Solving For X In 28x + 8y = 164 A Step-by-Step Guide

Hey guys! Ever stumbled upon an equation that looks like a tangled mess of numbers and letters? Well, don't sweat it! We're going to break down one of those equations today and make it super easy to solve. We're diving into the world of algebra to find the value of 'x' in the equation 28x + 8y = 164, but here's the cool part: we already know that y = 3! Yes, you read that right, 3 with an exclamation mark! That's factorial notation, and it's our first step to cracking this puzzle.

So, what exactly is a factorial? It sounds intimidating, but it's actually quite simple. A factorial (represented by the "!" symbol) means you multiply a number by every positive whole number less than it. For example, 3! means 3 × 2 × 1, which equals 6. Now we know that y isn't just 3; it's actually 6! This is a crucial piece of information because it allows us to simplify our original equation. We can substitute 'y' with '6' in the equation 28x + 8y = 164. This substitution transforms the equation from having two variables (x and y) to just one variable (x), making it much easier to solve. By replacing 'y' with '6', we get 28x + 8(6) = 164. See how we're making progress already? We've turned a seemingly complex equation into something much more manageable. Now, let's keep going and unlock the value of 'x'. We need to simplify further, so the next step is to perform the multiplication: 8 multiplied by 6. What does that give us? It's 48! So, our equation now looks like this: 28x + 48 = 164. We're getting closer to isolating 'x' and figuring out its value. The key to solving for 'x' is to isolate it on one side of the equation. This means we need to get rid of the '+ 48' on the left side. How do we do that? We use the magic of inverse operations! Since we're adding 48, we need to subtract 48 from both sides of the equation. Remember, whatever you do to one side of the equation, you must do to the other side to keep things balanced. Subtracting 48 from both sides gives us: 28x + 48 - 48 = 164 - 48. The +48 and -48 on the left side cancel each other out, leaving us with 28x on the left. On the right side, 164 minus 48 equals 116. So, our equation is now simplified to 28x = 116. We're in the home stretch now! We're almost there! The final step in isolating 'x' is to get rid of the '28' that's multiplying it. Again, we use the inverse operation. Since 'x' is being multiplied by 28, we need to divide both sides of the equation by 28. This gives us: (28x) / 28 = 116 / 28. On the left side, the 28s cancel out, leaving us with just 'x'. On the right side, 116 divided by 28... well, that's where a calculator or some long division skills come in handy! If you divide 116 by 28, you get approximately 4.14 (rounded to two decimal places). So, the value of x is approximately 4.14.

Alright, let's break down the solution step-by-step, just to make sure we've got everything crystal clear. Sometimes seeing the whole process laid out can really help solidify your understanding. We're going to revisit the equation 28x + 8y = 164, where we know y = 3!, and walk through each step to find the value of 'x'. Think of this as a detailed roadmap to solving the puzzle. We'll start from the very beginning and make sure no step is missed. This way, you can follow along easily and apply the same logic to similar problems in the future. Solving equations can feel like detective work, and these steps are our clues!

Step 1: Understand Factorials. The very first thing we encountered was the factorial notation, 3!. Remember, a factorial means multiplying a number by every positive whole number less than it. So, 3! is the same as 3 × 2 × 1. This is a foundational concept, so it's crucial to get it right. If you're ever unsure, just write out the multiplication sequence. It's a simple but powerful tool! Now, let's calculate 3!. Multiplying 3 × 2 gives us 6, and then multiplying 6 × 1 still gives us 6. So, 3! = 6. This means that the value of 'y' in our equation is 6. We've successfully deciphered the factorial part of the problem. This was our first clue, and it's a big one! Knowing the value of 'y' allows us to move forward and simplify the equation.

Step 2: Substitute the Value of y. Now that we know y = 6, we can substitute this value into the original equation. This is a key step in solving for 'x' because it reduces the number of unknowns in the equation. Replacing 'y' with '6' in the equation 28x + 8y = 164 gives us: 28x + 8(6) = 164. Notice how we've replaced the 'y' with the number '6'. This is the magic of substitution! It allows us to rewrite the equation in a simpler form that we can work with more easily. We've gone from having two variables (x and y) to just one variable (x). This makes the equation much more manageable and brings us closer to finding the solution. Substitution is a powerful technique in algebra, and it's one you'll use again and again.

