Solving X + 2y - Z = 4, 2x + 4y - Z = 5, 5x - 4y + Z = 2 A Step-by-Step Guide

Hey guys! Ever stumbled upon a system of linear equations that looks like a tangled mess? You're not alone! Many students find these types of problems challenging, but don't worry, we're here to break it down and make it super easy to understand. In this article, we're going to tackle a specific system of three equations with three variables: x + 2y - z = 4, 2x + 4y - z = 5, and 5x - 4y + z = 2. We'll go through each step in detail so you can confidently solve similar problems in the future. Get ready to become a master of linear equations!

Understanding the Problem: What Are We Solving For?

Before we dive into the nitty-gritty, let's make sure we understand what we're trying to achieve. We have a system of three linear equations:

1. x + 2y - z = 4
2. 2x + 4y - z = 5
3. 5x - 4y + z = 2

Our goal is to find the values of x, y, and z that satisfy all three equations simultaneously. Think of it like finding the single point where three planes intersect in 3D space. Each equation represents a plane, and the solution is the common point they share. This is a fundamental concept in algebra and has applications in various fields like engineering, economics, and computer science. We will explore how different methods can help us pinpoint this intersection, making sure you grasp the underlying logic every step of the way.

The key to solving such systems lies in systematically eliminating variables until we can isolate one variable and find its value. Once we know one variable, we can substitute it back into the equations to find the others. There are several methods to achieve this, such as substitution, elimination, and matrix methods. In this guide, we will focus on the elimination method, which is a powerful and often efficient way to solve systems of linear equations. This method will allow us to strategically combine equations to cancel out variables, simplifying the problem step-by-step. We'll walk through the process carefully, showing you how to choose the right steps and avoid common pitfalls. Remember, the goal is not just to get the right answer, but to understand why the method works. This understanding will empower you to tackle any system of linear equations, no matter how complex it may seem at first glance. By the end of this guide, you’ll not only be able to solve this specific problem but also have a solid foundation for handling similar challenges in your studies and beyond. So, let's get started and unlock the secrets of solving linear equations!

Step 1: Choosing Which Variables to Eliminate

Okay, the first step in the elimination method is deciding which variable we want to get rid of first. Looking at our equations:

1. x + 2y - z = 4
2. 2x + 4y - z = 5
3. 5x - 4y + z = 2

We can see that the variable z appears with coefficients of -1, -1, and +1. This is great news because it means we can easily eliminate z by adding equations together. The goal here is to look for variables that have either the same coefficient or coefficients that are easy to make the same (or opposite) by multiplying one or more equations by a constant. Eliminating z looks promising because the coefficients are already close to being opposites. This strategic choice will simplify our calculations and reduce the chances of making errors.

Why is eliminating a variable so important? It's because each time we eliminate a variable, we reduce the complexity of the system. We go from three equations with three unknowns to two equations with two unknowns, which is significantly easier to solve. This process of simplification is at the heart of the elimination method. Think of it like peeling away layers of an onion – each layer we remove brings us closer to the core. In our case, the core is the solution – the values of x, y, and z that satisfy all equations. Choosing the right variable to eliminate first can save you a lot of time and effort. Sometimes, a little bit of planning at the beginning can make the entire process smoother and more efficient. So, let's stick with eliminating z for now and move on to the next step, where we'll actually perform the elimination.

Step 2: Eliminating z from Equations 1 and 3

Alright, let's start eliminating z! We'll begin by focusing on Equations 1 and 3:

1. x + 2y - z = 4
2. 5x - 4y + z = 2

Notice that the coefficients of z are -1 and +1. This makes our job super easy! All we need to do is add these two equations together. When we add the left-hand sides, the -z in Equation 1 will cancel out the +z in Equation 3. This is the magic of the elimination method in action! By carefully choosing which equations to combine, we can strategically eliminate variables and simplify the system. Let's go ahead and perform the addition:

(x + 2y - z) + (5x - 4y + z) = 4 + 2

Now, let's combine like terms:

x + 5x + 2y - 4y - z + z = 6

This simplifies to:

6x - 2y = 6

Great! We've successfully eliminated z and obtained a new equation with only x and y. Let's call this Equation 4:

4. 6x - 2y = 6

This equation is a significant step forward because it reduces the problem to a system of two variables. We're one step closer to finding the values of x, y, and z. This process highlights the power of the elimination method: by strategically combining equations, we can systematically reduce the complexity of the problem. Now, we need to eliminate z from another pair of equations to get another equation in terms of x and y. This will allow us to create a system of two equations with two unknowns, which we can then solve using various techniques. So, let's move on to the next step and eliminate z again!

Step 3: Eliminating z from Equations 2 and 3

Okay, we've eliminated z from Equations 1 and 3. Now, let's eliminate z again, this time using Equations 2 and 3:

1. 2x + 4y - z = 5
2. 5x - 4y + z = 2

Just like before, the coefficients of z are -1 and +1, so we can simply add the equations together:

(2x + 4y - z) + (5x - 4y + z) = 5 + 2

Combining like terms, we get:

2x + 5x + 4y - 4y - z + z = 7

Simplifying, we have:

7x = 7

Awesome! We've eliminated z and y in one go, leaving us with a very simple equation. Let's call this Equation 5:

5. 7x = 7

This is a huge breakthrough! We've isolated x and can easily solve for its value. This demonstrates the elegance of the elimination method when things work out nicely. Sometimes, strategic choices and a bit of luck can lead to significant simplifications. Now, we're in a fantastic position to start unraveling the solution. We know the value of x, and we can use this information to find the values of y and z. This is like a domino effect – once we knock down the first domino (find the value of x), the rest will follow. So, let's move on to the next step and find the value of x!

