Solving Systems Of Equations Elimination Method

Jul 11, 2025 by ADMIN 48 views

#h1 Solve the System of Equations

Solving systems of equations is a fundamental concept in algebra, with applications across various fields like engineering, economics, and computer science. This article will guide you through the process of solving the given system of equations, focusing on the elimination method. We'll break down each step, providing clear explanations and insights to enhance your understanding. Understanding how to solve systems of equations is crucial for various mathematical and real-world applications, making this a valuable skill to master. Let's dive into the process of effectively solving this system using the elimination method, ensuring a comprehensive understanding along the way.

The Given System of Equations

The system of equations we're going to solve is:

6x - 3y = 3
-2x + 6y = 14

Our goal is to find the values of x and y that satisfy both equations simultaneously. There are several methods to solve systems of equations, including substitution, elimination, and graphing. In this article, we will focus on the elimination method, which is particularly effective for systems like this one. The elimination method involves manipulating the equations so that when they are added together, one of the variables is eliminated. This simplifies the system, allowing us to solve for the remaining variable. By carefully choosing the multipliers, we can strategically eliminate either x or y, making the process straightforward and efficient. Let's proceed with identifying the appropriate multiplier to eliminate the x terms in our given system.

Identifying the Multiplier for Elimination

The question asks: What number would you multiply the second equation by in order to eliminate the x-terms when adding to the first equation?

To eliminate the x-terms, we need to make the coefficients of x in both equations opposites of each other. The coefficient of x in the first equation is 6, and in the second equation, it is -2. To make the x coefficients opposites, we can multiply the second equation by a number that will turn -2 into -6. Multiplying the second equation by 3 achieves this, since 3 * (-2) = -6. Consequently, when added to the first equation, the x terms will cancel out. This step is crucial in the elimination method as it simplifies the system to a single variable equation. Choosing the correct multiplier is essential for successful elimination, and it often involves identifying the least common multiple of the coefficients of the variable you wish to eliminate. In our case, multiplying the second equation by 3 will effectively set up the system for the elimination of the x terms, paving the way for solving the system.

Step-by-Step Solution

Multiply the second equation by 3:
```
3 * (-2x + 6y) = 3 * 14
-6x + 18y = 42
```
This step is a pivotal part of the elimination method. By multiplying the entire equation by 3, we ensure that the equation remains balanced while strategically altering the coefficients. The goal here is to manipulate the coefficients of one variable so that they are additive inverses of the corresponding coefficients in the other equation. This allows for the elimination of that variable when the equations are added together. In our specific case, multiplying the second equation by 3 results in a new equation, -6x + 18y = 42, where the coefficient of x is -6, which is the additive inverse of the x coefficient in the first equation. This prepares us perfectly for the next step, where we will add the two equations to eliminate x and solve for y.
Add the modified second equation to the first equation:
```
(6x - 3y) + (-6x + 18y) = 3 + 42
```
Adding the equations is a core step in the elimination method. By carefully aligning the terms and adding the left-hand sides and the right-hand sides separately, we can eliminate one of the variables. In this case, adding the first equation (6x - 3y = 3) to the modified second equation (-6x + 18y = 42) results in the x terms canceling each other out. This is because 6x and -6x are additive inverses, summing to zero. The resulting equation will only contain the y variable, making it straightforward to solve. This simplification is the key advantage of the elimination method, allowing us to reduce a two-variable system into a single-variable equation. Let's continue to simplify the equation to isolate y.
Simplify the resulting equation:
```
15y = 45
```
This simplified equation is the direct result of adding the two equations together, which led to the elimination of the x variable. Simplifying the equation is a crucial step in isolating the variable we want to solve for. In this case, -3y + 18y combines to 15y, and 3 + 42 equals 45. The equation 15y = 45 is now a simple linear equation in one variable, which can easily be solved for y. By dividing both sides of the equation by 15, we can find the value of y. This step demonstrates the power of the elimination method in transforming a system of equations into a much more manageable form, highlighting its efficiency and effectiveness in solving for the variables.
Solve for y:
```
y = 45 / 15
y = 3
```
Solving for y involves isolating it on one side of the equation. This step is a fundamental algebraic operation that directly leads us to one of the solutions of the system. In our case, dividing both sides of the equation 15y = 45 by 15 gives us y = 3. This means that the y-coordinate of the solution to the system of equations is 3. Finding the value of one variable is a significant milestone in solving a system of equations, as it allows us to substitute this value back into one of the original equations to find the value of the other variable. Let's proceed with substituting y = 3 into one of the original equations to solve for x.
Substitute y = 3 into one of the original equations to solve for x. Let's use the first equation:
```
6x - 3(3) = 3
6x - 9 = 3
```
Substituting the value of y back into one of the original equations is a critical step in finding the value of x. This substitution allows us to transform the equation into one with a single variable, which we can then solve using basic algebraic techniques. By replacing y with 3 in the first equation (6x - 3y = 3), we get 6x - 3(3) = 3, which simplifies to 6x - 9 = 3. This new equation only involves x, making it straightforward to isolate and solve for. This step highlights the interconnectedness of the variables in a system of equations and how solving for one variable provides valuable information for finding the others. Now, let's proceed with solving for x in this equation.
Continue solving for x:
```
6x = 3 + 9
6x = 12
x = 12 / 6
x = 2
```
Continuing to solve for x involves isolating x on one side of the equation. This process typically includes adding or subtracting terms to both sides and then dividing by the coefficient of x. In our case, we start with 6x - 9 = 3. Adding 9 to both sides gives us 6x = 12. Then, dividing both sides by 6 yields x = 2. This completes the solution for x, giving us the x-coordinate of the solution to the system of equations. Now that we have found both x and y, we have a complete solution to the system. The next step is to verify that these values satisfy both original equations to ensure the accuracy of our solution.

The Solution

The solution to the system of equations is x = 2 and y = 3. This can be written as an ordered pair (2, 3).

To be absolutely sure of our solution, it's always a good practice to check if these values satisfy both original equations. Verification is a crucial step in solving systems of equations, as it confirms that the solution we found is indeed correct and avoids any potential errors in our calculations. Let's substitute x = 2 and y = 3 into both original equations to verify our solution:

For the first equation, 6x - 3y = 3:
```
6(2) - 3(3) = 12 - 9 = 3
```
The values satisfy the first equation.
For the second equation, -2x + 6y = 14:
```
-2(2) + 6(3) = -4 + 18 = 14
```
The values satisfy the second equation as well. Thus, our solution (2, 3) is correct. This verification step provides confidence in our solution and highlights the importance of checking our work in mathematical problem-solving.

Answer to the Question

The number you would multiply the second equation by in order to eliminate the x-terms when adding to the first equation is 3.

Understanding the underlying principles of solving systems of equations is essential for various applications in mathematics and beyond. Mastering the elimination method provides a powerful tool for tackling such problems efficiently. In this article, we've walked through a detailed step-by-step solution, emphasizing the logic behind each step and the importance of checking our work. By understanding the concepts and practicing the techniques, you can confidently approach and solve a wide range of systems of equations. The ability to solve systems of equations is not just a mathematical skill; it's a problem-solving skill that can be applied in many real-world scenarios, making it a valuable asset in your academic and professional pursuits. This concludes our exploration of solving the given system of equations.