Direct and inverse variation problems become manageable once the relationship is written as a formula. Find the constant of proportionality from one supplied pair of values, write the complete model, and only then use new data. These direct and inverse variation worked examples show that sequence and explain how to test whether the answer behaves as it should.
This lesson also includes variation with a square and a combined model. Formula rearrangement appears where it is needed to find an unknown variable. The examples are mathematical models with stated assumptions, rather than claims that every real situation follows an exact proportional relationship.
Which relationship does the statement describe?
If y varies directly as x, the ratio y/x stays constant wherever x is nonzero. Write y = kx, where k is the constant of proportionality. Doubling x doubles y when k is fixed.
If y varies inversely as x, their product xy stays constant. Write y = k/x, with x ≠ 0. In a positive-valued model, doubling x halves y. The constant may have units, so write the quantities in consistent units before calculating it.
| Statement | Model | Constant to test |
|---|---|---|
| y varies directly as x | y = kx | y/x |
| y varies inversely as x | y = k/x | xy |
| y varies directly as x squared | y = kx2 | y/x2 |
| y varies directly as x and inversely as z | y = kx/z | yz/x |
The proportionality symbol ∝ states the relationship but is not yet a complete equation for calculation. Replace it with an equals sign and a constant: y ∝ x becomes y = kx. Do not assume k = 1 unless the data show that it is.
Worked example 1: construct a direct-variation formula
Question: The mass m of a uniform wire varies directly as its length L. A 2.4 m length has mass 18 g. Find the mass of 7 m of the same wire.
- Write the model: m = kL.
- Use the known pair: 18 = k × 2.4.
- Solve for k: k = 18/2.4 = 7.5 g/m.
- Write the complete formula: m = 7.5L, when m is in grams and L is in metres.
- For L = 7, m = 7.5 × 7 = 52.5 g.
The unit g/m means that each metre contributes 7.5 g under the uniform-wire assumption. As a check, the length scale factor is 7/2.4. Multiplying the original 18 g by that same factor also gives 52.5 g.
Changing the length unit would change the numerical value of k. With length in centimetres, the formula would be m = 0.075L. Both models describe the same wire; the units determine which coefficient is correct.
Worked example 2: use the fixed product in inverse variation
Question: In a model, n identical outlets each deliver water at the same constant rate. Together they fill a fixed tank in time t, and t varies inversely as n. Five outlets take 36 minutes. How long would eight outlets take?
The stated relationship gives t = k/n. Substituting the known values, 36 = k/5, so k = 180. The complete formula is t = 180/n, with time in minutes.
For n = 8, t = 180/8 = 22.5 minutes. The fixed product check is 5 × 36 = 8 × 22.5 = 180.
A direct-proportion calculation would give 36 × 8/5 = 57.6 minutes, but that would make more equal-rate outlets take longer. Under the stated idealised model, this fails the direction check.
The inverse model depends on the outlets maintaining their individual rates. A real shared water supply might not do that. In an examination problem, use the given assumption; outside that model, obtain evidence before treating the relationship as exact.
Worked example 3: “as the square” changes the scale factor
Question: A quantity y varies directly as x2. When x = 4, y = 28. Find y when x = 6, then find positive x when y = 112.
Write y = kx2. Using the known pair, 28 = 16k, so k = 7/4. The model is y = (7/4)x2.
When x = 6, y = (7/4) × 36 = 63. The increase in x is a factor of 6/4, so y increases by (6/4)2 = 9/4. Indeed, 28 × 9/4 = 63.
To find x from y, rearrange the model: x2 = 4y/7. Substituting y = 112 gives x2 = 64. The positive solution is x = 8. Without the positive restriction, the algebraic equation would also allow x = −8.
The phrase “directly as the square” does not mean “directly as x”. Doubling x in this model multiplies y by four rather than by two.
Worked example 4: combine direct and inverse relationships
Question: A positive quantity q varies directly as a and inversely as b2. When a = 12 and b = 3, q = 20. Find q when a = 18 and b = 6.
Translate both parts into one formula: q = ka/b2. The known values give 20 = 12k/9. Multiplying by 9 and dividing by 12 gives k = 15.
Now substitute the new values into q = 15a/b2:
q = (15 × 18)/36 = 7.5.
Check with scale factors. The variable a increases by 18/12 = 1.5, tending to multiply q by 1.5. The variable b doubles, so b2 increases fourfold and divides q by 4. The combined factor is 1.5/4 = 0.375, giving 20 × 0.375 = 7.5.
Test a proposed model instead of guessing
Two quantities both increasing does not establish direct variation. Consider the pairs (x, y) = (2, 7), (4, 11) and (6, 15). The ratios y/x are different, so these data do not fit one model y = kx. They instead satisfy y = 2x + 3.
