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

Hey guys! Let's dive into the fascinating world of solving systems of equations. In this article, we're going to tackle a specific system and break down the process step-by-step. We'll focus on how to solve the system:

$\left\{\begin{aligned}8 x-6 z & =14 \\3 x-2 y+4 z & =44 \\2 x+2 y & =-74\end{aligned}\right.$

Systems of equations might seem daunting at first, but trust me, with the right approach, they're totally manageable. We'll use a combination of techniques to find the values of x, y, and z that satisfy all three equations simultaneously. So, grab your pencils and let's get started!

Understanding Systems of Equations

Before we jump into solving this specific system, let's make sure we're all on the same page about what a system of equations actually is. Basically, it's a set of two or more equations that share variables. Our goal is to find the values for these variables that make all the equations true at the same time. Think of it as a puzzle where each equation is a piece, and we need to fit them together to find the solution.

In our case, we have three equations with three variables (x, y, and z). This means we're looking for a single set of values for x, y, and z that works in all three equations. There are several methods we can use to solve systems of equations, including substitution, elimination, and matrix methods. We'll primarily use elimination and substitution in this example, as these methods are particularly effective for this type of system. The key idea behind elimination is to manipulate the equations so that when we add or subtract them, one of the variables cancels out. This simplifies the system and allows us to solve for the remaining variables. Substitution involves solving one equation for one variable and then substituting that expression into another equation.

This might sound a bit abstract, but don't worry! As we work through the example, it'll become much clearer. Just remember that our aim is to find the values of x, y, and z that satisfy all three equations. Keep in mind that a system of equations can have one solution, no solutions, or infinitely many solutions. The nature of the solution depends on the relationships between the equations. For instance, if two equations represent the same line, there will be infinitely many solutions. If the equations represent parallel lines, there will be no solutions. In our case, we expect to find a unique solution, meaning one specific set of values for x, y, and z.

Step 1: Simplifying the First Equation

Okay, let's start by simplifying the first equation: 8x - 6z = 14. We can make things a bit easier by dividing both sides of the equation by 2. This gives us a simpler equivalent equation: 4x - 3z = 7. Simplifying equations whenever possible is a great strategy because it reduces the size of the numbers we're dealing with, which makes calculations less prone to errors. This might seem like a small step, but it's these small steps that add up to make the overall process smoother.

Now, let's rewrite our system with this simplified equation:

$\left\{\begin{aligned}4x - 3z & = 7 \\3 x-2 y+4 z & =44 \\2 x+2 y & =-74\end{aligned}\right.$

You'll notice that the first equation now has smaller coefficients, which is always a good thing. This simple division step sets the stage for the next steps in our solution. We've essentially made the first piece of our puzzle a little easier to handle. Remember, the goal is to strategically manipulate the equations to eliminate variables. By simplifying the first equation, we've positioned ourselves to do just that. The process of simplifying equations often involves looking for common factors or multiples that can be divided out. In this case, the greatest common divisor of 8, 6, and 14 is 2, which allowed us to divide the entire equation by 2 without changing its meaning. This type of simplification is a fundamental technique in algebra and is used extensively in solving various types of equations.

Step 2: Eliminating 'y' from Equations 2 and 3

The next thing we're going to do is eliminate the variable 'y' from equations 2 and 3. Why 'y'? Well, if you look closely, you'll see that the 'y' terms in equations 2 and 3 have opposite signs (-2y and +2y). This makes them perfect candidates for elimination! To eliminate 'y', we can simply add equation 2 and equation 3 together. This is a classic move in solving systems of equations – looking for opportunities to add or subtract equations to cancel out variables. This technique is based on the principle that if a = b and c = d, then a + c = b + d. By adding the left-hand sides and the right-hand sides of the equations, we maintain the equality while eliminating one of the variables.

So, let's add equations 2 and 3:

(3x - 2y + 4z) + (2x + 2y) = 44 + (-74)

Simplifying this, we get:

5x + 4z = -30

Now we have a new equation with only 'x' and 'z'. Let's call this equation 4. This is a major step because we've reduced the number of variables in one of our equations. Remember, our goal is to get down to a point where we can solve for one variable at a time. By eliminating 'y', we're one step closer to achieving that goal. This process of elimination is a powerful technique because it systematically reduces the complexity of the system. It allows us to work with smaller, more manageable equations, making the solution process much easier. The key is to identify which variables can be easily eliminated and then strategically add or subtract the equations to make it happen.