Step 3: Simplify the Equation. We've substituted 'y', and now it's time to simplify things further. We need to perform the multiplication: 8 multiplied by 6. This is a straightforward calculation, but it's important to get it right. 8 multiplied by 6 equals 48. So, we can replace 8(6) with 48 in our equation. This gives us: 28x + 48 = 164. We've made the equation even simpler! By performing the multiplication, we've reduced the number of terms and made the equation easier to visualize. Simplifying equations is all about making them more manageable, and we're doing a great job so far. Each step we take brings us closer to isolating 'x' and finding its value. Keep going, guys! We're on the right track!

Step 4: Isolate the Term with x. The goal now is to get the term with 'x' (which is 28x) all by itself on one side of the equation. This is a crucial step in solving for 'x'. To do this, we need to get rid of the '+ 48' that's on the same side as 28x. How do we do that? We use the inverse operation! Since we're adding 48, the inverse operation is subtraction. We need to subtract 48 from both sides of the equation. Remember, whatever we do to one side, we must do to the other to keep the equation balanced. Subtracting 48 from both sides gives us: 28x + 48 - 48 = 164 - 48. Now, let's simplify. On the left side, the +48 and -48 cancel each other out, leaving us with just 28x. On the right side, 164 minus 48 equals 116. So, our equation is now: 28x = 116. Look how far we've come! We've successfully isolated the term with 'x'. We're one step closer to finding the value of 'x'. The power of inverse operations is amazing, isn't it?

Step 5: Solve for x. We're in the final stretch! We have 28x = 116, and we need to find 'x'. 'x' is being multiplied by 28, so to isolate 'x', we need to use the inverse operation of multiplication, which is division. We need to divide both sides of the equation by 28. This gives us: (28x) / 28 = 116 / 28. On the left side, the 28s cancel out, leaving us with just 'x'. On the right side, we have 116 divided by 28. This is where we'll need a calculator or some long division skills. 116 divided by 28 is approximately 4.14 (rounded to two decimal places). So, the value of x is approximately 4.14. We did it! We've successfully solved for 'x'! It might have seemed like a long journey, but we broke it down step-by-step and conquered it. Remember, solving equations is like building something – each step is a brick that gets us closer to the final result.

Okay, so we've tackled this equation head-on using a classic algebraic approach. But guess what? There's often more than one way to skin a cat, as they say! Let's explore some alternative approaches to solving for x in the equation 28x + 8y = 164, given that y = 3!. These different perspectives can not only help you double-check your answer but also deepen your understanding of algebraic principles. Think of it as having multiple tools in your toolbox – the more tools you have, the better equipped you are to handle any situation!

Method 1: Graphical Approach. Sometimes, visualizing an equation can make it easier to understand and solve. In this case, we can think of 28x + 8y = 164 as a linear equation, which represents a straight line on a graph. While we ultimately want to find the value of 'x' when y = 6, we can graph the entire line and then pinpoint the specific point where y is 6. To graph the line, we need to find at least two points that satisfy the equation. One way to do this is to set x = 0 and solve for y, and then set y = 0 and solve for x. When x = 0, the equation becomes 8y = 164, so y = 164 / 8 = 20.5. This gives us the point (0, 20.5). When y = 0, the equation becomes 28x = 164, so x = 164 / 28 ≈ 5.86. This gives us the point (5.86, 0). We can plot these two points on a graph and draw a line through them. Now, we know that y = 3! = 6. To find the corresponding x-value, we can look at the graph and find the point on the line where y is approximately 6. The x-coordinate of that point will be our solution for 'x'. While this method isn't as precise as the algebraic method, it provides a visual confirmation of our answer. It's a great way to check if your answer is in the right ballpark! Plus, it helps you understand the relationship between equations and their graphical representations. Graphing is a powerful tool in mathematics, and it's always good to have it in your arsenal.

