Direct proportion equations are a Higher-tier topic that links algebra with ratio and proportion. When two quantities are directly proportional, one is always a constant multiple of the other. Understanding how to set up, find, and use the equation connecting them is essential for securing marks on this topic.
What Is Direct Proportion?
Two quantities are in direct proportion when one quantity is always the same multiple of the other. If y is directly proportional to x, then doubling x also doubles y, tripling x triples y, and so on.
The proportionality symbol ∝ is used to express this relationship. Writing y ∝ x means "y is directly proportional to x". To turn this into an equation, introduce a constant of proportionality, k, giving y = kx. Once you know k, you can find y for any value of x, or x for any value of y.
Direct proportion does not only apply to y ∝ x. At Higher tier you must also handle y ∝ x², y ∝ x³, y ∝ √x, and other variations. In each case, the method is the same: replace ∝ with = k, use a given pair of values to find k, then write the final equation.
Key Formulas
Step-by-Step Method
- Write the proportionality statement using ∝ (e.g., y ∝ x²).
- Replace ∝ with = k to form the equation (e.g., y = kx²).
- Substitute the given pair of values to find k.
- Rewrite the equation with the value of k.
- Use the equation to answer the question.
Worked Example 1 — Foundation Level
Question: y is directly proportional to x. When x = 4, y = 20. Find y when x = 7.
Working: y ∝ x, so y = kx. Substitute x = 4, y = 20: 20 = k × 4, so k = 5. Equation: y = 5x. When x = 7: y = 5 × 7 = 35.
Answer: y = 35.
Worked Example 2 — Higher Level
Question: y is directly proportional to x². When x = 3, y = 36. Find y when x = 5.
Working: y ∝ x², so y = kx². Substitute x = 3, y = 36: 36 = k × 9, so k = 4. Equation: y = 4x². When x = 5: y = 4 × 25 = 100.
Answer: y = 100.
Worked Example 3 — Exam Style
Question: The cost C of a diamond is directly proportional to the square of its weight w grams. A diamond weighing 2 g costs £600. Find the cost of a diamond weighing 3.5 g.
Working: C ∝ w², so C = kw². Substitute w = 2, C = 600: 600 = k × 4, so k = 150. Equation: C = 150w². When w = 3.5: C = 150 × 12.25 = £1,837.50.
Answer: The diamond costs £1,837.50.
Common Mistakes
- Forgetting to square (or square-root) x before substituting. If y ∝ x², the equation is y = kx², not y = kx. Read the proportionality statement carefully.
- Using the wrong type of proportion. Direct proportion means y = kx^n; inverse proportion means y = k/x^n. Check whether y increases or decreases as x increases.
- Not finding k first. Some students try to jump directly to the answer. Always find k explicitly — it earns method marks and prevents errors.
Exam Tips
- Always write the proportionality statement and the equation with k as separate lines. Examiners award marks for each step.
- If the question says "y is proportional to the square root of x", write y ∝ √x, then y = k√x.
- Check your answer makes sense: if y is proportional to x² and x increases, y should increase by a larger factor.
Practice Questions
Q1 (Foundation): y is directly proportional to x. When x = 6, y = 15. Find y when x = 10.
Q2 (Foundation): y ∝ x. When x = 8, y = 24. Find x when y = 42.
Q3 (Higher): y is directly proportional to the cube of x. When x = 2, y = 48. Find y when x = 3.
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Related Topics
Summary
- Direct proportion means one quantity is a constant multiple of another.
- Replace ∝ with = k to form the equation, then find k using given values.
- Common forms include y = kx, y = kx², y = kx³, and y = k√x.
- Always find k first, write the full equation, then substitute to answer the question.
- Check your answer is consistent with the type of proportion described.
Test your understanding
5 quick MCQs to identify any misconceptions on this topic.
Further reading from leading academic institutions — free and open-access.
Cambridge challenges on forming and solving equations.
University of Cambridge · Free · Open AccessStep-by-step methods for linear and more complex equations.
Corbett Maths · Free · Open AccessCambridge problem-solving with ratio and proportion.
University of Cambridge · Free · Open AccessSimplifying, sharing in a ratio, and proportion problems.
Corbett Maths · Free · Open Access