Inverse Variation

Inverse variation occurs when two quantities have a constant product. For a fixed rectangular area, increasing the width requires decreasing the length. For a fixed journey distance, increasing a constant travel speed reduces the travel time.

An inverse-variation model requires a constant product; one quantity decreasing as another increases is not sufficient.

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A constant product

The standard equation is

y=\frac{k}{x},\qquad xy=k,\qquad k\ne0,\quad x\ne0.

The constant $k$ has the product units of $x$ and $y$ . If they are rectangle side lengths in metres, $k$ is an area in square metres. If they are speed and time, $k$ is a distance.

The non-zero constant excludes the degenerate zero-output case. For $k>0$ , the full real graph lies in the first and third quadrants. Positive physical quantities normally use only its first-quadrant branch. For $k<0$ , its branches lie in the second and fourth quadrants. The algebraic definition does not require both quantities to be positive.

Find the rule from a known pair

If inverse variation is given and $y=12$ at $x=6$ , multiply rather than divide:

k=xy=6\times12=72.

The rule is $y=72/x$ . At $x=9$ , $y=8$ . At $x=4.5$ , $y=16$ . In each case the product is 72.

Find the rule from a known pair: reference table
$x$	$y=72/x$	Product $xy$
3	24	72
6	12	72
9	8	72
12	6	72

These outputs do not decrease by equal amounts as $x$ increases by equal steps. An inverse-variation graph is not a straight line.

The reciprocal scaling rule

For two allowed inputs $x_1,x_2$ and outputs $y_1,y_2$ ,

x_1y_1=x_2y_2,\qquad \frac{y_2}{y_1}=\frac{x_1}{x_2}.

Multiplying the input by a non-zero factor $s$ divides the output by $s$ . Doubling the input halves the output; multiplying it by $3/2$ multiplies the output by $2/3$ .

A 20% increase in a positive input multiplies it by 1.2. The new output is $1/1.2=5/6$ of the old output, a decrease of $1/6$ , or $16\frac23\%$ . It is not a 20% decrease. Percentage changes do not reverse through simple subtraction.

Fixed area: different rectangles

A rectangle has area 72 m². Its side lengths $w$ and $l$ satisfy $wl=72$ , hence $l=72/w$ for $w>0$ .

Rectangles with the same area of 72 m²

Widening a rectangle from 6 to 9 while holding area fixed reduces its height from 12 to 8.

Diagram explanation and exact data

Both rectangles use the same drawing scale. The second is 1.5 times as wide but two thirds as high. Equal area does not imply equal perimeter or similar shape. This is inverse variation between width and height, not uniform enlargement.

Rectangles with the same area of 72 m²: exact reference data
Width (m)	Height (m)	Area (square m)	Perimeter (m)
6	12	72	36
9	8	72	34

A 6 m by 12 m rectangle and a 9 m by 8 m rectangle have equal areas. They are not similar: their side ratios are different. Their perimeters also differ, at 36 m and 34 m. A constant product does not preserve every geometric measurement.

There is no rectangle with a zero side length and area 72 m². Although the graph approaches an axis as a side becomes very large or very small, it never reaches that axis. In an application, available space would impose additional restrictions before either dimension became extreme.

Fixed distance: time and speed

For an idealized 120 km journey at constant speed $v$ km/h with no stops,

t=\frac{120}{v}\text{ hours},\qquad v>0.

At 60 km/h the time is 2 h; at 80 km/h it is 1.5 h. The distance is the fixed product $vt$ . This is a mathematical comparison of idealized journeys, not a recommendation about travel speed.

If a 30 min stop is added independently of travel speed, total time becomes $T=120/v+0.5$ . The moving time still varies inversely with speed, but total time no longer has a constant product with speed. Identify the quantity being modelled before applying the rule.

Fixed work: state the assumptions

Suppose a divisible task requires 60 worker-hours under a simplified model. Six equally productive workers need 10 h. Four workers need $60/4=15$ h; twelve need 5 h.

This assumes equal constant productivity, simultaneous work and no interference or fixed-duration step. Real tasks may violate those assumptions. A task that must happen in sequence cannot necessarily be completed twice as quickly by doubling the workforce.

An equivalent formulation uses rates: each worker completes $1/60$ of the task per hour, so $n$ workers complete $n/60$ per hour and take $60/n$ hours. Here $n$ must be a positive integer even though the algebraic formula accepts other positive inputs.

Distinguish three relationships

In direct variation, $y/x$ is constant. In inverse variation, $xy$ is constant. In a fixed-sum relation such as $y=20-x$ , it is $x+y$ that stays constant.

At $x=4$ and $x=8$ , the fixed-sum outputs are 16 and 12. Their products are 64 and 96, so the decreasing line is not inverse variation. “One goes up while the other goes down” is an observation, not a defining test.

Likewise, $y=k/x^2$ is inverse-square variation, not simple inverse variation between $x$ and $y$ . Doubling $x$ quarters the output instead of halving it.

Finally, inverse variation and an inverse function are different ideas. An inverse function reverses an input-output rule. Inverse variation names a particular reciprocal relationship. Inverse Functions treats the former.

Practice questions

Given inverse variation and $(x,y)=(6,12)$ , find $y$ when $x=9$ .
In a positive inverse-variation model, the input triples. What happens to the output?
Is $y=20-x$ inverse variation? Use two inputs to justify your answer.

Check your answers

The fixed product is 72. Thus $y=72/9=8$ , and $9\times8=72$ checks the answer.
The output becomes one third as large because $(3x)(y/3)=xy$ .
No. At $x=4$ the product is $4\times16=64$ , while at $x=8$ it is $8\times12=96$ . The rule preserves a sum, not a product.

Sources and further reading

OpenStax, Elementary Algebra 2e, §8.9: Use Direct and Inverse Variation. Constant-product models and restrictions.
OpenStax, Intermediate Algebra 2e, §7.5: Solve Applications with Rational Equations. Motion and work models.