Solving For C In The Equation A = (2/3)(B + C)

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In the realm of mathematics and algebra, manipulating equations to isolate specific variables is a fundamental skill. This article delves into the process of solving for C in the equation A = (2/3)(B + C). We will embark on a step-by-step journey, demystifying each stage and illuminating the underlying principles. Our goal is to provide a comprehensive understanding that empowers you to confidently tackle similar algebraic challenges.

Understanding the Equation

Before we dive into the solution, let's first dissect the equation A = (2/3)(B + C). This equation expresses a relationship between four variables: A, B, and C. A is equal to two-thirds of the sum of B and C. Our mission is to isolate C on one side of the equation, thereby expressing C in terms of A and B. This process involves applying algebraic operations strategically to both sides of the equation while maintaining equality. The beauty of algebra lies in its ability to transform equations without altering their fundamental meaning, allowing us to unveil hidden relationships between variables.

Step-by-Step Solution

  1. Eliminate the Fraction: The presence of the fraction 2/3 can initially appear daunting. However, we can readily eliminate it by multiplying both sides of the equation by its reciprocal, 3/2. This operation effectively cancels out the fraction on the right side, simplifying the equation. Remember, the golden rule of algebra dictates that any operation performed on one side of the equation must be mirrored on the other side to preserve equality. Multiplying both sides by 3/2 yields:

    (3/2) * A = (3/2) * (2/3)(B + C)

    This simplifies to:

    (3/2)A = B + C

  2. Isolate C: Now, our equation has transformed into a more manageable form. To isolate C, we need to eliminate B from the right side. This can be achieved by subtracting B from both sides of the equation. This operation effectively moves B to the left side, leaving C isolated on the right. Subtracting B from both sides, we get:

    (3/2)A - B = B + C - B

    This simplifies to:

    (3/2)A - B = C

  3. Final Solution: We have successfully isolated C! The equation C = (3/2)A - B expresses C in terms of A and B. This is our final solution. We can confidently state that C is equal to three-halves of A minus B. This solution provides a direct formula for calculating C given the values of A and B.

Alternative Representation

While C = (3/2)A - B is a perfectly valid solution, we can further refine it by combining the terms on the left side into a single fraction. This involves finding a common denominator for (3/2)A and B. Since B can be expressed as (2/2)B, we can rewrite the equation as:

C = (3/2)A - (2/2)B

Now, we can combine the terms over the common denominator of 2:

C = (3A - 2B) / 2

This alternative representation, C = (3A - 2B) / 2, is equally valid and may be preferred in certain contexts. It presents C as a single fraction, which can be advantageous for further calculations or analysis.

Practical Applications and Examples

The ability to solve for a specific variable in an equation has wide-ranging applications across various fields. Let's explore some practical scenarios where solving for C in the equation A = (2/3)(B + C) might prove invaluable.

1. Geometry

Imagine a scenario where A represents the area of a trapezoid, B represents the length of one of the parallel sides, and C represents the length of the other parallel side. The equation A = (2/3)(B + C) could be a simplified representation of the formula for the area of a trapezoid. If we know the area (A) and the length of one parallel side (B), we can use our derived formula, C = (3/2)A - B, to calculate the length of the other parallel side (C). This is a concrete example of how algebraic manipulation can be applied to solve geometric problems.

2. Physics

In physics, equations often describe relationships between physical quantities. Suppose A represents the final velocity of an object, B represents its initial velocity, and C is related to the acceleration and time. The equation A = (2/3)(B + C) might arise as a simplified model of a physical system. If we know the final velocity (A) and the initial velocity (B), we can use the formula C = (3/2)A - B to determine the value of C, which could then be used to infer information about the acceleration or time involved in the object's motion. This highlights the versatility of algebraic techniques in analyzing physical phenomena.

3. Everyday Life

Even in everyday situations, the ability to manipulate equations can be surprisingly useful. Consider a scenario where A represents the total cost of a project, B represents the cost of materials, and C represents the cost of labor. The equation A = (2/3)(B + C) might be a simplified representation of how the total cost is calculated. If you know the total cost (A) and the cost of materials (B), you can use the formula C = (3/2)A - B to determine the cost of labor (C). This simple example demonstrates how algebraic skills can be applied to personal finance and budgeting.

