Year 8 - Linear relations in real-life problems - Free practice

Year 8 Linear Relations in Real-Life Problems Worksheets & Practice

Q: What kinds of real-life problems use linear relations in Year 8?

Common contexts include mobile phone plans, hourly wages, taxi fares, savings accounts and distance-time problems. Students write an equation, solve it, and interpret the answer in plain language.

Q: How is this different from just solving linear equations?

Solving equations is one step in the process. In real-life problems students also need to define variables, build the equation from a description, and explain what the answer means in context. That full cycle is what these worksheets practise.

Q: What does it mean to compare two linear models?

Students might compare two phone plans, say Plan A costs $20 plus $0.10 per text and Plan B costs $5 plus $0.30 per text. They find the number of texts where both plans cost the same, then decide which is cheaper above or below that point.

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Quick answer

How do you use linear equations to solve real-life problems?

Write a linear equation by identifying a starting value (the intercept) and a constant rate of change (the slope). Then substitute known values to solve for the unknown. For example, if a plumber charges a 60 dollar call-out fee plus 80 dollars per hour, the total cost is C = 80h + 60, so 3 hours costs 300 dollars.

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Six questions across Foundation, Standard and Extension, with full worked solutions, written to the Australian Curriculum.

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Worked examples

Year 8 linear relations in real-life problems: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Writing a linear equation from a word problem
A plumber charges a call-out fee of $80 plus $60 for each hour worked. Write a linear equation for the total cost $C$ (in dollars) in terms of the number of hours worked $h$ .
Show worked solution
Key idea: A linear equation has the form $C = mh + b$ , where $m$ is the rate of change (cost per hour) and $b$ is the fixed starting value (call-out fee).
1. Identify the fixed cost (intercept): the call-out fee is $80, so $b = 80$ .
2. Identify the rate of change (slope): the cost increases by $60 per hour, so $m = 60$ .
3. Write the equation:
  $C = 60h + 80$
4. Check: for $h = 1$ , $C = 60(1) + 80 = 140$ . One hour should cost $80 + $60 = $140. ✓
Answer: $C = 60h + 80$
Example 2 · Interpreting slope and intercept in context
Maya is saving money. She already has $150 in her bank account and saves $25 each week. Her savings $S$ (in dollars) after $w$ weeks is modelled by $S = 25w + 150$ . How many weeks will it take Maya to save $500?
Show worked solution
Key idea: Substitute the target value for $S$ and solve the linear equation for $w$ .
1. The slope $25$ means savings grow by $25 per week; the intercept $150$ is the amount she starts with.
2. Set $S = 500$ :
  $500 = 25w + 150$
3. Subtract 150 from both sides:
  $350 = 25w$
4. Divide both sides by 25:
  $w = 14$
5. Check: $S = 25(14) + 150 = 350 + 150 = 500$ . ✓
Answer: 14 weeks
Example 3 · Comparing two plans to find the best-value point
A school is hiring a bus for an excursion. Company A charges a flat fee of $320 plus $4 per student. Company B charges a flat fee of $200 plus $6 per student. Write a linear equation for the total cost of each company in terms of the number of students $n$ , find the number of students at which both companies charge the same amount, and state which company is cheaper when 80 students attend.
Show worked solution
Key idea: Write an equation for each option, set them equal to find the break-even point, then compare for the given number of students.
1. Write the equations:
  $A = 4n + 320 B = 6n + 200$
2. Set $A = B$ to find the break-even point:
  $4n + 320 = 6n + 200$
3. Subtract $4n$ from both sides:
  $320 = 2n + 200$
4. Subtract 200, then divide by 2:
  $\begin{aligned} 120 &= 2n \\ n &= 60 \end{aligned}$
5. For $n = 80$ : $A = 4(80) + 320 = 640$ ; $B = 6(80) + 200 = 680$ . Since $640 < 680$ , Company A is cheaper.
6. Check break-even: $A = 4(60) + 320 = 560$ ; $B = 6(60) + 200 = 560$ . ✓
Answer: Both companies charge the same for 60 students. For 80 students, Company A is cheaper at $640.

Common mistakes

Where students usually lose marks

Swapping the slope and intercept

The plumber charges $60 call-out and $80 per hour, so $C = 80h + 60$ .

Always match each number to its meaning in the problem. The flat fee that is charged regardless of time is the intercept $b$ , and the amount charged per unit of time is the slope $m$ . Re-read the question and label each value before writing the equation.

Forgetting to include the fixed cost

The plumber charges $60 per hour, so $C = 60h$ .

Check whether the problem mentions any starting amount, joining fee, or flat charge. If it does, that value must be added as the intercept. A linear model with no fixed cost only applies when the total is zero when the variable is zero.

Setting up the break-even equation incorrectly

To compare Company A and Company B, add them: $(4n + 320) + (6n + 200) = 0$ .

To find when two options cost the same, set the two expressions equal to each other: $4n + 320 = 6n + 200$ . You are looking for the value of $n$ that makes both costs identical, not a sum.

Misreading which model is cheaper after the break-even point

Company B has a smaller flat fee so it is always cheaper.

A smaller flat fee does not mean a plan is always cheaper. After the break-even point, the plan with the higher slope (rate per unit) becomes more expensive. Always substitute the actual number into both equations and compare the results.

Try one yourself

Your turn

A mobile plan charges a flat monthly fee of $15 and $0.08 per text message sent. In a month where Maya sends $n$ messages, her bill is given by $C = 0.08n + 15$ . How many text messages did Maya send if her bill was $36.20? Give your answer as a whole number.

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About this worksheet

Why we built it

Linear relations in real-life problems is where Year 8 students see algebra start to matter outside the classroom. These worksheets build the skill of setting up and solving linear models from everyday and financial contexts, such as phone plans, wages, savings goals and distance-time situations, with worked solutions for every question so students can check exactly how to move from the written problem to an equation and back to a real-world answer.

What's covered

Sub-skills your student will practise

✓Identifying the variables and writing a linear equation from a word problem
✓Interpreting the slope and intercept in a real-world context
✓Solving linear equations to answer applied questions
✓Comparing two linear models to find a break-even or best-value point
✓Working with financial contexts such as wages, costs and savings
✓Reading and drawing graphs that represent real-life linear relationships

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Frequently asked

Questions parents ask about linear relations in real-life problems

What kinds of real-life problems use linear relations in Year 8?