Solving For Variables And Equations A Mathematical Exploration

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In the realm of mathematics, the ability to manipulate formulas and solve equations is paramount. These skills not only form the bedrock of various mathematical disciplines but also find extensive applications in diverse fields like physics, engineering, and economics. This article delves into two fundamental mathematical problems: making a variable the subject of a formula and solving a linear equation. Through a detailed exploration of these problems, we aim to enhance your understanding of algebraic manipulation and problem-solving strategies.

(a) Making 'n' the Subject of the Formula: I = (nε) / (R + nr)

In this section, we embark on a journey to isolate the variable 'n' in the given formula, I = (nε) / (R + nr). This process involves a series of algebraic manipulations designed to rearrange the equation such that 'n' stands alone on one side. The formula represents a common scenario in electrical circuits, where 'I' denotes current, 'ε' represents electromotive force (EMF), 'R' is the external resistance, 'r' is the internal resistance, and 'n' signifies the number of cells.

Step-by-Step Solution

  1. Eliminate the Fraction: To begin, we aim to eliminate the fraction by multiplying both sides of the equation by the denominator (R + nr). This yields:

    I * (R + nr) = nε

  2. Expand the Left Side: Next, we distribute 'I' on the left side of the equation to remove the parentheses:

    IR + Inr = nε

  3. Gather Terms with 'n': Our objective is to isolate 'n', so we need to gather all terms containing 'n' on one side of the equation. Let's subtract 'Inr' from both sides:

    IR = nε - Inr

  4. Factor out 'n': Now, we can factor out 'n' from the right side of the equation:

    IR = n(ε - Ir)

  5. Isolate 'n': Finally, to make 'n' the subject, we divide both sides by (ε - Ir):

    n = IR / (ε - Ir)

Therefore, we have successfully made 'n' the subject of the formula. This transformed equation allows us to calculate the number of cells ('n') required in a circuit, given the current (I), external resistance (R), internal resistance (r), and electromotive force (ε). Understanding the manipulation of formulas is crucial for solving real-world problems in various scientific and engineering disciplines.

Importance of Subject Manipulation

Making a variable the subject of a formula is a fundamental skill in algebra. It allows us to rearrange equations to solve for a specific variable of interest. This is particularly useful in physics, engineering, and other sciences where formulas are used to model real-world phenomena. By manipulating formulas, we can gain insights into the relationships between different variables and make predictions about the behavior of systems. In the context of the given formula, making 'n' the subject allows us to determine the number of cells needed to achieve a desired current in a circuit, given the other parameters. This practical application highlights the importance of mastering this algebraic technique.

(b) Solving the Linear Equation: 2(2x - 5) = 3(x - 6)

In this section, we tackle the task of solving a linear equation. Linear equations are fundamental to algebra and have wide-ranging applications. Our goal is to find the value of 'x' that satisfies the equation 2(2x - 5) = 3(x - 6). This involves a series of algebraic steps to isolate 'x' on one side of the equation.

Step-by-Step Solution

  1. Expand Both Sides: The first step is to eliminate the parentheses by distributing the numbers on both sides of the equation:

    4x - 10 = 3x - 18

  2. Gather 'x' Terms: Next, we want to collect all terms containing 'x' on one side of the equation. Let's subtract '3x' from both sides:

    4x - 3x - 10 = 3x - 3x - 18

x - 10 = -18

  1. Isolate 'x': To isolate 'x', we add 10 to both sides of the equation:

    x - 10 + 10 = -18 + 10

x = -8

Therefore, the solution to the equation 2(2x - 5) = 3(x - 6) is x = -8. This means that when we substitute -8 for 'x' in the original equation, both sides will be equal. Mastering the process of solving linear equations is a crucial step in developing algebraic proficiency.

Importance of Solving Linear Equations

Solving linear equations is a cornerstone of algebra. These equations appear in numerous mathematical contexts and are used to model a wide variety of real-world situations. From determining the break-even point in business to calculating the trajectory of a projectile in physics, linear equations provide a powerful tool for problem-solving. The ability to solve these equations accurately and efficiently is essential for success in mathematics and related fields. The step-by-step approach outlined above provides a systematic method for tackling linear equations, ensuring that each step is logically sound and leads to the correct solution.

Conclusion: Mastering Algebraic Techniques

In this article, we have explored two fundamental algebraic techniques: making a variable the subject of a formula and solving a linear equation. These skills are not only essential for success in mathematics but also have broad applications in various scientific, engineering, and economic disciplines. By understanding the underlying principles and mastering the step-by-step procedures, you can confidently tackle a wide range of mathematical problems. The ability to manipulate formulas and solve equations is a powerful tool that will serve you well in your academic and professional pursuits. Remember, practice is key to mastering these techniques, so continue to challenge yourself with new problems and explore the fascinating world of mathematics.

By delving into these problems, we have reinforced the importance of algebraic manipulation and problem-solving strategies. These skills are not just confined to the classroom; they are applicable in countless real-world scenarios, making them indispensable tools in any analytical toolkit. Keep practicing, keep exploring, and keep pushing the boundaries of your mathematical understanding.