Eliminating X-terms In Systems Of Equations A Step-by-step Guide
3x + (1/2)y = 3
6x - y = 2
Which of the following operations would eliminate the x-terms if the two equations were added together afterward? A. Multiply the first equation by -6.
Understanding the Elimination Method
In mathematics, particularly in algebra, solving systems of equations is a fundamental skill. One of the most effective methods for tackling these systems is the elimination method, also known as the addition method. This technique focuses on manipulating the equations in a system to eliminate one variable, thereby simplifying the problem and allowing us to solve for the remaining variable. This article delves into the intricacies of the elimination method, using the given system of equations as a practical example.
The core idea behind the elimination method is to make the coefficients of one variable in the two equations opposites of each other. When we add the equations together, this variable will cancel out, leaving us with a single equation in one variable. This simplified equation can then be easily solved. The solution for this variable can be substituted back into either of the original equations to find the value of the other variable. This process transforms a complex system of equations into a series of simpler, manageable steps.
To effectively use the elimination method, we often need to multiply one or both equations by a constant. The choice of this constant is crucial. It should be selected in such a way that the coefficients of the variable we want to eliminate become opposites. For instance, if we want to eliminate x, we look for a multiplier that will make the x coefficients in the two equations have the same magnitude but opposite signs. This may involve multiplying one equation by a negative number or finding the least common multiple of the coefficients to guide our choice of multiplier. By mastering this strategic approach, we can efficiently solve a wide range of systems of equations.
Analyzing the Given System of Equations
Let's consider the given system of equations:
3x + (1/2)y = 3
6x - y = 2
Our goal is to determine which operation will eliminate the x-terms when the two equations are added together. To achieve this, we need to manipulate the equations so that the coefficients of x are additive inverses (i.e., they have the same magnitude but opposite signs). The x coefficients in the given equations are 3 and 6. To eliminate x, we need to make these coefficients opposites. This means we want one equation to have a coefficient of -6 for x.
Looking at the first equation, 3x + (1/2)y = 3, we can see that multiplying the entire equation by -2 will result in a -6x term. This is because -2 * 3x = -6x. The modified equation would then be:
-6x - y = -6
Now, if we add this modified equation to the second equation in the original system, which is 6x - y = 2, the x-terms will indeed be eliminated:
(-6x - y) + (6x - y) = -6 + 2
This simplifies to:
-2y = -4
From this, we can easily solve for y. However, the question asks specifically which operation would eliminate the x-terms. We've determined that multiplying the first equation by -2 achieves this. This is a crucial step in the elimination method, as it sets the stage for solving the system. By carefully analyzing the coefficients and choosing the appropriate multiplier, we can effectively eliminate a variable and simplify the system.
Evaluating the Proposed Operation
The question presents a specific operation: Multiply the first equation by -6. Let's examine how this operation would affect the system of equations.
If we multiply the first equation, 3x + (1/2)y = 3, by -6, we get:
-6 * (3x + (1/2)y) = -6 * 3
This simplifies to:
-18x - 3y = -18
Now, let's add this modified equation to the second equation in the original system, 6x - y = 2:
(-18x - 3y) + (6x - y) = -18 + 2
Combining like terms, we get:
-12x - 4y = -16
As we can see, the x-terms were not eliminated in this case. Instead, we have a new equation with both x and y terms. This result highlights the importance of choosing the correct multiplier to achieve the desired elimination. Multiplying the first equation by -6 did not lead to the elimination of x because the resulting coefficient of x (-18) was not the additive inverse of the x coefficient in the second equation (6).
This exercise demonstrates a key aspect of the elimination method: the multiplier must be chosen carefully to ensure that the coefficients of the variable being eliminated become opposites. Simply multiplying by any number will not guarantee elimination. The goal is to create a situation where adding the equations will cancel out one of the variables, making the system easier to solve. In this instance, multiplying by -6 did not achieve this goal, emphasizing the need for a more strategic approach.
Determining the Correct Operation for Elimination
To correctly eliminate the x-terms in the given system of equations, we need to identify the operation that will make the x coefficients additive inverses. We have the system:
3x + (1/2)y = 3
6x - y = 2
We want to manipulate the first equation so that when it's added to the second equation, the x-terms cancel out. The x coefficient in the second equation is 6. Therefore, we need to make the x coefficient in the first equation -6.
To achieve this, we need to multiply the first equation by a factor that, when multiplied by 3 (the current x coefficient), results in -6. This factor is -2. Let's multiply the first equation by -2:
-2 * (3x + (1/2)y) = -2 * 3
This simplifies to:
-6x - y = -6
Now, we can add this modified equation to the second equation in the original system:
(-6x - y) + (6x - y) = -6 + 2
Adding the equations, we get:
-2y = -4
As we can see, the x-terms have been successfully eliminated, leaving us with an equation in just y. This demonstrates that multiplying the first equation by -2 is the correct operation to eliminate x. This result confirms the importance of identifying the appropriate multiplier to achieve the desired elimination.
The key to successful elimination is to focus on making the coefficients of the variable you want to eliminate additive inverses. This often involves multiplying one or both equations by a carefully chosen constant. By understanding this principle, you can effectively solve a wide range of systems of equations using the elimination method.
Conclusion
In conclusion, solving systems of equations using the elimination method requires careful consideration of the coefficients and strategic multiplication to eliminate variables. In the given system:
3x + (1/2)y = 3
6x - y = 2
the operation that would eliminate the x-terms if the two equations were added together afterward is not multiplying the first equation by -6. Instead, the correct operation is to multiply the first equation by -2. This is because multiplying by -2 results in an x coefficient of -6 in the first equation, which is the additive inverse of the x coefficient (6) in the second equation. When the equations are added, the x-terms cancel out, simplifying the system and allowing us to solve for y.
This example illustrates the importance of understanding the underlying principles of the elimination method. By choosing the correct multiplier, we can effectively eliminate variables and solve complex systems of equations. The process involves analyzing the coefficients, identifying the target variable for elimination, and selecting the appropriate multiplier to create additive inverses. Mastering this technique is crucial for success in algebra and beyond. The ability to solve systems of equations is a fundamental skill with applications in various fields, including science, engineering, and economics. By practicing and applying the elimination method, we can develop a strong foundation for tackling more advanced mathematical problems.
