Problem 4 - Algebra 1: Concepts and Skills

Question

Find the constant of variation. \(y\) varies directly with \(x,\) and \(y=8\) when \(x=32\)

Step-by-Step Solution

Answer

The constant of variation \(k\) is \(0.25\).

Step 1: Interpret the problem

Given the equation of direct variation \(y = kx\), where \(k\) is the constant of variation. We also have that \(y=8\) and \(x=32\).

Step 2: Substitute the given values of y and x into the formula

We can substitute \(y = 8\) and \(x = 32\) into the formula, we get \(8 = k \times 32\).

Step 3: Solve for k

To solve for \(k\), we then divide both sides of the equation by 32, which gives us \(k = \frac{8}{32}\).

Key Concepts

Constant of Variation

In a direct variation, when one variable changes, the other variable changes at a constant rate. This characteristic is defined by the constant of variation, often denoted as \(k\). The relationship is expressed algebraically as \(y = kx\). In this formula:

• \(y\) is the output or dependent variable.
• \(x\) is the input or independent variable.
• \(k\) is the constant of variation, describing how much \(y\) changes with \(x\).

It's essential to understand that the constant \(k\) remains unchanged throughout the equation, irrespective of the specific values \(x\) and \(y\) take, provided they maintain a direct variation relationship. For example, if \(y = 8\) when \(x = 32\), like in the given exercise, it means \(k\) will maintain this proportional relationship between \(y\) and \(x\). Finding \(k\) ensures we can predict \(y\) for any \(x\) using the same formula.

Algebraic Equation

An algebraic equation in the context of direct variation is a fundamental mathematical expression that defines the relationship between variables. In a direct variation scenario, this equation is expressed as \(y = kx\). Understanding the components of this equation:

• **Equation Structure:** The algebraic form \(y = kx\) indicates a direct proportionality.
• **Variables and Constants:** \(y\) and \(x\) are the variables, and \(k\) is the constant that links them.

Algebraic equations like this one allow us to establish meaningful connections between numbers and enable us to solve for unknowns. It lays the groundwork for algebra, helping solve real-world problems where proportional relationships exist. The equation \(8 = k \times 32\) derived from substituting values illustrates how algebraic manipulation helps us find unknown constants in a given situation.

Solving Equations

Solving equations involves finding the value of unknown variables. For direct variation problems, this often means isolating the constant of variation \(k\). Here's how you can solve for an unknown in the equation \(y = kx\):1. **Substitute Known Values:** Replace \(y\) and \(x\) with the known quantities in your equation. In our example, this gave us \(8 = k \times 32\).2. **Isolate the Unknown:** To find \(k\), the equation must be manipulated so that \(k\) is alone on one side of the equation. This is usually done by performing opposite operations.3. **Perform Operations:** Here, we divided both sides by 32, simplifying the equation to \(k = \frac{8}{32}\), which equals \(\frac{1}{4}\).By completing these steps, we determine the constant \(k\), unlocking the ability to make further calculations and predictions. This method of solving equations is crucial not only in mathematics but also in science and engineering, where understanding relationships and predicting outcomes is key.