Solving Systems Of Linear Equations A Step-by-Step Guide
In the realm of mathematics, particularly in algebra, solving systems of linear equations is a fundamental skill. These systems arise in various applications, from modeling real-world scenarios to solving complex problems in engineering and economics. A system of linear equations is a set of two or more linear equations involving the same variables. The solution to such a system is a set of values for the variables that satisfy all equations simultaneously. This article delves into a step-by-step approach to solving a system of three linear equations with three unknowns, using the method of elimination. We will illustrate this with a specific example and then explore the broader implications and applications of these techniques.
Understanding Linear Equations and Systems
Before diving into the solution, it's crucial to understand what linear equations are. A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. There are no exponents or other non-linear operations on the variables. When we have more than one of these equations considered together, we have a system of linear equations. Solving such a system means finding the values for the variables that make all the equations true at the same time. This can be visualized geometrically as finding the point(s) where the lines or planes represented by the equations intersect. The number of equations and variables can vary, but in this article, we will focus on systems with three equations and three variables, a common scenario in many applications. These systems can represent a variety of real-world situations, such as determining the flow of traffic in a network, balancing chemical equations, or modeling economic relationships. Understanding the underlying concepts of linear equations and systems is essential for mastering the techniques used to solve them.
The Elimination Method: A Step-by-Step Approach
The elimination method is a powerful technique for solving systems of linear equations. The core idea behind this method is to systematically eliminate variables by adding or subtracting multiples of equations. This process reduces the system to a simpler form, eventually allowing us to solve for one variable at a time. The method involves several steps, which we will outline and then apply to our example. First, we choose two equations and manipulate them so that the coefficients of one variable are opposites. This is often achieved by multiplying one or both equations by a suitable constant. Next, we add the equations together, which eliminates the chosen variable, resulting in a new equation with fewer variables. We repeat this process with a different pair of equations, eliminating the same variable. This gives us a second new equation with the same reduced set of variables. Now we have a system of two equations with two variables, which can be solved using similar elimination techniques or substitution. Once we have the values for these two variables, we substitute them back into one of the original equations to solve for the remaining variable. This step-by-step approach ensures that we systematically reduce the complexity of the system until we can easily find the solution. The elimination method is particularly effective for systems with many variables, as it provides a structured way to simplify the problem.
Step 1: Setting up the Equations
Our system of linear equations is as follows:
x + y + z = -7
4x - y - 9z = -29
-x - y - 3z = 3
We have three equations and three unknowns (x, y, and z). The goal is to find the values of x, y, and z that satisfy all three equations simultaneously. Before we start eliminating variables, it's helpful to label the equations for easy reference. Let's call the first equation (1), the second equation (2), and the third equation (3). This will make it easier to keep track of our steps as we manipulate the equations. We can rewrite the system with these labels:
(1) x + y + z = -7
(2) 4x - y - 9z = -29
(3) -x - y - 3z = 3
Now we are ready to apply the elimination method. We will begin by choosing a variable to eliminate and then strategically combining equations to achieve this. The choice of which variable to eliminate first can sometimes depend on the specific equations, but in this case, the coefficients of y are particularly convenient for elimination, as we will see in the next step. Setting up the equations clearly is a crucial first step in solving any system of linear equations, as it provides a solid foundation for the subsequent steps.
Step 2: Eliminating 'y' from Equations (1) and (2)
To eliminate 'y' from equations (1) and (2), we observe that the coefficients of 'y' are already opposites (+1 and -1). This makes the elimination process straightforward. We can simply add equations (1) and (2) together. Adding the left-hand sides of the equations, we get:
(x + y + z) + (4x - y - 9z) = x + 4x + y - y + z - 9z = 5x - 8z
Adding the right-hand sides, we get:
-7 + (-29) = -36
So, the new equation we obtain is:
(4) 5x - 8z = -36
This equation no longer contains 'y', effectively eliminating it from the system. We have reduced the number of variables in this equation, bringing us closer to solving the system. This process of adding equations to eliminate a variable is a key step in the elimination method. By carefully choosing which equations to combine, we can systematically reduce the complexity of the system. In this case, the coefficients of 'y' made the elimination particularly easy. Now we need to eliminate 'y' from another pair of equations to further simplify the system.
Step 3: Eliminating 'y' from Equations (1) and (3)
Next, we eliminate 'y' from equations (1) and (3). Again, the coefficients of 'y' are opposites (+1 and -1), which simplifies the process. We add equations (1) and (3) together:
(x + y + z) + (-x - y - 3z) = x - x + y - y + z - 3z = -2z
Adding the right-hand sides, we get:
-7 + 3 = -4
So, the new equation is:
(5) -2z = -4
This equation is particularly simple as it only involves one variable, 'z'. This is a significant step forward, as we can now easily solve for 'z'. Eliminating 'y' from another pair of equations has given us an equation that is much easier to handle. The strategic elimination of variables is what makes the elimination method so powerful. By reducing the number of variables in each equation, we gradually simplify the system until we can solve for each variable one by one. Now that we have an equation with only 'z', we can proceed to find its value.
