Year 9 Maths Linear Relations

Linear Modelling

Q: What does the gradient mean in a real-life problem?

The gradient is the rate of change: how much y changes for each one-unit increase in x, such as dollars per hour or litres per kilometre. A negative gradient means the quantity goes down.

Q: What does the y-intercept mean in a real-life problem?

It is the value when x is zero: the fixed fee, the starting amount or the value at the start. For example, in C = 8n + 75 the 75 is a fixed cost paid even when nothing is made.

Q: How do you find a linear equation from two points?

Divide the change in y by the change in x to get the gradient. Then substitute one of the points into y = mx + c to find c. Check the rule with the other point.

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Theory

A linear model describes something that changes by the same amount each step: a fixed amount plus a rate times the number of steps. This Year 9 maths page shows how to build the rule from words, a table, a graph or two data points, solve it for an unknown, find the break-even point of a profit model, treat simple interest as a linear model, and say when the model stops working.

A linear model has the form $y = mx + c$. The graph is a straight line because $y$ changes by the same amount, $m$, every time $x$ goes up by $1$.

The gradient $m$ is the rate of change: dollars per hour, litres per minute, centimetres per kilogram. A positive gradient means the quantity grows; a negative gradient means it falls.

The intercept $c$ is the value when $x = 0$: a fixed fee, a starting amount or a full tank. In $C = 8n + 75$, the $75$ is paid even when nothing is made.

When $y$ is directly proportional to $x$ there is no fixed part: $y = kx$, the line passes through the origin, and doubling $x$ doubles $y$. Currency conversions work this way.

A profit model breaks even where the profit is $0$: money in equals money out. Every model also has a domain: the values of $x$ for which it describes the real situation.

In $y = mx + c$ the gradient $m$ is the rate (rise over run) and $c$ is the starting or fixed amount.

A quantity that starts at $c$ and changes at a rate of $m$ per unit:

\[y = mx + c\]

y = m x + c

The 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}}

Direct proportion (no fixed part):

\[y = kx \qquad k = \dfrac{y}{x}\]

y = k x

An allowance or a starting threshold. If a phone plan includes the first \(50\) GB and each extra gigabyte costs \(3\) dollars, only the data past \(50\) GB is charged: \(C = 40 + 3(g - 50)\), for \(g \ge 50\).

Simple interest is a linear model. Simple interest pays the same amount every year: \(I = Prn\), with the rate \(r\) written as a decimal. $2000 invested at \(4\%\) p.a. earns \(2000 \times 0.04 = 80\) dollars a year, so after \(n\) years it is worth \(A = 2000 + 80n\). The gradient \(80\) is the interest per year and the intercept \(2000\) is the amount invested. After \(5\) years, \(A = 2400\), so it is worth $2400.

Net rate. When one thing adds and another removes at the same time, combine the rates first: filling at \(300\) L/h while draining at \(120\) L/h is a net \(+180\) L/h.

How to apply a linear model

Name the variables and their units. Say what $x = 0$ means.
Find the rate $m$: read it from the words, or divide the change in $y$ by the change in $x$ from a table, a graph or two data points. Make it negative if the quantity falls.
Find the starting value $c$: read it off, or substitute one known point into $y = mx + c$.
Substitute and solve: put in the known value and solve the linear equation for the unknown.
Interpret: answer in the units of the question, convert part of an hour to minutes, and check the answer lies inside the model's domain.

Example 1 — A rule from two points

An electrician charges a call-out fee plus an hourly rate. A $2$-hour job costs $250 and a $5$-hour job costs $505.
(i) Find the hourly rate and the call-out fee.
(ii) Write a formula for the charge $C$ dollars for a job of $h$ hours.
(iii) A bill was $760. How long was the job?

Solution

(i) The extra $3$ hours cost the difference in price. The fee is what is left once the hours are paid for.

