x-y=32x-2y=kThe linear system cannot have a unique solution, regardless of the value of k.
To determine the uniqueness of the solution, we need to analyze the system of equations
To determine the uniqueness of the solution, we need to analyze the system of equations.
The given system of equations is:
Equation 1: x – y = 32x – 2y
Equation 2: k
The second equation, k, is not a traditional equation as it does not involve any variables (x or y). It is simply a constant value, denoted by the variable k.
Now, let’s consider the first equation. In order for the system to have a unique solution, the two equations must be independent, meaning they do not represent the same line. One way to determine this is by checking the slopes of the two lines represented by the equations.
Let’s rewrite Equation 1 in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept:
x – y = 32x – 2y
y = (31/2)x
From this equation, we can see that the slope is 31/2.
Now, let’s consider Equation 2. Since it is not a traditional equation, we cannot determine its slope or any other information about it.
Since we have no information about the slope or any other characteristic of Equation 2, we cannot definitively say whether it is the same line as Equation 1 or not. However, regardless of the value of k, Equation 2 does not provide any additional information to the system of equations. It is essentially a separate equation unrelated to the first equation.
Therefore, the given system of equations cannot have a unique solution, regardless of the value of k, because we have insufficient information to determine whether Equation 2 is the same line or not.
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