Eliminating Y From Equations A Step By Step Guide
In mathematics, particularly in algebra, solving systems of equations is a fundamental skill. One common method for solving these systems involves eliminating one of the variables to simplify the problem. This article delves into the specific scenario of eliminating the variable y from a given system of two linear equations. We will explore the steps involved, the reasoning behind the method, and why a particular operation is needed. By the end of this guide, you will have a solid understanding of how to eliminate a variable in a system of equations, and you'll be able to apply this knowledge to various similar problems.
Understanding the Problem: The Equations at Hand
Before diving into the solution, let's first clearly state the problem. We are given the following two equations:
-3x - 3y = 3 (E1)
-x - 2y = 1 (E2)
The goal is to determine which operation, from the given options, will eliminate the variable y when applied to these equations. The options are:
A. 3E1 + 2E2 B. 3E1 - 2E2 C. 2E1 - 3E2 D. 2E1 + 3E2
Each option represents a linear combination of the two equations, where E1 and E2 are multiplied by constants and then either added or subtracted. The key to eliminating y lies in choosing the correct multipliers so that the y terms cancel each other out. This requires a strategic approach, focusing on the coefficients of y in both equations.
The Strategy Behind Elimination
The core idea behind eliminating a variable is to manipulate the equations so that the coefficients of that variable are either equal in magnitude but opposite in sign (so they cancel out when added) or equal in magnitude and the same sign (so they cancel out when subtracted). This manipulation is achieved by multiplying one or both equations by suitable constants. The choice of these constants is crucial for the success of the elimination process. By carefully selecting the constants, we can transform the system into a simpler form where one variable is immediately solved or can be easily solved.
In our case, we focus on the y terms. In equation E1, the coefficient of y is -3, and in equation E2, it is -2. To eliminate y, we need to find multipliers that will make these coefficients equal in magnitude but with opposite signs or equal in magnitude and same signs. This involves finding the least common multiple (LCM) of the absolute values of the coefficients and then determining the appropriate multipliers. This method ensures that the y terms will cancel out when the equations are combined, leaving us with an equation in just x, which can then be solved directly.
Detailed Analysis of the Options
To identify the correct operation, we need to examine how each option affects the coefficients of y in the resulting equation. Let's break down each option step by step:
Option A: 3E1 + 2E2
Applying this operation means multiplying equation E1 by 3 and equation E2 by 2, then adding the resulting equations:
3 * (-3x - 3y) = 3 * 3 => -9x - 9y = 9
2 * (-x - 2y) = 2 * 1 => -2x - 4y = 2
Adding these two equations:
(-9x - 9y) + (-2x - 4y) = 9 + 2
-11x - 13y = 11
In this case, the y terms do not cancel out (-9y - 4y = -13y). Therefore, option A is not the correct choice.
Option B: 3E1 - 2E2
This operation involves multiplying equation E1 by 3 and equation E2 by 2, then subtracting the second resulting equation from the first:
3 * (-3x - 3y) = 3 * 3 => -9x - 9y = 9
2 * (-x - 2y) = 2 * 1 => -2x - 4y = 2
Subtracting the second equation from the first:
(-9x - 9y) - (-2x - 4y) = 9 - 2
-9x - 9y + 2x + 4y = 7
-7x - 5y = 7
Again, the y terms do not cancel out (-9y + 4y = -5y). So, option B is also incorrect.
Option C: 2E1 - 3E2
Here, we multiply equation E1 by 2 and equation E2 by 3, then subtract the second resulting equation from the first:
2 * (-3x - 3y) = 2 * 3 => -6x - 6y = 6
3 * (-x - 2y) = 3 * 1 => -3x - 6y = 3
Subtracting the second equation from the first:
(-6x - 6y) - (-3x - 6y) = 6 - 3
-6x - 6y + 3x + 6y = 3
-3x = 3
In this case, the y terms do cancel out (-6y + 6y = 0), leaving us with an equation involving only x. This indicates that option C is a potential solution.
