Solving 4x + 3y = 9 And 2x + 5y = 15 A Step-by-Step Guide

Introduction: Understanding Systems of Linear Equations

In the realm of mathematics, particularly in algebra, systems of linear equations form a foundational concept with wide-ranging applications. A system of linear equations is a collection of two or more linear equations involving the same set of variables. The goal is to find values for these variables that satisfy all equations simultaneously. These systems pop up in various fields, from economics and engineering to computer science and data analysis. Understanding how to solve them is a crucial skill for anyone delving into quantitative disciplines. At its core, solving a system of linear equations involves finding the point(s) where the lines represented by the equations intersect. In a two-variable system, like the one we'll explore, each equation represents a line on a graph. The solution to the system is the coordinate point where these lines cross. However, not all systems have a unique solution. Some systems might have infinitely many solutions (if the equations represent the same line), while others might have no solution at all (if the lines are parallel). This introduction sets the stage for a deeper dive into various methods for tackling systems of linear equations and highlights the practical significance of this mathematical concept. Whether you're a student grappling with algebra or a professional using mathematical models, mastering systems of linear equations is an invaluable asset.

This article delves into the process of solving a specific system of linear equations: $\begin{array}{l}4 x+3 y=9 \ 2 x+5 y=15\end{array}$. We will explore different methods to find the solution, offering a step-by-step guide to make the process clear and understandable. Understanding how to solve such systems is a fundamental skill in algebra and has applications in various fields, including engineering, economics, and computer science. This initial system serves as an excellent case study to illustrate the common techniques used to solve systems of linear equations. By the end of this article, you'll not only understand the solution to this particular system but also gain a broader understanding of how to approach similar problems. The concepts discussed here form the bedrock for more advanced topics in linear algebra and mathematical modeling.

Linear equations, at their heart, represent relationships between variables in a straightforward, predictable manner. When we talk about a 'system' of these equations, we're essentially looking at how multiple such relationships interact. Imagine you're trying to balance a budget, considering income and expenses, or perhaps optimizing a production process with various constraints. These real-world scenarios often translate directly into systems of linear equations. Each equation captures a specific condition or constraint, and the solution to the system represents a scenario where all conditions are met simultaneously. This is why the ability to solve these systems is so powerful. It allows us to model and analyze complex situations, make informed decisions, and predict outcomes based on mathematical models. The equations might represent supply and demand curves, the forces acting on a structure, or even the flow of information in a network. The beauty of linear equations lies in their simplicity and the wide array of techniques available to solve them, making them a versatile tool for problem-solving across various disciplines.

Method 1: Substitution Method

The substitution method is a powerful technique for solving systems of linear equations. The core idea behind this method is to solve one equation for one variable and then substitute that expression into the other equation. This substitution results in a single equation with a single variable, which can then be easily solved. Once you've found the value of one variable, you can substitute it back into either of the original equations to find the value of the other variable. This method is particularly effective when one of the equations can be easily solved for one variable in terms of the other. For instance, if one equation has a variable with a coefficient of 1, isolating that variable is a straightforward process. The substitution method is a cornerstone of algebraic problem-solving and is applicable to a wide range of systems of linear equations. Its elegance lies in its ability to reduce a complex system into simpler, solvable parts.

Let's apply the substitution method to our system: $\begin{array}{l}4 x+3 y=9 \ 2 x+5 y=15\end{array}$.

1. Solve the second equation for $x$: $2x = 15 - 5y \Rightarrow x = \frac{15 - 5y}{2}$.
2. Substitute this expression for $x$ into the first equation: $4(\frac{15 - 5y}{2}) + 3y = 9$.
3. Simplify and solve for $y$: $2(15 - 5y) + 3y = 9 \Rightarrow 30 - 10y + 3y = 9 \Rightarrow -7y = -21 \Rightarrow y = 3$.
4. Substitute $y = 3$ back into the expression for $x$: $x = \frac{15 - 5(3)}{2} = \frac{0}{2} = 0$.

Therefore, the solution using the substitution method is $x = 0$ and $y = 3$.

