How To Prove That A Linear System Cannot Have A Unique Solution: An Elimination Method For Math Enthusiasts

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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