Problem 61 - Algebra 2

Question

If \(p\) and \(q\) vary inversely, and \(p=10\) when \(q=-4,\) what is \(q\) when \(p=-2 ?\) $$ \begin{array}{lllll}{\text { F. } 20} & {\text { G. } \frac{4}{5}} & {\text { H. }-\frac{4}{5}} & {\text { 1. }-20}\end{array} $$

Step-by-Step Solution

Answer

F.20

Step 1: Determine the constant of variation.

We are given that \(p = 10\) when \(q = -4\). We can use the formula for inverse variation \(pq = k\) to determine the constant \(k\). Thus, \(10*(-4) = k\), which yields \(k = -40\).

Step 2: Solve for \(q\)

We're asked to find the value of \(q\) when \(p = -2\). For this, we can rearrange the formula to \(q = k/p\) and substitute \(k = -40\) and \(p = -2\). Thus we get \(q = -40/-2\), and on simplification, we find that \(q = 20\).

Key Concepts

Constant of Variation

When two variables vary inversely, their product remains constant. This constant is known as the "Constant of Variation." In an inverse variation, like the problem provided, the relationship can be expressed as \( p \times q = k \), where \( k \) is the constant of variation.
Let's break down how to determine \( k \). Given \( p = 10 \) and \( q = -4 \), we use the formula:

• Multiply \( p \) and \( q \) to find \( k \).
• Here, \( 10 \times (-4) = -40 \).
• Thus, \( k = -40 \).

Knowing \( k \) is crucial because it allows us to find other unknowns when one variable changes while keeping the inverse relationship intact.

Inverse Proportionality

Inverse proportionality occurs when an increase in one variable results in a proportional decrease in the other, so their product remains constant. In math terms, if \( p \) and \( q \) are inversely proportional, then \( pq = k \) holds true, where \( k \) is a constant.

• This means if \( p \) increases, \( q \) must decrease to keep \( pq \) the same.
• Conversely, if \( p \) decreases, \( q \) increases, maintaining the relationship.

In our exercise, this concept shows when \( p \) changed from \( 10 \) to \( -2 \), \( q \) was required to adjust to maintain the constant product \( k = -40 \). Thus, such relationships help predict how changes impact each variable.

Algebraic Manipulation

Algebraic manipulation involves using algebraic techniques to solve for variables. When dealing with inverse variations, manipulating formulas allows us to find unknown quantities effectively. Let's consider how this is applied in the problem:1. **Rearranging the Formula:** - Begin with the known formula of inverse variation, \( pq = k \). - To solve for \( q \), rearrange it as \( q = \frac{k}{p} \).2. **Substituting Known Values:** - With \( k = -40 \) and replacing \( p = -2 \), plug these values into the rearranged equation. - This gives \( q = \frac{-40}{-2} = 20 \).By systematically isolating and substituting the variables, algebraic manipulation provides a powerful way to find unknowns, validating the inverse relationship each step of the way.