Solving For X In The Equation 2.5(6x - 4) = 10 + 4(1.5 + 0.5x)

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Introduction

In this article, we will delve into the process of solving for the value of x in the equation 2.5(6x - 4) = 10 + 4(1.5 + 0.5x). This equation involves algebraic manipulation, distribution, and combining like terms to isolate the variable x. Understanding how to solve such equations is a fundamental skill in mathematics, applicable in various fields, including physics, engineering, and economics. This comprehensive guide will walk you through each step, ensuring clarity and understanding, making it easier to tackle similar problems in the future. We will break down the equation, explain the logic behind each operation, and provide insights into why these steps are necessary. By the end of this article, you should have a solid grasp of how to solve this specific equation and be well-equipped to handle more complex algebraic problems.

Step-by-Step Solution

1. Distribute the Constants

The first step in solving the equation is to distribute the constants outside the parentheses to the terms inside. This means multiplying 2.5 by both 6x and -4 on the left side of the equation, and multiplying 4 by both 1.5 and 0.5x on the right side. This process, known as the distributive property, is a cornerstone of algebraic simplification. It allows us to eliminate parentheses and rearrange terms, making the equation easier to manage. By distributing, we effectively break down the complex terms into simpler components, paving the way for combining like terms and isolating the variable. The careful application of the distributive property is crucial for maintaining the balance of the equation and arriving at the correct solution.

  • Left side: 2.5 * (6x - 4) = (2.5 * 6x) - (2.5 * 4) = 15x - 10
  • Right side: 4 * (1.5 + 0.5x) = (4 * 1.5) + (4 * 0.5x) = 6 + 2x

Our equation now becomes:

15x - 10 = 10 + 6 + 2x

2. Combine Like Terms on the Right Side

Next, we simplify the right side of the equation by combining the constant terms. This involves adding 10 and 6 together. Combining like terms is a fundamental algebraic technique that simplifies equations by grouping similar elements. In this case, we are adding the constant terms to reduce the number of individual terms, making the equation more manageable. This step is essential for efficiently isolating the variable and solving for x. By combining like terms, we streamline the equation, making it easier to perform subsequent operations and arrive at the solution. The accuracy in this step is paramount, as any error here can propagate through the rest of the solution process.

  • 10 + 6 = 16

So, the equation is now:

15x - 10 = 16 + 2x

3. Move the Variable Term to One Side

To isolate the variable x, we need to move all terms containing x to one side of the equation. In this case, we can subtract 2x from both sides. This maintains the equality of the equation while consolidating the x terms. Moving the variable term is a critical step in solving for x, as it groups all instances of x together, allowing us to eventually isolate x on one side. This process is governed by the basic principles of algebra, which ensure that performing the same operation on both sides of the equation does not alter the solution. By subtracting 2x, we are one step closer to having x by itself, making the subsequent steps more straightforward.

  • 15x - 2x - 10 = 16 + 2x - 2x
  • 13x - 10 = 16

4. Move the Constant Term to the Other Side

Now, we need to move the constant term (-10) to the right side of the equation. We can do this by adding 10 to both sides. This step isolates the term with x on the left side, further simplifying the equation. Moving the constant term is a crucial maneuver in the process of solving for x, as it separates the variable term from the constant terms, setting the stage for the final isolation of x. The principle behind this step is the same as before: maintaining the balance of the equation by performing the same operation on both sides. By adding 10, we are effectively undoing the subtraction of 10 on the left side, bringing us closer to a solution.

  • 13x - 10 + 10 = 16 + 10
  • 13x = 26

5. Isolate x

Finally, to isolate x, we divide both sides of the equation by the coefficient of x, which is 13. This will give us the value of x. Isolating the variable is the ultimate goal in solving an equation, and this step is where we achieve that. By dividing both sides by the coefficient of x, we are effectively scaling down the equation until x stands alone. This process is the inverse operation of multiplication, and it is the final step in unraveling the equation to reveal the value of x. The precision in this step is critical, as it directly determines the accuracy of the solution.

  • (13x) / 13 = 26 / 13
  • x = 2

Solution

Therefore, the value of x in the equation 2.5(6x - 4) = 10 + 4(1.5 + 0.5x) is x = 2. This solution represents the point at which the left and right sides of the original equation are equal. The careful execution of each step, from distribution to isolation, ensures the accuracy of this result. This value can be substituted back into the original equation to verify its correctness, a practice that is highly recommended in mathematics to confirm the solution and ensure no errors were made during the process. The solution not only answers the question but also solidifies the understanding of the algebraic principles involved.

Verification

To ensure the correctness of our solution, we can substitute x = 2 back into the original equation: 2.5(6x - 4) = 10 + 4(1.5 + 0.5x). This process, known as verification, is a crucial step in problem-solving, as it provides a check against errors that may have occurred during the solution process. By substituting the value back into the original equation, we are essentially testing whether our solution satisfies the conditions set forth by the equation. This step not only validates the solution but also reinforces the understanding of the equation's structure and the relationships between its components. A successful verification instills confidence in the solution and confirms the mastery of the algebraic techniques used.

  • Left side: 2.5(6(2) - 4) = 2.5(12 - 4) = 2.5(8) = 20
  • Right side: 10 + 4(1.5 + 0.5(2)) = 10 + 4(1.5 + 1) = 10 + 4(2.5) = 10 + 10 = 20

Since both sides of the equation are equal when x = 2, our solution is correct.

Common Mistakes to Avoid

When solving algebraic equations, several common mistakes can lead to incorrect solutions. Awareness of these pitfalls is crucial for students to avoid them and improve their problem-solving accuracy. One frequent error is the incorrect distribution of constants across parentheses. It's essential to ensure that the constant outside the parentheses is multiplied by every term inside. Another common mistake is the incorrect combination of like terms, such as adding or subtracting terms that are not similar. Sign errors are also prevalent, particularly when dealing with negative numbers. Keeping track of signs during each step is vital for maintaining the integrity of the equation. Additionally, arithmetic errors in basic operations can derail the entire solution process. Double-checking calculations, especially in multi-step problems, can prevent these errors. Finally, forgetting to perform the same operation on both sides of the equation is a fundamental mistake that violates the principle of equality. Always remember that any operation performed on one side must be mirrored on the other side to maintain balance.

Conclusion

In conclusion, solving the equation 2.5(6x - 4) = 10 + 4(1.5 + 0.5x) involves a series of algebraic steps, including distribution, combining like terms, and isolating the variable. The solution we found, x = 2, satisfies the equation, as verified by substituting it back into the original equation. Mastering these steps is fundamental to success in algebra and other mathematical disciplines. The process of solving algebraic equations is not just about finding the right answer; it's about developing a logical and systematic approach to problem-solving. The skills acquired in solving equations like this one are transferable to many other areas of mathematics and beyond, making it a crucial skill for students to develop. By understanding the underlying principles and practicing regularly, students can build confidence and proficiency in solving algebraic equations, paving the way for more advanced mathematical concepts.