Direct and Inverse Proportion

Intuition

The ratio and percent lessons kept circling one idea without ever naming it. Buy twice as many oranges and you pay twice as much; feed twice as many people and a recipe needs twice the flour. Some pairs of quantities are locked together so that their ratio never changes. For those pairs a single fixed number, the amount of one for each unit of the other, controls the whole relationship. Other pairs pull the opposite way: put twice as many workers on a job and it takes half as long. In that kind of pair it is the product that stays fixed. This lesson names both patterns, direct and inverse variation, and pins down the one constant that governs each. For those two patterns, the lesson hands you a single reliable method: use one known pair of values to find the constant, then let the constant answer every other question. The same idea returns later for relationships that depend on more than one quantity, where it takes a complete set of matching values, not just a pair, to pin down the constant.

Learning goals

Recognize direct variation as $y = kx$ with a constant ratio
Identify inverse variation as $y = \dfrac{k}{x}$ with a constant product
Find $k$ from one known pair for direct or inverse variation, then reuse it
Read a table by testing whether $y/x$ or $xy$ repeats
Extend to joint $y = kxz$ and combined variation

Direct variation

Two quantities are in direct variation when one is always a fixed multiple of the other. We say $y$ varies directly with $x$ , or that $y$ is directly proportional to $x$ , when there is a constant $k$ with

y = kx.

The number $k$ is the constant of proportionality, the single value that turns each $x$ into its matching $y$ . Dividing both sides by $x$ (for $x \neq 0$ ) shows what stays fixed:

k = \frac{y}{x}.

The ratio $y/x$ is the same for every matching pair, and that shared ratio is exactly $k$ . Because $y$ is always the same multiple of $x$ , doubling $x$ doubles $y$ , tripling $x$ triples $y$ , and halving $x$ halves $y$ . In every example in this lesson the quantities are positive counts or amounts, apples, dollars, weight, so more of one always means more of the other: the two rise and fall together, locked in step.

Here is the cost of apples sold at a fixed price. Watch the ratio column hold steady:

$x$ (apples)	$y$ (dollars)	$y/x$
$2$	$3$	$1.5$
$4$	$6$	$1.5$
$6$	$9$	$1.5$
$10$	$15$	$1.5$

Every row has $y/x = 1.5$ , so the cost varies directly with the number of apples, and the constant of proportionality is $k = 1.5$ dollars per apple. Plot those pairs on the coordinate plane and they line up: the points $(2, 3)$ , $(4, 6)$ , $(6, 9)$ , and $(10, 15)$ all sit on one straight line. That line runs through the origin, because $x = 0$ forces $y = k \cdot 0 = 0$ . A smaller example makes the shape plain.

When a relationship is direct, the matching pairs (x, y) fall on a straight line through the origin. Reading two points off the line, as x rises from 1 to 2 the value y rises from 2 to 4, so doubling x doubles y.

The line through the origin is the visual signature of direct variation. Whatever the value of $k$ , a direct relationship always graphs as a straight line through the origin, and $k$ sets how steep that line is.

Find the constant, then solve

The reason variation is so useful is that a single matching pair (or, for joint and combined variation met later in this lesson, one complete matching set) unlocks the whole relationship. You do not need a table; you need one complete measurement. The four-step method never changes, only what you substitute does:

Write the variation equation with $k$ still unknown ( $y = kx$ for direct variation).
Substitute the known values, one matching pair for direct or inverse variation, one complete matching set for joint or combined variation, and solve for $k$ .
Rewrite the equation with $k$ filled in.
Substitute the new values and compute the answer.

Finding $k$ first turns the problem into simple arithmetic: once $k$ is known, the equation produces every other answer directly.

Worked example 1 A spring stretch (direct variation)

The distance a spring stretches varies directly with the weight hung from it. An $8$ kg weight stretches the spring $12$ cm. How far will a $10$ kg weight stretch it?

Let $y$ be the stretch and $x$ the weight. Direct variation means $y = kx$ . Substitute the known pair, $y = 12$ when $x = 8$ , and solve for the constant:

12 = k \times 8, \qquad k = \frac{12}{8} = 1.5.

The constant of proportionality is $1.5$ cm per kg. Put it back into the equation:

y = 1.5x.

Now answer the question with $x = 10$ :

y = 1.5 \times 10 = 15 \text{ cm}.

