Problem 38 - College Algebra with Corequisite Support

Question

For the following exercises, use the given information to find the unknown value. \(y\) varies jointly as \(x\) and \(z\) and inversely as \(w\). When \(x=5, z=2,\) and \(w=20,\) then \(y=4\). Find \(y\) when \(x=3\) and \(z=8,\) and \(w=48\).

Step-by-Step Solution

Answer

For the new conditions, \( y = 4 \).

Step 1: Understand the Relationship

Since the problem states that \( y \) varies jointly as \( x \) and \( z \) and inversely as \( w \), we can write this relationship as: \( y = k \frac{xz}{w} \), where \( k \) is the constant of variation.

Step 2: Find the Constant of Variation, k

Use the given values to find \( k \). Plug in \( x = 5 \), \( z = 2 \), \( w = 20 \), and \( y = 4 \) into the equation: \[ 4 = k \frac{5 \, \cdot \, 2}{20} \]. Simplify to get \[ 4 = k \frac{10}{20} \]. Therefore, \[ 4 = k \frac{1}{2} \]. Solving for \( k \), we get \( k = 8 \).

Step 3: Use the Constant to Find the New y

Now, use the constant \( k = 8 \) and the new values \( x = 3 \), \( z = 8 \), \( w = 48 \) to find \( y \). Plug these into the equation: \[ y = 8 \frac{3 \, \cdot \, 8}{48} \]. Simplify to get \[ y = 8 \frac{24}{48} \].

Step 4: Simplify to Find y

Simplify \( \frac{24}{48} \) to get \( \frac{1}{2} \). Multiply by \( 8 \): \[ y = 8 \times \frac{1}{2} = 4 \].

Key Concepts

Constant of Variation

In mathematical relationships, a constant of variation is a crucial factor that describes how one variable depends on others. It determines the proportionality in a joint variation or an inverse relationship. The concept might often seem abstract, but it's essential to clarify these relationships.
In the context of the given problem, the constant of variation, denoted as \( k \), solidifies the relationship among \( y, x, z, \) and \( w \) as:

• \( y = k \frac{xz}{w} \)

By setting specific values for \( x, z, w, \) and solving for \( y \), we can figure out the value of \( k \). After finding \( k \), it remains constant as long as the relationship holds true, serving as a key to solve for other variables.
Think of \( k \) as a binding factor, which helps translate one set of variable conditions into another. Even when variable values change, \( k \) allows us to transfer these changes consistently to find our new outcome.

Inverse Variation

Inverse variation is a fascinating concept where two variables change in opposite directions. This means when one variable increases, the other decreases, maintaining a constant product. It is represented as \( y = \frac{k}{x} \), where \( k \) is a constant. In the exercise, the relationship includes inverse variation, coupled with joint variation, captured as:

• \( y = k \frac{xz}{w} \)

Here, \( y \) varies inversely with \( w \). When \( w \) escalates, \( y \) diminishes if \( x \) and \( z \) remain the same.
Understanding inverse variation offers substantial insights into real-world dynamics. Consider how speed and time are inversely related in travel – as speed increases, travel time decreases, holding the distance constant. Recognizing these patterns empowers us to predict, calculate, and make informed decisions in various algebraic problems.

Algebraic Problem Solving

Algebraic problem solving is the heart of tackling mathematical challenges like joint and inverse variations. It's all about finding unknown values given specific scenarios, which involve forming equations, identifying relationships, and solving for unknowns systematically.
In the provided problem, the strategy included:

• Recognizing the relationship as \( y = k \frac{xz}{w} \)
• Identifying \( k \) by plugging in known values and solving the equation
• Using the derived \( k \) to find new values of \( y \) under different conditions

These steps are vital because they provide a clear direction to solve complex algebraic questions. The successful application of algebraic problem solving transforms abstract ideas into practical outcomes, making it a fundamental skill. Always remember, patience and practice enhance proficiency in this area.