Problem 62 - Algebra 2

Question

Which equation shows that \(z\) varies directly with the square of \(x\) and inversely with the cube of \(y ?\) $$ \begin{array}{llll}{\text { A. } z=\frac{x^{2}}{y^{3}}} & {\text { B. } z=\frac{x^{3}}{y^{2}}} & {\text { C. } z=\frac{y^{2}}{x^{3}}} & {\text { D. } z=\frac{y^{3}}{x^{2}}}\end{array} $$

Step-by-Step Solution

Answer

The correct equation is Option A:\(z=\frac{x^{2}}{y^{3}}\).

Step 1: Identifying the correct formula

In direct variation, the variable in question is in the numerator of the right side of the equation, and in inverse variation, the variable in question is in the denominator. Hence, an equation which exhibits \(z\) varying directly as the square of \(x\) and inversely as the cube of \(y\) would look like \(z = \frac{x^2}{y^3}\).

Step 2: Comparing the deduced formula with the given options

Having determined the correct form of the equation, we now compare this with the four given options. Only Option A matches the deduced formula, thus Option A is the correct choice.

Key Concepts

Quadratic Relationship

A quadratic relationship refers to a situation where one variable, say \( z \), is proportional to the square of another variable, like \( x^2 \). In this exercise, we are looking at how \( z \) depends directly on the square of \( x \). When a variable is said to vary directly with the square of another, it means it increases with the square of the other variable. Therefore, if \( x \) doubles, the output \( z \) becomes four times greater because the square impacts the rate of increase. Understanding quadratic relationships is crucial in many fields such as physics and engineering, where acceleration (a square of velocity) plays a significant role. Remember, in direct variation, \( z = k \, x^2 \), where \( k \) is a constant. In this given problem, the constant is represented by \( 1 \), which is implicit.

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are both polynomials. They are prevalent in algebra especially when dealing with direct and inverse variations. In problem-solving, especially in this exercise, it’s crucial to understand how to manipulate these expressions. Here, the correct equation \( z = \frac{x^2}{y^3} \) is a rational expression implying that \( z \) is affected by changes in both \( x \) and \( y \) through division. The numerator \( x^2 \) indicates a direct relationship, while the denominator \( y^3 \) shows an inverse relation with \( z \). This means that as \( y \) increases (while \( x \) remains constant), \( z \) will decrease due to the increase in the denominator of the fraction. Understanding rational expressions is essential in algebra as it helps in simplifying complex equations and solving for unknowns effectively.

Problem Solving in Algebra

Problem-solving in algebra often involves finding patterns, deducing relationships, and sometimes solving equations. In exercises like this one, understanding the terms "directly" and "inversely" forms the foundation of accurately forming and identifying the right equations. To solve problems effectively:

• Identify the relation type: Direct or Inverse.
• Translate these relations into mathematical expressions. For instance, direct relations typically place the variable in the numerator, and inverse relations in the denominator.
• Compare your expression with given options or form an equation if needed.

Solving algebraic problems requires a strategic approach. Break down the question—what variable impacts what? Examine constant terms and verify relationships through calculations or substitutions. Practice applying these steps to improve problem-solving skills in algebraic contexts.