Understand · explore · practise
Build straight-line models from data, interpret gradient and intercept with units, assess fit and distinguish interpolation, extrapolation and direct proportion.
Before you startGradients, straight-line equations and simultaneous equations
01 / Proportion
Two quantities are directly proportional when y = kx for a fixed k. For x ≠ 0, the ratio y/x is constant. Doubling x doubles y.
Direct proportion: y = kx
General linear model: y = ax + b
A general straight-line model has a constant rate of change, but a non-zero intercept means it is not direct proportion.
If printing costs £0.08 per page with no fixed fee, C = 0.08n is direct proportion. With a £2 setup charge, C = 0.08n + 2 is linear but not directly proportional. The model can be restricted to positive integer page counts even though its algebraic line extends further.
A uniform cable has mass 1.8 kg per metre
M = 1.8L
Here M is in kg and L in metres.
At L = 2: M = 3.6
At L = 4: M = 7.2
The length doubles and so does the mass.
M/L = 1.8 for L > 0
The gradient has units kg per metre. This assumes the cable is uniform.
02 / Build a model
Define the variables and their units first. For an exact linear relationship through (x₁,y₁) and (x₂,y₂), find a = (y₂ − y₁)/(x₂ − x₁), then substitute either point to find b.
If you are given a drawn line of best fit, choose two well-separated readable points on that line. They need not be measured data points.
Two points alone always determine a line when their inputs differ. They do not prove that the wider real-world relationship is linear.
A hire costs £38 for 2 hours
and £65 for 5 hours
Let C be cost in pounds and t be hours. Assume C = at + b.
a = (65 − 38)/(5 − 2) = 9
The rate is £9 per hour.
38 = 9(2) + b ⇒ b = 20
The fixed charge is £20.
C = 9t + 20
Check t = 5 gives £65.
03 / Coefficients
In y = ax + b, the gradient is the change in y for one unit increase in x. Its units are “y units per x unit”. The intercept is the model’s output when x = 0.
The meaning of zero depends on the variable definition. If t counts years after 2025, t = 0 means 2025. An intercept is not automatically a meaningful prediction if zero lies outside the model’s valid range.
V = 72 − 4t
If V is liquid volume in litres and t is minutes since draining started, 72 represents the starting volume and −4 represents a decrease of 4 litres per minute. A natural model domain is 0 ≤ t ≤ 18, assuming a constant rate until empty.
Uniform volume loss need not imply uniform depth loss in a container whose cross-sectional area changes. Say which physical quantity your rate describes.
04 / Assess the fit
Real measurements need not lie exactly on a line. A linear approximation may be useful when the points remain close to it without a clear curved pattern.
Switch between these illustrative datasets. The blue candidate line stays fixed. You can reveal the vertical differences between observed and predicted outputs.
In the curved dataset, both endpoints lie on the candidate line, but the intermediate points sit systematically below it. Joining endpoints hides that mismatch.
“Close” depends on the units, measurement uncertainty and the purpose of the prediction. This activity compares patterns; it is not a statistical fitting algorithm or a universal goodness-of-fit test.
| x | Observed y | Line y |
|---|---|---|
| 0 | 8 | 8 |
| 2 | 15 | 14 |
| 4 | 19 | 20 |
| 6 | 27 | 26 |
| 8 | 31 | 32 |
| 10 | 39 | 38 |
The observations are close to the line, with small differences on both sides. A linear approximation is plausible over the observed interval; it is not exact.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 / Predictions
For the hire model C = 9t + 20, four hours cost £56. If the charge is £92, solve 92 = 9t + 20 to obtain t = 8 hours.
Respect how the situation charges for time: a model for continuously measured hours differs from a tariff that rounds up to whole hours. State the assumption.
Interpolation predicts within the observed input range. Extrapolation predicts outside it and depends more strongly on the relationship continuing.
An instrument reads R = 14 at T = 0
and R = 74 at T = 30
Assume a linear relationship between reading R and temperature T in °C.
a = 60/30 = 2
R = 2T + 14
At T = 12, the model predicts R = 38.
