Direct Variation Revision Notes for HSC SSCE Mathematics Standard

Direct Variation

What is direct variation?

Direct variation (also called direct proportion) happens when one quantity depends directly on another quantity. When one value increases or decreases, the other value changes at the same rate in the same direction.

Think about a simple example: your wage depends on how many hours you work. If you work more hours, you earn more money. If you work fewer hours, you earn less money. The relationship between hours worked and wage is a direct variation.

infoNote

Real-World Connection: Hourly Wages

In the wage example, if you earn $20 per hour:

Work 5 hours → earn $100
Work 10 hours → earn $200
Work 15 hours → earn $300

Notice how doubling your hours doubles your wage. This is the essence of direct variation!

Key characteristics of direct variation:

Both quantities change in the same direction (both increase or both decrease)
The rate of change stays constant
When graphed, direct variation creates a straight line through the origin

The constant of variation

Every direct variation relationship includes a constant of variation, represented by the letter $k$ . This constant tells us the rate at which the quantities change together.

In the wage example above, if you earn $20 per hour, then $k = 20$ . This means:

For every additional hour worked, you earn an extra $20
The constant stays the same no matter how many hours you work

The constant of variation is the multiplier that connects the two variables. Finding the value of $k$ is an essential step in solving direct variation problems.

chatImportant

Understanding the Constant

The constant of variation ( $k$ ) is NOT just any number—it has meaning! It represents the actual rate of change:

In hourly wages: $k$ is the dollars per hour
In speed problems: $k$ is the distance per unit of time
In currency conversion: $k$ is the exchange rate

Always think about what $k$ represents in your specific problem!

The direct variation formula

When a quantity $y$ varies directly with quantity $x$ , we can write this relationship as an equation:

$y = kx$

Where:

$y$ is the dependent variable (the quantity that depends on the other)
$x$ is the independent variable (the quantity that controls the relationship)
$k$ is the constant of variation (the rate of change)

This formula tells us that $y$ equals $k$ times $x$ . Once we know the value of $k$ , we can calculate $y$ for any value of $x$ , or calculate $x$ for any value of $y$ .

infoNote

Formula Flexibility

The direct variation formula can be rearranged depending on what you need to find:

To find $y$ : $y = kx$
To find $k$ : $k = \frac{y}{x}$
To find $x$ : $x = \frac{y}{k}$

All three forms represent the same relationship!

Solving direct variation problems

Follow these three steps to solve any direct variation problem:

Step 1: Write an equation relating the two variables

Use the formula $y = kx$ , replacing $y$ and $x$ with the actual variables from your problem.

Step 2: Find the constant of variation $k$

Substitute the given values for $x$ and $y$ into your equation, then solve for $k$ .

Step 3: Write the complete equation and solve

Replace $k$ in your equation with the value you found, then substitute the new given value to answer the question.

lightbulbExample

Worked Example: Photocopier Costs

Problem: The cost of a photocopier ( $C$ , in dollars) varies directly with its speed ( $s$ , in pages per minute). A photocopier with a speed of 50 pages per minute costs $700.

a) What is the constant of variation ( $k$ )?

b) Write an equation connecting $C$ and $s$ .

c) How much does a photocopier with a speed of 60 pages per minute cost?

Solution:

Part a) Finding the constant of variation:

Cost is directly proportional to speed, so we write: $C = ks$
We know that when $s = 50$ , $C = 700$
Substitute these values into the formula:

$700 = k \times 50$
Solve for $k$ :

$k = \frac{700}{50}$

$k = 14$

Part b) Writing the equation:

Now that we know $k = 14$ , we can write the complete equation:

$C = 14s$

Part c) Finding the cost for 60 pages per minute:

Use the equation $C = 14s$ with $s = 60$ :

$C = 14 \times 60$

$C = 840$
Answer: A photocopier with a speed of 60 pages per minute costs $840.

Using graphs with direct variation

Direct variation relationships can be shown on graphs. These graphs are always straight lines that pass through the origin (0, 0). We can use these graphs to:

Convert between different quantities
Find values that aren't in a table
Visualise the relationship between variables

infoNote

Why Through the Origin?

Direct variation graphs always pass through (0, 0) because when one quantity is zero, the other must also be zero. Think about it: if you work 0 hours, you earn $0. If you travel at any speed for 0 time, you travel 0 distance. This is a key feature that distinguishes direct variation from other linear relationships!

Worked example: Currency conversion

lightbulbExample

Worked Example: Currency Conversion

Problem: The graph below shows the conversion between Australian dollars and New Zealand dollars.

Use the graph to convert:

a) 40 Australian dollars to New Zealand dollars

b) 25 NZD to AUD

Solution:

To read values from a conversion graph:

Find the known value on the appropriate axis
Draw a vertical or horizontal line to the graph
Draw a line from the graph to the other axis
Read the converted value

Part a) Converting 40 AUD to NZD:

Find 40 on the horizontal axis (AUD)
Read across to the graph line
Read the value on the vertical axis (NZD)
Answer: 40 AUD = 50 NZD

Part b) Converting 25 NZD to AUD:

Find 25 on the vertical axis (NZD)
Read across to the graph line
Read the value on the horizontal axis (AUD)
Answer: 25 NZD = 20 AUD

Worked example: Creating a graph from a table

lightbulbExample

Worked Example: Creating a Graph from a Table

Problem: The table below shows the cost of postage ( $c$ ) as a function of the weight of the parcel ( $w$ ).

Weight ( $w$ ) kg	1	2	3	4	5
Cost ( $c$ ) $	1.2	2.4	3.6	4.8	6.0

a) Draw a graph of cost ( $c$ ) against the weight of the parcel ( $w$ )

b) Use the graph to determine the cost of a parcel if the weight is 2.5 kg

Solution:

Part a) Drawing the graph:

Set up a number plane with weight ( $w$ ) on the horizontal axis and cost ( $c$ ) on the vertical axis
Plot the points from the table:
- (1, 1.2)
- (2, 2.4)
- (3, 3.6)
- (4, 4.8)
- (5, 6.0)
Join the points with a straight line

Part b) Finding the cost for 2.5 kg:

Locate 2.5 kg on the horizontal axis
Draw a vertical line upward until it meets the graph
From that point, draw a horizontal line across to the vertical axis
Read the cost value
Answer: A parcel weighing 2.5 kg would cost $3

chatImportant

Exam Tip: Accurate Graph Reading

When reading values from a graph, always use a ruler to draw straight lines. This helps you:

Read accurate values from the axes
Show clear working in your solution
Avoid making careless errors

Examiners look for neat, precise work—using a ruler demonstrates good mathematical practice!

Remember!

bookmarkSummary

Key Points to Remember:

Direct variation occurs when two quantities change at the same rate in the same direction
The constant of variation ( $k$ ) represents the rate of change between the two quantities
The formula for direct variation is $y = kx$ , where $k$ is the constant of variation
To solve direct variation problems: (1) write the equation, (2) find $k$ using given values, (3) use the complete equation to answer the question
Graphs of direct variation are always straight lines passing through the origin

Direct Variation (HSC SSCE Mathematics Standard): Revision Notes