Solving Systems Of Equations: A Step-by-Step Guide

Hey guys! Let's dive into the fascinating world of solving systems of equations. This is a crucial skill in mathematics, and we're going to break it down step by step, making it super easy to understand. We'll be tackling a system of three equations, which might seem daunting at first, but trust me, with the right approach, it's totally manageable. We will use the elimination method, a classic technique that simplifies the problem by strategically eliminating variables. We'll start by focusing on eliminating the variable 'z', then we will move to solve the resulting equations. This article will also show you how to check your work and ensure you've found the correct solution. Ready to get started? Let's go!

Understanding the Problem: The System of Equations

Alright, so we're starting with a system of three equations. That means we have three equations, and we need to find values for x, y, and z that satisfy all three equations simultaneously. Think of it like a puzzle where the solution has to fit perfectly into each piece. Here's our system:

1. 1. 2x + y + 3z = 12
2. 2. x - 2y + z = -5
3. 3. 5x - y + 2z = 5

Each equation represents a plane in 3D space. The solution to the system is the point (or points) where all three planes intersect. In this case, we're hoping for a single point of intersection, which will give us a unique solution for x, y, and z. The equations are linear, meaning the variables are raised to the power of 1. This makes the system easier to solve compared to non-linear equations. We can use methods like substitution, elimination, or matrix operations to find the solution. The elimination method is particularly efficient when we can easily manipulate the coefficients of the variables to cancel them out. It involves multiplying equations by constants and then adding or subtracting them to eliminate one variable at a time. The goal is to reduce the system to a simpler form that we can solve more readily. By carefully choosing which variables to eliminate and the order in which we do so, we can systematically work towards finding the values of x, y, and z that satisfy all three equations. Remember, the key is to stay organized and patient, as it might take a few steps to reach the final answer. The ability to solve systems of equations is essential in various fields, including physics, engineering, and economics, where you often encounter problems that involve multiple variables and constraints.

Eliminating 'z' Using Equations 2 and 3

Okay, let's start eliminating variables. We're going to focus on getting rid of 'z' first. To do this effectively, we'll use equations 2 and 3: x - 2y + z = -5 and 5x - y + 2z = 5. The goal is to manipulate these equations so that when we add or subtract them, the 'z' terms cancel out. Take a look at the coefficients of z. In equation 2, it's 1, and in equation 3, it's 2. To make them cancel, we can multiply equation 2 by -2. This changes the z coefficient to -2, and then, when we add the modified equation to equation 3, the z terms will vanish.

So, let's multiply equation 2 by -2:

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

This simplifies to:

-2x + 4y - 2z = 10

Now, we add this new equation to equation 3 (5x - y + 2z = 5):

  -2x + 4y - 2z = 10
+ 5x - y + 2z = 5

Adding these together, we get:

3x + 3y = 15

We have successfully eliminated 'z' and created a new equation with only x and y. Let's call this equation 4. This is a critical step because it simplifies our system, reducing the number of variables in one of the equations. The strategy here is to eliminate one variable at a time, eventually arriving at a single equation with a single variable that can be solved directly. Now, we have a simpler equation to work with, which is a significant progress towards finding the solution for x, y, and z. Keep in mind that you could choose to eliminate another variable first, but eliminating 'z' as our first step is an arbitrary but reasonable decision. The important thing is to be consistent with the calculations and to keep track of the equations you are creating. Remember, each time you eliminate a variable, you simplify the system and make it easier to solve.

Creating Another Equation Without 'z'

We're making great progress! We've eliminated 'z' once, but we need another equation without 'z' to further simplify our system. Now, let's use equations 1 (1.2x + y + 3z = 12) and 2 (x - 2y + z = -5) to eliminate 'z' again. The goal is to manipulate these equations so that we can cancel out the 'z' terms when we add or subtract them. Notice the coefficients of 'z' are 3 and 1. To make them cancel, we can multiply equation 2 by -3. This will change the 'z' coefficient to -3. This way, when we add the modified equation 2 to equation 1, the 'z' terms will vanish.

So, let's multiply equation 2 by -3:

-3(x - 2y + z) = -3(-5)

This simplifies to:

-3x + 6y - 3z = 15

Now, we add this new equation to equation 1 (1.2x + y + 3z = 12):

  1.  2x + y + 3z = 12
+ -3x + 6y - 3z = 15

Adding these together, we get:

-1.8x + 7y = 27

Let's call this equation 5. Now, we have two new equations, equation 4 (3x + 3y = 15) and equation 5 (-1.8x + 7y = 27), both without 'z'. This is exactly what we wanted: two equations with two unknowns. From these two equations, we can solve for x and y. By repeatedly eliminating variables, we have reduced the complexity of the original system of equations. Our next task is to use equations 4 and 5 to find the values of x and y. Remember, each step brings us closer to the solution.

