Linear Combination In Systems With No Solution A Detailed Explanation

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In the realm of mathematics, particularly when dealing with systems of equations, a fascinating scenario arises when a system has no solution. This often occurs when the equations represent lines that are parallel and never intersect. This article dives deep into the concept of linear combinations within systems of equations, focusing on a specific example where no solution exists. We'll explore how to identify a linear combination and understand the underlying principles that lead to such outcomes. Specifically, we will analyze the given system of equations:

{23x+52y=154x+15y=12\left\{ \begin{array}{l} \frac{2}{3} x+\frac{5}{2} y=15 \\ 4 x+15 y=12 \end{array} \right.

Our goal is to determine which equation could represent a linear combination of this system. This involves manipulating the original equations through multiplication and addition to arrive at a new equation that reveals the system's inherent nature of having no solution. Understanding this process is crucial for solving various problems in algebra and linear algebra. This article will guide you through the steps, providing a clear explanation and the rationale behind each action. By the end of this discussion, you'll have a solid grasp of how linear combinations work and how they help determine the solvability of a system of equations. We will examine the given options and deduce which one is a legitimate linear combination derived from the initial equations. The approach will involve multiplying each equation by suitable constants and then adding them together to form a new equation. This new equation will shed light on the relationship between the original equations and confirm that the system is indeed inconsistent, meaning it has no solutions. Moreover, this analysis strengthens your problem-solving skills in algebra and provides you with the tools to solve similar mathematical puzzles confidently.

Understanding Linear Combinations

To fully grasp the problem, let’s first define what a linear combination is in the context of systems of equations. A linear combination of two equations is created by multiplying each equation by a constant and then adding the resulting equations together. This technique is a powerful tool in algebra, allowing us to manipulate equations in a way that can simplify the system or reveal important properties, such as whether the system has a unique solution, infinitely many solutions, or no solution. In essence, a linear combination can help transform the original system into an equivalent system that is easier to solve or analyze. When we create a linear combination, we are essentially forming a new equation that lies within the same solution space as the original equations, provided that the constants we multiply by are not both zero. The significance of linear combinations is particularly evident when dealing with systems that might have inconsistencies, such as parallel lines in a two-variable system. By carefully choosing the constants, we can potentially eliminate one variable, which allows us to determine if the system leads to a contradiction, indicating no solution, or if it results in an identity, suggesting infinitely many solutions. This method is not just a mathematical trick; it is based on fundamental algebraic principles that ensure the resulting equation is a valid representation of the relationships within the original system. Thus, mastering linear combinations is crucial for anyone delving into the intricacies of algebra and equation solving.

Applying Linear Combinations to the System

Now, let’s apply the concept of linear combinations to our specific system of equations:

{23x+52y=154x+15y=12\left\{ \begin{array}{l} \frac{2}{3} x+\frac{5}{2} y=15 \\ 4 x+15 y=12 \end{array} \right.

To find a linear combination, we'll multiply each equation by a constant and add them together. The key here is to choose constants that might eliminate one of the variables, making it easier to analyze the relationship between x and y. We can rewrite the first equation to eliminate fractions, which will simplify the subsequent calculations. Multiplying the first equation by 6 gives us:

4x+15y=904x + 15y = 90

Now, we have the following system:

{4x+15y=904x+15y=12\left\{ \begin{array}{l} 4x + 15y = 90 \\ 4 x+15 y=12 \end{array} \right.

Notice that the left-hand sides of both equations are now identical (4x+15y4x + 15y). This is a crucial observation. If we subtract the second equation from the first, we get:

(4x + 15y) - (4x + 15y) = 90 - 12$ $0 = 78

This resulting equation, 0=780 = 78, is a clear contradiction. This contradiction implies that the original system of equations has no solution. The fact that we arrived at a false statement through legitimate algebraic manipulation confirms that the lines represented by these equations are parallel and never intersect. The beauty of linear combinations is evident here: by carefully choosing our constants, we were able to transform the system into a form that immediately reveals its inconsistency. This technique is widely used in higher mathematics, particularly in linear algebra, where it forms the basis for solving complex systems of linear equations and understanding the properties of vector spaces and matrices. In our case, the contradiction we found directly answers the question of why this system lacks a solution, highlighting the power and elegance of the linear combination method.

