x-y=32x-2y=kThe linear system cannot have a unique solution, regardless of the value of k.
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To show that the linear system cannot have a unique solution, regardless of the value of k, we can use the method of elimination to solve the system of equations.
x – y = 32 (Equation 1)
32x – 2y = k (Equation 2)
Multiplying Equation 1 by 2, we get:
2x – 2y = 64 (Equation 3)
Now, we can eliminate y by adding Equation 2 and Equation 3:
32x – 2y + 2x – 2y = k + 64
Grouping like terms:
34x – 4y = k + 64
Dividing both sides by 2, we get:
17x – 2y = k/2 + 32
Now, we have an equation with two variables, which means that there are an infinite number of solutions. To see this, we can solve for y in terms of x:
y = (17/2)x – k/4 – 16
So, for any value of k, there is a corresponding line in the x-y plane that satisfies the equation 17x-2y=k/2+32. Therefore, there are an infinite number of solutions to the system of equations, and it cannot have a unique solution.
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