Equations With Shared Solution For X As (3/5)(30x - 15) = 72

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In the realm of algebraic equations, finding solutions that satisfy a given equation is a fundamental skill. This article delves into the process of identifying equations that share the same solution for x as the equation (3/5)(30x - 15) = 72. We will explore various techniques for simplifying and manipulating equations, ultimately leading to the selection of three equivalent options. This exploration will not only enhance your understanding of equation solving but also demonstrate the interconnectedness of algebraic expressions. Let's embark on this mathematical journey together, unraveling the intricacies of equation equivalence and solidifying your problem-solving prowess. Through detailed explanations and step-by-step solutions, we aim to provide a comprehensive guide that empowers you to confidently tackle similar challenges in the future. This article is designed to be a valuable resource for students, educators, and anyone with a passion for mathematics. We encourage you to engage with the content, ask questions, and explore the fascinating world of algebra.

Understanding the Core Equation: (3/5)(30x - 15) = 72

Before we begin to solve, let's deeply understand the core equation: (3/5)(30x - 15) = 72. This equation presents a linear relationship involving a variable x. Our primary goal is to isolate x and determine its value. To achieve this, we'll systematically simplify the equation using algebraic principles. The equation involves a fraction, parentheses, and constants, each requiring careful manipulation. By understanding the order of operations and applying the distributive property, we can effectively simplify the equation. This process not only helps us find the solution for x but also provides valuable insights into the structure of the equation. Understanding the core equation is paramount as it serves as the foundation for comparing and contrasting with other equations. We will break down the equation into smaller, manageable steps, ensuring clarity and accuracy in our solution. This approach will also help in identifying equivalent equations later on in the article. Remember, a strong grasp of the initial equation is crucial for successful problem-solving.

Step-by-Step Simplification

To begin simplifying the equation, we first focus on the distributive property. The fraction 3/5 is multiplied by the expression within the parentheses (30x - 15). This means we multiply 3/5 by both 30x and -15. This step is crucial as it eliminates the parentheses and allows us to work with individual terms. The distributive property is a fundamental concept in algebra, and its correct application is essential for solving equations. Let's break down this step further: (3/5) * 30x equals 18x, and (3/5) * -15 equals -9. Therefore, the equation now becomes 18x - 9 = 72. This simplified form is much easier to work with. We have successfully removed the parentheses and reduced the complexity of the equation. This step-by-step approach ensures accuracy and clarity in the simplification process. Now, we can proceed with isolating x by further algebraic manipulations. This methodical approach is key to solving complex equations effectively. By understanding each step, we build a strong foundation for more advanced problem-solving.

Isolating x: The Final Solution

Having simplified the equation to 18x - 9 = 72, our next objective is to isolate x. This involves performing inverse operations to move constants to the right side of the equation. First, we add 9 to both sides of the equation. This cancels out the -9 on the left side, leaving us with 18x = 81. Adding the same value to both sides maintains the equality of the equation, a fundamental principle in algebra. Now, to completely isolate x, we divide both sides of the equation by 18. This results in x = 81/18, which simplifies to x = 4.5. This is the solution to the core equation. We have successfully isolated x and determined its value. This process demonstrates the power of algebraic manipulation in solving equations. The solution x = 4.5 is crucial as we will use it to identify equivalent equations later on. Understanding how to isolate variables is a core skill in algebra, and this step-by-step process provides a clear illustration of the technique. This foundational knowledge will empower you to solve a wide range of algebraic problems with confidence.

Analyzing the Options: Identifying Equivalent Equations

Now that we know the solution to the original equation is x = 4.5, we can analyze the provided options to identify which equations have the same solution. This involves either solving each equation individually or substituting x = 4.5 into each equation to see if it holds true. This process highlights the concept of equivalent equations, which are equations that have the same solution set. Identifying equivalent equations is a valuable skill in algebra, as it allows us to simplify problems and choose the most convenient form for solving. We will systematically examine each option, applying algebraic principles and substitution techniques. This analysis will not only help us identify the correct answers but also deepen our understanding of equation equivalence. The ability to recognize equivalent equations is crucial for efficient problem-solving in mathematics. By comparing and contrasting the different equations, we gain a deeper appreciation for the flexibility and power of algebraic manipulation. This section will provide a practical demonstration of how to apply the solution we found earlier to solve a larger problem.