So a $10$ kg weight stretches the spring $15$ cm. Check the ratio: $\frac{12}{8} = \frac{15}{10} = 1.5$ , so both pairs share the same constant, exactly as direct variation demands.

Check your understanding

A recipe uses sugar in direct proportion to flour. $6$ cups of flour need $4$ cups of sugar. How much sugar do $15$ cups of flour need?

10

cups B

12

cups C

9

cups D

13

cups

Inverse variation

Not every pair of quantities rises together. Sometimes making one larger makes the other smaller in exact proportion. Add more workers and a job finishes sooner; drive faster and a fixed trip takes less time. We say $y$ varies inversely with $x$ , or that $y$ is inversely proportional to $x$ , when there is a constant $k$ with

y = \frac{k}{x}, \qquad \text{equivalently} \qquad xy = k \qquad (x \neq 0).

Now it is the product $xy$ that stays fixed, not the ratio. Because $y = k/x$ , doubling $x$ halves $y$ , tripling $x$ cuts $y$ to a third, and halving $x$ doubles $y$ . The quantities move in opposite directions, but not by adding and subtracting: they move so that their product never budges.

Here is a fixed trip covered at different speeds, so the distance stays the same. Watch the product column:

$x$ (mph)	$y$ (hours)	$xy$
$10$	$12$	$120$
$20$	$6$	$120$
$30$	$4$	$120$
$40$	$3$	$120$

Every row has $xy = 120$ , so the travel time varies inversely with the speed, and the constant is $k = 120$ (the fixed $120$ -mile distance). The find-the-constant method works just as before, only now the equation is $y = k/x$ .

The difference between the two kinds of variation is which thing you are holding still, and a rectangle built one square at a time can show both, as long as you are careful about which two quantities are playing the roles of $x$ and $y$ in each demonstration. A rectangle’s area is the product of its two side lengths, and its perimeter is not.

Start with inverse variation. Here $x$ is the width and $y$ is the height. Keep the area reading $24$ : build $3$ by $8$ , then $4$ by $6$ , then $6$ by $4$ , then $8$ by $3$ . Doubling the width from $3$ to $6$ halves the height from $8$ to $4$ , precisely what $y = 24/x$ demands, and the area never moves off $24$ . Watch the perimeter while you do it. It reads $22$ , $20$ , $20$ , $22$ , so the sum of the sides is emphatically not what is being held constant here. Only the product, width times height, is, which is the whole content of $xy = k$ .

Now direct variation, from the same figure, but pair the width with a different quantity this time: the area itself, not the height. Set the height to $3$ and leave it alone, then step the width up from $1$ : the area runs $3$ , $6$ , $9$ , $12$ , and so on. Divide each area by its width and you get $3$ every time, so with $x$ the width and $y$ the area, $y = 3x$ with a constant of proportionality $k = 3$ , which is just the height you fixed. Width and height are not the variation pair this time, width and area are, and holding the height still is exactly what keeps that ratio constant. The same rectangle can demonstrate either pattern, but only one pairing of quantities at a time; deciding what is playing the role of $x$ and what is playing the role of $y$ is part of reading any variation problem.

Rectangle explorer

Width Height

A rectangle 3 units wide and 8 units tall. Perimeter 22 units. Area 24 square units.

A rectangle whose width and height you set a square at a time, with the perimeter and the area both reported. The area is the product of the two sides, so a set of rectangles that all report one area is a set of number pairs sharing one product.

Worked example 2 Workers and time (inverse variation)

Assume every worker paves at the same steady rate and none gets in another’s way. Under that assumption the time to pave a road varies inversely with the number of workers assigned. With $4$ workers the job takes $9$ days. How long will it take with $6$ workers?

Let $y$ be the days and $x$ the number of workers. Inverse variation means $xy = k$ , so find the constant by multiplying the known pair:

k = 4 \times 9 = 36.

The fixed amount of work is $36$ worker-days. Write the equation for the time:

y = \frac{36}{x}.

Now set $x = 6$ :

y = \frac{36}{6} = 6 \text{ days}.

So $6$ workers finish in $6$ days. Notice the time fell from $9$ days to $6$ , not by subtracting the two extra workers but because the product of workers and days must stay $36$ . The tempting wrong move is to subtract; the correct relationship divides. Headcount alone predicts the time only because every worker paves at the same rate; with workers of different speeds it is their combined rate, not their count, that the time varies against.