$\text{rate}$	$=$	$\dfrac{505 - 250}{5 - 2}$
	$=$	$\dfrac{255}{3}$
	$=$	$\mathbf{\$85} \text{ per hour}$
$\text{fee}$	$=$	$250 - 2 \times 85 = \mathbf{\$80}$

(ii) The rate is the gradient and the fee is the starting value.

$C$

$=$

$85h + 80$

(iii) Substitute $C = 760$ and solve.

$85h + 80$	$=$	$760$
$85h$	$=$	$680$
$h$	$=$	$\mathbf{8}\text{ hours}$

C = 85 h + 80

Check with the other data point: $85 \times 5 + 80 = 505$. The rule fits both.

Example 2 — A falling model and its limits

A candle is $24$ cm tall when it is lit at 7:15 pm. It burns down at a steady rate, and after $1.5$ hours it is $18.6$ cm tall.
(i) Find a rule for the length $L$ cm after $t$ hours.
(ii) The candle goes out when it reaches the holder, $3$ cm from the bottom. At what time does it go out?
(iii) For what values of $t$ does the rule make sense?

Solution

(i) The length falls, so the gradient is negative.

$\text{rate}$	$=$	$\dfrac{24 - 18.6}{1.5}$
	$=$	$3.6 \text{ cm per hour}$
$L$	$=$	$24 - 3.6t$

(ii) Set $L = 3$.

$24 - 3.6t$	$=$	$3$
$3.6t$	$=$	$21$
$t$	$=$	$5.8333\ldots$
$0.8333\ldots \times 60$	$=$	$50 \text{ minutes}$

L = 24 - 3.6 t

$5$ h $50$ min after 7:15 pm is 1:05 am.

(iii) The rule only describes the candle while it burns: $0 \le t \le 5\dfrac{5}{6}$. Past that, it would give lengths the candle never reaches.

Example 3 — Profit and break-even

A school market stall makes candles. Its costs are a fixed $45 for the stall plus $1.80 to make each candle. Each candle sells for $4.80.
(i) Write a rule for the profit $P$ dollars when $n$ candles are made and sold.
(ii) How many candles must be sold to break even?
(iii) How many candles must be sold to make a profit of $120?

Solution

(i) Profit is money in minus money out. Each candle adds $4.80 - 1.80 = 3$ dollars, and the $45$ is paid whatever happens.

$P$	$=$	$4.8n - (45 + 1.8n)$
	$=$	$\mathbf{3n - 45}$

(ii) Break even means a profit of $0$.

$3n - 45$	$=$	$0$
$3n$	$=$	$45$
$n$	$=$	$\mathbf{15}\text{ candles}$

3 n - 45 = 0

(iii) Put $P = 120$:

$3n - 45$	$=$	$120$
$3n$	$=$	$165$
$n$	$=$	$\mathbf{55}\text{ candles}$

Check: $55 \times 4.80 = 264$ and $45 + 55 \times 1.80 = 144$, so the profit is $264 - 144 = 120$.

Example 4 — Total cost or average cost?

A bakery's daily cost is $240 in fixed costs plus $1.80 for each loaf baked.
(i) Write a formula for the total daily cost $C$ dollars when $n$ loaves are baked.
(ii) Find the average cost per loaf when $150$ loaves are baked.
(iii) How many loaves must be baked to bring the average cost down to $2.60 a loaf?

Solution

$C$

$=$

$240 + 1.8n$

(ii) Average cost is the total cost shared over the loaves.

$C$	$=$	$240 + 1.8 \times 150$
	$=$	$510$
$\text{average}$	$=$	$\dfrac{510}{150} = \mathbf{\$3.40}$

(iii) The straight-line rule is for the total. Write the total two ways and solve.

$240 + 1.8n$	$=$	$2.6n$
$240$	$=$	$0.8n$
$n$	$=$	$\mathbf{300}\text{ loaves}$

240 + 1.8 n = 2.6 n

The average falls as $n$ grows because the fixed $240 is shared over more loaves. Doubling the loaves does not double the total cost: only the $1.80-a-loaf part doubles.

