Solving For 'a': A Step-by-Step Guide

Hey guys! Let's dive into solving this equation: $rac{49}{7-a}=rac{2}{a+5}$. This might look a little intimidating at first, but trust me, we can break it down into manageable steps. The key here is to isolate the variable 'a' and find its value. This falls squarely into the realm of algebra, specifically dealing with fractional equations. We'll use cross-multiplication, simplification, and a bit of rearranging to get to the solution. So, grab your pencils, and let's get started. Understanding how to solve such equations is super important, not just for your math class, but also because the problem-solving skills you gain here are applicable in tons of real-world scenarios. We'll make sure to go through each step carefully, explaining the 'why' behind each move, so you can totally grasp the concept. Are you ready to crack this equation? Let’s get to it and find that value for a!

Step 1: Cross-Multiplication - The Magic Wand

Okay, the first thing we're going to do is get rid of those fractions. And the easiest way to do that is by using cross-multiplication. Think of it like this: you're multiplying the numerator of the first fraction by the denominator of the second fraction, and vice-versa. So, we'll multiply 49 by (a+5) and 2 by (7-a). This gives us the equation: $49(a+5) = 2(7-a)$. See? No more fractions! This is a much cleaner equation to work with. Cross-multiplication is a fundamental technique when dealing with equations involving fractions. It allows us to transform a complex fractional equation into a linear equation, making it easier to solve. The concept is straightforward: you're essentially multiplying both sides of the equation by the denominators of the fractions. This cancels out the denominators on both sides, leaving you with a simplified expression. This is where we start building a strong foundation for tackling these types of problems. Remember, the core idea is to eliminate the fractions, and cross-multiplication is our go-to tool for this purpose. So, make sure you understand this step well. It's the foundation of the rest of the solution!

Cross-multiplication is a fundamental technique in algebra that simplifies equations involving fractions. It essentially eliminates the fractions by multiplying the numerator of one fraction by the denominator of the other. For instance, in our original equation, we multiply 49 by (a + 5) and 2 by (7 - a). This process allows us to transform a fractional equation into a linear equation, which is much easier to solve. The effectiveness of cross-multiplication lies in its ability to eliminate the denominators, which are often the main source of complexity in such equations. By applying this technique, you can quickly simplify the equation and move towards isolating the variable. This step is crucial for transforming the equation into a more manageable form.

Step 2: Expanding the Parentheses - Time to Distribute!

Now, we're going to expand the parentheses using the distributive property. This means multiplying the number outside the parentheses by each term inside the parentheses. So, for the left side, we'll multiply 49 by a and 49 by 5. For the right side, we'll multiply 2 by 7 and 2 by -a. This results in: $49a + 245 = 14 - 2a$. See how we've expanded the equation? Now, it's starting to look a lot more like a standard algebraic equation that we can easily solve. The distributive property is a key concept here, so make sure you understand how it works. Essentially, you're distributing the multiplication across the terms inside the parentheses. It's important to be accurate with your multiplication and pay attention to the signs. A small mistake here can throw off your entire answer! Carefully expanding the parentheses ensures that we don't miss any terms and that we maintain the correct mathematical relationships. We've got to make sure every term is accounted for, and by distributing correctly, we are one step closer to isolating a. This is another critical step, so don't rush through it.

Expanding the parentheses is a crucial step in simplifying the equation. It involves applying the distributive property, which means multiplying the term outside the parentheses by each term inside the parentheses. This process transforms the equation into a more manageable form. In our case, expanding the parentheses gives us a clearer view of the equation's components. Remember to be meticulous with the multiplication and pay attention to the signs. This step ensures that all terms are accounted for correctly, setting the stage for the next steps in solving the equation. The distributive property allows us to eliminate the parentheses and combine like terms, moving us closer to isolating the variable.

Step 3: Isolating the Variable - Getting 'a' Alone!

Alright, it's time to get a all by itself. We'll do this by moving all the terms with a to one side of the equation and all the constant terms (the numbers without a) to the other side. First, let's add 2a to both sides of the equation. This gets rid of the -2a on the right side. The equation becomes: $49a + 2a + 245 = 14$. Then, let’s subtract 245 from both sides to get the constant terms on the right side. The equation now looks like this: $51a = 14 - 245$. We're making progress! By adding and subtracting the same values from both sides, we maintain the equality of the equation. The goal is to collect all the a terms together and all the constant terms together. This process may seem simple, but it's fundamentally important in solving algebraic equations. Remember, the core principle is to perform the same operation on both sides of the equation to keep it balanced. Isolating the variable is like giving a its own space, so we can see its exact value clearly. By moving terms strategically, we bring a into focus.

