Problem 5 - Introductory Algebra for College Students

Question

In Exercises \(5-8,\) determine the constant of variation for each stated condition. \(y\) varies directly as \(x,\) and \(y=80\) when \(x=4\)

Step-by-Step Solution

Answer

The constant of variation, \(k\), is 20.

Step 1: Understand the Problem

In this problem, \(y\) is said to vary directly as \(x\). This is represented mathematically as \(y=kx\), where \(k\) is the constant of variation. We are given that \(y=80\) when \(x=4\), which we can use to find \(k\).

Step 2: Substitute the Given Values

Substitute the values of \(y\) and \(x\) into the equation \(y=kx\) to form the equation \(80=k \cdot 4\).

Step 3: Solve for the Constant of Variation

Now, divide both sides of the equation \(80=k \cdot 4\) by 4 to solve for \(k\), obtaining \(k=80 \div 4\).

Step 4: Calculate the Result

Calculate the value of \(k\) by performing the division \(80 \div 4\) to get \(k=20\).

Key Concepts

Direct Variation

Understanding direct variation is critical for grasping the connection between two variables that change together in a specific way. In the realm of algebra, direct variation describes a situation where as one quantity increases, another quantity increases at a constant rate, and vice versa.

For instance, when we say that \( y \) varies directly as \( x \), this relationship can be represented by the equation \( y = kx \), where \( k \) is known as the constant of variation. This constant determines how quickly \( y \) changes in relation to \( x \). A key characteristic here is proportionality—wherever \( x \) goes, \( y \) follows, maintaining a consistent ratio that is defined by \( k \).

Solving for k in Direct Variation

When you come across a problem involving direct variation, and you're tasked with finding the constant of variation, \( k \), there's a simple process to follow.

Firstly, remember the direct variation equation \( y = kx \). When given specific values for \( y \) and \( x \), you substitute them into the equation to form a solvable expression for \( k \). For example, if \( y = 80 \) when \( x = 4 \), by substituting these values into the equation, you get \( 80 = k \cdot 4 \). To solve for \( k \) you simply divide both sides by \( 4 \) to isolate \( k \), leading you to \( k = 20 \). This value of \( k \) is what maintains the direct proportionality between \( x \) and \( y \).

Proportional Relationships

Proportional relationships are the backbone of direct variation and involve two quantities that are in proportion to one another. Basically, if one quantity is doubled, the other is doubled as well; if one is halved, so is the other. This indicates a consistent ratio or fraction between them, which is where the constant of variation plays a pivotal role.

Such relationships are not only vital in mathematics but also in real-life situations. Consider the speed of a car (distance/time): if the speed is constant, the distance covered varies directly with the time traveled. Proportional relationships help predict one quantity when the other is known, making it a powerful tool for modeling and solving problems across various fields.