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When two quantities are connected by a proportion, they are described as being **proportional** to one another. An equivalent way of characterizing this type of relationship is through the language of **variation**. Whereas a proportion focuses on the equality of two ratios, variation emphasizes how one quantity changes in response to changes in the other. In algebra, two principal forms of variation are studied: **direct variation**, in which both quantities increase or decrease together, and **inverse variation**, in which one quantity increases as the other decreases.

Proportional Quantities and Variation

For any two variables $$x$$ and $$y$$, the variable $$y$$ **varies directly** with $$x$$ if their relationship can be expressed as

$$y = kx, \quad \text{where } k \neq 0$$

The nonzero constant $$k$$ is called the **constant of variation** (sometimes called the constant of proportionality). It represents the fixed multiplier that connects the two quantities: whenever one variable is known, the other is found by multiplying by $$k$$. Because $$k$$ is constant, doubling $$x$$ will always double $$y$$, tripling $$x$$ will triple $$y$$, and so on.

For example, if Lindsay earns \$15 per hour, her salary $$s$$ and hours worked $$h$$ satisfy $$s = 15h$$. Here $$k = 15$$, and her salary varies directly with the number of hours she works — the salary is always the product of the constant $$15$$ and the hours.

Direct Variation

To find the equation that relates two variables in a direct variation relationship when one pair of values is known, follow this four-step procedure:

1. **Write the formula for direct variation.** Begin with the general equation $$y = kx$$.
2. **Substitute the given values for the variables.** Replace $$x$$ and $$y$$ with the known pair of values.
3. **Solve for the constant of variation.** Divide both sides by the known value of $$x$$ to isolate $$k$$.
4. **Write the equation that relates $$x$$ and $$y$$.** Substitute the value of $$k$$ back into $$y = kx$$ to obtain the specific equation.

Once this equation is established, it can be used to determine the value of $$y$$ for any other value of $$x$$.

Solving Direct Variation Problems

**Problem:** If $$y$$ varies directly with $$x$$ and $$y = 20$$ when $$x = 8$$, find the equation that relates $$x$$ and $$y$$.

**Step 1 — Write the direct variation formula:**
$$y = kx$$

**Step 2 — Substitute the given values:** Replace $$y$$ with $$20$$ and $$x$$ with $$8$$:
$$20 = k \cdot 8$$

**Step 3 — Solve for the constant of variation:** Divide both sides by $$8$$:
$$\frac{20}{8} = k$$
$$k = 2.5$$

**Step 4 — Write the equation:** Substitute $$k = 2.5$$ back into the general formula:
$$y = 2.5x$$

The equation relating $$x$$ and $$y$$ is $$y = 2.5x$$.

Finding the Direct Variation Equation Given $$y = 20$$ When $$x = 8$$

**Problem:** When Raoul runs on the treadmill at the gym, the number of calories $$c$$ he burns varies directly with the number of minutes $$m$$ he uses it. He burned 315 calories in 18 minutes.

**(a) Find the equation relating $$c$$ and $$m$$.**

**Step 1 — Write the direct variation formula** using $$c$$ in place of $$y$$ and $$m$$ in place of $$x$$:
$$c = km$$

**Step 2 — Substitute the given values** ($$c = 315$$, $$m = 18$$):
$$315 = k \cdot 18$$

**Step 3 — Solve for $$k$$:** Divide both sides by $$18$$:
$$k = \frac{315}{18} = 17.5$$

**Step 4 — Write the equation:**
$$c = 17.5m$$

**(b) Find the calories burned in 25 minutes.**

Substitute $$m = 25$$ into the equation:
$$c = 17.5(25) = 437.5$$

Raoul would burn 437.5 calories if he used the treadmill for 25 minutes.

Calories Burned on a Treadmill: A Direct Variation Application

For any two variables $$x$$ and $$y$$, the variable $$y$$ **varies inversely** with $$x$$ if their relationship can be expressed as

$$y = \frac{k}{x}, \quad \text{where } k \neq 0$$

The nonzero constant $$k$$ is called the **constant of variation**. In an inverse variation, as one variable increases the other decreases, and vice versa. The term "inverse" refers to the multiplicative inverse (reciprocal): since the multiplicative inverse of $$x$$ is $$\frac{1}{x}$$, the inverse variation formula can be understood as $$y$$ equaling a constant times the reciprocal of $$x$$. This stands in contrast to direct variation, where $$y = kx$$ and both variables increase or decrease together.

