Question

Please answer the above question with steps...I will mark as brainliest..​

Answer 1

Answer:

k = -6

Step-by-step explanation:

(i)

kx + 3y = k - 3

It can be written as,

kx + 3y - (k - 3) = 0

Here, a₁ = k, b₁ = 3, c₁ = -(k - 3)

(ii)

12x + ky = k

It can be written as,

12x + ky - k = 0

Here, a₂ = 12, b₂ = k, c₂ = -k

Since equation has no solution,

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

⇒ k/12 = 3/k ≠ -(k - 3)/-k

On taking first two parts, we get

k/12 = 3/k

⇒ k² = 36

⇒ k = ±6

On taking last two parts, we get

3/k ≠ -(k - 3)/-k

⇒ 3/k ≠ (k - 3)/k

⇒ 3k ≠ k² - 3k

⇒ k² - 6k ≠ 0

⇒ k ≠ 0,6

Therefore, the value of k = -6

Hope it helps!

Answer 2

If  a1 /  a2 ≠  b1 /  b2 then the pair of linear equations a1x + b1y + c1 = 0

a2x+ b2y + c2 = 0  has a unique solution.

Given pair of  equations K x + 3y = K-3, 12x + Ky = K.

Here a1 = k ,b1 = 3 , a2 = 12 , b2 = k .

k / 12 ≠ 3 / k

k2 ≠ 36.

k ≠ ± 6.