Step 3: Solving for 'x' and 'z'

Now we have two equations with two variables (x and z):

$\left\{\begin{aligned}4x - 3z & = 7 \\5x + 4z & = -30\end{aligned}\right.$

To solve for 'x' and 'z', we can use the elimination method again. This time, we need to manipulate the equations so that either the 'x' or the 'z' coefficients are opposites. Let's choose to eliminate 'z'. To do this, we can multiply the first equation by 4 and the second equation by 3. This will give us 'z' coefficients of -12 and +12, which will cancel out when we add the equations. This is a common strategy when using the elimination method. The idea is to find the least common multiple of the coefficients you want to eliminate and then multiply the equations accordingly. This ensures that the coefficients become opposites, allowing for easy cancellation when the equations are added.

Multiplying the equations, we get:

$\left\{\begin{aligned}16x - 12z & = 28 \\15x + 12z & = -90\end{aligned}\right.$

Now, we add these equations together:

(16x - 12z) + (15x + 12z) = 28 + (-90)

Simplifying, we get:

31x = -62

Now we can solve for 'x' by dividing both sides by 31:

x = -2

Awesome! We've found the value of 'x'. Now that we know 'x', we can substitute it back into one of the equations with 'x' and 'z' to solve for 'z'. Let's use the first equation (4x - 3z = 7):

4(-2) - 3z = 7

-8 - 3z = 7

-3z = 15

z = -5

So, we've found that x = -2 and z = -5. We're making great progress! By systematically eliminating variables, we've narrowed down the problem to a simple equation that we can easily solve. The key to success with systems of equations is to be organized and methodical. Keep track of your equations and the steps you've taken, and you'll be able to tackle even the most complex systems.

Step 4: Solving for 'y'

Now that we know x = -2 and z = -5, we can substitute these values into any of the original equations to solve for 'y'. Let's use the third equation (2x + 2y = -74) because it looks the simplest:

2(-2) + 2y = -74

-4 + 2y = -74

2y = -70

y = -35

Excellent! We've found the value of 'y'. So, we have x = -2, y = -35, and z = -5. This means we have a complete solution to our system of equations. Remember, the solution to a system of equations is a set of values for the variables that make all the equations true simultaneously. We've used a combination of elimination and substitution to systematically find these values. The process of substituting the known values back into the equations is crucial for solving for the remaining variables. It's like a chain reaction – once you find one variable, you can use it to find the others. This step-by-step approach is what makes solving systems of equations manageable. By breaking down the problem into smaller parts, we can solve each part individually and then put the pieces together to find the complete solution.

Step 5: Verifying the Solution

It's always a good idea to verify our solution by plugging the values of x, y, and z back into the original equations. This helps us catch any mistakes we might have made along the way. It's like a final check to make sure everything adds up correctly. Verification is a critical step in any mathematical problem-solving process. It gives us confidence in our solution and helps us avoid errors. By plugging the values back into the original equations, we're essentially testing whether our solution satisfies all the conditions of the problem.

Let's plug x = -2, y = -35, and z = -5 into the original equations:

• Equation 1: 8x - 6z = 14

  8(-2) - 6(-5) = -16 + 30 = 14 (Correct!)

• Equation 2: 3x - 2y + 4z = 44

  3(-2) - 2(-35) + 4(-5) = -6 + 70 - 20 = 44 (Correct!)

• Equation 3: 2x + 2y = -74

  2(-2) + 2(-35) = -4 - 70 = -74 (Correct!)

Since our values satisfy all three equations, we can be confident that our solution is correct. This final verification step provides a sense of closure and confirms that we've successfully solved the system of equations. In this case, the verification process confirms that our calculations were accurate and that the values x = -2, y = -35, and z = -5 are indeed the solution to the system.

Conclusion

So, there you have it! We've successfully solved the system of equations. The solution is x = -2, y = -35, and z = -5. We used a combination of simplification, elimination, and substitution to find these values. Remember, the key to solving systems of equations is to be organized, methodical, and to look for opportunities to eliminate variables. With practice, you'll become a pro at solving these types of problems!

Solving systems of equations is a fundamental skill in mathematics and has applications in various fields, including engineering, physics, and economics. The techniques we've used in this article can be applied to a wide range of systems, regardless of the number of equations or variables. The process of breaking down the problem into smaller, manageable steps is a valuable problem-solving strategy that can be applied to many different types of challenges. So, keep practicing, and you'll be well-equipped to tackle any system of equations that comes your way! Great job, guys!