91Ó°ÊÓ

Direct variation describes a relationship where if one variable increases, the other variable also increases proportionally. The formula for direct variation is expressed as:
\( y = kx \)
Here, \( y \) is the dependent variable (the amount of medicine), \( x \) is the independent variable (the weight of the child), and \( k \) is the constant of variation. Whenever both variables change at the same rate, you’re dealing with a direct variation.
To understand this better, think of the physician's prescription of medicine. If the child’s weight increases, the amount of prescribed medicine increases. This straightforward relationship can be calculated using the direct variation formula once you know the constant of variation.

The constant of variation, \( k \), is a crucial part of the direct variation equation \( y = kx \). To determine \( k \), you need one set of corresponding values for \( x \) and \( y \). For instance, we are given that 180 mg of medicine is prescribed for a 40-lb child.
Substituting these values into the formula, we get:
\( 180 = k \times 40 \).
Solving for \( k \), we divide both sides by 40:
\( k = \frac{180}{40} = 4.5 \).
So, the constant of variation \( k \) is 4.5. Now we can use this value to determine the amount of medicine for any given weight.

Using algebraic equations, we can solve for unknown values once we have the constant of variation. The algebraic equation of direct variation is \( y = kx \).
Let’s use this equation to solve the problems:

• For a 50-lb child: \( y = 4.5 \times 50 = 225 \text{ mg} \).
• For a 60-lb child: \( y = 4.5 \times 60 = 270 \text{ mg} \).
• For a 70-lb child: \( y = 4.5 \times 70 = 315 \text{ mg} \).

These equations help us find the correct amount of medicine based on the child’s weight by simply substituting the weight into the equation.
If asked for the weight of a child given an amount of medicine, we rearrange the equation to solve for \( x \):
\( y = kx \)
\( x = \frac{y}{k} \).
For example, if 135 mg of medicine is prescribed:
\( x = \frac{135}{4.5} = 30 \text{ lbs} \).

To solve a direct variation problem, follow these steps:

• Understand the concept of direct variation and the formula \( y = kx \).
• Find the constant of variation, \( k \), using given values.
• Use the equation to solve for the unknown variable by substituting known values.

For our exercise:

1. Identify the relationship: The amount of medicine varies directly with the weight of the child.
2. Calculate \( k \) with given values: \( 180 = k \times 40 \), so \( k = 4.5 \).
3. Solve for the new values using the equation \( y = kx \). For a 50-lb child: \( y = 4.5 \times 50 = 225 \text{ mg} \).

Repeat similar steps for different weights to find corresponding amounts of medicine. If the amount of medicine is given, rearrange the formula to find the weight. For example, finding weight for 135 mg of medicine:
\( x = \frac{135}{4.5} = 30 \text{ lbs} \).
Following these steps ensures you systematically address direct variation problems and find accurate solutions.