Problem 26 - Algebra 2

Question

If \(y\) varies directly as \(x\) and \(x=6\) when \(y=0.5,\) find \(y\) when \(x=10\)

Step-by-Step Solution

Answer

When \(x=10\), \(y=\frac{5}{6}\).

Step 1: Understand the Direct Variation

If a variable y varies directly as another variable x, we can express this relationship using the equation \( y = kx \), where \( k \) is a constant. We are given that \( y = 0.5 \) when \( x = 6 \). Our first task is to find the constant \( k \).

Step 2: Solve for the Constant k

To find \( k \), substitute the given values into the direct variation equation. We have \( 0.5 = k \times 6 \). Solving for \( k \) gives \( k = \frac{0.5}{6} = \frac{1}{12} \).

Step 3: Use the Constant to Find y

Now that we have the constant \( k = \frac{1}{12} \), we can use it to find \( y \) when \( x = 10 \). Substitute \( x = 10 \) into the equation \( y = kx \) to get \( y = \frac{1}{12} \times 10 \).

Step 4: Calculate the Value of y

Perform the multiplication to find \( y \). We have \( y = \frac{10}{12} = \frac{5}{6} \). Thus, \( y \) is \( \frac{5}{6} \) when \( x = 10 \).

Key Concepts

Constant of Variation

In direct variation problems, the constant of variation, often denoted by the letter \( k \), plays a crucial role. It describes the factor by which one variable is multiplied to get another variable in a direct variation relationship. This is expressed mathematically as \( y = kx \), meaning \( y \) is a product of \( x \) multiplied by \( k \).

Finding the constant of variation is necessary to understand how changes in one variable affect the other. Once you know \( k \), you can predict one variable's value when you change the other. In our example, we are given that \( y = 0.5 \) when \( x = 6 \). To find \( k \), we substitute these values into the equation, yielding \( 0.5 = k \times 6 \). By solving for \( k \), we determine \( k = \frac{1}{12} \). This value defines how \( y \) changes in response to changes in \( x \).

Remember, the constant remains the same for all pairs of \( x \) and \( y \) in a given direct variation relationship, making it a fundamental component of solving these problems.

Solve for a Constant

To solve for the constant in a direct variation, you'll apply some basic algebra. The idea is to isolate \( k \) in the direct variation equation \( y = kx \).

Here's how you can solve for \( k \):

• Start by substituting the given values for \( x \) and \( y \) into the equation. For instance, with \( y = 0.5 \) and \( x = 6 \), the equation becomes \( 0.5 = k \times 6 \).
• To isolate \( k \), divide both sides of the equation by \( x \). This yields \( k = \frac{y}{x} \).
• In our example, \( k = \frac{0.5}{6} = \frac{1}{12} \).

When you solve for the constant, you unlock the equation necessary to calculate \( y \) for any value of \( x \) in this relationship. It simplifies further interactions with the equation, letting you find unknowns easily as they arise in similar variations.

Proportional Relationships

Understanding proportional relationships is key to direct variation problems. A proportional relationship means that two quantities increase or decrease at the same rate. In terms of algebra, if \( y \) varies directly as \( x \), then the ratio \( \frac{y}{x} \) remains constant, represented by \( k \).

Direct variation is a special kind of proportional relationship where the equation \( y = kx \) perfectly describes how \( y \) changes concerning \( x \). This constant ratio (\( k \)) tells us that for every unit increase in \( x \), \( y \) changes by \( k \) units.

Let's break it down:

• At \( x = 6 \), \( y \) was found to be \( 0.5 \), giving us \( k = \frac{1}{12} \).
• With this constant, you can predict that when \( x = 10 \), substituting in \( y = kx \), the result is \( y = \frac{10}{12} = \frac{5}{6} \).
• This consistent variation supports how \( y \) and \( x \) are directly proportional for any value.

Recognizing these relationships helps predict changes and find solutions to real-world problems modeled by direct variations.