For which value (s) of k will the pair of equations kx + 3y = k -3, 12x+ ky = k have no solution?
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Step-by-step explanation:
KX + 3Y = K -3
KX + 3Y - ( K -3) = 0 ------------(1)
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 +3
And,
A2 = 12 , B2 = K and C2 = -K
For no solution we must have,
A1/A2 = B1/B2 # C1/C2 [ Where # stand for not equal]
K / 12 = 3/K
K² = 12 × 3
K² = 36
K = ✓36 = 6
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