Problem 26 - Beginning and Intermediate Algebra

Question

Solve. If \(q\) varies directly as \(p,\) and \(q=10\) when \(p=4,\) find \(q\) when \(p=10\)

Step-by-Step Solution

Answer

When \(p = 10\), the value of \(q\) is \(25\).

Step 1: Write the direct variation formula

We are given that \(q\) varies directly as \(p\). The formula for direct variation is: \[q = k \times p\] where k is the constant of variation.

Step 2: Determine the constant of variation (k)

We are given that \(q = 10\) when \(p = 4\). We can plug these values into the direct variation formula and solve for k: \[10 = k \times 4\] Now, solve for k: \[k = \frac{10}{4} = \frac{5}{2}\] So, the constant of variation (k) is \(\frac{5}{2}\).

Step 3: Find the value of q when p = 10

We now have the constant of variation, k, and are given that p = 10. We can use the direct variation formula to find q: \[q = \frac{5}{2} \times 10\] Multiply: \[q = 5 \times 5 = 25\] So, when \(p = 10\), the value of \(q\) is \(25\).

Key Concepts

Constant of Variation

In mathematics, when we talk about direct variation, the **constant of variation** is a key concept. It represents a fixed number that relates two directly proportional variables. When one variable changes, the other changes in a predictable way.
In our problem, we are given that \( q \) varies directly as \( p \). The direct variation formula is expressed as:

• \( q = k \times p \)

Here, \( k \) is the constant of variation. It's the factor by which \( p \) is multiplied to get \( q \). To find \( k \), you use known values of \( p \) and \( q \) from a given situation. By substituting \( q = 10 \) and \( p = 4 \), we found that \( k = \frac{5}{2} \). Once you have \( k \), you can use it to determine the relationship between \( q \) and \( p \) for different values, ensuring the variables increase together in a consistent way.

Algebraic Equations

**Algebraic equations** are mathematical statements that express the equality between two expressions. In our problem, we start with the direct variation formula \( q = k \times p \), which is an algebraic equation involving the variables \( q \) and \( p \).
Understanding algebraic equations involves knowing how to manipulate these equations to find unknown values. When given \( q = 10 \) and \( p = 4 \) to find the constant \( k \), we set up the equation:

• \( 10 = k \times 4 \)

To solve for \( k \), algebraic manipulation leads us to:

• \( k = \frac{10}{4} = \frac{5}{2} \)

Once \( k \) is determined, the equation can be used to solve for other values, such as finding \( q \) when \( p = 10 \). This practical use of algebraic equations helps solve real-world problems by establishing and manipulating relationships through equations.

Proportional Relationships

A **proportional relationship** is when two quantities maintain a constant ratio. In the case of direct variation, \( q \) and \( p \) are proportional, meaning as one changes, the other changes at a constant rate.
When exploring proportional relationships, the constant of variation \( k \) indicates how \( q \) changes with respect to \( p \). For instance, the relationship \( q = \frac{5}{2} \times p \) shows that for every unit increase in \( p \), \( q \) increases by \( \frac{5}{2} \). This is a linear relationship, and can be visualized as a straight line graph passing through the origin.

• When \( p = 4 \), \( q = 10 \)
• When \( p = 10 \), \( q = 25 \)

These examples demonstrate how proportional relationships through direct variation provide a predictable and reliable way to explore changes in related variables. Understanding this concept gives insight into many natural and mathematical phenomena where one quantity depends directly on another.