Algebra — Worked Solutions (Preliminary Maths Standard)
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Worked examples for Preliminary Maths Standard 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 Preliminary Maths Standard — 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 — Substituting into a formula
Question
The stopping distance of a car is given by $d = 0.4v + \dfrac{v^2}{20}$, where $d$ is the distance in metres and $v$ is the speed in metres per second. Find the stopping distance when the car is travelling at $v = 30$ m/s.
Solution
Substitute $v = 30$ straight into the formula and evaluate each term.
$d = 0.4(30) + \dfrac{30^2}{20}$.
First term: $0.4 \times 30 = 12$. Second term: $\dfrac{900}{20} = 45$.
$d = 12 + 45 = 57$ m.
Write the units. A stopping distance with no metres attached loses the final mark.
The formula is already done for us — our job is just to put the right number in the right place. We're told $v = 30$, so everywhere we see $v$ we write $30$.
$d = 0.4(30) + \dfrac{30^2}{20}$.
Take it one term at a time so nothing slips. The first term is $0.4 \times 30 = 12$. For the second, square first ($30^2 = 900$) then divide ($900 \div 20 = 45$).
Adding them: $d = 12 + 45 = 57$ metres.
Doing it term-by-term and keeping the units the whole way through is what stops careless slips on a question that's really just careful arithmetic.
Substitute $v = 30$.
- $d = 0.4(30) + \dfrac{30^2}{20}$
- $0.4 \times 30 = 12$
- $\dfrac{900}{20} = 45$
- $d = 12 + 45 = 57$ m
Where the marks go
- 1 mark: Correct substitution of $v = 30$ into the formula
- 1 mark: Correct evaluation of both terms ($12$ and $45$)
- 1 mark: Correct answer with units ($57$ m)
Key idea
Substitution means replacing the variable with its value, then evaluating each term carefully — and always include units in the final answer.
Example 2 — Linear equation from a model
Question
A plumber charges a $\$80$ call-out fee plus $\$65$ per hour. The total cost $C$ for a job of $h$ hours is $C = 80 + 65h$. A customer is charged $\$340$. Set up and solve a linear equation to find how many hours the plumber worked.
Solution
The total is $\$340$, so set the cost formula equal to $340$ and solve for $h$.
$80 + 65h = 340$.
Subtract the call-out fee: $65h = 260$.
Divide: $h = \dfrac{260}{65} = 4$.
The plumber worked $4$ hours. Undo the constant first, then the coefficient — that order keeps the algebra clean.
We want to find $h$, the number of hours. The clue is that the total charge equals $\$340$, so we set the whole cost expression equal to that total.
$80 + 65h = 340$.
To get $h$ on its own, we peel away everything attached to it. First the $\$80$ call-out fee, which is added on, so we subtract it from both sides: $65h = 340 - 80 = 260$.
Now $h$ is only multiplied by $65$, so we divide both sides by $65$: $h = \dfrac{260}{65} = 4$.
So the plumber worked $4$ hours. The reason we subtract before we divide is that we reverse the operations in the opposite order they were applied — like taking off your shoes before your socks.
Set cost $= 340$.
- $80 + 65h = 340$
- $65h = 260$
- $h = \dfrac{260}{65} = 4$
Plumber worked $4$ hours.
Where the marks go
- 1 mark: Sets up the correct equation $80 + 65h = 340$
- 1 mark: Subtracts the call-out fee to get $65h = 260$
- 1 mark: Divides to solve $h = 4$
- 1 mark: States the answer in context (4 hours)
Key idea
To solve a linear model, set the formula equal to the given value, then undo the operations in reverse — subtract the constant, then divide by the coefficient.
Frequently asked questions
Step-by-step solutions to Algebra questions in Preliminary Maths Standard, 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 Preliminary Maths Standard, using the methods and notation expected in exams and assessments.