Question

what is the value of K so that the pair of linear equation kx-y=2 and 6x-2y=3 has a unique solution???
a) k=3
b) k is not equal to 3
c) k=0
d) k is not equal to 0​

Answer 1

Step-by-step explanation:

if pair of linear equations

$a_1x + b_1y + c_1 = 0 \\ \\ a_2x + b_2y + c_2 = 0$

has a unique solution,then

$\frac{a_1}{a_2} \not = \frac{b_1}{b_2}$

So,here in the question

$kx - y - 2 = 0 \\ 6x - 2y - 3 = 0$

take the coefficient of x and y from both and put in the relation stated above

$a_1 = k \: \: b_1 = - 1 \\ a_2 = 6 \: \: b_2 = - 2$

$\frac{k}{ - 1} \not= \frac{6}{ - 2}$

$k \not = 3$

For equation to be have unique solutions k≠3,Option b is correct.

Hope it helps you.

Answer 2

Answer:

Step-by-step explanation:

The given system is

Kx - y - 2 =0

6x - 2y -3 = 0

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

For the system to have a unique solution we must have

a₁/a₂ ≠ b₁/b₂

k /6  ≠ -1 /-2

-2k  ≠ -6

k  ≠ 3

Hence K  ≠ 3