Option 1: 18x - 15 = 72

Let's analyze the first option: 18x - 15 = 72. To determine if this equation is equivalent to the original, we can substitute x = 4.5 into the equation. If the equation holds true, then it shares the same solution. Substituting x = 4.5, we get 18(4.5) - 15 = 81 - 15 = 66. Since 66 ≠ 72, this equation does not have the same solution as the original equation. This process of substitution is a powerful tool for verifying solutions and identifying equivalent equations. By performing this check, we can confidently eliminate this option. This example demonstrates the importance of careful calculation and attention to detail in algebraic problem-solving. Understanding how to substitute values and verify solutions is a fundamental skill in mathematics. This step-by-step analysis helps to build a strong foundation for more complex equation solving.

Option 2: 50x - 25 = 72

Next, we consider the second option: 50x - 25 = 72. Again, we substitute x = 4.5 into the equation to check for equivalence. Substituting, we get 50(4.5) - 25 = 225 - 25 = 200. Since 200 ≠ 72, this equation does not have the same solution as the original equation. This further reinforces the importance of verifying solutions through substitution. By systematically checking each option, we can narrow down the possibilities and identify the correct answers. This process also highlights the diversity of algebraic expressions and how different equations can lead to different solutions. The ability to quickly evaluate equations and compare results is a valuable skill in mathematics. This step-by-step approach ensures that we accurately assess each option and make informed decisions about their equivalence to the original equation.

Option 3: 18x - 9 = 72

Now, let's examine the third option: 18x - 9 = 72. We substitute x = 4.5 into the equation: 18(4.5) - 9 = 81 - 9 = 72. This equation holds true when x = 4.5, indicating that it shares the same solution as the original equation. This is our first equivalent equation. This process of verification through substitution is a fundamental technique in algebra. It allows us to confidently identify equations that are mathematically equivalent. This example demonstrates how seemingly different equations can have the same solution set. Recognizing this equivalence is a key skill in simplifying and solving algebraic problems. By carefully evaluating each option, we are building a strong understanding of equation relationships and problem-solving strategies.

Option 4: 3(6x - 3) = 72

Moving on to the fourth option: 3(6x - 3) = 72. We substitute x = 4.5 into the equation: 3(6(4.5) - 3) = 3(27 - 3) = 3(24) = 72. This equation also holds true when x = 4.5, confirming that it is equivalent to the original equation. This is our second equivalent equation. This reinforces the concept that multiple equations can have the same solution. The distributive property plays a key role in this equation, and its correct application is crucial for verification. This example highlights the interconnectedness of algebraic concepts and the importance of mastering fundamental skills. By systematically checking each option, we are developing a comprehensive understanding of equation equivalence and problem-solving techniques.

Option 5: x = 4.5

The final option is x = 4.5. This directly states the solution we found for the original equation. Therefore, this is our third equivalent option. This option serves as a clear confirmation of our solution and provides a direct link to the original equation. This highlights the ultimate goal of solving equations: to find the value of the variable that satisfies the equation. This simple option underscores the power of algebraic manipulation in isolating the variable and determining its value. By including this option, the problem emphasizes the importance of understanding what a solution represents in the context of an equation. This comprehensive analysis of all options ensures a thorough understanding of equation equivalence and solution verification.

Conclusion: The Three Equivalent Equations

In conclusion, after carefully analyzing each option, we have identified three equations that share the same solution for x as the original equation (3/5)(30x - 15) = 72. These equations are:

  • 18x - 9 = 72
  • 3(6x - 3) = 72
  • x = 4.5

This exercise has demonstrated the importance of understanding equation equivalence and the techniques for verifying solutions. By simplifying equations, substituting values, and applying algebraic principles, we can confidently identify equations that share the same solution set. This skill is crucial for success in algebra and beyond. The ability to recognize and manipulate equivalent equations allows for efficient problem-solving and a deeper understanding of mathematical relationships. We hope this comprehensive guide has provided valuable insights and empowered you to tackle similar challenges with confidence. Remember, practice and a strong foundation in algebraic principles are key to mastering equation solving. This article serves as a valuable resource for students, educators, and anyone seeking to enhance their mathematical skills. We encourage you to continue exploring the fascinating world of algebra and its applications in various fields.

Keywords: Algebraic equations, equation equivalence, solving equations, substituting values, verifying solutions, distributive property, isolating variables, linear relationships, mathematical principles, problem-solving techniques.