Question

The value of k for which system of equations x+ky=0,2x-y=0 has a unique solution is________________​

Answer 1

GIVEN :

Equations :

x + ky = 0 and 2x - y = 0

We know that,

For unique solution, a1/a2 ≠ b1/b2

From the equations,

a1 = 1 b1 = k

a2 = 2 b2 = -1

1/2 ≠ k/-1

2k ≠ -1

k ≠ -1/2

Therefore, the value of k ≠ -1/2.

Answer 2

Answer :-

$k \neq \dfrac{-1}{2}$

Given :-

x + ky = 0

2x - y = 0

To find :-

The value of k such that the system of equations has unique solution.

Solution:-

we have ,

$a_1 = 1 , a_2 = 2\\ b_1 = k ,b_2 = -1$

For unique solution we have ,

$\huge \boxed{\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}}$

Now , put the given value

→$\dfrac{1}{2}\neq \dfrac{k}{-1}$

By cross multipication

→$-1 \neq 2k$

→$k \neq \dfrac{-1}{2}$

hence, the value of k for which the system of equation has infinity solution is $k \neq \dfrac{-1}{2}$