even when the working looks tidy.

A common trap

Expanding $- 2 (x - 3)$ as $- 2 x - 6$ .

Instead: Multiply both terms by $- 2$ : the result is $- 2 x + 6$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Unknowns on both sides

Solve $7 x - 8 = 3 x + 12$ .

Subtract $3 x$ from each side: $4 x - 8 = 12$ .
Add 8: $4 x = 20$ .
Divide by 4: $x = 5$ .
Check: $7 (5) - 8 = 27$ and $3 (5) + 12 = 27$ .

x = 5

Example 2

A bracket and a fraction

Solve $\frac{3 (x - 2)}{2} = 9$ .

Multiply both sides by 2: $3 (x - 2) = 18$ .
Divide by 3: $x - 2 = 6$ .
Add 2: $x = 8$ .
Check: $3 (8 - 2) / 2 = 9$ .

x = 8

Check your understanding

Solve $4 x + 3 = 2 x + 11$ .

Show answer

$x = 4$

Subtract $2 x$ and then 3: $2 x = 8$ , so $x = 4$ . Both original sides equal 19.

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $6 x + 5 = 29$ .
2 marks
Solve $8 x - 7 = 3 x + 18$ .
2 marks
Solve $3 (x + 4) = 27$ .
2 marks
Solve $5 - 2 (x - 1) = 13$ .
2 marks
Solve $\frac{x + 5}{3} = 4$ .
2 marks
A number is multiplied by 4 and then 9 is subtracted. The result equals twice the number plus 7. Find the number.
3 marks

Read the worked solutions

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

Solve $6 x + 5 = 29$ .
2 marks
Subtract 5: $6 x = 24$ .
Divide by 6: $x = 4$ .
$x = 4$
Solve $8 x - 7 = 3 x + 18$ .
2 marks
Subtract $3 x$ : $5 x - 7 = 18$ .
Add 7 and divide by 5.
$x = 5$
Solve $3 (x + 4) = 27$ .
2 marks
Divide by 3: $x + 4 = 9$ .
Subtract 4.
$x = 5$
Solve $5 - 2 (x - 1) = 13$ .
2 marks
Expand: $7 - 2 x = 13$ .
Subtract 7: $- 2 x = 6$ .
Divide by $- 2$ .
$x = - 3$
Solve $\frac{x + 5}{3} = 4$ .
2 marks
Multiply by 3: $x + 5 = 12$ .
Subtract 5.
$x = 7$
A number is multiplied by 4 and then 9 is subtracted. The result equals twice the number plus 7. Find the number.
3 marks
Let the number be $x$ . Then $4 x - 9 = 2 x + 7$ .
Subtract $2 x$ and add 9: $2 x = 16$ .
Thus $x = 8$ .
8

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Gradient, midpoint and distance; algebraic and graphical straight-line relationships; interpreting models and testing parameters. Parallel and perpendicular lines are included as a natural school-course extension. All explanations and questions are original.

Official curriculum source ↗

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