Identifying the Correct Linear Combination

Having established that the system has no solution, we need to identify which of the given options represents a linear combination of the original equations. The process we undertook in the previous section – multiplying the first equation by 6 and then subtracting the second equation – provides us with a clear path to the solution. We arrived at the equation 0=780 = 78, which is a direct result of combining the equations in a specific way. Now let's re-examine the steps to explicitly show the linear combination:

  1. Multiply the first original equation by 6: $6 \left( \frac{2}{3} x+\frac{5}{2} y \right) = 6(15)$

This simplifies to: $4x + 15y = 90$

  1. The second original equation remains unchanged: $4x + 15y = 12$

  2. Subtract the second equation from the modified first equation: $(4x + 15y) - (4x + 15y) = 90 - 12$ $0 = 78$

Now, we need to relate this process back to the provided options. We are looking for an equation that could have resulted from a similar manipulation. Option A states: $ \frac{4}{3} x=42

This equation doesn't immediately appear to be a direct result of our linear combination. However, we must consider whether a different combination could lead to this equation. To assess this, we need to determine if there exist constants that, when multiplied by the original equations and added, will yield $\frac{4}{3}x = 42$. Let's consider multiplying the first original equation by a constant `a` and the second original equation by a constant `b`: $a \left( \frac{2}{3} x+\frac{5}{2} y \right) + b(4x + 15y) = 15a + 12b

We want this combination to result in an equation of the form 43x=42\frac{4}{3}x = 42. This means the coefficient of y must be zero, and the coefficient of x must be 43\frac{4}{3}. Let's set up a system of equations based on these conditions:

  • Coefficient of x: 23a+4b=43\frac{2}{3}a + 4b = \frac{4}{3}
  • Coefficient of y: 52a+15b=0\frac{5}{2}a + 15b = 0
  • Constant term: 15a+12b=4215a + 12b = 42

Solving this system of equations will tell us if such constants a and b exist. If they do, then option A is a valid linear combination. This approach allows us to rigorously check whether the given option is indeed a possible outcome of combining the original equations, thus ensuring we choose the correct answer.

Detailed Analysis of Option A

To determine if option A, 43x=42\frac{4}{3} x = 42, represents a linear combination of the original system, we need to solve the system of equations we derived in the previous section. This involves finding constants a and b that satisfy the following conditions:

  • Coefficient of x: 23a+4b=43\frac{2}{3}a + 4b = \frac{4}{3}
  • Coefficient of y: 52a+15b=0\frac{5}{2}a + 15b = 0
  • Constant term: 15a+12b=4215a + 12b = 42

Let's start by solving the second equation for a:

\frac{5}{2}a = -15b$ $a = -15b \cdot \frac{2}{5}$ $a = -6b

Now, substitute this expression for a into the first equation:

\frac{2}{3}(-6b) + 4b = \frac{4}{3}$ $-4b + 4b = \frac{4}{3}$ $0 = \frac{4}{3}

This result is a contradiction. It indicates that there are no values of a and b that can simultaneously satisfy the first two conditions. Therefore, option A, 43x=42\frac{4}{3} x = 42, cannot be a linear combination of the given system of equations. This outcome is crucial because it demonstrates that not every equation that appears related to the system is a valid linear combination. The stringent requirements for the coefficients and constant terms to match up, as dictated by the linear combination process, must be met. Our analysis has revealed that these requirements are not met for option A, confirming its incorrectness. This process underscores the importance of a systematic approach when dealing with linear combinations, ensuring that each candidate equation is rigorously tested against the fundamental principles of linear algebra.

Conclusion

In conclusion, analyzing the given system of equations $ \left{ \begin{array}{l} \frac{2}{3} x+\frac{5}{2} y=15 \ 4 x+15 y=12 \end{array} \right.

reveals that it has no solution due to the contradiction arising from the **_linear combination_** process. By multiplying the first equation by 6 and subtracting the second equation, we arrived at the false statement $0 = 78$, confirming the system's inconsistency. Furthermore, our detailed analysis of option A, $\frac{4}{3} x = 42$, demonstrated that it cannot be a linear combination of the original equations. We systematically attempted to find constants that would satisfy the conditions for a linear combination but encountered a contradiction, thereby disproving option A. This exploration highlights the power and precision of linear combinations in determining the solvability of systems of equations. It also underscores the importance of rigorous mathematical analysis when evaluating potential solutions. The principles discussed here are fundamental to algebra and linear algebra, providing essential tools for solving and understanding various mathematical problems. By mastering these concepts, one can confidently tackle complex systems of equations and gain deeper insights into the relationships between variables and equations.