Solve This System Of Linear Equations

Hey there, math enthusiasts! Today, we're diving into the fascinating world of systems of linear equations. These are sets of equations with multiple variables, and our goal is to find the values of those variables that satisfy all the equations simultaneously. It's like solving a puzzle where each equation gives you a crucial clue. The system we'll be tackling today is:

$$\left\{\begin{array}{l} 3 x+3 y+4 z=6 \\ 5 x-y+4 z=10 \\ -4 x+2 y-3 z=-8 \end{array}\right.$$

Don't let the three variables (x, y, and z) and three equations intimidate you! We've got a solid strategy to break this down. Our main objective is to solve this system of equations, meaning we want to find a unique set of values for x, y, and y that makes all three equations true. If we find such a set, the system is considered to have a solution. If, however, we reach a contradiction, it means there's no single solution that works for all equations, and the system is deemed impossible to solve. We'll explore the methods to determine this, focusing on clarity and providing a step-by-step approach to make the process as understandable as possible. Let's get started on unraveling this mathematical challenge!

Understanding Systems of Linear Equations

Before we jump into solving our specific problem, let's take a moment to appreciate what systems of linear equations represent. In essence, each linear equation with variables like x, y, and z can be thought of as defining a plane in three-dimensional space. When we have a system of these equations, we're looking for the point (or points) where all these planes intersect. If there's a single point where all three planes meet, the system has a unique solution. If the planes intersect along a line, there are infinitely many solutions. And if the planes never intersect at a common point (perhaps they are parallel or intersect in pairs but not all together), then the system has no solution. Our goal today is to solve the system of equations and determine which of these scenarios applies. The methods we commonly use for solving such systems include substitution, elimination, and matrix methods (like Gaussian elimination or Cramer's Rule). Each method has its strengths, but they all aim to systematically reduce the system to a form where the solution is easily identifiable. For our given system, we will likely use a combination of substitution and elimination to simplify the problem step-by-step. It's crucial to be meticulous with arithmetic and algebraic manipulations, as a small error can lead to an incorrect final answer. Remember, the beauty of mathematics lies in its logical progression, and by following these established procedures, we can confidently navigate through the complexities of systems of linear equations.

Method 1: Elimination Strategy

One of the most effective ways to solve a system of equations like the one we have is through the elimination method. The core idea here is to strategically add or subtract multiples of one equation from another to eliminate one of the variables. We'll repeat this process to reduce the system from three equations with three variables to two equations with two variables, and then finally to a single equation with one variable. Let's label our equations for clarity:

(1) $3x + 3y + 4z = 6$ (2) $5x - y + 4z = 10$ (3) $-4x + 2y - 3z = -8$

Our first step is to eliminate one variable, say 'y', from two pairs of equations. Let's combine equations (1) and (2). Notice that the 'y' terms have coefficients +3 and -1. To eliminate 'y', we can multiply equation (2) by 3 and then add it to equation (1):

$3 * (5x - y + 4z = 10) \implies 15x - 3y + 12z = 30$

Now, add this modified equation (2) to equation (1):

$(3x + 3y + 4z) + (15x - 3y + 12z) = 6 + 30$ $18x + 16z = 36$

Let's call this new equation (4). Now, we need to eliminate 'y' from another pair of equations. Let's use equations (2) and (3). The 'y' coefficients are -1 and +2. We can multiply equation (2) by 2 and add it to equation (3):

$2 * (5x - y + 4z = 10) \implies 10x - 2y + 8z = 20$

Now, add this modified equation (2) to equation (3):

$(-4x + 2y - 3z) + (10x - 2y + 8z) = -8 + 20$ $6x + 5z = 12$

Let's call this new equation (5). We have successfully reduced our system to two equations with two variables:

(4) $18x + 16z = 36$ (5) $6x + 5z = 12$

We can simplify equation (4) by dividing by 2: $9x + 8z = 18$. Now we have:

(4') $9x + 8z = 18$ (5) $6x + 5z = 12$

Our next step is to eliminate either 'x' or 'z' from these two equations. Let's eliminate 'x'. We can multiply equation (4') by 2 and equation (5) by -3 (or multiply (5) by 3 and subtract):

$2 * (9x + 8z = 18) \implies 18x + 16z = 36$ $-3 * (6x + 5z = 12) \implies -18x - 15z = -36$

Now, add these two resulting equations:

$(18x + 16z) + (-18x - 15z) = 36 + (-36)$ $z = 0$

Fantastic! We've found the value of one variable. This is a significant breakthrough in our quest to solve the system of equations. The elimination method has systematically brought us closer to the solution by progressively simplifying the problem. We've successfully eliminated variables, reducing the complexity of the system at each stage. The fact that we arrived at a specific numerical value for 'z' is a good sign that our system likely has a unique solution. Now, we need to use this value of z to find x and y.

