Question

Find the value of k for which the pair of linear equation kx+3y=k-2 and 12 x+ky=k has no solution

Answer 1

In equation 1, kx + 3y =2k

here a1=k b1=3 & c1=2k

And equation is 12x + ky = k

here a2=12 b2=k and c2=k

For a equation to have no solutions,

a1/a2 = b1/b2 not equal to c1/c2

Now a1/a2 = b1/b2

so k/12 = 3/k

=> k^2=3×12

=>k = √(12×3)=6

Now b1/b2 not equal to c1/c2

=> 3/k not =2k/k

=>3/k not=2

=>k not=3/2

=> k not equal to 1.5

So the value of k for which the pair of equations have no soln is 6.

hope this answer helpful u

Answer 2

Answer:

k/k=-2-3y-x

Step-by-step explanation:

The answer can be

kx+3y=k-2

kx=k-2-3y

k=k-2-3y-x

k/k=-2-3y-x