Step 4: Solving for x

This step is super straightforward. We have Equation 5:

7x = 7

To solve for x, we simply divide both sides of the equation by 7:

7x / 7 = 7 / 7

This gives us:

x = 1

Woohoo! We've found the value of x! This is a major milestone in solving the system of equations. Knowing the value of one variable opens the door to finding the others. Think of it as having one piece of a puzzle – now we can start fitting the other pieces together. This step highlights the importance of simplification in problem-solving. By systematically eliminating variables, we've reduced the complex system to a simple equation that we can easily solve. Now that we know x = 1, we can substitute this value back into one of our earlier equations to solve for y. This process of substitution is a key technique in solving systems of equations, and it allows us to leverage the information we've already obtained. So, let's move on to the next step and use the value of x to find y!

Step 5: Solving for y

Now that we know x = 1, we can substitute this value into Equation 4 to solve for y. Remember Equation 4? It was:

6x - 2y = 6

Let's substitute x = 1 into this equation:

6(1) - 2y = 6

This simplifies to:

6 - 2y = 6

Now, we want to isolate y. First, let's subtract 6 from both sides:

6 - 2y - 6 = 6 - 6

This gives us:

-2y = 0

Finally, we divide both sides by -2:

-2y / -2 = 0 / -2

So, we get:

y = 0

Excellent! We've found the value of y! We're now two-thirds of the way to solving the entire system. This step demonstrates the power of substitution – using the values we've already found to uncover more information. By substituting x = 1 into Equation 4, we were able to transform an equation with two unknowns into an equation with just one unknown, which we could easily solve. This is a common strategy in algebra and a valuable tool for tackling complex problems. Now that we know both x = 1 and y = 0, we can move on to the final step: finding the value of z. We'll use these values and substitute them into one of the original equations to solve for z. So, let's finish this puzzle and find the last piece!

Step 6: Solving for z

We're in the home stretch! We know x = 1 and y = 0. Now, let's find z. We can substitute these values into any of the original equations. Let's use Equation 1, as it looks the simplest:

x + 2y - z = 4

Substituting x = 1 and y = 0, we get:

1 + 2(0) - z = 4

This simplifies to:

1 + 0 - z = 4

1 - z = 4

Now, let's isolate z. Subtract 1 from both sides:

1 - z - 1 = 4 - 1

-z = 3

Finally, multiply both sides by -1:

(-1)(-z) = (-1)(3)

So, we get:

z = -3

We did it! We've found the value of z! We now have the complete solution to the system of equations. This final step highlights the importance of carefully tracking your progress and using the information you've already gathered. By substituting the values of x and y into one of the original equations, we were able to isolate z and find its value. This is a common strategy in solving multi-step problems – break the problem down into smaller steps, solve each step, and then use the results to move forward. Now that we have the values of x, y, and z, let's write down our final solution and then verify that it satisfies all three original equations.

Step 7: The Solution and Verification

Okay, we've found our solution! We have:

• x = 1
• y = 0
• z = -3

So, the solution to the system of equations is the ordered triple (1, 0, -3). But before we celebrate, let's make sure we're right! It's crucial to verify our solution by substituting these values back into the original equations. This is a simple but essential step that can catch any errors we might have made along the way. Let's start with Equation 1:

1. x + 2y - z = 4

Substituting our values, we get:

1 + 2(0) - (-3) = 4

1 + 0 + 3 = 4

4 = 4

Great! It checks out for Equation 1. Now let's try Equation 2:

2. 2x + 4y - z = 5

Substituting, we get:

2(1) + 4(0) - (-3) = 5

2 + 0 + 3 = 5

5 = 5

Perfect! It works for Equation 2 as well. Finally, let's check Equation 3:

3. 5x - 4y + z = 2

Substituting, we get:

5(1) - 4(0) + (-3) = 2

5 - 0 - 3 = 2

2 = 2

Awesome! Our solution satisfies all three original equations. This verification step gives us confidence that we've solved the system correctly. It's like getting a gold star on your math homework! Now we can confidently say that the solution to the system of equations is indeed (1, 0, -3). This entire process demonstrates the power of systematic problem-solving. By breaking down the problem into smaller, manageable steps, and by using techniques like elimination and substitution, we were able to successfully find the solution. So, congratulations! You've mastered solving this system of linear equations, and you're well-equipped to tackle similar problems in the future!

Conclusion: You've Cracked the Code!

Alright, guys, we've done it! We successfully solved the system of equations x + 2y - z = 4, 2x + 4y - z = 5, and 5x - 4y + z = 2. We walked through the entire process step-by-step, from choosing which variable to eliminate first to verifying our final solution. You now have a solid understanding of how to use the elimination method to solve systems of linear equations. Remember, the key is to be systematic, break the problem down into smaller steps, and double-check your work along the way. Solving systems of linear equations might seem daunting at first, but with practice and a clear understanding of the methods involved, you can confidently tackle these problems. This skill is not just useful for exams; it's a fundamental tool in many areas of science, engineering, and mathematics. So, keep practicing, and you'll become a pro in no time! Now go forth and conquer those equations!