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