For what value of k the equation will have no solution ?

Question

For what value of k the equation will have no solution ?

kx - 3y = 1

(4-k)y - x +1 = 0 solve without using the concept of Coordinate Geometry.. .Thanks in Advance .........

Answer 1

Find the rank of the matrix with first row [k  -3  -1] and the second row [(4-k)  -1  1]

on applying row operation R2 <-- R1-3R2

the rows become [k  -3  -1] and [4k-12  0   -4]

in order to get the inconsistent system, 4k-12=0

Hence, k=3

Answer 2

kx - 3y = 1

kx - 3y - 1 = 0________(equation 1)

(4 -k)y - x + 1= 0

- x + 1 + y(4 - k) = 0

-x + 1+ 4y - ky = 0

-x + 4y - ky + 1 = 0________(equation 2)

The equations will have no solution if a1/a2 = b1/b2 not equal to c1/c2

k/ -1  = -3/4 - k not equal to -1/1

now cross multiply any 2

u will get the answer

I hope this has helped u!

Answer 3

kx -3y = 1 ans: x =0;

(4-k)y - x=0 ans;y =0;

because at this value k becomes infinite n taht is not possible