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

Hey guys! Let's dive into the world of solving systems of equations. This is a fundamental concept in mathematics, and it's super useful for all sorts of real-world problems. Today, we'll tackle a specific system and break down the process step-by-step. Get ready to flex those math muscles! We'll be solving the following system of equations:

egin{array}{r} x+y+z=7 \\ 2 x+3 y-z=5 \\ 2 x+3 y-2 z=6 \end{array}

Understanding Systems of Equations

Alright, before we jump into the nitty-gritty, let's make sure we're all on the same page. What exactly is a system of equations? Well, it's simply a collection of two or more equations, each containing one or more variables. Our goal is to find the values of these variables that satisfy all the equations in the system simultaneously. Think of it like a puzzle where you need to find the specific values that make all the pieces fit perfectly. The solution to a system of equations is the set of values for the variables that makes each equation true. These values represent the point(s) where the graphs of the equations intersect (if the equations are linear, like the ones we are dealing with). There are several methods for solving systems of equations, and the best method to use depends on the specific system.

For instance, if we're dealing with a system of linear equations, as we are here, we can often use methods like substitution, elimination, or even graphing. For more complex systems, you might need to use more advanced techniques. But don't worry, we'll focus on the basics for now! So, why is solving systems of equations important? Well, it crops up everywhere! From calculating the best price for items to determine the optimal production levels in a business. It's used in science, engineering, and of course, in everyday life when trying to solve a set of problems. It’s a core skill that builds the foundation for more advanced math concepts. This helps to develop logical reasoning and problem-solving skills, which are transferable to all kinds of scenarios in life. The first step to become a math wizard is to understand the basics!

The Elimination Method: Our Weapon of Choice

For this particular system, the elimination method is a great approach. The elimination method involves manipulating the equations in such a way that when you add or subtract them, one or more of the variables cancels out. This simplifies the system, allowing you to solve for the remaining variables. This is a very effective strategy when solving linear equations. Before we go any further, let me show you how it works in the simplest format. First, let's take a look at our equations:

• Equation 1: $x + y + z = 7$
• Equation 2: $2x + 3y - z = 5$
• Equation 3: $2x + 3y - 2z = 6$

Notice that the 'z' variable in equation 2 and equation 3 have opposite signs (-z and -2z). This is an excellent opportunity for elimination! The goal is to eliminate one variable at a time until you're left with just one variable. With just a little bit of algebraic manipulation, you'll be able to quickly solve any kind of equation!

Eliminating 'z' in Equations 1 and 2

Let's start by eliminating 'z' using Equation 1 and Equation 2. To do this, we need to make the coefficients of 'z' opposites. In this case, we can simply add Equation 1 and Equation 2 together because the 'z' values will cancel out. So, let's do it!

• Equation 1: $x + y + z = 7$
• Equation 2: $2x + 3y - z = 5$

Adding these two equations gives us:

$(x + 2x) + (y + 3y) + (z - z) = 7 + 5$

Which simplifies to:

$3x + 4y = 12$

We'll call this new equation Equation 4. It's a key step in simplifying the system of equations. We've successfully eliminated 'z' from the equation. High five!

Eliminating 'z' in Equations 2 and 3

Now, let's eliminate 'z' again, this time using Equation 2 and Equation 3. But this is the tricky part, so pay close attention! We need to manipulate the equations so that the 'z' terms cancel out when we add or subtract them. To do this, let's multiply Equation 2 by -2. This will make the coefficient of 'z' in Equation 2 become +2, which is the opposite of the -2z in Equation 3. So now our equations look like this:

• Equation 2 (multiplied by -1): $-2x - 3y + z = -5$
• Equation 3: $2x + 3y - 2z = 6$

Adding these two equations gives us:

$(-2x + 2x) + (-3y + 3y) + (z - 2z) = -5 + 6$

Which simplifies to:

$-z = 1$

Solving for z, we get:

$z = -1$

Awesome, we now have the value for 'z'! Great work! Now that we know 'z', it's easier to find the values of 'x' and 'y'. Keep going!

Solving for x and y

Now that we know the value of z, we can substitute it into our equations. Let's substitute $z = -1$ into Equation 1:

$x + y + (-1) = 7$

Which simplifies to:

$x + y = 8$ (Equation 5)

We also have Equation 4: $3x + 4y = 12$

Now, we have a system of two equations with two variables (x and y). We can solve this using either substitution or elimination. Let's use elimination again. Multiply Equation 5 by -3:

$-3x - 3y = -24$

Now, add this modified equation to Equation 4:

$3x + 4y = 12$

Adding these gives us:

$(3x - 3x) + (4y - 3y) = 12 - 24$

This simplifies to:

$y = -12$

We've found the value of 'y'! Yay!

To find the value of x, substitute $y = -12$ into Equation 5:

$x + (-12) = 8$

So,

$x = 20$

The Solution

So guys, we did it! We solved the system of equations! The solution is:

• $x = 20$
• $y = -12$
• $z = -1$

We can write this as an ordered triple: $(20, -12, -1)$. This means that when we plug these values back into the original equations, they all hold true. That's the beauty of solving systems of equations; the final answer gives a set of values that satisfy all the equations simultaneously. You have now successfully mastered solving a system of equations. Keep practicing, and it will become second nature to you in no time.

Verification

It's always a good idea to verify your solution! Let's substitute these values back into our original equations to make sure they're correct:

• Equation 1: $x + y + z = 7$ -> $20 + (-12) + (-1) = 7$. Correct!
• Equation 2: $2x + 3y - z = 5$ -> $2(20) + 3(-12) - (-1) = 5$. Correct!
• Equation 3: $2x + 3y - 2z = 6$ -> $2(20) + 3(-12) - 2(-1) = 6$. Correct!

Since all three equations hold true with these values, we know our solution is correct! Great job, everyone!

Conclusion

Solving systems of equations can seem a bit daunting at first, but with a systematic approach and practice, you can master it. We've walked through the elimination method today, showing you how to systematically eliminate variables to arrive at a solution. Remember to always double-check your work to ensure accuracy. Keep practicing, and you'll be solving complex systems of equations in no time! Keep in mind that math is not a spectator sport, so the more you work on solving these kinds of problems, the easier it becomes. You've got this! And, as always, don't be afraid to ask for help if you get stuck. Keep up the awesome work!