Problem 14 - Algebra 2

Question

Suppose that \(x\) and \(y\) vary inversely. Write a function that models each inverse variation and find \(y\) when \(x=10 .\) $$ x=20 \text { when } y=-4 $$

Step-by-Step Solution

Answer

The equation modeling the inverse variation between x and y is \(y = -80 / x\). When \(x = 10\), \(y = -8\).

Step 1: Understand Inverse Variation

An inverse variation between two parameters \(x\) and \(y\) is defined by \(xy = k\), where \(k\) is a constant. This relationship indicates that as \(x\) increases, \(y\) decreases proportionally, and vice versa.

Step 2: Determination of the constant \(k\)

Using the given relationship \(x = 20\) when \(y = -4\), we substitute into the inverse variation equation: \[20 (-4) = k\] This will result in \(k = -80\).

Step 3: Formulation of the Inverse Variation Function

Since we have found value of \(k\), we can write down the inverse function: \[f(x) = -80 / x \] This function models the given inverse variation between \(x\) and \(y\).

Step 4: Find y when \(x = 10\)

Now that we have the function, we can substitute \(x = 10\) into the function to find the corresponding value of \(y\): \[f(10) = -80 / 10 = -8\]

Key Concepts

Constant of Variation

In mathematical relationships, especially in inverse variations, the "constant of variation" plays a significant role. This constant, often denoted as \(k\), is a key value that defines how one variable changes relative to another. For inverse variations, the relation is represented as \(xy = k\). Here, the product of \(x\) and \(y\) is always equal to \(k\).
For example, consider the relationship we were given: when \(x = 20\) and \(y = -4\), we substitute these values into the equation to find \(k\):

• \(20 \times (-4) = k\), which leads to \(k = -80\).

This constant value means that no matter what values \(x\) and \(y\) take, their product will always be \(-80\). Recognizing \(k\) helps us build the inverse function and understand the nature of the inverse relationship between \(x\) and \(y\).

Inverse Function

An inverse function is designed to illustrate the inverse relationship between two variables. For our example, once we determine the constant of variation \(k\), we can formulate this inverse function. In our case, the inverse function is:

• \(f(x) = \frac{-80}{x}\).

This function represents the relationship where as one variable, \(x\), increases, the other, \(y\), decreases so that their product equals the constant \(-80\).
Working with this function is simple:

• To find \(y\) for any given \(x\), just divide the constant \(-80\) by \(x\).
• In our problem, when \(x = 10\), \(y\) is calculated as \(f(10) = \frac{-80}{10} = -8\).

Thus, inverse functions make it easier to determine how variations in one quantity affect another while preserving the underlying constant relationship.

Algebraic Equations

Algebraic equations form the backbone of relationships in mathematics, including inverse variations. They are powerful tools that help express how different quantities relate to each other mathematically. In the case of inverse variation, the relationship \(xy = k\) is a simple yet profound algebraic equation.

• In our specific example, this equation is \(xy = -80\), indicating that regardless of \(x\) or \(y\) values they must multiply to \(-80\).

Solving algebraic equations involves applying mathematical operations to find unknown values. For instance, if given one variable, such as \(x = 10\), we can rearrange our inverse equation to find \(y\):

• Substitute \(x = 10\) into the equation: \(10y = -80\).
• Solve for \(y\) by dividing both sides by 10, resulting in \(y = -8\).

This process highlights how algebraic equations allow us to explore and solve real-world problems by expressing relationships numerically.