Option D: 2E1 + 3E2
Finally, let's examine multiplying equation E1 by 2 and equation E2 by 3, then adding the resulting equations:
2 * (-3x - 3y) = 2 * 3 => -6x - 6y = 6
3 * (-x - 2y) = 3 * 1 => -3x - 6y = 3
Adding these two equations:
(-6x - 6y) + (-3x - 6y) = 6 + 3
-9x - 12y = 9
The y terms do not cancel out (-6y - 6y = -12y), making option D incorrect.
The Correct Operation: Why 2E1 - 3E2 Works
Based on our analysis, option C (2E1 - 3E2) is the correct operation to eliminate y. The reason this works lies in the careful selection of the multipliers 2 and 3. By multiplying equation E1 by 2 and equation E2 by 3, we create y terms with the same coefficient magnitude (-6y) but with the same sign. Subtracting the equations then results in the y terms canceling each other out.
Step-by-Step Breakdown
- Multiply E1 by 2: This gives us -6x - 6y = 6.
- Multiply E2 by 3: This gives us -3x - 6y = 3.
- Subtract the new E2 from the new E1: (-6x - 6y) - (-3x - 6y) = 6 - 3.
- Simplify: -6x - 6y + 3x + 6y = 3, which simplifies further to -3x = 3.
As we can see, the y terms have been successfully eliminated, leaving us with a simple equation in x that can be easily solved. This method demonstrates the power of strategic manipulation in solving systems of equations. The choice of multipliers is not arbitrary; it is a deliberate process aimed at achieving a specific outcome—in this case, the elimination of the variable y.
Solving for x and y: Completing the Solution
While the original problem only asked for the operation needed to eliminate y, let's take it a step further and solve for both x and y to provide a complete solution. We've already found that -3x = 3 after eliminating y. Now, we can solve for x:
-3x = 3
x = 3 / -3
x = -1
Now that we have the value of x, we can substitute it back into either equation E1 or E2 to solve for y. Let's use equation E2:
-x - 2y = 1
-(-1) - 2y = 1
1 - 2y = 1
-2y = 0
y = 0
Therefore, the solution to the system of equations is x = -1 and y = 0. This comprehensive solution not only answers the original question but also demonstrates how to proceed once the elimination is achieved.
Alternative Methods and Considerations
While the elimination method is effective, it is worth noting that there are other methods for solving systems of equations, such as substitution and matrix methods. The choice of method often depends on the specific structure of the equations and the solver's preference. However, the elimination method is particularly useful when the coefficients of one variable are multiples of each other, as it allows for straightforward elimination with minimal algebraic manipulation.
Another consideration is the potential for inconsistent or dependent systems. Inconsistent systems have no solution, while dependent systems have infinitely many solutions. When using the elimination method, these situations can be identified by either all variables being eliminated, resulting in a contradiction (inconsistent system), or all variables being eliminated, resulting in a true statement (dependent system). Understanding these possibilities is crucial for a complete grasp of solving systems of equations.
Conclusion: Mastering Elimination Techniques
In conclusion, determining the correct operation to eliminate a variable in a system of equations is a critical skill in algebra. In the given problem, the correct operation to eliminate y is 2E1 - 3E2. This choice is based on a strategic analysis of the coefficients of y and the application of the principle that terms with equal magnitude and same sign will cancel out when subtracted. We have also shown how to proceed to solve for both variables once the elimination is achieved, and we've briefly discussed alternative methods and considerations.
By mastering these techniques, you'll be well-equipped to tackle a wide range of problems involving systems of equations. Remember, the key is to understand the underlying principles and apply them systematically. With practice, you'll become proficient at identifying the most efficient method for solving any given system, making complex algebraic problems much more manageable. This skill is not only valuable in academic settings but also in various real-world applications where mathematical modeling is required.