Check your understanding

For a fixed trip, travel time varies inversely with speed. At $45$ mph the trip takes $8$ hours. How long does it take at $60$ mph?

5

hours B

6

hours C

7

hours D

10

hours

Why a single constant is enough

For direct and inverse variation, one data pair fixes the whole relationship#

Take direct variation first. To say $y$ varies directly with $x$ is to say $y = kx$ for some fixed number $k$ , and that single number is the only thing the relationship leaves open: the equation, the table, and the graph are all settled the moment $k$ is known. So how much information does it take to find $k$ ? Exactly one pair of matching values. Suppose $y$ equals $y_0$ when $x$ equals $x_0$ , with $x_0 \neq 0$ . Substituting gives $y_0 = k x_0$ , and dividing by $x_0$ gives $k = y_0 / x_0$ . One pair, one division, and $k$ is known; from there $y = kx$ produces $y$ for any $x$ you like.

Inverse variation runs the same way with one change. Here $y = k/x$ , so a known pair $(x_0, y_0)$ gives $y_0 = k / x_0$ , and multiplying by $x_0$ gives $k = x_0 y_0$ . Again a single pair pins down $k$ , and again the whole relationship follows. This is why every direct or inverse variation problem needs just one complete pair of values to begin: that pair is enough to find the constant, and the constant answers everything else. Joint and combined variation, met later in this lesson, follow the identical logic with more quantities: instead of one pair, it takes one complete set of matching values, one value for every quantity in the equation, to pin down $k$ .

Reading a table: ratio or product?

When a problem hands you a table and does not say which kind of variation it is, let the numbers decide. Test the ratio $y/x$ down the rows first: if it repeats, the relationship is direct and that repeated value is $k$ . If the ratio drifts, test the product $xy$ : if it repeats, the relationship is inverse and that repeated value is $k$ . If neither holds steady, the relationship is something else (a quantity that grows with the square of another varies by a power, and those wait for later chapters).

The two variations differ in what stays constant. In direct variation the ratio y over x repeats down the table; in inverse variation the product x times y repeats. Testing which one is steady tells you which kind of variation you have.

Worked example 3 Which kind is it? Reading a table

A table pairs these values: $(x, y) = (3, 8),\ (4, 6),\ (6, 4),\ (8, 3)$ . Decide whether the variation is direct or inverse, find the constant, and predict $y$ when $x = 12$ .

First test the ratio $y/x$ for direct variation. The first two ratios are $\frac{8}{3}$ and $\frac{6}{4} = 1.5$ , which already disagree, so the ratio is not constant and the variation is not direct.

Now test the product $xy$ for inverse variation:

3 \times 8 = 24, \quad 4 \times 6 = 24, \quad 6 \times 4 = 24, \quad 8 \times 3 = 24.

The product is $24$ in every column, so the variation is inverse with $k = 24$ , and the equation is $y = \frac{24}{x}$ . Predict the value at $x = 12$ :

y = \frac{24}{12} = 2.

So when $x = 12$ , $y = 2$ . A steady product, not a steady ratio, is what marks the table as inverse.

Check your understanding

A table pairs these values: $(x, y) = (3, 21),\ (5, 35),\ (8, 56)$ . Test the ratio and the product. Which kind of variation is this, and what is $k$ ?

A Direct variation,

k = 7

B Inverse variation,

k = 63

C Direct variation,

k = 63

D Neither

Joint and combined variation

Real quantities often depend on several others at once, and the constant-of-proportionality idea stretches to cover them. When $y$ varies directly with two quantities together, we say $y$ varies jointly with $x$ and $z$ :

y = kxz.

When $y$ varies directly with one quantity and inversely with another, the relationship is called combined variation:

y = \frac{kx}{z}.

A quantity in the numerator drives $y$ up as it grows; a quantity in the denominator drives $y$ down. In both cases the method is unchanged: substitute one complete set of matching values to find $k$ , then use the finished equation. The only new care is placing each quantity on the correct side of the fraction.

Worked example 4 A banner's cost (joint variation)

The cost of a printed banner varies jointly with its width and its height. A banner $3$ ft wide and $4$ ft tall costs $18$ dollars. What does a banner $5$ ft wide and $6$ ft tall cost?

Let $c$ be the cost, $w$ the width, and $h$ the height. Joint variation means $c = kwh$ . Substitute the known banner to find the constant:

18 = k \times 3 \times 4 = 12k, \qquad k = \frac{18}{12} = 1.5.

