Worked Solutions
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.
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.