Question

22. For what value of k will the following pair of linear equations have no solutions?
3x+y=1
(2k-1)x + (k-1) y = 2k +1​

Answer 1

Answer:

If we consider a pair of linear equations in two variables, say:

a1x + b1y + c1 = 0 and

a2x + b2y + c2 = 0

Then, the condition for no solution is:

a1/a2 = b1/b2 ≠ c1/c2

Here,

The given pair of linear equations is;

=> 3x + y = 1

=> 3x + y - 1 = 0

and

=> (2k - 1)x + (k - 1)y = 2k + 1

=> (2k - 1)x + (k - 1)y - (2k + 1) = 0

Clearly, here we have;

a1 = 3

a2 = 2k - 1

b1 = 1

b2 = k - 1

c1 = -1

c2 = - (2k +1)

Thus, the given equations will have no solution, if and only if,

=> a1/a2 = b1/b2

=> 3/(2k - 1) = 1/(k - 1)

=> 3(k - 1) = 2k - 1

=> 3k - 3 = 2k - 1

=> 3k - 2k = 3 - 1

=> k = 2

Hence, the required value of k is 2.

Answer 2

Answer:

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