Solving For X And Y In Linear Equations A Step By Step Guide

In mathematics, we often encounter situations where we need to solve for multiple unknowns. One common scenario is when we have a system of linear equations. These equations represent relationships between variables, and our goal is to find the values of those variables that satisfy all equations simultaneously. This article delves into the process of solving a system of linear equations, using a specific example to illustrate the steps involved. We'll explore the concepts, methods, and nuances of solving these problems, providing a comprehensive understanding for anyone interested in mathematics and its applications.

Introduction to Systems of Linear Equations

A system of linear equations is a set of two or more linear equations containing the same variables. A linear equation is an equation in which the highest power of any variable is 1. For example, x + y = 5000 and 0.15x + 0.10y = 80 are linear equations. Solving a system of linear equations means finding the values for the variables that make all the equations true. These values represent the point(s) where the lines represented by the equations intersect on a graph. There are several methods to solve such systems, including substitution, elimination, and matrix methods. Understanding these methods is crucial for various applications in fields like economics, engineering, and computer science.

The system of equations we will focus on in this article is:

1. x + y = 5000
2. 0.15x + 0.10y = 80

This system has two equations and two unknowns (x and y), making it a classic example of a linear system that can be solved using various algebraic techniques. Before we dive into the solution, it's essential to understand why solving such systems is important and how these solutions can be interpreted in real-world contexts. For instance, in economics, 'x' and 'y' might represent the quantities of two different goods, and the equations could represent budget constraints and market demands. The solution would then give the optimal quantities of each good to maximize utility within the given constraints.

Methods for Solving Systems of Linear Equations

There are several methods to solve systems of linear equations, each with its own advantages and disadvantages. The most common methods include:

• Substitution Method: This involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be solved. The value obtained is substituted back into one of the original equations to find the value of the other variable.
• Elimination Method: This involves manipulating the equations so that the coefficients of one of the variables are the same (or additive inverses). By adding or subtracting the equations, one variable is eliminated, resulting in a single equation with one variable. This equation is then solved, and the value is substituted back to find the other variable.
• Graphical Method: This involves plotting the equations on a graph and finding the point of intersection, which represents the solution. This method is particularly useful for visualizing the solution but may not be accurate for complex systems.
• Matrix Methods: These involve using matrices and matrix operations to solve the system. Methods like Gaussian elimination, Gauss-Jordan elimination, and using the inverse of a matrix are part of this category. Matrix methods are especially efficient for solving large systems of equations.

For the system we are considering, both the substitution and elimination methods are viable options. The choice often depends on personal preference and the specific structure of the equations. In this article, we will primarily use the elimination method, as it can be quite efficient for this particular system. However, we will also briefly touch upon how the substitution method could be applied.

Solving the System Using the Elimination Method

The elimination method involves manipulating the equations to eliminate one of the variables. To do this, we need to make the coefficients of one of the variables the same (or additive inverses) in both equations. Let's consider our system again:

1. x + y = 5000
2. 0.15x + 0.10y = 80

To eliminate y, we can multiply the first equation by 0.10, which will make the coefficient of y in the first equation equal to the coefficient of y in the second equation. This gives us:

• 0.10(x + y) = 0.10(5000), which simplifies to 0.10x + 0.10y = 500

Now, we have the following system:

1. 0.10x + 0.10y = 500
2. 0.15x + 0.10y = 80

Next, we subtract the first equation from the second equation to eliminate y:

• (0.15x + 0.10y) - (0.10x + 0.10y) = 80 - 500

This simplifies to:

• 0.05x = -420

Now, we can solve for x by dividing both sides by 0.05:

• x = -420 / 0.05 = -8400

We have found the value of x. Now, we can substitute this value into one of the original equations to find y. Let's use the first equation:

• x + y = 5000
• -8400 + y = 5000

Add 8400 to both sides:

• y = 5000 + 8400 = 13400

So, the solution to the system of equations is x = -8400 and y = 13400.