Illustrative Examples

To solidify your understanding, let's work through a few numerical examples:

Example 1:

If A = 10 and B = 4, find C.

Using the formula C = (3/2)A - B:

C = (3/2) * 10 - 4

C = 15 - 4

C = 11

Example 2:

If A = 20 and B = 8, find C.

Using the formula C = (3/2)A - B:

C = (3/2) * 20 - 8

C = 30 - 8

C = 22

Example 3:

If A = 15 and B = 3, find C.

Using the formula C = (3/2)A - B:

C = (3/2) * 15 - 3

C = 22.5 - 3

C = 19.5

These examples demonstrate how the derived formula can be readily applied to calculate the value of C for different values of A and B. By substituting the given values into the formula, we can quickly and accurately determine the corresponding value of C.

Common Mistakes and Pitfalls

When solving algebraic equations, it's crucial to be mindful of common mistakes and pitfalls that can lead to incorrect solutions. Let's examine some frequent errors that students often encounter when solving for C in the equation A = (2/3)(B + C), and how to avoid them.

1. Incorrectly Distributing the Fraction

A common mistake is to misapply the distributive property when dealing with the fraction 2/3. Remember, the fraction must be multiplied by both B and C inside the parentheses. A frequent error is to only multiply the fraction by B or C, but not both. To avoid this, always ensure that you distribute the fraction correctly:

(2/3)(B + C) = (2/3)B + (2/3)C

2. Forgetting to Perform the Same Operation on Both Sides

The fundamental principle of algebra is that any operation performed on one side of the equation must be mirrored on the other side to maintain equality. A common mistake is to perform an operation, such as multiplying by 3/2 or subtracting B, on only one side of the equation. This leads to an imbalance and an incorrect solution. Always remember to apply the same operation to both sides to preserve the equation's integrity.

3. Errors in Arithmetic

Simple arithmetic errors, such as incorrect multiplication or subtraction, can derail the entire solution process. It's essential to double-check your calculations at each step to ensure accuracy. Pay close attention to signs (positive and negative) and the order of operations (PEMDAS/BODMAS). A small arithmetic error can cascade through the solution, leading to a significantly wrong answer.

4. Incorrectly Isolating C

The goal is to isolate C on one side of the equation. A common mistake is to stop the solution process prematurely or to perform the isolation steps incorrectly. For instance, students might subtract B from only the left side or fail to completely eliminate the terms surrounding C. Ensure that you meticulously follow the steps outlined in the solution process to isolate C effectively.

5. Not Simplifying the Final Answer

While obtaining a correct solution is paramount, it's also important to simplify the final answer whenever possible. This might involve combining like terms, reducing fractions, or expressing the answer in an alternative form. Leaving the answer in an unsimplified form can sometimes obscure the underlying relationship between the variables. In our case, we saw how the solution C = (3/2)A - B could be further simplified to C = (3A - 2B) / 2.

Tips to Avoid Mistakes

  1. Write Clearly: Maintain a neat and organized solution process. This makes it easier to track your steps and identify potential errors.
  2. Show All Steps: Don't skip steps, even if they seem trivial. Writing out each step explicitly reduces the chances of making a mistake.
  3. Double-Check Your Work: After completing the solution, review each step carefully to ensure accuracy.
  4. Substitute the Solution: Substitute your final solution back into the original equation to verify that it holds true. This is a powerful way to catch errors.
  5. Practice Regularly: Consistent practice is the key to mastering algebraic manipulation. The more you practice, the more comfortable and confident you will become.

By being aware of these common mistakes and adhering to the tips provided, you can significantly improve your accuracy and proficiency in solving algebraic equations.

Conclusion

Solving for C in the equation A = (2/3)(B + C) is a fundamental exercise in algebraic manipulation. By systematically applying algebraic operations, we successfully isolated C and derived the formula C = (3/2)A - B (or its equivalent form, C = (3A - 2B) / 2). This process highlights the power of algebra in transforming equations and revealing relationships between variables. Furthermore, we explored practical applications of this equation in geometry, physics, and everyday scenarios, demonstrating the versatility of algebraic skills. By understanding the step-by-step solution, recognizing common mistakes, and practicing regularly, you can confidently tackle similar algebraic challenges and unlock the beauty of mathematical problem-solving.