Question

Find the value of k so that the system of equations has no solution:

3x-y-5=0; 6x-2y-k=0
samaj seva answer what do you want ​

Answer 1

Step-by-step explanation:

3x/6x = 1/2

-y/-2y = 1/2

-5/-k = ?

let put the value of k =10

-5/-10 = 1/2...

so the value of k will be 10....

hope it helps....

Answer 2

I am give you 3-4 answers

Choose by yourself :

1.

Answer:

The value of k = 10.

Step-by-step explanation:

Given two equations, 3x – y – 5 = 0 (i.e) 3x - y = 5    …. (1)

6x - 2y – k = 0 (i.e) 6x - 2y = k    …. (2)

Divide equation (2) by 2, it becomes, 3x – y =  … (3)

From equation (1) and (3) LHS are equal, hence consider RHS

= 5, k = 10 for the problem to have solution,  

Condition for no solution be k ≠ 10.  

2.

k can be any value because in no solution condition

a1/a2=b1/b2 is not equal to c1/c2

3.

To have no solution the two equations should have the form - a1/ a2 = b1/b2 not equals to c1/c2

So the value of k is any no. except 10 .

4.

Given Equation is 3x - y - 5 = 0  = > a1 = 3, b1 = -1, c1 = -5.

Given Equation is 6x - 2y - k = 0 = > a2 = 6, b2 = -2, c2 = -k.

Now,

Given that Equation has no solutions.

= > (a1/a2) = (b1/b2) ≠ (c1/c2)

= > (3/6) = (-1/-2) ≠ (-5/-k)

=. (1/2) = (1/2) ≠ (5/k)

= > (1/2) ≠ (5/k)

= > k ≠ 10.

Therefore, k can take all real values except 10

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