The constant is $1.5$ dollars per square foot. Write the finished equation and apply it to the new banner:

c = 1.5wh = 1.5 \times 5 \times 6 = 45 \text{ dollars}.

So the larger banner costs $45$ dollars. Joint variation is just direct variation with two partners at once, and one complete measurement, cost together with both side lengths, fixes the constant.

Check your understanding

The volume of a box varies jointly with its length and its width, for a fixed height. A box $4$ in long and $3$ in wide holds $24$ in $^3$ . How much does a box $6$ in long and $5$ in wide hold?

60

^3

54

^3

48

^3

66

^3

Worked example 5 A beam's load (combined variation)

The safe load a wooden beam can carry varies directly with its width and inversely with its length. A beam $4$ inches wide and $10$ feet long safely carries $800$ pounds. How much can a beam $6$ inches wide and $8$ feet long carry?

Let $L$ be the safe load, $w$ the width, and $\ell$ the length. Combined variation means

L = \frac{kw}{\ell}.

Substitute the first beam, $L = 800$ when $w = 4$ and $\ell = 10$ :

800 = \frac{k \times 4}{10} = \frac{2k}{5}, \qquad k = 800 \times \frac{5}{2} = 2{,}000.

Write the finished equation and apply it to the second beam, $w = 6$ and $\ell = 8$ :

L = \frac{2{,}000 \times 6}{8} = \frac{12{,}000}{8} = 1{,}500 \text{ pounds}.

So the second beam safely carries $1{,}500$ pounds. This beam is both wider and shorter. Width sits in the numerator, so more width raises the load. Length sits in the denominator, so less length also raises it, and both changes push the safe load up from $800$ to $1{,}500$ pounds.

Check your understanding

The variable $y$ varies directly with $x$ and inversely with $z$ . When $x = 6$ and $z = 2$ , $y = 15$ . Find $y$ when $x = 8$ and $z = 4$ .

20

10

40

5

Common mistakes

Confusing direct with inverse. Direct variation keeps the ratio $y/x$ constant and uses $y = kx$ ; inverse variation keeps the product $xy$ constant and uses $y = k/x$ . One quantity growing while the other shrinks is a clue for inverse variation, not proof of it. Before solving, test the ratio and the product and decide which one actually stays fixed; forcing an inverse relationship into the direct formula makes the answer grow when it should shrink.
Adding or subtracting instead of scaling. With $4$ workers a job takes $9$ days, so $8$ workers do not take $9 - 4 = 5$ days. They take half the time, $4.5$ days, because doubling the workers halves the time. The direct, inverse, joint, and combined patterns in this lesson all multiply and divide; none of them add or subtract.
Skipping the constant. Do not leap from the given values straight to the answer without a method behind it. Scaling directly between two known values works when the numbers line up evenly for direct or inverse variation, but finding $k$ first works every time, for all four patterns in this lesson: write the equation, solve for $k$ , then use it.
Misplacing a variable in combined variation. A quantity you vary directly with belongs in the numerator; one you vary inversely with belongs in the denominator. Swapping them turns $y = kx/z$ into the wrong relationship.

Key takeaways

Direct variation: $y = kx$ , so the ratio $y/x = k$ stays constant; doubling $x$ doubles $y$ , and the graph is a straight line through the origin.
Inverse variation: $y = \dfrac{k}{x}$ , so the product $xy = k$ stays constant; doubling $x$ halves $y$ .
$k$ is the constant of proportionality, and a relationship has exactly one, so one matching set of values determines it, as long as every directly varying quantity in that set is nonzero (for direct and inverse variation, any $(x, y)$ pair with $x \neq 0$ will do; for joint and combined variation, every directly varying variable in the set must be nonzero).
The method: write the variation equation, then find $k$ , using one known pair for direct or inverse variation, or one complete matching set for joint or combined variation; rewrite the equation with $k$ filled in, then substitute the new values.
Recognize from a table: if $y/x$ repeats, the variation is direct; if $xy$ repeats, it is inverse.
Joint variation is $y = kxz$ (direct with two quantities); combined variation is $y = \dfrac{kx}{z}$ (direct with one, inverse with another).

Direct and Inverse Proportion

Learning goals

Direct variation

Find the constant, then solve

Inverse variation

Why a single constant is enough

Reading a table: ratio or product?

Joint and combined variation

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice