Solving A System Of Linear Equations: A Step-by-Step Guide

Hey guys! Ever stumbled upon a set of equations that look like a jumbled mess? Don't worry, we've all been there. In this article, we're going to break down how to solve a system of linear equations. We'll use a real example to make sure you get the hang of it. Let's dive in!

Understanding the Problem

So, what are we even trying to do? When we have a system of linear equations, we're looking for the values of the variables (usually x, y, and z) that make all the equations true at the same time. Think of it like finding the exact spot where all the lines (or planes, in 3D) intersect. In our case, we have these equations:

1. 2x - y + 2z = 8
2. x + 2y - z = -1
3. x + y + z = 1

Our mission is to find the values of x, y, and z that satisfy all three of these equations. There are a few ways to tackle this, but we're going to use the substitution or elimination method. Let's get started!

Choosing a Method: Elimination

When solving systems of equations, you've got a few cool tools in your arsenal. Substitution and elimination are the big ones. Sometimes, one method is clearly easier than the other. Looking at our equations, elimination seems like a solid choice because we can easily add or subtract equations to get rid of one variable. The elimination method involves strategically adding or subtracting multiples of the equations to eliminate one variable at a time. This simplifies the system until we can solve for the remaining variables. Let's start by eliminating x from the second and third equations. We can achieve this by subtracting the third equation from the second equation.

Step 1: Eliminate x from Equations 2 and 3

Let's subtract equation (3) from equation (2):

(x + 2y - z) - (x + y + z) = -1 - 1 x + 2y - z - x - y - z = -2 y - 2z = -2

Now we have a new equation:

4. y - 2z = -2

Next, we want to eliminate x from equation (1) as well, using equation (3). To do this, we'll subtract twice equation (3) from equation (1):

(2x - y + 2z) - 2(x + y + z) = 8 - 2(1) 2x - y + 2z - 2x - 2y - 2z = 8 - 2 -3y = 6 y = -2

Great! We've found the value of y! Now we can use this value to find z using equation (4).

Step 2: Solve for z

Plug the value of y into equation (4):

y - 2z = -2 -2 - 2z = -2 -2z = 0 z = 0

Alright, we've got z = 0. Now we just need to find x.

Step 3: Solve for x

We can use any of the original equations to solve for x. Let's use equation (3) since it looks the simplest:

x + y + z = 1 x + (-2) + 0 = 1 x - 2 = 1 x = 3

So, we've found that x = 3.

Step 4: Verify the Solution

Before we celebrate, let's make sure our solution actually works. Plug the values of x, y, and z into all three original equations to see if they hold true:

Equation 1: 2x - y + 2z = 8 2(3) - (-2) + 2(0) = 6 + 2 + 0 = 8 (Correct!)

Equation 2: x + 2y - z = -1 3 + 2(-2) - 0 = 3 - 4 - 0 = -1 (Correct!)

Equation 3: x + y + z = 1 3 + (-2) + 0 = 3 - 2 + 0 = 1 (Correct!)

Our solution works for all three equations. Awesome!

Alternative Method: Substitution

Now, let's explore how we could solve the same system of equations using the substitution method. This approach involves solving one equation for one variable and then substituting that expression into the other equations. While elimination was quite efficient for this particular system, substitution can be more straightforward in other cases.

Step 1: Solve One Equation for One Variable

Look at equation (3): x + y + z = 1. It seems easiest to solve for x:

x = 1 - y - z

Now we have an expression for x in terms of y and z.

Step 2: Substitute into the Other Equations

Substitute this expression for x into equations (1) and (2):

Equation 1: 2x - y + 2z = 8 2(1 - y - z) - y + 2z = 8 2 - 2y - 2z - y + 2z = 8 2 - 3y = 8 -3y = 6 y = -2

Equation 2: x + 2y - z = -1 (1 - y - z) + 2y - z = -1 1 + y - 2z = -1

Step 3: Solve for the Remaining Variables

From the substitution into equation (1), we found y = -2. Now plug this value into the modified equation (2):

1 + y - 2z = -1 1 + (-2) - 2z = -1 -1 - 2z = -1 -2z = 0 z = 0

Now we have y = -2 and z = 0. Plug these values back into the expression for x:

x = 1 - y - z x = 1 - (-2) - 0 x = 1 + 2 x = 3

Step 4: Verify the Solution

As before, let's verify our solution by plugging the values of x, y, and z into all three original equations:

Equation 1: 2x - y + 2z = 8 2(3) - (-2) + 2(0) = 6 + 2 + 0 = 8 (Correct!)

Equation 2: x + 2y - z = -1 3 + 2(-2) - 0 = 3 - 4 - 0 = -1 (Correct!)

Equation 3: x + y + z = 1 3 + (-2) + 0 = 3 - 2 + 0 = 1 (Correct!)

The solution x = 3, y = -2, and z = 0 works for all three equations. Both methods yield the same correct solution!

Solution Set

Alright, after all that work, we've found our solution. The solution set for the system of equations is:

• x = 3
• y = -2
• z = 0

We can write this as an ordered triple: (3, -2, 0).

Conclusion

Solving systems of linear equations might seem tough at first, but with a bit of practice, you'll get the hang of it. Whether you prefer the elimination method or the substitution method, the key is to stay organized and double-check your work. Now go out there and conquer those equations, you've got this!