Each step in the substitution method is designed to break down the complexity of the system into manageable pieces. By isolating one variable and substituting its expression into the other equation, we eliminate one variable and create a simpler equation to solve. This process of simplification is a key strategy in algebra and is used in various problem-solving scenarios. The ability to manipulate equations and isolate variables is a fundamental skill that underpins much of mathematical reasoning. Once we have solved for one variable, the substitution back into the original equation allows us to find the value of the other variable, completing the solution to the system. This methodical approach not only provides the answer but also enhances our understanding of the relationships between the variables in the system. The substitution method, therefore, is more than just a technique; it's a powerful tool for unraveling the intricacies of linear systems.

Method 2: Elimination Method

The elimination method, also known as the addition method, offers another powerful approach to solving systems of linear equations. Unlike the substitution method, which focuses on isolating one variable, the elimination method aims to eliminate one variable by manipulating the equations so that either the coefficients of $x$ or the coefficients of $y$ are opposites. When you add the equations together, the variable with opposite coefficients will cancel out, leaving you with a single equation in one variable. This method is particularly useful when the equations are in standard form (Ax + By = C) and neither variable is easily isolated. The elimination method relies on the principle that adding equal quantities to both sides of an equation does not change its solution, allowing us to manipulate the equations without altering their fundamental relationships. Its strength lies in its ability to handle systems with more complex coefficients and structures, making it a versatile tool in the algebraic toolbox.

Let's apply the elimination method to the system: $\begin{array}{l}4 x+3 y=9 \ 2 x+5 y=15\end{array}$.

1. Multiply the second equation by -2 to make the coefficients of $x$ opposites: $-2(2x + 5y) = -2(15) \Rightarrow -4x - 10y = -30$.
2. Add the modified second equation to the first equation: $(4x + 3y) + (-4x - 10y) = 9 + (-30)$.
3. Simplify and solve for $y$: $-7y = -21 \Rightarrow y = 3$.
4. Substitute $y = 3$ back into either of the original equations to solve for $x$. Let's use the first equation: $4x + 3(3) = 9 \Rightarrow 4x + 9 = 9 \Rightarrow 4x = 0 \Rightarrow x = 0$.

Thus, the solution using the elimination method is $x = 0$ and $y = 3$.

At the heart of the elimination method is the strategic manipulation of equations to create a situation where adding them together eliminates one variable. This is achieved by multiplying one or both equations by a constant that will make the coefficients of one of the variables opposites. The beauty of this approach is that it simplifies the system by reducing it to a single equation with one unknown. Once we solve for that unknown, we can substitute the value back into one of the original equations to find the value of the other variable. The elimination method is a testament to the power of algebraic manipulation and demonstrates how seemingly complex problems can be broken down into simpler steps. It's a valuable skill for anyone working with systems of equations, offering an alternative approach to substitution that can be more efficient in certain situations. The method's strength lies in its ability to handle equations with various coefficients and its clear, methodical approach to problem-solving.

Comparison of Methods

Both the substitution and elimination methods are effective for solving systems of linear equations, but they differ in their approach and may be more suitable for different types of systems. The substitution method is particularly useful when one of the equations can be easily solved for one variable in terms of the other. This makes the substitution process straightforward and efficient. However, if neither equation has a variable that is easily isolated, the substitution method can become cumbersome and involve fractions, potentially increasing the risk of errors. The elimination method, on the other hand, shines when the equations are in standard form (Ax + By = C) or when the coefficients of one variable are easily made opposites by multiplication. This method avoids the fractions that can sometimes arise in substitution, making it a clean and often faster approach for certain systems. In the example we explored, both methods led to the same solution, demonstrating their equivalence. The choice between the methods often comes down to personal preference and the specific characteristics of the system at hand. Being proficient in both methods provides a versatile toolkit for tackling a wide range of linear systems.

