Question

find the value of k so that the following system of equation has no solution
3x-y-5=0
6x-2y+k=0

Answer 1

3x - y - 5 = 0

6x - 2y + k = 0

Here, a1 = 3, a2 = 6, b1 = - 1, b2 = - 2, c1 = - 5, c2 = k

If these equations has no zeroes, then

a1 / a2 = b1 / b2 ≠ c1 / c2

Now,

b1 / b2 ≠ c1 / c2

- 1 / - 2 ≠ - 5 / k

1 / 2 ≠ - 5 / k

$<b>k ≠ - 10</b>$

Hence, k has a values except - 10.

Answer 2

Solution :- 3x-y-5 = 0
6x-2y+k = 0

Here, A1 = 3 A2 =6.
B1 = -1 B2 =-2
C1 = -5. C3= k

Condition for not solution :-

A1 / A2 = B1/B2 is not equal to C1/C2
So,
3/6 = -1/-2 is not equal to -5/k
1/2=1/2 is not equal to -5/k
1/2 is not equal to -5/k
k is not equal to -10

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