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Chapter 1: Problem 69

Solve for the specified variable. $$ L=2 d+3.25(r+R) \quad \text { for } R $$

Short Answer

Expert verified
\( R = \frac{L - 2d - 3.25r}{3.25} \)

Step by step solution

01

Simplify the Equation

The equation given is \( L = 2d + 3.25(r + R) \). First, distribute \(3.25\) over the terms inside the parentheses: \[ L = 2d + 3.25r + 3.25R \]
02

Isolate Terms Involving R

Next, we want to isolate the term involving \( R \). Subtract \( 2d + 3.25r \) from both sides of the equation to get:\[ L - 2d - 3.25r = 3.25R \]
03

Solve for R

Now, divide both sides of the equation by \(3.25\) to solve for \( R \):\[ R = \frac{L - 2d - 3.25r}{3.25} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isolating Variables
When solving an equation for a specific variable, such as finding \( R \) in the equation \( L = 2d + 3.25(r + R) \), we are performing a task called isolating the variable. This process involves manipulating the equation until the variable of interest is by itself on one side.

Here鈥檚 how you can approach it:
  • First, identify the variable that you need to isolate; in our case, it鈥檚 \( R \).
  • Use various algebraic operations such as addition, subtraction, multiplication, or division to move other terms to the opposite side of the equation.
  • Our goal is to express \( R \) in terms of the other variables and constants. This makes it easier to evaluate \( R \) if the other variable values are known.

In the provided exercise, we need to carefully move other terms involving other variables or constants away from \( 3.25R \). It鈥檚 important to perform the same operation on both sides of the equation to maintain equality, a key principle in algebra.
Distributive Property
Applying the distributive property is crucial when you encounter brackets or parentheses with terms inside them in an equation. The distributive property states that \( a(b+c) = ab + ac \).

In our original equation \( L = 2d + 3.25(r + R) \), we apply the distributive property to eliminate the parentheses, thereby allowing us to simplify the equation.

Here鈥檚 a step-by-step way to use the distributive property:
  • Multiply each term inside the parentheses by the factor outside. In the equation, \( 3.25 \) multiplies both \( r \) and \( R \).
  • Rewrite the equation as \( L = 2d + 3.25r + 3.25R \). Now, the equation does not contain parentheses, and we can proceed to isolate \( R \).

The distributive property simplifies the equation, making it easier to manipulate, and eventually isolate the desired variable.
Algebraic Manipulation
Algebraic manipulation involves using various algebraic techniques to transform an equation into a more useful or simpler form. This is essential when solving for a specific variable.

In the exercise, algebraic manipulation is used in several steps:
  • First, after distributing, you identify terms to move. Here, terms not involving \( R \) need to be transferred to the other side.
  • Subtract \( 2d + 3.25r \) from both sides to isolate terms containing \( R \). This leads to \( L - 2d - 3.25r = 3.25R \).
  • Finally, divide both sides by \( 3.25 \) to completely solve for \( R \), resulting in: \[ R = \frac{L - 2d - 3.25r}{3.25} \]

Each step of the manipulation is aimed at getting closer to having \( R \) completely isolated. With practice, these steps become intuitive, simplifying complex equations into understandable forms.

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