Question

Value of 'k' in which system of equations kx-y=2 and 6x-2y=3 has a unique solutions.
a)=3
b)not equal to 3
c),not equal to 0
d)=0

Answer 1

$\bold{\huge{Answer:}}$

Hello here is your Answer!!!!!

Given Equation:-

Kx-y=2

Here,
a1=k
b1=-1

c1=3

Another equation:;

6x-2y=3

Here
a2=6

b2=-2

c2=3

For Unique solution we must be have;

$\bold{ \frac{a1}{a2} = \frac{b1}{b2} } \\ \\ \\ \bold{ \frac{k}{6} = \frac{ \cancel- 1}{ \cancel- 2} } \\ \\ \bold{ \frac{k}{6} = \frac{ 1}{ 2} } \\ \\ \bold{2k = 6} \\ \\ \\ \\ \\ \bold{k =) \frac{6}{2} = 3} \\ \\ \bold{so \: k = 3}$

Solution:- k=3 and k≠3

Answer 2

Answer:

The given equation is..kx-y=2 and 6x-2y=3

so, let's take the first equation,

kx-y=2

a1=k

b2=-1 and c3=2...

Now, let's take the other equation,

6x-2y=3

a2=6,

b2=-2 and c3=3...

SO, for UNIQUE SOLUTION, WE KNOW...

a1/a2=! b1/b2

=》k/6=!-1/-2

=》k/6=! 1/2

=》k=!3

Therefore, the answer is k is not equals to 3