Hey guys! Ever found yourself staring at a system of equations and feeling totally lost? Don't worry, you're not alone! Solving systems of equations can seem daunting, but with the right approach, it becomes a manageable and even kinda fun puzzle. In this guide, we'll break down a specific system step-by-step and provide you with the tools to tackle similar problems on your own. So, let's dive in and conquer those equations!
The Challenge: A Three-Variable System
Our mission today is to solve the following system of three linear equations with three unknowns (x, y, and z):
3x + 2y - 5z = 54
-x + 4y + 8z = 5
7x - 2y + 8z = 7
This might look intimidating at first glance, but fear not! We'll use a systematic method called elimination to find the values of x, y, and z that satisfy all three equations simultaneously. Think of it as a detective game where we eliminate clues (variables) one by one until we uncover the solution.
The Elimination Method: Our Detective Tool
The elimination method is a powerful technique for solving systems of equations. The basic idea is to manipulate the equations (by multiplying them by constants and adding or subtracting them) to eliminate one variable at a time. This reduces the system to a smaller one, which is easier to solve. We'll repeat this process until we're left with a single equation in a single variable, which we can then solve directly. Once we know the value of one variable, we can substitute it back into the previous equations to find the values of the other variables. Let's get started with our system!
Step 1: Eliminating 'y' from Equations 1 and 3
Notice that the coefficients of 'y' in the first and third equations are +2 and -2, respectively. This is perfect for elimination! If we simply add these two equations together, the 'y' terms will cancel out. Let's do it:
(3x + 2y - 5z) + (7x - 2y + 8z) = 54 + 7
Simplifying, we get:
10x + 3z = 61 (Equation 4)
Great! We've eliminated 'y' and now have an equation with only 'x' and 'z'.
Step 2: Eliminating 'y' from Equations 1 and 2
Now, we need to eliminate 'y' from a different pair of equations. Let's choose equations 1 and 2. To eliminate 'y', we need to make the coefficients of 'y' opposites. We can achieve this by multiplying equation 1 by 2:
2 * (3x + 2y - 5z) = 2 * 54
This gives us:
6x + 4y - 10z = 108 (Modified Equation 1)
Now we can add this modified equation to equation 2:
(6x + 4y - 10z) + (-x + 4y + 8z) = 108 + 5
Wait a minute! We made a mistake. We wanted to eliminate y, but by adding the equations directly, we ended up with 8y. Our goal is to have the 'y' terms cancel out. So, instead of adding the equations, we need to subtract equation 2 from the modified equation 1:
(6x + 4y - 10z) - (-x + 4y + 8z) = 108 - 5
Simplifying, we get:
7x - 18z = 103 (Equation 5)
Awesome! We've successfully eliminated 'y' again and now have another equation with only 'x' and 'z'.
Step 3: Solving for 'x' and 'z'
We now have two equations (Equation 4 and Equation 5) with two unknowns ('x' and 'z'):
10x + 3z = 61
7x - 18z = 103
To solve this system, we'll use the elimination method again. Let's eliminate 'z' this time. To do this, we'll multiply Equation 4 by 6:
6 * (10x + 3z) = 6 * 61
Which gives us:
60x + 18z = 366 (Modified Equation 4)
Now we can add the modified Equation 4 to Equation 5:
(60x + 18z) + (7x - 18z) = 366 + 103
Simplifying, we get:
67x = 469
Dividing both sides by 67, we find:
x = 7
Hallelujah! We've found the value of 'x'! Now we can substitute this value back into either Equation 4 or Equation 5 to solve for 'z'. Let's use Equation 4:
10 * 7 + 3z = 61
Simplifying:
70 + 3z = 61
3z = -9
z = -3
We've got 'z' too!
Step 4: Solving for 'y'
Now that we know 'x' and 'z', we can substitute their values back into any of the original three equations to solve for 'y'. Let's use Equation 1:
3 * 7 + 2y - 5 * (-3) = 54
Simplifying:
21 + 2y + 15 = 54
2y + 36 = 54
2y = 18
y = 9
And there it is! We've found the value of 'y'.
The Solution: Unveiling the Mystery
We've successfully solved the system of equations! The solution is:
(x, y, z) = (7, 9, -3)
This means that the values x = 7, y = 9, and z = -3 satisfy all three original equations simultaneously. You can verify this by plugging these values back into the equations and checking if they hold true. Go ahead, give it a try!
Key Takeaways: Mastering the Art of Solving Systems
- The elimination method is a powerful tool for solving systems of equations. It involves strategically manipulating equations to eliminate variables one by one.
- Careful arithmetic is crucial! A small mistake can throw off the entire solution.
- Practice makes perfect! The more you practice solving systems of equations, the more comfortable and confident you'll become.
Beyond the Basics: Exploring Other Methods
While the elimination method is great, it's not the only way to solve systems of equations. Other popular methods include:
- Substitution: Solving one equation for one variable and substituting that expression into the other equations.
- Matrices: Using matrix operations to solve systems, especially useful for larger systems.
Practice Problems: Sharpen Your Skills
Ready to put your newfound skills to the test? Try solving these systems of equations using the elimination method (or any method you prefer!):
-
2x + y = 7 x - y = 2 -
x + 2y - z = 1 2x - y + z = 3 x + y + z = 2
Conclusion: You've Got This!
Solving systems of equations can be a fun and rewarding challenge. By mastering the elimination method and other techniques, you'll be well-equipped to tackle a wide range of mathematical problems. Remember, practice is key, so keep solving and keep learning! And if you ever get stuck, don't hesitate to ask for help. We're all in this together!
So, that's it for this guide. I hope you found it helpful. Now go out there and conquer those equations, guys!