Solving Systems Of Equations By Elimination A Step-by-Step Guide
In the realm of mathematics, solving systems of equations is a fundamental skill. Among the various methods available, elimination stands out as a powerful technique. This article will guide you through the process of solving a system of equations using the elimination method, providing a step-by-step approach with a focus on clarity and understanding. We will delve into the intricacies of this method, ensuring you grasp the underlying concepts and can confidently apply them to various problems. Our example will be the system of equations:
8x + 7y = 39
4x - 14y = -68
We will explore how to manipulate these equations to eliminate one variable, allowing us to solve for the other. This involves strategic multiplication and addition, techniques that we will break down in detail. By the end of this guide, you'll have a solid grasp of the elimination method and its applications.
1. Understanding the Elimination Method
The elimination method, a cornerstone of algebraic problem-solving, offers a systematic approach to solving systems of equations. The core principle behind this method is to manipulate the equations in such a way that when they are added together, one of the variables cancels out, leaving us with a single equation in one variable. This simplification allows us to easily solve for the remaining variable. To effectively utilize the elimination method, a deep understanding of its underlying principles is crucial. This involves recognizing how multiplying equations by constants can alter their coefficients without changing their fundamental solutions. It also requires a keen eye for identifying which variable is easiest to eliminate, often based on the coefficients present in the equations. By mastering these fundamental aspects, you'll be well-equipped to tackle a wide range of systems of equations. The beauty of the elimination method lies in its versatility and efficiency. It can be applied to systems with two or more variables, making it a powerful tool in various mathematical contexts. Furthermore, the method encourages strategic thinking and problem-solving skills, as it requires careful planning and execution. By understanding the theoretical foundations and practicing the steps involved, you can confidently use the elimination method to solve complex problems and gain a deeper appreciation for the elegance of algebra.
2. Multiplying the First Equation for y-Term Elimination
The primary goal in this step is to strategically manipulate the equations so that the y-terms have opposite coefficients. This will allow them to cancel each other out when the equations are added together. In our given system:
8x + 7y = 39
4x - 14y = -68
We observe that the coefficients of the y-terms are 7 and -14. To eliminate y, we need to make the coefficient of y in the first equation the opposite of the coefficient in the second equation. Since -14 is twice 7 with a negative sign, we can multiply the first equation by 2. This will result in the y-term in the first equation becoming 14y, which is the opposite of -14y in the second equation. Multiplying the entire first equation (8x + 7y = 39) by 2, we get:
2 * (8x + 7y) = 2 * 39
16x + 14y = 78
This new equation is equivalent to the original first equation, but it now has a y-term that will conveniently cancel with the y-term in the second equation. This step is crucial for setting up the elimination process. By carefully choosing the multiplier, we ensure that the elimination will be successful and lead us closer to the solution. The ability to strategically manipulate equations in this way is a key skill in algebra and is essential for mastering the elimination method. Furthermore, it's important to remember that we must multiply the entire equation, both sides, to maintain the equality. This ensures that we are not changing the fundamental relationship expressed by the equation. By paying attention to detail and applying the principles of algebraic manipulation, we can confidently proceed to the next step in the elimination process.
3. Adding the Equations to Eliminate a Variable
Now that we've strategically multiplied the first equation, we have the following system:
16x + 14y = 78
4x - 14y = -68
The coefficients of the y-terms are now opposites (14 and -14), which is exactly what we aimed for. The next step is to add the two equations together. This is where the elimination magic happens. When we add the left-hand sides of the equations, we add the x-terms together (16x + 4x) and the y-terms together (14y + (-14y)). Similarly, we add the right-hand sides of the equations (78 + (-68)). This process can be written as follows:
(16x + 14y) + (4x - 14y) = 78 + (-68)
Simplifying the equation, we get:
20x = 10
Notice that the y-terms have completely canceled each other out, leaving us with a simple equation in just one variable, x. This is the essence of the elimination method – reducing the system to a single equation that can be easily solved. This step is crucial because it allows us to isolate one variable and determine its value. Once we know the value of one variable, we can substitute it back into one of the original equations to find the value of the other variable. The addition step requires careful attention to signs and coefficients. It's important to add like terms correctly and ensure that the cancellation of the targeted variable occurs as planned. By performing this step accurately, we pave the way for solving for the remaining variable and ultimately finding the solution to the system of equations. The elimination of a variable is a powerful technique that simplifies the problem and makes it much easier to solve.
