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Ch. 3 - Polynomial and Rational Functions
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 4, Problem 13

Write an equation that expresses each relationship. Then solve the equation for y. x varies directly as the cube of z and inversely as y.

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Identify the given variation relationship: "x varies directly as the cube of z and inversely as y." This means x is proportional to \( z^3 \) and inversely proportional to \( y \).
Write the variation equation using a constant of proportionality \( k \): \[ x = \frac{k \cdot z^3}{y} \]
To solve for \( y \), multiply both sides of the equation by \( y \) to get rid of the denominator: \[ x \cdot y = k \cdot z^3 \]
Next, isolate \( y \) by dividing both sides by \( x \): \[ y = \frac{k \cdot z^3}{x} \]
The equation is now solved for \( y \) in terms of \( x \), \( z \), and the constant \( k \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Direct Variation

Direct variation describes a relationship where one variable is proportional to another. If x varies directly as the cube of z, it means x = k * z³ for some constant k, indicating that as z increases, x changes proportionally to z cubed.
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Inverse Variation

Inverse variation means one variable changes in the opposite way to another. Here, x varies inversely as y, so x = k / y, implying that as y increases, x decreases proportionally, and vice versa.
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Solving for a Variable in an Equation

Solving for y involves manipulating the equation algebraically to isolate y on one side. This requires understanding how to rearrange terms, use inverse operations, and simplify expressions to express y explicitly in terms of x and z.
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