Solving For 'a': Math Problems Explained Simply

Hey guys! Let's dive into some math problems that might seem a little tricky at first, but trust me, they're totally manageable. We're going to figure out how to find the value of a number, represented by the letter 'a,' in two different scenarios. These problems involve some basic algebra, but don't worry, I'll break it down step-by-step so it's super clear. The core concept here is understanding how to manipulate equations and use the given information to isolate the variable we're trying to find. Think of it like a puzzle – we have all the pieces, and our job is to put them together correctly to reveal the answer. We'll start with the first problem, where we have two equations with 'a', 'x', and 'y' involved. Don't worry, you don't need to solve for x and y individually, the key is to use what we know about them to find 'a'. So, grab your pencils and let's get started! We'll go through each step carefully, explaining the logic behind it. This way, you'll not only get the answers but also understand the why behind the how. Understanding the underlying principles is crucial for tackling similar problems in the future, and that's exactly what we're aiming for. We'll also talk about the importance of checking our work, which is a good habit to develop in mathematics. It helps to ensure that the solution makes sense in the context of the problem. Keep an open mind, ask questions if something isn't clear, and most importantly, have fun with it! Math can be a blast when you approach it the right way. Ready to get started? Let's jump right into it!

Problem 1: Unveiling 'a' with ax + ay = 45 and x + y = 9

Alright, let's tackle the first problem. We're given two equations: ax + ay = 45 and x + y = 9. Our mission? To find the value of 'a'. The first thing we should notice is that in the equation ax + ay = 45, the variable 'a' is multiplied by both 'x' and 'y'. This is a super important clue! Think about how we can simplify this equation. Remember, our goal is to isolate 'a'. The expression 'ax + ay' has a common factor, which is 'a'. We can factor out 'a' from both terms on the left side of the equation. So, let's rewrite ax + ay = 45 by factoring out 'a'. This gives us: a(x + y) = 45. See how we've simplified things already? Now, let's bring in the second equation, x + y = 9. We already know the value of x + y! That means we can substitute the value of x + y from the second equation into the first equation. Substitute '9' for x + y in the equation a(x + y) = 45. This gives us a * 9 = 45, or simply 9a = 45. To find 'a', we need to isolate it. Right now, 'a' is multiplied by 9. To undo this, we need to divide both sides of the equation by 9. So, divide both sides by 9: 9a / 9 = 45 / 9. This simplifies to a = 5. And there you have it, guys! The value of 'a' in this problem is 5. We used the power of factoring and substitution to solve this one. See? Not so hard, right? We took the information we were given, simplified and rearranged it to unveil the value of 'a'.

Now, to be absolutely sure, it's always a good idea to double-check our answer. We can plug the value of 'a' back into the original equations to see if they hold true. If a = 5, then the first equation becomes 5x + 5y = 45. Factoring out the 5, we get 5(x + y) = 45. And since we know that x + y = 9, we have 5 * 9 = 45, which is true! So, our solution of a = 5 is correct. Always check your work when you get the solution so you can make sure you did everything right! It's a great habit to form, and it'll save you from mistakes in the future. This process not only confirms our answer but also reinforces our understanding of the problem. We're not just finding a number; we're using a logical approach to solve it. Next, we will tackle the second problem, which will help you solidify your understanding and apply the same concept in a different context.

Problem 2: Finding 'a' when ax + ay = 471 and x + y = 3

Let's move on to the second part of the problem. This one is very similar to the first, which is great because you should already be starting to feel comfortable with this type of question. We're given two equations again: ax + ay = 471 and x + y = 3. Our aim, as always, is to find the value of 'a'. Just like before, let's start with the first equation, ax + ay = 471. We see that the variable 'a' is present in both terms on the left side. What should we do first? That’s right! We can factor out 'a'. Factoring out 'a' from the left side gives us: a(x + y) = 471. So, we have simplified our first equation by factoring. Now, look at our second equation. We have x + y = 3. Guess what? This equation gives us the value of x + y! We can now substitute the value of x + y from the second equation into the factored form of the first equation. We replace x + y with 3, so we have: a * 3 = 471, or simply 3a = 471. To isolate 'a,' we must divide both sides of the equation by 3. This gives us 3a / 3 = 471 / 3. Now, do the math! If you divide 471 by 3, you get 157. So, a = 157. There we have it! The value of 'a' in this second problem is 157. This time we didn't need to make many steps and we can find the answer quickly. We followed the same logic as before: factoring, substitution, and solving for 'a'. Remember to always stay organized, especially when you have to do multiple steps. It keeps your work neat, and it makes it easier to find any errors.

Now, let's check our answer! If a = 157, the first equation becomes 157x + 157y = 471. Factoring out 157, we get 157(x + y) = 471. Since x + y = 3, we can verify that 157 * 3 = 471, which is indeed true. Thus, our solution, a = 157, is correct! This is a good way to check to make sure that you did not make any mistakes. This practice not only builds confidence but also helps you avoid careless errors.

Key Takeaways and Tips for Success

So, what have we learned, guys? These two problems highlight a few key concepts that are super important in algebra. First, factoring. Recognizing common factors and being able to factor them out is a crucial skill. It simplifies the equations and allows you to see hidden relationships. Secondly, substitution. Using the information we have to replace a part of the equation with its value is also key. This technique helps us to get rid of variables and get closer to the solution. Finally, isolating the variable. After substituting the known values, we needed to isolate 'a'. We did this by performing the opposite operation (division, in our case) on both sides of the equation.

Here are some tips for success:

• Practice, practice, practice! The more problems you solve, the more comfortable you'll become with the concepts. Try different examples and variations of these problems.
• Always read the problem carefully. Understand what's being asked and what information you have.
• Show your work step-by-step. This makes it easier to follow your logic and spot any mistakes.
• Double-check your answers! Always substitute your solution back into the original equations to make sure everything is correct. This step is super important!
• Don't be afraid to ask for help! If you get stuck, ask your teacher, a friend, or look online for help.

Math might seem intimidating at first, but it is absolutely achievable with the right approach and a little bit of practice. Remember to take your time, stay organized, and never give up. Every problem is an opportunity to learn and grow. Keep practicing and you'll get the hang of it! Good job today guys! You have solved two math problems. Now you can solve problems like this by yourself. Keep up the great work! Remember, math is all about logical thinking and problem-solving. Keep practicing, and you'll find it becomes easier and more enjoyable over time!