Inverse variation

Inverse variation is a special kind of relationship between two quantities where one goes up exactly as the other goes down, in such a way that their product stays the same.

In the broader chapter on ratios and proportions, you have already seen how two quantities can change together “in the same direction” (direct variation). Here we look at the opposite behavior.

What inverse variation means

Two variables $x$ and $y$ are said to vary inversely if:

• When $x$ increases, $y$ decreases.
• When $x$ decreases, $y$ increases.
• The product $x \cdot y$ is constant for all corresponding pairs of values.

We write this relationship as:

$$
y \text{ varies inversely as } x
$$

or more simply:

$$
y \propto \frac{1}{x}
$$

This is read as “$y$ is proportional to $1/x$.”

The key algebraic form of an inverse variation is:

$$
y = \frac{k}{x}
$$

where $k$ is a constant called the constant of variation (or constant of proportionality).

• If $k$ is positive, then $y$ is positive when $x$ is positive and negative when $x$ is negative.
• The larger $k$ is, the larger the product $x \cdot y$ is for any pair of values.

For every pair $(x, y)$ that follows this relationship,

$$
x \cdot y = k.
$$

This product being the same every time is the hallmark of inverse variation.

Recognizing inverse variation from data

Suppose you have a table of values for $x$ and $y$:

• If the ratio $\dfrac{y}{x}$ stays the same, that suggests direct variation.
• If the product $x \cdot y$ stays the same, that suggests inverse variation.

So to test for inverse variation:

1. Compute $x \cdot y$ for each row of a table.
2. If each product is the same number, the relationship is an inverse variation.

Example (inverse variation):

• $(x, y) = (2, 6)$ gives $2 \cdot 6 = 12$
• $(x, y) = (3, 4)$ gives $3 \cdot 4 = 12$
• $(x, y) = (4, 3)$ gives $4 \cdot 3 = 12$

Since the product is always $12$, we have $y = \dfrac{12}{x}$.

Compare with a non-example:

• $(x, y) = (2, 7)$ gives $2 \cdot 7 = 14$
• $(x, y) = (3, 4)$ gives $3 \cdot 4 = 12$

The products differ, so these do not follow a single inverse variation.

Using the constant of variation

In most problems, you will be given one pair of values and asked to find the constant $k$, then use it to find other values.

The steps are:

1. Start with the general form $y = \dfrac{k}{x}$.
2. Plug in a known pair $(x, y)$ to find $k$.
3. Use that $k$ to find additional values.

Example:

“$y$ varies inversely as $x$, and $y = 8$ when $x = 5$. Find $y$ when $x = 10$.”

1. General form: $y = \dfrac{k}{x}$.
2. Use $x = 5$, $y = 8$ to find $k$:

$$
8 = \frac{k}{5} \quad \Rightarrow \quad k = 8 \cdot 5 = 40.
$$

3. Now $y = \dfrac{40}{x}$. For $x = 10$,

$$
y = \frac{40}{10} = 4.
$$

Notice how doubling $x$ from $5$ to $10$ cut $y$ in half from $8$ to $4$. That “multiply one by 2, divide the other by 2” behavior is typical of inverse variation.

Solving inverse variation word problems

You have already seen how to translate verbal descriptions into proportions for direct variation and general ratios. For inverse variation, the sentences look a bit different and lead to $y = \dfrac{k}{x}$ rather than $y = kx$.

Common phrases that signal inverse variation:

• “$y$ varies inversely as $x$.”
• “$y$ is inversely proportional to $x$.”
• “$y$ is proportional to $\dfrac{1}{x}$.”
• “$y$ and $x$ are in inverse proportion.”

Once you see one of these phrases, you can immediately write the equation $y = \dfrac{k}{x}$.

General approach

When reading a word problem about inverse variation:

1. Identify the two quantities that change together.
2. Look for wording that indicates inverse variation.
3. Write $y = \dfrac{k}{x}$ (choose names that match the context: $t = \dfrac{k}{n}$, $v = \dfrac{k}{d}$, etc.).
4. Use one known pair to find $k$.
5. Use the equation with that $k$ to find the unknown quantity.

