Solving The Equation 945/450 = (P[(1 + R/100)^2 - 1]) / (P[(1 + R)/100 - 1]) For R

Jul 13, 2025 by ADMIN 83 views

$Unveiling the Solution to \\frac{945}{450}=\\frac{P[ (1+\\frac{R}{100})^2-1 ]}{P[ (\\frac{1+R}{100})-1 ]} - A Comprehensive Guide$

Introduction

In this comprehensive guide, we embark on a detailed journey to solve the mathematical equation \frac{945}{450}=\frac{P[ (1+\frac{R}{100})^2-1 ]}{P[ (\frac{1+R}{100})-1 ]}. This equation, which may initially appear daunting, unravels into a fascinating exploration of algebraic manipulation and problem-solving techniques. Our goal is to dissect each component of the equation, simplify its structure, and ultimately, determine the value of the unknown variable, R. This process will not only provide a solution to this specific problem but also enhance your understanding of mathematical principles applicable to a wide range of scenarios. We will guide you through each step, ensuring clarity and fostering a deeper appreciation for the elegance of mathematical solutions.

Step-by-Step Solution

Simplifying the Equation

Let's begin by simplifying the given equation: \frac{945}{450}=\frac{P[ (1+\frac{R}{100})^2-1 ]}{P[ (\frac{1+R}{100})-1 ]}. The first step involves reducing the fraction on the left-hand side to its simplest form. Both 945 and 450 are divisible by 45, which simplifies the fraction to \frac{21}{10}. This seemingly small step significantly reduces the complexity of the equation, making it more manageable for subsequent operations.

Now, let's focus on the right-hand side of the equation. Notice that the variable P appears in both the numerator and the denominator. This allows us to cancel out P, further simplifying the equation. By canceling out the common factor P, we eliminate a variable from the equation, making it easier to isolate and solve for R. This simplification is a crucial step in our quest to find the value of R.

Expanding and Rearranging

After canceling out P, our equation transforms into \frac{21}{10} = \frac{(1+\frac{R}{100})^2-1}{(\frac{1+R}{100})-1}. The next step involves expanding the squared term in the numerator. Expanding (1+\frac{R}{100})^2 gives us 1 + \frac{2R}{100} + \frac{R^2}{10000}. This expansion is a critical step as it allows us to work with the individual terms and further simplify the equation. By expanding the squared term, we unveil the quadratic nature of the equation, paving the way for applying appropriate solution methods.

Substituting the expanded term back into the equation, we get \frac{21}{10} = \frac{1 + \frac{2R}{100} + \frac{R^2}{10000} - 1}{\frac{R}{100}}. Notice that the '1' and '-1' in the numerator cancel each other out, which further simplifies the equation. This cancellation is a prime example of how careful observation and simplification can significantly reduce the complexity of a mathematical problem.

Isolating R and Solving

With the equation simplified, we now have \frac{21}{10} = \frac{\frac{2R}{100} + \frac{R^2}{10000}}{\frac{R}{100}}. To eliminate the fractions within the fraction, we can multiply both the numerator and the denominator of the right-hand side by 10000. This operation clears the denominators and transforms the equation into a more workable form. By multiplying by 10000, we effectively scale the equation, making it easier to manipulate and solve.

Multiplying by 10000, we get \frac{21}{10} = \frac{200R + R^2}{100R}. Now, we can further simplify by multiplying both sides of the equation by 100R, which eliminates the denominator on the right-hand side. This step is crucial in isolating the terms containing R and setting up the equation for solving. Multiplying both sides by 100R allows us to work with a simpler, linear equation.

After multiplying by 100R, we have \frac{21}{10} * 100R = 200R + R^2, which simplifies to 210R = 200R + R^2. Rearranging the terms, we obtain a quadratic equation: R^2 - 10R = 0. This quadratic equation is a pivotal point in our solution as it allows us to employ standard methods for finding the roots of a quadratic equation.

Solving the Quadratic Equation

The quadratic equation R^2 - 10R = 0 can be easily solved by factoring out R. Factoring out R gives us R(R - 10) = 0. This factored form provides us with two possible solutions for R: R = 0 or R = 10. These solutions represent the values of R that satisfy the original equation.

However, we must consider the context of the original equation. If R = 0, the denominator of the original equation becomes zero, which is undefined in mathematics. Therefore, R = 0 is an extraneous solution and must be discarded. This step highlights the importance of verifying solutions within the context of the original problem.

Therefore, the only valid solution for R is R = 10. This is the value of R that satisfies the original equation without leading to any mathematical inconsistencies. This solution represents the culmination of our step-by-step simplification and algebraic manipulation.

Alternative Approaches

While the step-by-step solution provides a clear and methodical approach, it's worth noting that alternative methods can also be employed to solve the equation \frac{945}{450}=\frac{P[ (1+\frac{R}{100})^2-1 ]}{P[ (\frac{1+R}{100})-1 ]}. One such approach involves recognizing the structure of the numerator and denominator on the right-hand side as related to compound interest or financial mathematics. This recognition can lead to a more intuitive understanding of the equation and potentially offer a shortcut to the solution.

For instance, the expression (1+\frac{R}{100})^2-1 can be interpreted as the effective interest rate over two periods, while \frac{R}{100} represents the interest rate for a single period. This perspective can guide the simplification process and provide insights into the nature of the problem.

Another approach involves using numerical methods or computer algebra systems to find the solution. These tools can be particularly helpful for more complex equations where algebraic manipulation becomes cumbersome. Numerical methods provide approximate solutions, while computer algebra systems can provide exact solutions. This alternative approach highlights the power of technology in solving mathematical problems.

Practical Applications and Relevance

The equation \frac{945}{450}=\frac{P[ (1+\frac{R}{100})^2-1 ]}{P[ (\frac{1+R}{100})-1 ]} , while presented in a purely mathematical context, has practical applications in various fields, particularly in finance and economics. The structure of the equation closely resembles formulas used in calculating compound interest rates and investment returns. Understanding the solution to this equation can provide valuable insights into financial calculations and decision-making.

For example, the equation can be used to determine the interest rate R that would yield a specific return over a two-year period. This is a common problem in investment analysis, where investors seek to understand the potential returns of different investment options. By solving for R, individuals and organizations can make informed decisions about where to allocate their capital.

Furthermore, the equation illustrates the power of algebraic manipulation in solving real-world problems. The steps involved in simplifying and solving the equation demonstrate how complex problems can be broken down into smaller, more manageable parts. This approach is applicable not only to mathematics but also to problem-solving in general.

Conclusion

In conclusion, solving the equation \frac{945}{450}=\frac{P[ (1+\frac{R}{100})^2-1 ]}{P[ (\frac{1+R}{100})-1 ]} has been a rewarding journey through the realm of algebraic manipulation and problem-solving. By systematically simplifying the equation, expanding terms, and isolating the variable R, we successfully determined that R = 10. This solution not only provides a numerical answer but also enhances our understanding of mathematical principles and their practical applications.

Throughout this guide, we have emphasized the importance of step-by-step problem-solving, careful simplification, and verification of solutions. These principles are fundamental to success in mathematics and other disciplines. By mastering these techniques, you can confidently tackle a wide range of mathematical challenges.

Moreover, we have explored alternative approaches and the relevance of the equation to real-world scenarios, particularly in finance and economics. This broader perspective highlights the interconnectedness of mathematics with other fields and its importance in everyday life. The skills and knowledge gained from solving this equation will undoubtedly serve you well in your future endeavors.