Problem 2 - Algebra A Combined Function

Question

If y varies directly as \(x\), find the constant of variation and the direct variation equation for each situation. \(y=6\) when \(x=30\)

Step-by-Step Solution

Answer

The constant of variation is \(\frac{1}{5}\), and the direct variation equation is \(y = \frac{1}{5}x\).

Step 1: Understand Direct Variation

In a direct variation, if variable \(y\) varies directly as variable \(x\), then \(y = kx\). Here, \(k\) is the constant of variation. Given \(y=6\) when \(x=30\), we use this information to find \(k\).

Step 2: Find the Constant of Variation

To find the constant \(k\), substitute the given values of \(y\) and \(x\) into the equation \(y = kx\). We have: \\[ 6 = k \cdot 30 \] \Solve for \(k\) by dividing both sides by 30, \\[ k = \frac{6}{30} = \frac{1}{5} \]

Step 3: Write the Direct Variation Equation

Now that we have \(k = \frac{1}{5}\), substitute it back into the direct variation formula \(y = kx\). The equation becomes: \\[ y = \frac{1}{5}x \]

Key Concepts

Constant of Variation

The "Constant of Variation" is a key element in understanding direct variation problems in algebra. When two variables are linked by direct variation, this constant is the fixed number that defines that relationship.
Imagine it as the multiplier that scales one variable to get the other. In simpler terms, it's the factor that helps us understand how much one quantity changes with another.
To find the constant of variation, we use the basic direct variation equation:

• Identify known values for both variables.
• Substitute these known values into the equation to solve for the constant.
• In our example, by substituting the given values, we determined that the constant is \( k = \frac{1}{5} \), meaning for every unit change in \( x \), \( y \) changes by one-fifth of that.

When solving such problems, always ensure units and proportions make sense to maintain consistency between the variables.

Direct Variation Equation

The "Direct Variation Equation" is fundamentally about creating a simple relationship between two variables, commonly represented by the formula \( y = kx \).
This equation tells us that the variable \( y \) changes at a consistent rate with regard to changes in the variable \( x \).
Here's how you can understand and use the direct variation equation effectively:

• The variable \( y \) represents the dependent variable, which changes based on changes in \( x \).
• \( k \), the constant of variation, quantifies the relationship between \( y \) and \( x \).
• To write the direct variation equation, simply substitute the constant of variation back into \( y = kx \).

In our case, substituting \( k = \frac{1}{5} \) yields the equation \( y = \frac{1}{5}x \).
This equation expresses how \( y \) directly depends on \( x \) through a simple multiplier. Such equations are fundamental in expressing proportional relationships.

Algebraic Equations

"Algebraic Equations" like the ones encountered in direct variation problems are essential tools in mathematics. They represent relationships using symbols and numbers, allowing us to solve real-world problems by defining relationships between quantities.
Specifically, in the context of direct variation problems:

• Algebraic equations describe how one quantity changes with another, using variables like \( x \) and \( y \).
• These equations can be manipulated by adding, subtracting, multiplying, or dividing both sides to solve for unknowns.
• Equations like \( y = kx \) in direct variation give us a formula to predict the value of \( y \) for any given \( x \).

Algebraic equations help break down complex problems into manageable parts. When faced with solving a problem, divide it into steps, applying known values and rules until an unknown is revealed. This step-by-step process simplifies even daunting math challenges into clear, logical steps, making algebra an invaluable tool for solving a wide range of issues.