Problem 27 - Algebra 2

Question

Suppose \(y\) varies jointly as \(x\) and \(z .\) Find \(y\) when \(x=\frac{1}{2}\) and \(z=6,\) if \(y=45\) when \(x=6\) and \(z=10\).

Step-by-Step Solution

Answer

The value of \( y \) is 2.25.

Step 1: Understanding Joint Variation

When we say that a variable varies jointly as two others, we mean that there is a constant of proportionality, such that the variable equals this constant multiplied by the two others. Here, that means there is a constant, say \( k \), where \( y = kxz \). We need to find this constant based on the information given.

Step 2: Finding the Constant of Variation

Given that \( y = 45 \) when \( x = 6 \) and \( z = 10 \), we plug these values into the equation for joint variation: \( 45 = k imes 6 imes 10 \). Simplifying gives \( 45 = 60k \), which means \( k = \frac{45}{60} = \frac{3}{4} \).

Step 3: Substitute and Solve for New Conditions

Now that we know \( k = \frac{3}{4} \), we use it to find \( y \) when \( x = \frac{1}{2} \) and \( z = 6 \). Substitute these into the joint variation formula: \( y = \frac{3}{4} \times \frac{1}{2} \times 6 \).

Step 4: Calculate the Value of y

Simplify the equation to find \( y \): \( y = \frac{3}{4} \times \frac{1}{2} \times 6 \). First, multiply the fractions: \( \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} \). Then multiply by 6: \( \frac{3}{8} \times 6 = \frac{18}{8} = \frac{9}{4} = 2.25 \).

Key Concepts

Proportionality Constant

When dealing with joint variation, grasping the idea of the proportionality constant is crucial. Joint variation describes a scenario where a variable depends on two or more other variables. This relationship can often be expressed as a product of these variables with a constant factor, also known as the proportionality constant.

• In our problem, we express joint variation as: \( y = kxz \)
• The goal is to identify the constant \( k \)

Given the data \( y = 45 \), \( x = 6 \), and \( z = 10 \), we identify \( k \) by substituting these values into the equation.
Solving \( 45 = k \times 6 \times 10 \) simplifies to \( 45 = 60k \), and thus \( k = \frac{3}{4} \).
The constant \( k \) ensures that any set of \( x \) and \( z \) values maintains consistent proportionality, keeping \( y \) sensible and linear to changes in \( x \) and \( z \). Understanding how to calculate and use this constant is vital for accurately predicting the value of \( y \) under different conditions.

Solving Equations

Solving equations is necessary to understand variations like joint variation. Once you establish a relationship, the next step is to manipulate the equation to find unknowns by using algebraic techniques.
In this example, after establishing \( 45 = 60k \), solving for \( k \) involves simple division; \( k = \frac{45}{60} = \frac{3}{4} \).

• Key steps include isolating variables and simplifying expressions.
• When substituting new values, follow the same process to keep consistency.

To find \( y \) for a new set of values \( x = \frac{1}{2} \) and \( z = 6 \), substitute into \( y = \frac{3}{4} \times \frac{1}{2} \times 6 \).
Simplifying the equation involves multiplying the fractions and whole numbers, following steps like:
1. \( \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} \)
2. \( \frac{3}{8} \times 6 = \frac{18}{8} = \frac{9}{4} = 2.25 \)
This results in \( y \)'s new value, after correctly solving the equation.

Variables

Variables are fundamental to formulating any equation involving joint variation. They represent quantities that can change and are governed by certain mathematical relationships.
In this example:

• \( y \) is the dependent variable; its value varies based on \( x \) and \( z \).
• \( x \) and \( z \) are independent variables; they influence but are not affected directly by \( y \).

By understanding how these variables interact through the proportionality constant \( k \), we predict how changes in \( x \) and \( z \) affect \( y \).
This relationship highlights the significance of identifying and manipulating variables correctly. Remember:
• Independent variables like \( x \) and \( z \) are inputs.
• The dependent variable \( y \) is the output we aim to discover.
Mastering the understanding of these types of variables in equations is crucial for solving real-world problems where predicting outcomes is essential.