4. Solving for x
After adding the equations, we arrived at the simplified equation:
20x = 10
This equation is a straightforward linear equation in one variable, x. To solve for x, we need to isolate it on one side of the equation. This is achieved by performing the inverse operation of multiplication, which is division. We divide both sides of the equation by the coefficient of x, which is 20. This maintains the equality and allows us to isolate x:
20x / 20 = 10 / 20
Simplifying, we get:
x = 1/2
Therefore, the value of x is 1/2 or 0.5. This is a significant step in solving the system of equations. We have successfully determined the value of one of the variables. Now that we know the value of x, we can proceed to find the value of y. This is typically done by substituting the value of x back into one of the original equations. Solving for x is a fundamental algebraic skill that is essential for various mathematical problems. It involves understanding the properties of equality and applying inverse operations to isolate the variable of interest. In this case, dividing both sides of the equation by the coefficient of x was the key to finding its value. This process highlights the importance of maintaining balance in equations and performing the same operations on both sides to preserve the equality. By mastering these techniques, you'll be well-equipped to solve a wide range of linear equations and apply them to more complex problems.
5. Substituting x to Solve for y
Now that we have found the value of x (x = 1/2), the next step is to substitute this value into one of the original equations to solve for y. We can choose either of the original equations:
8x + 7y = 39
4x - 14y = -68
Let's choose the first equation (8x + 7y = 39) for this substitution. Substituting x = 1/2 into this equation, we get:
8 * (1/2) + 7y = 39
Simplifying the equation:
4 + 7y = 39
Now, we need to isolate the y-term. To do this, we subtract 4 from both sides of the equation:
7y = 39 - 4
7y = 35
Finally, to solve for y, we divide both sides of the equation by 7:
y = 35 / 7
y = 5
Therefore, the value of y is 5. This step demonstrates the power of substitution in solving systems of equations. By replacing a variable with its known value, we can transform an equation with two variables into a simpler equation with only one variable, which can then be easily solved. The choice of which equation to substitute into is often a matter of convenience. In this case, the first equation was chosen, but the second equation would have yielded the same result. Substituting the value of x and solving for y completes the process of finding the solution to the system of equations. We now know the values of both x and y, which allows us to express the solution as an ordered pair.
6. The Solution
We have successfully solved for both variables in the system of equations. We found that:
x = 1/2
y = 5
Therefore, the solution to the system is the ordered pair (1/2, 5). This means that the point (1/2, 5) is the intersection of the two lines represented by the original equations. It is the unique point that satisfies both equations simultaneously. The solution to a system of equations is a fundamental concept in algebra and represents the point where all the equations in the system are true. In the case of linear equations, the solution corresponds to the point where the lines intersect on a graph. This graphical interpretation provides a visual understanding of the solution. The process of finding the solution involves manipulating the equations in a way that isolates the variables and allows us to determine their values. In this case, we used the elimination method, which is a powerful technique for solving systems of equations. The solution (1/2, 5) can be verified by substituting these values back into the original equations. If the equations hold true with these values, then we have confirmed that the solution is correct. This verification step is a good practice to ensure the accuracy of the solution. Understanding the concept of a solution and how to find it is essential for various applications in mathematics, science, and engineering. Systems of equations arise in many real-world problems, and the ability to solve them is a valuable skill.
7. Verifying the Solution
To ensure the accuracy of our solution (x = 1/2, y = 5), it's crucial to verify it by substituting these values back into the original equations. This process confirms that the solution satisfies both equations simultaneously. Let's start with the first equation:
8x + 7y = 39
Substituting x = 1/2 and y = 5, we get:
8 * (1/2) + 7 * 5 = 39
4 + 35 = 39
39 = 39
The equation holds true. Now, let's check the second equation:
4x - 14y = -68
Substituting x = 1/2 and y = 5, we get:
4 * (1/2) - 14 * 5 = -68
2 - 70 = -68
-68 = -68
This equation also holds true. Since the solution (x = 1/2, y = 5) satisfies both original equations, we can confidently conclude that it is the correct solution to the system of equations. Verification is a critical step in problem-solving. It provides a check against potential errors and ensures that the solution is accurate. This process is particularly important in mathematics, where a small mistake can lead to an incorrect answer. By substituting the solution back into the original equations, we can confirm that the values of the variables are consistent with the relationships expressed by the equations. This step not only validates the solution but also reinforces our understanding of the problem and the solution process. Furthermore, verification helps to develop a habit of carefulness and attention to detail, which are valuable skills in any field. In the context of systems of equations, verification ensures that we have found the point that simultaneously satisfies all the equations in the system.
By following these steps, you can confidently solve systems of equations using the elimination method. Remember to practice regularly to solidify your understanding and improve your problem-solving skills.
