Worked Solutions
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.
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.