Question

find the value of k for which the given linear pair has a unique solution 2x+3y-5=0 ,kx-6y-8=0​

Answer 1

Answer:

K can be any value from 0 to infinity in both negative and positive except (-4)

Step-by-step explanation:

For unique solution a1/b1 ≠ a2/b3

Here, a1 = 2

b1 = 3

a2= k

b2= - 6

As, a1/b1 ≠ a2/b2

2/3 ≠ k/-6

If k = - 4 then k/-6 will be 2/3 which will be = a1/b1 which should not happen

Therefore k can be any value except (-4)

Answer 2

Answer:

if you want a unique solution then the lines of these equations in a graph should be intersecting

∴the value of k can be anything other than -4

hope it helps

Step-by-step explanation: