Question

Find the Value of K for which bthe pair of linear equations KX+3y = K-2 and 12X+KY-K = 0 has no solution.

Answer 1

Hey

Here is your answer,

Answer 2

Hii friend,

The given equation are

KX + 3Y = K-2

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

And,

12X +KY -K = 0

These equation are in the form of

A1X + B1Y +C1 = 0 and A2X+B2Y+C2 = 0

Where,

A1 = K , B1 = 3 , C1 = (2-K) and A2= 12 , B2= K , C2 = -K

Therefore,

A1/A2= K/12 , B1/B2= 3/K and C1/C2 = (2-K)/-K = (K-2)/K

The given system of equation has no solution

Then,

A1/A2 = B1/B2 is not equal to C1/C2

=> K/12 = 3/K not equal to (K-2)/K

=> K/12 = 3/K and 3/K not equal to (K-2)/K

=> K² = 36 and K²-2K not equal to 3K

=> K²= 36 and K²-5K not equal to 0

=> K = ✓36 and K(K-5) not equal to 0

=>K = 6 or K=-6

When,

K = 6

Then,

=> K(K-5) = 6(6-5) = 6 × 1 = 6 not equal to 0

When,

K = -6

Then,

=> K(K-5) = (-6)(-6-5) = (-6) × (-11) = 66 not equal to 0 .

Thus,

In each case , the given system has no solution.

Hence,

K = 6 or K = -6


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