Method 2: Back-Substitution to Find Remaining Variables

Now that we have found $z=0$, we can use the back-substitution technique to find the values of x and y. This means we'll substitute the known value of z back into one of the simpler equations we derived, and then use that result to find the other variables. Let's use equation (5) (or our simplified version of (4') if we prefer, but (5) looks simpler): $6x + 5z = 12$. Substitute $z=0$ into this equation:

$6x + 5(0) = 12$ $6x = 12$ $x = 2$

Excellent! We've found the value of x. So far, we have $x=2$ and $z=0$. The final step is to substitute these values into one of the original equations to find y. Let's use equation (1): $3x + 3y + 4z = 6$. Substitute $x=2$ and $z=0$:

$3(2) + 3y + 4(0) = 6$ $6 + 3y + 0 = 6$ $3y = 6 - 6$ $3y = 0$ $y = 0$

So, we have found our potential solution: $x=2$, $y=0$, and $z=0$. This is a crucial part of the process of solving the system of equations – systematically working backward from a known variable to find the others. Back-substitution is a powerful tool because it leverages the values you've already discovered, simplifying subsequent calculations. It's like putting the final pieces of a puzzle together once you've identified a few key elements. The clarity of this method hinges on accurately calculating the values of the variables in the preceding steps. Any arithmetic error in the elimination phase would have propagated through to this stage, leading to incorrect values for x, y, and z. Therefore, double-checking calculations is always a wise practice when dealing with systems of equations.

Verification of the Solution

We've arrived at a potential solution: $x=2$, $y=0$, and $z=0$. However, to be absolutely sure that we have indeed managed to solve the system of equations correctly, we must verify our solution by substituting these values back into all three of the original equations. This step is non-negotiable and serves as a final check to ensure consistency.

Let's check equation (1): $3x + 3y + 4z = 6$ Substitute $x=2, y=0, z=0$: $3(2) + 3(0) + 4(0) = 6$ $6 + 0 + 0 = 6$ $6 = 6$

Equation (1) holds true. Great!

Now, let's check equation (2): $5x - y + 4z = 10$ Substitute $x=2, y=0, z=0$: $5(2) - (0) + 4(0) = 10$ $10 - 0 + 0 = 10$ $10 = 10$

Equation (2) also holds true. We're on the right track!

Finally, let's check equation (3): $-4x + 2y - 3z = -8$ Substitute $x=2, y=0, z=0$: $-4(2) + 2(0) - 3(0) = -8$ $-8 + 0 - 0 = -8$ $-8 = -8$

Equation (3) is also satisfied. Since our values $x=2, y=0, z=0$ satisfy all three original equations, we can confidently conclude that this is the unique solution to the system. The verification step is paramount in confirming the accuracy of our work when we solve the system of equations. It's the final stamp of approval that confirms our calculated values indeed make all statements true simultaneously. This systematic process ensures the integrity of our mathematical findings.

Conclusion: The Solution and Its Implications

We have successfully navigated the challenge of solving the system of equations $\left\{\begin{array}{l} 3 x+3 y+4 z=6 \\ 5 x-y+4 z=10 \\ -4 x+2 y-3 z=-8 \end{array}\right.$ using the elimination and back-substitution methods. Our rigorous step-by-step process led us to the unique solution: $x=2$, $y=0$, and $z=0$. This means that the point $(2, 0, 0)$ is the single point in three-dimensional space where the three planes represented by these equations intersect. The fact that we found a unique solution indicates that the system is consistent and independent. This contrasts with systems that might have no solution (inconsistent) or infinitely many solutions (dependent). The successful verification of our solution in all three original equations confirms the accuracy of our calculations and the validity of the solution. Understanding how to solve systems of linear equations is a fundamental skill in mathematics with broad applications in various fields, including engineering, economics, computer science, and physics. Whether it's optimizing resource allocation, analyzing circuits, or predicting trends, these mathematical tools provide a powerful framework for modeling and solving complex real-world problems. The methods we employed, like elimination and substitution, are versatile and can be adapted to solve systems of varying sizes and complexities. Always remember the importance of careful calculation and verification to ensure the reliability of your results when you solve the system of equations.

For further exploration into the methods and applications of solving systems of linear equations, you can visit Khan Academy, a fantastic resource for learning and practicing mathematics.