Number & Algebra — Worked Solutions (Year 7 Maths)
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Worked examples for Year 7 Maths Number & Algebra. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Year 7 Maths — Number & Algebra. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Order of operations with integers and fractions
Question
Evaluate $-12 + 3 \times (5 - 8) + \dfrac{1}{2}$, giving your answer as a fraction.
Solution
Work through the order of operations: brackets, then multiplication, then add left to right.
Brackets first: $5 - 8 = -3$.
Multiply: $3 \times (-3) = -9$.
Now add: $-12 + (-9) = -21$, then $-21 + \dfrac{1}{2}$.
Write $-21$ as $-\dfrac{42}{2}$, so $-\dfrac{42}{2} + \dfrac{1}{2} = -\dfrac{41}{2}$.
The answer is $-\dfrac{41}{2}$. Don't reach for the half until you've cleared the brackets and the multiply — that's where marks get lost.
Let's follow the order of operations — the rule that tells us what to do first so everyone gets the same answer. Brackets come before multiplying, and multiplying comes before adding.
Start inside the brackets: $5 - 8 = -3$ (we go three below zero).
Next the multiplication: $3 \times (-3) = -9$. A positive times a negative is negative.
Now we add from left to right. $-12 + (-9)$ means we go nine further down, giving $-21$.
Finally add the half. To add a fraction we need the same denominator, so write $-21$ as $-\dfrac{42}{2}$. Then $-\dfrac{42}{2} + \dfrac{1}{2} = -\dfrac{41}{2}$.
So the answer is $-\dfrac{41}{2}$. The reason we keep to the order is that it stops the half being added too early, which would change the result.
Order of operations: brackets, $\times$, then $+$.
- Brackets: $5 - 8 = -3$
- Multiply: $3 \times (-3) = -9$
- Add: $-12 + (-9) = -21$
- Half: $-21 + \dfrac{1}{2} = -\dfrac{42}{2} + \dfrac{1}{2} = -\dfrac{41}{2}$
Answer: $-\dfrac{41}{2}$.
Where the marks go
- 1 mark: Evaluates the bracket correctly: $5 - 8 = -3$
- 1 mark: Correct multiplication and integer addition: $-12 + (-9) = -21$
- 1 mark: Correct final answer including the fraction: $-\dfrac{41}{2}$
Key idea
Follow the order of operations — brackets, then multiplication, then addition — and add fractions only after giving them a common denominator.
Example 2 — Solving a linear equation and substitution
Question
Solve $3x + 7 = 22$ for $x$. Then use your value of $x$ to find the value of $2x - 5$.
Solution
Get $x$ on its own by undoing the operations in reverse.
Subtract 7 from both sides: $3x = 15$.
Divide both sides by 3: $x = 5$.
Now substitute into $2x - 5$: $2(5) - 5 = 10 - 5 = 5$.
So $x = 5$ and $2x - 5 = 5$. Keep both sides balanced at every step — do the same thing to each side or the equation breaks.
An equation is like a balanced scale, so whatever we do to one side we must do to the other. We want $x$ by itself.
First remove the $+7$ by subtracting 7 from both sides: $3x + 7 - 7 = 22 - 7$, which gives $3x = 15$.
The $3x$ means "3 lots of $x$", so we divide both sides by 3: $x = \dfrac{15}{3} = 5$.
Now the second part asks us to substitute, which just means replacing $x$ with 5. $2x - 5 = 2(5) - 5 = 10 - 5 = 5$.
So $x = 5$ and $2x - 5 = 5$. Keeping the scale balanced is what makes sure the value of $x$ we find really is the solution.
Solve, then substitute.
- $3x + 7 = 22$
- Subtract 7: $3x = 15$
- Divide by 3: $x = 5$
- Substitute: $2x - 5 = 2(5) - 5 = 5$
$x = 5$, $\;2x - 5 = 5$.
Where the marks go
- 1 mark: Subtracts 7 from both sides to get $3x = 15$
- 1 mark: Divides by 3 to find $x = 5$
- 1 mark: Substitutes correctly to get $2x - 5 = 5$
Key idea
Solve a linear equation by doing the same operation to both sides until $x$ is alone, then substitute that value wherever $x$ appears.
Frequently asked questions
Step-by-step solutions to Number & Algebra questions in Year 7 Maths, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Year 7 Maths, using the methods and notation expected in exams and assessments.