Problem 41 - Algebra 1: Concepts and Skills

Question

The variables \(x\) and \(y\) vary directly. Use the given values to write an equation that relates x and y . \(\text { (Lesson } 4.6)\) $$ x=6, y=42 $$

Step-by-Step Solution

Answer

The equation that describes the directly proportional relationship between \(x\) and \(y\) is \(y = 7x\).

Step 1: Understanding the equation of direct variation

The equation of direct variation is presented as \(y=kx\), where \(k\) is the constant of variation. The given values in the exercise, \(x=6, y=42\), will be plugged into this equation to find the value of \(k\).

Step 2: Substituting the given values into the equation

Substitute the given values into the equation: \(42=k * 6\).

Step 3: Solving for \(k\)

To solve for \(k\), you divide 42 by 6. So, \(k = 42 / 6 = 7\).

Step 4: Writing the final equation

Now that we know that \(k = 7\), we can write the final equation that describes the relationship between \(x\) and \(y\) as \(y = 7x\).

Key Concepts

Equation of Direct Variation

In mathematics, when we talk about direct variation, we mean a relationship between two variables in which one is a constant multiple of the other. That constant is what we call the "constant of variation."
The basic formula that represents this relationship is given by the equation \( y = kx \). Here, \( y \) and \( x \) are the variables that vary directly with each other, and \( k \) is the constant of variation that remains consistent as changes occur between \( x \) and \( y \).
Direct variation tells us that as one variable increases or decreases, the other does as well, in proportion to the constant \( k \). Understanding this linear relationship is crucial when dealing with real-world scenarios that involve proportional relationships.
Some properties of direct variation include:

• The graph of the equation \( y = kx \) is a straight line that passes through the origin (0,0).
• If \( k > 0 \), both \( y \) and \( x \) increase together, showing a positive relationship.
• If \( k < 0 \), \( y \) decreases while \( x \) increases, indicating a negative relationship.

Constant of Variation

The constant of variation, \( k \), plays a vital role in the equation \( y = kx \). It's the value that determines how much \( y \) will change for a given change in \( x \).
To find \( k \), we use specific given values of \( x \) and \( y \), as shown in the exercise:
For example, when \( x = 6 \) and \( y = 42 \), we can determine \( k \) by substituting these values into the direct variation equation:\[ 42 = k \times 6 \]
The solution to \( k \) involves isolating it by performing the operation: \( k = \frac{42}{6} = 7 \).
With this constant of variation, we can tell:

• For every unit increase in \( x \), \( y \) increases by 7 times that unit.
• Since \( k \) is positive, we know the relationship is direct and positive; \( y \) increases as \( x \) increases.

Solving Equations

Solving equations in the context of direct variation involves finding the value of the constant of variation \( k \) and then using it to understand the relationship between the variables.
The step-by-step approach is as follows:
1. **Identify Known Values**: Start with the known values of the variables, such as \( x = 6 \) and \( y = 42 \).2. **Direct Variation Formula**: Apply these values to the equation of direct variation \( y = kx \) to get a specific equation.3. **Isolate \( k \)**: Perform algebraic operations to solve for \( k \). In our example, the equation \( 42 = 6k \) simplifies by dividing both sides by 6, yielding \( k = 7 \).4. **Final Equation**: With \( k \) known, write the final equation that shows the relationship, which would be \( y = 7x \). This shows that \( y \) is seven times \( x \), summarizing the direct relationship.
Solving these equations helps in predicting the behavior of one variable based on another in situations governed by direct variation.