System Of Equations No Solution Linear Combination

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In the realm of linear algebra, a system of equations is a collection of two or more equations that share the same variables. Solving a system of equations involves finding values for the variables that satisfy all equations simultaneously. However, not all systems have solutions. Some systems have a unique solution, others have infinitely many solutions, and some have no solution at all. In this article, we will delve into a specific case of a system of equations with no solution and explore how to identify such systems through linear combinations.

The system of equations presented is:

{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.

This system consists of two linear equations in two variables, x and y. Our goal is to determine why this system has no solution and to identify a linear combination that demonstrates this property. A linear combination of equations is formed by multiplying each equation by a constant and then adding the resulting equations together. The key to understanding systems with no solution lies in recognizing when a linear combination leads to a contradiction, such as an equality that is always false.

Understanding Systems with No Solution

To grasp why a system of equations might have no solution, let's consider the geometric interpretation of linear equations. Each equation in the system represents a line in the xy-plane. A solution to the system corresponds to a point where the lines intersect. If the lines intersect at a single point, the system has a unique solution. If the lines coincide (are the same line), the system has infinitely many solutions. However, if the lines are parallel and distinct, they never intersect, and the system has no solution.

In the given system:

{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.

We can rewrite the equations in slope-intercept form (y = mx + b) to better visualize their graphical representation. The first equation becomes:

52y=−23x+15y=−415x+6\frac{5}{2}y = -\frac{2}{3}x + 15\\ y = -\frac{4}{15}x + 6

The second equation becomes:

15y=−4x+12y=−415x+4515y = -4x + 12\\ y = -\frac{4}{15}x + \frac{4}{5}

Notice that both lines have the same slope (-4/15) but different y-intercepts (6 and 4/5). This indicates that the lines are parallel but not identical. Therefore, they will never intersect, and the system has no solution. This is a crucial concept in understanding why certain systems of equations have no solution.

Identifying Linear Combinations that Lead to Contradictions

Now, let's explore how to find a linear combination of the given equations that demonstrates the absence of a solution. The goal is to manipulate the equations in such a way that we arrive at a contradiction, such as an equation of the form 0 = constant, where the constant is non-zero. This would definitively prove that the system cannot have any solutions.

Consider the given system again:

{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 create a linear combination, we can multiply the first equation by a constant, say A, and the second equation by another constant, say B, and then add the resulting equations. Our aim is to choose A and B such that the x and y terms cancel out, leaving us with a constant on one side and zero on the other side, indicating a contradiction.

Let's multiply the first equation by 6 and the second equation by -1/4:

6â‹…(23x+52y)=6â‹…154x+15y=906 \cdot \left(\frac{2}{3} x+\frac{5}{2} y\right) = 6 \cdot 15\\ 4x + 15y = 90

−14⋅(4x+15y)=−14⋅12−x−154y=−3-\frac{1}{4} \cdot (4 x+15 y) = -\frac{1}{4} \cdot 12\\ -x - \frac{15}{4}y = -3

Multiplying the equations, we get:

6â‹…(23x+52y)=6â‹…154x+15y=90\begin{aligned} 6 \cdot \left(\frac{2}{3} x+\frac{5}{2} y\right) &= 6 \cdot 15 \\ 4x + 15y &= 90 \end{aligned}

−14⋅(4x+15y)=−14⋅12−x−154y=−3\begin{aligned} -\frac{1}{4} \cdot (4 x+15 y) &= -\frac{1}{4} \cdot 12 \\ -x - \frac{15}{4}y &= -3 \end{aligned}

To eliminate the x term, we can multiply the first equation by -6:

−6⋅(23x+52y)=−6⋅15−4x−15y=−90-6 \cdot \left(\frac{2}{3} x+\frac{5}{2} y\right) = -6 \cdot 15\\ -4x - 15y = -90

Now, add this modified first equation to the second equation:

(−4x−15y)+(4x+15y)=−90+120=−78\begin{aligned} (-4x - 15y) + (4x + 15y) &= -90 + 12 \\ 0 &= -78 \end{aligned}

This result, 0 = -78, is a clear contradiction. It demonstrates that there are no values of x and y that can simultaneously satisfy both equations in the system. Therefore, the system has no solution. The equation 0 = -78 represents a linear combination of the original equations that highlights this contradiction.

Analyzing the Given Options

Now, let's examine the given options to see which one represents a linear combination of the system:

  1. 43x=42\frac{4}{3} x=42
  2. 0=−780=-78
  3. 152y=−102\frac{15}{2}y=-102

We have already derived the linear combination 0 = -78, which directly demonstrates the inconsistency of the system. This matches option 2. Let's analyze why the other options might not be as direct or conclusive.

  • Option 1: 43x=42\frac{4}{3} x=42

    This equation involves only the variable x. While it might be part of an intermediate step in solving the system, it doesn't directly show the contradiction. To obtain this equation, we would need to manipulate the original equations in a way that eliminates y. However, simply having an equation with only x doesn't guarantee that the system has no solution. We would need to find a corresponding equation for y or a contradiction to definitively conclude that there is no solution.

  • Option 3: 152y=−102\frac{15}{2}y=-102

    Similar to option 1, this equation involves only the variable y. It doesn't, on its own, reveal the contradiction that proves the system has no solution. We would need to find a corresponding inconsistent equation involving x or a direct contradiction to confirm the absence of solutions.

Therefore, the most direct and conclusive representation of a linear combination demonstrating the system's lack of solution is option 2: 0 = -78. This equation explicitly shows the contradiction that arises from the system, proving that no values of x and y can satisfy both equations simultaneously.

Conclusion

In summary, the 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.

has no solution because the equations represent parallel lines that never intersect. This can be demonstrated by transforming the equations into slope-intercept form and observing that they have the same slope but different y-intercepts. Furthermore, a linear combination of the equations can lead to a contradiction, such as 0 = -78, which definitively proves the absence of a solution. Among the given options, 0 = -78 is the most direct representation of this contradiction and therefore the correct answer. Understanding how to identify and create such linear combinations is a valuable skill in solving and analyzing systems of equations in linear algebra.

Given the 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.

which has no solution, which of the following equations represents a linear combination of the system that demonstrates this?

43x=42\frac{4}{3} x=42 0=−780=-78 152y=−102\frac{15}{2}y=-102

System of Equations with No Solution Identifying Linear Combinations