Direct & Inverse ProportionEdexcel IGCSE Maths: Revision notes

Section 1

What does it mean for two quantities to be directly proportional?

Two variables $y$ and $x$ are directly proportional (written $y \propto x$ ) if $y$ increases at the same rate as $x$ increases, so their ratio stays constant.

The equation is $y = kx$ , where $k$ is a fixed number called the constant of proportionality.
The graph of $y$ against $x$ is a straight line through the origin with gradient $k$ .
Doubling $x$ doubles $y$ ; trebling $x$ trebles $y$ .
To find $k$ , substitute one known pair of values $(x, y)$ into $y = kx$ and solve.

$x$	$y$ (if $y = 3x$ )
1	3
2	6
4	12

Key termsdirectly proportionalconstant of proportionality

Example

If $y = kx$ and $y = 20$ when $x = 4$ , then $k = 5$ , so $y = 5x$ . When $x = 7$ , $y = 35$ .

Exam tip

Always find $k$ first using given values before answering the rest of the question.

Section 2

What does it mean for two quantities to be inversely proportional?

Two variables $y$ and $x$ are inversely proportional (written $y \propto \dfrac{1}{x}$ ) if $y$ decreases as $x$ increases, such that their product stays constant.

The equation is $y = \dfrac{k}{x}$ , equivalent to $xy = k$ .
The graph of $y$ against $x$ is a curve (a reciprocal graph) that never touches either axis.
Doubling $x$ halves $y$ ; trebling $x$ makes $y$ one-third of its value.
To find $k$ , substitute a known pair of values into $y = \dfrac{k}{x}$ , or simply multiply $x$ and $y$ together.

Key termsinversely proportionalreciprocal graph

Example

If $y = \dfrac{k}{x}$ and $y = 6$ when $x = 5$ , then $k = 30$ , so $y = \dfrac{30}{x}$ . When $x = 10$ , $y = 3$ .

Common mistake

Do not confuse inverse proportion with a straight line with negative gradient — $y = \dfrac{k}{x}$ is always a curve, never a straight line.

Section 3

How do power and root variants of proportion work?

Proportion is not limited to $x$ itself — it can involve a power or root of $x$ . The method for solving is identical: replace $x$ with the given expression.

Direct variants:

$y \propto x^2 \Rightarrow y = kx^2$
$y \propto x^3 \Rightarrow y = kx^3$
$y \propto \sqrt{x} \Rightarrow y = k\sqrt{x}$

Inverse variants:

$y \propto \dfrac{1}{x^2} \Rightarrow y = \dfrac{k}{x^2}$
$y \propto \dfrac{1}{\sqrt{x}} \Rightarrow y = \dfrac{k}{\sqrt{x}}$

In every case: substitute the given $(x, y)$ pair to find $k$ , then use the completed formula to answer the question.

Key termspower variantroot variant

Example

$y \propto x^2$ and $y = 18$ when $x = 3$ . Then $18 = k(3)^2 = 9k$ , so $k = 2$ and $y = 2x^2$ . When $x = 5$ , $y = 50$ .

Think of it like this

Think of $y = kx^2$ like the area of a square scaling with side length — double the side length and the area (not just the perimeter) quadruples.

Section 4

How do you set up and solve a proportion problem step by step?

Follow the same four steps regardless of the type of proportion:

Write the proportionality statement, e.g. $y \propto x^2$ .
Convert to an equation using $k$ , e.g. $y = kx^2$ .
Substitute known values to find $k$ .
Rewrite the formula with $k$ and substitute the new value to find the unknown.

This method works whether the relationship is direct, inverse, or involves a power/root — only the substitution step changes.

Exam tip

Underline the word 'proportional' or 'varies' in the question and check immediately whether it says 'directly' or 'inversely' — this decides whether $k$ multiplies or divides.

Section 5

How do you spot proportion from a table of values?

Exam questions sometimes give a table and ask which type of proportion (if any) fits.

Direct proportion: $\dfrac{y}{x}$ is the same for every pair (constant ratio).
Inverse proportion: $xy$ is the same for every pair (constant product).
If neither ratio nor product is constant, the data may fit a power/root variant — try $\dfrac{y}{x^2}$ , $\dfrac{y}{x^3}$ , or $y\sqrt{x}$ instead.

Check at least two pairs of values before concluding, since one matching pair could be coincidence.

Key termsconstant ratioconstant product

Common mistake

Testing only one pair of values and assuming the relationship holds for the whole table — always check a second pair.

Must Know

Direct proportion: $y = kx$ — straight line through the origin; ratio $\dfrac{y}{x}$ is constant.
Inverse proportion: $y = \dfrac{k}{x}$ — reciprocal curve; product $xy$ is constant.
Power/root variants replace $x$ with $x^2$ , $x^3$ , $\sqrt{x}$ , etc., but the solving method is identical.
Always find $k$ first by substituting a given pair of values into the equation.
Doubling $x$ : direct proportion doubles $y$ ; inverse proportion halves $y$ ; $y = kx^2$ quadruples $y$ .
Always state the equation with $k$ found before calculating the final answer — method marks depend on it.

That's the notes covered.

Carry on to the next subtopic.