Problem 15 - Algebra Form and Function

Question

In Exercises 14-17, assume the two quantities are directly proportional to each other. If \(p=24\) when \(q=6,\) find \(q\) when \(p\) is 32 .

Step-by-Step Solution

Answer

Answer: When p = 32, q = 8.

Step 1: Find the constant of proportionality (k)

Since we know that \(p=24\) when \(q=6\), we can use the equation \(p=kq\) to find the constant of proportionality (k): \(24 = k \cdot 6\) To solve for k, we can divide both sides of the equation by 6: \(k = \frac{24}{6}\) \(k = 4\)

Step 2: Substitute the value of k back into the equation \(p=kq\)

Now that we have the value of the constant of proportionality (k = 4), we can rewrite the equation as: \(p = 4q\)

Step 3: Solve for q when p is 32

To find the value of q when p is 32, we can substitute 32 for p in the equation and solve for q: \(32 = 4q\) To solve for q, we can divide both sides of the equation by 4: \(q = \frac{32}{4}\) \(q = 8\) So when \(p = 32\), the value of \(q = 8\).

Key Concepts

Proportionality Constant

When two quantities are directly proportional, this means that as one quantity increases, the other increases at a certain constant rate, and vice versa. This constant rate is known as the **proportionality constant** or constant of variation, denoted as \(k\). The relationship can be expressed mathematically as \(p = kq\), where \(p\) and \(q\) are the quantities.

• This constant \(k\) acts as a multiplier that scales one variable in relation to the other.
• Direct proportionality can be visualized as a straight line passing through the origin on a graph of \(p\) versus \(q\).

To find \(k\) from numerical data, you simply divide one variable by the other, provided one of them isn't zero. In our exercise, \(p = 24\) and \(q = 6\). Using the formula \(p = kq\), we calculate \(k\) by dividing 24 by 6, obtaining \(k = 4\). This \(k\) will be used for any \(p\), \(q\) pair obeying this direct relationship.

Linear Equations

Linear equations are foundational to understanding relationships between quantities. A linear equation establishes a straight-line relationship between variables when plotted on a graph. For direct proportionality, our equation \(p = kq\) is a simple form of a linear equation.

• The equation can be rearranged into a familiar format \(y = mx\), where \(y\) and \(x\) are variables and \(m\) represents the slope of the line. In our case, \(p = 4q\) has a slope of 4.
• Linear relationships do not account for constant terms added or subtracted, which is why a direct proportion always passes through the origin (0,0).

Understanding linear equations helps envision how the changes in one variable necessitate changes in the other, maintaining the direct proportion specified by \(k\). This balance maintains a straight path, with the slope being the constant of proportionality \(k\).

Solving Algebraic Equations

One of the crucial aspects of mathematics is being able to solve algebraic equations accurately and swiftly. This means finding the value of unknown variables. Here, our main task is to find \(q\) when \(p\) is given.

• You start with the direct variation equation you've established, such as \(p = 4q\).
• By substituting the known value of \(p = 32\) into the equation, you solve for \(q\) by isolating it on one side of the equation.

The steps to solve for \(q\) are straightforward and involve basic algebra:1. Substitute \(p = 32\) into \(p = 4q\), resulting in the equation \(32 = 4q\).2. To find \(q\), divide both sides by the constant 4, yielding \(q = 8\).Once solved, these algebraic solutions validate the constant relationship between the variables. Solving such equations allows a clear understanding of how quantities intertwine in direct relationships.