Linear modelling - Year 9 Mathematics

What you will learn

build a linear model $y = mx + c$ from a real-world description,
interpret $m$ as the rate of change and $c$ as the starting value,
calculate simple interest using $I = PrT$ and total amount $A = P + I$ ,
compare two linear models graphically and algebraically to decide which is better under given conditions,
determine when two linear models give the same value (break-even point).

Worked example 0 Real-world example: choosing a streaming plan

Plan A costs $12 per month with no per-movie charge. Plan B costs $4 per month plus $2 per movie. Which plan is cheaper if you watch $5$ movies a month?

Plan A: $C_A = 12$ (constant, no matter how many movies).
Plan B: $C_B = 2n + 4$ , where $n$ is the number of movies.
At $n = 5$ : $C_B = 2(5) + 4 = 14$ .
Plan A ($12) is cheaper than Plan B ($14) at $5$ movies.
Break-even: $12 = 2n + 4$ , so $2n = 8$ , $n = 4$ . Below $4$ movies, Plan B is cheaper.

Key idea: the gradient ($2 per movie) is the variable cost; the $y$ -intercept ($4) is the fixed cost. Comparing both tells you which plan wins at a given usage level.

1. Building linear models

A linear model takes the form $y = mx + c$ , where:

$m$ = rate of change (how much $y$ changes for each unit increase in $x$ ),
$c$ = starting value (the value of $y$ when $x = 0$ ).

To build a model from words: identify the fixed amount (that is $c$ ) and the per-unit amount (that is $m$ ).

Worked example 1 Writing a linear model from a description

A plumber charges a $80 call-out fee plus $65 per hour. Write a linear model for the total cost $C$ after $h$ hours.

Fixed cost (call-out fee): $c = 80$ .
Rate: $m = 65$ dollars per hour.
Model: $C = 65h + 80$ .
For a $3$ -hour job: $C = 65(3) + 80 = 195 + 80 = 275$ dollars.

Worked example 2 Model from two data points

A candle is $30$ cm tall. After burning for $2$ hours it is $24$ cm tall. Assuming the height decreases at a constant rate, find a linear model for the height $h$ after $t$ hours.

When $t = 0$ , $h = 30$ , so $c = 30$ .
When $t = 2$ , $h = 24$ . Gradient: $m = \dfrac{24 - 30}{2 - 0} = \dfrac{-6}{2} = -3$ .
Model: $h = -3t + 30$ .
Interpret: the candle burns down $3$ cm per hour and will reach $0$ cm when $t = 10$ hours.

2. Simple interest

Simple interest is a linear model. The interest earned or charged is the same amount each period.

Formula reference

I = PrT

$I$ = interest, $P$ = principal (initial amount), $r$ = annual interest rate (as a decimal), $T$ = time in years.

A = P + I = P + PrT = P(1 + rT)

$A$ = total amount after $T$ years.

Worked example 3 Calculating simple interest

Mia deposits $2000 into an account paying $4\%$ p.a. simple interest. How much interest does she earn in $3$ years, and what is her total balance?

$P = 2000$ , $r = 0.04$ , $T = 3$ .
$I = PrT = 2000 \times 0.04 \times 3 = 240$ .
$A = P + I = 2000 + 240 = 2240$ dollars.

As a linear model: $A = 80T + 2000$ , where the gradient $80$ represents $80 of interest earned each year.

Worked example 4 Finding the rate or time

Liam borrows $5000 at simple interest. After $4$ years he owes $6200 in total. What is the annual interest rate?

$I = A - P = 6200 - 5000 = 1200$ .
$I = PrT$ : $1200 = 5000 \times r \times 4$ .
$1200 = 20000r$ , so $r = 0.06 = 6\%$ p.a.

3. Comparing linear models

When two situations are modelled by different linear equations, you can find the break-even point by setting them equal and solving for $x$ .

Worked example 5 Comparing two gym memberships

Gym A charges $50 per month plus $5 per visit. Gym B charges $20 per month plus $10 per visit.

Gym A: $C_A = 5v + 50$ .
Gym B: $C_B = 10v + 20$ .
Break-even: $5v + 50 = 10v + 20$ , so $30 = 5v$ , $v = 6$ visits.
At $6$ visits both cost $80. Below $6$ visits Gym B is cheaper; above $6$ visits Gym A is cheaper.

Gym A (C = 5v + 50) vs Gym B (C = 10v + 20). The lines cross at v = 6 visits, where both cost $80. Below 6 visits Gym B is cheaper; above 6 Gym A is cheaper.

