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

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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 .........

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  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


  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!


  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

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