Solving Equations A Step By Step Guide For $\frac{1}{6}x + \frac{3}{2} = -\frac{1}{3}$

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Hey guys! Ever find yourself staring at an equation, feeling like you're trying to decipher an ancient code? Well, you're not alone! Solving for variables is a fundamental skill in mathematics, and once you get the hang of it, it's like unlocking a superpower. Let's break down how to tackle the equation 16x+32=βˆ’13\frac{1}{6}x + \frac{3}{2} = -\frac{1}{3} and, more importantly, how to check your answer to make sure you've nailed it. This comprehensive guide will walk you through each step, making the process crystal clear. So, grab your pencil and paper, and let's dive in!

Understanding the Basics of Solving Equations

Before we jump into the specifics of this equation, let's cover some ground rules. Solving for a variable means isolating it on one side of the equation. Think of it like a balancing act: whatever you do to one side, you must do to the other to keep things balanced. The main goal is to get 'x' (our variable) all by itself on one side, so we know its value. We achieve this by using inverse operations – operations that "undo" each other. For example, addition and subtraction are inverse operations, as are multiplication and division. Mastering these basics is crucial for tackling more complex equations later on. Think of inverse operations as the keys to unlocking the solution. To solve for the variable, use inverse operations to isolate the variable. Remember to perform the same operation on both sides of the equation to maintain balance. For instance, if we see addition, we'll use subtraction to cancel it out. If we see multiplication, we'll use division. This balancing act is the cornerstone of equation solving, ensuring that we maintain equality while manipulating the terms. The inverse operation concept is the fundamental key to solving equations effectively. Understanding this concept thoroughly will empower you to tackle a wide range of mathematical problems. With a solid grasp of inverse operations, you'll be well-equipped to navigate the intricacies of algebraic manipulations and confidently arrive at accurate solutions. So, always remember to maintain the balance by applying the same operation to both sides, and let inverse operations be your guide in isolating the variable and unveiling its value. Equations can seem daunting at first glance, but by breaking them down into manageable steps and applying the principles of inverse operations, you can systematically unravel the solution. Each step you take brings you closer to isolating the variable and revealing its numerical value. It's a process of carefully undoing the operations that surround the variable, ensuring that the equation remains balanced and the solution remains accurate. So, embrace the challenge, trust your understanding of inverse operations, and approach each equation with confidence, knowing that you have the tools to solve it effectively.

Step-by-Step Solution for 16x+32=βˆ’13\frac{1}{6}x + \frac{3}{2} = -\frac{1}{3}

Okay, let's get down to business and solve this equation step by step. Our equation is 16x+32=βˆ’13\frac{1}{6}x + \frac{3}{2} = -\frac{1}{3}. The first thing we want to do is get rid of that 32\frac{3}{2} on the left side. Since it's being added, we'll use its inverse operation: subtraction. We'll subtract 32\frac{3}{2} from both sides of the equation. This gives us: 16x+32βˆ’32=βˆ’13βˆ’32\frac{1}{6}x + \frac{3}{2} - \frac{3}{2} = -\frac{1}{3} - \frac{3}{2}. Simplifying the left side, the 32\frac{3}{2} terms cancel each other out, leaving us with 16x\frac{1}{6}x. Now, we need to simplify the right side. To subtract fractions, we need a common denominator. The least common multiple of 3 and 2 is 6, so we'll convert both fractions to have a denominator of 6. βˆ’13-\frac{1}{3} becomes βˆ’26-\frac{2}{6} (multiply numerator and denominator by 2), and βˆ’32-\frac{3}{2} becomes βˆ’96-\frac{9}{6} (multiply numerator and denominator by 3). Now we have: βˆ’26βˆ’96-\frac{2}{6} - \frac{9}{6}. Subtracting these, we get βˆ’116-\frac{11}{6}. So, our equation now looks like this: 16x=βˆ’116\frac{1}{6}x = -\frac{11}{6}. We're almost there! Now, we need to get rid of the 16\frac{1}{6} that's multiplying 'x'. The inverse operation of multiplication is division, but dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of 16\frac{1}{6} is 6 (or 61\frac{6}{1}). So, we'll multiply both sides of the equation by 6. This gives us: 6βˆ—16x=6βˆ—βˆ’1166 * \frac{1}{6}x = 6 * -\frac{11}{6}. On the left side, the 6 and 16\frac{1}{6} cancel out, leaving us with just 'x'. On the right side, one of the 6's cancels out, leaving us with -11. Therefore, our solution is x=βˆ’11x = -11. This methodical approach, breaking down the problem into smaller, manageable steps, makes even complex equations feel less intimidating. Each step builds upon the previous one, leading you closer to the final solution. The key is to remain focused, apply the appropriate inverse operations, and maintain the balance of the equation throughout the process. By meticulously addressing each step, you can confidently navigate the equation and arrive at the accurate value of the variable. This systematic approach not only helps you solve the equation but also enhances your understanding of the underlying mathematical principles. It's a testament to the power of breaking down complex problems into simpler components, making the journey to the solution more accessible and the final answer more satisfying.

Checking Your Answer Is Key!

