Unraveling The Equation: $6(3-5w) = 5(4-2w) - 20w$

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Decoding the Equation: $6(3-5w) = 5(4-2w) - 20w$

Hey math enthusiasts! Today, we're diving headfirst into the world of algebraic equations. Specifically, we'll be breaking down how to solve the equation: 6(3βˆ’5w)=5(4βˆ’2w)βˆ’20w6(3-5w) = 5(4-2w) - 20w. Don't worry if it looks a bit intimidating at first; we'll go through it step by step, making sure everyone's on the same page. The goal? To find the value of 'w' that makes this equation true. So, grab your pencils, open your minds, and let's get started!

Step-by-Step Solution: A Friendly Approach

Step 1: The Distributive Property – Clearing the Parentheses

Alright, guys, our first move is to tackle those pesky parentheses. Remember the distributive property? It's like spreading the love (or in this case, the multiplication) to everything inside the parentheses. On the left side, we have 6(3βˆ’5w)6(3-5w). We multiply the 6 by both the 3 and the -5w. So, 6βˆ—3=186 * 3 = 18 and 6βˆ—βˆ’5w=βˆ’30w6 * -5w = -30w. This gives us 18βˆ’30w18 - 30w. On the right side, we have 5(4βˆ’2w)βˆ’20w5(4-2w) - 20w. Similarly, we distribute the 5 to both terms inside the parentheses: 5βˆ—4=205 * 4 = 20 and 5βˆ—βˆ’2w=βˆ’10w5 * -2w = -10w. This gives us 20βˆ’10w20 - 10w. Don't forget the βˆ’20w-20w at the end! So now our equation looks like this: 18βˆ’30w=20βˆ’10wβˆ’20w18 - 30w = 20 - 10w - 20w. Pretty neat, huh?

This first step, using the distributive property, is crucial. It simplifies the equation, allowing us to isolate the variable 'w'. Think of it as opening up the puzzle box. Without this, the pieces (terms) inside would remain hidden, making it impossible to solve. By distributing, we reveal those hidden pieces and make them accessible for rearrangement and simplification. The distributive property is one of the fundamental pillars of algebra, so understanding it well is key to solving more complex equations down the road. It essentially says that multiplying a number by a sum or difference inside parentheses is the same as multiplying that number by each term individually and then adding or subtracting the results. It's like having a magical key that unlocks the equation, allowing you to manipulate it in ways that ultimately lead to the solution. The core of this step is to transform an equation containing parentheses into an equivalent one without them, making subsequent steps easier. This transformation is not just a cosmetic change; it's a fundamental shift in the way we view and approach the problem. It streamlines the equation, laying the groundwork for the next steps in our solution. This simplification is the cornerstone of algebraic manipulation, essential for simplifying and solving equations across diverse mathematical problems. So, always remember: distribute first!

Step 2: Combining Like Terms – Tidying Up

Now that we've cleared the parentheses, it's time to tidy things up a bit. We're going to combine 'like terms'. Like terms are terms that have the same variable raised to the same power. In our equation, 18βˆ’30w=20βˆ’10wβˆ’20w18 - 30w = 20 - 10w - 20w, we have '-10w' and '-20w' on the right side. Combining these gives us βˆ’10wβˆ’20w=βˆ’30w-10w - 20w = -30w. So, our equation now simplifies to 18βˆ’30w=20βˆ’30w18 - 30w = 20 - 30w. See how much cleaner that looks? Combining like terms is all about simplifying the equation to make it easier to solve. It's like organizing your desk before you start a project; it makes everything more manageable. This step ensures that we have the fewest possible terms, reducing the chance of errors in subsequent calculations. It's a fundamental principle of algebra, streamlining equations to facilitate finding solutions. It's like grouping similar objects together to make them easier to count and manipulate. This simplifies the equation, preparing it for the next steps toward the final answer. Therefore, understanding and implementing the concept of combining like terms is essential for solving equations.

Step 3: Isolating the Variable – Getting 'w' Alone

This is where we start to isolate 'w'. Our goal is to get all the terms with 'w' on one side of the equation and the constants (numbers without variables) on the other. Currently, we have βˆ’30w-30w on both sides of the equation: 18βˆ’30w=20βˆ’30w18 - 30w = 20 - 30w. Let's try to get all 'w' terms to the left side. To do this, we can add 30w30w to both sides. This gives us: 18βˆ’30w+30w=20βˆ’30w+30w18 - 30w + 30w = 20 - 30w + 30w. Simplifying, we get 18=2018 = 20. Uh oh! Does this mean something went wrong?