Solving the System Using the Substitution Method

Although we solved the system using the elimination method, it's also beneficial to see how the substitution method could be applied. This provides a different perspective and reinforces the understanding of solving systems of equations. Again, our system is:

1. x + y = 5000
2. 0.15x + 0.10y = 80

We can start by solving the first equation for y:

• y = 5000 - x

Now, substitute this expression for y into the second equation:

• 0. 15x + 0.10(5000 - x) = 80

Distribute the 0.10:

• 0. 15x + 500 - 0.10x = 80

Combine like terms:

• 0. 05x + 500 = 80

Subtract 500 from both sides:

• 0. 05x = -420

Divide by 0.05:

• x = -420 / 0.05 = -8400

Now that we have x, we can substitute it back into the expression for y:

• y = 5000 - (-8400) = 5000 + 8400 = 13400

As we can see, the substitution method leads to the same solution: x = -8400 and y = 13400. This confirms the consistency of the methods and the accuracy of our solution.

Verification of the Solution

To ensure that our solution is correct, it's crucial to verify it by substituting the values of x and y back into the original equations. This step helps to catch any arithmetic errors or inconsistencies in the solution process. Let's substitute x = -8400 and y = 13400 into the original equations:

1. x + y = 5000

   • -8400 + 13400 = 5000
   • 5000 = 5000 (True)

2. 0.15x + 0.10y = 80

   • 0. 15(-8400) + 0.10(13400) = 80
   • -1260 + 1340 = 80
   • 80 = 80 (True)

Since the values of x and y satisfy both equations, our solution is correct. This verification step is essential in mathematics and provides confidence in the accuracy of the results. It also highlights the importance of careful and methodical problem-solving.

Interpreting the Solution

Now that we have the solution x = -8400 and y = 13400, it's important to understand what these values mean. In a mathematical context, these values represent the point of intersection of the two lines defined by the equations. However, in a real-world scenario, the interpretation can be more nuanced and depends on the context of the problem. For instance, if x and y represent quantities of goods, a negative value for x might indicate a debt or a borrowing situation, while a large positive value for y could represent a significant amount of the other good.

The interpretation of a solution is crucial for applying mathematical results to practical problems. Without a proper understanding of what the variables represent, the numerical values alone are meaningless. In the absence of a specific context, we can state that the solution x = -8400 and y = 13400 satisfies the given system of linear equations. However, to provide a meaningful interpretation, we would need additional information about the problem's context.

Common Mistakes and How to Avoid Them

When solving systems of linear equations, several common mistakes can occur. Being aware of these pitfalls can help in avoiding them and ensuring accurate solutions. Some common mistakes include:

• Arithmetic Errors: Miscalculations while performing arithmetic operations (addition, subtraction, multiplication, division) are a frequent source of errors. It's important to double-check each step and use a calculator when necessary.
• Incorrect Substitution: When using the substitution method, it's crucial to substitute the expression correctly and consistently. Substituting into the same equation from which the expression was derived is a common mistake.
• Sign Errors: Errors involving positive and negative signs can easily occur, especially when dealing with the elimination method. Paying close attention to signs and double-checking them is essential.
• Misunderstanding the Problem: Failing to fully understand the problem's context can lead to incorrect interpretations of the solution. Make sure to read the problem carefully and identify what the variables represent.

To avoid these mistakes, it's helpful to follow a systematic approach, show all steps clearly, and verify the solution by substituting the values back into the original equations. Regular practice and a solid understanding of the underlying concepts are also key to improving accuracy and avoiding errors.

Conclusion

Solving systems of linear equations is a fundamental skill in mathematics with wide-ranging applications. In this article, we have explored the process of solving a system of two linear equations with two unknowns using both the elimination and substitution methods. We have also emphasized the importance of verifying the solution and interpreting it in the context of the problem. By understanding the methods, avoiding common mistakes, and practicing regularly, you can master the art of solving linear systems and apply this skill to various mathematical and real-world problems. The system of equations x + y = 5000 and 0.15x + 0.10y = 80 provided a concrete example for illustrating the steps and nuances involved in solving such problems. Remember, the key to success in mathematics is not just memorizing formulas and procedures, but also developing a deep understanding of the concepts and their applications.

By mastering these techniques, you'll be well-equipped to tackle more complex mathematical problems and apply these skills in various fields, including engineering, economics, and computer science. The journey of learning mathematics is a continuous process, and each solved problem adds to your understanding and expertise.