MathematicsFind 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)  

Find the value of k for which the following pair of equation has no solution.


x+2y=3 (k1)x+(k+1)y=(k+2)  


  1. A
    1
  2. B
    2
  3. C
    3
  4. D
    4 

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    Solution:

    Given pair of linear equations are: x+2y=3   and (k1)x+(k+1)y=(k+2)  . We need to find the value of k, for these equations have no solution.
    A system of linear equations a 1 x+ b 1 y+ c 1 =0   and a 2 x+ b 2 y+ c 2 =0   will have no solutions when a 1 a 2 = b 1 b 2 c 1 c 2  .
    Now from x+2y=3   and (k1)x+(k+1)y=(k+2)  , we have,
    a 1 =1, b 1 =2, c 1 =3   and a 2 =k1, b 2 =k+1, c 2 =k2  
    Now, substituting the values in the condition we have,
    1 k1 = 2 k+1 3 k2  
    On comparing, we have,
    1 k1 = 2 k+1 k+1=2(k1) k+1=2k2 2kk=2+1 k=3  
    Thus, the value of k is 3.
    Therefore, option 3 is correct.
     
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