A nonzero added constant prevents direct proportionality. A direct-variation graph y = kx passes through the origin. For an inverse model with positive k and positive x, the graph curves downward rather than forming a straight line through the origin.
One known pair determines k only after the form of the relationship is given. That pair alone cannot tell you whether the intended model is direct, inverse or something else.
Three practice problems with explained answers
Practice 1: direct variation
y varies directly as x. When x = 8, y = 30. Find y when x = 14.
Solution: From y = kx, k = 30/8 = 15/4. Therefore y = (15/4) × 14 = 52.5. The ratio y/x remains 15/4 in both pairs.
Practice 2: inverse variation
p varies inversely as r. When r = 6, p = 14. Find r when p = 8.
Solution: The constant product is pr = 14 × 6 = 84. Thus 8r = 84 and r = 10.5. Since p decreases, r increases, as a positive inverse model requires.
Practice 3: square and reciprocal together
z varies directly as x2 and inversely as y. When x = 3 and y = 2, z = 27. Find z when x = 5 and y = 10.
Solution: Write z = kx2/y. Substitution gives 27 = 9k/2, so k = 6. Then z = 6 × 25/10 = 15. The scale-factor check is 27 × (5/3)2 × (2/10) = 15.
Joint variation: two independent inputs multiply
The direct, inverse, square and combined cases above remain useful. An explicit joint-product model adds a different statement: y varies jointly as x and z means y = kxz. Here x and z are independent inputs that may be changed separately; y is the dependent quantity. The model holds only over the stated domain, with one fixed constant k. Do not replace the product xz with the sum x + z.
Worked model and rearrangement
A paper model states that positive, dimensionless y varies jointly as positive, dimensionless x and z. When x = 4 and z = 3, y = 30. Find the model, predict y when x = 6 and z = 5, and find z when y = 90 and x = 8.
- Translate the joint statement: y = kxz.
- Use all three calibration values: 30 = k × 4 × 3. Thus k = 30/12 = 2.5.
- The complete model is y = 2.5xz. At x = 6 and z = 5, y = 2.5 × 6 × 5 = 75.
- For the inverse calculation, divide y = 2.5xz by 2.5x. Since x is positive, z = y/(2.5x). With y = 90 and x = 8, z = 90/20 = 4.5.
Check the prediction independently by scale factors: 30 × (6/4) × (5/3) = 75. The changes multiply; adding their scale factors would describe no stated model. Substituting the recovered z gives 2.5 × 8 × 4.5 = 90. If x doubles while z triples, y becomes six times its original value. If x doubles while z halves, the product and y are unchanged.
Independent practice
A new model states that positive p varies jointly as positive a and b. All quantities are dimensionless. When a = 5 and b = 2, p = 18. Find the model and p when a = 4 and b = 7. Then find b when a = 6 and p = 54. Explain whether p must increase whenever a increases.
Checked answer: k = 18/(5 × 2) = 1.8, so p = 1.8ab. The requested value is 1.8 × 4 × 7 = 50.4. Rearranging gives b = p/(1.8a) = 54/10.8 = 5. Increasing a increases p only if b is held fixed and k remains positive. If a doubles and b falls to one quarter, p halves. One calibration triple fixes k after the model is supplied; it does not prove that an unknown real system follows joint variation.
Background definition: OpenStax, modelling with variation. These values and worked checks are original.
Frequently asked variation questions
Must I find the constant every time?
A ratio method can compare two cases without displaying k. Finding it explicitly is a reliable approach when the question asks for a formula or several predictions.
Does inverse variation mean subtracting?
No. It means reciprocal dependence, such as y = k/x. Subtracting x from a constant produces a different relationship.
Can the constant of proportionality have units?
Yes. Its units come from the model. For mass directly proportional to length, k has units of mass per unit length.
What if the units change halfway through a question?
Convert them consistently before substituting, or rewrite the coefficient for the new units. Mixing metres and centimetres with one unchanged numerical coefficient gives incorrect results.
Is y = 4x + 2 direct variation?
No. The added 2 means y/x is not constant. A direct-variation formula has the form y = kx without an added term.
Can an inverse-variation input equal zero?
No. In y = k/x, zero would make the denominator undefined. State any further restrictions required by the context.
Why does doubling both variables not always leave the answer unchanged?
It depends on their powers. In q = ka/b, doubling a and b cancels. In q = ka/b2, the same change halves q.
Continue your Mathematics practice
Use the existing Mathematics notes hub and Mathematics topical-test collection to find related revision resources. Both currently display a login requirement. All worked models and practice solutions for this lesson are included above.
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