Question

the sstem of equation 6x+2y=2 and kx+y=1 has many solution find the value of K​

Answer 1

Step-by-step explanation:

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.

Answer 2

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For a system of equations

a¹x+b¹y+c¹

=0;a²x+b²y+c²

=0 to have unique solution, the condition to be satisfied is

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

⇒6/k not equal to −2/−1 not equals to 3/2

∴ For unique solution of system of equations, k not equals to 3,4