Solving Systems Of Equations By Elimination A Step-by-Step Guide

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In mathematics, particularly in algebra, solving systems of equations is a fundamental skill. One powerful method for tackling these systems is the elimination method. This approach is especially effective when dealing with linear equations, where the goal is to find the values of the variables that satisfy all equations simultaneously. In this article, we will delve into the elimination method, providing a detailed explanation and step-by-step instructions, along with a practical example to illustrate the process. We will solve the system of equations:

{4x+2y=63x−2y=8{ \begin{cases} 4x + 2y = 6 \\ 3x - 2y = 8 \end{cases} }

to determine the correct solution from the given options.

Understanding the Elimination Method

The elimination method, also known as the addition method, involves manipulating the equations in the system so that when they are added together, one of the variables is eliminated. This leaves us with a single equation in one variable, which can be easily solved. The solution for this variable can then be substituted back into one of the original equations to find the value of the other variable. This method is particularly useful when the coefficients of one of the variables are opposites or can be easily made opposites by multiplying one or both equations by a constant.

To effectively employ the elimination method, it's crucial to understand its underlying principles and the steps involved. The primary goal is to manipulate the equations in such a way that when they are combined, one variable cancels out, leaving us with a single equation containing only one unknown. This is achieved by ensuring that the coefficients of one of the variables are either the same or additive inverses (opposites) in the two equations. Once we have a single equation with one variable, we can easily solve for that variable. The solution obtained is then substituted back into one of the original equations to determine the value of the other variable, thus providing the complete solution to the system of equations.

Step 1: Align the Equations

The first step in the elimination method is to ensure that the equations are properly aligned. This means that the like terms (terms with the same variable) should be in the same columns. For example, the x terms should be aligned, the y terms should be aligned, and the constant terms should be aligned. Proper alignment is crucial for the subsequent steps, as it allows for the easy addition or subtraction of the equations without making errors. Misalignment can lead to incorrect cancellation of variables and, consequently, an incorrect solution. Therefore, taking the time to ensure the equations are correctly aligned is a fundamental step in the elimination method.

In our example, the equations are already aligned:

{4x+2y=63x−2y=8{ \begin{cases} 4x + 2y = 6 \\ 3x - 2y = 8 \end{cases} }

The x terms (4x and 3x), the y terms (2y and -2y), and the constant terms (6 and 8) are all aligned in their respective columns. This proper alignment sets the stage for the next steps in the elimination process.

Step 2: Identify a Variable to Eliminate

Next, we need to identify which variable we want to eliminate. Look for variables that have coefficients that are either the same or opposites. If the coefficients are opposites, we can simply add the equations together to eliminate the variable. If the coefficients are the same, we can subtract one equation from the other. Sometimes, it may be necessary to multiply one or both equations by a constant to make the coefficients of one of the variables the same or opposites. This multiplication ensures that when the equations are added or subtracted, the chosen variable will be eliminated, simplifying the system into a single equation with one unknown.

In our system:

{4x+2y=63x−2y=8{ \begin{cases} 4x + 2y = 6 \\ 3x - 2y = 8 \end{cases} }

we can see that the coefficients of the y terms are 2 and -2. These are opposites, which means we can eliminate the y variable by adding the two equations together. This direct cancellation simplifies the process, as we don't need to multiply any equations by constants. Identifying such opportunities for direct elimination is a key step in efficiently solving systems of equations using this method.

Step 3: Eliminate the Variable

Now, we perform the elimination. Since the coefficients of y are opposites (2 and -2), we add the two equations together:

(4x+2y)+(3x−2y)=6+8{ (4x + 2y) + (3x - 2y) = 6 + 8 }

Adding the left-hand sides and the right-hand sides separately, we get:

7x=14{ 7x = 14 }

The y terms have been eliminated, leaving us with a simple equation in terms of x. This is the crucial step where the elimination method simplifies the system, allowing us to solve for one variable directly. The elimination of the y variable has transformed the system into a single equation, which is much easier to solve.

Step 4: Solve for the Remaining Variable

Now we solve the resulting equation for x:

7x=14{ 7x = 14 }

Divide both sides by 7:

x=147{ x = \frac{14}{7} }

x=2{ x = 2 }

So, we have found the value of x. This step is a straightforward algebraic manipulation, isolating the variable x to determine its value. The result, x = 2, is a significant milestone in solving the system, as it provides one half of the solution. We now know the x-coordinate of the point where the two lines represented by the equations intersect.

Step 5: Substitute to Find the Other Variable

Substitute the value of x (which is 2) into either of the original equations to solve for y. Let's use the first equation:

4x+2y=6{ 4x + 2y = 6 }

Substitute x = 2:

4(2)+2y=6{ 4(2) + 2y = 6 }

8+2y=6{ 8 + 2y = 6 }

Subtract 8 from both sides:

2y=−2{ 2y = -2 }

Divide by 2:

y=−1{ y = -1 }

Thus, we have found the value of y. This substitution step is essential for completing the solution, as it uses the known value of x to determine the corresponding value of y. By substituting x = 2 into the first equation, we've successfully solved for y, obtaining y = -1. This completes the solution process, giving us both the x and y coordinates of the intersection point.

Step 6: Write the Solution as an Ordered Pair

The solution to the system of equations is the ordered pair (x, y), which in this case is (2, -1). This ordered pair represents the point where the two lines intersect on the coordinate plane. It's the unique pair of values for x and y that satisfies both equations simultaneously. Writing the solution as an ordered pair is the standard convention, clearly indicating the values of both variables in the solution.

Checking the Solution

To ensure our solution is correct, we can substitute the values of x and y back into both original equations:

For the first equation:

4x+2y=6{ 4x + 2y = 6 }

Substitute x = 2 and y = -1:

4(2)+2(−1)=6{ 4(2) + 2(-1) = 6 }

8−2=6{ 8 - 2 = 6 }

6=6{ 6 = 6 }

The first equation is satisfied.

For the second equation:

3x−2y=8{ 3x - 2y = 8 }

Substitute x = 2 and y = -1:

3(2)−2(−1)=8{ 3(2) - 2(-1) = 8 }

6+2=8{ 6 + 2 = 8 }

8=8{ 8 = 8 }

The second equation is also satisfied. This verification step is crucial to confirm the accuracy of the solution. By substituting the values of x and y back into the original equations, we can check that both equations hold true. This ensures that the solution (2, -1) is indeed the correct solution to the system of equations.

Conclusion

Therefore, the solution to the system of equations

{4x+2y=63x−2y=8{ \begin{cases} 4x + 2y = 6 \\ 3x - 2y = 8 \end{cases} }

is (2, -1), which corresponds to option D. The elimination method is a powerful tool for solving systems of equations, particularly when the coefficients of one of the variables are easily made opposites. By following the steps outlined in this article, you can confidently solve a wide range of systems of equations. The key is to carefully align the equations, identify a variable to eliminate, perform the elimination by adding or subtracting the equations, solve for the remaining variable, and then substitute back to find the value of the eliminated variable. Finally, always check your solution by plugging the values back into the original equations to ensure accuracy.