This step is all about getting to the heart of the matter – the value of 'w'. We do this by manipulating the equation in such a way that 'w' is alone on one side. This process involves using inverse operations – doing the opposite of whatever operation is being performed on 'w'. For example, if 'w' is being multiplied by a number, we divide both sides by that number. If a number is being added to 'w', we subtract that number from both sides. The key is to keep the equation balanced – whatever we do to one side, we must do to the other. This ensures that the equality remains true throughout the process. Isolating the variable is like slowly peeling away layers to reveal the answer. It requires a clear understanding of the properties of equality and the ability to apply inverse operations strategically. It is a fundamental process in algebra, essential for uncovering the unknown values in various mathematical contexts. This crucial step moves us closer to finding the value of 'w'. Think of it like a game of hide-and-seek, where the aim is to bring the hidden 'w' into the open. By correctly isolating the variable, you prepare the equation for the final reveal of the solution.

Step 4: Interpreting the Result

In the previous step, we ended up with 18=2018 = 20. This is a contradiction, meaning it's not a true statement. This indicates that the original equation has no solution. There's no value of 'w' that, when plugged back into the original equation, would make both sides equal. In other words, there is no value of w that can satisfy this equation.

This final step is crucial. The work you've put in thus far culminates in an interpretation of the outcome. Whether you arrive at a specific value for the variable or, as in this case, discover a contradiction, understanding the meaning behind the result is what truly completes the problem-solving process. This step is about drawing conclusions based on your calculations. It ensures that your answer is valid. It's not just about crunching numbers; it's about making sense of the solution and understanding what it tells you about the equation. Interpreting the result is like the final piece of a puzzle; it provides closure and helps you understand the overall picture. It’s where the abstract symbols transform into meaningful insights. Recognizing a contradiction highlights the importance of checking your work and understanding the implications of your findings. It's an essential element of mathematical problem-solving, making sure that your conclusion is grounded in reason. The goal is to provide a comprehensive view of the problem and its solution, demonstrating a deep understanding of mathematical concepts. This means that the equation doesn't have a solution. It's a very common outcome in algebra, and it doesn't mean you've made a mistake. It just means the equation doesn't have a value for 'w' that makes it true.

Summary: Putting It All Together

Let's recap what we've learned. We started with the equation 6(3βˆ’5w)=5(4βˆ’2w)βˆ’20w6(3-5w) = 5(4-2w) - 20w. We used the distributive property to eliminate the parentheses, then combined like terms. Next, we attempted to isolate 'w', but we arrived at a contradiction, 18=2018 = 20, showing that the equation has no solution. It's important to remember that not all equations have solutions, and it's essential to understand how to interpret the results. So, when you're faced with an equation, don't be afraid to take it step by step, and always keep an eye on what your results mean. You've got this!

Common Pitfalls and How to Avoid Them

  • Forgetting the Distributive Property: This is the most common mistake. Always remember to multiply the term outside the parentheses by every term inside. Forgetting even one term can throw off the entire solution. Double-check your distribution to avoid this.
  • Incorrectly Combining Like Terms: Ensure you're only combining terms that have the same variable raised to the same power. For instance, you can combine 2x2x and 3x3x to get 5x5x, but you can't combine 2x2x and 3x23x^2.
  • Sign Errors: Be extra careful with negative signs! A simple mistake with a negative sign can drastically change the outcome. Take your time and double-check your signs.
  • Not Recognizing No Solution: Some equations, like this one, have no solution. Don't assume you've made a mistake if you end up with a contradiction. Understand what the result means.

By keeping these common pitfalls in mind, you'll be well-equipped to tackle similar equations with confidence. Practice makes perfect, so keep practicing, and you'll become a pro at solving these types of problems in no time!

Conclusion: Mastering the Math

Guys, congratulations! You've successfully navigated the equation 6(3βˆ’5w)=5(4βˆ’2w)βˆ’20w6(3-5w) = 5(4-2w) - 20w, even if we didn't get a specific value for 'w'. This is a valuable lesson. Remember that solving equations is like a puzzle, requiring a systematic approach, a strong understanding of mathematical principles, and careful attention to detail. Keep practicing, keep learning, and keep asking questions. Mathematics is all about exploring and understanding the world around us, and with each equation you solve, you're becoming a more skilled explorer. Keep up the excellent work, and always remember to enjoy the journey. And that's a wrap on this equation! Keep practicing, and you'll conquer more math problems in the future. Cheers!