find k:

a). 2x + ky = 1 ( for unique solution)

b). 2x - 3y = 7

(k + 2) x - (2k + 1)y = 3(2k -1) (for infinte number of solutions)

c). kx + 3y = 3

12x + ky = 6(for no solution)

Part (a) of the question is incomplete. Recheck your question and do get back to us.

Here is the answer for part (b) and part (c).

(b).  The given equations are  

 2x - 3y -7 = 0  ...............(1)

 and  (k + 2)x  - (2k + 1)y - 3 (2k - 1) = 0  ..............(2)

comparing above two equations with standard equations that are,

 a1x + b1y + c1 = 0  and  a2x + b2y + c2 = 0  , we get 

 a1 = 2 ; b1 = -3 ; c1 = -7 and  a2 = k + 2 ; b2 = - (2k + 1) ; c2 = -3(2k - 1)

For infinitely many solutions, we have

So the given system of equations has infinitely many solutions for k = 4

(c).  The given equations are  

 kx + 3y = 3  ............(1)

12x + ky = 6  ............(2)

comparing above two equations with standard equations that are,

 a1x + b1y + c1 = 0  and  a2x + b2y + c2 = 0  , we get 

 a1 = k ; b1 = 3 ; c1 = -3 and  a2 = 12 ; b2 = k ; c2 = -6

For no solution, we have 


So the given system of equations has no solution for k = -6  

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