Question

Find k so that the following pair of linear
equations has no solution.
3x + y = 1; (2k-1)x + (k-1)y = 2k + 1.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

For no solution

a1/a2 = b1/b2 ≠ c1/c2

Here, a1 is 3, b1 is 1 and c1 is -1

a2 is (2k-1), b2 is (k-1) and c2 is 2k+1

CASE I

3/2k - 1    =     1/k-1

by cross multiplication,

3k-3 = 2k - 1

k=2

CASE II

1/ k-1 ≠ -1/-(2k+1)

1/(k-1) ≠ 1/(2k+1)

Again by cross multiplication,

2k+1 ≠ k-1

k≠-2

So, at last we conclude that k is never equal equal to -2 and is equal to 2

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