Question

Find the value(s) of k for which the pair of linear equations kx + 3y = k – 2 and 12x + ky = k,has no solution.

Answer 1

Step-by-step explanation:

kx+3y-k-2=0 ____________eq1

12x+ky-k=0______________eq2

a1\b1=a2\b2 is not equal to a3\b3

k\12=3\k is not equal to -2\-k

by cross multiplying we get

1. -k^2 ( is not equal to)sign -12k-24

by solving it you got the value of k

2. -3k is not equal to -k^2-2k

by solving it you got the value of k