Question

For what value of k the system of equations has a unique solution? x - ky = 2 and 3x + 2y = -5

Answer 1

The given system of equations:

3x + y = 1

⇒ 3x + y − 1 = 0 ...(i)

And, (2k − 1)x + (k − 1)y = (2k + 1)

⇒ (2k − 1)x + (k − 1)y − (2k + 1) = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Here, a1 = 3, b1= 1, c1 = −1 and a2 = (2k − 1), b2 = (k − 1), c2 = −(2k + 1)

In order that the given system of equations has no solution, we must have:

i.e.

and

⇒ 3(k − 1) = 2k − 1 and −(2k + 1) ≠ −(k − 1)

⇒ 3k − 3 = 2k − 1 and −2k − 1 ≠ −k + 1

⇒ k = 2 and k ≠ −2

Thus, holds when k is equal to 2.

Hence, the given system of equations has no solution when the value of k is 2.

Answer 2

Answer:

The value of the k is 2

Step-by-step explanation:

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