Question

find the value of k if the pair of equations x minus y equal to 2 and 6 x minus 2 y equal to 3 has a unique solution​

Answer 1

Answer:

Given system of equations are

6x - 2y = 3

6x - 2y - 3 = 0 ----( 1 )

kx - y = 2

kx - y - 2 = 0 ----( 2 )

Compare above equations with

a1 x + b1 y + c1 = 0 and

a2 x + b2 y + c2 = 0 , we get

a1 = 6 , b1 = -2 , c1 = -3 ;

a2 = k , b2 = -1 , c2 = -2 ;

Now ,

a1/a2 ≠ b1/b2

[ Given they have Unique solution ]

6/k ≠ ( -2 )/( -1 )

6/k ≠ 2

k/6 ≠ 1/2

k ≠ 6/2

k ≠ 3

Therefore ,

For all real values of k , except k≠ 3,

Above equations has unique solution.