Problem 53 - Elementary Algebra

Question

For exercises 53-56, the formula \(F=\frac{100 S_{u} C_{p}}{S_{p} C_{u}}\) describes the fractional excretion of sodium, \(F\). Is the relationship of the given variables a direct variation or an inverse variation? $$ S_{u}, S_{p} \text {, and } C_{u} \text { are constant; the relationship of } F \text { and } C_{p} \text {. } $$

Step-by-Step Solution

Answer

Direct variation

Step 1: Identify Constants and Variables

Identify the constants and the variables in the given formula. According to the exercise, the constants are: - Constants: - \(S_{u}\) - \(S_{p}\) - \(C_{u}\) - Variables: - \(F\) - \(C_{p}\)

Step 2: Write Down the Given Formula

The formula given is: \[ F = \frac{100 S_{u} C_{p}}{S_{p} C_{u}} \]

Step 3: Simplify the Formula

Since \(S_{u}, S_{p}\), and \(C_{u}\) are constants, simplify the equation to show the relationship between the variables \(F\) and \(C_{p}\): \[ F = \frac{k C_{p}}{1} \] where \(k = \frac{100 S_{u}}{S_{p} C_{u}}\).

Step 4: Determine the Type of Variation

Since \(F\) is proportional to \(C_{p}\) multiplied by a constant \(k\), the relationship can be classified as direct variation. In direct variation, when one variable increases, the other variable also increases proportionally.

Key Concepts

Constant and Variable Identification

In algebra, identifying constants and variables is crucial for understanding and simplifying expressions. Constants are values that do not change. Variables, on the other hand, represent values that can change.

For instance, in the given exercise, the formula is:
\[ F = \frac{100 S_{u} C_{p}}{S_{p} C_{u}} \]
Here, we need to identify which terms are constants and which are variables. According to the exercise:

• Constants: \( S_{u} \), \( S_{p} \), \( C_{u} \)
• Variables: \( F \), \( C_{p} \)

Once the constants and variables are identified, it becomes easier to understand their roles in the equation.

Direct Variation

Direct variation occurs when two variables maintain a consistent ratio. That is, as one variable increases, the other variable increases proportionally. Conversely, if one decreases, the other decreases proportionally.

In the formula \[ F = \frac{100 S_{u} C_{p}}{S_{p} C_{u}} \]
since \( S_{u}, S_{p} \), and \( C_{u} \) are constants, we can simplify the formula by combining these constants into one constant \( k \).
This results in:
\[ F = k C_{p} \]
where \[ k = \frac{100 S_{u}}{S_{p} C_{u}} \]
This simplified equation shows that \( F \) and \( C_{p} \) maintain a direct variation. When \( C_{p} \) increases, \( F \) increases proportionally and vice versa. Understanding direct variation helps in predicting the behavior of variables in algebraic expressions.

Algebraic Simplification

Simplifying algebraic expressions is a fundamental process in algebra. It involves reducing an expression to its simplest form to make it easier to work with.
For the given formula, it starts as:
\[ F = \frac{100 S_{u} C_{p}}{S_{p} C_{u}} \]
Given \( S_{u}, S_{p} \), and \( C_{u} \) are constants, we combined them into a single constant \( k \). This transforms the formula to:
\[ F = k C_{p} \]
where \[ k = \frac{100 S_{u}}{S_{p} C_{u}} \]
By simplifying the formula, you can more easily observe the relationship between the variables. Instead of dealing with multiple constants, you're left with a clear, direct variation equation:

• \( F \) depends directly on \( C_{p} \)
• Changes in \( C_{p} \) lead to proportional changes in \( F \)

Simplification in algebra helps to clarify complex relationships and make analytical work more straightforward.