Comprehensive Guide to Direct Variation Equations

Interpreting Lexical Cues: In algebraic word problems, the phrase "varies directly" serves as a specific linguistic trigger. As soon as this phrase is identified, it indicates a direct proportional relationship involving a constant.
The Meaning of the Constant $k$ : The transcript explains that the phrase "varies directly" effectively means "equals $k$ " within the context of the relationship between variables.
The Relationship of Variables: When the prompt states " $y$ varies directly as $x$ ," it establishes the fundamental starting point for the mathematical model where $x$ and $y$ are related by the constant $k$ .
The Relationship as an Equation: The conceptual phrasing " $y$ varies directly as $x$ " is mathematically translated into the formula $y = kx$ .

Initial Data Requirement: Direct variation problems typically begin by providing two specific pieces of information: an initial value for $y$ and its corresponding value for $x$ .
Step 1: Solving for the Constant of Variation: Once the initial pair of values ( $x$ and $y$ ) is identified, the first objective is to solve for the unknown constant, $k$ .
Step 2: Establishing the Specific Model: After calculating the value of $k$ , it is inserted back into the general direct variation equation. This results in an equation that is specifically tailored to the unique parameters of the problem.
Step 3: Final Computation: With the specific equation established, the student can solve for any given scenario. If a new value for $x$ is provided, the equation can be used to determine the new value for $y$ . Conversely, if a value for $y$ is provided, the student can solve for $x$ .

Scenario Description: The task is to calculate the value of $y$ given that $y$ varies directly as $x$ .
Given Initial Conditions:
- $y = 15$
- $x = -5$
Objective: Find the value of $y$ when $x = 7$ .
The Solution Process:
- First, the known values are substituted into the direct variation template ( $y = kx$ ) to isolate the constant. This creates the equation: $15 = k \times (-5)$ .
- To solve for $k$ , the equation is simplified by dividing both sides by $-5$ : $\frac{15}{-5} = k$ .
- The resulting value for the constant is $k = -3$ .
- Next, the specific equation for this problem is defined: $y = -3x$ .
- Finally, to find the specific answer for the goal condition ( $x = 7$ ), the value is substituted into the specific equation: $y = -3 \times 7$ .
- The final calculation yields: $y = -21$ .

Instructional Continuity: Following the step-by-step walkthrough of the first problem, the instructor transitions to student-led practice.
Audience Prompt: The instructor asks the students directly, "Do you guys wanna try number two and number three?" indicating that the next phase of the lesson involve independent or guided practice on subsequent problems of the same nature.