Solving Systems Of Equations 6x - Y = 12 And 3x + Y = 6

JU07/13/2025 08, 2025 by THE IDEN 56 views

Solving Systems of Equations A Comprehensive Guide to 6x - y = 12 and 3x + y = 6

In the realm of mathematics, solving systems of equations is a fundamental skill with applications spanning various fields, from engineering to economics. This article delves into the intricacies of solving a specific system of linear equations: 6x - y = 12 and 3x + y = 6. We will explore different methods, providing a step-by-step guide to arrive at the solution. Understanding how to solve systems of equations is crucial for anyone seeking to master algebra and its practical applications. Our focus will be on providing a clear, comprehensive approach, ensuring that both beginners and those with some experience can benefit from this guide.

Understanding Systems of Equations

Before diving into the solution, it's essential to understand what a system of equations represents. A system of equations is a set of two or more equations containing the same variables. The solution to a system of equations is the set of values for the variables that satisfy all equations simultaneously. In our case, we have two linear equations with two variables, x and y. The goal is to find the values of x and y that make both equations true.

Linear equations, which are at the heart of this system, represent straight lines when graphed on a coordinate plane. The solution to a system of two linear equations corresponds to the point where the two lines intersect. This intersection point represents the (x, y) values that satisfy both equations. When we talk about solving the system 6x - y = 12 and 3x + y = 6, we're essentially looking for the coordinates of this intersection point. There are several methods to find this point, each with its own advantages and suitability for different types of systems. We will explore these methods in detail, ensuring a thorough understanding of the underlying principles.

Methods for Solving Systems of Equations

There are several methods for solving systems of equations, including:

Substitution Method: This method involves solving one equation for one variable and substituting that expression into the other equation.
Elimination Method: This method involves adding or subtracting the equations to eliminate one variable.
Graphical Method: This method involves graphing the equations and finding the point of intersection.

In the context of solving 6x - y = 12 and 3x + y = 6, we will primarily focus on the substitution and elimination methods, as they are the most efficient for this particular system. The graphical method, while useful for visualization, may not always provide precise solutions, especially when the intersection point has non-integer coordinates. Each method offers a unique approach to tackling the problem, and understanding their nuances is key to becoming proficient in solving systems of equations. We will break down each method into manageable steps, illustrating how they lead to the same solution.

Solving the System Using the Elimination Method

The elimination method is particularly well-suited for this system because the coefficients of y in the two equations are opposites (-1 and 1). This allows us to eliminate y by simply adding the two equations together. Here’s how it works:

Write down the equations:

$6x - y = 12$

$3x + y = 6$
Add the equations together:

$(6x - y) + (3x + y) = 12 + 6$

This simplifies to:

$9x = 18$
Solve for x:

Divide both sides by 9:

$x = 2$

As you can see, the elimination method allows us to efficiently isolate one variable by strategically manipulating the equations. The beauty of this method lies in its simplicity and directness, especially when the coefficients of one variable are opposites or can be easily made opposites through multiplication. In this case, the opposite coefficients of y made the elimination straightforward. This step-by-step process demonstrates the elegance of algebraic manipulation in solving complex problems. Now that we have found the value of x, we can substitute it back into either of the original equations to find the value of y.

Finding the Value of y

Now that we have found the value of x (x = 2), we can substitute it back into either of the original equations to solve for y. Let's use the second equation, 3x + y = 6, for this step:

Substitute x = 2 into the equation:

$3(2) + y = 6$
Simplify:

$6 + y = 6$
Solve for y:

Subtract 6 from both sides:

$y = 0$

This substitution process highlights the interconnectedness of the variables in a system of equations. Once we determine the value of one variable, it becomes a crucial piece in unlocking the value of the other. By carefully substituting the known value into one of the equations, we can reduce the problem to a simple one-variable equation, making it easy to isolate and solve for the remaining unknown. In this instance, we found that y = 0, completing the solution to our system. Now, we have both x and y values, allowing us to state the solution as an ordered pair.

The Solution

The solution to the system of equations is the ordered pair (x, y) = (2, 0). This means that the point (2, 0) is the intersection point of the two lines represented by the equations 6x - y = 12 and 3x + y = 6. To ensure the accuracy of our solution, we can verify it by substituting these values back into both original equations.

