Comparing Solutions Of Two Systems Of Equations System A And System B

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Introduction

In the realm of mathematics, systems of equations play a pivotal role in solving real-world problems. Understanding how to solve and compare different systems is a fundamental skill. In this article, we delve into two distinct systems of equations, System A and System B, to determine which statements hold true regarding their solutions. We will explore the intricacies of solving these systems using various methods, compare their properties, and ultimately arrive at a definitive conclusion. Our focus will be on providing a comprehensive analysis that not only answers the question but also enhances your understanding of linear systems.

Problem Statement

We are presented with two systems of equations:

System A

3x + 2y = 3
-2x - 8y = -1

System B

-x - 14y = 1
-2x - 8y = -1

Our objective is to determine which statement accurately compares the solutions of these two systems. This involves solving both systems and comparing their x and y values to identify any relationships or differences.

Solving System A

To solve System A, we can employ several methods, including substitution, elimination, and matrix operations. Here, we will use the elimination method, which is particularly effective for systems with two variables.

System A Equations

  1. 3x + 2y = 3
  2. -2x - 8y = -1

To eliminate x, we can multiply the first equation by 2 and the second equation by 3. This will give us coefficients of x that are opposites, allowing us to eliminate x by adding the equations.

Modified Equations

  1. 2 * (3x + 2y) = 2 * 3 -> 6x + 4y = 6
  2. 3 * (-2x - 8y) = 3 * (-1) -> -6x - 24y = -3

Now, add the modified equations:

(6x + 4y) + (-6x - 24y) = 6 + (-3)

This simplifies to:

-20y = 3

Solving for y:

y = 3 / -20 y = -3/20

Now that we have the value of y, we can substitute it back into one of the original equations to solve for x. Let's use the first equation, 3x + 2y = 3.

3x + 2(-3/20) = 3

3x - 3/10 = 3

Add 3/10 to both sides:

3x = 3 + 3/10

3x = 30/10 + 3/10

3x = 33/10

Divide by 3:

x = (33/10) / 3 x = 33/30 x = 11/10

Thus, the solution for System A is x = 11/10 and y = -3/20.

Solving System B

Now, let's solve System B using the elimination method again.

System B Equations

  1. -x - 14y = 1
  2. -2x - 8y = -1

To eliminate x, we can multiply the first equation by -2. This will make the coefficients of x opposites, allowing us to eliminate x by adding the equations.

Modified Equation

  1. -2 * (-x - 14y) = -2 * 1 -> 2x + 28y = -2
  2. -2x - 8y = -1

Now, add the modified equation 1 to equation 2:

(2x + 28y) + (-2x - 8y) = -2 + (-1)

This simplifies to:

20y = -3

Solving for y:

y = -3/20

Now that we have the value of y, we can substitute it back into one of the original equations to solve for x. Let's use the first equation, -x - 14y = 1.

-x - 14(-3/20) = 1

-x + 42/20 = 1

-x + 21/10 = 1

Subtract 21/10 from both sides:

-x = 1 - 21/10

-x = 10/10 - 21/10

-x = -11/10

Multiply by -1:

x = 11/10

Therefore, the solution for System B is x = 11/10 and y = -3/20.

Comparing Solutions

After solving both systems, we find the following solutions:

  • System A: x = 11/10, y = -3/20
  • System B: x = 11/10, y = -3/20

Upon comparing the solutions, it is evident that the values of x and y are identical for both System A and System B. This means that the systems have the same solution set.

Detailed Analysis and Discussion

When dealing with systems of equations, it is crucial to understand that the solution represents the point(s) where the lines (in the case of two-variable linear equations) intersect. In this scenario, since both systems have the same solution (x = 11/10, y = -3/20), it implies that the lines represented by the equations in both systems intersect at the same point. This can occur if the equations in one system are linear combinations of the equations in the other system, or if the systems are essentially equivalent.

Another important aspect to consider is the consistency and dependency of the systems. A system is consistent if it has at least one solution, and it is independent if it has a unique solution. In this case, both systems are consistent and independent since they have exactly one solution. If the systems had infinitely many solutions, they would be consistent but dependent, indicating that the equations represent the same line. If the systems had no solution, they would be inconsistent, indicating that the lines are parallel and do not intersect.

Furthermore, understanding the graphical representation of these systems can provide additional insights. Each equation in a system of two variables represents a line in a two-dimensional plane. The solution to the system is the point where these lines intersect. If the lines are parallel, there is no solution; if they coincide, there are infinitely many solutions; and if they intersect at a single point, there is a unique solution. In this case, the graphical representation would show that the lines in both systems intersect at the same point (11/10, -3/20).

Key Concepts Revisited

  1. Systems of Equations: A set of two or more equations with the same variables.
  2. Solution of a System: The set of values for the variables that satisfy all equations in the system.
  3. Elimination Method: A method to solve systems of equations by eliminating one variable through addition or subtraction.
  4. Substitution Method: A method to solve systems of equations by solving one equation for one variable and substituting that expression into the other equation.
  5. Consistency and Dependency: A system is consistent if it has at least one solution, and it is independent if it has a unique solution.

Practical Applications

The ability to solve and compare systems of equations has numerous practical applications in various fields, including:

  • Engineering: Designing structures and circuits involves solving systems of equations to ensure stability and functionality.
  • Economics: Modeling supply and demand, predicting market trends, and optimizing resource allocation often require solving systems of equations.
  • Computer Science: Solving linear systems is essential in areas such as computer graphics, machine learning, and network analysis.
  • Physics: Analyzing motion, forces, and energy often involves solving systems of equations to describe physical phenomena.

Conclusion

In conclusion, after solving both System A and System B, we have determined that they share the same solution. Both systems have x = 11/10 and y = -3/20 as their solution. This detailed analysis not only provides the answer but also deepens our understanding of systems of equations and their applications. By mastering these concepts, we are better equipped to tackle a wide range of mathematical and real-world problems. The methodical approach of solving each system independently and then comparing the results ensures accuracy and reinforces the fundamental principles of algebra. This exercise highlights the importance of critical thinking and problem-solving skills in mathematics, which are essential for success in STEM fields and beyond.

By applying the elimination method, we systematically reduced the equations to solve for each variable. The careful steps of multiplying equations by constants and adding or subtracting them allowed us to isolate the variables and find their values. This process underscores the power of algebraic manipulation in simplifying complex problems and arriving at precise solutions.

Moreover, the comparison of the solutions reinforces the concept that different sets of equations can lead to the same solution. This is a crucial insight in mathematics, as it allows us to approach problems from multiple angles and choose the most efficient method for solving them. The ability to recognize equivalent systems of equations is a valuable skill that can save time and effort in mathematical problem-solving.

In summary, the comparison of System A and System B demonstrates the fundamental principles of solving and analyzing systems of equations. The shared solution highlights the interconnectedness of mathematical concepts and the importance of a thorough understanding of algebraic techniques. This exercise serves as a valuable learning opportunity, reinforcing the skills needed to tackle more complex problems in mathematics and its applications.