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 for finding solutions efficiently. This article provides a comprehensive guide on how to solve a system of equations using the elimination method, with a specific focus on the example you provided. We will walk through each step, ensuring a clear understanding of the process. This method is particularly useful when dealing with linear equations, where the goal is to find the values of the variables that satisfy all equations simultaneously. By mastering this technique, you will be well-equipped to tackle a wide range of mathematical problems, from simple algebraic expressions to more complex real-world applications. So, let’s dive in and explore the intricacies of solving systems of equations by elimination.

Understanding the Elimination Method

The elimination method hinges on manipulating the equations in a system to eliminate one variable, making it easier to solve for the other. The core principle involves adding or subtracting the equations in such a way that one of the variables cancels out. This is achieved by multiplying one or both equations by a constant, ensuring that the coefficients of one variable are opposites. Once this is accomplished, the equations can be added together, eliminating one variable and leaving a single equation with one unknown. This resulting equation can then be easily solved, and the value of the solved variable can be substituted back into one of the original equations to find the value of the other variable. The beauty of the elimination method lies in its systematic approach, which minimizes the chances of errors and provides a clear path to the solution. It is particularly effective when dealing with systems of linear equations, where the variables are raised to the power of one. By understanding the underlying principles and practicing the steps involved, you can become proficient in using the elimination method to solve a variety of mathematical problems.

Step 1 Multiplying the First Equation

The first crucial step in solving the system of equations using the elimination method involves strategically multiplying one or both equations. The goal here is to make the coefficients of one of the variables opposites, setting the stage for elimination. In the given system:

8x + 7y = 39
4x - 14y = -68

We focus on the y-term. To eliminate y, we need to make the coefficients of y in both equations opposites. The coefficients are 7 and -14. A simple way to achieve this is to multiply the first equation by 2. This will make the coefficient of y in the first equation 14, which is the opposite of -14 in the second equation. The multiplication ensures that when we add the equations together, the y-terms will cancel out, leaving us with an equation in terms of x only. This step is a critical setup for the rest of the solution, as it simplifies the system and allows us to isolate one variable. Understanding how to choose the appropriate multiplier is key to efficiently applying the elimination method. It often involves identifying the least common multiple of the coefficients of the variable you want to eliminate.

Multiplying the first equation (8x + 7y = 39) by 2, we get:

2 * (8x + 7y) = 2 * 39
16x + 14y = 78

Now, our system of equations looks like this:

16x + 14y = 78
4x - 14y = -68

Step 2 Adding the Equations

With the coefficients of the y-terms now opposites, the next step is to add the two equations together. This is where the elimination method truly shines, as the y-terms will cancel each other out, leaving us with a single equation in terms of x. This step is a direct application of the principle that adding equal quantities to both sides of an equation does not change its validity. By adding the equations:

16x + 14y = 78
4x - 14y = -68

We combine the like terms: (16x + 4x) and (14y - 14y), and also add the constants (78 + (-68)). This process simplifies the system significantly, transforming two equations with two variables into a single equation with one variable. This resulting equation can then be easily solved for x, providing us with one piece of the solution to the original system. The addition step is a powerful tool in the elimination method, as it streamlines the process and brings us closer to finding the values of both variables. It's a clear demonstration of how strategic manipulation of equations can lead to a more manageable form.

Adding the equations:

(16x + 14y) + (4x - 14y) = 78 + (-68)

Simplifying, we get:

20x = 10

Solving for x

Now that we have a simplified equation, 20x = 10, solving for x is a straightforward algebraic step. The goal is to isolate x on one side of the equation. This is achieved by performing the inverse operation of multiplication, which is division. Dividing both sides of the equation by 20 ensures that the equality is maintained while isolating x. This step is a fundamental principle in solving equations, as it allows us to determine the value of the unknown variable. The result of this division will give us the numerical value of x, which is one half of the solution to the system of equations. This value can then be used in subsequent steps to find the value of the other variable, y. The process of isolating a variable is a cornerstone of algebra, and mastering this step is essential for solving a wide range of mathematical problems.

Dividing both sides by 20:

x = 10 / 20
x = 1/2

Substituting x to Find y

Having found the value of x, the next crucial step is to substitute this value back into one of the original equations to solve for y. This process is a fundamental aspect of solving systems of equations, as it allows us to determine the value of the remaining unknown variable. We can choose either of the original equations, but it's often strategic to select the one that appears simpler or easier to work with. In this case, let's use the first original equation:

8x + 7y = 39

Substituting x = 1/2 into this equation, we get:

8 * (1/2) + 7y = 39

Simplifying this equation will allow us to isolate y and solve for its value. This substitution method is a powerful technique in algebra, as it bridges the gap between the known and the unknown, allowing us to systematically uncover the solution. The accuracy of the substitution is paramount, as any error in this step will propagate through the rest of the solution. Therefore, careful attention to detail is essential when performing this substitution and simplification.

Substituting x = 1/2 into the equation 8x + 7y = 39:

8 * (1/2) + 7y = 39
4 + 7y = 39

Now, subtract 4 from both sides:

7y = 35

Finally, divide by 7:

y = 5

Solution

Therefore, the solution to the system of equations is x = 1/2 and y = 5. This solution represents the point where the two lines represented by the equations intersect on a graph. To verify the solution, we can substitute these values back into both original equations and confirm that they hold true. This is a crucial step in ensuring the accuracy of the solution and catching any potential errors. The solution (x = 1/2, y = 5) is the unique pair of values that satisfies both equations simultaneously. It's a testament to the power of the elimination method, which allows us to systematically solve systems of equations by manipulating them in a way that eliminates one variable at a time. Understanding how to find and verify solutions is a fundamental skill in algebra and has wide-ranging applications in various mathematical and real-world problems. The ability to accurately solve systems of equations is a valuable asset in many fields, from engineering to economics.

Conclusion

In conclusion, solving systems of equations by elimination is a powerful and versatile method in mathematics. By strategically manipulating equations, we can eliminate variables and simplify the system, leading to a clear path to the solution. The example we walked through demonstrates the step-by-step process, from multiplying equations to eliminate a variable, to adding the equations, solving for one variable, and finally, substituting to find the other. The solution, x = 1/2 and y = 5, represents the point of intersection of the two lines described by the equations. This method is not only effective but also provides a systematic approach to solving problems, minimizing the risk of errors. Mastering the elimination method equips you with a valuable tool for tackling a wide range of mathematical challenges. The ability to solve systems of equations is a fundamental skill that underpins many areas of mathematics and its applications in science, engineering, economics, and beyond. As you continue your mathematical journey, the elimination method will undoubtedly prove to be a valuable asset in your problem-solving toolkit. Remember to practice and apply this method to various problems to solidify your understanding and proficiency.