Question

find the value of k for which the pair of leniar equation kx+3y=k-2 and 12x+ky=k has no solutions

Answer 1

therefore , k = 6 .
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Answer 2

Heya !!!



KX + 3Y = K -2




KX + 3Y - (K -2 ) = 0




And,




12X + KY = K


12X - KY - K = 0



These equations are of the form of A1X + B1Y + C1 = 0 and A2X + B2Y + C2 = 0




Where,



A1 = K , B1 = 3 and C1 = -K +2



And,


A2 = 12 , B2 = K and C2 = -K





For no solution we must have,



A1/A2 = B1/B2 # C1/C2 [ Where # stands for not equal ]




K / 12 = 3 / K




K² = 12 × 3




K² = 36



K = ✓36



K = 6






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