Question

for what values of k will the pair of linear equations kx+3y-(k-1)=0 and 12 x+ky=k wi;; have unique solution

Answer 1

Heya !!!


KX + 3Y - (K-1) = 0--------(1)


12X + KY - K = 0----------(2)


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

Where,


A1 = K , B1 = 3 and C1 = -(K-1)

And,


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


A1/A2 = K/12 , B1/B2 = 3/K and C1/C2 = -K+1/-K



For unique solution we must have;


A1/A2= B1/B2 not equal C1/C2.


K/12 = 3/K


K² = 36


K = ✓36 = 6




HOPE IT WILL HELP YOU...... ;-)