Problem 30 - Algebra 2

Question

If \(y\) varies inversely as \(x\) and \(y=4\) when \(x=12,\) find \(y\) when \(x=5\)

Step-by-Step Solution

Answer

When \( x = 5 \), \( y = 9.6 \).

Step 1: Understand the Inverse Variation

When one variable varies inversely as another, it means their product is constant. Mathematically, if \( y \) varies inversely as \( x \), then \( y = \frac{k}{x} \) for some constant \( k \).

Step 2: Determine the Constant of Variation

Using the given values, \( y = 4 \) when \( x = 12 \), substitute these into the equation \( yx = k \) to find \( k \):\[ k = y \cdot x = 4 \cdot 12 = 48 \]Thus, the constant \( k \) is 48.

Step 3: Set Up the Inverse Variation Equation

Having determined \( k = 48 \), the relationship between \( y \) and \( x \) is given by:\[ y = \frac{48}{x} \]

Step 4: Calculate the New Value of \( y \)

To find \( y \) when \( x = 5 \), substitute \( x = 5 \) into the inverse variation equation:\[ y = \frac{48}{5} = 9.6 \]

Step 5: Conclusion

Thus, when \( x = 5 \), the value of \( y \) is 9.6.

Key Concepts

Constant of Variation

The constant of variation is a key element in understanding inverse relationships between two variables. If a relationship is described as "inverse variation"—which means as one value increases, the other decreases at a proportional rate—the constant of variation remains unchanged.

To find this constant, known as "\(k\)," you multiply the values of the variables when they are first given. In our example from the exercise, we know that when \( y = 4 \) and \( x = 12 \), we find \( k \) using the equation \( y \times x = k \), resulting in \( 48 \).

• This means \( y \times x = 48 \) no matter the values of \( x \) and \( y \) as long as they follow the inverse relationship condition.
• This constant helps us keep the relationship between both variables intact.

Inverse Relationship

Inverse relationships describe how two variables are connected in such a way that their product remains constant. This implies that as one variable increases, the other must decrease in a way that maintains the same product.

For instance, if \( y \) varies inversely as \( x \), then the equation that describes this relationship is \( y = \frac{k}{x} \), where \( k \) is the constant of variation previously found.

• When \( x \) increases, \( y \) must decrease to keep the product \( y \times x \) equal to \( k \).
• An example of this can be seen when moving from \( x = 12 \) (where \( y = 4 \)) to \( x = 5 \). For \( k \) to remain \( 48 \), we calculate \( y \) using the equation \( y = \frac{48}{x} \).
• This results in \( y = 9.6 \) when \( x = 5 \), showing the inverse relationship in action.

Algebraic Equations

Algebraic equations are the backbone of describing inverse variations mathematically. In our exercise, the main algebraic expression used is \( y = \frac{k}{x} \).

This equation helps solve any unknowns when given certain values in inverse variation problems:

• First, identify the constant of variation \( k \) using initial values. Here, \( 4 \times 12 = 48 \).
• Next, apply this constant to find new variable values. For example, when \( x = 5 \), substitute into the equation to find \( y = \frac{48}{5} \).
• This results in \( y = 9.6 \), making it easy to complete and understand how changes in \( x \) affect \( y \).

The flexibility of algebraic equations allows for clear problem-solving paths, making these types of equations indispensable when dealing with inverse relationships.