Isolating the variable is a fundamental step in solving any algebraic equation. It involves rearranging the equation to have all terms containing the variable on one side and all the constant terms on the other. This process requires adding or subtracting terms from both sides of the equation to maintain the balance. For example, to isolate 'a' in our equation, we first add 2a to both sides and then subtract 245 from both sides. The goal is to group like terms together, making it easier to solve for the variable. This step is about organizing the equation in a way that allows us to see the variable in isolation, revealing its value. Understanding how to isolate the variable is crucial for successfully solving a wide range of algebraic problems.

Step 4: Simplifying and Solving - The Final Countdown

Now, let's simplify our equation. We'll combine the terms with a on the left side: $51a = -231$. Next, we'll divide both sides by 51 to isolate a: $a = rac{-231}{51}$. Finally, let's simplify that fraction. -231 divided by 51 is -4.529, or approximately -4.53. Boom! We've found the value of a. The last step is all about making things look as neat and clear as possible. Simplifying the equation and performing the final calculations bring us to the solution. Always double-check your arithmetic to make sure you didn’t make any simple mistakes. This is the moment of truth! We are trying to find the single number that fits a, and we have succeeded. Always check your work – this will help you to build confidence in your math skills! You can check your answer by plugging it back into the original equation to see if it makes the equation true. So, there you have it, folks! We've successfully solved for a. This entire process underscores the power of algebraic techniques for problem-solving. It's a journey from a complex fractional equation to a simple, clean solution, all made possible by understanding and applying these fundamental principles.

In this final step, we simplify the equation and solve for the variable. This involves performing the necessary calculations to isolate the variable and determine its value. For example, in our equation, we divide both sides by 51 to isolate 'a'. Always simplify fractions to their simplest form. Then, we find the solution, which is the value of 'a' that satisfies the original equation. We then arrive at the final answer. Double-check your arithmetic to avoid any errors. This step is about bringing the solution to light. This process highlights the practical application of algebraic principles.

Step 5: Verification (Important!) - Checking Your Work

Always, always, always check your answer! Substitute the value of a back into the original equation: $rac49}{7-(-4.53)}=rac{2}{-4.53+5}$. Let's simplify the denominator on both sides $rac{49{11.53}=rac{2}{0.47}$. Now, let's use a calculator to get the decimal values: 49/11.53 is approximately 4.25, and 2/0.47 is approximately 4.25. Since both sides are equal, we can confirm that our solution is correct! This is a crucial step! Verifying your solution ensures that you have found the correct answer and reinforces your understanding of the equation. It's easy to make a small calculation error, so checking your work is super important. This helps build confidence and ensures accuracy. Verification is not just about confirming the answer; it's about solidifying the entire problem-solving process. This step is about closing the loop and confirming the accuracy of your answer.

Verification is an essential step in solving any mathematical equation. It involves substituting the solution back into the original equation to ensure that it satisfies the equation. This step helps to identify any errors that may have occurred during the solving process. In our case, we substitute the value of 'a' back into the original equation and verify that both sides are equal. Verification ensures the accuracy of the solution and strengthens the understanding of the equation-solving process. This step is about checking and confirming the accuracy of the answer.

Conclusion: You Did It!

Congratulations, you guys! We successfully solved for a in our equation. We started with a fractional equation and, through a series of steps (cross-multiplication, expanding, isolating, and simplifying), we arrived at a solution. Remember, practice makes perfect! The more you practice these types of problems, the easier they will become. Don’t be afraid to try different problems and to learn from your mistakes. Algebra is a powerful tool, and with a little bit of effort, you can master it. Keep up the great work, and keep practicing! I hope you found this guide helpful. If you have any questions, feel free to ask! See you in the next problem!

Solving for a variable in an equation, such as $rac{49}{7-a}=rac{2}{a+5}$, involves a systematic approach that utilizes several key algebraic techniques. Initially, cross-multiplication is used to eliminate the fractions, converting the equation into a more manageable linear form. Expanding the parentheses, using the distributive property, is the next crucial step. This clears the parentheses and prepares the equation for further simplification. Then, isolating the variable involves rearranging the equation to place the variable terms on one side and the constant terms on the other. This is done by adding or subtracting terms from both sides of the equation to maintain balance. Finally, simplifying and solving the equation, which includes combining like terms and performing basic arithmetic operations, results in determining the value of the variable. Checking the solution by substituting it back into the original equation verifies the accuracy of the solution and ensures that the answer satisfies the original equation. Each step is essential for obtaining a correct and reliable solution, demonstrating the practicality and utility of algebraic principles in problem-solving.