Typical everyday contexts

Here are some common real-world situations that can often be modeled (at least approximately) by inverse variation:

• Speed and travel time (for the same distance):

For a fixed distance, time $t$ varies inversely as speed $v$:
$$
t = \frac{k}{v}.
$$
Faster speed means less time; slower speed means more time.

• Number of workers and time to finish a job (with all workers working at the same rate and no interference):

For a fixed amount of work, time $T$ varies inversely as number of workers $n$:
$$
T = \frac{k}{n}.
$$

• Intensity (or brightness) and distance in some simplified models:

For certain types of light or sound models, intensity $I$ can vary inversely with a power of distance (though that becomes more advanced). A simplified version for learning purposes may use $I = \dfrac{k}{d}$.

These situations are not always perfectly inverse in real life, but they capture the basic “more of one, less of the other” pattern in a clean mathematical way.

Example: workers and time

“A group of 4 workers can paint a wall in 6 hours. Assuming all workers work at the same rate, how long will it take 8 workers to paint the same wall?”

1. Quantities: number of workers $n$ and time $T$.
2. As $n$ goes up, $T$ goes down. Assume inverse variation: $T = \dfrac{k}{n}$.
3. Use the known values: $T = 6$ when $n = 4$:

$$
6 = \frac{k}{4} \quad \Rightarrow \quad k = 24.
$$

4. With 8 workers:

$$
T = \frac{24}{8} = 3 \text{ hours}.
$$

Doubling the workers from 4 to 8 cut the time in half from 6 to 3.

Example: speed and time

“A driver travels a fixed distance. At 60 km/h, the trip takes 3 hours. How long would the trip take at 90 km/h (same route, same conditions)?”

Let $t$ be time, $v$ be speed. For fixed distance, $t$ varies inversely as $v$:

1. $t = \dfrac{k}{v}$.
2. Use $t = 3$ when $v = 60$:

$$
3 = \frac{k}{60} \quad \Rightarrow \quad k = 180.
$$

3. For $v = 90$:

$$
t = \frac{180}{90} = 2 \text{ hours}.
$$

Increasing speed from 60 to 90 (a factor of $\tfrac{3}{2}$) reduces time from 3 to 2 (a factor of $\tfrac{2}{3}$). The changes are in opposite directions.

Comparing inverse and direct variation

You have already seen direct variation in the “Direct variation” part of the parent chapter. Here, it is useful only to highlight the key differences without re-explaining the entire topic.

• Direct variation: $y = kx$; the ratio $\dfrac{y}{x}$ is constant.
• Inverse variation: $y = \dfrac{k}{x}$; the product $x \cdot y$ is constant.

So to quickly tell them apart in simple data:

• Check if $\dfrac{y}{x}$ is constant → suggests direct variation.
• Check if $x \cdot y$ is constant → suggests inverse variation.

Graph of an inverse variation

While a full study of graphs belongs mainly to later chapters, it is useful here to note the general shape.

For a positive constant $k > 0$, the graph of

$$
y = \frac{k}{x}
$$

has these features:

• For positive $x$, $y$ is positive and decreases as $x$ increases.
• As $x$ gets very large, $y$ gets closer and closer to $0$ (but never reaches it).
• As $x$ gets close to $0$ (from the positive side), $y$ gets very large.
• The graph never crosses the axes: $x = 0$ is not allowed (division by zero), and $y$ never equals $0$ for any finite $x$.

This kind of curve is called a hyperbola. The exact study of its properties will appear in later chapters, but you should recognize its general “bending” shape and the idea that points are arranged so that $x \cdot y$ is always the same fixed number.

Summary

• Inverse variation describes a relationship where $y$ is proportional to $\dfrac{1}{x}$.
• The equation for inverse variation is $y = \dfrac{k}{x}$, with $x \cdot y = k$ constant.
• To check for inverse variation, see whether the products $x \cdot y$ stay the same.
• Common contexts: speed and time for a fixed distance; number of workers and time for a fixed job.
• In word problems, phrases like “varies inversely” or “is inversely proportional” tell you to use $y = \dfrac{k}{x}$.
• Once you find $k$ from one known pair, you can use the same equation to find any other value in the relationship.