Question 126

The value of k for which the system ky + 3y = k - 3 and 12x + ky = k of equations has infinitely many solutions is:

Solution

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

$$K^2 = 12 × 3$$

$$K^2 = 36$$

$$K = \sqrt{36} = 6$$

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