Solve -3 = 9(5-2k)/5: A Step-by-Step Guide

Introduction

Hey guys! Let's dive into solving a mathematical equation today. We're tackling the equation -3 = 9(5-2k)/5. This type of problem falls under basic algebra and involves simplifying expressions, distributing values, and isolating the variable k to find its value. This guide will break down the solution step-by-step, making it super easy to follow along. Whether you're a student brushing up on algebra or just someone who enjoys a good math challenge, you've come to the right place. We'll not only solve the equation but also understand the underlying principles so you can confidently tackle similar problems in the future. So, let's jump right in and unravel this equation together!

Breaking Down the Equation

Our equation is -3 = 9(5-2k)/5. The first thing we want to do is get rid of that pesky fraction. Fractions can make equations look intimidating, but don't worry, we've got this! To eliminate the fraction, we need to multiply both sides of the equation by the denominator, which in this case is 5. By multiplying both sides by 5, we maintain the balance of the equation – think of it like a seesaw; whatever you do to one side, you have to do to the other to keep it level. So, multiplying both sides by 5, we get: -3 * 5 = [9(5-2k)/5] * 5. The 5 in the denominator on the right side cancels out with the 5 we're multiplying by, simplifying things nicely. This leaves us with -15 = 9(5-2k). Now, the equation looks a lot cleaner and more manageable, doesn't it? We've successfully cleared the fraction and are ready for the next step in solving for k. This initial step of eliminating the fraction is a crucial one, as it sets the stage for easier simplification and solving of the equation.

Distributing and Simplifying

Now that we have -15 = 9(5-2k), the next step is to distribute the 9 across the terms inside the parentheses. Remember the distributive property? It's like sharing the 9 with both the 5 and the -2k. So, we multiply 9 by both 5 and -2k. 9 multiplied by 5 is 45, and 9 multiplied by -2k is -18k. This transforms our equation into -15 = 45 - 18k. See how we're making progress? We've removed the parentheses and now have a simpler equation to work with. The next part of simplifying involves getting the terms with k on one side and the constants on the other. To do this, we'll subtract 45 from both sides of the equation. This is another way we keep the equation balanced – by performing the same operation on both sides. Subtracting 45 from both sides gives us -15 - 45 = 45 - 18k - 45. This simplifies to -60 = -18k. We're getting closer and closer to isolating k and finding its value. Each step we take is a strategic move to peel away the layers and reveal the solution.

Isolating the Variable

We've reached the point where our equation is -60 = -18k. Now, the goal is to get k all by itself on one side of the equation. To do this, we need to undo the multiplication that's happening between -18 and k. The opposite of multiplication is division, so we'll divide both sides of the equation by -18. Dividing both sides by -18 keeps the equation balanced, just like we've been doing all along. So, we have -60 / -18 = (-18k) / -18. When we perform the division, we get k isolated on the right side, and on the left side, we have -60 divided by -18. A negative number divided by a negative number results in a positive number. So, -60 / -18 simplifies to 60/18. But we're not quite done yet! We can simplify this fraction further. Both 60 and 18 are divisible by 6. Dividing 60 by 6 gives us 10, and dividing 18 by 6 gives us 3. So, the simplified fraction is 10/3. Therefore, our solution is k = 10/3. We've successfully isolated k and found its value! This final step is the culmination of all the previous steps, bringing us to the answer we were seeking.

Checking the Solution

Alright, we've found that k = 10/3, but how do we know if we're right? This is where checking our solution comes in super handy. It's like a safety net, ensuring we haven't made any mistakes along the way. To check our solution, we'll substitute k = 10/3 back into the original equation: -3 = 9(5-2k)/5. So, we replace k with 10/3, which gives us -3 = 9(5 - 2(10/3))/5. Now, we need to simplify this expression and see if both sides of the equation are equal. First, let's tackle the term inside the parentheses: 2(10/3) is 20/3. So, we have -3 = 9(5 - 20/3)/5. Next, we need to subtract 20/3 from 5. To do this, we need a common denominator, so we rewrite 5 as 15/3. Now we have -3 = 9(15/3 - 20/3)/5, which simplifies to -3 = 9(-5/3)/5. Now, let's multiply 9 by -5/3. This gives us -45/3, which simplifies to -15. So, our equation becomes -3 = -15/5. Finally, we divide -15 by 5, which gives us -3. So, we have -3 = -3. Yay! Both sides of the equation are equal, which means our solution k = 10/3 is correct. Checking our solution not only gives us confidence in our answer but also reinforces our understanding of the equation-solving process.

Alternative Methods for Solving

While we've solved the equation -3 = 9(5-2k)/5 using a step-by-step approach, it's always good to know there are often multiple paths to the same destination in mathematics. This not only broadens our problem-solving toolkit but also deepens our understanding of the underlying concepts. One alternative method we could have used involves cross-multiplication. Remember, cross-multiplication is a handy shortcut when you have a fraction equal to another fraction or a whole number (which can be thought of as a fraction with a denominator of 1). In our case, we can rewrite the equation as -3/1 = 9(5-2k)/5. Now, we can cross-multiply, which means multiplying the numerator of the first fraction by the denominator of the second fraction, and vice versa. This gives us -3 * 5 = 1 * 9(5-2k), which simplifies to -15 = 9(5-2k). Notice that this is the same equation we arrived at after our first step in the original method! From here, we would proceed with distributing and isolating k as before. This alternative method demonstrates how a different initial approach can lead us to the same point in the solution process. Exploring different methods not only makes math more interesting but also helps us develop a more flexible and intuitive understanding of algebraic manipulations. Another approach could involve dividing both sides by 9 right after clearing the fraction. This would lead to smaller numbers and might simplify the arithmetic slightly. The key takeaway is that understanding the properties of equality allows us to manipulate equations in various ways to reach the solution.

Common Mistakes to Avoid

When tackling equations like -3 = 9(5-2k)/5, it's easy to make a few common mistakes if you're not careful. Identifying these pitfalls can help you steer clear of them and boost your accuracy. One frequent error is forgetting to distribute properly. When we have a term multiplied by an expression in parentheses, like 9(5-2k), we need to multiply the 9 by both terms inside the parentheses. Some people might just multiply 9 by 5 but forget to multiply it by -2k, which would lead to an incorrect equation and, ultimately, a wrong answer. Another common mistake is mishandling negative signs. Negative signs are like tiny mathematical landmines – one wrong move and you can end up with the wrong sign on your answer. For example, when we had -15 = 45 - 18k, we subtracted 45 from both sides to get -60 = -18k. A mistake here could be adding 45 instead of subtracting, or forgetting the negative sign on the -18k. These seemingly small errors can throw off the entire solution. Another area where mistakes often occur is with fractions. When clearing fractions, it's crucial to multiply every term on both sides of the equation by the common denominator. Forgetting to multiply even one term can throw off the balance of the equation. Finally, it's always a good idea to double-check your work, especially the arithmetic. Simple calculation errors can happen to anyone, but they're easily caught with a quick review. By being aware of these common pitfalls, you can approach equation-solving with more confidence and precision.

Real-World Applications of Solving Equations

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