Solving Linear Equations: Find The Value Of 'w'
Hey math enthusiasts! Let's dive into the world of linear equations and figure out how to find the value of a specific variable. Today, we're tackling a system of two equations with two unknowns, and our mission is to pinpoint the value of 'w'. This is a classic problem in algebra, and understanding the solution process is key to mastering more complex mathematical concepts. So, buckle up, grab your pencils, and let's get started!
Understanding the Problem: The System of Equations
First things first, let's take a look at the system of equations we're dealing with. We have:
- 5v + 4w = 1
- 3v - 6w = 2
These equations represent two straight lines on a graph. The solution to the system is the point where these two lines intersect. This point gives us the values of 'v' and 'w' that satisfy both equations simultaneously. Our goal is laser-focused: to find the specific value of 'w' at this intersection point. No sweat, right? Let's break it down into manageable steps.
To solve this, we can use a variety of methods, but we'll focus on the elimination method here. Why elimination? Because it's often a straightforward and efficient way to solve systems of equations like these. The elimination method involves manipulating the equations to eliminate one of the variables, allowing us to solve for the other. Then, we can substitute the known value back into one of the original equations to find the value of the eliminated variable. Sounds good, yeah?
This problem is a cornerstone in algebra, providing a solid foundation for more advanced topics. It's a fantastic example of how mathematics provides tools to solve real-world problems. For instance, these equations could represent relationships in economics, physics, or any other field where two or more variables interact. So, understanding the how and why behind the solution is super important! The ability to manipulate and solve these equations is a core skill for anyone venturing into the world of STEM or any field that requires analytical thinking.
The Elimination Method: A Step-by-Step Guide
Alright, guys, let's get down to the nitty-gritty. Our goal is to eliminate either 'v' or 'w' to solve for the other. Looking at our equations, it seems easier to eliminate 'v' first. Here's how we'll do it. Notice that the coefficients of 'v' are 5 and 3. To eliminate 'v', we want these coefficients to be opposites. So, we'll multiply the first equation by 3 and the second equation by 5. This gives us:
- 3 * (5v + 4w) = 3 * 1 => 15v + 12w = 3
- 5 * (3v - 6w) = 5 * 2 => 15v - 30w = 10
Now we have two new equations. Notice that the 'v' terms have the same coefficient. Subtract the second new equation from the first new equation to eliminate 'v':
- (15v + 12w) - (15v - 30w) = 3 - 10
- 15v + 12w - 15v + 30w = -7
- 42w = -7
Now, we've successfully eliminated 'v'! We can now easily solve for 'w'. Divide both sides of the equation by 42:
- w = -7 / 42
- w = -1/6
And there you have it! The value of 'w' is -1/6. We've done it! By strategically manipulating the equations, we've isolated 'w' and found its value. Isn't math awesome?
This method is a powerful tool in your math toolbox. It's not just about solving this particular problem; it's about developing a strategic approach to problem-solving. It teaches you to break down complex problems into smaller, more manageable steps, and to apply logical reasoning to find solutions. This skill is invaluable in many areas of life, not just in mathematics. Learning the elimination method is a game changer for tackling systems of equations, and it can be applied to problems of varying complexity.
Verifying the Solution: Checking Our Work
We've found our answer, but it's always a good idea to check our work. To do this, we'll substitute the value of 'w' (-1/6) back into one of the original equations and solve for 'v'. Let's use the first equation:
- 5v + 4w = 1
- 5v + 4(-1/6) = 1
- 5v - 4/6 = 1
- 5v - 2/3 = 1
- 5v = 1 + 2/3
- 5v = 5/3
- v = (5/3) / 5
- v = 1/3
So, we found that v = 1/3. Now, let's substitute both v = 1/3 and w = -1/6 back into the second original equation to verify that our solution is correct:
- 3v - 6w = 2
- 3(1/3) - 6(-1/6) = 2
- 1 + 1 = 2
- 2 = 2
Since the equation holds true, our solution is correct! We can be confident that w = -1/6 and v = 1/3 is the correct solution to the system of equations. Always double-check your work; it's a super-important habit to cultivate in mathematics.
This verification step is crucial. It not only confirms the accuracy of your answer, but it also reinforces your understanding of the concepts. It helps to catch any potential errors early on and builds your confidence in your problem-solving abilities. It shows that both equations are satisfied by the values of v and w, which confirms that our solution is correct. This is a very important step to build in your approach to problem-solving.
Conclusion: Mastering Linear Equations
So, there you have it, folks! We've successfully solved a system of linear equations and found the value of 'w'. We used the elimination method, a powerful tool for solving these types of problems. Remember, the key is to understand the concepts, break down the problem into manageable steps, and always check your work.
This whole process of solving linear equations builds a foundation for more advanced topics in algebra and other areas of mathematics. The skills you develop here – like algebraic manipulation, problem-solving strategies, and logical reasoning – are invaluable in all sorts of fields. Keep practicing, keep exploring, and keep the math adventure going! You've got this!
Solving linear equations is a fundamental skill in mathematics, with applications in various fields like physics, engineering, and computer science. By mastering this concept, you not only improve your math skills but also gain a valuable tool for tackling real-world problems. The ability to understand and manipulate equations is a stepping stone to higher-level concepts and applications. Always remember that practice and a solid understanding of the basics are key to success.