Common pitfalls

Watch the sign of the gradient. A quantity that falls, such as fuel, value or water in a draining tank, has a negative gradient: $V = 40 - 1.5t$, not $V = 40 + 1.5t$.

Watch the step size in a table. If the table goes up in steps of $2$ hours, divide the change by $2$ to get the rate per hour.

Watch totals and averages. The straight-line rule is for the total. Turn an average cost per item back into a total (average $\times$ number) before you use it.

Watch parts of an hour. $6.4$ hours is $6$ h $24$ min, not $6$ h $40$ min. Multiply the decimal part by $60$.

Watch the domain. A model can give impossible answers outside its range: a negative length, a spring stretched past breaking, or a charge for kilometres that were already included.

Frequently asked questions

How do you write a linear equation from a word problem?

Find the fixed or starting amount and the rate of change. The rule is $y = mx + c$, with the rate as $m$ and the starting amount as $c$. Make the rate negative if the quantity decreases.

What does the gradient mean in a real-life problem?

The gradient is the rate of change: how much $y$ changes for each one-unit increase in $x$, such as dollars per hour or litres per kilometre. A negative gradient means the quantity goes down.

What does the y-intercept mean in a real-life problem?

It is the value when $x = 0$: the fixed fee, the starting amount or the value at the start. In $C = 8n + 75$, the $75$ is a fixed cost paid even when nothing is made.

How do you find a linear equation from two points?

Divide the change in $y$ by the change in $x$ to get the gradient $m$. Then substitute one of the points into $y = mx + c$ to find $c$. Check the rule with the other point.

What does break even mean?

It means the profit is zero: the money coming in equals the money going out. Write the profit rule, set it equal to $0$ and solve. For example, $P = 5n - 80$ breaks even when $n = 16$.

Is a linear model always accurate?

No. A linear model only works over a sensible range of values. Outside that range it can give impossible answers, such as a negative amount of fuel or a spring stretching forever.

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\(\text{rate}\)	\(=\)	\(\dfrac{505 - 250}{5 - 2}\)
\(\)	\(=\)	\(\dfrac{255}{3}\)
\(\)	\(=\)	\(\mathbf{\$85} \text{ per hour}\)
\(\text{fee}\)	\(=\)	\(250 - 2 \times 85 = \mathbf{\$80}\)

\(85h + 80\)	\(=\)	\(760\)
\(85h\)	\(=\)	\(680\)
\(h\)	\(=\)	\(\mathbf{8}\text{ hours}\)

\(\text{rate}\)	\(=\)	\(\dfrac{24 - 18.6}{1.5}\)
\(\)	\(=\)	\(3.6 \text{ cm per hour}\)
\(L\)	\(=\)	\(24 - 3.6t\)

\(24 - 3.6t\)	\(=\)	\(3\)
\(3.6t\)	\(=\)	\(21\)
\(t\)	\(=\)	\(5.8333\ldots\)
\(0.8333\ldots \times 60\)	\(=\)	\(50 \text{ minutes}\)

\(P\)	\(=\)	\(4.8n - (45 + 1.8n)\)
\(\)	\(=\)	\(\mathbf{3n - 45}\)

\(3n - 45\)	\(=\)	\(0\)
\(3n\)	\(=\)	\(45\)
\(n\)	\(=\)	\(\mathbf{15}\text{ candles}\)

\(3n - 45\)	\(=\)	\(120\)
\(3n\)	\(=\)	\(165\)
\(n\)	\(=\)	\(\mathbf{55}\text{ candles}\)

\(C\)	\(=\)	\(240 + 1.8 \times 150\)
\(\)	\(=\)	\(510\)
\(\text{average}\)	\(=\)	\(\dfrac{510}{150} = \mathbf{\$3.40}\)

\(240 + 1.8n\)	\(=\)	\(2.6n\)
\(240\)	\(=\)	\(0.8n\)
\(n\)	\(=\)	\(\mathbf{300}\text{ loaves}\)

Linear Modelling

📖 Theory

How to apply a linear model

Common pitfalls

Frequently asked questions

Theory