Inverse Variation

To find the equation that relates two variables in an inverse variation relationship when one pair of values is known, follow this four-step procedure:

1. **Write the formula for inverse variation.** Begin with the general equation $$y = \frac{k}{x}$$.
2. **Substitute the given values for the variables.** Replace $$x$$ and $$y$$ with the known pair of values.
3. **Solve for the constant of variation.** Multiply both sides by the known value of $$x$$ to isolate $$k$$.
4. **Write the equation that relates $$x$$ and $$y$$.** Substitute the value of $$k$$ back into $$y = \frac{k}{x}$$ to obtain the specific equation.

This procedure mirrors the one used for direct variation problems — only the starting formula changes from $$y = kx$$ to $$y = \frac{k}{x}$$. Once the equation is established, it can be used to determine the value of $$y$$ for any other value of $$x$$.

Solving Inverse Variation Problems

**Problem:** If $$y$$ varies inversely with $$x$$ and $$y = 20$$ when $$x = 8$$, find the equation that relates $$x$$ and $$y$$.

**Step 1 — Write the inverse variation formula:**
$$y = \frac{k}{x}$$

**Step 2 — Substitute the given values:** Replace $$y$$ with $$20$$ and $$x$$ with $$8$$:
$$20 = \frac{k}{8}$$

**Step 3 — Solve for the constant of variation:** Multiply both sides by $$8$$:
$$8 \cdot 20 = 8 \cdot \frac{k}{8}$$
$$160 = k$$

**Step 4 — Write the equation:** Substitute $$k = 160$$ back into the general formula:
$$y = \frac{160}{x}$$

The equation relating $$x$$ and $$y$$ is $$y = \frac{160}{x}$$.

Finding the Inverse Variation Equation Given $$y = 20$$ When $$x = 8$$

A **proportion** is an equation of the form $$\frac{a}{b} = \frac{c}{d}$$, where $$b \neq 0$$ and $$d \neq 0$$, asserting that two rational expressions are equal. This equation is read as "$$a$$ is to $$b$$, as $$c$$ is to $$d$$." For instance, $$\frac{1}{2} = \frac{4}{8}$$ is a proportion — read "1 is to 2 as 4 is to 8" — because both fractions represent the same value. Proportions frequently appear in applications that involve scaling a known ratio to a larger or smaller context; when an unknown quantity occupies any of the four positions, it can be found by solving the equation. Because a proportion is a rational equation, it is solved by multiplying both sides by the LCD to clear the fractions and then solving the resulting equation. As with all rational expressions, any variable values that would make either denominator equal zero must be excluded.

Proportion

Two figures are **similar** if they satisfy two conditions simultaneously: the measures of all their corresponding angles are equal, and the lengths of their corresponding sides are in the same ratio. In other words, similar figures have exactly the same shape but may differ in size — every angle in one figure has an identical match in the other, and the side lengths are uniformly scaled by a single constant factor.

Similar Figures

A supervisor at a manufacturing plant notes that as the speed of a conveyor belt increases, the time it takes to move a part from one end to the other decreases. Which algebraic term best describes the relationship between the speed and the time?

Match each workplace scenario or mathematical term to the type of variation it represents.

In a professional procurement context, when the total cost of an order increases at a constant rate as the quantity of items purchased increases, the relationship between cost and quantity is specifically described as ____ variation.

True or False: In a relationship of inverse variation, an increase in one quantity results in a decrease in the other quantity.

In a professional reporting environment, we use the language of variation to describe how proportional quantities are connected. Briefly define 'direct variation' and 'inverse variation' by explaining the direction of change for the second quantity when the first quantity increases.

Defining Variation in Professional Metrics

A retail manager is analyzing store performance metrics to understand how different factors influence one another. Based on the manager's observations provided in the case study, identify the two specific algebraic terms used to describe these types of variation. Additionally, provide the formal definition for each term, explaining the characteristic relationship between the quantities involved.

Classifying Variation in Retail Operations

In professional environments, we often use the language of 'variation' to describe how one metric changes in response to another. Define the terms 'direct variation' and 'inverse variation.' In your response, clearly state what happens to the second quantity in each type of relationship when the first quantity increases.

Defining Direct and Inverse Variation in Professional Metrics

In a workplace productivity analysis, arrange the logical steps used to transition from identifying a general proportion to the specific classification of direct variation.

In professional data modeling, while a **proportion** focuses on the equality of two ratios, which algebraic term is used to emphasize how one quantity changes in response to changes in another?

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