4. Interpreting gradient and intercept in context

The power of a linear model is in the meaning of $m$ and $c$ :

Component	Mathematical meaning	Real-world meaning
Gradient $m$	Rate of change of $y$ per unit of $x$	Cost per item, speed, burn rate, interest per year
$y$ -intercept $c$	Value of $y$ when $x = 0$	Fixed fee, starting height, initial deposit

Worked example 6 Interpreting from a graph description

A water tank is being filled. The amount of water (litres) is modelled by $W = 15t + 40$ , where $t$ is time in minutes.

Gradient $= 15$ : the tank fills at $15$ litres per minute.
$y$ -intercept $= 40$ : the tank already had $40$ litres when filling started.
After $10$ minutes: $W = 15(10) + 40 = 190$ litres.
To reach $250$ litres: $250 = 15t + 40$ , so $t = 14$ minutes.

Practice

Fluency

Tier 1: basic skills

A taxi charges a $4.50 flag-fall plus $2.20 per km. Write a linear model for the fare $F$ after $d$ km.
Using your model from Q1, find the fare for a $12$ km trip.
Calculate the simple interest on $3000 at $5\%$ p.a. for $2$ years.
Find the total amount when $8000 is invested at $3.5\%$ p.a. simple interest for $4$ years.
A phone battery starts at $100\%$ and drains at $8\%$ per hour. Write a linear model for battery percentage $B$ after $t$ hours.
Using your model from Q5, after how many hours will the battery reach $20\%$ ?
A pool contains $12\,000$ litres. Water drains at $500$ litres per hour. Write a model for the volume $V$ after $t$ hours.
Gemma earns $18.50 per hour plus a $25 daily transport allowance. Write a model for her daily earnings $E$ after $h$ hours.

Reasoning

Tier 2: mixed practice

Plan X charges $30 per month plus $0.10 per text. Plan Y charges $15 per month plus $0.40 per text. Find the break-even number of texts and state which plan is cheaper for $60$ texts per month.
Omar invests $4000 at simple interest and after $5$ years has $4800. Find the annual interest rate.
A car is worth $25000 new and depreciates by $3000 per year (linear model). Write the model for its value $V$ after $t$ years and find when it will be worth $7000.
Two runners start a race. Runner A starts $20$ m ahead and runs at $6$ m/s. Runner B starts at the start line and runs at $8$ m/s. Write models for their positions and find when Runner B catches Runner A.
The cost of hiring a marquee is modelled by $C = 150 + 45h$ , where $h$ is the number of hours. Interpret the $150$ and the $45$ in context.
A swimming pool is being filled. After $3$ hours it has $2700$ litres; after $7$ hours it has $5100$ litres. Find the linear model and state how much water was in the pool initially.

Reasoning

Tier 3: explain and apply

Explain why the simple interest formula $A = P(1 + rT)$ is linear in $T$ . What is the gradient of the graph of $A$ vs $T$ ?
Maria has $10000 to invest. Bank A offers $4\%$ p.a. simple interest. Bank B offers $3\%$ p.a. simple interest plus a one-off $200 bonus at account opening. Which bank gives more money after $8$ years?
A hire company charges $120 for the first day and $80 for each additional day. Is this a linear model from day $1$ ? Write the model for total cost $C$ after $d$ days ( $d \geq 1$ ) and explain.
Two towns are connected by road ( $120$ km). Car A leaves Town X at $60$ km/h. Car B leaves Town Y at the same time at $40$ km/h, driving toward Town X. Write position models for each car (from Town X) and find when and where they meet.

Challenge

Reasoning

Harder reasoning

Aisha borrows $15000 at $6\%$ p.a. simple interest. She repays $500 at the end of each year (after interest is calculated on the original principal). Write a model for the amount still owed after $n$ years and find after how many years the loan is fully repaid.
A phone company offers three plans. Plan A: $0 monthly, $0.80 per minute. Plan B: $20 monthly, $0.30 per minute. Plan C: $50 monthly, unlimited calls. Find the usage ranges (in minutes per month) for which each plan is cheapest.
A candle and a sparkler are lit at the same time. The candle is $20$ cm tall and burns down $2$ cm per minute. The sparkler is $12$ cm tall and burns down $4$ cm per minute. Find when they are the same height and when each burns out completely.