To verify the solution, we substitute x = 2 and y = 0 into both equations:

For the first equation, 6x - y = 12:

$6(2) - 0 = 12$

$12 = 12$ (This is true)
For the second equation, 3x + y = 6:

$3(2) + 0 = 6$

$6 = 6$ (This is also true)

Since the values x = 2 and y = 0 satisfy both equations, we can confidently conclude that this is indeed the correct solution. This verification step is a crucial part of the problem-solving process, as it provides a final check to catch any potential errors. The solution (2, 0) represents the only point that lies on both lines simultaneously, emphasizing the unique nature of the solution to this system of equations.

Solving the System Using the Substitution Method

Another effective method for solving this system is the substitution method. This method involves solving one equation for one variable and then substituting that expression into the other equation. Let's walk through the steps:

Choose an equation and solve for one variable:

Let's choose the second equation, 3x + y = 6, and solve for y:

$y = 6 - 3x$
Substitute the expression for y into the other equation:

Substitute $y = 6 - 3x$ into the first equation, 6x - y = 12:

$6x - (6 - 3x) = 12$
Simplify and solve for x:

$6x - 6 + 3x = 12$

$9x - 6 = 12$

$9x = 18$

$x = 2$

The substitution method provides an alternative pathway to the same solution, showcasing the versatility of algebraic techniques. By isolating one variable in one equation, we create an expression that can be seamlessly integrated into the other equation, effectively reducing the system to a single-variable equation. This method is particularly useful when one equation can be easily solved for one of the variables, as was the case in our example. The ability to choose the most convenient method for a given system of equations is a hallmark of a skilled problem solver. As we found with the elimination method, the substitution method also leads us to the value x = 2, further reinforcing the correctness of this partial solution.

Finding the Value of y Using Substitution

Now that we have found the value of x (x = 2) using the substitution method, we need to find the value of y. We can do this by substituting the value of x back into the expression we found for y in the first step:

Substitute x = 2 into the expression for y:

$y = 6 - 3x$

$y = 6 - 3(2)$
Simplify:

$y = 6 - 6$

$y = 0$

This step mirrors the process we undertook in the elimination method, highlighting the consistency of mathematical solutions regardless of the method employed. By plugging the x-value back into the expression we derived earlier, we effortlessly arrive at the y-value. This underscores the importance of maintaining accuracy and clarity throughout the problem-solving process. The consistent result of y = 0, obtained through both the elimination and substitution methods, strengthens our confidence in the overall solution to the system of equations. We have now determined both x and y values using the substitution method, allowing us to confirm the complete solution.

Confirming the Solution

Again, we arrive at the solution (x, y) = (2, 0), which matches the solution we found using the elimination method. This consistency reinforces the correctness of our answer. To be absolutely sure, we should still verify this solution by substituting it back into the original equations, as we did previously.

To ensure the accuracy of our solution, we substitute x = 2 and y = 0 into both original equations:

For the first equation, 6x - y = 12:

$6(2) - 0 = 12$

$12 = 12$ (This holds true)
For the second equation, 3x + y = 6:

$3(2) + 0 = 6$

$6 = 6$ (This also holds true)

The fact that the values x = 2 and y = 0 satisfy both equations definitively confirms that (2, 0) is the correct solution to the system. This verification process is a critical step in mathematical problem-solving, as it provides a robust check against errors and solidifies our understanding of the solution. By confirming the solution through both elimination and substitution methods, we have demonstrated a comprehensive and rigorous approach to solving systems of equations.

Conclusion

In this article, we have thoroughly explored the process of solving the system of equations 6x - y = 12 and 3x + y = 6. We have demonstrated two primary methods: the elimination method and the substitution method. Both methods lead us to the same solution, (x, y) = (2, 0). This solution represents the point where the two lines represented by the equations intersect on a graph. The ability to solve systems of equations is a valuable skill in mathematics and has numerous applications in real-world problems.

Understanding these methods not only allows you to solve specific problems but also enhances your overall problem-solving skills in mathematics. The elimination method excels when the coefficients of one variable are opposites or can be easily made so, while the substitution method shines when one variable can be readily isolated in one of the equations. Mastering both methods provides a versatile toolkit for tackling a wide range of systems of equations. This comprehensive guide has provided a clear and detailed explanation of each step involved, empowering you to confidently solve similar problems in the future. Remember to always verify your solutions to ensure accuracy and deepen your understanding of the underlying concepts.