Year 10 Maths Advanced Straight Line Graphs

Linear Models

Q: What is a linear model in maths?

It is a rule of the form y = mx + c that describes a real situation where one quantity changes at a constant rate as another changes. Its graph is a straight line.

20 practice questions 2 video lessons Theory + worked examples

A linear model uses a straight-line rule such as $C=70t+120$ to describe a real situation. The gradient is the rate of change, such as a cost per hour, and the vertical intercept is the starting value, such as a call-out fee. In NSW Year 10 Advanced maths you build models from words, tables and graphs, and use them to compare options (NESA outcome MA5‑LIN‑C‑02).

You will explain what the gradient and intercept mean, find a rule from two data points, and use it to predict values. You will find break-even points and when two plans charge the same, from removalists and bike hire to jumping castles and a school camp. Other models describe a falling dam level, the speed of sound and two burning candles. You will also judge when a prediction is not sensible. It is part of Year 10 Advanced Straight Line Graphs.

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Theory

A linear model describes a quantity that changes by the same amount for each unit of another: $y = mx + c$, with $m$ the rate and $c$ the fixed or starting amount. This Year 10 Advanced maths page covers writing rules from words, tables, graphs and two facts, break-even problems, comparing plans, and the limits of a model.

A linear model is a rule of the form $y = mx + c$ used to describe a real situation. Its graph is a straight line.

The gradient $m$ is the rate of change: how much $y$ changes for each one unit of $x$, such as dollars per hour or centimetres per day. A negative gradient means the quantity is falling.

The intercept $c$ is the value of $y$ when $x = 0$: a fixed fee, a starting height or an initial amount.

A model only works over a sensible range of values. Using it far outside the data, called extrapolation, can give silly answers, such as a negative height or an adult mass for a small child.

In $y = mx + c$, the gradient $m$ is the rate (e.g. dollars per hour) and $c$ is the starting or fixed amount.

Profit $=$ income $-$ cost. Where the two lines cross, profit is zero: the break-even point.

\[y = mx + c \qquad \text{total} = \text{rate} \times \text{amount} + \text{fixed amount}\]

y = m x + c

Rate from two data points $(x_1,\ y_1)$ and $(x_2,\ y_2)$:

\[m = \dfrac{y_2 - y_1}{x_2 - x_1}\]

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

\[\text{profit} = \text{income} - \text{costs} \qquad \text{break-even: profit} = 0\]

P = I - C

Percentages and units. A commission of \(3\%\) of sales \(s\) is \(0.03s\). A charge of \(25\) cents per km is \(0.25\) dollars per km. Keep every term in the same units.

A late start. If something starts \(1.5\) hours after time zero, its time is \(t - 1.5\). A car leaving \(600\) km away at \(90\) km/h towards you is at distance \(600 - 90(t - 1.5)\).

How to build and use a linear model

Define the variables with units, e.g. $C$ dollars after $t$ hours.
Find the rate: given in words, from a table (change in $y$ $\div$ change in $x$), from a graph, or from two facts.
Find the fixed amount: the value when $x = 0$, or substitute a known pair into $y = mx + c$.
Use the rule: substitute to predict, or set equal to a value and solve.
Interpret the answer in words, round sensibly (whole people or items), and check it makes sense.

Common types of problem

Two facts, unknown rule: the rate is the difference in totals over the difference in amounts (Example 2). If the rule is $C = a + bh$, two facts give two equations to solve simultaneously.
Break-even and minimum numbers: solve profit $= 0$ or payments $\ge$ costs, then round up to the next whole item (Example 3).
When does it pass a level? Solve an inequality, or solve the equation and test whole numbers either side.
Comparing two models: set them equal to find the crossover, then test a value to see which is cheaper on each side (Example 4).
Meeting problems: two moving objects meet when their distances from the same starting point are equal.
Using a formula backwards: e.g. if $v = 331 + 0.6T$ and the speed is known, solve for $T$.
Validity: ask whether the input is within a sensible range and whether a steady rate is realistic for that long.

Example 1 — Writing and using a rule

A plumber charges a $\$85$ call-out fee plus $\$110$ per hour.
(i) Write a rule for the cost $C$ dollars of a job lasting $t$ hours, and say what each number means.
(ii) Find the cost of a $2.5$-hour job.
(iii) A bill is $\$580$. How long was the job?

Solution

(i) Cost $=$ rate $\times$ time $+$ fixed amount.

$C$

$=$

$110t + 85$

C = 110 t + 85

The gradient $110$ is the cost per hour; the intercept $85$ is the call-out fee, charged even for a very short job.

(ii)

$C$	$=$	$110(2.5) + 85$
	$=$	$275 + 85 = \mathbf{\$360}$

(iii) Now $C$ is known; solve for $t$.

$110t + 85$	$=$	$580$
$110t$	$=$	$495$
$t$	$=$	$\mathbf{4.5}\text{ hours}$

110 t + 85 = 580, t = 4.5

Example 2 — A rule from two facts

A gym charges a joining fee plus a fixed amount each month. Three months cost $\$305$ in total and eight months cost $\$630$.
(i) Find the monthly charge and the joining fee.
(ii) Write a rule for the total cost $C$ of $m$ months, and find the cost of a year.

Solution

(i) The extra $5$ months cost the extra amount, so the rate is the gradient between the two facts.