Method 2: Using Online Calculators or Solvers. In today's digital age, we have access to a plethora of online tools that can help us solve mathematical problems. There are many websites and apps that offer equation solvers. These tools can be incredibly helpful for checking your work or for solving more complex equations. To use an online calculator or solver, you simply input the equation and the value of y, and the tool will calculate the value of x for you. This is a quick and easy way to verify your answer. However, it's important to remember that these tools are meant to be aids, not replacements for understanding the underlying concepts. It's crucial to learn how to solve equations manually so that you can tackle problems even when you don't have access to a calculator. Think of online solvers as a safety net – they're there to catch you if you make a mistake, but you should always strive to solve the problem yourself first. Using these tools wisely can enhance your learning and boost your confidence.

Method 3: Estimation and Approximation. Sometimes, especially in real-world situations, you don't need an exact answer. An estimate or approximation can be good enough. Let's see how we can estimate the value of 'x' in our equation. We know that 28x + 8y = 164 and y = 6. So, the equation becomes 28x + 48 = 164. We can round the numbers to make the calculations easier. Let's round 28 to 30 and 164 to 160. Our equation now becomes approximately 30x + 50 = 160. Subtracting 50 from both sides gives us 30x = 110. Dividing both sides by 30 gives us x ≈ 110 / 30 ≈ 3.67. This is an approximate value of 'x'. It's not the exact answer, but it gives us a rough idea of what to expect. Estimation is a valuable skill in mathematics and in life. It allows you to quickly check if your answer is reasonable and to make informed decisions even when you don't have all the information. By practicing estimation, you can develop a better number sense and improve your problem-solving abilities. So, the next time you encounter an equation, don't be afraid to try estimating the answer first!

Alright guys, now that we've mastered the art of solving for 'x' in our equation, let's take a step back and think about the bigger picture. Why is this important? Where do these skills come in handy in the real world? You might be surprised to learn that solving equations is a fundamental skill that has applications in a wide range of fields and everyday situations. It's not just about abstract numbers and symbols; it's about understanding relationships and using logic to find solutions. Let's explore some real-world scenarios where solving equations can be a game-changer. Think of this as your guide to unlocking the practical power of algebra!

1. Budgeting and Finance. Imagine you're planning a party, and you have a budget of $200. You know that the venue rental costs $50, and each guest will cost you $10 for food and drinks. You want to figure out how many guests you can invite without exceeding your budget. This is a classic equation-solving problem! You can represent the situation with an equation: 50 + 10x = 200, where 'x' is the number of guests. Solving for 'x' will tell you the maximum number of guests you can invite. Similarly, in personal finance, you might want to calculate how long it will take to pay off a loan or how much you need to save each month to reach a financial goal. These calculations often involve setting up and solving equations. Whether you're managing your personal finances or running a business, understanding how to solve equations is essential for making informed financial decisions. It's about taking control of your money and planning for the future. So, the next time you're budgeting, remember the power of equations!

2. Cooking and Baking. Believe it or not, cooking and baking are full of mathematical ratios and proportions. If a recipe calls for certain amounts of ingredients to serve a specific number of people, you might need to adjust the quantities to serve a different number of people. This often involves setting up proportions, which are essentially equations. For example, if a recipe for 4 people calls for 2 cups of flour, you can set up a proportion to find out how much flour you need for 6 people. You can also use equations to convert between different units of measurement, such as cups and milliliters. Baking, in particular, requires precision, and solving equations can help you ensure that your recipes turn out perfectly every time. So, the next time you're in the kitchen, remember that you're also practicing your equation-solving skills!

3. Science and Engineering. In science and engineering, equations are the language of the universe. Physicists use equations to describe the motion of objects, the behavior of light and electricity, and the forces that govern the cosmos. Engineers use equations to design bridges, buildings, and machines. Whether you're calculating the trajectory of a rocket, designing an electrical circuit, or determining the strength of a material, you'll be using equations. Solving equations is a fundamental skill for anyone working in a STEM field. It's about understanding the laws of nature and using them to create and innovate. So, if you're interested in science or engineering, mastering equation-solving is a must!

