Direct Variation Equation

What is a direct variation equation?

A direct variation equation represents a relationship where the quotient of two variables remains completely constant. As the input variable increases, the output variable increases or decreases by a consistent multiple.

If two quantities vary directly, their graph always forms a perfectly straight line that passes exactly through the origin, which is the coordinate point $(0, 0)$. The relationship never has a starting value other than zero.

Identify the form y equals kx

The standard formula for a direct variation equation is written as $y = kx$.

In this formula, the variables $x$ and $y$ represent the changing quantities. The letter $k$ represents the constant of proportionality, which is the fixed multiplier connecting the two variables.

The constant $k$ can be any real number except zero. If a linear equation includes an added or subtracted value at the end, such as $y = 3x + 2$, it is not a direct variation equation because its graph does not pass through the origin.

Find the constant of variation

To write a direct variation equation, you must first find the numerical value of $k$. Because $y = kx$, you can isolate $k$ by dividing both sides of the equation by $x$.

The constant of variation is found by dividing the output by the input.

The formula to find the constant of variation is $k = \dfrac{y}{x}$.

Substitute any valid pair of $x$ and $y$ values into this formula. For example, if a bicycle travels $45$ kilometers in $3$ hours, the constant $k$ is $45$ divided by $3$, which equals $15$. In word problems, this constant often represents a unit rate, such as $15$ kilometers per hour.

Write an equation from values

When given a data table or a set of paired values, you can write the direct variation equation by calculating the constant multiplier connecting each pair.

Follow these exact steps to build the equation:

1. Select one pair of non-zero values for $x$ and $y$.
2. Divide the $y$ value by the $x$ value to calculate $k$.
3. Check another pair of values from the table to confirm the ratio remains identical.
4. Substitute your verified value of $k$ into the standard equation form $y = kx$.

Write an equation from a graph

You can also determine the direct variation equation by analyzing its visual representation on a coordinate grid. Directly proportional graphs are always straight lines extending from the origin, $(0, 0)$.

1. Verify that the graphed line is straight and intersects the origin.
2. Identify the exact coordinates of another clear point on the line, written as $(x, y)$.
3. Divide the $y$-coordinate by the $x$-coordinate to find the constant $k$.
4. Write the final formula by substituting $k$ into the equation $y = kx$.

Worked examples

Review these step-by-step examples to see how the direct variation equation is built in different mathematical contexts.

Example 1: Finding an equation from a word problem

Question: A painter earns $240$ dollars for $8$ hours of work. The total pay $y$ varies directly with the number of hours worked $x$. Write the direct variation equation for this relationship.

Method:

1. Identify the given values: $y = 240$ and $x = 8$.
2. Calculate the constant of variation using $k = \dfrac{y}{x}$.
3. Substitute the values: $k = \dfrac{240}{8} = 30$.
4. Write the standard equation: $y = kx$.
5. Substitute the constant: $y = 30x$.

Answer: The direct variation equation is $y = 30x$.

Check: Multiply $30$ by $8$ hours to confirm it produces exactly $240$ dollars.

Example 2: Writing an equation with a fractional constant

Question: If $y$ varies directly with $x$, and $y = 6$ when $x = 24$, what is the direct variation equation?

Method:

1. Identify the values: $y = 6$ and $x = 24$.
2. Find the constant of variation: $k = \dfrac{y}{x}$.
3. Substitute the numbers: $k = \dfrac{6}{24}$.
4. Simplify the fraction by dividing the numerator and denominator by $6$. The result is $k = \dfrac{1}{4}$.
5. Insert $k$ into $y = kx$.

Answer: The direct variation equation is $y = \dfrac{1}{4}x$.

Check: Substitute $x = 24$ into the equation. $\dfrac{1}{4} \times 24 = 6$, which matches the original given value.

Example 3: Checking if a mixed equation is a direct variation

Question: Does the equation $5y = 35x$ represent direct variation? If yes, find the constant of variation.

Method:

1. A direct variation must be written in the form $y = kx$.
2. Isolate $y$ in the given equation by dividing both sides by $5$.
3. Calculate the right side: $\dfrac{35x}{5} = 7x$.
4. The simplified equation is $y = 7x$. This strictly matches the form $y = kx$ with no added terms.

Answer: Yes, it is a direct variation equation. The constant of variation is $7$.

Check: The equation $y = 7x$ has no $y$-intercept other than zero, meaning it passes directly through the origin.

Common mistakes

Dividing in the wrong order. Students frequently calculate $k = \dfrac{x}{y}$ instead of $k = \dfrac{y}{x}$. Always divide the dependent output variable by the independent input variable.

Adding a constant value. A direct variation equation cannot have a non-zero $y$-intercept. An equation like $y = 4x + 1$ is a valid linear equation, but it is not a direct variation because the line would cross the $y$-axis at $1$, missing the origin completely.

Frequently asked questions

Can the constant of variation be a fraction or a decimal?

Yes, $k$ can be any non-zero real number. Equations like $y = \dfrac{2}{3}x$ or $y = 1.25x$ are completely valid direct variation equations.

Can the constant of variation be negative?

Yes. A negative constant means that as the $x$ value increases, the $y$ value decreases at a constant rate. The graph of an equation like $y = -5x$ still forms a straight line passing through the origin, sloping downward.

What is the difference between direct and inverse variation?

In direct variation, the two variables increase or decrease together while maintaining a constant ratio. In direct and inverse variation, an inverse variation means one variable decreases proportionately as the other increases, keeping a constant product where $x \times y = k$.