Question

5. The system Kx - y = 2 and 6x-2y = 3 has a unique solution only when
a) K=0
b) K+ 0
K=3
d) K+ 3

Answer 1

Answer:

Kx - y = 2 .......(i)

6x -2y = 3 ......(ii)

now compairing both the equation to standard form , we get

kx - y -2 = 0

6x -2y -3 = 0

a1 = k , b1 = -1 , c1 = -2

a2 = 6 , b2 = -2 ,c2 = -3

As it is unique equation , so

$\dfrac{a1}{a2} ≠ \dfrac{b1}{b2}$

$\dfrac{k}{6} ≠ \dfrac{-1}{-2}$

$\dfrac{k}{6} ≠ \dfrac{1}{2}$

Now , by cross multiplication,

2k ≠ 6

k ≠$\dfrac{6}{2}$

k ≠ 3

Therefore option "c" is correct.

Answer 2

Answer:

c is the correct option