Solving System Of Equations Find P And Q In 0.5x - 0.3y = 2.3 And 0.4x + 0.7y = 0.9

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Hey everyone! Today, we're diving into the exciting world of simultaneous equations, specifically focusing on how to solve for two unknowns, which in this case, are p and q. While the original title mentions x and y, let's clarify that the process remains the same regardless of the variables used. So, whether you're dealing with x and y, p and q, or any other pair of variables, the fundamental techniques we'll explore here will be your trusty sidekicks. We'll tackle a system of equations: 0.5x - 0.3y = 2.3 and 0.4x + 0.7y = 0.9. Get ready to sharpen your pencils (or fire up your favorite equation solver) as we unravel this mathematical puzzle step by step! Solving systems of equations might seem daunting at first, but trust me, with a bit of practice and the right strategies, you'll be cracking these problems like a pro. Think of it like learning a new language – once you grasp the grammar and vocabulary, you can start stringing together sentences and expressing yourself fluently. Similarly, in the world of math, mastering techniques like substitution and elimination opens up a whole new realm of problem-solving possibilities. We'll break down each method, showing you how to apply them to this specific problem and giving you the confidence to tackle similar challenges in the future. So, stick with me, and let's transform those equation-solving anxieties into triumphant “Aha!” moments!

Understanding Systems of Equations

Before we jump into the solution, let's take a moment to understand what a system of equations actually is. Simply put, it's a set of two or more equations that share the same variables. The goal is to find the values of those variables that satisfy all the equations simultaneously. Think of it like a detective trying to solve a mystery with multiple clues – each equation is a clue, and the solution is the set of values that fits all the clues perfectly. In our case, we have two equations, and we're looking for the values of x and y that make both equations true. There are several methods for solving systems of equations, but the most common ones are the substitution method and the elimination method. We'll explore both of these in detail as we work through our example problem. Understanding the underlying concepts is crucial, guys, because it's not just about memorizing steps; it's about developing a deep understanding that allows you to adapt and apply these techniques to a wide range of problems. So, let's ditch the rote learning and embrace the power of conceptual understanding! When you truly grasp the 'why' behind each step, the 'how' becomes so much easier. This approach not only helps you solve problems more efficiently but also builds a solid foundation for tackling more advanced mathematical concepts in the future. Remember, mathematics is a journey, not just a destination. So, let's embark on this journey together, exploring the fascinating world of equations and uncovering the beauty of logical problem-solving.

Method 1: The Elimination Method

The elimination method, also known as the addition method, is a fantastic technique for solving systems of equations. The core idea is to manipulate the equations so that when you add them together, one of the variables cancels out, leaving you with a single equation in one variable. This simplified equation is then easy to solve, and once you know the value of one variable, you can plug it back into either of the original equations to find the value of the other variable. Let's apply this to our system:

    1. 5x - 0.3y = 2.3
  2. 4x + 0.7y = 0.9

Our goal is to make either the x coefficients or the y coefficients opposites of each other. To eliminate x, we can multiply the first equation by -0.4 and the second equation by 0.5. This will give us -0.2x in the first equation and 0.2x in the second equation. Alternatively, to eliminate y, we can multiply the first equation by 0.7 and the second equation by 0.3. This will give us -0.21y in the first equation and 0.21y in the second equation. Let's choose to eliminate x in this case. Here's what happens when we multiply:

  • Equation 1 multiplied by -0.4: -0.2x + 0.12y = -0.92
  • Equation 2 multiplied by 0.5: 0.2x + 0.35y = 0.45

Now, we add the two modified equations together:

(-0.2x + 0.12y) + (0.2x + 0.35y) = -0.92 + 0.45

The x terms cancel out, leaving us with:

  1. 47y = -0.47

Now, we can easily solve for y by dividing both sides by 0.47:

y = -1

Awesome! We've found the value of y. Now, let's substitute this value back into either of the original equations to find x. Let's use the first equation:

  1. 5x - 0.3(-1) = 2.3
  2. 5x + 0.3 = 2.3
  3. 5x = 2

x = 4

And there you have it! We've solved for x and y using the elimination method. The solution is x = 4 and y = -1. The elimination method is particularly useful when the coefficients of one of the variables are already opposites or can be easily made opposites by multiplying one or both equations by a constant. This method streamlines the process and can save you time and effort. But remember, guys, the key is understanding the 'why' behind each step. When you grasp the underlying logic, you can confidently adapt this technique to a variety of equation-solving scenarios. So, keep practicing, and you'll become an elimination method master in no time!

