Problem 53 - Algebra 2

Question

Suppose that \(y\) varies directly with \(x\) and inversely with \(z^{2},\) and \(x=48\) when \(y=8\) and \(z=3 .\) Find \(x\) when \(y=12\) and \(z=2\) .

Step-by-Step Solution

Answer

The value of \(x\) when \(y = 12\) and \(z = 2\) is 24.

Step 1: Translate the problem into equation form

The direct and inverse variation statement can be written as: \(y = k \times \frac{x}{z^2}\)

Step 2: Find the constant \(k\)

We are given a scenario where \(x = 48\), \(y = 8\), and \(z = 3\). Substitute these values into the equation and solve for \(k\). \[8 = k \times \frac{48}{3^2} \rightarrow k = \frac{8 \times 3^2}{48} \rightarrow k = 2\] So, the constant \(k\) is found to be 2.

Step 3: Find \(x\) when \(y = 12\) and \(z = 2\)

Now, we are asked to find \(x\) when \(y = 12\) and \(z = 2\). Substitute \(k = 2\), \(y = 12\), and \(z = 2\) into the equation and solve for \(x\). \[12 = 2 \times \frac{x}{2^2} \rightarrow x = \frac{12 \times 2^2}{2} \rightarrow x = 24\] Thus, the value of \(x\) when \(y = 12\) and \(z = 2\) is found to be 24.

Key Concepts

Variation Constant

In problems involving both direct and inverse variation, the key is finding a special number called the 'variation constant', represented as \(k\). This number forms the bridge between the variables in the equation. In the problem we are addressing, the relationship between \(y\), \(x\), and \(z\) can be modeled by the equation \(y = k \times \frac{x}{z^2}\). Here, \(k\) remains unchanged irrespective of the values of \(x\), \(y\), and \(z\).
By substituting known values into the equation, it becomes possible to solve for \(k\). In our example, with \(x = 48\), \(y = 8\), and \(z = 3\), once these values are plugged into the equation, we calculate \(k\) to be 2. Knowing \(k\) is crucial because it allows us to explore new scenarios and solve for other unknowns efficiently.

Solving Equations

Solving equations in this context involves substituting known quantities into the equation and systematically deducing unknown quantities. We are often given some values for \(x\), \(y\), and \(z\), along with a relationship that connects these through the variation constant \(k\).
To find an unknown, we input the known values into our equation, like \(y = k \times \frac{x}{z^2}\), and rearrange to solve for the unknown. For example, if you need to find \(x\) when \(y = 12\), \(z = 2\), and knowing \(k = 2\), you substitute into the equation and solve for \(x\). This systematic approach ensures accuracy and helps manage complex relationships easily.

Mathematical Modeling

Mathematical modeling is about creating equations that accurately represent real-world relationships. In our problem, the relationship is dictated by both direct and inverse variation, represented as \(y = k \times \frac{x}{z^2}\). This equation simplifies complex interactions between \(y\), \(x\), and \(z\) into a manageable form, allowing for accurate predictions and calculations.
By understanding how to construct and interpret such models, you open doors to solving a vast array of problems in physics, engineering, and economics. This model, which encapsulates varying influences on \(y\) from \(x\) and \(z\), is both precise and versatile, offering insights into proportional reasoning across different disciplines.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate and solve for unknown values. In our case, once we have the equation \(y = k \times \frac{x}{z^2}\), we need to solve for \(x\).
This involves rearranging terms and performing operations to isolate \(x\) on one side of the equation. For instance, for \(12 = 2 \times \frac{x}{2^2}\), we clear the fraction by multiplying both sides by \(2^2\), then divide by 2 to solve for \(x\). Such manipulations always adhere to algebraic principles, maintaining equation balance. This skill is essential to solving many different types of problems and falls under fundamental algebra skills required in further mathematical studies.