Year 11 General Linear Equations And Their Graphs

Linear Modelling

10 practice questions 2 video lessons Theory + worked examples

Theory

A linear model uses the equation $y = m x + c$ to describe a real-world relationship with a constant rate of change. The slope $m$ is the rate (dollars per month, km per hour, depreciation per year) and the intercept $c$ is the starting value (joining fee, initial distance, purchase price). Use models to find values, compare plans (break-even) and track depreciation.

A linear model is an equation of the form $y = m x + c$ used to describe a real-world situation where one quantity changes at a constant rate with respect to another. The slope $m$ and the intercept $c$ each carry real-world meaning.

There are three core skills with linear models:

Build a model from a description, two readings, or a graph.
Use the model by substituting a value of $x$ (or $y$ ) and solving.
Interpret the slope and intercept in the real-world context.

When two plans meet at the same cost, this is called a break-even point. Set the two cost expressions equal and solve. When the slope is negative, the quantity is decreasing — common in depreciation, water draining and fuel burning.

The first diagram shows a typical linear model: a fixed monthly fee plus a per-minute rate. The intercept is the joining fee and the slope is the rate. The second shows two competing plans meeting at a break-even point — the usage at which both plans cost the same.

A linear model:

C = 0.10 t + 25

. The intercept is the fixed fee, the slope is the per-minute rate.

Plans

C_{A} = 50 m + 200

and

C_{B} = 70 m

break even at

m = 10

months,

C = $ 700

A linear model uses the standard form:

y = m x + c

y = m x + c

For break-even between two plans $C_{A} = m_{A} x + c_{A}$ and $C_{B} = m_{B} x + c_{B}$ , set them equal and solve:

C_{A} = C_{B}

Interpreting $m$ and $c$ in context

Situation	Slope $m$	Intercept $c$
Cost vs. months	$ per month	joining fee / setup
Distance vs. time	speed (km/h)	starting distance
Water bill vs. usage	$ per kL	fixed monthly fee
Asset value vs. years	$ per year (negative for depreciation)	purchase price

Reading "until reaches zero". A question like "how long until the value reaches zero" means set y=0 and solve for x — not use the y-intercept.

How to work with a linear model

Identify what each variable represents (with units) and what's being asked.
Build or use the model. To build from two readings, treat each as a point and use the gradient formula then point-slope form. To use the model, substitute the known value and solve for the unknown.
Interpret the answer in context. Include units, and check the answer is reasonable for the situation.

EXAMPLE 1 — USE A MODEL

A taxi charges

$ 5

flagfall plus

$ 3.50

per km. Find the cost

C

for a

20

km trip, and the distance covered for

$ 110

SOLUTION

Set up the model with $d$ for distance in km. Flagfall is the fixed cost, the per-km rate is the slope.

C

=

3.50 d + 5

Substitute $d = 20$ for the cost of a 20 km trip.

$C$	$=$	$3.50 (20) + 5$
$C$	$=$	$$ 75$

For $$ 110$ , set $C = 110$ and solve for $d$ .

$110$	$=$	$3.50 d + 5$
$105$	$=$	$3.50 d$
$d$	$=$	$30 km$

C = 75, d = 30

EXAMPLE 2 — BUILD FROM TWO POINTS

A young tree is

80

cm tall after

6

months and

200

cm tall after

18

months. Find a linear model

H = m t + c

for height

H

in cm after

t

months.

SOLUTION

Use the points $(6, 80)$ and $(18, 200)$ . The slope is the growth rate in cm per month.

$m$	$=$	$\frac{200 - 80}{18 - 6}$
$m$	$=$	$\frac{120}{12} = 10$

Substitute $m = 10$ and the point $(6, 80)$ into the point-slope form.

$H - 80$	$=$	$10 (t - 6)$
$H$	$=$	$10 t + 20$

The tree grows at $10$ cm/month and was $20$ cm tall at $t = 0$ .

H = 10 t + 20

EXAMPLE 3 — BREAK-EVEN

Plan A:

C_{A} = 50 m + 200

. Plan B:

C_{B} = 70 m

. Find the number of months at which both plans cost the same.

SOLUTION

Set the two cost expressions equal and solve.

$50 m + 200$	$=$	$70 m$
$200$	$=$	$20 m$
$m$	$=$	$10 months$

At $m = 10$ both plans cost $70 \times 10 = $ 700$ .

m = 10

EXAMPLE 4 — DEPRECIATION

A laptop's value is modelled by

V = - 120 t + 1800

(in dollars, after

t

years). Find the time until

V = 0

SOLUTION

The slope $- 120$ means the laptop loses $$ 120$ of value each year. Set $V = 0$ and solve.

$0$	$=$	$- 120 t + 1800$
$120 t$	$=$	$1800$
$t$	$=$	$15 years$

The model predicts the laptop reaches zero value after $15$ years (though this is a model — in reality it would still have some scrap value).

t = 15

Common pitfalls

Negative slope means a decreasing quantity. Depreciation, draining water, fuel burning — the slope is negative because the quantity is going down over time, not because of a sign error.

"Reaches zero" means set $y = 0$ . A question like "how long until the value reaches

$ 0

" means substitute

0

for the dependent variable and solve for the independent variable — not read off the

y

-intercept.

State the answer in context. An answer of

m = 10

is incomplete — say "the two plans cost the same after 10 months." Always include units and explain what the number means.

Models have limits. Linear models are good approximations in the short term but unrealistic over long ranges. Trees don't grow at the same rate forever; cars don't reach negative dollar values. Sanity-check the model for the input you're using.

Frequently asked questions

What is a linear model?

A linear model is an equation of the form y equals m x plus c used to describe a real-world situation where one quantity changes at a constant rate with respect to another. The slope m is the rate of change and the y-intercept c is the starting value.

What does the slope mean in a linear model?

The slope is the rate of change in real-world units. For cost versus months, it is dollars per month. For distance versus time, it is speed. For asset value over years, a negative slope is the rate of depreciation in dollars per year.

What does the y-intercept mean in a linear model?

The y-intercept c is the value of the dependent variable when x equals 0 — the starting point. It might be the joining fee, the fixed monthly charge, the purchase price of an asset, or the starting distance.

What is a break-even point and how do I find it?

A break-even point is the value of x at which two cost (or revenue) plans are equal. Set the two cost expressions equal to each other and solve for x. Then substitute back into either model to find the matching cost.

How do I build a linear model from two data points?

Treat each piece of data as an ordered pair. Find the slope using m equals (y2 minus y1) divided by (x2 minus x1), then substitute the slope and either point into the point-slope form y minus y1 equals m times (x minus x1) and rearrange.

When is a linear model not appropriate?

Linear models assume a constant rate of change. They work well in the short term but become unrealistic over time. A tree does not keep growing 10 cm per month forever, and a car's value cannot drop below zero. Always check the model is reasonable for the input value.