Table of Contents
Understanding Direct Variation
In this chapter, we focus on a special kind of proportional relationship between two quantities called direct variation. You will see how to recognize it, how to write equations for it, and how to use it to solve simple problems.
What Direct Variation Means
Two quantities $x$ and $y$ are in direct variation if:
- When $x$ increases, $y$ increases in a consistent way.
- When $x$ decreases, $y$ decreases in the same consistent way.
- Doubling $x$ doubles $y$, tripling $x$ triples $y$, and so on.
The key idea: the ratio $\dfrac{y}{x}$ stays the same every time (as long as $x \neq 0$).
So if $y$ varies directly with $x$, then
$$
\frac{y}{x} = \text{constant}.
$$
This constant is called the constant of variation or constant of proportionality.
The Equation of Direct Variation
The most important way to describe a direct variation is with an equation of the form:
$$
y = kx
$$
where:
- $y$ is the dependent quantity (it changes when $x$ changes),
- $x$ is the independent quantity,
- $k$ is the constant of variation (a fixed number).
If $y$ varies directly as $x$, we always can write $y = kx$ for some number $k$.
Finding the Constant of Variation
If you know one pair of values $(x, y)$ for a direct variation, you can find $k$:
From $y = kx$, solve for $k$:
$$
k = \frac{y}{x}, \quad x \neq 0.
$$
Example (just to illustrate the calculation):
If $y = 18$ when $x = 6$, then:
$$
k = \frac{y}{x} = \frac{18}{6} = 3,
$$
so the direct variation equation is $y = 3x$.
Recognizing Direct Variation from Pairs of Values
You might be given a small table or list of corresponding $x$ and $y$ values and asked whether $y$ varies directly with $x$.
To check:
- Compute $\dfrac{y}{x}$ for each pair (with $x \neq 0$).
- If all the ratios $\dfrac{y}{x}$ are the same, it is direct variation.
- That common ratio is $k$.
If the ratios are not the same, then $y$ does not vary directly as $x$.
For example:
- If $(x, y)$ pairs are $(2, 10)$, $(4, 20)$, $(6, 30)$, then
$$
\frac{10}{2} = 5,\quad \frac{20}{4} = 5,\quad \frac{30}{6} = 5,
$$
so $k = 5$ and $y = 5x$ is a direct variation. - If $(x, y)$ are $(1, 4)$ and $(2, 9)$, then
$$
\frac{4}{1} = 4,\quad \frac{9}{2} = 4.5,
$$
the ratios are different, so this is not a direct variation.
Recognizing Direct Variation from an Equation
You can also be asked whether a given equation represents direct variation.
An equation does represent direct variation if it can be written in the form:
$$
y = kx
$$
with $k$ a constant and no extra terms.
Some typical cases:
- $y = 7x$ is direct variation ($k = 7$).
- $y = \frac{1}{2}x$ is direct variation ($k = \frac{1}{2}$).
- $y = -4x$ is direct variation ($k = -4$).
But these are not direct variation:
- $y = 3x + 2$ (extra $+2$ term),
- $y = x^2$ (not just $kx$; $x$ is squared),
- $2y = 5x$ (this can be rewritten as $y = \frac{5}{2}x$, which is direct variation),
- $x = 4$ (this doesn’t define $y$ as $kx$).
So if needed, first solve the equation for $y$ and see if you get $y = kx$ with no additional terms.
Graph of a Direct Variation
Although the details of graphing are covered elsewhere, it helps to know what the graph of a direct variation looks like:
- It is a straight line.
- It passes through the origin $(0, 0)$.
- Its equation is $y = kx$.
The fact that it passes through $(0, 0)$ reflects that if $x = 0$, then $y = 0$ in a direct variation.
If a straight-line graph does not go through the origin, it is not a direct variation.
Solving Problems with Direct Variation
Problems involving direct variation usually give you one pair of related values, tell you that the relationship is direct variation, and then ask you to find another missing value.
The general steps:
- Use the given pair $(x, y)$ to find $k$:
$$
k = \frac{y}{x}.
$$ - Write the direct variation equation:
$$
y = kx.
$$ - Use the equation to find the unknown value.
Example Pattern
Problem pattern:
“$y$ varies directly as $x$. If $y = 12$ when $x = 3$, find $y$ when $x = 10$.”
Step 1: Find $k$:
$$
k = \frac{12}{3} = 4.
$$
Step 2: Equation:
$$
y = 4x.
$$
Step 3: Find $y$ when $x = 10$:
$$
y = 4 \cdot 10 = 40.
$$
You can solve any basic direct variation word problem by following this pattern.
Interpreting $k$: “How Many per One”
In many practical situations, $k$ is a rate: how much $y$ changes for each one unit of $x$.
- If $y$ is cost in dollars and $x$ is number of items, then $k$ is “dollars per item.”
- If $y$ is distance and $x$ is time, then $k$ is “distance per unit time” (like speed).
So when you calculate $k = \dfrac{y}{x}$, you are finding the “per one” amount that stays the same each time.
Common Mistakes to Avoid
- Forgetting that in $y = kx$, you must have no extra constant term.
- Calling a relationship direct variation just because $x$ and $y$ both increase. The test is that $\dfrac{y}{x}$ is the same constant each time.
- Mixing up direct variation with other types of variation (such as inverse variation) which have different forms and behavior.
Summary
- Direct variation means $y$ is proportional to $x$ with a constant ratio:
$$
y = kx \quad \text{and} \quad \frac{y}{x} = k.
$$ - To test for direct variation from data, check whether $\dfrac{y}{x}$ is the same for all pairs.
- To test from an equation, solve for $y$ and see if the form is exactly $y = kx$.
- In problems, find $k$ using given values, then use $y = kx$ to find the unknown.