Problem 12 - Introductory Algebra for College Students

Question

Use the four-step procedure for solving variation problems given on page 551 to solve Exercises \(9-12\). \(y\) varies inversely as \(x . y=5\) when \(x=3 .\) Find \(y\) when \(x=9\)

Step-by-Step Solution

Answer

When \( x = 9 \), \( y = 5/3 \).

Step 1: Find the constant of variation

First, substitute the given values of \( x \) and \( y \) into the formula for inverse variation \( y = k/x \). Thus, when \( y = 5 \) and \( x = 3 \), we get \( 5 = k/3 \). Solving this for \( k \), we find that \( k = 15 \). This is the constant of variation.

Step 2: Find y when x = 9

Next, substitute the constant of variation (\( k = 15 \)) and given \( x = 9 \) into the formula for inverse variation. Thus, \( y = 15 / 9 \). Simplifying this fraction reveals \( y = 5/3 \).

Step 3: State the final solution

Given the inverse variation relationship, when \( x = 9 \), \( y = 5/3 \)

Key Concepts

Understanding the Constant of Variation

The constant of variation plays a critical role in solving problems involving inverse variation. This constant, often denoted by the symbol k, remains unchanged when two variables change at an inverse rate. In other words, as one value increases, the other decreases proportionally, ensuring the product of the two variables is always equal to the constant of variation.

To find the constant, we use the basic inverse variation formula, which is \( y = \frac{k}{x} \) . By substituting the given values of x and y, we solve for k to establish this unchanging relationship. For instance, in the provided exercise, the calculation \( 5 = k/3 \) reveals that \( k = 15 \) , firmly establishing the basis upon which we can solve various problems across different values.

Inverse Variation Formula and Its Application

The inverse variation formula is an equation that states one variable is equal to the constant of variation divided by the other variable, and is expressed as \( y = \frac{k}{x} \) . It eloquently describes a relationship where one quantity decreases as another increases.

Applying this formula enables us to find the value of one variable when the other is known. For example, with the constant of variation, k, determined to be 15, and when x is given as 9, we substitute these values into the formula to find y, resulting in \( y = \frac{15}{9} \) . Through this application, the formula becomes a powerful tool in inverse variation problems.

Solving Variation Equations

Solving variation equations involves a systematic approach to unravel the relationships between varying quantities. The process typically includes isolating variables, substituting known values, and solving for the unknowns.

Once the constant of variation is identified, it becomes simpler to navigate through inverse variation problems. In the textbook's step-by-step example, the second step after finding the constant is to apply it to find a new value of y for a different x. By maintaining the inverse relationship, as in \( y = \frac{15}{9} \) which simplifies to \( y = \frac{5}{3} \) , we effectively use the equation to find the corresponding variation of one variable as the other changes.

Algebraic Problem Solving

Algebraic problem solving is a critical skill for interpreting and solving various math problems, including those involving inverse variation. This skill set combines recognizing patterns, understanding mathematical properties, and manipulating algebraic expressions to arrive at a solution.

By breaking down problems into smaller, manageable parts, such as identifying the constant of variation before applying it to find unknown values, we follow a logical progression that simplifies complex problems. The four-step procedure for solving variation problems is an excellent example of strategic algebraic problem solving, which can be applied to a wide array of mathematical challenges.