Step 4: Solving for 'z'
From equation (5), we have:
-2z = -4
To solve for 'z', we divide both sides of the equation by -2:
z = (-4) / (-2) = 2
So, we have found that z = 2. This is a crucial step in solving the system. Now that we have the value of one variable, we can substitute it back into other equations to find the values of the remaining variables. Finding the value of 'z' is a significant milestone, as it allows us to reduce the number of unknowns in the other equations. This process of back-substitution is a common technique in solving systems of equations. By substituting the known value of 'z' into an equation with two variables, we can solve for one of those variables. The value of z = 2 will be used in the next step to solve for 'x'.
Step 5: Solving for 'x'
Now that we know z = 2, we can substitute this value into equation (4) to solve for 'x'. Equation (4) is:
5x - 8z = -36
Substituting z = 2, we get:
5x - 8(2) = -36
5x - 16 = -36
Adding 16 to both sides:
5x = -36 + 16
5x = -20
Dividing both sides by 5:
x = -20 / 5
x = -4
So, we have found that x = -4. Now we have the values for both 'x' and 'z'. This means we are just one step away from finding the complete solution to the system. The process of substituting the known value of 'z' into equation (4) allowed us to isolate 'x' and solve for its value. This is a common strategy in solving systems of equations: use known values to reduce the number of unknowns in other equations. Now that we have 'x' and 'z', we can substitute both of these values into one of the original equations to solve for 'y'.
Step 6: Solving for 'y'
We now know that x = -4 and z = 2. We can substitute these values into any of the original equations to solve for 'y'. Let's use equation (1):
x + y + z = -7
Substituting the values of 'x' and 'z', we get:
(-4) + y + (2) = -7
Simplifying:
y - 2 = -7
Adding 2 to both sides:
y = -7 + 2
y = -5
So, we have found that y = -5. Now we have the values for all three variables: x, y, and z. We have successfully solved the system of linear equations. The final step of substituting the known values of 'x' and 'z' into equation (1) allowed us to isolate 'y' and find its value. This completes the solution process. We now have a set of values for x, y, and z that satisfy all three equations simultaneously. However, it's always a good practice to verify our solution to ensure accuracy.
Step 7: Verifying the Solution
To verify our solution, we substitute the values x = -4, y = -5, and z = 2 into all three original equations:
Equation (1):
x + y + z = -7
(-4) + (-5) + (2) = -7
-7 = -7 (True)
Equation (2):
4x - y - 9z = -29
4(-4) - (-5) - 9(2) = -29
-16 + 5 - 18 = -29
-29 = -29 (True)
Equation (3):
-x - y - 3z = 3
-(-4) - (-5) - 3(2) = 3
4 + 5 - 6 = 3
3 = 3 (True)
Since the values satisfy all three equations, our solution is correct. We have successfully solved the system of linear equations and verified our result. Verifying the solution is an important final step in the process. It ensures that we have not made any errors in our calculations and that our solution is accurate. By substituting the values back into the original equations, we can confirm that they satisfy all the equations simultaneously. This gives us confidence in our solution and ensures that we have correctly solved the system.
The Solution
The solution to the system of linear equations is:
(x, y, z) = (-4, -5, 2)
This corresponds to option C. (-4, -5, 2). We have successfully solved the system using the elimination method and verified our solution. The solution represents the point in three-dimensional space where the three planes represented by the equations intersect. This point is the unique solution to the system, meaning that there are no other values of x, y, and z that satisfy all three equations simultaneously. Understanding the geometric interpretation of the solution can provide further insight into the nature of systems of linear equations.
Applications of Solving Systems of Linear Equations
Solving systems of linear equations has numerous applications in various fields. In engineering, these systems are used to analyze circuits, determine forces in structures, and model fluid flow. In economics, they are used to model supply and demand, analyze market equilibrium, and perform input-output analysis. In computer science, they are used in computer graphics, cryptography, and optimization problems. Systems of linear equations also play a crucial role in data analysis and machine learning, where they are used in regression analysis, classification algorithms, and dimensionality reduction techniques. The ability to solve these systems is a fundamental skill for anyone working in these fields. The applications of linear systems extend beyond these specific areas. They are used in chemistry to balance chemical equations, in physics to solve problems in mechanics and electromagnetism, and in operations research to optimize resource allocation. The versatility of these techniques makes them an essential tool in many areas of science and technology. Mastering the methods for solving systems of linear equations is therefore a valuable asset for students and professionals alike.
Conclusion: Mastering Linear Systems
In this comprehensive guide, we have explored the process of solving systems of linear equations using the elimination method. We have demonstrated a step-by-step approach, from setting up the equations to verifying the solution. The example provided illustrates how to systematically eliminate variables and solve for the unknowns. We have also highlighted the wide range of applications of these techniques in various fields. Mastering the methods for solving linear systems is essential for anyone studying mathematics, science, or engineering. These techniques provide a powerful tool for modeling and solving real-world problems. The elimination method, in particular, is a versatile and efficient approach for solving systems with multiple variables. By understanding the underlying principles and practicing the steps involved, you can develop the skills necessary to tackle a wide range of problems involving linear systems. The ability to solve these systems is not only a valuable mathematical skill but also a crucial asset for success in many professional fields. Continuous practice and application of these techniques will further enhance your understanding and proficiency in this important area of mathematics.