Algebra — Worked Solutions (HSC Maths Standard 2)
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Worked examples for HSC Maths Standard 2 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 HSC Maths Standard 2 — Algebra. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach 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 — Linear and non-linear modelling
Question
A drone hire company charges a fixed booking fee plus a constant rate per hour. Hiring the drone for 2 hours costs \$95, and hiring it for 5 hours costs \$200.
(a) Write a linear equation for the total cost $C$ dollars in terms of the number of hours $h$.
(b) The company also offers a maintenance plan whose cost follows $M = 4h^2 + 20$. For how many whole hours is the hire cost $C$ cheaper than the maintenance plan $M$?
Solution
Find the rate first. Cost rises by $200 - 95 = 105$ over $5 - 2 = 3$ hours, so the rate is $\dfrac{105}{3} = 35$ per hour.
Back-substitute to get the fixed fee: at $h = 2$, $95 = 35(2) + b$, so $b = 25$. Hence $C = 35h + 25$.
(b) Want $C < M$: $35h + 25 < 4h^2 + 20$, i.e. $4h^2 - 35h - 5 > 0$. The positive root is $h = \dfrac{35 + \sqrt{1225 + 80}}{8} = \dfrac{35 + \sqrt{1305}}{8} \approx 8.89$.
So $C < M$ once $h > 8.89$, meaning from 9 whole hours onward. Read the inequality direction carefully — that's where marks are lost.
Let's build the linear model. A linear cost has the form $C = (\text{rate})h + (\text{fixed fee})$. The rate is how fast the cost changes, which is the change in cost divided by the change in hours: $\dfrac{200 - 95}{5 - 2} = \dfrac{105}{3} = 35$ dollars per hour.
Now we need the fixed fee. Use a known point — at $h = 2$ the cost is \$95: $95 = 35 \times 2 + b = 70 + b$, so $b = 25$. That gives $C = 35h + 25$.
(b) "$C$ cheaper than $M$" means $C < M$, so $35h + 25 < 4h^2 + 20$. Rearranging into the usual form: $4h^2 - 35h - 5 > 0$. Solving the quadratic, the positive crossing point is $h = \dfrac{35 + \sqrt{1305}}{8} \approx 8.89$. The $h^2$ term grows fastest, so beyond that point $M$ overtakes $C$ — the hire is cheaper from 9 hours onward.
(a) Linear model $C = mh + b$.
- Rate: $m = \dfrac{200 - 95}{5 - 2} = 35$
- Fee: $95 = 35(2) + b \Rightarrow b = 25$
- $C = 35h + 25$
(b) $C < M$:
- $35h + 25 < 4h^2 + 20$
- $4h^2 - 35h - 5 > 0$
- Positive root $h = \dfrac{35 + \sqrt{1305}}{8} \approx 8.89$
- Cheaper from $h = 9$ hours onward
Where the marks go
- 1 mark: Correct rate of \$35 per hour
- 1 mark: Correct linear equation $C = 35h + 25$
- 1 mark: Solves $C < M$ and states 9 whole hours onward
Key idea
A linear model is rate × quantity + fixed amount; comparing it with a non-linear model means solving an inequality and checking which side satisfies it.
Example 2 — Simultaneous equations
Question
A school canteen sells two snack packs. On Monday it sells 12 fruit packs and 8 muesli packs for total takings of \$76. On Tuesday it sells 9 fruit packs and 14 muesli packs for total takings of \$83.
Let $f$ be the price of a fruit pack and $m$ the price of a muesli pack. Set up and solve a pair of simultaneous equations to find the price of each pack.
Solution
Set up the two equations straight from the takings:
$12f + 8m = 76$ … (1) $9f + 14m = 83$ … (2)
Use elimination. Multiply (1) by 3 and (2) by 4 to match the $f$ terms: $36f + 24m = 228$ and $36f + 56m = 332$. Subtract: $32m = 104$, so $m = 3.25$.
Back-substitute into (1): $12f + 8(3.25) = 76 \Rightarrow 12f = 50 \Rightarrow f = 4.166\ldots$ — check that: $12f = 76 - 26 = 50$, $f \approx 4.17$.
So a fruit pack is about \$4.17 and a muesli pack \$3.25. Always verify both numbers in the other equation before you commit.
First, turn the words into equations. Each day's takings are (number of fruit packs)$\times f$ plus (number of muesli packs)$\times m$:
$12f + 8m = 76$ … (1) $9f + 14m = 83$ … (2)
To solve them together we eliminate one variable. I'll line up the $f$ terms: multiply (1) by 3 to get $36f + 24m = 228$, and (2) by 4 to get $36f + 56m = 332$. Subtracting removes $f$: $32m = 104$, so $m = 3.25$.
Now substitute back to find $f$. Using (1): $12f + 8(3.25) = 76$, so $12f + 26 = 76$, giving $12f = 50$ and $f \approx 4.17$. The reason elimination works is that making the coefficients equal lets us cancel that variable cleanly. So fruit packs are \$4.17 and muesli packs \$3.25.
Equations:
- $12f + 8m = 76$ … (1)
- $9f + 14m = 83$ … (2)
Eliminate $f$:
- (1)$\times 3$: $36f + 24m = 228$
- (2)$\times 4$: $36f + 56m = 332$
- Subtract: $32m = 104 \Rightarrow m = 3.25$
Back-substitute:
- $12f + 8(3.25) = 76 \Rightarrow 12f = 50 \Rightarrow f \approx 4.17$
Fruit \$4.17, muesli \$3.25.
Where the marks go
- 1 mark: Correct pair of simultaneous equations from the context
- 1 mark: Valid elimination/substitution step matching coefficients
- 1 mark: Correct muesli price $m = 3.25$
- 1 mark: Correct fruit price $f \approx 4.17$
Key idea
Two unknowns need two equations; matching coefficients lets you eliminate one variable, then back-substitute for the other.
Frequently asked questions
Step-by-step solutions to exam-style questions on Algebra in HSC Maths Standard 2, 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 in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Standard 2 syllabus for Algebra, using the methods and notation expected in the exam.