Question

fint the value of k for which the following pair of equation has no solution x+2y=3;(k-1)x+(k+1)y=(k+2)​

Answer 1

Answer:

k = 3

k ≠ 1

Step-by-step explanation:

Given a pair of linear equations such that,

x + 2y = 3 .......(1)

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

To find the value of k for which it has no solution.

We know that,

For given pair of linear equations, there will be no solution iff ,

• a1/a2 = b1/b2 ≠ c1/c2

Therefore, we will get,

=> 1/(k-1) = 2/(k+1) ≠ 3/(k+2)

Thus, we will get,

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

=> k + 1 = 2k - 2

=> 2k - k = 2+1

=> k = 3

And

=> 2(k+2)≠ 3(k+1)

=> 2k + 4 ≠ 3k + 3

=> 3k-2k = 4-3

=> k ≠1

Hence, the required values of k= 3 and k≠1.