Now, before you shout "Eureka!" and move on, there's a crucial step we need to take: checking our answer. It's like the golden rule of equation solving! Plugging our solution back into the original equation is the ultimate test to make sure we haven't made any sneaky mistakes along the way. It's like having a built-in error detector! So, let's take our solution, x=βˆ’11x = -11, and substitute it back into the original equation: 16x+32=βˆ’13\frac{1}{6}x + \frac{3}{2} = -\frac{1}{3}. Replacing 'x' with -11, we get: 16(βˆ’11)+32=βˆ’13\frac{1}{6}(-11) + \frac{3}{2} = -\frac{1}{3}. Now, let's simplify. 16(βˆ’11)\frac{1}{6}(-11) is βˆ’116-\frac{11}{6}. So, our equation becomes: βˆ’116+32=βˆ’13-\frac{11}{6} + \frac{3}{2} = -\frac{1}{3}. Again, we need a common denominator to add these fractions. We already know the least common multiple of 6 and 2 is 6, so let's convert 32\frac{3}{2} to have a denominator of 6. Multiplying the numerator and denominator by 3, we get 96\frac{9}{6}. Now our equation looks like this: βˆ’116+96=βˆ’13-\frac{11}{6} + \frac{9}{6} = -\frac{1}{3}. Adding these fractions, we get βˆ’26-\frac{2}{6}. Now, let's simplify βˆ’26-\frac{2}{6}. Both 2 and 6 are divisible by 2, so we can reduce this fraction to βˆ’13-\frac{1}{3}. So, our equation now reads: βˆ’13=βˆ’13-\frac{1}{3} = -\frac{1}{3}. Woohoo! The left side equals the right side. This means our solution, x=βˆ’11x = -11, is correct! Checking your answer is not just a formality; it's an essential part of the problem-solving process. It provides you with the confidence that your solution is accurate and allows you to identify and correct any errors you might have made. Think of it as the final seal of approval on your mathematical journey. This step is important to check if the answer is correct. Verifying the solution by substituting it back into the original equation serves as a powerful validation of your work. It's the ultimate safeguard against potential errors and ensures that your final answer is accurate and reliable. The process of checking your answer not only confirms the correctness of your solution but also reinforces your understanding of the equation-solving process. It provides a tangible demonstration of how the solution satisfies the original equation, solidifying your grasp of the underlying mathematical principles. So, embrace the habit of checking your answers, and let it be a cornerstone of your mathematical practice. It's the key to building confidence and ensuring accuracy in your problem-solving endeavors. Don't skip this important step!

Common Mistakes to Avoid

Alright, let's talk about some common pitfalls people often stumble into when solving equations. Knowing these beforehand can help you steer clear of them! One big mistake is not performing the same operation on both sides of the equation. Remember, it's all about balance! If you subtract something from one side, you must subtract it from the other. Another common error is messing up the order of operations. Remember PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction)? Make sure you're applying the correct order when simplifying expressions. Also, be super careful with negative signs! They can be tricky little devils. Double-check your signs at each step to avoid making errors. Another pitfall is not simplifying fractions correctly. Always reduce fractions to their simplest form. And lastly, the biggest mistake of all: not checking your answer! We've already hammered this one home, but it's worth repeating. Checking your answer is your safety net. Avoiding these common mistakes is crucial for solving equations accurately and efficiently. Understanding where errors typically occur can help you develop strategies to prevent them. One effective approach is to work through equations systematically, paying close attention to each step and double-checking your calculations. Another helpful technique is to break down complex equations into smaller, more manageable parts, addressing each part individually before combining them. By being mindful of these common mistakes and implementing preventative measures, you can significantly reduce the likelihood of errors and enhance your equation-solving skills. Remember, practice makes perfect, and the more you solve equations, the more adept you'll become at identifying and avoiding these pitfalls. So, keep honing your skills, stay vigilant, and approach each equation with confidence and precision. With a combination of knowledge, carefulness, and practice, you can conquer the challenges of equation solving and achieve accurate results every time. Steer clear of these errors, and you'll be an equation-solving pro in no time!

Conclusion: You've Got This!

Solving for variables might seem daunting at first, but with a clear understanding of the steps and some practice, you can conquer any equation that comes your way. Remember the key takeaways: use inverse operations to isolate the variable, maintain balance by performing the same operation on both sides, simplify fractions, and always check your answer! We tackled the equation 16x+32=βˆ’13\frac{1}{6}x + \frac{3}{2} = -\frac{1}{3}, found that x=βˆ’11x = -11, and verified our solution. You've got this! Keep practicing, and soon you'll be solving equations like a mathematical rockstar. Now you’re equipped with the knowledge and strategies to confidently approach and solve similar equations. The journey of mastering equation solving is a continuous one, filled with opportunities for growth and refinement. Embrace the challenges, celebrate your successes, and never stop exploring the fascinating world of mathematics. Each equation you solve is a step forward in your mathematical journey, building your skills, confidence, and understanding. So, keep practicing, keep learning, and keep pushing your boundaries, and you'll be amazed at what you can achieve. You've unlocked the power to solve equations, and now it's time to unleash that power and conquer any mathematical challenge that comes your way. The world of mathematics is vast and exciting, and with your newfound skills, you're ready to explore it with confidence and enthusiasm. Keep solving, keep growing, and keep shining as a mathematical problem-solver!

So the final answer is:

x=βˆ’11x = -11