Solving for 'x' and 'y'

Alright, we now have two equations with two variables: equation 4 (3x + 3y = 15) and equation 5 (-1.8x + 7y = 27). Our next task is to solve for x and y. Let's use the elimination method again, but this time to eliminate either x or y. To do this, we can multiply one or both equations by constants so that the coefficients of either x or y are opposites. Let's aim to eliminate 'x'. To make the x coefficients opposites, we can multiply equation 4 by 0.6. This gives us 1.8x, and we can add the modified equation 4 to equation 5, and the x term will disappear.

So, multiply equation 4 by 0.6:

0. 6(3x + 3y) = 0.6(15)

This simplifies to:

1. 8x + 1.8y = 9

Now, add this new equation to equation 5 (-1.8x + 7y = 27):

  1.  8x + 1.8y = 9
+ -1.8x + 7y = 27

Adding these together, we get:

8. 8y = 36

Now, divide both sides by 8.8 to solve for y:

y = 36 / 8.8 y = 4.09

Well, let's round that to 4. We now have a value for y. We can substitute this value back into either equation 4 or equation 5 to find x. Let's use equation 4 (3x + 3y = 15) and substitute y = 4:

3x + 3(4) = 15 3x + 12 = 15 3x = 3 x = 1

So, we've found that x = 1 and y = 4. We can now use these values to solve for z. Using the elimination method allows us to systematically reduce the number of variables until we can isolate and solve for each one. This approach is highly effective in solving systems of linear equations.

Solving for 'z'

Fantastic! Now that we have the values for x (x=1) and y (y=4), we can easily solve for z. We can substitute these values into any of the original three equations. Let's choose equation 2 (x - 2y + z = -5) because it looks simple to work with. Remember, we found x = 1 and y = 4, so let's plug these values into the equation:

1 - 2(4) + z = -5 1 - 8 + z = -5 -7 + z = -5

Now, solve for z by adding 7 to both sides: z = -5 + 7 z = 2

So, z = 2. We now have our solution: x = 1, y = 4, and z = 2. That means our three planes intersect at the point (1, 4, 2). The method of substitution is a powerful tool to solve for the remaining variables once the others have been found. By plugging in the known values, we can isolate and solve for the unknown variable. The solution represents the point in 3D space where all three equations are satisfied simultaneously. The strategy of substitution provides a direct path to the solution once some of the variables are known.

Checking Your Work

Alright, we've got our solution, but it's always a good idea to check our work. It's easy to make a small calculation error, so let's plug our values (x = 1, y = 4, z = 2) back into the original three equations to make sure everything holds up. This is a crucial step to ensure the accuracy of the result.

Equation 1: 1.2x + y + 3z = 12

1. 2(1) + 4 + 3(2) = 12
2. 2 + 4 + 6 = 12 12 = 12 (This checks out!)

Equation 2: x - 2y + z = -5

1 - 2(4) + 2 = -5 1 - 8 + 2 = -5 -5 = -5 (This checks out!)

Equation 3: 5x - y + 2z = 5

5(1) - 4 + 2(2) = 5 5 - 4 + 4 = 5 5 = 5 (This checks out!)

Since all three equations hold true with our values for x, y, and z, we can be confident that our solution is correct. Checking the solution is a vital step in solving systems of equations because it helps to identify any mistakes and ensures that the final result is accurate. By carefully substituting the values back into the original equations, you can verify that the solution satisfies all the conditions of the system. This step gives you confidence in the correctness of your work and allows you to catch any errors that might have occurred during the solving process. Always remember to check your work; it's a critical part of the problem-solving process and helps build your confidence.

Conclusion

We did it, guys! We successfully solved a system of three equations using the elimination method. We systematically eliminated variables, creating new equations until we could isolate and solve for each variable. We then checked our work to make sure our solution was correct. This method is incredibly useful, and with practice, you'll become more comfortable and confident in solving these types of problems. Remember the key steps: eliminate a variable, create new equations, and substitute the values to find the final solution. Keep practicing, and you'll be solving these equations in no time. If you're looking for more practice, try solving other systems of equations, or maybe even explore the substitution method. Keep up the great work, and happy solving! By mastering this method, you've gained a valuable tool that can be used to solve a wide range of mathematical problems. Keep practicing, and you'll become proficient in solving systems of equations with ease.