Method 2: The Substitution Method

Another powerful technique for tackling systems of equations is the substitution method. This method involves solving one equation for one variable and then substituting that expression into the other equation. This effectively eliminates one variable, leaving you with a single equation in one variable that you can easily solve. Once you've found the value of one variable, you can substitute it back into either of the original equations (or the expression you derived earlier) to find the value of the other variable. Let's see how this works with our system of equations:

    1. 5x - 0.3y = 2.3
  2. 4x + 0.7y = 0.9

First, we need to choose one equation and solve it for one variable. Let's choose the first equation and solve for x. To do this, we'll isolate x on one side of the equation:

  1. 5x = 0.3y + 2.3

Now, divide both sides by 0.5:

x = (0.3y + 2.3) / 0.5 x = 0.6y + 4.6

Great! We've now expressed x in terms of y. Next, we substitute this expression for x into the second equation:

  1. 4(0.6y + 4.6) + 0.7y = 0.9

Now, we have an equation with only y as the variable. Let's simplify and solve for y:

  1. 24y + 1.84 + 0.7y = 0.9
  2. 94y + 1.84 = 0.9
  3. 94y = -0.94

y = -1

Fantastic! We've found the value of y, which is -1. Now, let's substitute this value back into the expression we derived for x:

x = 0.6(-1) + 4.6 x = -0.6 + 4.6 x = 4

And there you have it! Using the substitution method, we've found that x = 4 and y = -1. Notice that we arrived at the same solution as with the elimination method, which confirms the accuracy of our calculations. The substitution method shines when one of the equations is already solved for one variable or can be easily solved for one variable. It's a versatile technique that empowers you to tackle a wide range of systems of equations. Just like with the elimination method, understanding the logic behind each step is key. When you grasp the 'why,' the 'how' becomes second nature. So, keep practicing, experiment with different equations, and you'll master the art of substitution in no time!

Solution Verification

It's always a good idea, guys, to verify your solution to ensure accuracy. This is especially important in exams or when working on complex problems. To verify our solution, we simply substitute the values we found for x and y (x = 4 and y = -1) back into the original equations and check if they hold true.

Let's start with the first equation:

  1. 5x - 0.3y = 2.3

Substitute x = 4 and y = -1:

  1. 5(4) - 0.3(-1) = 2.3
  2. 0 + 0.3 = 2.3
  3. 3 = 2.3 (This is true!)

Now, let's check the second equation:

  1. 4x + 0.7y = 0.9

Substitute x = 4 and y = -1:

  1. 4(4) + 0.7(-1) = 0.9
  2. 6 - 0.7 = 0.9
  3. 9 = 0.9 (This is also true!)

Since our solution satisfies both equations, we can confidently say that our solution (x = 4, y = -1) is correct! Verification is a crucial step in the problem-solving process. It's like proofreading your writing before submitting it – it helps you catch any errors and ensures that your answer is accurate. Don't skip this step, guys! It can save you valuable points and boost your confidence in your problem-solving abilities. So, always take a moment to plug your solution back into the original equations and give yourself that extra peace of mind. Think of it as the final flourish on a masterpiece – the perfect way to complete your mathematical creation!

Conclusion

Alright, guys, we've successfully navigated the world of systems of equations and conquered the challenge of solving for x and y in the equations 0.5x - 0.3y = 2.3 and 0.4x + 0.7y = 0.9. We explored two powerful methods – the elimination method and the substitution method – and saw how each can be used to arrive at the same solution. We also emphasized the importance of solution verification to ensure accuracy. Remember, the key to mastering these techniques is practice, practice, practice! The more you work with systems of equations, the more comfortable and confident you'll become in solving them. Don't be afraid to experiment with different methods and see which one works best for you in different situations. And most importantly, remember to break down complex problems into smaller, more manageable steps. This approach not only makes the problem-solving process less daunting but also helps you develop a deeper understanding of the underlying concepts. Whether you're tackling algebra problems, calculus conundrums, or any other mathematical challenge, the skills you've learned here – problem-solving strategies, logical thinking, and attention to detail – will serve you well. So, go forth and conquer, guys! Embrace the beauty of mathematics and enjoy the thrill of solving those equations! You've got this! Keep practicing, keep exploring, and keep those mathematical gears turning!