Topic 6

Direct and Inverse Variation

A direct variation is a proportional relationship of the form

y = kx,

where

k

is the constant of variation. As

x

increases,

y

increases proportionally. Graph: a straight line through the origin with slope

k

An inverse variation is a relationship where one quantity increases as the other decreases, given by

y = \frac{k}{x},

where

k

is the constant of variation. The product

xy = k

is constant. Graph: a hyperbola in the first and third quadrants when

k > 0

Core Skills

Identify whether a situation or equation shows direct or inverse variation.
Find the constant $k$ from given data and use it to make predictions.
Recognize $y/x = \text{constant}$ as direct variation and $xy = \text{constant}$ as inverse variation.
Graph and interpret the meaning of $k$ in context.

y

varies directly with

x

and

y=12

when

x=4

, find the equation.

y = kx \Rightarrow 12 = 4k \Rightarrow k = 3 \Rightarrow \boxed{y = 3x}.

y

varies inversely with

x

and

y=8

when

x=2

, find the equation.

y = \frac{k}{x} \Rightarrow 8 = \frac{k}{2} \Rightarrow k = 16 \Rightarrow \boxed{y = \frac{16}{x}}.

Determine the type of variation:

y = 5x + 2.

This is not a variation because it does not pass through the origin.

y = 10x \quad \text{(direct)} \qquad y = \frac{20}{x} \quad \text{(inverse)}.

y = 7x

and

x=12

, find

y

y = 7(12) = \boxed{84}.

y = \frac{24}{x}

and

x=6

, find

y

y = \frac{24}{6} = \boxed{4}.

The time

t

to complete a job varies inversely with the number of workers

n

. If 4 workers take 18 hours, find how long 6 workers will take.

t = \frac{k}{n} \Rightarrow 18 = \frac{k}{4} \Rightarrow k = 72.

t = \frac{72}{6} = \boxed{12\text{ hours}}.

Direct variation: $y = kx$ . The ratio $y/x$ is constant.
Inverse variation: $y = k/x$ . The product $xy$ is constant.
Graphs of direct variation are lines through the origin; inverse variation graphs are hyperbolas.
Use variation equations to find missing quantities and interpret proportional relationships.