Question

6. Find the value of k so that the following system of equations has 3x-y-5 = 0.6x-2y + k = 0​

Answer 1

(4 – 5) – (13 – 18 + 2).

= -1-(13+2-18).

= -1-(15-18).

= -1-(-3).

= -1+3.

= 2

Thanks

Answer 2

Answer:

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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Step-by-step explanation:

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