For R = 50:
T = (50 − 14)/2 = 18°C
Reverse the model to recover temperature.
R = T ⇒ 2T + 14 = T
T = −14
Numerical equality is an algebraic result. It lies outside the 0–30°C calibration range, so physical validity needs more evidence.
06 / Model limits
A model simplifies reality. State the assumption that makes a constant gradient reasonable, then identify where it might fail.
For a hypothetical club with 480 members in 2025, N = 480 + 35t models an increase of 35 members per year. It is an approximation, and t must be measured from the chosen starting year.
If a fitted line predicts a negative length or count at small inputs, that signals a domain limitation. It does not make a negative physical quantity possible.
07 / Compare models
Two straight-line models can describe competing tariffs, costs and revenue, or supply and demand. Equate their outputs and interpret the intersection in the original units.
For tariffs C₁ = 8t + 24 and C₂ = 12t + 8, equality gives t = 4 and C = £56. Below four hours, tariff 2 is cheaper; above four hours, tariff 1 is cheaper, assuming these formulas remain valid.
Demand: P = 50 − 2Q
Supply: P = 8 + Q
Let Q be quantity in hundreds and P be price in pounds. These are illustrative models.
50 − 2Q = 8 + Q
Q = 14
Substitute into either model.
P = 22
The equilibrium prediction is 1,400 units at £22 each.
Check the model domain
Negative quantity or price would not be meaningful in this setting.
08 / Your turn
Explain the units of coefficients and the assumptions behind predictions.
Classify M = 2.4L and C = 2.4L + 6.
Direct proportion requires a zero intercept.
M is directly proportional to L, with ratio M/L = 2.4 for L ≠ 0. C is linear but not directly proportional to L; the fixed 6 prevents the ratio from staying constant.
A service costs £31 for 3 hours and £55 for 7 hours. Find C = at + b and interpret both constants.
Use change in cost divided by change in time.
a = (55 − 31)/(7 − 3) = 6
b = 31 − 6(3) = 13
C = 6t + 13
£6 per hour plus a £13 fixed charge, assuming the tariff remains unchanged.
Using C = 6t + 13, predict the charge for 5 hours and the time corresponding to £73.
Substitute t = 5; then solve 73 = 6t + 13.
C = 6(5) + 13 = £43
t = (73 − 13)/6 = 10 hours
Relative to the observations at 3 and 7 hours, 5 hours is interpolation and 10 hours is extrapolation.
V = 90 − 6t models volume in litres t minutes after draining begins. Interpret its coefficients and give a sensible time domain.
Find when the model predicts zero volume.
Starting volume: 90 litres. Change: −6 litres per minute. The container empties at t = 15 minutes, so use 0 ≤ t ≤ 15. The model assumes a constant loss rate.
For x = 0, 1, 2, 3, the outputs are 2, 3, 6, 11. Is one constant-gradient line an exact model?
Compare successive output changes over equal input intervals.
No. The output changes are 1, 3 and 5, so the gradient is not constant. The values follow y = x² + 2; this curved pattern cannot be represented exactly by one straight line.
A model predicts 820 subscribers at the start of 2026 and 45 additional subscribers each year. Write N in terms of t years after that date, and predict the start-of-2029 count.
At the start of 2029, t = 3.
N = 820 + 45t
N(3) = 955
This assumes a constant net increase of 45 per year; it need not remain realistic indefinitely.
Two charges are C₁ = 5t + 30 and C₂ = 8t + 12. Find when they agree and which is cheaper for t = 4.
Set the charges equal, then compare at t = 4.
5t + 30 = 8t + 12
t = 6, C = £60
At t = 4: C₁ = £50, C₂ = £44
The second tariff is cheaper at four hours.
A line fitted to observations for 10 ≤ x ≤ 30 is y = 4x − 12, where y is a physical length. A student predicts y = −8 at x = 1. Comment.
Separate correct substitution from a meaningful model prediction.
The substitution 4(1) − 12 = −8 is algebraically correct. But x = 1 is outside the observed range and a physical length cannot be negative. The fitted model should not be used there without further evidence or a different model.
09 / Recap
Section 1 of 9 · Proportion