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karansingh
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Find the values of k for which the pair of linear equations kx + 3y = k-2 and 12x + ky = k has no solution.

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  1. The given pair of linear equations is

    kx + 3y = k – 3 …(i)

    12x + ky = k …(ii)

    On comparing the equations (i) and (ii) with ax + by = c = 0,

    We get,

    a1 = k, b1 = 3, c1 = -(k – 3)

    a2 = 12, b2 = k, c2 = – k

    Then,

    a1 /a2 = k/12

    b1 /b2 = 3/k

    c1 /c2 = (k-3)/k

    For no solution of the pair of linear equations,

    a1/a2 = b1/b2≠ c1/c2

    k/12 = 3/k ≠ (k-3)/k

    Taking first two parts, we get

    k/12 = 3/k

    k2 = 36

    k = + 6

    Taking last two parts, we get

    3/k ≠ (k-3)/k

    3k ≠ k(k – 3)

    k2 – 6k ≠ 0

    so, k ≠ 0,6

    since k is not 6 it must be -6.

    k=-6

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