Problem 40 - Algebra 2

Question

If \(x\) varies directly as \(y\) and \(y=\frac{1}{5}\) when \(x=11,\) find \(x\) when \(y=\frac{2}{5}\)

Step-by-Step Solution

Answer

The value of \( x \) is 22 when \( y = \frac{2}{5} \).

Step 1: Understanding Direct Variation

When a variable varies directly with another, it means that there is a constant ratio between them. The direct variation equation can be expressed as \( x = ky \), where \( k \) is the constant of variation.

Step 2: Determine the Constant of Variation

Given that \( x = 11 \) when \( y = \frac{1}{5} \), substitute these values into the direct variation formula to find \( k \): \[ 11 = k \times \frac{1}{5} \] Multiply both sides by 5 to solve for \( k \): \[ k = 55 \].

Step 3: Substitute to Find New Value of x

Now that we know \( k = 55 \), use it to find \( x \) when \( y = \frac{2}{5} \). Substitute \( y = \frac{2}{5} \) and \( k = 55 \) into the variation equation: \[ x = 55 \times \frac{2}{5} \].

Step 4: Solve for x

Calculate the expression: \[ x = rac{55 imes 2}{5} = rac{110}{5} = 22 \]. So, \( x = 22 \) when \( y = \frac{2}{5} \).

Key Concepts

Constant of Variation

When we talk about direct variation, one of the most important terms you will come across is the **constant of variation**. This is the constant value, often represented by the letter \( k \), that relates two variables which change proportionally. In simpler terms, it tells us how much one variable changes in relation to another.
If you have two variables, say \( x \) and \( y \), which are directly proportional, you can always express their relationship as \( x = ky \). Here, \( k \) remains constant throughout, even as \( x \) and \( y \) take on different values.

• This constant is found by dividing one variable by the other, given that they vary directly. For example, if we know \( x = 11 \) when \( y = \frac{1}{5} \), \( k \) can be found using the equation: \( k = \frac{x}{y} \).
• In this particular example, substituting the known values gives \( k = \frac{11}{\frac{1}{5}} = 11 \times 5 = 55 \).

By understanding this constant, you can predict how the variables will behave with changes, making it incredibly useful in algebraic calculations.

Proportionality

Proportionality is a fundamental concept in mathematics, especially useful for understanding how two quantities relate to each other. When two variables are said to have direct proportionality, a change in one variable always causes a proportional change in the other. But what does that mean?
In the context of our problem, since \( x \) varies directly as \( y \), it means:- For any change in \( y \), there is an exact, constant change in \( x \), governed by our constant of variation, \( k \).- The relationship is linear, represented by the equation \( x = ky \).
This idea facilitates predictions and calculations. Knowing proportional relationships allows us to solve for unknown variables efficiently. For instance, if \( y \) changes in such a way that it is now \( \frac{2}{5} \), we can determine \( x \) as it also changes proportionally: - Given \( k = 55 \), plug \( y = \frac{2}{5} \) back into the equation: \( x = 55 \times \frac{2}{5} \).- Simplifying, we find \( x = \frac{110}{5} = 22 \).
That's the essence of proportionality—consistency in the way one quantity responds to changes in another.

Algebraic Equation

An **algebraic equation** is a mathematical statement that uses letters to represent numbers, showing relationships between quantities. In the case of direct variation, the equation \( x = ky \) is key.
This equation is straightforward in its structure:

• \( x \) is a dependent variable or the output, changing as \( y \) changes.
• \( y \) is an independent variable or the input, which you control by assigning it different values.
• \( k \) is the constant of variation, determining how significantly \( x \) reacts to changes in \( y \).

The beauty of algebraic equations lies in their ability to model real-world situations and solve problems. With the knowledge of \( k \) and one value, finding the other becomes a matter of substituting and solving the equation. In our exercise, knowing that \( k = 55 \) allowed us to compute the new value of \( x \) easily when \( y = \frac{2}{5} \).
By practicing these equations, you sharpen your problem-solving skills and become proficient in predicting outcomes based on an established relationship.