$\text{Monthly charge}$	$=$	$\dfrac{630 - 305}{8 - 3} = \dfrac{325}{5} = \$65$
$\text{Joining fee}$	$=$	$305 - 65 \times 3 = \$110$

m = \frac{325}{5} = 65, c = 110

(ii)

$C$	$=$	$65m + 110$
$m = 12:\quad C$	$=$	$65(12) + 110 = \mathbf{\$890}$

C = 65 m + 110, C (12) = 890

Check with the other fact: $65(8) + 110 = 630\ \checkmark$

Example 3 — Profit and break-even

A market stall sells candles for $\$14$ each. Its costs are $\$260$ for the stall plus $\$5.50$ per candle.
(i) Write a rule for the profit $P$ dollars when $n$ candles are sold.
(ii) How many must be sold to make a profit?
(iii) Find the profit when $80$ are sold.

Solution

(i) Profit is income minus costs. Brackets keep the subtraction right.

$P$	$=$	$14n - (260 + 5.5n)$
	$=$	$8.5n - 260$

P = 8.5 n - 260

(ii) Find where $P = 0$, the break-even point.

$8.5n - 260$	$=$	$0$
$n$	$=$	$\dfrac{260}{8.5} = 30.58\ldots$

n = \frac{260}{8.5} \approx 30.6

Candles are sold whole. $30$ candles give $P = -\$5$, a loss; $31$ give $P = \$3.50$. At least $\mathbf{31}$ candles must be sold.

(iii)

$P$

$=$

$8.5(80) - 260 = \mathbf{\$420}$

Example 4 — Comparing two plans

Car hire A costs $\$60$ a day plus $25$ cents per kilometre. Car hire B costs $\$45$ a day plus $40$ cents per kilometre. For a one-day hire, for what distances is B cheaper?

Solution

Write both rules in dollars, with $d$ kilometres.

$A$	$=$	$60 + 0.25d$
$B$	$=$	$45 + 0.4d$

Find where the costs are equal, then decide which side B is cheaper.

$45 + 0.4d$	$=$	$60 + 0.25d$
$0.15d$	$=$	$15$
$d$	$=$	$100$

45 + 0.4 d = 60 + 0.25 d, d = 100

B starts cheaper ($\$45$ against $\$60$) but rises faster. Check $d = 50$: A costs $\$72.50$ and B costs $\$65$. So B is cheaper for trips of less than $\mathbf{100}$ km; at $100$ km both cost $\$85$.

Common pitfalls

Rate and fixed amount swapped. In $C = 110t + 85$ the $110$ multiplies the time, so it is the hourly rate. Writing $C = 85t + 110$ swaps their meanings.

Bracket the costs. Profit $= 14n - (260 + 5.5n)$. Without brackets the $5.5n$ is added instead of subtracted.

Round to suit the situation. $30.6$ candles to break even means $31$ must be sold. But the number of full buses you can fill rounds down.

Mixed units. A rate of $40$ cents per km with a fee in dollars must be written $0.4$ dollars per km.

Do not trust a model too far. A steady rate fitted over a few weeks or months may be useless years later. Check that a prediction is realistic.

Frequently asked questions

What is a linear model in maths?

It is a rule of the form $y = mx + c$ that describes a real situation where one quantity changes at a constant rate as another changes. Its graph is a straight line.

What do the gradient and the y-intercept mean in a real-life problem?

The gradient is the rate, such as dollars per hour or litres per minute. The $y$-intercept is the starting or fixed amount, such as a call-out fee or an initial height.

How do you find a linear rule from two data points?

Divide the change in the output by the change in the input to get the rate. Then substitute one data point into $y = mx + c$ to find the fixed amount $c$.

What is the break-even point?

It is where income equals costs, so the profit is zero. Below it the business makes a loss; above it, a profit. On a graph it is where the income and cost lines cross.

How do you compare two pricing plans?

Write a linear rule for each, set the rules equal and solve to find where they cost the same. Then test a value on either side to see which plan is cheaper there.

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\(C\)	\(=\)	\(110(2.5) + 85\)
\(\)	\(=\)	\(275 + 85 = \mathbf{\$360}\)

\(110t + 85\)	\(=\)	\(580\)
\(110t\)	\(=\)	\(495\)
\(t\)	\(=\)	\(\mathbf{4.5}\text{ hours}\)

\(\text{Monthly charge}\)	\(=\)	\(\dfrac{630 - 305}{8 - 3} = \dfrac{325}{5} = \$65\)
\(\text{Joining fee}\)	\(=\)	\(305 - 65 \times 3 = \$110\)

\(C\)	\(=\)	\(65m + 110\)
\(m = 12:\quad C\)	\(=\)	\(65(12) + 110 = \mathbf{\$890}\)

\(P\)	\(=\)	\(14n - (260 + 5.5n)\)
\(\)	\(=\)	\(8.5n - 260\)

\(8.5n - 260\)	\(=\)	\(0\)
\(n\)	\(=\)	\(\dfrac{260}{8.5} = 30.58\ldots\)

\(45 + 0.4d\)	\(=\)	\(60 + 0.25d\)
\(0.15d\)	\(=\)	\(15\)
\(d\)	\(=\)	\(100\)

Linear Models

📖 Theory

How to build and use a linear model

Common types of problem

Common pitfalls

Frequently asked questions

Theory