Number & Algebra — Worked Solutions (Year 9 Maths)
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Worked examples for Year 9 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 9 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 — Index laws
Question
Simplify $\dfrac{6a^5 b^3 \times 2a^2 b}{4a^3 b^2}$, writing your answer with positive indices.
Solution
Deal with the numbers and each letter separately. Multiply out the top first, then divide.
Numbers: $\dfrac{6 \times 2}{4} = 3$.
The $a$ terms: $\dfrac{a^5 \times a^2}{a^3} = \dfrac{a^7}{a^3} = a^4$ — add the indices on top, then subtract.
The $b$ terms: $\dfrac{b^3 \times b}{b^2} = \dfrac{b^4}{b^2} = b^2$.
Answer: $3a^4 b^2$. Keep numbers, $a$'s and $b$'s in separate columns and you won't lose a mark.
Index laws let us multiply and divide powers by just adding or subtracting the indices, so let's take it one base at a time.
Start with the plain numbers (the coefficients): $6 \times 2 = 12$, and $12 \div 4 = 3$.
Now the $a$'s. When we multiply we add indices, so $a^5 \times a^2 = a^{5+2} = a^7$. When we divide we subtract, so $a^7 \div a^3 = a^{7-3} = a^4$.
The $b$'s work the same way: $b^3 \times b^1 = b^4$, then $b^4 \div b^2 = b^2$.
Putting the pieces back together gives $3a^4 b^2$. The reason we add when multiplying is that we're just counting how many times the base appears all together.
Coefficients, then each base.
- Numbers: $\frac{6 \times 2}{4} = 3$
- $a$: $\frac{a^5 a^2}{a^3} = a^{5+2-3} = a^4$
- $b$: $\frac{b^3 b}{b^2} = b^{3+1-2} = b^2$
Answer: $3a^4 b^2$.
Where the marks go
- 1 mark: Correct coefficient $3$
- 1 mark: Correct $a$ power $a^4$ using add/subtract of indices
- 1 mark: Correct $b$ power $b^2$ and fully simplified answer $3a^4 b^2$
Key idea
Multiplying powers of the same base adds the indices; dividing subtracts them. Handle the coefficients and each base separately.
Example 2 — Linear equations and graphing
Question
A straight line has equation $2x + y = 6$. Find the gradient and the $y$-intercept, then find where the line crosses the $x$-axis.
Solution
Rearrange into $y = mx + c$ form first — then the gradient and $y$-intercept just read off.
$2x + y = 6 \Rightarrow y = -2x + 6$. So the gradient is $m = -2$ and the $y$-intercept is $c = 6$, i.e. the point $(0, 6)$.
The $x$-intercept is where $y = 0$: $0 = -2x + 6 \Rightarrow 2x = 6 \Rightarrow x = 3$, so $(3, 0)$.
Always convert to $y = mx + c$ before reading off the gradient — guessing from the original form loses marks.
The easiest way to see the gradient and intercept is to get $y$ by itself, because the form $y = mx + c$ tells us both directly.
Subtract $2x$ from both sides of $2x + y = 6$ to get $y = -2x + 6$. Now compare with $y = mx + c$: the number in front of $x$ is the gradient, so $m = -2$, and the constant is the $y$-intercept, so $c = 6$. That intercept is the point $(0, 6)$.
To find where the line meets the $x$-axis, remember that every point on the $x$-axis has $y = 0$. Setting $y = 0$ gives $0 = -2x + 6$, and solving, $2x = 6$ so $x = 3$. The line crosses at $(3, 0)$.
Thinking about which axis means which variable is zero is what makes intercept questions click.
Rearrange to $y = mx + c$.
- $2x + y = 6 \Rightarrow y = -2x + 6$
- Gradient $m = -2$; $y$-intercept $c = 6$, point $(0, 6)$
$x$-intercept ($y = 0$):
- $0 = -2x + 6 \Rightarrow x = 3$, point $(3, 0)$
Where the marks go
- 1 mark: Rearranges to $y = -2x + 6$
- 1 mark: States gradient $m = -2$
- 1 mark: States $y$-intercept $6$ (point $(0, 6)$)
- 1 mark: Finds $x$-intercept $(3, 0)$ by setting $y = 0$
Key idea
Rearrange a linear equation into $y = mx + c$ to read off the gradient $m$ and $y$-intercept $c$; set $y = 0$ for the $x$-intercept.
Frequently asked questions
Step-by-step solutions to Number & Algebra questions in Year 9 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 9 Maths, using the methods and notation expected in exams and assessments.