Solving SPLDV 0.2x + Y = 11 And X - Y = 1 A Step-by-Step Guide

Hey guys! Are you wrestling with systems of linear equations? Don't worry, you're not alone! Many students find these problems a bit tricky at first, but with a little guidance and practice, you'll be solving them like a pro in no time. In this article, we're going to break down one such system step by step: 0. 2x + y = 11 and x - y = 1. We'll explore different methods and show you how to arrive at the solution. So, grab your pencils and let's dive in!

Understanding Systems of Linear Equations

Before we jump into solving the specific system, let's make sure we're all on the same page about what systems of linear equations actually are. Simply put, a system of linear equations is a set of two or more linear equations that involve the same variables. The solution to the system is the set of values for the variables that make all the equations true simultaneously. Think of it like finding the sweet spot where all the lines intersect on a graph. There are several methods to solve these systems, including substitution, elimination, and graphing. Each method has its own advantages and disadvantages, and the best method to use often depends on the specific equations in the system. In our case, we have two equations:

1. 2x + y = 11
2. x - y = 1

Our goal is to find the values of 'x' and 'y' that satisfy both of these equations. We'll walk through the process using the elimination method, which is particularly handy when you notice that the coefficients of one of the variables are opposites (like the 'y' terms in our equations!). This makes it easy to eliminate one variable and solve for the other. Remember, the key is to manipulate the equations in a way that allows us to isolate one variable and then substitute that value back into one of the original equations to find the other variable. Stay with me, and we'll get through this together!

Solving with the Elimination Method

The elimination method is our weapon of choice for this system, and it's a fantastic tool to have in your math arsenal! The beauty of this method lies in its ability to cancel out one variable by adding or subtracting the equations. Looking at our system:

1. 2x + y = 11
2. x - y = 1

You'll notice something brilliant: the 'y' terms have opposite signs! This is exactly what we want. The 'y' in the first equation is positive, and the 'y' in the second equation is negative. This means that if we add the two equations together, the 'y' terms will magically disappear, leaving us with an equation that only involves 'x'. How cool is that?

So, let's add the equations together:

(0.2x + y) + (x - y) = 11 + 1

Simplifying this, we get:

1. 2x + x = 12
2. 2x = 12

Now we have a simple equation with just one variable, 'x'. To solve for 'x', we divide both sides of the equation by 1.2:

x = 12 / 1.2 x = 10

Boom! We've found the value of 'x'. Now we know that x = 10. But we're not done yet! We still need to find the value of 'y'. This is where the second part of the elimination method comes into play: substitution. We'll take this value of 'x' that we just found and plug it back into one of our original equations to solve for 'y'. Ready to see how it's done?

Finding 'y' with Substitution

Now that we've heroically rescued the value of 'x' (x = 10), it's time to bring 'y' into the spotlight. We'll use the substitution method to find 'y'. This involves plugging the value of 'x' that we just found into one of our original equations. It doesn't matter which equation we choose; we'll get the same answer for 'y' either way. For simplicity's sake, let's use the second equation:

x - y = 1

Now, we substitute x = 10 into this equation:

10 - y = 1

Our mission now is to isolate 'y'. To do this, we can subtract 10 from both sides of the equation:

-y = 1 - 10 -y = -9

But wait! We're not quite there yet. We have '-y' equals -9, but we want to know what 'y' equals. To get rid of the negative sign, we can multiply both sides of the equation by -1:

y = 9

Ta-da! We've found the value of 'y'. So, we now know that y = 9. We've successfully navigated the substitution process and unearthed the value of our second variable. But before we declare victory, let's do one final check to make sure our solution is rock solid.

Verifying the Solution

Before we do a victory dance, it's crucial to verify our solution. We've found that x = 10 and y = 9, but we need to make sure these values actually work in both of our original equations. This is a super important step because it helps us catch any mistakes we might have made along the way. So, let's plug these values back into our equations and see if they hold true.

First, let's check the first equation:

1. 2x + y = 11

Substitute x = 10 and y = 9:

2. 2(10) + 9 = 11
3. 2 + 9 = 11
4. 1 = 11

Hooray! The first equation checks out. Now, let's move on to the second equation:

x - y = 1

Substitute x = 10 and y = 9:

10 - 9 = 1 1 = 1

Yes! The second equation also holds true. This means that our solution, x = 10 and y = 9, is indeed the correct solution to the system of equations. We've successfully verified our answer, and we can now confidently say that we've solved the system. Give yourselves a pat on the back, guys! You've earned it.

Alternative Methods: Substitution Method

While we conquered this system using the elimination method, it's worth noting that the substitution method could also have been used. The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This can be particularly useful when one of the equations is already solved for a variable or can be easily solved for a variable. Let's briefly explore how we might have approached this system using substitution.

Looking at our equations:

1. 2x + y = 11
2. x - y = 1

We can easily solve the second equation for 'x':

x = y + 1

Now, we substitute this expression for 'x' into the first equation:

3. 2(y + 1) + y = 11

Now we have an equation with only 'y'. We can solve for 'y':

4. 2y + 0.2 + y = 11
5. 2y = 10.8
6. y = 9

And just like before, we find that y = 9. We can then substitute this value back into either equation to find 'x'. If we use x = y + 1:

x = 9 + 1 x = 10

So, we arrive at the same solution, x = 10 and y = 9, using the substitution method. This highlights that there's often more than one way to solve a math problem, and choosing the method that feels most comfortable and efficient for you is key. Now that we've explored both elimination and substitution, you have a broader toolkit for tackling systems of linear equations!

Conclusion: Mastering SPLDV

So, there you have it, guys! We've successfully solved the system of linear equations 0.2x + y = 11 and x - y = 1 using the elimination method and explored how the substitution method could also be applied. Remember, the key to mastering these types of problems is practice, practice, practice! The more you work through different systems, the more comfortable you'll become with the various methods and strategies. Don't be afraid to experiment with different approaches and find what works best for you.

Systems of linear equations are a fundamental concept in algebra, and they pop up in all sorts of real-world applications, from engineering and economics to computer science and even everyday problem-solving. So, the skills you're developing here are incredibly valuable. Keep practicing, keep exploring, and keep challenging yourselves. You've got this! And remember, if you ever get stuck, there are tons of resources available to help, including online tutorials, textbooks, and, of course, awesome articles like this one! Keep up the great work, and happy problem-solving!