Problem 10 - Algebra 1: Concepts and Skills

Question

Suppose \(y=6\) when \(x=4 .\) For the given type of variation, find an equation that relates \(x\) and \(y .\) $$\text{\(x\) and \(y\) vary directly.}$$

Step-by-Step Solution

Answer

The equation that relates \(x\) and \(y\) is \(y = 1.5x\).

Step 1: Identification of Known Values

The problem states that for certain \(x\) and \(y\), specifically when \(x = 4\), \(y = 6\).

Step 2: Calculate the Constant of Variation

This can be done by re-arranging the direct variation equation \(y = kx\) to \(k = \frac{y}{x}\), and then substituting our known values. So we get, \(k = \frac{6}{4} = 1.5\)

Step 3: Formulate the equation

Substitute the constant of variation \(k = 1.5\) into the equation of direct variation, yielding the final equation \(y = 1.5x\).

Key Concepts

Constant of Variation

The concept of the *Constant of Variation* refers to a specific value, represented by the letter \(k\), in a direct variation equation. In a direct variation, as one variable changes, the other changes in a specific, consistent way, maintained by this constant. For instance, with the relationship between \(x\) and \(y\) expressed as \(y = kx\), the constant \(k\) determines how much \(y\) changes in response to changes in \(x\).

To find this constant, use known values of \(x\) and \(y\). In our example, when \(x=4\) and \(y=6\), \(k\) is calculated by rearranging the equation to \(k = \frac{y}{x}\). Plugging in these values, we discover that \(k = \frac{6}{4} = 1.5\). This tells us that for every unit change in \(x\), \(y\) changes by 1.5 times that amount.

Equation Formulation

After discovering the constant of variation, \(k\), the next crucial step is *Equation Formulation*. This involves putting together the pieces of the direct variation equation. With a known \(k\), the task is to create a mathematical representation linking \(x\) and \(y\).

In our scenario, we already calculated \(k = 1.5\), so we substitute this into the standard form \(y = kx\). The resulting equation, \(y = 1.5x\), gives a complete description of the relationship. This new formula indicates that \(y\) is directly proportional to \(x\), with a factor of 1.5.

This equation is powerful as it allows predicting \(y\) for any \(x\), offering a clear framework to understand how these variables interact.

Mathematical Relationship

A *Mathematical Relationship* in direct variation reveals how two variables interact in a predictable pattern. In our context, the relationship between \(x\) and \(y\) shows that as \(x\) changes, \(y\) changes proportionally due to the constant \(k\).

With the equation \(y = 1.5x\), we capture this relationship explicitly:

• If \(x\) is doubled, \(y\) doubles as well because \(y\) is directly proportional to \(x\).
• If \(x\) is halved, \(y\) also halves, maintaining the stable ratio dictated by \(k = 1.5\).

This consistent pattern makes direct variation a valuable tool in many fields, allowing predictions and scaling. The mathematical relationship is concise but robust, providing a clear connection between input and output variables, underpinning the seamless flow of predictable change.