Answers

Answer key

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Year 9 core - answers

Fluency

Tier 1: basic skills

$F = 2.2d + 4.5$ .
$F = 2.2(12) + 4.5 = 26.4 + 4.5 = 30.90$ dollars.
$I = 3000 \times 0.05 \times 2 = 300$ dollars.
$I = 8000 \times 0.035 \times 4 = 1120$ dollars. Total: $A = 8000 + 1120 = 9120$ dollars.
$B = -8t + 100$ .
$20 = -8t + 100$ , so $8t = 80$ , $t = 10$ hours.
$V = -500t + 12000$ .
$E = 18.5h + 25$ .

Reasoning

Tier 2: mixed practice

$30 + 0.1n = 15 + 0.4n$ , so $15 = 0.3n$ , $n = 50$ texts. At $60$ texts: Plan X $= 36$ , Plan Y $= 39$ . Plan X is cheaper.
$I = 4800 - 4000 = 800$ . $800 = 4000 \times r \times 5$ . $r = 0.04 = 4\%$ p.a.
$V = -3000t + 25000$ . Set $V = 7000$ : $7000 = -3000t + 25000$ , $3000t = 18000$ , $t = 6$ years.
Runner A: $d_A = 6t + 20$ . Runner B: $d_B = 8t$ . Set equal: $8t = 6t + 20$ , $2t = 20$ , $t = 10$ s. Position: $80$ m from the start.
The $150 is the fixed hire cost (charged regardless of time). The $45 is the hourly rate (cost per additional hour).
Gradient: $m = \dfrac{5100 - 2700}{7 - 3} = \dfrac{2400}{4} = 600$ L/h. Using $(3, 2700)$ : $2700 = 600(3) + c$ , $c = 900$ . Model: $W = 600t + 900$ . The pool initially had $900$ litres.

Reasoning

Tier 3: explain and apply

$A = P(1 + rT) = PrT + P$ . This has the form $y = mx + c$ with $m = Pr$ (constant interest per year) and $c = P$ (initial deposit). The gradient is $Pr$ .
Bank A after $8$ years: $A = 10000(1 + 0.04 \times 8) = 10000 \times 1.32 = 13200$ dollars. Bank B after $8$ years: $A = 10000(1 + 0.03 \times 8) + 200 = 10000 \times 1.24 + 200 = 12600$ dollars. Bank A gives $600 more.
For $d \geq 1$ : first day costs $120, each additional day costs $80. Total: $C = 80(d - 1) + 120 = 80d + 40$ . Yes, this is linear in $d$ : gradient $= 80$ (daily rate), $y$ -intercept $= 40$ (but note the model only applies for $d \geq 1$ , so the actual starting cost is $C(1) = 120$ ).
Car A from Town X: $d_A = 60t$ . Car B from Town X: $d_B = 120 - 40t$ . Set equal: $60t = 120 - 40t$ , $100t = 120$ , $t = 1.2$ hours ( $1$ h $12$ min). Position: $60 \times 1.2 = 72$ km from Town X.

Reasoning

Challenge

Interest per year: $15000 \times 0.06 = 900$ dollars. Net reduction per year: $500 - 900 = -400$ dollars (the debt grows by $400 each year because repayments do not cover interest). Actually, since the question states interest is on the original principal: amount owed after $n$ years $= 15000 + 900n - 500n = 15000 + 400n$ . The debt increases, so the loan is never repaid under these terms. Alternatively, if we interpret “repays $500 after interest” as reducing the balance: total interest over $n$ years is $900n$ . Total repaid: $500n$ . Amount owed: $15000 + 900n - 500n = 15000 + 400n$ . The repayments ($500/year) are less than the annual interest ($900), so the loan balance grows. The repayment amount would need to exceed $900 per year to reduce the debt.
Plan A vs B: $0.8m = 20 + 0.3m$ , $0.5m = 20$ , $m = 40$ min. Plan B vs C: $20 + 0.3m = 50$ , $0.3m = 30$ , $m = 100$ min. Plan A cheapest: $0$ to $40$ min. Plan B cheapest: $40$ to $100$ min. Plan C cheapest: over $100$ min.
Candle: $h_c = -2t + 20$ , burns out at $t = 10$ min. Sparkler: $h_s = -4t + 12$ , burns out at $t = 3$ min. Same height: $-2t + 20 = -4t + 12$ , $2t = -8$ , $t = -4$ . Since $t = -4$ is negative, they are never the same height during the time both are burning (the sparkler starts shorter and burns faster, so the candle is always taller while both exist).

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