Choosing between the substitution and elimination methods is not just about finding a solution; it's also about efficiency and minimizing the chances of making errors. The substitution method is conceptually simple: isolate one variable and plug its expression into the other equation. This is intuitive and easy to remember. However, it can lead to more complex algebra if the coefficients are not simple, potentially resulting in fractions and increased computational effort. The elimination method, in contrast, focuses on manipulating the equations as a whole to eliminate a variable. This often involves multiplying equations by constants, which can be a more direct path to a solution when the equations are already in a suitable form. The key is to recognize the structure of the equations and choose the method that will lead to the fewest algebraic steps. Some systems lend themselves more naturally to one method over the other, and experience in solving these systems helps in developing this intuition. Ultimately, mastering both methods provides a flexible and powerful approach to solving linear equations, allowing you to select the most efficient path to the solution.

In practical applications, the choice between the substitution and elimination methods can have a significant impact on the efficiency of the problem-solving process. Imagine dealing with a complex system of equations that represents a real-world scenario, such as optimizing a business process or designing an engineering structure. Choosing the right method can save valuable time and effort, especially when dealing with a large number of equations or variables. The substitution method can be particularly advantageous when one variable is already expressed in terms of the others, perhaps as a result of a previous step in the analysis. This allows for a direct substitution, simplifying the system quickly. The elimination method, on the other hand, can be more robust when dealing with equations that have similar coefficients or when the goal is to find the relationship between variables rather than their specific values. In these cases, the elimination method can provide a clearer path to understanding the underlying structure of the system. By understanding the strengths and weaknesses of each method, you can develop a strategic approach to solving systems of equations, choosing the method that best fits the specific problem at hand.

Conclusion: Solution and Implications

In conclusion, by applying both the substitution and elimination methods, we have found that the solution to the system of equations $\begin{array}{l}4 x+3 y=9 \ 2 x+5 y=15\end{array}$ is $x = 0$ and $y = 3$. This means that the point (0, 3) is the intersection of the two lines represented by these equations. This solution satisfies both equations simultaneously, confirming that it is the unique solution to this system. Understanding how to solve systems of linear equations is a fundamental skill in mathematics with broad applications in various fields. From modeling real-world scenarios to solving complex problems in science and engineering, the ability to find solutions to these systems is crucial. The techniques discussed in this article provide a solid foundation for further exploration of linear algebra and its applications. The solution not only provides the values of $x$ and $y$ but also gives us insights into the relationship between the two equations and their graphical representation.

This solution, $x = 0$ and $y = 3$, has a specific geometric interpretation. Each linear equation represents a line on a two-dimensional graph. The point (0, 3) is the exact location where these two lines intersect. This intersection point is the only coordinate pair that lies on both lines, thus satisfying both equations simultaneously. In a broader context, understanding the geometric interpretation of solutions to systems of linear equations helps to visualize and analyze more complex systems. For instance, in a system of three equations with three variables, each equation represents a plane in three-dimensional space, and the solution represents the point where all three planes intersect. This geometric perspective is invaluable in fields like computer graphics, where linear equations are used to model and manipulate objects in space. The concept of intersection also extends to higher dimensions, where solutions represent the points where hyperplanes intersect. Therefore, mastering the basics of solving systems of linear equations not only provides algebraic skills but also develops geometric intuition, which is crucial for tackling more advanced mathematical concepts.

Beyond the specific solution, the process of solving this system of equations highlights several important principles in mathematics and problem-solving. First, it demonstrates the power of algebraic manipulation in simplifying complex problems. By strategically applying techniques like substitution and elimination, we can transform a system of equations into a more manageable form, leading to a clear solution. Second, it illustrates the importance of having multiple methods for solving the same problem. Both substitution and elimination are valid approaches, and the choice between them often depends on the specific characteristics of the equations. Being proficient in both methods provides a flexible toolkit for tackling a wider range of problems. Third, the process underscores the interconnectedness of mathematical concepts. Solving systems of linear equations draws upon skills in algebra, arithmetic, and logical reasoning. It also lays the foundation for more advanced topics in linear algebra, calculus, and differential equations. Ultimately, the ability to solve systems of linear equations is not just a mathematical skill; it's a valuable tool for critical thinking and problem-solving in various aspects of life.