Question

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Heya friend ✋✋ can anybody help me in this question.....


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❇ Find the value of 'k' for which the system of equation kx-y=2 and 6x-2y=3 will have:-
✴unique solution
✴Infinite solution
✴no solution

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Answer 1

kx - y = 2..........1st eqn
6x - 2y = 3........2nd eqn


Comparing the two equations by the form
a1x + b1y - c1 = 0
& a2x + b2y + c2 = 0

We get
a1 = k. b1 = -1 & c1 = -2
a2 = 6. b2 = 2. & c2 = -3

Now,
When the two equns have a unique solution then
a1 / a2 is not equal to b1 / b2
k/6 not = -1/2
k not= -6/2
K not equal to -3
Therefore
The Two equations will have a unique solution for all values of k except -3.

Now
If the two equations have no solution
a1 / a2 = b1 / b2 not = c1 / c2
a1 / a2 = b1 / b2
k / 6 = -1 / 2
k = -6/2
k = -3