Algebraic Equations: Solving For 'a'

by Alex Johnson 37 views

Welcome, math enthusiasts, to a deep dive into the fascinating world of algebraic equations! Today, we're going to tackle a specific problem: solving for 'a' in the equation 3aβˆ’5(aβˆ’2)=343a - 5(a - 2) = 34. This might seem a bit daunting at first glance, especially with those parentheses throwing a curveball, but fear not! By breaking it down step-by-step, we'll unravel this mystery together. Algebra is like a puzzle, and each step we take brings us closer to the solution. Understanding how to manipulate these equations is a fundamental skill in mathematics, opening doors to more complex problem-solving in various scientific and engineering fields. The beauty of algebra lies in its universality; the principles we use here apply to countless other problems, making it a powerful tool in our intellectual arsenal. So, let's put on our thinking caps and get ready to explore the elegant process of isolating our variable, 'a', and discovering its true value.

Understanding the Equation Structure

Before we jump into the actual solving process for the equation 3aβˆ’5(aβˆ’2)=343a - 5(a - 2) = 34, it's crucial to understand the components at play. We have terms involving our variable 'a', constant numbers, and an expression enclosed in parentheses. The presence of parentheses, like (aβˆ’2)(a - 2), signifies that the operation outside them (in this case, multiplication by -5) applies to everything inside. This is a key concept in the order of operations, often remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). In our equation, the βˆ’5-5 needs to be distributed to both the 'a' and the '-2' within the parentheses. Ignoring this distribution step is a common pitfall, leading to incorrect solutions. Furthermore, recognizing like terms is another vital aspect. 'Like terms' are terms that have the same variable raised to the same power. In our equation, 3a3a and βˆ’5a-5a (after distribution) are like terms because they both involve 'a' to the power of 1. Combining like terms is essential for simplifying the equation and making it easier to solve. The ultimate goal in solving for 'a' is to isolate it on one side of the equals sign, meaning we want to get it to a state where it looks like 'a=extsomenumbera = ext{some number}'. This involves a series of inverse operations – performing the opposite of what is currently being done to 'a' to both sides of the equation to maintain balance.

Step-by-Step Solution

Let's now embark on the journey of solving for 'a' in the equation 3aβˆ’5(aβˆ’2)=343a - 5(a - 2) = 34! Our first mission is to eliminate those pesky parentheses. We do this by distributing the βˆ’5-5 to both terms inside: βˆ’5imesa=βˆ’5a-5 imes a = -5a and βˆ’5imesβˆ’2=+10-5 imes -2 = +10. So, our equation transforms into: 3aβˆ’5a+10=343a - 5a + 10 = 34. See? Already looking a bit friendlier! Now, we identify and combine our like terms. The like terms here are 3a3a and βˆ’5a-5a. When we combine them, we get 3aβˆ’5a=βˆ’2a3a - 5a = -2a. So, the equation becomes: βˆ’2a+10=34-2a + 10 = 34. Our next goal is to isolate the term with 'a' (which is βˆ’2a-2a). To do this, we need to move the constant term, +10+10, to the other side of the equation. We achieve this by performing the inverse operation: subtracting 10 from both sides. So, βˆ’2a+10βˆ’10=34βˆ’10-2a + 10 - 10 = 34 - 10, which simplifies to βˆ’2a=24-2a = 24. We're so close! The variable 'a' is currently being multiplied by βˆ’2-2. To isolate 'a', we perform the inverse operation: divide both sides by βˆ’2-2. Therefore, rac{-2a}{-2} = rac{24}{-2}. This gives us our final answer: a=βˆ’12a = -12. It's always a good practice to check your solution by substituting this value back into the original equation to ensure it holds true. Let's do that: 3(βˆ’12)βˆ’5(βˆ’12βˆ’2)=343(-12) - 5(-12 - 2) = 34. First, βˆ’36βˆ’5(βˆ’14)=34-36 - 5(-14) = 34. Then, βˆ’36+70=34-36 + 70 = 34. Finally, 34=3434 = 34. Success! Our solution is correct.

Why This Matters: Algebraic Thinking

The process of solving for 'a' in the equation 3aβˆ’5(aβˆ’2)=343a - 5(a - 2) = 34 is more than just an arithmetic exercise; it's a fundamental demonstration of algebraic thinking. Algebra, at its core, is about representing unknown quantities with variables and then using logical steps and rules of mathematics to find the values of those unknowns. This equation, while relatively simple, involves several key algebraic principles: the distributive property, combining like terms, and the concept of inverse operations to isolate a variable. These principles are the building blocks for understanding more complex mathematical concepts, from quadratic equations and calculus to statistics and data analysis. In the real world, algebraic thinking is indispensable. Scientists use it to model phenomena, engineers to design structures, economists to predict market trends, and computer programmers to develop algorithms. Even in everyday life, we implicitly use algebraic reasoning when budgeting, planning, or problem-solving. The ability to break down a problem into smaller, manageable steps, to manipulate symbols and equations logically, and to arrive at a verifiable solution are skills that transcend the classroom and are vital for critical thinking and effective decision-making. So, the next time you encounter an equation, remember that you're not just crunching numbers; you're honing a powerful cognitive skill.

Conclusion: Mastering Algebraic Solutions

In conclusion, we have successfully navigated the process of solving for 'a' in the equation 3aβˆ’5(aβˆ’2)=343a - 5(a - 2) = 34, arriving at the answer a=βˆ’12a = -12. We've seen how the distributive property is key to simplifying the expression, how combining like terms streamlines the equation, and how inverse operations are used to isolate the variable. Mastering these techniques is not just about getting the right answer; it's about developing a robust understanding of algebraic principles that will serve you well in all your mathematical endeavors. Remember to always perform operations on both sides of the equation to maintain balance, and don't shy away from checking your solutions. This systematic approach to problem-solving is a transferable skill, applicable to countless situations beyond mathematics. Keep practicing, and you'll find that algebraic equations become increasingly intuitive.

For further exploration into the foundational concepts of algebra, I recommend visiting the Khan Academy mathematics section, a fantastic resource for learning and reinforcing mathematical skills.