4. Everyday Problem Solving. Even in everyday situations, we often encounter problems that can be solved using equations. For example, you might want to calculate the distance you can travel on a certain amount of gas, given the gas mileage of your car. Or you might want to figure out how much time it will take you to drive to a destination, given your speed and the distance. These calculations involve setting up and solving simple equations. Equation-solving is a powerful tool for making informed decisions and navigating the world around us. It's about breaking down complex problems into smaller, manageable steps and using logic to find solutions. So, the next time you're faced with a challenge, remember the power of equations!

Okay, guys, we've covered a lot of ground in this discussion! We've tackled the equation 28x + 8y = 164, where y = 3!, and we've explored various methods for solving it. We've also delved into the real-world applications of equation-solving. Before we wrap things up, let's recap some key takeaways and discuss resources for further learning. Think of this as your cheat sheet and roadmap for continued success in algebra! We'll highlight the essential concepts and skills we've learned and point you in the direction of more resources to deepen your understanding.

1. Factorials. Remember, a factorial (denoted by the "!" symbol) means multiplying a number by every positive whole number less than it. 3! = 3 × 2 × 1 = 6. Understanding factorials is crucial for simplifying expressions and solving equations that involve them. They pop up in various areas of mathematics, including combinatorics and probability, so mastering them is a worthwhile investment.

2. Substitution. Substitution is a powerful technique for solving equations with multiple variables. By replacing a variable with its known value, we can simplify the equation and solve for the remaining variable. In our case, we substituted y = 6 into the equation 28x + 8y = 164, which allowed us to solve for x. Substitution is a fundamental skill in algebra, and it's one you'll use frequently.

3. Inverse Operations. To isolate a variable and solve for it, we use inverse operations. Addition and subtraction are inverse operations, as are multiplication and division. We used inverse operations to get rid of the +48 and the 28 in our equation, ultimately isolating x. Mastering inverse operations is key to solving equations efficiently.

4. Solving for x. We successfully found that x ≈ 4.14 in the equation 28x + 8y = 164, where y = 3!. We used a step-by-step approach, breaking down the problem into smaller, manageable steps. This systematic approach is essential for tackling more complex equations.

5. Real-World Applications. We explored how equation-solving is used in budgeting, cooking, science, and everyday problem-solving. This highlights the practical importance of algebra and its relevance to our lives. Understanding the real-world applications of mathematics can make learning more engaging and meaningful.

Further Learning Resources. If you're eager to delve deeper into algebra and equation-solving, here are some resources you might find helpful:

• Khan Academy: This free online platform offers a wealth of math tutorials and exercises, covering everything from basic algebra to advanced calculus. It's a fantastic resource for self-paced learning.
• Mathway: This website provides a powerful equation solver and calculator. You can input equations and see step-by-step solutions, which can be helpful for checking your work or understanding different approaches.
• Textbooks and Workbooks: Traditional textbooks and workbooks can provide a structured approach to learning algebra. Look for resources that include plenty of practice problems and clear explanations.

By continuing to practice and explore, you can build your confidence and competence in algebra. Remember, math is a journey, not a destination. Enjoy the process of learning and discovering!

So, there you have it, guys! We've successfully navigated the equation 28x + 8y = 164, given that y = 3!. We've not only found the value of 'x' but also explored different methods for solving it and discussed its real-world applications. Hopefully, this journey has demystified the process of equation-solving and shown you how powerful and versatile these skills can be. Remember, math isn't just about numbers and formulas; it's about problem-solving, logical thinking, and understanding the world around us. By mastering equation-solving, you're equipping yourself with a valuable tool that can help you in countless situations.

Keep practicing, keep exploring, and keep challenging yourselves. The world of mathematics is vast and fascinating, and there's always something new to learn. And remember, if you ever get stuck, don't hesitate to break the problem down into smaller steps, use the resources available to you, and ask for help when you need it. You've got this!