Question

If the system of linear equation 6.x - 2y = 3 and kx - y - 2 = 0 has
unique solution, then find the value of k.​

Answer 1

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.

I hope this helps you.

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Answer 2

Answer:

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Step-by-step explanation: